logarithms
inequalities
calculus
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ \sqrt{\frac{a^3\plus{}abc}{(b\plus{}c)^3}}\plus{}\sqrt{\frac{b^3\plus{}abc}{(c\plus{}a)^3}}\plus{}\sqrt{\frac{c^3\plus{}abc}{(a\plus{}b)^3}} \ge \frac{3}{2} \cdot \left( \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\right)^{2\minus{}\log_2 3}\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3 \plus{} abc}{(c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3 \plus{} abc}{(a \plus{} b)^3}} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
:)[/quote]
I have a solution with no SOS, EV, pqr, ... :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3 \plus{} abc}{(c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3 \plus{} abc}{(a \plus{} b)^3}} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
:)[/quote]

Nice, my friend :) . I think
\[ \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \ge \sum \frac a{b\plus{}c}\ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]

[quote="dduclam"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3 \plus{} abc}{(c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3 \plus{} abc}{(a \plus{} b)^3}} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
:)[/quote]

Nice, my friend :) . I think
\[ \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \ge \sum \frac a{b \plus{} c}\ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
[/quote]
Yes, that's it, my friend. :)
But do you have a nice solution for
\[ \sum \frac a{b \plus{} c}\ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}\]
?
I meant with no SOS, EV, pqr, ... :)

Did you meant:
$ (\sum\ \frac{a}{b\plus{}c})^2 \geq\ \frac{9(a^2\plus{}b^2\plus{}c^2)}{4(ab\plus{}bc\plus{}ca)}$
 :wink:

[quote="nguoivn"]Did you meant:
$ (\sum\ \frac {a}{b \plus{} c})^2 \geq\ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}$
 :wink:[/quote]

Try with $ a \equal{} b \equal{} 1,1$ and $ c \equal{} 0,8$.

Especially

[quote="can_hang2007"]
\[ \sum \frac a{b \plus{} c}\ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
[/quote]

Equality holds if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$.  :wink:

[quote="manlio"]$ \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \ge \sum \frac a{b \plus{} c}$

Is it an old inequality? I don'tm remember it, sorry. :huh:[/quote]
Maybe Can means that it's obvious for him?
By Holder we obtain:
$ \left(\sum_{cyc} \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}}\right)^2\sum_{cyc}\frac{a^2}{a^2\plus{}bc}\geq\left(\sum_{cyc}\frac{a}{b\plus{}c}\right)^3.$
Thus, it remains to prove that $ \sum_{cyc}\frac{a}{b\plus{}c}\geq\sum_{cyc}\frac{a^2}{a^2\plus{}bc},$ which is equivalent to
$ \sum_{cyc}\frac{a(a\minus{}b)(a\minus{}c)}{(b\plus{}c)(a^2\plus{}bc)}\geq0,$ which is obvious even for me. :D

[quote="arqady"][quote="manlio"]$ \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \ge \sum \frac a{b \plus{} c}$

Is it an old inequality? I don'tm remember it, sorry. :huh:[/quote]
Maybe Can means that it's obvious for him?
By Holder we obtain:
$ \left(\sum_{cyc} \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}}\right)^2\sum_{cyc}\frac {a^2}{a^2 \plus{} bc}\geq\left(\sum_{cyc}\frac {a}{b \plus{} c}\right)^3.$
Thus, it remains to prove that $ \sum_{cyc}\frac {a}{b \plus{} c}\geq\sum_{cyc}\frac {a^2}{a^2 \plus{} bc},$ which is equivalent to
$ \sum_{cyc}\frac {a(a \minus{} b)(a \minus{} c)}{(b \plus{} c)(a^2 \plus{} bc)}\geq0,$ which is obvious even for me. :D[/quote]
That's it, arqady. :)

[quote="can_hang2007"][quote="dduclam"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3 \plus{} abc}{(c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3 \plus{} abc}{(a \plus{} b)^3}} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
:)[/quote]

Nice, my friend :) . I think
\[ \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \ge \sum \frac a{b \plus{} c}\ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
[/quote]
Yes, that's it, my friend. :)
But do you have a nice solution for
\[ \sum \frac a{b \plus{} c}\ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
?
I meant with no SOS, EV, pqr, ... :)[/quote]
Actually, I have proved this inequality by dividing it into 2 cases
Case 1. If $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge 2$, then we will prove that
\[ \sum \frac {a}{b \plus{} c} \ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
Case 2. If $ 1 \le \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \le 2$, then we will prove that
\[ \sum \frac {a}{b \plus{} c} \ge \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{2} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
For cases 2, I would like to see a nice proof without full expansion, SOS, ... for
\[ \sum \frac {a}{b \plus{} c} \ge \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{2}
\]
:)
[quote="arqady"][quote="manlio"]$ \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}} \ge \sum \frac a{b \plus{} c}$

Is it an old inequality? I don'tm remember it, sorry. :huh:[/quote]
Maybe Can means that it's obvious for him?
By Holder we obtain:
$ \left(\sum_{cyc} \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}}\right)^2\sum_{cyc}\frac {a^2}{a^2 \plus{} bc}\geq\left(\sum_{cyc}\frac {a}{b \plus{} c}\right)^3.$
Thus, it remains to prove that $ \sum_{cyc}\frac {a}{b \plus{} c}\geq\sum_{cyc}\frac {a^2}{a^2 \plus{} bc},$ which is equivalent to
$ \sum_{cyc}\frac {a(a \minus{} b)(a \minus{} c)}{(b \plus{} c)(a^2 \plus{} bc)}\geq0,$ which is obvious even for me. :D[/quote]
We can also prove it by taking notice that
\[ (a^2 \plus{} bc)(b^2 \plus{} ca) \minus{} ab(a \plus{} c)(b \plus{} c) \equal{} c(a \plus{} b)(a \minus{} b)^2 \ge 0
\]
Therefore
\[ \left( \sum \sqrt {\frac {a^3 \plus{} abc}{(b \plus{} c)^3}}\right)^2 \equal{} \sum \frac {a^3 \plus{} abc}{(b \plus{} c)^3} \plus{} 2\sum \sqrt {\frac {ab(a^2 \plus{} bc)(b^2 \plus{} ca)}{(a \plus{} c)^3(b \plus{} c)^3}} \ge \sum \frac {a^3 \plus{} abc}{(b \plus{} c)^3} \plus{} 2\sum \frac {ab}{(a \plus{} c)(b \plus{} c)} \equal{} \sum \frac {a(a \minus{} b)(a \minus{} c)}{(b \plus{} c)^3} \plus{} \left( \sum \frac {a}{b \plus{} c}\right)^2 \ge \left(\sum \frac {a}{b \plus{} c}\right)^2
\]
:)

[quote="manlio"]$ \sum \sqrt {\frac {a^3 + abc}{(b + c)^3}} \ge \sum \frac a{b + c}$

Is it an old inequality? I don'tm remember it, sorry. :huh:[/quote]

$ \sum \sqrt {\frac {a^3 + abc}{(b + c)^3}} \ge \sum \frac a{b + c}$

$ \Leftrightarrow \sum \frac{\sqt a(a-b)(a-c)}{\sqrt{(b+c)^3}(\sqrt{a^2+bc}+\sqrt{ab+ac})}\ge0$ which true by Vornicu-Schur  :)

Dear can_hang2007,

why

[quote="can_hang2007"]

Case 2. If $ 1 \le \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \le 2$, then we will prove that
\[ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{2} \ge \frac {3}{2} \cdot \left( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^{2 \minus{} \log_2 3}
\]
[/quote]

It is hard for me. How do you prove it without calculus? Thankyou very much. :)

As you used the words "very easy" I think you have a nice prove of this fact without calculus, but I cannot see it.

Please, a hint. :)


factorial
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logarithms
inequalities
integration
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function

Let $ \nu_{p}(n!)=\sum_{k=1}^{\infty}\left\lfloor{\frac{n}{p^{k}\right\rfloor}}$, be the exponent of $ p$ in the prime factorization of $ n!$.
For which $ \alpha$ and $ \beta$, positive real numbers, does the following series converge:
\[ \sum_{n=1}^{\infty}\sum_{p~ prime}\frac{(\nu_{p}(n!))^{\beta}}{n^{\alpha}}\]
This is a generalization to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159572]didilica's question[/url]; don't go to it unless you want a hint! :lol:

Let $ \mathbb{N}= \{1,2, \ldots\}$, $ \mathbb{P}= \{p \in \mathbb{N}: p \mbox{ is prime}\}$ and $ \mathbb{P}(n) = \{p \in \mathbb{P}: p \le n\}$, for any $ n \in \mathbb{N}$. Then
\[ \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)^\beta}{n^\alpha}= \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}(n)}\frac{v_{p}(n!)^\beta}{n^\alpha},\quad\qquad (1) \]
as obviously $ v_{p}(n!) = 0$, whenever $ n \in \mathbb{N}$ and $ p \in \mathbb{P}$ is greater than $ n$. Now consider that
\[ \frac{n}{2p}\le \left\lfloor \frac{n}{p}\right\rfloor \le v_{p}(n!) = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^{k}}\right\rfloor \le \sum_{k=1}^\infty \frac{n}{p^{k}}= \frac{n}{p-1}\le \frac{2n}{p}.\quad\qquad (2) \]
for any $ n \in \mathbb{N}$ and any $ p \in \mathbb{P}(n)$. Hence it’s proved there exist absolute positive constants $ c_{1}, c_{2}\in \mathbb{R}$ such that
\[ c_{1}\sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right) \le \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)^\beta}{n^\alpha}\le c_{2}\sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right). \]
Then it’s pretty clear the series diverges if $ \alpha-\beta \le 1$ and converges if $ \alpha > 1+\beta$ and $ \beta > 1$. So we have to find just a way to deal with the case when $ \alpha > 1+\beta$ and $ 0 < \beta \le 1$. Any ideas?

[quote="s.tringali"]
Then it’s pretty clear the series in hand with the problem converges iff $ \beta > 1$ and $ \alpha > \beta+1$, while diverging otherwise.[/quote]

Everything was elegant, but I think you mean "if" and not "iff".  Careful, the series even converges for $ \beta>0$.

s.tringali, I believe you settled the case when $ \beta>1$.  Lets look at the other possibilities for $ \beta$.  Correct me if I am wrong, but for $ 0< \beta <1$, $ \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\approx_{\beta}\frac{n^{1-\beta}}{\log n}$ where the constant depends on $ \beta$.  So that for any such fixed $ \beta$
\[ \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)^\beta}{n^\alpha}\approx \sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right) \approx \sum_{n = 1}^\infty \frac{1}{n^{\alpha-1}\log n}\]
which converges iff $ \alpha>1$.  Lastly, the case where $ \beta=1$ is not too bad.

[quote="s.tringali"][...] it’s proved there exist absolute positive constants $ c_{1}, c_{2}\in \mathbb{R}$ such that
\[ c_{1}\sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right) \le \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)^\beta}{n^\alpha}\le c_{2}\sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right). \]
[/quote]
Hi, baz. :) Well, do we agree on the fact that the inequality above holds for any positive $ \alpha, \beta \in \mathbb{R}$? If we do, then note that
\[ \sum_{n = 2}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right) \ge \frac{1}{2^{\beta}}\cdot \sum_{n = 2}^\infty \frac{1}{n^{\alpha-\beta}}, \]
which suggests the series in hand with the problem is divergent, whenever $ \alpha-\beta \le 1$. What's missing? :maybe: Besides, you're right when saying $ \alpha > \beta+1$ and $ \beta > 1$ is just a [i]sufficient[/i] condition for convergence. It's my fault, I should seriously work to improve my English... :oops: So actually I'm editing my previous post to make plain what I meant to say.

[quote="baz"]Correct me if I am wrong, but for $ 0 < \beta <1$, $ \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\approx_{\beta}\frac{n^{1-\beta}}{\log n}$ where the constant depends on $ \beta$.[/quote]
I think the problem with your reasoning is in the asymptotic estimation you're exploiting to handle the sum $ \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}$, when $ 0 < \beta \le 1$. Indeed, how do you manage to get such bounds?

s.tringali!  :lol: Our interests seem quite similar.  I am glad you are on this forum.  :)
Now, I think that you have a good point.  I am not sure what happened.  Here is what I thought to be true for $ 0 < \beta <1$:
\[ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}=\int_{2}^{n}\frac{d(\pi(x))}{x^\beta}= \frac{\pi(x)}{x^{\beta}}\bigg|_{2}^{n}+\beta \int_{2}^{n}\frac{\pi(x) dx}{x^{1+\beta}}\approx \frac{n^{1-\beta}}{\log n}+\beta \int_{2}^{n}\frac{dx}{x^{\beta}\log x}\]
where breaking the last integral into two parts $ \int_{2}^{n^{1-\beta}/\log n}$ and $ \int_{n^{1-\beta}/\log n}^{n}$ yields that $ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}\approx_\beta \frac{n^{1-\beta}}{\log n}$.

[quote="baz"]Our interests seem quite similar. I am glad you are on this forum.[/quote]
I'm glad you are on the forum too. I find your problems to be quite fine and challenging. :) But now I have two questions.

[quote="baz"]I am not sure what happened. Here is what I thought to be true for $ 0 < \beta <1$:
\[ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}= \int_{2}^{n}\frac{d(\pi(x))}{x^\beta}\]
[/quote]
How would you formalize this? :|

[quote="baz"]
\[ \int_{2}^{n}\frac{d(\pi(x))}{x^\beta}\approx \int_{2}^{n}\frac{d(x/\log x)}{x^\beta}\]
[/quote]
And this? :huh: I know $ \pi(x) \approx \frac{x}{\ln(x)}$, when $ x \in [2,+\infty[$, that meaning (at least, in my notations) there are absolute constants $ c_{1}, c_{2}\in \mathbb{R}$ such that $ c_{1}\frac{x}{\ln(x)}\le \pi(x) \le c_{2}\frac{x}{\ln(x)}$. But then!? If $ f, g: [0,+\infty[ \to \mathbb{R}$ are defined taking $ f(x) = \frac{x}{2}$ and $ g(x) = f(x)+\sin(x)$, as $ x$ ranges over $ [0,+\infty[$, then clearly $ g(x) \approx f(x)$. Yet $ g^\prime(x) \not\approx f^\prime(x)$. :wink:

I think I have a proof to get off the case when $ \beta = 1$ and $ \alpha > 2$. It isn't much, I know, but it's even better than nothing. Furthermore, it could suggest anyone a way to conclude.

[quote="s.tringali"]Hence it’s proved there exist absolute positive constants $ c_{1}, c_{2}\in \mathbb{R}$ such that
\[ c_{1}\sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right) \le \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)^\beta}{n^\alpha}\le c_{2}\sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right). \]
Then it’s pretty clear the series diverges if $ \alpha-\beta \le 1$ and converges if $ \alpha > 1+\beta$ and $ \beta > 1$. So we have to find just a way to deal with the case when $ \alpha > 1+\beta$ and $ 0 < \beta \le 1$.[/quote]
I'll restart from here. It's known (Mertens' second theorem) there exist absolute positive constants $ b_{1}, b_{2}\in \mathbb{R}$ such that
\[ \forall n \in \mathbb{N}, n \ge 2: b_{1}\cdot\ln(\ln n) \le \sum_{p \in \mathbb{P}(n)}\frac{1}{p}\le b_{2}\cdot\ln(\ln n). \]
So there are two more absolute constants $ a_{1}, a_{2}\in \mathbb{R}^{+}$ such that
\[ a_{1}\sum_{n=2}^\infty \frac{\ln(\ln n)}{n^{\alpha-1}}\le \sum_{n = 2}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)}{n^\alpha}\le a_{2}\sum_{n = 2}^\infty \frac{\ln(\ln n)}{n^{\alpha-1}}\]
This proves the claim. So summarizing, we have the series in hand is convergent if $ \beta \ge 1$ and $ \alpha > 1+\beta$, whereas divergent if $ \alpha \le 1+\beta$. Now we need a way to tackle the case when $ \alpha > 1+\beta$ and $ 0 < \beta < 1$.

[quote="s.tringali"]How would you formalize this? [/quote]
See [url=http://mathworld.wolfram.com/StieltjesIntegral.html]Stieltjes Integral[/url]

[quote="s.tringali"] And this? :huh: I know $ \pi(x) \approx \frac{x}{\ln(x)}$, when $ x \in [2,+\infty[$, that meaning (at least, in my notations) there are absolute constants $ c_{1}, c_{2}\in \mathbb{R}$ such that $ c_{1}\frac{x}{\ln(x)}\le \pi(x) \le c_{2}\frac{x}{\ln(x)}$. But then!? If $ f, g: [0,+\infty[ \to \mathbb{R}$ are defined taking $ f(x) = \frac{x}{2}$ and $ g(x) = f(x)+\sin(x)$, as $ x$ ranges over $ [0,+\infty[$, then clearly $ g(x) \approx f(x)$. Yet $ g^\prime(x) \not\approx f^\prime(x)$. :wink:[/quote]
You are absolutely correct!  I found it easier to (hopefully) correct the earlier post above.  Please tell me your thoughts.  Also for $ \beta=1$, I had the same proof in mind.  Thanks again!

[quote="baz"]For $ 0< \beta <1$, so that for any such fixed $ \beta$
$ \sum_{n = 1}^\infty \sum_{p \in \mathbb{P}}\frac{v_{p}(n!)^\beta}{n^\alpha}\approx \sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right) \approx \sum_{n = 1}^\infty \frac{1}{n^{\alpha-1}\log n}$
which converges iff $ \alpha>1$. [/quote]
I think I should have been more careful and said iff $ \alpha>1+\beta$ (as you pointed out).

[quote="baz"]
\[ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}=\int_{2}^{n}\frac{d(\pi(x))}{x^\beta}\]
[/quote]
Yet I don't realize how you can settle this! :?:

But actually I'm quoting from your last edit, and despite of your corrections I can't find out a way to prove the first equality you set, when writing - but consider here in Italy it's a quarter past one in the morning, so I'm not even aware of myself :lol: - that
\[ \forall \beta \in \,]0, 1[ \;\wedge\;\forall n \in \mathbb{N}: \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}= \int_{2}^{n}\frac{\pi(x)}{x^\beta}, \]
which in fact is apparently false if $ n = 2$. Let me know, I'm just a little in confusion.

s.tringali,

I am not sure if you saw this: [url=http://mathworld.wolfram.com/StieltjesIntegral.html]Stieltjes Integral[/url].  But, if you are not familiar with this type of integral (ubiquitous in analytic number theory) (ubiquitous=shows up everywhere  :lol: cool english word for you, feel free to share italian), then all I can say is that in general the integral takes values only at its jumps.  I.e. notice that $ \pi(x)$ is a step function which jumps up 1 at each prime number.  Essentially is whats happening is the Stieltjes integral below only takes values at the jumps, so that in general for a real-valued, bounded function $ f(x)$:
\[ \sum_{p \in \mathbb{P}(n)}f(p) = \int_{2}^{n}f(x) d(\pi(x)). \] Notice we are differentiating with respect to $ \pi(x)$ not $ x$ (not a Riemann integral).  Get some sleep and grazie.

[quote="baz"]s.tringali,

I am not sure if you saw this: [url=http://mathworld.wolfram.com/StieltjesIntegral.html]Stieltjes Integral[/url].  But, if you are not familiar with this type of integral (ubiquitous in analytic number theory) (ubiquitous=shows up everywhere  :lol: cool english word for you, feel free to share italian), then all I can say is that in general the integral takes values only at its jumps.[/quote]
O.K., I'll try to go deeper in the topic of Stieltjes integral and come back to talk again over the problem, thank you so much. :) Now, just as a margin note, "ubiquitous" is a word of Latin derivation, so you can imagine we have got it in our dictionary... :D In fact, we say "ubiquo", which sounds quite similar to the English. So said, goodnight. :)

[quote="s.tringali"]Now, just as a margin note, "ubiquitous" is a word of Latin derivation, so you can imagine we have got it in our dictionary...In fact, we say "ubiquo", which sounds quite similar to the English. So said, goodnight.[/quote]  :blush: Thanks and goodnight.

[b]MILD CORRECTION[/b]
Again see [url=http://mathworld.wolfram.com/StieltjesIntegral.html]Stieltjes Integral[/url].  Notice that $ \pi(x)$ is a step function which jumps up 1 at each prime number.  Essentially whats happening is the Stieltjes integral below only takes values at the jumps, so that in general for a real-valued, bounded function $ f(x)$:
\[ \sum_{p \le n}f(p) = \int_{1}^{n}f(x) d(\pi(x)). \] or \[ \sum_{p \le n}f(p) = \int_{2}^{n}f(x) d(\pi(x))+f(2) \]
(as above where $ f(x)=1/x^\beta$, I should have added $ 1/2^\beta$ to obtain
\[ \sum_{p \le n}\frac{1}{p^\beta}= \int_{2}^{n}\frac{d\pi(x)}{x^\beta}+1/2^\beta~ ). \]
Either way this is a mere constant, yet this also concedes that we need $ \alpha\ge \beta+1$.  Indeed, recall $ \sum_{n = 1}^\infty \left(\frac{1}{n^{\alpha-\beta}}\cdot \sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}\right)$. The inner sum is what I claimed plus a constant, hence for that constant we require $ C\sum_{n = 1}^\infty \frac{1}{n^{\alpha-\beta}}$ to converge.

[quote="baz"][...] in general the integral takes values only at its jumps.  I.e. notice that $ \pi(x)$ is a step function which jumps up 1 at each prime number. Essentially what's happening is the Stieltjes integral below only takes values at the jumps, so that in general for a real-valued, bounded function $ f(x)$:
\[ \sum_{p \in \mathbb{P}(n)}f(p) = \int_{2}^{n}f(x) d(\pi(x)). \]
Notice we are differentiating with respect to $ \pi(x)$ not $ x$ (not a Riemann integral).  Get some sleep and grazie.[/quote]
I've just finished to read a simple good survey about Riemann-Stieltjes integrals, and actually (almost) everything you wrote in the previous posts is plain. :) Indeed, I have a couple of remarks for you: firstly, I think it should be $ \sum_{p \in \mathbb{P}(n)}f(p) = \int_{2}^{n+\epsilon}f(x) d(\pi(x))$, where $ 0 < \epsilon < 1$, rather than $ \sum_{p \in \mathbb{P}(n)}f(p) = \int_{2}^{n}f(x) d(\pi(x))$ (in fact, the last would entail an absurd, whenever $ n$ is a prime and $ f: [2, n] \mapsto \mathbb{R}$ is bounded and [i]positive[/i], that is exactly the case in hand with us). In the second place, I think you dropped some constants somewhere, which on the contrary it's necessary to take in account.

[quote="baz"]Here is what I thought to be true for $ 0 < \beta <1$:
\[ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}=\int_{2}^{n}\frac{d(\pi(x))}{x^\beta}\]
[/quote]
Indeed, I think the correct form to fix your arguments is to write that
\[ \sum_{p \in \mathbb{P}}\frac{1}{p^\beta}= \int_{2}^{n+\epsilon}\frac{d(\pi(x))}{x^\beta}. \]
where $ \epsilon \in \;]0, 1[$ and $ n\ge 2$ ($ n \in \mathbb{N}$) are fixed. Btw, note the presence of $ \epsilon$ is not really significant, since $ \pi(\cdot)$ is right continuous: actually, it serves only to formalize everything. :wink: Now we come to the constant I was talking over a few lines above. Indeed, integrating by parts gives
\[ \int_{2}^{n+\epsilon}\frac{d(\pi(x))}{x^\beta}= \frac{\pi(n+\epsilon)}{(n+\epsilon)^\beta}-\frac{1}{2^\beta}+\beta \int_{2}^{n+\epsilon}\frac{\pi(x)}{x^{\beta+1}}dx. \]
So recalling the prime number theorem, there exists positive constants $ \kappa,\kappa_{1}, \kappa_{2}\in \mathbb{R}$ (depending uniquely on $ \beta$) such that
\[ \kappa_{1}\left(\frac{(n+\epsilon)^{1-\beta}}{\ln(n+\epsilon)}+\beta \int_{2}^{n+\epsilon}\frac{dx}{x^\beta \ln(x)}\right) \le \kappa+\sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}+\le \kappa_{2}\left(\frac{(n+\epsilon)^{1-\beta}}{\ln(n+\epsilon)}+\beta \int_{2}^{n+\epsilon}\frac{dx}{x^\beta \ln(x)}\right), \]
and consequently (just taking the limit with $ \epsilon \to 0^{+}$):
\[ \kappa_{1}\left(\frac{n^{1-\beta}}{\ln n}+\beta \cdot J(n) \right) \le \kappa+\sum_{p \in \mathbb{P}(n)}\frac{1}{p^\beta}+\le \kappa_{2}\left(\frac{n^{1-\beta}}{\ln n}+\beta \cdot J(n)\right), \]
where (for future reference) $ J(n) = \int_{2}^{n}\frac{dx}{x^\beta \ln(x)}$. Now I can figure out the way neither to kill that $ \kappa$ (which would be useful) nor to give an asymptotic estimation of $ J(n)$ as you did.

[quote="baz"]
\[ \int_{2}^{n}\frac{d(\pi(x))}{x^\beta}= \frac{\pi(x)}{x^{\beta}}\bigg|_{2}^{n}+\beta \int_{2}^{n}\frac{\pi(x) dx}{x^{1+\beta}}\approx \frac{n^{1-\beta}}{\log n}+\beta \int_{2}^{n}\frac{dx}{x^{\beta}\log x}\]
where breaking the last integral into two parts $ \int_{2}^{n^{1-\beta}/\log n}$ and $ \int_{n^{1-\beta}/\log n}^{n}$ yields that $ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}\approx_\beta \frac{n^{1-\beta}}{\log n}$.[/quote]
In particular, I don't understand this part of your reasoning. It's clear we need somehow a second constant to balance the effect of $ \kappa$, and I believe it's only the foregoing integral that can provide it for sure. Well, now I'm going to bed, as I feel like I'm saying a lot of nonsense... :D So goodnight and let me know. :wink:

It took me some time to write the previous post... :blush: Maybe one day my English won't be so bad and then... ta-dah! :D In the meanwhile...

[quote="baz"]Essentially whats happening is the Stieltjes integral below only takes values at the jumps, so that in general for a real-valued, bounded function $ f(x)$: [...]
\[ \sum_{p \le n}f(p) = \int_{2}^{n}f(x) d(\pi(x))+f(2) \]
[/quote]
Well, adding up that constant has no apparent explanation and doesn't seem even to make much sense. E.g., it would entail $ \sum_{p \le 4}f(p) = 2f(2)+f(3)$, that is generally false in total evidence. :wink:

[quote="s.tringali"] Indeed, I have a couple of remarks for you: firstly, I think it should be $ \sum_{p \in \mathbb{P}(n)}f(p) = \int_{2}^{n+\epsilon}f(x) d(\pi(x))$, where $ 0 < \epsilon < 1$, rather than $ \sum_{p \in \mathbb{P}(n)}f(p) = \int_{2}^{n}f(x) d(\pi(x))$[/quote]
I think there is some confusion with the upper and lower limits. I think it should be
\[ \sum_{p \le n}f(p) = \int_{2^{-}}^{n}f(x) d(\pi(x)) = \int_{2}^{n}f(x) d(\pi(x))+f(2) \]  I also addressed this issue in my previous post (maybe you did not see it?).

[quote] Btw, note the presence of $ \epsilon$ is not really significant, since $ \pi(\cdot)$ is right continuous[/quote] Exactly!  If you plot this graph, notice that the integral does not take value at 2 but does take value at the upper limit, b/c I think (this is why this integral is weird) if you don't jump then you don't take a value.  I am not sure how to formalize this other than the equality above.  This may be something good to ask in the analysis forum to make clear this issue.

[quote="baz"]
\[ \int_{2}^{n}\frac{d(\pi(x))}{x^\beta}= \frac{\pi(x)}{x^{\beta}}\bigg|_{2}^{n}+\beta \int_{2}^{n}\frac{\pi(x) dx}{x^{1+\beta}}\approx \frac{n^{1-\beta}}{\log n}+\beta \int_{2}^{n}\frac{dx}{x^{\beta}\log x}\]
where breaking the last integral into two parts $ \int_{2}^{n^{1-\beta}/\log n}$ and $ \int_{n^{1-\beta}/\log n}^{n}$ yields that $ \sum_{p\in\mathbb{P}(n)}\frac{1}{p^\beta}\approx_\beta \frac{n^{1-\beta}}{\log n}$.[/quote]
[quote="s.tringali"]In particular, I don't understand this part of your reasoning. It's clear we need somehow a second constant to balance the effect of $ \kappa$, and I believe it's only the foregoing integral that can provide it for sure. [/quote]
I may have to write something up or scan it.  Then, I can send it to you for your scrutinizing.

[quote="baz"]
I think there is some confusion with the upper and lower limits. I think it should be
\[ \sum_{p \le n}f(p) = \int_{2^{-}}^{n}f(x) d(\pi(x)) = \int_{2}^{n}f(x) d(\pi(x))+f(2) \]
I also addressed this issue in my previous post (maybe you did not see it?).
[/quote]
Oh, yes, I did! :) But in general this even lacks of consistency (as I tried to point out), since yielding
\[ f(2)+f(3) = \sum_{p \in \mathbb{P}(4)}f(p) = 2f(2)+f(3). \]
But I think all that mess with $ \epsilon$ is just a simple trick to take away any kind of troubles.

[quote="baz"]This may be something good to ask in the analysis forum to make clear this issue.[/quote]
It's not a bad idea at all! :D

[quote="baz"]
I may have to write something up or scan it.  Then, I can send it to you for your scrutinizing.[/quote]
Really, it would be better if you can write up everything on the forum. Then everyone interested could possibly give his own contribution or simply share his own advice. :wink:

[quote="baz"] I think there is some confusion with the upper and lower limits. I think it should be
\[ \sum_{p \le n}f(p) = \int_{2^{-}}^{n}f(x) d(\pi(x)) = \int_{2}^{n}f(x) d(\pi(x))+f(2)\][/quote]

[quote="s.tringali"]  Oh, yes, I did! Smile But in general this even lacks of consistency (as I tried to point out), since yielding...[/quote]
I think it should be
\[ f(2)+f(3) = \sum_{p \in \mathbb{P}(4)}f(p) = \int_{2}^{3}f(x) d(\pi(x))+f(2) = f(3)+f(2)\]
mainly because  $ \int_{2}^{3}f(x) d(\pi(x))=f(3)$ and [b]not [/b]$ f(2)+f(3)$.  So, how we define this integral is where the problem lies.  Try posting some type of question (in the forum) regarding the definition.  :) I hope this makes sense.

[quote="baz"]
I think it should be
\[ f(2)+f(3) = \sum_{p \in \mathbb{P}(4)}f(p) = \int_{2}^{3}f(x) d(\pi(x))+f(2) = f(3)+f(2) \]
mainly because  $ \int_{2}^{3}f(x) d(\pi(x))=f(3)$ and [b]not [/b]$ f(2)+f(3)$.  So, how we define this integral is where the problem lies.  Try posting some type of question (in the forum) regarding the definition.  :) I hope this makes sense.[/quote]
Ah, O.K. Finally I get your point! :) So we just write $ \sum_{p \in \mathbb{P}(n)}= \int_{2}^{n^{+}}\frac{d(\pi(x))}{x^\beta}+\frac{1}{2^\beta}$ and everything is in its place. Then
\[ \forall n \in \mathbb{N}, n \ge 2: \sum_{n \in \mathbb{P}(n)}\frac{1}{p^\beta}= \int_{2}^{n^{+}}\frac{d(\pi(x))}{x^\beta}+\frac{1}{2^\beta}= \frac{\pi(n)}{n^\beta}+\beta \int_{2}^{n^{+}}\frac{\pi(x)}{x^{\beta+1}}dx, \]
and hence there exist positive constants $ \kappa_{1}, \kappa_{2}\in \mathbb{R}$ such that
\[ \forall n \in \mathbb{N}, n \ge 2: \kappa_{1}\left(\frac{n^{1-\beta}}{\ln n}+\beta J(n) \right) \le \sum_{n \in \mathbb{P}(n)}\frac{1}{p^\beta}\le \kappa_{2}\left(\frac{n^{1-\beta}}{\ln n}+\beta J(n) \right), \]
where $ J(n) = \int_{2}^{n}\frac{dx}{x^\beta \ln x}$. Indeed, thinking of a way to avoid troubles stemming from the integral upper limit made me blind to the problems coming out from the lower one. I'm a real donkey... :P Now we need a way to deal with $ J(n)$. So what?

---------------------------------------

I made some progress. :) In fact, by an easy induction, one proves that
\[ \forall m \in \mathbb{N}: J(n) = \sum_{k=1}^{m-1}\frac{(k-1)!}{(1-\beta)^{k}}\left(\frac{n^{1-\beta}}{\ln^{k}(n)}-\frac{2^{1-\beta}}{\ln^{k}(2)}\right)+\frac{(m-1)!}{(1-\beta)^{m-1}}\int_{2}^{n}\frac{dx}{x^\beta \ln^{m}(x)}. \]
I hope this can help to get further advance.

How to deal with $ J(n) = \int_{2}^{n}\frac{dx}{x^\beta \log x}$ (I use $ \log$ to of course mean $ \ln$):
Let $ U=U(n)=\frac{n^{1-\beta}}{\log n}$, then $ J(n)=\int_{2}^{U}+\int_{U}^{n}.$
Now, first remember that $ U$ is the size of our main term.  So, we need only show that each of these error terms (each integral mentioned above) has size no bigger than a constant times $ U$.
The first satisfies $ 0<\int_{2}^{U}\le U\cdot\max\bigg\{ \frac{1}{x^\beta \ln x}\bigg\}\ll U$.
While the latter integral satisfies $ 0<\int_{U}^{n}\frac{dx}{x^\beta \ln x}\le \frac{1}{\log U}\int_{U}^{n}\frac{dx}{x^\beta}\le \frac{n^{1-\beta}}{(1-\beta)\log U}\ll_\beta U$ since $ \log U = (1-\beta)\log n-\log\log n \approx_\beta \log n$.
Note the dependence of $ \beta$ in the constant which is denoted by $ \ll_\beta$.  Moreover, we could have bound these integrals by something asymptotically smaller with a better choice of $ U$.   :wink:


Lines AB and CD are parallel to each other and 8 inches apart.
Point P is located BETWEEN the two parallel lines and 2 inches from line AB.

a) Describe the locus of points that are:

1) equidistant from lines AB and CD

2) 3 inches from point P

b) How many points satisfy BOTH conditions in part a above?

1)  the line that is four inches from each of the two lines
2)  the circle of radius 3, centered at point p
b)  Because point p is two inches from line AB, the line from part one is 2 inches from point p.  Because this is within the circle from part 2, they have two points of intersection.

wow, that was easy!(I mean the solution.)

It seems like I've seen this problem before.  (Or maybe it was just a similar one)  Do you know anything about this?

I think Interval posted something like 10 problems in this section and in HSB, all regarding this stuff... I'd suggest him not to propose problems which vary just for the numbers involved... it's a waste of time...
Then, if you can't figure out the analogies among these problems, then try to generalise them, as I already told you in a past post...  :roll:


geometry
3D geometry
sphere
algebra
function
domain
vector

I need help to show that any bounded domain $U\subset\mathbb{R}^{n}$ such that $\partial U$ is $C^{2}$ satisfies an external sphere condition.

Let $\vec n(x)$ be the exterior unit normal vector at $x\in\partial U$. Since $\partial U$ is $C^{2}$, the function $x\mapsto \vec n(x)$ is Lipschitz. Let $L$ be its Lipschitz constant. Set $\epsilon=1/(2L)$ and consider the sphere of radius $\epsilon$ centered at $a=x+\epsilon\vec n(x)$.


calculus derivative quadratics trigonometry parameterization inequalities algebra

I'm not sure i can post here, but i'd like to have alternative solutions:

A ball is trown from a hill of height h with an initial speed v(0) and an angle t relative to the horizontal.
Find the angle t which maximizes the range.

I'm wondering if there's a solution not implying difficult derivatives...

Just a question: isn't that angle the same that maximizes the range when the ball is thrown from the ground?

No, it's not 45...For example, it has been calculated that from the height of 1 m the angle which maximazes the range is 42.

The derivative method is standard and ugly. (First you solve a quadratic to get an expression for time, multiply by horizontal component of velocity and then maximise this expression). I really want to see if anyone has a non-derivative method.

[hide="hmmm....."]
the given problem is equivalent to maximising the distance a ball travels after it has gained a height $ h$ when thrown with initial speed $ v_{i} \equal{} \sqrt {v_{o}^{2} \plus{} 2gh} \equal{} \sqrt {\beta}$ by changing the angle from ground
clearly the ball traces back the same path when thrown from the the point where it lands
also for a given initial speed the total range of a ball when it ia at a certain height is maximised when it is thrown with an angle $ 45^{\circ}$ so the ball must be thrown to the left with the speed $ v_{i}$ and angle $ 45^{\circ}$.
so when it reaches height $ h$ the angle $ t$ which is the required angle is
$ \tan t \equal{} 1 \minus{} \frac {2gx}{\beta}$ where $ x$ is the horizontal distance covered
which is given by
$ x \equal{} \frac {\beta \minus{} \sqrt {\beta^{2} \minus{} 4\beta h g}}{2g}$
thus $ t \equal{} \tan^{ \minus{} 1}(\sqrt {1 \minus{} \frac {4hg}{\beta}})$
thus $ \boxed{t \equal{} \tan^{ \minus{} 1}(\sqrt {1 \minus{} \frac {4hg}{v_{o}^{2} \plus{} 2gh}})}$
clearly this holds true iff $ v_{o} \geq \sqrt {2gh}$
EDIT:i changed $ gh \rightarrow 2gh$ and added the last condition[/hide]

That's what i thought..but you can't be sure the ball reaches a certain height h with speed v(0) after being thrown frome the ground with an angle of 45.i
In few words, your result doesn't work if under the sqrt there's a negative value (try for h=30m, v(0)=20).
Anyway it would be a nice solution if we find the relation between the initial angle from the ground and the angle t at the height h. Maybe that's not possible..

well i was going to correct my solution
see the edited version :wink:
also what is the answer

ok..so we have a partial solution. I don't want to give up yet...maybe we should observe what happen changing the direction of the speed at the height h...we could point out a geometric figure..it's just a suggestion...i don't know if it works

If you use lagrange multipliers, you can avoid having to take derivatives of square roots =)
[hide]
Let $ x$ be the range. We know that it takes $ \frac {x}{v_0 \cos{t}}$ time to get there.

Therefore we have
\[ \minus{} \frac {g}{2} \left(\frac {x}{v_0 \cos{t}}\right)^2 \plus{} v_0 \sin{(t)} \left(\frac {x}{v_0 \cos{t}}\right) \plus{} h \equal{} 0
\]
or
\[ g(t,x) \equal{} \minus{} g x^2 \plus{} 2 x v_0^2 \sin{t} \cos{t} \plus{} 2 h v_0^2 \cos^2 {t} \equal{} \minus{} g x^2 \plus{} x v_0^2 \sin{2t} \plus{} h v_0^2 (\cos{2t} \plus{} 1) \equal{} 0
\]
We are thus trying to maximize $ f(t,x) \equal{} x$ under the condition $ g(t,x) \equal{} 0$. We have that $ [\frac {\partial f}{\partial t} \; \frac {\partial f}{\partial x}] \equal{} [0 \; 1]$, so we want $ \frac {\partial g}{\partial t} \equal{} 2 x v_0^2 \cos{2t} \minus{} 2 h v_0^2 \sin{2t} \equal{} 0$ or $ x \equal{} h \tan{2t}$.

Substituting this back into $ g(t,x)$, we have \[ \minus{} g h^2 \tan^2{2t} \plus{} h v_0^2 \sin{2t} \tan{2t} \plus{} h v_0^2 (\cos{2t} \plus{} 1) \equal{} \minus{} g h \tan^2{2t} \plus{} v_0^2 \sin{2t} \tan{2t} \plus{} v_0^2 (\cos{2t} \plus{} 1) \equal{} 0\] or
\[ g h \tan^2{2t} \equal{} v_0^2 \sin{2t} \tan{2t} \plus{} v_0^2 (\cos{2t} \plus{} 1)
\]
or
\[ \frac {gh}{v_0^2} \equal{} \cos{2t} \plus{} \frac {\cos^2{2t} \plus{} \cos{2t}}{\sin{2t}}
\]
Simplifying this, we get $ \frac {\cot^2{t} \minus{} 1}{2} \equal{} \frac {gh}{v_0^2}$.[/hide]

[b]@calc rulz[/b] by no calculus i think ummagumma meant a more physical solution for ex. see the case for which i solved above

Before closing the thread (there's no evident way to solve the general case), i suggest a quick solution for v>=sqrt(2gh). Basically, i use the same method pardesi showed. At the height h, the ball has a speed v(0), with an angle t relative to x axis. We can think the ball has been trown from the ground with a speed v(1) and an angle w relative to x (we can assume it if the parameters respect a condition).
Along the x axis, the motion is uniform, so:
v(1) cosw = v(0) cost                                                                       [i](a)[/i]
Along the y axis the motion is uniformly accelerated, i can use the following equation:
2gh = (v(1)*sinw)^2 - (v(0)*sint)^2
2gh = v(1)^2 - (v(1)*cosw)^2 - (v(0)*sint)^2
Using [i](a)[/i]:
2gh = v(1)^2 - v(0)^2
v(1)= sqrt(v(0)^2 + 2gh)
The maxime range for a point thrown from the ground is reached for an angle of 45.
I rewrite the [i](a)[/i], assuming w=45
sqrt(0.5*(v(0)^2 + 2gh)) = v(0)*cost
cost=sqrt(0.5*v(0) + gh/v(0))

t = arccos sqrt(0.5*v(0) + gh/v(0))                                                     [i](b)[/i]

It's possible calculate the angle t in this way when 0.5*v(0) + gh/v(0)<=1
We have the condition v(0)>=sqrt(2gh).

A non- calculus solution. Setting of the problem and notation slightly changed to avoid confusion. (If projection point is on a hill there is the complication  that particle can land on the hill. Use t to denote time)

Let a particle be projected from the edge of a vertical cliff height h with Speed v at an angle a to the horizontal and let the particle reach the ground at a horizontal distance x from the foot of the cliff after a time t.

If v is fixed and a is variable we want to find the value of a which maximises x.

Let the point of projection be at the origin of X-Y axes. Then the particle hits the ground at a point (x,-h). and

x=vtcosa  ..........(1)

-h =vtsina-$ \frac{gt^2}{2}$ ...............(2)

Elimination of a between these two equations leads to the quadratic equation in $ t^2$
$ \frac{g^2t^4}{4}$ - (gh+$ v^2$)$ t^2$ +$ h^2 + x^2$ =0 ...............(3)
This equation must have real roots. Therefore its discriminant must be $ \geq$ 0
That is   $ (gh + v^2)^2$ $ \geq$ $ g^2(h^2 + x^2)$ which leads to $ v^4 + 2v^2gh + g^2h^2$ $ \geq$ $ g^2h^2 + g^2x^2$ from which inequality we deduce that x cannot exceed $ x_{max}$ = $ \frac {{v(v^2 + 2gh)^{\frac{1}{2}}}$/g...........(4)

For this value of x the equation ...(3) has equal roots in $ t^2$ namely $ t^2$ =$ \frac {2(v^2 + gh)}_{g^2}$ giving

t=$ \frac {(\sqrt(2v^2 + 2gh))}_{g}}$ ........(5) as the time of flight when x is greatest.

Now substituting values of t and x from equations (4) and (5) into equation (1) we can find the   value of the angle  a  giving $ x_{max}$:

$ \frac {v\sqrt(v^2 + 2gh)}_{g}}$ =$ \frac {v\sqrt(2v^2 + 2gh)}_{g}}cosa$...........

I know of two other non-calculus solutions to this problem.Both methods give $ x_{max}$ fairly quickly  but determination of the corresponding value of a is a bit tricky.


Dear all!

I've searched this forum and the web, but with no luck!
I'm looking for a way to type a piecewise function in Latex, where each line is given a number, so I can give it a label and then make references to the exact line. I've tried arrays inside align, dcases and aligned, but I haven't succeded.

Is there someone, who might have a suggestion?

Cheers, Sabrina

When I search for "piecewise" in this forum, I find the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904834273&t=289984
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904834273&t=217238
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904834273&t=159550
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904834273&t=157124
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904834273&t=15365

I don't understand the rest of your question, though.

[code]\usepackage{empheq}
...
\begin{document}
\begin{empheq}[left={|x|= \empheqlbrace}]{align}
x &\text{ if } x \geq 0  \\
-x & \text{ if } x<0
\end{empheq}[/code]

Thank you very much!

I just have one more question that's based on the previous. I have four equations. I want to have a left bracket that spans the first three lines. I can do this using the code you just wrote. But then I want another bracket placed to the left of the other bracket. This bracket should span all four lines. I've tried to nest empheq inside each other, but that doesn't work?
I know that it's probably difficult :S

As empheq won't nest you will need to use the methods in [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] section 61.1 pages 103-104.

For example (you will need to adjust things consderably)
[code]\usepackage{bigdelim}
\usepackage{multirow}
...
\begin{document}
\newcommand{\addeqno}{\refstepcounter{equation}(\theequation)}
\begin{tabular*}{\linewidth-2cm}{llll@{\extracolsep{\fill}}l}
\ldelim\{{4}{0cm}[] &
 \ldelim\{{3}{0cm}[$|x|=$]
 & $x$ & if  $x > 0$  & \addeqno \\
 && $-x$  & if $x<0$  &\addeqno \\
 && $0$ & if $x=0$&\addeqno \\
 & $x=1$ &&& \addeqno \\
\end{tabular*}[/code]
gives

Thank you very much!!!!  :lol:

I modified the snip to the following:

[code]\begin{tabular*}{\linewidth-2cm}{llllr@{\extracolsep{\fill}}l}
\ldelim\{{5}{1cm}[{$(D_1)$}] &
 \ldelim\{{4}{0cm}[$(D'_1)$]
 & \hspace{1cm}$\frac{ \p^2 u }{ \p x^2 }+\frac{ \p^2 u }{ \p y^2 }=0$ & for\enskip  $0<x<a,\enskip 0<y<b$  && \hspace{1.55cm}\addeqno \\
 &&\hspace{1cm} $u(x,b) = 0$  & for\enskip $0\leq x\leq a$  &&\hspace{1.55cm}\addeqno \\
 &&\hspace{1cm} $u(0,y) =u(a,y)=0$ & for\enskip $0<y<b$&&\hspace{1.55cm}\addeqno \\
 &&\hspace{1cm} $u(x,0) = f(x)$ & for\enskip $0\leq x \leq a.$ && \hspace{1.55cm}\addeqno \\
\end{tabular*}[/code]


Hey guys!

As you may know, last summer, after returning from the IMO, I began working a photo-diary documenting my trip.

I have finally finished putting everything together. The new addition is the third part, but I have also made small changes to the exisiting parts.

Read it here: http://web.mit.edu/yufeiz/www/imo2006

I hope this will give you an idea of what the IMO experience is like from the eyes of a [i]contestant[/i].

Enjoy!

[img]http://web.mit.edu/yufeiz/www/imo2006/imopics/cl26.jpg[/img]

(Disclaimer: in order to stay true to the spirit, I tried to be as frank as possible in my writing. This includes making fun of certain people at times. I hope no one is actually going to be insulted.)

Great job, Yufei!  :thumbup:  Really enjoyed it.

I read all 3 parts of ur journal..
it's quite long..

but it was so fun!
 i really enjoyed it!
thnx!!

I should confess, it made me jealous.

Yufei, Where did you choose as the center when you inverted your diagram for problem 1, Incenter?

I don't remember. Probably that point (P I think?) that was defined by angles. Inversion changes the location of those angles.

[quote="princevegetam"]how can the chinese guy take a nap?!

is it really that easy for him?

i can't imagine... i mean he could at least use the time to check answers...[/quote]

No... some people are just too smart. But anywho, congrats Yufei!!!! The journal is awesome!!!

Wow the IMO experience is truly much more then a math contest; it is clearly seen in Yufei's blog. I really enjoyed it, as it was well written. It's nice to see how Yufei did not base his IMO experience on his result, I think that we can all learn from this.

Congratulations Yufei on all the success you have had. Wish you the best in the future.

Victoria provided me with some more photos. So I've made some additions to the journal (e.g. look at the section about with sleepy students in Bohinj).

[quote="billzhao"]Peng emailed me with the following reponse to the journal. I think you'll enjoy reading it:

[quote="Peng Shi"]However, I have a small complaint to make. In part 3, you wrote "Peng told me that while they were partying, they heard a bunch of police sirens, so everyone began to scramble. He said that he lost Farzin on the way and was somewhat concerned for his safety…Peng's story-telling abilities deserve an applause." But I was telling the truth. After coming back from the student party, which we didn't like because it was just a bunch of people sitting in a park, speaking Spanish, and drinking, we came back. On our way back to the residence, we saw a party of the guides in some pit on campus. They were gambling, smoking, and drinking orange juice mixed with vodka. After chatting with them for a while, and after Farzin accidentally drank what he thought was orange juice, we prepared to leave. Just then, some police officers came, yelling in Slovenian and waving their arms threateningly. We walked more quickly, and once we turned a corner, we decided to run. I ran faster than Farzin, and when I got back to the residence and turned my head, Farzin was nowhere in sight. So my story is completely true.

...

PS: It turned out that Farzin went back to the party and found that the guides have skillfully wooed the cops away and are continuing the party. So he kept partying with them for several more hours. [/quote][/quote]

hehe nice skills

Thank you so much, Yufei, for writing the journal! :clap2:  It is really interesting and fun to read. I see that IMO is an extremely exciting and fun event! :thumbup:

Now I want to go to IMO sooooooooooooooooooooooooooo bad!  :wow: I am going to train extremely hard to make it this year.

yes same! tHANKS FOR MOTIVATION YUFEI!!!! :wow:


geometry
3D geometry
number theory unsolved
number theory

[b] Problem 1[/b]  Find all solutions of $(x+y)^2=xy$ in integers

[b] Problem 2 (open problem ?) [/b] Find all solutions of $(x+y+z)^3=xyz$ in integers

Sorry for not obeying the rules and posting two problems in one topic, but they're strictly connected, and I just asking for the solution of problem 1 wheres I just would like to know if the problem 2 is solved by now ? (W. Sierpinski listed it in 1960 as unsolved) Or maybe someone will give solution for it   :D

I've realized that problem 1 is trivial - sorry for posting problems without lookind at them  :oops:

Still waiting for something about the second


[quote="236factorial"]What is the least multiple of 72 that has exactly 16 positive factors?[/quote]

[hide="Hint"]
Consider that $72=2*6^2$ and the next few multiples of it are just multiplying 1, 2, 3, and so on to it.

$2*6^2$

$2^2*6^2$

$2*6^2*3$

$2^3*6^2$
[/hide]

[hide]$72 = 2^3 * 3^2$ which means it has 12 factors.
multiplying that by 3 to get $2^3 * 3^3$ you can see that this number has 16 factors.
Therefore, the number is $2^3 * 3^3$ or $216$ [/hide]


If s(r) = 3e^(2r) + 1, determine the inverse of this function.

here's what i got, but it's been awhile since i learned inverses.  i figured i'd give my input anyway though. btw--im not good at LaTeX form, so i don't put it in such--sorry

f(x)=3e^(2x) + 1

y = 3e^(2x) + 1

x = 3e^(2y) + 1

x - 1 = 3e^(2y)

[u]x - 1[/u] = e^(2y)
...3

log [(x-1)/3] = log(e^(2y))

log [(x-1)/3] = 2y

[u]log [(x-1)/3][/u] = f(x)^-1
..........2

so that's the inverse i arrive at

solution looks good. Instead of $ \log$ I would use $ \ln$ since its a natural log (base e)

Also, with the latex, you just have to stick what you have between dollar signs
It's a whole lot easier than plain text. If you click on other people's LaTeX, you can see how it works.

For example

f(x)=3e^(2x) + 1

is simply
[code]$f(x)=3e^{2x} + 1$[/code]

$ f(x)\equal{}3e^{2x} \plus{} 1$


For positive $ a$, $ b$ and $ c$ prove that

$ 2>\frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}a}>1$

 :)

[quote="ElChapin"]For positive $ a$, $ b$ and $ c$ prove that

$ 2 > \frac {a}{a + b} + \frac {b}{b + c} + \frac {c}{c + a} > 1$

 :)[/quote]

$ \sum_{\cyc} \frac{a}{a+b} > \sum_{\cyc} \frac{a}{a+b+c} = 1$

Similar idea for $ 2$

Moreover, the inequalities, after a permuatation of the variables, are equivalent:

$ \frac {a}{c \plus{} a} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} > 1$
$ \frac {a}{c \plus{} a} \minus{} 1 \plus{} \frac {b}{a \plus{} b} \minus{} 1 \plus{} \frac {c}{b \plus{} c} \minus{} 1 > 1 \minus{} 3$
$ \minus{} \frac {c}{c \plus{} a} \minus{} \frac {a}{a \plus{} b} \minus{} \frac {b}{b \plus{} c} > \minus{} 2$
$ \frac {c}{c \plus{} a} \plus{} \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} < 2$

Other similar and easy to:

For $ a$, $ b$ and $ c$ sides of triangle(may be degenerated) prove that

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}<2$

 :)


$ \frac{a\minus{}b}{b^2\plus{}c^2}\plus{}\frac{b\minus{}c}{c^2\plus{}a^2}\plus{}\frac{c\minus{}a}{a^2\plus{}b^2}\geq0 \Longleftrightarrow$

$ a^5\plus{}b^5\plus{}c^5 \geq a^4b\plus{}b^4c\plus{}c^4a$ which is a well-known inequality.

$ \frac{2a\minus{}b\minus{}c}{2a(b\plus{}c)}\plus{}\frac{2b\minus{}c\minus{}a}{2b(c\plus{}a)}\plus{}\frac{2c\minus{}a\minus{}b}{2c(a\plus{}b)} \leq 0 \Longleftrightarrow$

$ a^3b^2\plus{}a^2b^3\plus{}b^3c^2\plus{}b^2c^3\plus{}c^3a^2\plus{}c^2a^3 \geq 2abc(ab\plus{}bc\plus{}ca)$ which is obvious by Muirhead.

Hence

$ \frac{a\minus{}b}{b^2\plus{}c^2}\plus{}\frac{b\minus{}c}{c^2\plus{}a^2}\plus{}\frac{c\minus{}a}{a^2\plus{}b^2}\geq0\geq$ $ \frac{2a\minus{}b\minus{}c}{2a(b\plus{}c)}\plus{}\frac{2b\minus{}c\minus{}a}{2b(c\plus{}a)}\plus{}\frac{2c\minus{}a\minus{}b}{2c(a\plus{}b)} >$

$ \frac{1}{10}\left(\frac{8a\minus{}5b\minus{}5c}{a(b\plus{}c)}\plus{}\frac{8b\minus{}5c\minus{}5a}{b(c\plus{}a)}\plus{}\frac{8c\minus{}5a\minus{}5b}{c(a\plus{}b)}\right)$  :wink:


if n is a prime number, what is the smallest composite number produced by n^2-n-1

[hide]8[/hide]

don't ask me how i did it cause i have no clue it was the first number that came to mind and it worked.

i know you can factor this but so far i only have (n+1)(n-1)-n
i doubt that will do any good

Uh 8 is definitely not right.. its a lot higher than that.

much, much higher. all i know is that 271 doesn't work... and nothing before that...

is it 1331? i did it by guess and check...  :oops:

Uh firstly, the question didn't ask for the smallest n, it asks for the smallest composite result. Second, yes 55 is composite, but 8 is not prime :P. Its slightly smaller than 1331.

1331 is too big but it works if you substitute in 13 which gives you 155

the answer is slightly smaller than 1331?

dum de dum. 155 is divisible by 5. Yep. Thats right.

yup, i made the same mistake too... :oops:

any solutions other than guess and check or did i miss it

Its impossible to have a non-guess and check solution with finding the first nonprime in a sequence.. otherwise you'd have a few nice results about finding primes.


One of Goldbach's conjectures states that every odd number greater than 5 can be expressed as the sum of three primes. What is the greatest possible product of three primes whose sum is 11?

Odd numbers adding up to 11 must be less than 11. Listing, we get:
2
3
5
7
9

It cannot be 9, as 11-9=2, and we need 3 primes, not 2, and 0 is not prime.
Brute Force, and we get 5,3,3
5*3*3=15*3=45

[quote="007math"][quote="kabi"]Hello
$ 45(5*3*3)$
Thank u.[/quote]

$ 5 \times 3 \times 3 \boxed{ = } 45$

[color=blue]
Fixed your $ \text{\LaTeX}$.
-Izzy
[/color][/quote]

$ 5 \times 3 \times 3=\boxed{45}$

[color=blue]
Fixed your $ \text{\LaTeX}$.
[/color]


probability
symmetry
integration
ratio
probability and stats

In a population, 10% have 1 story home, 50% have 2 story home and 40% have 3 story home, after speaking with 300 people, what is the chance that more than 40% live in a 3 story homes?
the answer is 0.5 but how to get to this? thanks!

First off, this is just binomial - narrow it down to just two possibilities: 3-story home (probability $ 0.4$) and not 3-story home (probability $ 0.6$.)

$ X$ is the number of respondents that said 3-story, out of 300. $ X$ is binomially distributed, and the mean is $ E(X) \equal{} 120.$ That would be 40% of 300. I see three ways to estimate $ P(X > 120),$ which is what is being asked for.

[b]First method: normal approximation to the binomial, quick and sloppy.[/b]

This distribution is approximately normal. A normal is symmetric about its mean. We're asking for the probability that a normal is greater than its mean. By that symmetry, that should be $ \frac12.$

[b]Second method: normal approximation to the binomial, but carefully this time.[/b]

The distribution has mean $ 120$ and standard deviation $ \sqrt {300\cdot .4\cdot.6} \equal{} \sqrt {72} \equal{} 6\sqrt{2}.$ For using continuous distributions to approximate inter-valued discrete distributions, it's best practice to split halfway between the integers. Write $ P(X > 120) \equal{} P(X > 120.5).$ If $ Z \equal{} \frac {X \minus{} 120}{6\sqrt{2}},$ we're asking for $ P\left(Z > \frac1{12\sqrt{2}}\right).$ We can look that up in a normal table, but $ \frac1{12\sqrt{2}}$ is small enough that I'm willing to use a simple estimate. For $ Z$ standard normal

$ P\left(0 < Z < \frac1{12}\right) \equal{} \frac1{\sqrt {2\pi}}\int_0^{\frac1{12\sqrt{2}}}e^{ \minus{} \frac {z^2}2}\,dz\approxeq\frac1{\sqrt {2\pi}}\int_0^{\frac1{12\sqrt{2}}}1\,dx \equal{} \frac1{24\sqrt {\pi}}\approxeq 0.0235.$

So based on that, we estimate $ P(X > 120)\approxeq 0.5 \minus{} 0.0235 \equal{} .04765.$

[b]Third method: the binomial probabilities directly.[/b]

Just by direct computation, $ P(X > 120) \equal{} 1 \minus{} P(X\le 120) \equal{} 1 \minus{} \sum_{k \equal{} 0}^{120}\binom{300}{k}(.4)^k(.6)^{300 \minus{} k}.$

Can we do that? The answer is yes. It's only 120 terms, there's an efficient scheme to to compute the terms recursively, and we all have computers. Let me fire up the spreadsheet - this should take about 2 minutes.

I get $ P(X > 120)\approxeq 0.47495.$

[b]Fourth statement:[/b]

There's still one approximation in that last answer - that we considered the result to be binomially distributed. This is exactly true only if our sampling procedure was to choose 300 people to survey with replacement. That is, pick a number at random to call. Then forget that we ever called that person, and pick a number at random to call - which might be the same one we've already called. If that happens, we just count them twice.

If we make sure we don't call the same person twice, then we're sampling with replacement, and if we're sampling with replacement, then the distribution isn't binomial, it's hypergeometric - and to know exactly what hypergeometric, we need to know the size of the whole population. We weren't given that information here. One plausible approximation is to assume that the total population is "large" - or at least much larger than 300. The limit as the size of the total population goes to infinity of a hypergeometric distribution (holding the ratios constant) is a binomial distribution.

Let X=1 if a person poled lives in a 3 storey home, 0 otherwise. then X~Bin(300,0.4). Use Normal approximation Y~N(120,72). Then P(Z>0.059)=0.4765 which is about 0.5. Hope I understood your question correctly.

Oops, you found my mistake, JL. 72, not 36. I'll go back and edit what I called my second method. But the third method is more accurate than that.


parameterization
vector
calculus
calculus computations

A missile is traveling along the curve $ \vec{r}t \equal{} t\vec{i} \plus{} t^2\vec{j} \plus{} t^3\vec{k}$. Dylan wants to shoot a laser so that the laser will be tangent to the missile at the point (2,4,8). Where should Dylan position himself on the ground (xy-plane)? You may assume that the earth is flat.

We have that $ \vec{r}(t)\equal{}\left< t,t^2,t^3\right>$, the so-called twisted cubic, and $ P\left( 2,4,8\right)$ which is generated by $ t\equal{}2$.

A tangent vector to the trajectory of the missile is $ \vec{T}(t)\equal{}\vec{r}\, ^{\prime}(t)\equal{}\left< 1,2t,3t^2\right>$. At the point of interest, $ \vec{T}(2)\equal{}\left< 1,4,12\right>$ . To position Dylan, construct parametric line parallel to $ \left< 1,4,12\right>$ passing through the point $ P\left( 2,4,8\right)$ : $ \vec{\ell} (t)\equal{}\vec{P}\plus{}t\, \vec{T}(2)\equal{}\left< 2\plus{}t,4\plus{}4t,8\plus{}12t\right>$ which intersects the xy-plane when $ t\equal{}\minus{}\textstyle{\frac{2}{3}}$ at the position $ Q\left(\textstyle{ \frac{4}{3}, \frac{4}{3}},0 \right)$, where Dylan should stand.


LaTeX
trigonometry
trig identities
cosine double angle
sine double angle

evaluate
(sorry i dont know latex, so this will look pretty ugly)
 2+( :sqrt: 4+( :sqrt: 16+ :sqrt: (256+...

you may like this problem :
[hide="HiddenProblem"]
Show that : $ \sin \frac{\pi}{10} = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{ 2 - \sqrt {2 + \cdots }}}}$
$ \sin \frac{\pi}{18} = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{ 2 + \sqrt {2 - \sqrt{2 + \sqrt{2 + \sqrt {2 - \cdots }}}  }}}}$

$\sin \frac{\pi}{34} = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{ 2 + \sqrt {2 + \sqrt{2 - \sqrt{2 + \sqrt {2 + \sqrt {2 + \sqrt{2 - \cdots } } }}}  }}}}$

$\cdots \cdots \cdots$

Can you see general pattern. ?  :blush:[/hide]

[quote="Hexaditidom"]I'm wondering, how did you guys simplify 2,4,16,256, to 1,1,1,1? I've always wondered if it was possible to simplify the root of a sum or difference.[/quote]
well, it's not that hard :)

$\sqrt {4}=2\sqrt {1}
\\
\sqrt {\sqrt {16}}=\sqrt {4\sqrt {1}}=2\sqrt {\sqrt {1}}
\\
etc. $

I took a similar approach

x = 2 + sqrt( 4 + sqrt( 16 + ...

x-2 = sqrt(4 + sqrt( 16 + ....

square both sides to get

x^2 - 2x + 4 = 4 + sqrt( 16 +...

x^2 - 2x = 2*sqrt(4 + sqrt(16 + ...

x^2 - 2x = 2(x-2)

x^2 - 6x +4 = 0

This gives x = 3 + sqrt(5) and x = 3 - sqrt(5). Previous posts confirm x = 3 + sqrt(5), but how do I know to eliminate the alternate solution?

Um the question is 2 + {Positive expression.}, and $3-\sqrt{5} < 2$, so we know to eliminate it.

$\displaystyle x = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{ 2 - \sqrt {2 + \cdots }}}}$
$\displaystyle x = \frac{1}{2}\sqrt{2 - \sqrt{2 + 2x}}$
$\displaystyle 4x^2=2-\sqrt{2+2x}$
$\displaystyle 16x^4-16x^2+4=2+2x$
$\displaystyle 8x^4-8x^2-x+1=0$

Now let's play with $\sin{\frac{\pi}{10}}$.
$\displaystyle \sin{\frac{\pi}{10}}=\cos{\frac{2\pi}{5}}$ since $\sin\theta=\cos(\pi-\theta)$
$\displaystyle \sin{\frac{\pi}{10}}=1-2\sin^2{\frac{\pi}{5}}$ by the cosine double angle identity.
$\displaystyle 1=\sin{\frac{\pi}{10}}+2\left(\sin{\frac{\pi}{5}}\right)^2$
$\displaystyle 1=\sin{\frac{\pi}{10}}+2\left(2\sin{\frac{\pi}{10}}\cos{\frac{\pi}{10}}\right)^2$ by the sine double angle identity
$\displaystyle 1=\sin{\frac{\pi}{10}}+8\left(\sin{\frac{\pi}{10}}\right)^2\left(\cos^2{\frac{\pi}{10}}\right)$
$\displaystyle 1=\sin{\frac{\pi}{10}}+8\left(\sin{\frac{\pi}{10}}\right)^2\left(1-\sin^2{\frac{\pi}{10}}\right)$
Let $\displaystyle x=\sin{\frac{\pi}{10}}$ and the expression becomes $\displaystyle 8x^4-8x^2-x+1=0$, so it is a root of $\displaystyle\frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{ 2 - \sqrt {2 + \cdots }}}}$.
Thus, $\displaystyle\sin{\frac{\pi}{10}=\frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{ 2 - \sqrt {2 + \cdots }}}}}$.

So is there a better way to do it? This process gets ugly with the other ones.


induction
inequalities
function
linear algebra
combinatorics unsolved
combinatorics

Can some body give a combinatorial proof for this equality:

$\sum_{k=0}^{n}{{3n}\choose{3k}}=\frac{1}{3}(8^{n}+2(-1)^{n})$

Induction might work, but this combinatorical proof sure is pretty.  It might have been posted before, but since I just figured it out I'll write it.  It uses De Mouivre's Theorem too.

$\displaystyle (1+cis(\frac{\pi}{3}))^{3n}=\displaystyle\sum_{k=0}^{3n}{3n\choose k} (cis(\frac{\pi}{3}))^k$
$=\displaystyle\sum_{k=0}^{3n}{3n\choose k} (cis{\frac{k\pi}{3})}$
$=\displaystyle\sum_{k=0}^n {3n\choose 3k}+\displaystyle\sum_{k=0}^{n-1}{3n\choose 3k+1}cis (\frac{\pi}{3})+\displaystyle\sum_{k=0}^{n-1} {3n\choose 3k+2} cis(\frac{2\pi}{3})$

Now we have that $\displaystyle\sum_{k=0}^{n-1}{3n\choose 3k+1}=\displaystyle\sum_{k=0}^{n-1}{3n\choose 3k+2}$ because $\displaystyle {n\choose k}={n\choose n-k}$.  Then the imaginary parts cancel, leaving

$=\displaystyle\sum_{k=0}^n {3n\choose 3k}-\frac{1}{2}\displaystyle\sum_{k=0}^{n-1} {3n\choose 3k+1}-\frac{1}{2}\displaystyle\sum_{k=0}^{n-1} {3n\choose 3k+2}$

Now, notice for the LHS that
$\displaystyle (1+cis(\frac{\pi}{3}))^{3n}=(\frac{1}{2}+\frac{i\sqrt{3}}{2})^{3n}=\displaystyle (cis(\frac{\pi}{6}))^{3n}=cis(\frac{n\pi}{2})=(-1)^n$

Now multiplying the whole inequality by two yields

$\displaystyle 2(-1)^n=2\displaystyle\sum_{k=0}^n {3n\choose 3k}-\displaystyle\sum_{k=0}^{n-1}{3n\choose 3k+1}-\displaystyle\sum_{k=0}^{n-1}{3n\choose 3k+2}$
$=3\displaystyle\sum_{k=0}^n {3n\choose 3k}-\displaystyle\sum_{k=0}^{3n}{3n\choose k}$
$=3\displaystyle\sum_{k=0}^n {3n\choose 3k}-8^n$

This finally gives the result after adding over $8^n$ and dividing by $3$.

I'm sorry, but your proof is completely algebraic...

An ideal proof would be a function, defined on almost every subset of $\{1, 2, \dots, 3n\}$, that has the following properties:

$f(f(f(S)))=S$

$|f(S)|-|S| = 1 (mod 3)$

There would be a few sets it was not defined on, which would correspond to the $(-1)^{n}$ term.

any ideas for such a function?


induction
LaTeX
number theory unsolved
number theory

We will prove that $ \boxed{a_n \equal{} (n \minus{} 1)! \plus{} n!} \ \ (1)$

$ (1)$ is equivalent to $ \boxed{na_n \equal{} (n \plus{} 1)!} \ \ (2)$

Indeed, $ na_n \equal{} n(n \minus{} 1)! \plus{} nn! \equal{} n! \plus{} nn! \equal{} (n \plus{} 1)n! \equal{} (n \plus{} 1)!$

$ (1)$ holds for $ n \equal{} 3: \ \ a_3 \equal{} 8 \equal{} 3! \plus{} 2!$

$ \star$ Notice that $ (1)$ holds also for $ n \equal{} 1$ and $ n \equal{} 2$

Suppose that $ (1)$ holds for any natural number $ < n$. Then, $ (2)$ also holds for any natural number $ < n$

So we have $ \boxed{(n \minus{} 1)a_{n \minus{} 1} \equal{} n!}$ and $ \boxed{(n \minus{} 2)a_{n \minus{} 2} \equal{} (n \minus{} 1)!}$

adding these two equations we get $ a_n \equal{} n! \plus{} (n \minus{} 1)!$

Now $ a_{2008} \minus{} a_{2007} \plus{} \cdots \plus{} a_2 \minus{} a_1 \equal{}$

$ (2008! \plus{} 2007!) \minus{} (2007! \plus{} 2006!) \plus{} \cdots \plus{} (4! \plus{} 3!) \minus{} (3! \plus{} 2!) \plus{} (2! \plus{} 1!) \minus{} (1! \plus{} 0!)$

$ \equal{} 2008! \minus{} 1$

How do you know $ a_n = n! + (n - 1)!$ ?
And also, is the following proof valid?

We will prove $ a_n = n! + (n - 1)!$ by induction.
{induction steps}
Proved.

Is this valid?

And last question: how do you make a box outside the $ \text{\LaTeX}$?

[quote="wangsacl"]How do you know $ a_n = n! + (n - 1)!$ ?[/quote]Very often, [url=http://www.research.att.com/~njas/sequences/]this[/url] site helps a lot  :D

[quote="wangsacl"]And also, is the following proof valid?

We will prove $ a_n = n! + (n - 1)!$ by induction.
{induction steps}
Proved.

Is this valid?[/quote]I cannot understand you

[quote]And last question: how do you make a box outside the $ \text{\LaTeX}$?[/quote] I don't know either

[quote="wangsacl"]
And also, is the following proof valid?

We will prove $ a_n \equal{} n! \plus{} (n \minus{} 1)!$ by induction.
{induction steps}
Proved.

Is this valid?
[/quote]

It is valid wangsacl. Pontios I think that his question is whether we can make the conjecture that $ a_n\equal{}n!\plus{}(n\minus{}1)!$ and prove it using induction or not. Of course given the hypothesis of the problem we can make any conjecture which we prove in anyway we want.

Alexandros

So, we can say that $ a_n\equal{}n!\plus{}(n\minus{}1)!$, and prove it by induction, without explaining how to get $ a_n\equal{}n!\plus{}(n\minus{}1)!$ ?

Of course.  It's perfectly valid to do a few computations, guess at the general formula, and prove rigorously by induction that it must be the case.


$ x$ and $ y$ are randomly chosen such that $ 0<x,y<1$. What is the probability that the leading nonzero digit of $ \frac{x}{y}$ is 2?

[hide]The restriction means that $ \frac{1}{5}\le\frac{x}{y}<\frac{3}{10}$.

This simplifies to:

$ \frac{10}{3}<\frac{y}{x}\le5$
$ \frac{10}{3}x<y\le5x$

So,

$ y>\frac{10}{3}x$ and $ y\le5x$

With the restriction of $ x<1$ and $ y<1$, the area enclosed by these is,

$ \frac{3}{20}\minus{}\frac{1}{10}\equal{}\frac{1}{20}$[/hide]

[quote="Carbon57"][hide]The restriction means that $ \frac {1}{5}\le\frac {x}{y} < \frac {3}{10}$.

This simplifies to:

$ \frac {10}{3} < \frac {y}{x}\le5$
$ \frac {10}{3}x < y\le5x$

So,

$ y > \frac {10}{3}x$ and $ y\le5x$

With the restriction of $ x < 1$ and $ y < 1$, the area enclosed by these is,

$ \frac {3}{20} \minus{} \frac {1}{10} \equal{} \frac {1}{20}$[/hide][/quote]

what about 2.asdfasdfsadfsadfsadf and 20.asdfasdfsadfsdaf and 200.asdfasdfsadfsdafsad and 2000.asdfasdfsadf (and like 2001.asdfasdfads and so forth), you get my point i think.

yes sry but it is much tougher than that. I have the answer but none of my approaches quite match it. Here it is

[hide="Answer"]$ \frac {4}{27}$ [/hide]

maybe it has something to do with Benford's law?

So modify the approach a little to account for these cases: [hide]calculate the probability that $ 2\cdot 10^n \leq \frac {y}{x} < 3\cdot 10^n$ in terms of $ n$ (maybe handle the cases $ n \geq 0$ and $ n < 0$ separately) and sum the resulting series over all integers $ n$.  The computations are not much harder than what Carbon57 did.[/hide]

Using the same method, what happens if you replace "2" with "1" or any other digit?  What if you change to base $ b$?


calculus
function
trigonometry
derivative
probability
LaTeX
real analysis

An important part of knowing calculus (analysis) is knowing your examples. So I propose a topic in which we do things backward. In each case, the example is the answer - but if it is the answer, what was the question? That is, for what proposed statement is this function an example or (more likely) a counterexample?

I'll start with one example; there will be others. Feel free to contribute your own. My first example:

$f(x)=x^2\sin\left(\frac1x\right)+\frac x4$ (And of course, $f(0)=0.$)

This function has discontinuous derivate. What are additional properties, Kent?

I remember, you asked for differentiable function $f$ s.t. $|f'|$ is not a summable function. Did someone give an example?

What additional properties? How about this:

If $f$ is differentiable everywhere and $f'(a)>0,$ does there exist a neighborhood of $a$ on which $f$ is an increasing function? No; see above example.

The example Myth asked for: $g(x)=x^2\sin(x^{-10}).$ The exponent is overkill, but why not make what's going on unmistakeable?

Moving along: a classic that everyone should know:

If $x=\frac pq$ in lowest terms, $f(x)=\frac1q.$

If $x$ is irrational, $f(x)=0.$

And what things does Riemann's function provide an example or counterexample for?

Hint. I wrote in my previous reply: "Opposite question"  :D

A second thing the Riemann function is good for: It's a Riemann-integrable function with infinitely many discontinuities.

I know the answer to your "opposite question" but will leave it for someone else.

[quote="Kent Merryfield"]I know the answer to your "opposite question" but will leave it for someone else.[/quote]
Do you have famous book "Examples and Counterexamples"?
Unfortunately, I don't. :(
Is there some electronic version of this book? Please!

A set of classroom examples from a course I was just teaching this last term:

Differentiable everyhere (on the interval in question) with a bounded derivative implies Lipschitz.

Lipschitz implies uniformly continuous.

Uniformly continuous implies continuous.

What are the examples you would use to show that the converses of these statements are false?

We just need to slightly transformate your very first example.  :)
Right?

[quote="Myth"]Opposite question:
Does there exist function, which is continuous in all rational points and discontinuous in all irrational points?[/quote]

I think this has been discussed before: the set of continuity points of a function is a $G_{\delta}$ set (the intersection of a sequence of open sets), and it's not hard to show that $\mathbb Q$ is not a $G_{\delta}$ set.

Although I would go ahead and use $f(x)=\sin\left(\frac1x\right)$ on $(0,1)$ for the third example (continuous but not uniformly continuous), I had far simpler examples in mind for the first two.

Changing the subject slightly:

$a_{m,n}=\frac{m}{m+n}$

$f(x,y)=\frac{2xy}{x^2+y^2},$  $f(0,0)=0$

$f(x,y)$ is continuous in $(0,0)$ with respect to $x$ and with respect to $y$, but $f$ is discontinuous in $(0,0)$.

So... sorry then :D
 Let Kent explain it!!

A function of bounded variation can be written as the difference of two nondecreasing functions, and from that we may obtain that a function of bounded variation must be differentiable almost everywhere. You could probably directly show that the Weierstrass nowhere differentiable function is not of bounded variation, but in any case, it can't be of bounded variation because it's not differentiable almost everywhere.

Any good gallery of examples should have the Cantor-Lebesgue "devil's staircase" function. I initially thought that harazi was trying to give us this example, although that turned out to not be the case. One construction:

Let $f_0(x)=x\,\mathrm{on}\,[0,1].$
Let $f_1(x)$ be the piecewise linear continuous function with vertices $(0,0),\left(\frac13,\frac12\right),\left(\frac23,\frac12\right),(1,1).$
Let $f_2(x)$ be the piecewise linear continuous function with vertices $(0,0),\left(\frac19,\frac14\right), \left(\frac29,\frac14\right), \left(\frac13,\frac12\right), \left(\frac23,\frac12\right) \left(\frac79,\frac34\right), \left(\frac89,\frac34\right), (1,1).$

Continue this process inductively. In each step, replace each sloping segment by three segments, with the function constant on the middle third and with the height on that middle third the midpoint height of the previous segment.

Let $f(x)=\lim_{n\to\infty}f_n(x).$ (The sequence converges uniformly.)

What properties does this function have? Do you have your own variant on how you would describe the construction?

(Incidentally, this function is monotone and bounded, hence of bounded variation, and it is differentiable almost everywhere.)

[quote="Kent Merryfield"]

(Incidentally, this function is monotone and bounded, hence of bounded variation, and it is differentiable almost everywhere.)[/quote]

 But it fails the Newton - Leibnitz fomular , which is

\[\int_{0}^{1} f'(x)dx = f(1) - f(0)\]

 Right? :D

Well, since $f'(x)=0$ wherever it exists (which is almost everywhere), you would have to say that it fails that formula.

You can also say that $f$, restricted to the Cantor set, maps the Cantor set onto $[0,1].$

Aha!! :lol:
 If I remeber right, every compact subset of a metric space is infact a continuous image of Cantor set. Fascinating!! This is what the function look like after a few steps

[img]http://users.forthnet.gr/ath/jgal/math/figs/CantorF0.GIF[/img][img]http://users.forthnet.gr/ath/jgal/math/figs/CantorF1.GIF[/img]
[img]http://users.forthnet.gr/ath/jgal/math/figs/CantorF2.GIF[/img][img]http://users.forthnet.gr/ath/jgal/math/figs/CantorF3.GIF[/img]

[img]http://users.forthnet.gr/ath/jgal/math/figs/CantorF4.GIF[/img][img]http://users.forthnet.gr/ath/jgal/math/figs/CantorF5.GIF[/img]

Nice, dickchimney. Can you do any drawings of the Billingsley function in post #25 of this topic?

To be honest, those drawings are not mine :blush:
 And I haven't fully understood the Billingsley's function yet.
 But, anyway
 If $x = 0,a_1a_2...$ in binary expansion then
    \[g(x) = \sum_{i=0}^\infty{a_ip^{i-k(i)+1}(1-p)^{k(i)-1}}\]
 Where $k(i)$ is the number of $1'$s in the first $i$  digits
If the above fomular is correct then :
- $f$ is continous
- $f$ is not differentiable at points $\frac{k}{2^m}$
- $f$ is differentiable at points $\frac{1}{2^n-1}$ for sufficiently large $n$

[quote="dickchimney"]If the above fomular is correct then : [/quote]
It is correct.

[quote]- $f$ is continous [/quote]
Continuous and also [b]strictly[/b] increasing.

[quote]- $f$ is not differentiable at points $\frac{k}{2^m}$ [/quote]
True, but that is not the whole set of points at which $f$ is not differentiable. There is in fact an uncountable set of points of nondifferentiability. What your points do show is that this set is dense.

[quote]- $f$ is differentiable at points $\frac{1}{2^n-1}$ for sufficiently large $n$[/quote]
As a monotone bounded function, $f$ in fact must be differentiable almost everywhere.

There is a relationship between $\lim_{i\to\infty}\frac{k(i)}i$ (in your notation) and differentiability. Can you find that relationship, and can you say what the derivative would be if it does exist?

There is an $\epsilon(p)$ depends on $p$ s.t

- If $A =\lim\sup_{i}{\frac{k(i)}{i}}$ is smaller then $\epsilon(p)$  then at that point $f'$ exists and equals to $0$
- If $A = \epsilon(p)$ then $f'$ exists at that point iff $\lim_{i\rightarrow \infty}{\frac{k(i)}{i}} = \epsilon(p)$ and if does it must be $1$
- $f'$ doesn't exist at any other point.
Right?

 Do you have the formula for the Hilbert's curve?

[quote]There is an $\epsilon(p)$ depends on p s.t

- If $A =\lim\sup_{i}{\frac{k(i)}{i}}$ is smaller then $\epsilon(p)$ then at that point $f'$ exists and equals to 0 [/quote]
Think of the sequence $\{a_i\}$ of binary digits as a sequence of coin flips.

Of course, $\epsilon(p)>\frac12$ and the probability (that is, measure) of the set on which $A<\epsilon(p)$ equals 1 by the weak law of large numbers. (The strong law of large numbers says that $A=\frac12$ almost surely.)

The set on which $A>\epsilon(p)$ is uncountable and (since it contains all of of the dyadic rationals) dense.

However, there are no points at which the derivative equals 1. For each point $x$ we can find the sequences of dyadic rationals that approach $x$ from each side and we can compute the slope of secant line through those two points. The sequence of successive slopes is a sequence of partial products of an infinite product where each factor is either $2p$ (which is < 1) or $2(1-p)$ (which is > 1). If an infinite product converges to a number other than zero, the factors must tend to 1 - which they don't here. Thus the infinite product cannot converge and the only possible value of the derivative is zero.

In case you weren't following all of those details, we have this:

The Billingsley function is a strictly increasing continuous function from $[0,1]$ onto $[0,1]$ that is differentiable almost everywhere but the derivative is zero wherever it exists. The set of points at which the derivative does not exist is an uncountable dense set of measure zero.

The graph of the function is a self-similar fractal, as can be seen from the functional equations satisfied by $g(x):$

If $0\le x\le \frac12,\,g(2x)=\frac1pg(x).$

If $\frac12\le x\le1,\,g(2x-1)=\frac1{1-p}(g(x)-p).$

[quote="Kent Merryfield"]In case you weren't following all of those details, we have this:

[/quote]

 Of course I was following those!! ;) Just a little upset that I didn't think of the coin flips ( the fact that I had heard about )
 Talking about fractal, can you in some way present the Hilbert curve or Peano curve, which fills the plane ,here?I don't know much about those things!! :D

I'd like to introduce a book here, "COUNTEREXAMPLES IN ANALYISIS" by B. R. Gelbaum and J. M. H. Olmsted, Holden-Day Inc, 1964.

Now comes the Peano curve, which fills the plane!!

 Let $t= (t_1t_2t_3....)_3$ be a ternary representation of $t \in [0,1]$. Then the Peano's curve is expressed as :
\[P(t) = (x(t),y(t))\]\[ x(t) = ((t_1(k^{t_2}t_3)(k^{t_2+t_4}t_5)(k^{t_2+t_4+t_6}t_7)...)_3)\]\[ y(t) = ((k^{t_1}t_2)(k^{t_1+t_3}t_4)(k^{t_1+t_3+t_5}t_6)...)_3)\] where the operator $k$ is defined as \[kt_j=2-t_j,%Error. "blank" is a bad command.
t_j= 0,1,2   \]and $k^lt_j$ is the $l'$th element in the sequence ${kt_j, k(kt_j),k(k(kt_j)),...}$ ( with $k^0t_j = t_j$ )

 Here below is the geometric generation of the curve

[img]http://www.appmath.osaka-wu.ac.jp/~takashiw/mathematica/geometry/Peanocurve/Peano1c.GIF[/img][img]http://www.appmath.osaka-wu.ac.jp/~takashiw/mathematica/geometry/Peanocurve/Peano2c.GIF[/img]
[img]http://www.appmath.osaka-wu.ac.jp/~takashiw/mathematica/geometry/Peanocurve/Peano3c.GIF[/img][img]http://www.appmath.osaka-wu.ac.jp/~takashiw/mathematica/geometry/Peanocurve/Peano4c.GIF[/img]

 Can you recognize the pattern ?? ;)


limit
real analysis
real analysis solved

Consider $(a_n)$ satisfies $\exists M>0$ such that $|a_n|<M \forall n$

and $a_n+a_{n+1} \geq 2a_{n+2} \forall n$

Prove that: $(a_n)$ is converge

[quote="silviu"]To prove that $(a_n)$ converge we just need po prove the monotony of  $(a_n)$,because $a_n<M$.
Having $a_n+a_{n+1} \ge 2a_{n+2},$
and also $a_{n-1}+a_n \ge 2a_{n+1},$
And for $b_n=a_{n-1}+2a_n$ we have that $b_n \ge b_{n+1}$
From here is easy to prove that we want .[/quote]
From here is easy to prove that $b_n$ is converge;but... $a_n$???

You are right the probleme is not finish.
To prove that $(a_n)$ converge using that $(b_n)$ converge we must prove the next probleme.

Having $(a_n),(b_n)$ and $x \in (0,1)$ and $b_n=a_{n+1}+xa_n, \forall n \in N$.
Prove that: $(a_n)$ converge $ \Leftrightarrow $ $(b_n)$ converge.

I have a solution for this probleme but is very long and not nice,so [u]I would someone would give me a nother solution more beatiful[/u]

That's not difficult, $b_{n}=a_{n+1}+xa_{n}\Longrightarrow%
%TCIMACRO{\dsum \limits_{i=0}^{n}}%
%BeginExpansion
{\displaystyle\sum\limits_{i=0}^{n}}
%EndExpansion
\left(  -x\right)  ^{i}b_{n-i}=a_{n+1}-\left(  -x\right)  ^{n+1}a_{0}$, we have a more general conclusion: if $%
%TCIMACRO{\dsum \limits_{n=1}^{\infty}}%
%BeginExpansion
{\displaystyle\sum\limits_{n=1}^{\infty}}
%EndExpansion
A_{n}$ converges absolutely, $\lim B_{n}$ exists, then $\lim%
%TCIMACRO{\dsum \limits_{i=1}^{n}}%
%BeginExpansion
{\displaystyle\sum\limits_{i=1}^{n}}
%EndExpansion
A_{i}B_{n-i}\ $ exists.

We can aslo use the following approach:
From the problem condtion it easy to get the sequence $a_{n+2}-a_{n}$ converges, so we may write $a_{n+2}-a_{n}\to{A_1}$ (1), then using the fact that $a_n+2a_{n+1}$ converges we also may write $a_n+2a_{n+1}\to{A_2}$ (2). So summing up (1) and (2) we obtain $a_{n+2}+2a_{n+1}\to{A_1+A_2}$, but from the other side it is known that $a_{n+1}+2a_{n+2}\to{A_2}$, thu summing the last two limits we get $a_{n+1}+a_{n+2}\to\frac{2A_1+A_2}{3}$  (3), substracting the last two we also get $a_{n+1}-a_{n+2}\to{A_1}$ (4). So, summing (3) and (4) we obtaing our result $a_{n+1}\to\frac{5A_1+A_2}{6}$


function
induction
algebra unsolved
algebra

[quote="mps"]Find all functions $f : N0 \Rightarrow N0$ such that
$mf(n) + nf(m) = (m + n)f(m^2 + n^2)$; m; n $\in$ $N0$[/quote]
lemma 1: $f(2^k)=f(0)=c$
x is special if and only if $x_1, .....,x_k$ exist such that $\sqrt{{x_i}^2+{ {x_{i+1}}^2}}$ is integer( $x=x_0$ ) And $x_k$ is a power of 2.
lemma 2: If t is special then 2t too. It prove using the lemma 1.
lemma 3: if a and b are specials then ab too. It prove using the lemma 1.
by induction in n: f(4n+r)=c; $n \in N_0 \ and \ r=0,1,2,3$ using the lemmas and the Pythagorean triples.


I mess around with source codes. See if you can spot the difference!

u wrote that there were a [b][u]LOT[/b][/u] of users on and registered and u changed everything on the sidebar

nice

You also changed all the numbers where it showed how many guests, hidden, and registered were online.
Interesting. Wonder what you could do with other websites (http://www.usa.gov would be a great place to start).

It's a secret. If I told you, it wouldn't be a secret now would it. :wink:

(Go to View -> Source and change some things, then apply changes. I would use Opera for this.)

I would use opera. An older version of IE works too.

dude I wish someone would fix the default opera skin. It looks good, but it belittles the aops icon on the tabs.  :o


Solve :

$ 1\blacktriangleright$ $ 2^{2x}\plus{}2\equal{}3\cdot 2^{x^2}$ .

$ 2\blacktriangleright$ $ 2^{3x}\plus{}2^{x\plus{}1}\equal{}3\cdot 2^{2x^2}$ .

$ 3\blacktriangleright$ $ 2^{4x}\plus{}2^{2x\plus{}1}\equal{}3\cdot 2^{3x^2}$ .

My solution is certainly a very naive approach.
  1) for x=0 and x=1 the equation is satisfied.
  2)  for x=o the equation is satisfied.
For x>0  we can see that $ 3\mid \left ( 2^{3x} \plus{} 2^{x\plus{}1}\right )$ (it can be verified with modulo 3;($ x\plus{}1$ and $ 3x$ being of different eveness)
 but then$ 3\mid \left ( 2^{3x} \plus{} 2^{x\plus{}1}\right )  < 2^{2^x}$
    3) equation is satisfied for x=0 and x=1 .
Now, it is necessary to find some generalisation.


Besides the recent school round, is there any other way to retrieve practice school round tests?

Google is your friend. One place you could go is...well, just search Minnesota Mathcounts and go to their resources page. If you absolutely can't find them, you can pm me your e-mail and I can e-mail all of my tests to you.

edit: actually it says your e-mail...but anyway if you can't find any I can e-mail them to you (just tell me).

i think you can get it off the mathcounts website.

yeah on the MC website u can get all the tests with questions and answers
i bet ur coach has a copy 2

[quote="math154"]Google is your friend. One place you could go is...well, just search Minnesota Mathcounts and go to their resources page. quote]

I think the website is http://www.mnspe.org/mathcounts/coach.htm
 :maybe: ive been using it even though I dont think im supposed to but....

It's illegal to post these tests online, but not to download them. :P


Does there exist positive integers $a_{1}<a_{2}<\cdots<a_{100}$ such that for $2 \le k \le 100$, the greatest common divisor of $a_{k-1}$ and $a_{k}$ is greater than the greatest common divisor of $a_{k}$ and $a_{k+1}$?

Yes, even if we replace $ 100$ by any finite positive integer.

Let $ n$ be the length of the sequence (the problem has $ n \equal{} 100$). Let $ p_1, p_2, p_3 , p_4, \ldots p_n$ be any strictly increasing sequence of primes. Now define the sequence to be $ a_i \equal{} p_1^{n\minus{}i} p_i^i$ for all $ 1 \leq i \leq n$. Then the greatest common divisor of $ a_k$ and $ a_{k\plus{}1}$ is $ p_1^{n\minus{}k\minus{}1}$ which is decreasing, and $ a_i < a_{i\plus{}1}$ because $ p_i < p_{i\plus{}1}$.


number theory
relatively prime
combinatorics proposed
combinatorics

Let $ S$ be a set such that
(a) Each element of $ S$ is a positive integer no greater than 100;

(b) For any two  dinstinct element $ a$ and $ b$ in $ S$ , there exist an element $ c$ in $ S$ such that $ a$ and $ c$ , $ b$ and $ c$ are relatively prime; and

(c) For any dinstinct two element $ a$ and $ b$ in $ S$ there exist third element $ d$ in $ S$ such that $ a$ and $ d$, $ b$ and $ d$ are not relatively prime to each other

Determine the maximum number of elements $ S$ can have

maximum number of elements $ S$ is $ 100$!!!!!! because $ S\subset\{1,2,....,100\}$
i think you want the minmum.

@aviateurpilot: Did you mean "the maximum number of elements in $ S$ is $ 100$" ? The condition (c) is not satisfied for the pair $ 97$ and $ 89$.


This question was in the Classroom Marathon in this forum.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42194&postorder=asc&start=740]www.artofproblemsolving.com/Forum/viewtopic.php?t=42194&postorder=asc&start=740[/url]
It is question 249. No one has the answer for a day now, so I decided to post it in this thread. The problem is:
A square with the largest area possible is inscribed in an equilateral triangle. What is the ratio of the side of the square to the side of the triangle?

[hide]
Say that the side of the square is $1$.
It is the longer leg of a 30-60-90 triangle (all of the angles of the triangle are 60)
So the shorter leg is $\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$
For each side of the equilateral triangle, that $\frac{\sqrt{3}}{3}$ is there twice, and also the side of the square.
So it is $2+\frac{\sqrt{3}}{3}=\frac{6+\sqrt{3}}{3}$
The ratio is thus
$\frac{1}{\frac{6+\sqrt{3}}{3}}=\frac{3}{6+\sqrt{3}}$ and simplified that is
$\frac{6-\sqrt{3}}{11}$[/hide]

Assuming the square has a side a, then the side of the equilateral triangle is: a[1 + 2/ sqrt (3)]. Therefore, the ratio should be sqrt (3)/[sqrt (3) + 2].[/list][/code]

biffandoc, u said
[quote]For each side of the equilateral triangle, that$\frac{\sqrt{3}}{3}$  is there twice, and also the side of the square.[/quote]
the side of the square is only on one side.

[hide="i get"]$2\sqrt{3}-3$
im in hurry, so can't show process.[/hide]

You got exactly the same answer as I showed here. Actually, this problem is not very clear, because you can have only one inscribled square in an equilateral triangle.

u have the same answer, except it's not simplified. u can't have a radical in the denominator.

[quote="qibojing"]How can I delete a message I posted?[/quote]

you can only delete your message if no one's posted after you yet.


function
algebra
functional equation
absolute value
algebra open

Question 1: Function f is defined as following: f : R->R, $f^{2}(x+y) = f^{2}(x) + f^{2}(y)$ .
Find: f(2006) - f(2005) + f(2004) - .... + f(2) - f(1)

A. 1
B. $2006^{2} - 1$
C. 0
D. $2006^{2} - 2005$
Question 2: Evaluate: $lim( \frac{x-|tanx|}{|sinx|-x} ) when x\to 0^{+}$

A. -1
B. 1
C. -2
D. 2

Question 3: If x,y are positive integers, then the number of ordered pairs (x,y) which satisfy $3x + 5y = 501$ is 34 ?

Question 4: let n = ( 1+1376)(2+1376)(3+1376)...(1376+1376) what is the largest value of k such that:$2^{k}|n$

A. 688
B. 689
C. 1375
D. 1376

Eastlake Math Competition?   :D

I did not like the first question.  If you assume $f^{2} (x) = ( f(x) )^2$ then you will see that $f^{2} (x) = x, 0$, but $f(x) = \pm \sqrt{x}$ can't satisfy the equation regardless of how piecewise you define it (simply put $y = -x$ and the functional equation can't be true).  Hence the only solution is $f(x) = 0$ and the answer is $\boxed{C}$.  But I feel that Amir wasn't sufficiently clear in his notation  :(

Second question.  First note the absolute value signs do nothing because we are approaching from the positive direction.  Next, Taylor series or l'Hopital's rule gives $\frac{ \frac{x^3}{3} }{ - \frac{x^3}{3!} } = -2$, so the answer is $\boxed{C}$.  I think.

Third question.  The smallest $y$ that produces a solution is $y = 3, x = 162$.  New solutions can be generated by increasing $y$ by $3$ and decreasing $x$ by $5$.  Note that $162 = 5(32) + 2$, and rewriting $x = 5k + 2$ we see that we can have $0 \le k \le 32$ producing $\boxed{33}$ solutions.

Fourth question.  First, count how many multiples of $2$ there are.  There are $\frac{1376}{2} = 688$ of them.  Next, realize that the multiples of $4$ will be evenly distributed across the multiples of $2$ - that is, every other multiple of $2$ is also a multiple of $4$, so the number of multiples of $4$ (which add extra powers of $2$ to the product) is $\frac{688}{2} = 344$.  Repeating this process for higher powers, our answer is $688 + 344 + 172 + 86 + 43 + 21 + 10 + 5 + 2 + 1 = \boxed{1372}$... strange.  Was there a $NOTA$ option on this question?

For the last one...I'm too lazy, but what about image of svilupping the multiplication and choose the element with the least power of 2? Can we use log and means?


Let $ AA_1,BB_1,CC_1$ be bisectors in triangle $ ABC$, and let $ G_1,G_2,G_3$ be the centroids of triangles $ AB_1C_1,BC_1A_1,CA_1B_1$, respectively. Prove that the lines $ AG_1,BG_2,CG_3$ are concurrent.

Let $ M$, $ N$, $ P$ be the midpoints of the segments $ B_1C_1$, $ C_1A_1$, $ A_1B_1$. Then, the lines $ A_1M$, $ B_1N$, $ C_1P$ are the medians of triangle $ A_1B_1C_1$ and thus concur. The lines $ AA_1$, $ BB_1$, $ CC_1$ are the angle bisectors of triangle $ ABC$, and thus also concur. Hence, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]the Cevian Nests theorem[/url], the lines $ AM$, $ BN$, $ CP$ also concur. But the lines $ AM$, $ BN$, $ CP$ are the lines $ AG_1$, $ BG_2$, $ CG_3$ (because the line $ AM$ is the $ A$-median of triangle $ AB_1C_1$ (since $ M$ is the midpoint of $ B_1C_1$), and thus passes through the centroid $ G_1$ of this triangle, and thus coincides with the line $ AG_1$; similarly for the lines $ BN$ and $ CP$). Thus, the lines $ AG_1$, $ BG_2$, $ CG_3$ concur.

  darij


[quote="A+MATH"]1. Albatross is at a fair to see cows and chickens, he counts 6 heads and and 22 legs. How many of each animal were there?

2. Frankinfueter mutated sheep they now have 5 legs. Frankinfueter is counting mutated sheep and dogs. He counts 5 heads and 23 legs. How many of each animal where there?

3. There are three legged frogs and 6 legged cats. If Bobert counts 10 heads and 45 legs how many of each animal are there?

Bonus: Mutated cows have two heads and three leg. Eustine is counting mutated cow and normal horses. He counts 12 heads and 33 legs.

P.S. Frankinfueter is the only one that enjoys mutating animals  :D
P.P.S No animals were harmed in the makings of this problem.[/quote]

[hide="Solutions and Answers"]If all the animals stood on two feet, there would be 12 feet on the ground, since 2*6=12. That means that 10 feet are in the air, so five cows and one chicken are seen by albatross.

I'm guessing that "mutated sheep and dogs" means that only the sheep were mutated. If all the animals stood on four legs, there would be 20 legs on the ground, and three sheep legs in the air. So there are three sheep and two dogs.

If all the animals stood on three legs, there would be 30 legs on the ground and 15 in the air. So there are five cats and five frogs.

We chop every cow in half, producing cows with one head and 1.5 legs. If all the animals stood on 1.5 legs, there would be 18 legs on the ground and 15 in the air. Those 15 belong to 6 horses. So there are 6 half-cows and 3 whole cows.

No animals were harmed in production of the above solutions.[/hide]

Yea, your answers should be correct.  :)

*Looks at problem and throws away KFC chicken leg*

[hide="Solution"]c + 2d = 19      6c + 2d = 44    5c = 25   c = 5

5 chickens and 7 ducks[/hide]

[quote]1. Albatross is at a fair to see cows and chickens, he counts 6 heads and and 22 legs. How many of each animal were there? [/quote]

[hide]all of them have at least 2 legs.
2*6=12
22-12=10
10/2=5 cows
6-5=1 duck[/hide]

[quote]1. Albatross is at a fair to see cows and chickens, he counts 6 heads and and 22 legs. How many of each animal were there?

2. Frankinfueter mutated sheep they now have 5 legs. Frankinfueter is counting mutated sheep and dogs. He counts 5 heads and 23 legs. How many of each animal where there?

3. There are three legged frogs and 6 legged cats. If Bobert counts 10 heads and 45 legs how many of each animal are there?

Bonus: Mutated cows have two heads and three leg. Eustine is counting mutated cow and normal horses. He counts 12 heads and 33 legs.

P.S. Frankinfueter is the only one that enjoys mutating animals
P.P.S No animals were harmed in the makings of this problem.[/quote]

[hide="1"]5cows/1chicken[/hide]
[hide="2"]3mutated sheeps/2dogs[/hide]
[hide="3"]5three legged frogs/5, 6legged cows[/hide]
[hide="bonus"]3mutated cows/6 normal horses[/hide]

[hide]1. Let x be number of cows and y number of chickens. So x+y=6 and 4*x+2*y=22 then 2*x=22-12=10. x=5,y=1
2. Let x be number of sheep and y number of dogs. Then x+y=5 and 5*x+4*y=23. So x=3 and y+2
3. Let x be number of frogs and y number of cats. Then x+y=10, 3*x+6*y=45. 3*y=45-30=15. So y=5, x=5[/hide]

Yup those are the right equations and answers.

[hide]
1. 5 cows, 1 chicken
2. 3 "mutated" sheep, 2 dogs
3. 5 frogs, 5 cats
Bonus: 3 cows, 6 horses[/hide]

[quote="dk2"][quote="Treething"]On a visit to the KFC factory, Treething counted 19 chopped-off heads and 44 dismembered feet next to the bird baking machine.

KFC chickens have 1 head and 6 legs to make more drumsticks per chicken.
KFC ducks have 2 heads and 2 legs because Colonel Sanders was drunk.

How many chickens and ducks were cooked?[/quote]

Extension:
The next time dk2 comes to the KFC factory, the birds have mutated once more:
KFC chickens have 1 head and 3 legs
KFC ducks have 7 heads and 25 legs

dk2 counts 49 legs and 15 heads. How many of each type were cooked.

HARD BONUS:
By the time dk2 reaches his house, he only remembers that there were 49 legs. Can he figure out how many birds of each type were cooked. What if he only remembers that there are 15 heads?[/quote]
[hide]
8 chickens, 1 duck
[/hide]

KFC chickens have 1 head and 3 legs
KFC ducks have 7 heads and 25 legs

HARD BONUS:
By the time dk2 reaches his house, he only remembers that there were 49 legs. Can he figure out how many birds of each type were cooked. What if he only remembers that there are 15 heads?

[hide]
Well, if he only remembers that there were 49 legs, then he can figure out how many birds of each type were cooked. That's because 3 doesn't go into 49, so you know that only chickens could not have been cooked. Next, you try if you had one duck. Subtract 25 from 49 and you get 24, which is the total number of legs that the cooked chickens had to have had. 3 goes into 24, so you know that combination works. There could have been 8 chickens and 1 duck cooked for a total of 49 legs. Lastly, see if there could have been more than one duck or all ducks. Two ducks would make 50 legs, which is more than 49, so that wouldn't work. Hence, all ducks wouldn't work, either. Because there's only one combination that works, you can figure out how many of each animal were cooked.

However, if he only remembers that there were 15 heads, there could be multiple combinations of the number of chickens and ducks were cooked. There could be 15 chickens, 8 chickens and 1 duck, or 1 duck and two chickens. Therefore, you cannot figure out how many birds of each type were cooked if you only know that there were 15 heads.

That was a bit lengthy and confusing... ;)
[/hide]


thats basically asking 1/8 of 240, which is 30


calculus
integration
function
parameterization
trigonometry
complex analysis

Evalute $ \int_\gamma z sin(z^2) dz$ where $ \gamma$ is the unit circle.

Is $ f(z)\equal{}z\sin(z^2)$ analytic on the unit circle...actually in a larger region than the unit circle?

[quote="JRav"]Is $ z$ just a variable?  Because if so, it is a pretty simple integral.[/quote]

Would the answer be $ \frac{\minus{}cos(z^2)}{2}$ and how does the unit circle $ \gamma$ factor into the integral?

No, this is not the right answer. If this isn't obvious (but it should be because this function is analityc on $ |z|\equal{}1$ so the answer must be $ 0$) you can perform a direct calculation. The parametrization would be $ z\equal{}e^{it},\ t\in [0,2\pi]$ and we have

\[ \int_{\gamma} z\sin (z^2)dz \equal{} i\int_0^{2\pi} e^{2it}\sin(e^{2it})dt.\]

The last one you can calculate easly as a "normal" integral.

Are you sure that should read $ e^{2it}$ outside of the sin?  I believe it should just be $ e^{it}$.

I'm pretty sure, the parametrization formula is as follows:

If $ \varphi(t)$ is parametrization of $ \gamma$, $ t\in[a,b]$, then

\[ \int_{\gamma} f(z)dz \equal{} \int_a^b f(\varphi(t))\varphi'(t)dt\]

In our case $ f(z) \equal{} z\sin(z^2)$, and $ \varphi(t) \equal{} e^{it},\ \varphi'(t)\equal{}ie^{it}$. So $ f(e^{it}) \equal{} e^{it}\sin(e^{2it})$, and multiplied by $ ie^{it}$ we get what has been written.

Maybe.  I don't know complex analysis yet so I didn't know that formula so you very well might be right.

The function $ f(z) \equal{} z \sin(z^2)$ is an entire function. Therefore, for any simple closed contour $ \gamma$,

$ \int _ {\gamma} z \sin(z^2)\,dz \equal{} 0$,

according to Cauchy-Goursat theorem.

The same could be said about $ \int _ {\gamma} \frac{\sin(z^2)}{z}\,dz$. How about $ \int _ {\gamma} \frac{\sin(z)}{z^2}\,dz$?

You mentioned $ \minus{}\frac{\cos(z^2)}{2}$; that's the antiderivative. This is a definite integral, so we actually evaluate that antiderivative at the two endpoints... which are the same point, so the integral is zero.

That's the fundamental theorem for line integrals, expressed in the complex form.

[quote="jmerry"]You mentioned $ \minus{} \frac {\cos(z^2)}{2}$; that's the antiderivative. This is a definite integral, so we actually evaluate that antiderivative at the two endpoints... which are the same point, so the integral is zero.

That's the fundamental theorem for line integrals, expressed in the complex form.[/quote]

Are you sure about this, what about $ z^{\minus{}1}$ over the unit-circle?

As Carcul says, there are no poles of the function inside the contour hence the integral is zero.

$ z^{\minus{}1}$ doesn't have an antiderivative on the whole unit circle.

[quote="Carcul"]
The same could be said about $ \int _ {\gamma} \frac {\sin(z^2)}{z}\,dz$. How about $ \int _ {\gamma} \frac {\sin(z)}{z^2}\,dz$?[/quote]

It is true that $ \int _ {\gamma} \frac {\sin(z^2)}{z}\,dz\equal{}0$, but it is because of the Cauchy integral formula and not because of analyticity like before. (Well, break into the case where $ \gamma$ contains $ 0$ and where it doesn't it is analytic).

For the other let $ f(z)\equal{}\sin z$, then $ f'(z)\equal{}\cos(z)$ and so by Cauchy formula $ \int _ {\gamma} \frac {\sin(z)}{z^2}\,dz\equal{}2\pi if'(0)\equal{}2\pi i$ which is not zero (if $ \gamma$ contains $ 0$).

$ \frac{\sin(z^2)}{z}$ is analytic at zero, and in fact entire. You can easily write its power series centered at zero. One might as well treat removable singularities as removed.

That's right. Writing the Laurent series of $ f(z) \equal{} \frac{\sin(z^2)}{z}$ about z = 0 we get

$ f(z) \equal{} z \minus{} \frac{z^5}{3!} \plus{} \frac{z^9}{5!} \minus{} ...$,

where we can see that f(z) has a removable singularity at z = 0 - however, defining now f(0) = 0 the function f(z) is therefore analytic at zero and in fact entire.