calculus
integration
logarithms
function
calculus computations

Evaluate $ \int_{\minus{}a}^{a} \frac{1}{e^x \plus{} 1} \, dx$.

First, find:

$ \int \frac{1}{1\plus{}e^x}\,dx$

this can be done using the substitution:

$ t\equal{}e^x\plus{}1$

which gives us:

$ \int\frac{1}{t(t\minus{}1)}\,dx\equal{}\int\frac{1}{t\minus{}1}\,dx\minus{}\int\frac{1}{t}\,dx\equal{}\ln(t\minus{}1)\minus{}\ln(t)\equal{}$

$ \equal{}\ln(e^x)\minus{}\ln(1\plus{}e^x)\equal{}x\minus{}\ln(1\plus{}e^x)$

Now, plugging in the boundaries gives us:

$ \int^{a}_{\minus{}a} \frac{1}{1\plus{}e^x}\,dx\equal{}a\minus{}\ln(1\plus{}e^a)\minus{}(\minus{}a\minus{}\ln(1\plus{}e^{\minus{}a}))\equal{}2a\plus{}\ln\left(\frac{1\plus{}e^{\minus{}a}}{1\plus{}e^a}\right)\equal{}$

$ \equal{}2a\plus{}\ln(e^{\minus{}a})\equal{}a$

$ \frac {1}{1 \plus{} e^x} \plus{} \frac {e^x}{1 \plus{} e^x} \equal{} 1,$

$ \int \frac {1}{1 \plus{} e^x}dx \equal{} \int \left(1 \minus{} \frac {e^x}{1 \plus{} e^x}\right)dx \equal{} x \minus{} \int \frac {(1 \plus{} e^x)'}{1 \plus{} e^x}dx \plus{} C.$

or remark that $ y\equal{}1$ is the even function, ....

[quote="t0rajir0u"]Evaluate $ \int_{ \minus{} a}^{a} \frac {1}{e^x \plus{} 1} \, dx$.[/quote]
Let $ \tau$ be your integral. After substitution $ u\equal{}\minus{}x,$

$ \tau  \equal{} \int_{ \minus{} a}^a {\frac{1}
{{e^x  \plus{} 1}}\,dx}  \equal{} \int_{ \minus{} a}^a {\frac{{e^x }}
{{e^x  \plus{} 1}}\,dx} .$

So $ 2\tau  \equal{} \int_{ \minus{} a}^a {dx}  \equal{} 2a.$ The rest follows.

I have shown the following solution several times.

Let $ f(x)\equal{}\int_{\minus{}x}^x \frac{1}{e^t\plus{}1}\ dt,$ $ f'(x)\equal{}\frac{1}{e^x\plus{}1}\plus{}\frac{1}{e^{\minus{}x}\plus{}1}\equal{}1.$ Since $ f(0)\equal{}0,$ yielding $ f(x)\equal{}x.$

The answer is $ f(a)\equal{}a.$

[color=blue]Practice[/color]

$ \boxed{1}$ [url=http://www.mathlinks.ro/viewtopic.php?t=185941]created by kunny[/url]

Evaluate $ \int_{ \minus{} 2008}^{2008} \frac {f'(x) \plus{} f'( \minus{} x)}{2008^x \plus{} 1}\ dx.$

$ \boxed{2}$ [url=http://www.mathlinks.ro/viewtopic.php?t=185867]from Greece forum[/url]

Αφού προτείνατε κάποια ολοκληρώματα να βάλω και εγώ ένα που μου άρεσε.

Να υπολογιστεί το $ \int^1_{ \minus{} 1}\frac {x^2 \plus{} 1}{2^x \plus{} 1}\,dx.$

$ \boxed{3}$ [url=http://www.mathlinks.ro/viewtopic.php?t=18738&search_id=1710069534]1979 Gakusyuin University[/url]

Evaluate the following definite integral.

\[ \int^1_{ \minus{} 1} \frac {x^2}{1 \plus{} e^x} dx.
\]

This is my 4- th solution.

$ \int_{ \minus{} a}^a \frac {1}{1 \plus{} e^x}\ dx \equal{} \left[x\cdot \frac {1}{1 \plus{} e^x}\right]_{ \minus{} a}^a \plus{} \int_{ \minus{} a}^a x\cdot \frac {e^x}{(1 \plus{} e^x)^2}\ dx$

$ \equal{} a\left(\frac {1}{1 \plus{} e^a} \plus{} \frac {1}{1 \plus{} e^{ \minus{} a}}\right) \plus{} 0 \equal{} a.$

Because $ \frac {xe^x}{(1 \plus{} e^x)^2}$ is odd function. The justification is left for you.

[quote="milin"]First, find:

$ \int \frac {1}{1 \plus{} e^x}\,dx$

this can be done using the substitution:

$ t \equal{} e^x \plus{} 1$

which gives us:

$ \int\frac {1}{t(t \minus{} 1)}\,dx \equal{} \int\frac {1}{t \minus{} 1}\,dx \minus{} \int\frac {1}{t}\,dx \equal{} \ln(t \minus{} 1) \minus{} \ln(t) \equal{}$

$ \equal{} \ln(e^x) \minus{} \ln(1 \plus{} e^x) \equal{} x \minus{} \ln(1 \plus{} e^x)$

Now, plugging in the boundaries gives us:

$ \int^{a}_{ \minus{} a} \frac {1}{1 \plus{} e^x}\,dx \equal{} a \minus{} \ln(1 \plus{} e^a) \minus{} ( \minus{} a \minus{} \ln(1 \plus{} e^{ \minus{} a})) \equal{} 2a \plus{} \ln\left(\frac {1 \plus{} e^{ \minus{} a}}{1 \plus{} e^a}\right) \equal{}$

$ \equal{} 2a \plus{} \ln(e^{ \minus{} a}) \equal{} a$[/quote]
shouldn't those dxs be dts?


function
inequalities
limit
triangle inequality
real analysis
real analysis unsolved

Let $ m(E)>0$.  Prove that $ L^{\infty}(E)$ is not separable.

I assume that $ E$ is meant to be a subset of $ \mathbb{R},$ or $ \mathbb{R}^n$ or in any case a "non-atomic" measure space. After all if $ E\equal{}\{1,2,\dots,n\}$ with counting measure, then $ L^{\infty}(E)$ is just $ \mathbb{R}^n$ with one of its many equivalent norms.

The following lemma would help:

For any $ r$ such that $ 0<r<m(E),$ there exists a measurable set $ E_r\subset E$ such that $ m(E_r)\equal{}r.$

Then once we have this lemma, we just look at the functions $ \mathbf{1}_{E_r}.$

Isn't the proof something along the lines that $ \left\|\chi_{E_{r}}\minus{}\chi_{E_{s}}\right\|_{\infty}\equal{}1$?, or something like that?  But why does that necessarily help?

Do you have a proof of the lemma?

You can't have an uncountable discrete set and be separable; if the minimum distance is $ 1$, you can only approximate countably many of those points within $ \frac12$ by your proposed countable "dense" set.

[quote="JRav"]Isn't the proof something along the lines that $ \left\|\chi_{E_{r}} \minus{} \chi_{E_{s}}\right\|_{\infty} \equal{} 1$?, or something like that?  But why does that necessarily help?[/quote]

There are uncountably many $ E_\alpha$s.  There are only countably many functions in your hypothetical countable dense subset.  So one member of this countable dense subset must be close to at least two $ E_\alpha$s, say, with indices $ s$ and $ r$.  Now use the triangle inequality and the equality $ \left\|\chi_{E_{r}} \minus{} \chi_{E_{s}}\right\|_{\infty} \equal{} 1$ to obtain a contradiction if you choose $ \varepsilon$ too small.

[quote="JRav"]Do you have a proof of the lemma?[/quote]

It's a good exercise to attempt.

Let $ D$ be the countable dense set and let $ f\in D$.  Assume that $ f$ is close to $ \chi_{E_{r}}$, say within $ 1/2$.  Then
$ 1\equal{}\left\|\chi_{E_{r}}\minus{}\chi_{E_{s}}\right\|_{\infty}\leq\left\|\chi_{E_{r}}\minus{}f\right\|_{\infty}\plus{}\left\|f\minus{}\chi_{E_{s}}\right\|_{\infty}<\frac{1}{2}\plus{}\left\|f\minus{}\chi_{E_{s}}\right\|_{\infty}$.
Hence $ \left\|f\minus{}\chi_{E_{s}}\right\|_{\infty}>\frac{1}{2}$, which means that there can't be more than one $ f\in D$ that is close to a $ \chi_{E_{s}}$, a contradiction.  My only remaining question is why one member of $ D$ must be close to at least two $ E_{\alpha}$'s/how the uncountable part comes in.

As for the lemma, let $ m(E)\equal{}E$.  Let $ f(x)\equal{}m(E\cap (\minus{}\infty,x])$ and let $ \epsilon>0$.  Note that $ f(x)$ is defined for all $ x$ and $ f$ is increasing.  Suppose $ x<y$ and $ y\minus{}x<\epsilon$.  Then $ 0\leq f(y)\minus{}f(x)\equal{}m(E\cap (x,y])<\epsilon$, so $ f$ is continuous.  We know that $ \sum_{n\equal{}\minus{}\infty}^{\infty}m(E\cap (n,n\plus{}1])\equal{}E$, so $ \lim_{x\rightarrow\minus{}\infty}f(x)\equal{}0$ and $ \lim_{x\rightarrow\infty}f(x)\equal{}E$.  Thus there exists $ x_{0}$ such that $ f(x_{0})\equal{}r$.  Let $ F\equal{}E\cap (\minus{}\infty,x_{0}]$.  Then $ F\subset E$ and $ m(F)\equal{}r$.

[quote="JRav"]My only remaining question is why one member of $ D$ must be close to at least two $ E_{\alpha}$'s/how the uncountable part comes in.
[/quote]

If no member of $ D$ is close to multiple $ \chi_{E_r}$s, then there is an injection from $ \mathbb{R}_{\geq 0}$ to $ D$, which contradicts the countability of $ D$.


Let x be an positive integer with ten different digits  and can be divided by 11111.  Find the numbers of all x. :?

huh? I think he means how many ways are there to arrange the digits 0-9 exactly one time each and create a number divisible by 11111? I have been thinking about this for awhile. All I have figured out is that 11111=41*271. I don't know how helpful that is, though...

Oops yeah, I the answer I put was $\frac{x}{11111}$.  But I just realized that that number doesn't work either   :?

[quote="eryaman"]Well, [hide]11115[/hide] is one of them...not sure if there are more though.[/quote]
what???
it should be ten digit number with diferent digits... ;)

The first one I've found is

[hide]$1234987650$[/hide]

There are more, actually, such as:

[hide]
2694173058 and 6804231957

This is a really nice problem. We had it in the intermediate counting/probability class, so I had to look back there for the solution.

A good place to start is breaking the 10 digit number into two 5 digit segments, x and y.
Then 100000x+y=99999z.
[/hide]

You can permute the digits to get other multiples of 11111.
[hide="Got proof?"]Underlined variables mean the digit is that variable, not multiplication.
Suppose the number is [u]abcdefghij[/u]. If it's divisible by 11111, then so is [u]bcdefghija[/u] because [u]bcdefghija[/u]=10([u]abcdefghij[/u])-9999999999a :equiv: [u]abcdefghij[/u] (mod 11111).[/hide]
I guess I went overboard with the last one.

Well, the way we did it,
[hide]
100000x+y=99999z implies x+y is 0 mod 99999.
So x+y=99999, since 199998 is too big.
Then corresponding digits of x and y have to add up to 9.
That should bring us home. :)
[/hide]


Dear Arqady,

I have proved the inequality
$ a\plus{}b\plus{}c\equal{}1$  , $ a,b,c >0$

$ \frac{1}{a^3} \plus{}\frac{1}{b^3}\plus{}\frac{1}{c^3} \geq \frac{225}{1\plus{}48abc}$

But unsure if I make any mistake along the prof . Can you prove it ?

[quote="nguoivn"][quote="Utkirstudios"]
\[ \begin{array}{l}a,b,c > o \\
a \plus{} b \plus{} c \equal{} 1 \\
prove that \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge \frac {{25}}{{1 \plus{} 48abc}} \\
\end{array}
\]
[/quote]

It's very easy. See my proof here  :)[/quote]

use english please, I don't understand your proof.

[quote="dima ukraine"][quote="Gibbenergy"]
$ a \plus{} b \plus{} c \equal{} 1$  , $ a,b,c > 0$

$ \frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3} \geq \frac {225}{1 \plus{} 48abc}$

[/quote] By CS
$ (\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3})(a \plus{} b \plus{} c)\ge(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2\ge \frac {25^2}{(1 \plus{} 48abc)^2}$ but
$ abc\le \frac {1}{27}$so
$ \frac {25^2}{(1 \plus{} 48abc)^2}\ge\frac {25^2}{(1 \plus{} 48abc)(\frac {25}{9})} \equal{} \frac {225}{1 \plus{} 48abc}$[/quote]

Nice. My sols is purely SOS.
How's about

$ a,b,c,d >0$ and $ a\plus{}b\plus{}c\plus{}d \equal{} 1$   Prove that
$ \frac1a \plus{}\frac1b\plus{}\frac1c\plus{}\frac1d \geq \frac{19}{1\plus{}48abcd}$     :D

[quote="Gibbenergy"][quote="dima ukraine"][quote="Gibbenergy"]
$ a \plus{} b \plus{} c \equal{} 1$  , $ a,b,c > 0$

$ \frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3} \geq \frac {225}{1 \plus{} 48abc}$

[/quote] By CS
$ (\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3})(a \plus{} b \plus{} c)\ge(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2\ge \frac {25^2}{(1 \plus{} 48abc)^2}$ but
$ abc\le \frac {1}{27}$so
$ \frac {25^2}{(1 \plus{} 48abc)^2}\ge\frac {25^2}{(1 \plus{} 48abc)(\frac {25}{9})} \equal{} \frac {225}{1 \plus{} 48abc}$[/quote]

Nice. My sols is purely SOS.
How's about

$ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$   Prove that
$ \frac1a \plus{} \frac1b \plus{} \frac1c \plus{} \frac1d \geq \frac {19}{1 \plus{} 48abcd}$     :D[/quote]
 Let $ x \equal{} (abcd)^{\frac {1}{4}}\le \frac {1}{4}$ Then
$ \frac1a \plus{} \frac1b \plus{} \frac1c \plus{} \frac1d \geq \frac {4}{x}$ and we must prove $ \frac {4}{x}\ge \frac {19}{1 \plus{} 48x^4}$
or $ 192x^4 \minus{} 19x \plus{} 4 \equal{} (4x \minus{} 1)(48x^3 \plus{} 12x^2 \plus{} 3x \minus{} 4)\ge 0$ for $ 0 < x\le \frac {1}{4}$ and its obviously true!
 :)

[quote="wangsacl"]
use english please, I don't understand your proof.
[/quote]

Sorry, it's here  :oops:

[quote="dima ukraine"][quote="Gibbenergy"]Dear Arqady,

I have proved the inequality
$ a \plus{} b \plus{} c \equal{} 1$  , $ a,b,c > 0$

$ \frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3} \geq \frac {225}{1 \plus{} 48abc}$

But unsure if I make any mistake along the prof . Can you prove it ?[/quote] By CS
$ (\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3})(a \plus{} b \plus{} c)\ge(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2\ge \frac {25^2}{(1 \plus{} 48abc)^2}$ but
$ abc\le \frac {1}{27}$so
$ \frac {25^2}{(1 \plus{} 48abc)^2}\ge\frac {25^2}{(1 \plus{} 48abc)(\frac {25}{9})} \equal{} \frac {225}{1 \plus{} 48abc}$[/quote]

By CS
$ (\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3})(a \plus{} b \plus{} c)\ge\color{red}{(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2\ge \frac {25^2}{(1 \plus{} 48abc)^2}}$

I can't understand how it is?

[quote="arqady"][quote="Utkirstudios"]
\[\begin{array}{l}a,b,c > o \\ a+b+c = 1 \\ prove that \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{{25}}{{1+48abc}}\\ \end{array}\]
[/quote]
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{{25}}{{1+48abc}}\Leftrightarrow$
$\Leftrightarrow\sum_{sym}(a^{4}b+3a^{3}b^{2}-9a^{3}bc+5a^{2}b^{2}c)\geq0\Leftrightarrow\sum_{cyc}c(a-b)^{2}(a+b-3c)^{2}\geq0.$ :)[/quote]

why do you have this result ?

$\Leftrightarrow\sum_{sym}(a^{4}b+3a^{3}b^{2}-9a^{3}bc+5a^{2}b^{2}c)\geq0.$

[quote="lemonhuyenthu"]

why do you have this result ?

$\Leftrightarrow\sum_{sym}(a^{4}b+3a^{3}b^{2}-9a^{3}bc+5a^{2}b^{2}c)\geq0.$[/quote]
We can get it after expanding.

I have new solution:
Suppose that $a\ge b\ge c\ge 0.$
We will use from Generalization of Schur Inequality.
After homogenization, we get that  \[ \sum (a-b)(a-c)(c(a-2b)^2+b(a-2c)^2+4a(b-c)^2)\ge 0 \] and \[ \sum x(a-b)(a-c) \ge 0 \]
where, $x=c(a-2b)^2+b(a-2c)^2+4a(b-c)^2$, ...
It's true. Because,  $(a,b,c)$ and $(x,y,z)$ are monotonic. :lol:


The first is (half of) a symmetric sum.  Note that we have every possible pair of $4$ distinct variables.  It can be written $\frac{1}{2}\sum_{sym}p_{i}p_{j}$ or similarly.

The second one is neither cyclic nor symmetric, to the best of my knowledge.

The second one is equivalent (just the variables in different order) to

$p_{1}p_{2}p_{3}+p_{2}p_{3}p_{4}+p_{3}p_{4}p_{1}+p_{4}p_{1}p_{2}= \sum_{cyc}p_{1}p_{2}p_{3}$,

i.e. we are cycling through the variables for each term (in the order $1, 2, 3, 4$).

I think the first is actually 1/4 of the symmetric sum.
The second is a cyclic sum. $\sum_{cyc}p_{i}p_{j}p_{k}$

yes the first is a quarter because the symmetric gives all possible (ie 4!) terms, 4!/4=6 terms

also, I think the second is 1/6 of the symmetric sum
is it correct?

[quote="me@home"]
In general for n variables the ratio of symmetric to cyclic (starting with the cyclic) is (n-1)![/quote]

This is only true if you have $n$ variables and the terms have $n-1$ variables... for instance, with four variables $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, yes we do have

$\sum_{cyc}p_{1}p_{2}p_{3}= \frac{1}{6}\sum_{sym}p_{1}p_{2}p_{3}$,

but

$\sum_{cyc}p_{1}p_{2}\neq \frac{1}{6}\sum_{sym}p_{1}p_{2}$.

Your statement is correct if you said the ratio of the number of terms in the symmetric and cyclic sums, but not that actual sum of the terms.

[quote="me@home"]Yes, by definition this is true for all cyclic vs. symmetric sums in four variables: in a cyclic there are 4 options for the first term, no more freedom for the others. in a symmetric, there are 4 options for first, 3 for second, 2 for third, no more freedom... so 4*3*2/4=6
In general for n variables the ratio of symmetric to cyclic (starting with the cyclic) is (n-1)![/quote]

Does that mean order is significant in a symmetric sum?

@paladin sorry yes, that's what i meant (i had the example in mind)
no order isn't significant whatsoever in a symmetric sum because whatever the order, all combos will be added... but it does matter in some cases for the cyclic... $\sum_{c}a^{2}b\not =\sum_{c}a^{2}c$


[img]http://i5.photobucket.com/albums/y194/jelly119/g01.jpg[/img]
$\mbox{Given} \triangle ABC,\overline {AB}=\overline {AC},\angle A=20^\circ,\angle {ACD}=20^\circ,\angle {ABE}=30^\circ,\mbox{find}\ \angle{CDE}=?$

(It's said that there are  at least 11 defferent solutions to this question, maybe we can find more about this... :roll:  )

I posted this problem once a long time ago. It is in deed harder than it first appears. Its a very cool problem though. But possibly too hard for this board.

[quote="jelly119"]
(It's said that there are  at least 11 defferent solutions to this question, maybe we can find more about this... :roll:  )[/quote]

By the way, I have a PDF file that contains 8 solutions to this problem. It came from someone who gave a talk about it at the Berkeley Math Circle. Maybe I'll post it later. :)

[quote="gauss202"]By the way, I have a PDF file that contains 8 solutions to this problem. It came from someone who gave a talk about it at the Berkeley Math Circle. Maybe I'll post it later. :)[/quote]

Yeah, by Tom Rike. here is the link:

http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf


[quote="lena_siz"]I would like to join USAMTS for the 07 - 08 season, since I will be moving to California. And I have a bunch of questions about it:
1) Do we have to register during a certain time period, or we can start registering now?
2) If we miss a round, would we still be allowed to participate?
3)What is the level of difficulty/topics used in the USAMTS?
4) How many rounds are there?
5) Is there a break between each round, i.e. there is one in September and then one in November and so on, or something like that?[/quote]
1) The registration for 07-08 isn't out yet, I don't think. You can register at any time.
2) Yes.
3) See for yourself... http://www.usamts.org/Problems/U_Problems.php
4) There are 4 rounds, 5 problems each.
5) I believe there is a small break (a week to a month, I can't remember  :| ) between the due date of a round and the release of problems for the next round.

[quote="lena_siz"]I would like to join USAMTS for the 07 - 08 season, since I will be moving to California. And I have a bunch of questions about it:
1) Do we have to register during a certain time period, or we can start registering now?
2) If we miss a round, would we still be allowed to participate?
3)What is the level of difficulty/topics used in the USAMTS?
4) How many rounds are there?
5) Is there a break between each round, i.e. there is one in September and then one in November and so on, or something like that?[/quote]

1) Registration for 07-08 hasn't been out yet. You can register at any time, but it is better to register at the beginning of the year so that you can accumulate more points.

2) If you miss a round, it is going to affect your score a great deal, but you are still allowed to participate. In other words, you don't HAVE to do a round; but if you don't do a round, you are going to lose 25 points, which automatically puts you at silver medal. I don't recommend skipping a round; you have an entire month (or more) for each round.

3) These are relatively easy proof problems. You need to spend a great deal of time thinking about the problems, which is the reason that they allow an entire month for it.

4) 4 rounds, 5 problems per round, 5 points per problem. Do the math. You can get 100 points per year.

5) There is a break between each round, although I don't think there is a set break time (like 1 week or something).

Here is a chart:
100 - Perfect score (gold)
96-99 - Gold
75-95 - Silver
60-74 - Bronze
40-59 - Honorable Mention

Actually I believe you need a 41 for HM and a 76 for silver.

When will the USAMTS site be updated with new deadlines, problems, and such?

We'll release the first round in early August.

last year, I received the first round problems in the mail. will that happen again this year?


parameterization
logarithms
algebra unsolved
algebra

Solve the inequation for parameter $ a$:

$ \log_{\frac{1}{2}}(x^2\minus{}2x\plus{}a)>\minus{}3$

Here is solution, I hope this is correct.

In the PDF file, there is a part wrote in red as below in which I forgot to write.

"If there exists real number $ x$ satisfying $ (Star)$, then [color=red]the line y=a[/color] intersects the region $ (D)$, not including the boarder."

[quote="EastyMoryan"]$ 1/2^( \minus{} 3) > x^2 \minus{} 2x \plus{} a$
$ 8 > x^2 \minus{} 2x \plus{} a$
$ a > x^2 \minus{} 2x \minus{} 8$

Yes?[/quote]

The first line is correct, but the second line is incorrect. Try again! :)


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that if H is a subgroup of Q, different from Q, then Q/H is neither finite nor cyclic.

We assume that Q\H is finite. For a prime p, let A(p) = {k is a positive integer | there exists u/v in H with gcd(u,v)=1 and p^k divides v}. It's clear that A(p) must be unbounded for each prime p and also, we notice that if k is in A(p), then all numbers <=k are in A(p). Therefore for each p, A(p)=N. So this means that H=Q a contradiction.

Now if Q\H is cyclic, it's generated by a single element a. Therefore Q\H = aZ for some a. Then Q = H U (aZ), the reunion of 2 proper subgroups, impossible.

Oh, I'm soo sorry! What I meant was, in fact, prove that Q/H is neither finite nor cyclic :D:D

Sorry for the inconvenience. You found nice (and true) results, though :).

a)
Suppose that Q/H is finite. Let A_p(H) = {k in N | there exists u/v in H with gcd(u,v)=1 and p^k divides v}. Then there exists p0 such that A_p0(H) is bounded, or otherwise H=Q. Q/H is finite ===> Q = a1*H U  a2*H U ... U ak*H It's clear that A_p0(ai*H) is bounded but this means that A_p0(Q) is bounded, a contradiction.

b)
Suppose that Q/H is cyclic. Let Q/H = <a>. Then Q = U k*aH , over all k in Z. But max A_p0(k*aH)<=max A_p0(H), so A_p0(Q) is bounded, a contradiction.

So Q ? H is neither finite, nor cyclic, where ? is "\" or "/". :)


prove that , for any positive integer $n$, there exist a prime $p$  such that $n< p<2n$ :P

Bertrands Postulate, whose proof can be found in most good books about elementary properties of primes.
At least see http://www.mathlinks.ro/Forum/viewtopic.php?t=40143 and http://mathworld.wolfram.com/BertrandsPostulate.html


geometry
circumcircle
trigonometry
power of a point
radical axis
geometry solved

This is not such a hard problem. Invert around $P$ (the power is $k$). Because $PBCX$ is cyclic and so on we get that $X_0 \in B_0C_0$... We want to prove that. $X_0-Y_0-Z_0$ are colliniar. This is equivalent by Menelaos's theorem with :
$\prod \frac {B_0X_0}{X_0C_0}=1$ which is equivalent with :
$\prod \frac {BX}{XC}=1$ (1) (this is because $B_0X_0=\frac {kBX}{PB*PX}$ (exercise!)...
But (1) is equivalent with :
$\prod \frac {sin BPA'}{sin CPA'}=1$ (because of the sine theorem). Which again by the sine theorem is equivalent with :
$\prod \frac {BA'}{A'B} * \prod \frac {PC}{PB}=1$ which is obviously true by Menelaos theorem in $ABC$ with points $A'$, $B'$, $C'$.

[quote="hossein11652"]Let $P$ be a point inside of a triangle $ABC$, and let $L$ be a line which intersects the lines $BC$, $CA$, $AB$ at the points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$.
Let the line $PA^{\prime}$ intersect the circumcircle of triangle $PBC$ at $X$. Let the line $PB^{\prime}$ intersect the circumcircle of triangle $PCA$ at $Y$. Let the line $PC^{\prime}$ intersect the circumcircle of triangle $PAB$ at $Z$.
Prove that the quadrilateral $PXYZ$ is cyclic.[/quote]

Here is a proof by Nikolaos Dergiades:

The line AB is the radical axis of the circles ABC and APB. The line BC is the radical axis of the circles ABC and BPC. The line PX is the radical axis of the circles PXY and BPC.

But the pairwise radical axes of three circles always concur; hence, the line PX (= the radical axis of the circles PXY and BPC), the line BC (= the radical axis of the circles ABC and BPC) and the radical axis of the circles ABC and PXY concur. In other words, the point A', lying on the lines PX and BC, must also lie on the radical axis of the circles ABC and PXY. Similarly, B' lies on this radical axis. Hence, the line A'B' is the radical axis of the circles ABC and PXY.

Now, the point C' lies on the line A'B', i. e. on the radical axis of the circles ABC and PXY. But on the other hand, C' lies on the line AB, i. e. on the radical axis of the circles ABC and APB. Since the pairwise radical axes of any three circles concur, this implies that C' also lies on the radical axis of the circles PXY and APB. On the other hand, the point P obviously lies on this radical axis. Hence, the line PC' is the radical axis of the circles PXY and APB. Now, since the point Z lies on this line, it must have equal powers with respect to the two circles PXY and APB. But the power of Z with respect to the circle APB is 0; hence, the power of Z with respect to the circle PXY is also 0, and Z lies on the circle PXY. Hence, the quadrilateral PXYZ is cyclic. Qed..

Another nice proof of the result was given by the late Igor F. Sharygin; the idea of his proof is to generalize by letting P be an arbitrary point P in space. If P does not lie on the plane ABC, then the theorem is very easy. Now, take the limit case of P lying on the plane ABC and you are done.

  Darij

Also inversion is possible.(After taht menelaus)

Yes .S_Alex solved it with inversion.

And excuse me every body, it was not really a difficult one> :)


trigonometry
limit
calculus
calculus computations

$ a_1\equal{}1$
$ a_{n\plus{}1}\equal{}\sin{a_n}$

$ \sum_{n\equal{}1}^{\infty}a_n <\infty$ ?

One can prove that $ \lim_{n\to\infty} n a_n \equal{} \infty$. Then comparisn with the harmonic series shows that $ \sum_{n\equal{}1}^{\infty} a_n \equal{} \infty$.

For $ x>0,$ $ \sin x>x\minus{}\frac{x^3}6.$

Hence, $ a_{n\plus{}1}>a_n\minus{}\frac{a_n^3}6\,$ or $ \ \frac{a_{n\plus{}1}}{a_n}>1\minus{}\frac{a_n^2}6.$

So, $ a_n>\prod_{k\equal{}1}^{n\minus{}1}\left(1\minus{}\frac{a_k^2}6\right).$

We do have that $ \lim_{n\to\infty}a_n\equal{}0,$ since it is a decreasing positive sequence whose limit must be a solution of $ \sin x\equal{}x.$ Thus it follows that
\[ \prod_{k\equal{}1}^{\infty}\left(1\minus{}\frac{a_k^2}6\right)\equal{}0\]
The theory of infinite products then shows us that for this product to diverge to zero, we must have
\[ \sum_{k\equal{}1}^{\infty}\frac{a_k^2}6\equal{}\infty.\]
But $ \sum_ka_k^2$ is (for $ a_k<1$) smaller than $ \sum_ka_k.$ Hence $ \sum_ka_k\equal{}\infty.$ We have actually proved something stronger than asked for, and stronger than sos440's intermediate result. ($ na_n\to\infty$ is not strong enough by itself to argue that $ \sum a_n^2\equal{}\infty.$)

And what about $ \sum_{n\equal{}1}^{\infty}a_{n}^{\alpha}$ ? ($ \alpha>0$)


An insect sets off upwards from the shaft of the minute hand of a clock at 12 o'clock. Moving along the hand, it reaches the end of the hand in a quarter of an hour. When was it at the highest position?

$7mins$ and $30secs$ after $12'oclock$ is the answer probably but I think the solution is too advanced for me :blush:  but I did it by a average hit-and-trial way which never fails me in other problems similiar to this.

Nope, that won't be it (assuming it travels along the hand at a constant rate) -- try comparing the height at that time to, say, the height at 8 minutes.

[hide]
Think of height of the bug as a function an angle from 0 to pi. You can see that this equation is $H(\theta) = \frac{\theta}{\pi}\cos{\theta}$. Now just find the maximum of this on the interval and you will find he gets to his maximum height at 8 minutes and 12.93 seconds. Is that what the question really asks for? Or does it want the height in some units? Not sure.[/hide]


group theory
abstract algebra
vector
linear algebra
matrix
superior algebra
superior algebra unsolved

Let E a vectorial space of finite dimension and G a finite subgroup of automorphisms of G. Let F be any subspace that is stable by all elements of G. Then prove that F has a suplementary that is also stable by all elements of G.

This appears to be incorrect. Let $\mathbb{F}$ be a finite field, and consider the two dimensional vector space over that field. Consider the group generated by the matrix

\[\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}\]
The span of $\begin{bmatrix}1 \\ 0\end{bmatrix}$ is fixed by this group, but there is no other fixed subspace.

Sorry, I forgot to add that the field is of zero characteristic.  :blush:

Apparently, there's a standard way of solving such problems: we take an average over all the elements of the group of some sort. Here we do it like this:

Let $F'$ be any supplement of $F$, and let $p$ be the projection on $F'$. We define a new projection, $\tilde p$, by $\tilde p(x)=\frac 1{|G|}\cdot\sum_{g\in G}p(gx)$. It's easy to check that it is idempotent, and thus a projection. Its kernel includes $F$, and we must show that it is, in fact, equal to $F$. This is true because the dimension of the kernel of $\tilde p$ equals $n-tr(\tilde p)$, and $tr(\tilde p)=tr(p)$, as we can easily check. Since the kernel of $p$ is $F$, it means that $n-tr(\tilde p)=n-tr(p)=\dim F$. We can now take our subspace to be $Im(\tilde p)$.

The zero characteristic was needed for that $\frac 1{|G|}$ thing. In Kent Merryfield's example, the order of the group was equal to the characteristic. We only need a weaker condition, actually, namely that $|G|$ is not divisible by the characteristic of the field.


calculus
derivative
Princeton
college
geometry
geometric transformation

I know this isn't math related, but I've taken a look at some of the SAT verbal sections and found that I don't know some of the words in their analogies, sentence completions, etc, just that section in general.  What would be a good way to improve your vocabulary?  I still have a 2 years left till the SAT's.

[quote="Gyan"]If  you can, take Latin.[/quote]Or learn Romanian :).

I preped for like a month when I was a senior for SATI and I got 650 in Verbal (should have gotten more, but heck I didn't knew that in the last break they don't let you do the the restroom anymore :( - you can imagine what happened in the last two tests :D). I owned this score to the fact that 50% of SAT words have indentical spelling (well almost identical) and meaning in Romanian (which is a direct derivative from Latin - the language of Romans (that's why we're called Romanians :D)).

One quite useful resource for vocabulary development is [url=http://www.rfwp.com/wordwithin.shtml]The Word within the Word[/url], a book series about English word roots. You can order it from the author's publishing company, [url=http://www.rfwp.com/]Royal Fireworks Press[/url].

Hope this helps! Merry Christmas.

I just got this the other day, but I'm really loving it

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0375762183/qid=1104053918/sr=8-1/ref=pd_csp_1/104-7132705-6312755?v=glance&s=books&n=507846]Princeton Review's Word Smart[/url]

The definitions are clear and precise, and the periodic quizzes really help inforce them. Also, the cultural references and general wittiness used in the examples are quite hilarious.

Also, in the back of the book there are word roots, SAT "Hit List", and GRE "Hit List" (I don't know what the GRE is though  :P)

[quote="requiem"](I don't know what the GRE is though  :P)[/quote]

The [url=http://www.gre.org/]GRE[/url] (Graduate Record Examinations, but everyone refers to it by the initials) is the usual test for admission to graduate schools in the United States. It's kind of like a slightly harder SAT I. You can find free brochures about the GRE in most public libraries.

The best way is just to read a lot in ur free time.

but if you specifically want to do well on teh SAT, then work through "10 real SATs" (or whatever the new thing is, now)

Look up and learn all the vocab words you don't know...680 to 780 improvement for me, b/w frosh and jr years.

[quote="**********"]but if you specifically want to do well on teh SAT, then work through "10 real SATs" (or whatever the new thing is, now)[/quote]

The [url=http://www.collegeboard.com/newsat/hs/readiness/studyguide.html]new book from the College Board[/url] for preparing for the "new" SAT is called [url=http://www.amazon.com/exec/obidos/tg/detail/-/0874477182/learninfreed]The Official SAT Study Guide: For the New SAT[/url]. It has actual questions that have appeared on previous SATs or in the experimental sections of SATs recently.

trying to improve vocabulary specifically
is like trying to improve just your ability in calculus;
feasible to an extent but the real project would
be to improve your command of, and appreciation of,
language (resp. mathematics).

Reading helps more if you look at relatively learned
or recherche authors who use more complex
language.  19th and some early 20th century writing
is likelier to be useful than most of what came later.
Translations of non-English works with sophisticated
language will give you a feel for the different linguistic
register as well as upgrade your vocabulary (see e.g.
Thomas Mann's works).  Keep a dictionary at hand while
reading.

Learning some root languages such as German, French,
Latin or Old/Middle English helps, as does developing
some feeling for British vs American (vs Irish, etc) vocabulary
and English etymology, as does gleaning something of history
and culture of times and places other than late 20th century USA.

Also, learning words specifically for the SAT seems
misguided, but good luck if you're trying that approach.

[quote="fleeting_guest"]Learning some root languages such as German, French, Latin or Old/Middle English helps, as does developing
some feeling for British vs American (vs Irish, etc) vocabulary
and English etymology, as does gleaning something of history
and culture of times and places other than late 20th century USA.

Also, learning words specifically for the SAT seems
misguided, but good luck if you're trying that approach.[/quote]

I agree about learning root languages, and don't forget about Greek.  But for SAT training, good old cramming with flashcards does work too.  I would recommend looking up the word in a good dictionary that has etymology (word origins) and then writing some mnemonic tidbit from the etymology on the back of your flashcard, along with the definition and sample sentence, to help you remember.

Don't forget that for SAT analogies, you may need to know more than just the definition of the word, but the connotation of how it is used, which is harder to get from word lists and dictionaries, so do read too :)  But reading only helps a lot if you stop to look up unfamiliar words.  Otherwise you may infer a definition for a word that is not quite right.

Good luck!


calculus derivative conics parabola function

A cylindrical vessel is filled with water upto the brim to a
height $ H$ from the ground.A small hole is created at some height $ "h"$ from the ground .$ h\le\ H$ .     find the value of $ h$ for which the water coming out has the maximum horizontal displacement

The displacement $ x$ of the water jet is given by $ x \equal{} 2\sqrt {hd}$ where $ h$ is the height of the outlet (from base) and $ d\;$  the depth of outlet (measured from liquid surface).

[hide="Derivation:"]
$ v_x \equal{} \sqrt {2gd} \\
 \\
t \equal{} \sqrt {\frac {2h}{g}} \\
 \\
x \equal{} tv_x \\
 \\
\boxed{\therefore x \equal{} 2\sqrt {hd}}$
[/hide]

$ x \equal{} 2\sqrt {hd} \\
 \\
\;\; \equal{} 2\sqrt {h(H \minus{} h)} \\
 \\
f(h) \equal{} 2\sqrt {h(H \minus{} h)} \\
 \\
f'(h) \equal{} 0 \text{ at } h \equal{} \frac {H}{2}$ and $ h\equal{}0$. I leave it to you to decide which one is the suitable position for maximum displacement.

Where from comes $ f '(h) \equal{} 0 \text { at } h \equal{} 0 \;\;$ ?

Work out the differentiation, dude! You get a maxima [b]and[/b] a minima! The maxima is at $ h \equal{} \frac {H}{2}$ and minima at $ h \equal{} 0$.

actually you should also check endpoints and points where $ f'(h) \equal{}$ undefined.
this adds in another possibiliy, $ x\equal{}H$, which also happens to be a minimum, so $ x\equal{}H/2$ must be the absolute maximum over the relevant interval.

OK : from the condition for x to be real comes the minima $ h \equal{} 0 \text{ and } h \equal{}  H \;\;$,  but not from the differentiation .

You are right, Immanuel. But, I did not use differentiation, I instead plotted the graph (mentally) which is a parabola concave downwards but does not exist below X-Axis. The function here is $ f: [0, s_{max}]\to[0,H]$. In fact using differentiation, you get only the maxima but not minima since the function is not differentiable at $ (0,0)$ and $ (H,0)$ which is really bad...


function
induction
inequalities
algebra unsolved
algebra

if $f$ is a one to one function in R,and there exits an $x_1$,such that $f(x_1)=x_1$,
and for any $x$, we have $f(2x-f(x))=x$
prove $f(x)=x$

Hm... it seems to me that I have found a counterexamle :(

Simply take $f(x) = x$ for integers and $f(x) = x + 1$ for not-integers $x$.

my mistake.$f$ should also be sequence  :)

Is f defined on the whole set of integer numbers?
If so then we could set $f(x) = x$ for even and $f(x) = x+2$ for odd $x$ and would obtain a counterexample.
Otherwise if f is defined only on the set of positive integers we need some information about $2x-f(x)$ being positive.
If we changed the condition to $f+f^{-1}=2id$ (equivalent to $(f^n(x))_n$ is an arithmethic series) we would not need the existence of a fixpoint at all ...

sorry,this topic is gone a lot of time.
And I see the problem again,
the conditon is $f$ is a continuous contifunction in $R$

May I will rewrite the problem:
if $f$ is a bijection from $R$ to $R$,$f$  is continous,and there exist a $c \in R$ such that $f(c)=c$
and for any $x$,we have $f(2x-f(x))=x$
prove $f(x)=x$
can anybody help me now?

[quote="tµtµ"]Nothing like a proof, just some ideas :

Let's take $x$ such as $f(x) \neq x$ but such as $x$ is close to a fixed point so that $f(x)-x$ can be chosen as small as needed (by continuity)

$f(x) = x + f(x) - x = f(2x + 2(f(x)-x) - f(x + (f(x)-x))$ so by injectivity $f(x + (f(x)-x)) = x + 2(f(x)-x$ and similarily $f(x + n*(f(x)-x)) = x + (n+1)(f(x)-x), \forall n \in \mathbb{Z}$

It means that from non-fixed-point we can translate slowly to an other non-fixed-point with increased gap.
That way we can get close to a fixed point and get a contradiction.[/quote]
thanks.but I think I have thought your idea when I posted:(
when use the way can't slove the prove compeletly.
For example if $x$ such as $f(x) > x$
then we can only prove that
there is no fixed point in $(-\infty,x]$
but can't prove no fixed  point in $(x,+\infty )$
maybe I forget something...

Well, I think the main idea in here is to consider a function \[ g(x)=f(x)-x \] From our equation we can get\[ x=f(2x-f(x))=f(x-g(x))=g(x-g(x))+(x-g(x)) \] so we obtained \[ g(x-g(x))=g(x) \]  Our conditions on $f$ can be also written in form that for any $a$, our function $g$ has exactly one intersection point with line $a-x$. Of course $g$ is also continous...
and we know that there exists $x_0$ such that $g(x_0)=0$ Now all we have to do is to prove, that with these conditions it has to be $g(x)=0$ for all $x$. I believe I'm able to prove it, but the proof is rather complicated and difficult to present.Maybe some other day I'll do it.

well,I hope you can complete it...
And my book's hint is:"just notice that $f$ is monotonic..........."

[quote="jastrzab"]Well, sorry for that... now it seems to me that I've made a great mistake in my proof...  :oops: but sincerely speaking I've no idea why the fact that $f$ is monotonic is a real "hint"[/quote]
So am I...
can anybody helps us :(

Hooray!!! I believe I can finally solve this problem  :lol:  Well it's not really beautiful...but it seems, that the fact of being monotonic is really useful   As \[ f(2x-f(x))=x \] our function can't be decreasing...cos' if it would be like that, we would obtain for $y>x$ : $2y-f(y)>2x-f(x)$ which would mean that \[ x<y=f(2y-f(y))<f(2x-f(x))=x \]  which is a contradiction. But our function is monotonic , it has to be increasing!! Also it's rather easy to show, that in this case function $2x-f(x)$ is continous, increasing and bijective. We  will  use those facts further. Now another useful fact: if for some a, $f(a)>a$ then of course $2a-f(a)<a$
Now assume, that there exists $x_0$ such that $f(x_0)>x_0$ Then let us define $x_k=2x_{k-1}-f(x_{k-1})$This gives us $f(x_{k+1})=x_k$ Using this fact we may easily obtain (sooner or later...) that $x_k-x_{k+1}=x_{k+1}-x_{k+2}$ Now by induction we can prove that for every $k$ :$f(x_k)>x_k$ which means also that sequence$x_i$ is strictly decreasing with constant difference between to consecutive terms. This means that $x_k$ converges to minus infinity (it can be infinitely small) We can see now that as $f$ is increasing, it can be proved that ( it's not difficult, I can do if later if youwant) for any $x_k>x>x_{k+1}$ we have $f(x)>x$ This means that for every $x\leq x_0$ there is $f(x)>x$. Now we will show that that the same inequality holds when $x\geq x_0$. As I wrote ealier, function $2x-f(x)$ is bijective, and increasing which means that for every $\alpha$ there exists exactly one $x$ such that $2x-f(x)=\alpha$ Now we can define new terms of our sequence(with negatives indices) $x_0=2x_{-1}-f(x_{-1})$ and consecutively: $x_i=2x_{i-1}-f(x_{i-1})$. Now again as function $2x-f(x)$ is increasing, we may prove that that for decreasing indices our sequence is increasing with constant difference betweem consecutive terms,and again $f(x_i)>x_i$ and as $f$ is contious we are able to show that for every $x\geq x_0$ we have $f(x)>x$ so we obtained for every real x: $f(x)>x$ which is contradicton cos' there exists $q$ for which $f(q)=q$.
   Now we may assume that there exists $x_0$ such that $f(x_0)<x_0$ then in rather analogical way ( maybe not exactly the same, but using the sames ideas) we can obtain that for every $x$ : $f(x)<x$ which one more time gives us a contradiction. This means that $f(x)=x$ and we are done..  if any part of this proof in unclear just ask me... I'll try to explain it more precisely

@jastrzab:
One of the most beautiful proofs I've seen on this forum :) .  It contains very good (and also very strange) ideas !


That's pretty disgusting. You should probably try finding ab instead of a and b, if that's possible...I don't know.

[quote="hoshattack"]I am finishing the work I got to do before supper...
It wasn't that hard, just time consuming.
[hide]

Right, so we simplify to c = 9. :blush:

Then we can sub c into the equation $320a-6c = 3786$

$320a-6(9) = 3786$
$320a = 3840$
$a = 12$

Then you can plug both values into one of the original equations.
$2(12)+3b-5(9) = 0$
$24+3b-45 = 0$
$3b = 21$
$b = 7$

So the answer this all yields is the ordered triple (12, 7, 9)
To recap: A=12; B = 7; C = 9

[/hide][/quote]
That's all correct, but you haven't answered the question.

[quote="hoshattack"]Oh yes, forgot to say what $\frac{a}{bc}$ is.
[hide]
$\frac{12}{(7)(9)}= \boxed{\frac{12}{63}}$[/hide][/quote]
[quote="i_like_pie"]Express $\frac{a}{bc}$ [b]in simplified form.[/b][/quote]
This is hilarious... :rotfl:

[quote="hoshattack"][hide]

Fine.  I havn't gotten too much sleep lately.  Brain's tired.

$\boxed{\frac{4}{21}}$

THERE'S YOUR ANSWER![/hide][/quote]
Correct! :D

[quote="daermon"]do you know of a faster method?[/quote]
I don't. There's all primes (thanks to me), so it's hard to simplify anything.


function
algebra
domain
LaTeX
algebra unsolved

Find all function $f: R\to R$ satisfying:
\[f(x^{2}+y+f(y))=2y+[f(x)]^{2}\]

GREAT!! Your way to prove $f(0)=0$ is so BEAUTIFUL!!

sorry, but i need a bit of explanation...
how can you be sure that f is surjective?
by putting x=0 you get
f(y+f(y))=2y+c^2
but now how can you be sure that y+f(y) covers all the domain?
thanks...and sorry for not knowing latex :blush:

"Surjective" means that $2y+c^{2}$ covers all domain.

[quote="cefer"] If $3f(x)-x=0 \Longrightarrow f(x)=\frac{x}{3}\rightarrow x=0$ .contradiction.
[/quote]
How can you get this ,cefer?

He puts $f(x)=\frac{x}{3}$ in the initial condition :wink:

but $f(x)=\frac{x}{3}$ is only valid for $x=\sqrt{\frac{t}{2}}$ :P  :wink:

OK.I thought that it is easy,so i didn't give exlanation.Here is explanation.
In $(**)$ replace $y \rightarrow x$.Then
$f(x+f(x))=2x \Longrightarrow f(\frac{4x}{3})=2x$
Let's take $t_{1}=2(\frac{4x}{3})^{2}$
[color=red]Case1[/color].If $f(t_{1})=f(-t_{1})$
Then $f(\frac{4x}{3})=\frac{4x}{9}=2x \rightarrow x=0$
[color=red]Case2[/color] If $f(-t_{1})=-f(t_{1})$
Then $f(\frac{4x}{3})=\frac{4x}{3}=2x \rightarrow x=0$
Ok?  :P

Thanks very much.You are very good at function equation,cefer. :)
I always be late to post my solution.
To this problem I have a very long solution ,longer than any solutions given in the link of Albanian Eagle  :blush: Though it defers from yours, I won't post it..


calculus
integration
limit
calculus computations

Hi, a friend had shown me this problem today.  I do not know how to solve it.  I was wondering if anyone here knows.

Problem:  Find $ \displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{k\equal{}0}^{n\minus{}1}(n\minus{}k)\int_{\frac{k}{n}}^{\frac{k\plus{}1}{n}}\frac{1}{x^4\plus{}1}dx$.

First step: bring that $ \frac{n\minus{}k}{n}$ factor inside the integral. Since $ x$ is approximately $ \frac{k}{n}$, it's about $ 1\minus{}x$. The limit becomes $ \lim_{n\to\infty}\sum_{k\equal{}0}^{n\minus{}1}\int_{k/n}^{(k\plus{}1)/n}\frac{1\minus{}x}{x^4\plus{}1}\,dx\equal{}\int_0^1\frac{1\minus{}x}{x^4\plus{}1}\,dx$.
Now, use partial fractions.


inequalities
calculus
inequalities proposed

Let $x, y, z \ge 0$ be real numbers. Prove that: \[\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)| .\]
[hide="Additional task"]Find the maximal real constant $\alpha$ that can replace $\frac{3}{4}$ such that the inequality is still true for any non-negative $x,y,z$.[/hide]

Are you sure is $\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)|$?

I`m asking that because we have the inequality $x^{3}+y^{3}+z^{3}-3xyz\geq  4(x-y)(y-z)(z-x)$

I'm very sure! I don't see the contradiction with your inequality. Anyway, one can prove it easily using a brute-force method (which works fine here).

[quote="pohoatza"]I think the most of you contestants took $x\leq y\leq z$ and $y=x+a$, $z=x+a+b$, where $a,b \geq 0$  :D[/quote]

Well, that would be an easy way to do it.

As for the additional task, you need two things to complete it in a reasonable amount of time: calculus and a computer algebra program.  :wink:

[quote="PhilAndrew"]Let $x, y, z \ge 0$ be real numbers. Prove that:
\[\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)| . \]
[hide="Additional task"]Find the maximal real constant $\alpha$ that can replace $\frac{3}{4}$ such that the inequality is still true for any non-negative $x,y,z$.[/hide][/quote]
with chebyshev inequalities see here http://diendantoanhoc.net/forum/index.php?showtopic=31296

[quote=Zamfirmihai][hide=*]Are you sure is $\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)|$?

I`m asking that because we have the inequality [/hide]$x^{3}+y^{3}+z^{3}-3xyz\geq  4(x-y)(y-z)(z-x)$[/quote]


calculus
integration
geometry
circumcircle
quadratics
number theory
relatively prime

1. If $ f(N \plus{} 1) \equal{} ( \minus{} 1)^{N \plus{} 1}N \minus{} 2f(N)$ for integral $ N \ge 1$, and $ f(1) \equal{} f(1989)$, then find the
value of $ f(1) \plus{} f(2) \plus{} f(3) \plus{} ... \plus{} f(1988)$.
[hide="my answer"]I got $ \minus{} 994/3$. For those interested, I got $ f(1) \equal{} \frac {2^{1989} \minus{} 1989}{2^{1989} \minus{} 1}$[/hide]

2. Let x be a positive integer such that $ x^2 \equal{} AABBCC$ where A, B, C are digits. If S is
the sum of all possible x, and B = 4, then find the sum of the digits of 37 · S.
[hide="what I got"]I got no solutions for this problem... not sure if I've made any miscalculations[/hide]

3. $ \triangle DEF$ is inscribed inside $ \triangle ABC$ such that $ D,E,F$ lie on $ BC, AC, AB$, respectively. The [[circumcircle]]s of $ \triangle DEC, \triangle BFD, \triangle AFE$ have centers $ O_1,O_2,O_3$, respectively. Also, $ AB \equal{} 23, BC \equal{} 25, AC \equal{} 24$, and $ \stackrel{\frown}{BF} \equal{} \stackrel{\frown}{EC},\ \stackrel{\frown}{AF} \equal{} \stackrel{\frown}{CD},\ \stackrel{\frown}{AE} \equal{} \stackrel{\frown}{BD}$. The length of $ BD$ can be written in the form $ \frac mn$, where $ m$ and $ n$ are relatively prime integers. Find $ m \plus{} n$.

4. In $ \triangle ABC$, $ AB$, $ BC$, and $ CA$ have lengths $ 3$, $ 4$, and $ 5$, respectively. Let the incircle, circle $ I$, of $ \triangle ABC$ touch $ AB$, $ BC$, and $ CA$ at $ C'$, $ A'$, and $ B'$, respectively. Construct three circles, $ A''$, $ B''$, and $ C''$, externally tangent to the other two and circles $ A''$, $ B''$, and $ C''$ are internally tangent to the circle $ I$ at $ A'$, $ B'$, and $ C'$, respectively. Let circles $ A''$, $ B''$, $ C''$, and $ I$ have radii $ a$, $ b$, $ c$, and $ r$, respectively. If $ \frac {r}{a} \plus{} \frac {r}{b} \plus{} \frac {r}{c} \equal{} \frac {m}{n}$ where $ m$ and $ n$ are positive integers, find $ m \plus{} n$.

[hide="one"]
Since $ f(1) \equal{} f(1989)$

$ \sum_{k \equal{} 1}^{1988} f(k) \equal{} \sum_{k \equal{} 1}^{1988} f(k \plus{} 1) \equal{} \sum_{k \equal{} 1}^{1988} \minus{} 2f(k) \plus{} ( \minus{} 1)^{k \plus{} 1}\cdot k$

$ 3\sum_{k \equal{} 1}^{1988} f(k) \equal{} \sum_{k \equal{} 1}^{1988} ( \minus{} 1)^{k \plus{} 1}\cdot k \equal{} \minus{} 944$

$ \sum_{k \equal{} 1}^{1988} f(k) \equal{} \frac { \minus{} 994}{3}$[/hide]

[hide="two"]
$ x^2 \equal{} AABBCC \equal{} 11(A\cdot 10^4 \plus{} 400 \plus{} C)$,

since $ 11|x^2\Longrightarrow 121|x^2$
$ \Longrightarrow A\cdot 10^4 \plus{} 400 \plus{} C \equiv 0 \mod 11 \Rightarrow A \plus{} C\equiv 7 \mod 11$

Since $ A \plus{} C \le 18$ we have $ A \plus{} C \equal{} 7$

Now by checking the numbers $ 11,22,33,44,55,66$ against the quadratic residues $ \mod 4$ and $ \mod 5$ it is clear that only $ 44$ is a quadratic residue $ \mod 100$

Hence we only have to check $ x^2 \equal{} 774400$ and $ 3344444$, but only the first case works so $ x \equal{} 880$

So $ d(37\cdot S) \equal{} d(880\cdot 37 ) \equal{} d(32560) \equal{} \boxed{16}$[/hide]


calculus
function
vector
Functional Analysis
advanced fields
advanced fields unsolved

Is it true that if f is a positive strictly increasing function, A is bounded self adjoint positive operator on a Hilbert Space, then

$ \langle x_n, f(A)x_n\rangle \rightarrow 0$
iff
$ \langle x_n, Ax_n\rangle \rightarrow 0$

for a sequence $ x_n$ with $ \langle x_n,x_n\rangle\equal{}1$?

Do you assume that $ \langle x_n, Ax_n\rangle \rightarrow 0$ for any sequence of unit vectors, or for a particular one?
Suppose that $ A$ is a diagonal operator in an ON basis $ e_n$; that is, $ Ae_n\equal{}a_n e_n$. Let $ x_n\equal{}n^{\minus{}1/2}(e_1\plus{}\dots\plus{}e_n)$. Then $ \langle x_n, Ax_n\rangle\equal{}\frac{1}{n}\sum_{k\equal{}1}^n a_k$ while $ \langle x_n, f(A)x_n\rangle\equal{}\frac{1}{n}\sum_{k\equal{}1}^n f(a_k)$. It's not easy to find an example where one sequence goes to zero and the other does not, but I'm sure they exist.

only assume a particular sequence of unit vectors.

I think the fact that $ A$ is bounded is crucial to the estimation? and $ A$ is self adjoint means it is closed operator, so that $ f(A)$ is also bounded...

The assumption means that $ A$ has 0 in its spectrum (by Rayleigh quotient), so $ f$ (defined on $ \sigma(A)$) should be defined on 0 also, it cannot get too large for $ f(a_k)$?

Obviously I missed a condition...

I need $ f(0) \equal{} 0$

otherwise the question is trivial enough: take $ f(x)\equal{}1\plus{}x$

OK, my supposed counterexample above does not work.


calculus
integration
trigonometry
function
limit
probability
real analysis

hola  :D

can you solve

$ \int_0^\infty\;\left[\prod_{k\equal{}1}^\infty\;\cos\,\left(\dfrac{x}{k}\right)\right]\;\textbf dx$

I don't think that this integral exist .
it's looks like $ \int_{0}^{\infty} \cos x dx$ :lol: . since function
$ f(x) \equal{} \prod_{k}\cos x/k$ has zero at points $ x \equal{} k\frac {\pi}{2}$, $ \forall k \in \mathbb{N}$  and not only, Hence it has no limit at point $ \infty$, and if it has limit then $ f$  must be $ f \equiv 0$, but this number $ \prod _{k} \cos \frac {x}{k}$ at point $ x \equal{} 1$ is not equal to $ 0$.
I believe that such integrals $ \int_{0}^{\infty} \cos x dx$ only misan's students are able to calculate    :P

:))) I delete my previos post, it was stupid. Function $ f_{n}(x) \equal{} \prod_{k \equal{} 1}^{n}{\cos\left(\frac {x}{k}\right)}$ has one of the periods (not the smallest, but in some cases it is) $ T \equal{} 2n!\pi$. So, integral of a pereodic function (for some big integer $ N$) will be $ \lim_{N\to\infty}\int_{0}^{N\cdot T}{f_{n}(x)dx} \equal{} N\cdot \int_{0}^{T}{f_{n}(x)dx} \equal{} N\cdot 0 \equal{} 0$ because $ \int_{0}^{T}{f_{n}(x)dx} \equal{} 0$.

But if we change the way of finding limit: if $ 0 < M < T$ then $ \lim_{N\to\infty}\int_{0}^{N\cdot T \plus{} M}{f_{n}(x)dx} \equal{} N\cdot \int_{0}^{T}{f_{n}(x)dx} \plus{} \int_{0}^{M}{f_{n}(x)dx} \equal{} \int_{0}^{M}{f_{n}(x)dx}\ne 0$.

So, final answer will depends of the way of integration. This means that it does not exist. Correct me if i'm wrong. :) This is from one hand.

But from the other, when $ n\to\infty$ this period $ T$ is alsow tends to $ \infty$, so our function $ f_{\infty}(x)$ is not a pereodic and we should think one more about this.

yuriy, this is a   known integral amongst the mathematical community, one can actually show   it's value  lies between $ 0$ and $ \frac {\pi}{4}$, but here is the interesting thing ..... it is [b][u]extremely[/u][/b] close to $ \frac {\pi}{4}$, but mind you, it is [color=red][b]NOT[/b][/color] $ \frac {\pi}{4}$ ................  i am poor  :( , so i don't have access to sophisticated tools like mathematica and maple, but i wrote a crude software  program in C so i can decipher   the integral's value is  $ 0.78538 \ldots$ (don't know the rest of the digits as error creeps in)  :(  anyway,  i observe from the "CRC Standard Mathematical Tables and Formula" that the value  of $ \frac {\pi}{4}$ is $ 0.785398 \ldots$ , hence as you can  see, it is very, very close.

I calculate integral it alsow by computer.
$ \textbf{I}(m,n) \equal{} \int_{0}^{m}\;\left[\prod_{k \equal{} 1}^{n}\;\cos\,\left(\dfrac{x}{k}\right)\right]\;\textbf dx$
approximate it by sum
$ \textbf{I}(m,n)\approx\sum_{j \equal{} 0}^{\frac {m}{\delta x}}{\;\left[\prod_{k \equal{} 1}^{n}\;\cos\,\left(\dfrac{j\cdot \delta x}{k}\right)\right]\; \delta x}$

I calculated $ \textbf{I}(10,50)$ with $ \delta x \equal{} 10^{ \minus{} 7}$ and get

$ \boxed{\textbf{I}(10,50)\approx 0.78539734}$
and
$ \frac {\pi}{4}\approx 0.7853981633974483$

btw/ did you see this one...  :D

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202733

Playing with the Fourier transform a bit, one can quickly realize that the integral in question equals $ \frac \pi 4(1\minus{}2p)$ where $ p$ is the probability that the sum $ \sigma\equal{}\sum_{k\equal{}1}^\infty \frac{\xi_k}{2k\plus{}1}>1$ ($ \xi_k$ are independent random variables uniformly distributed on $ [\minus{}1,1]$). It is clear that $ p>0$, so the integral is not $ \frac \pi 4$ but a bit less. To estimate $ p$ from above, I used the computer to compute $ Ee^{a\sigma}$ (which is a fairly simple product) and made the rough estimate $ p\le e^{\minus{}a}Ee^{a\sigma}$. It turned out that the optimal $ a$ is around $ 30$, which gives $ p\le e^{\minus{}10}$. This more or less agrees with the difference in the 6th digit after the decimal point though it suggests that my rough estimate is about 50 times as large as the truth. It would be nice to estimate this probability more carefully :).

fedja, can you explain how you get probability and $ \frac{\pi}{4}$ from that product? :)

Certainly. Note that $ \cos x$ is the characteristic function of the random variable $ \zeta$ taking the values $ \pm 1$ with probability $ 1/2$ each. Thus, by the Fourier inversion formula, the integral in question is just $ \pi$ times the density of the distribution of the sum $ \sum_{k\ge 1}\frac{\zeta_k}k$ where $ \zeta_k$ are independent Bernoulli random variables. Note now that $ \sum_{\ell\ge 0}\frac{\zeta_{2^\ell}}{2^\ell}$ is uniformly distributed on $ [\minus{}2,2]$. So, the density in question is the value at $ 0$ of the infinite convolution $ *_{k\ge 0} \tfrac{2k\plus{}1}4\chi_{[\minus{}\frac 2{2k\plus{}1},\frac 2{2k\plus{}1}]}$. The first function in the convolution is $ \tfrac 14\chi_{[\minus{}2,2]}$ and, if we denote the rest by $ \Phi$, we'll get that the integral equals $ \frac \pi 4 \int_{\minus{}2}^2 \Phi(x)\,dx$. Noting that $ \Phi$ is even, the full integral of $ \Phi$ is $ 1$ and returning to the probabilistic interpretation, you get the formula I mentioned (up to rescaling).

Of course, you do not need to think in probability terms at all. You may just use the well-known formula $ \prod_{\ell\equal{}0}^\infty \cos\frac x{2^\ell}\equal{}\frac{\sin 2x}{2x}\equal{}F(x)$ to bring the infinite product to a more convenient form $ \prod_{k\ge 1}F\left(\frac x{2k\plus{}1}\right)$ for numerical computations and use some quadrature formulae to get a really decent approximation to the actual value of the integral.

In any case, I doubt there is a nice closed form expression for this value, so my answer to Misan's original question is "No, I cannot solve it explicitly".


USAMTS
AMC
AIME
USA(J)MO
USAMO
AIME II
AMC 10

How do USAMTS scores convert to USAMO indices? I don't think I qualified anyway, but I would like to know how close I came...

Anyway, USAMTS - 70 (after 3 rounds, 95 for the whole year), AMC 12A-102 and AIME II-6, would have been 8 without my computational mistakes, for an index of 162 not considering USAMTS.

I think they double the USAMTS score, since 75*2=150...so that would mean your index is 200...

A qualifying USAMTS score is the same as a 100 AMC 12 score when looking at USAMO index.

possible, but I doubt that because that would mean that 70 is equivalent to a 140 on the AMC, which (I think) is much harder to achieve because of time limitations.  If there were no time limits, I would have had no problem on the AMC.  USAMTS doesn't have comparable time limits.  Also, I believe that only 68+ were invited to take the AIME, and I highly doubt that everyone qualifying for AIME via USAMTS got at least a 136 from it.  I've heard that in the past they have given everyone who qualifies via USAMTS 100, but more recently they have been requesting the scores, but no one seems to know how they use the scores.

[quote="Carbon57"]Really, I thought the USAMTS was harder than the AMC, even with no time limits...My AMC 10 score was 135 and my USAMTS score for the first 3 rounds was...34 :([/quote]

the main thing is that the USAMTS is not part of the qualifying process for the AMC/AIME (not everyone can/will take it) so I doubt they would value it more than the AMC tests.  So its like 100.

it might be harder...but in my opinion i think the format of the USAMTS makes it easier.

Ouch, Carbon...

I know USAMTS is harder in a different way.  You have to write proofs, not just find an answer, which can be more difficult.  But the time limits on the AMC is really what screws me - I am fine with proof writing.  Anyway, I suppose my programming skills helped a bit last year when I wasn't good with combinatorics or didn't realize I could do a case study and just wrote programs to find the answers for me.  Definitely easier to do it mathematically (via sums, combinations, premutations, case studies, whatever) when possible than to program it.  This year, none of my solutions required programs, and I only used 1 to help me find the solution (#5 on round 4) so I could then work backwards to prove it.  But without these programming skills, I probably would have missed a few more problems and been somewhere around a 40 after 3 rounds my first year (last year, not this year).  Experience definitely helps with USAMTS, much less than it does for AMC.

I was never that great at AMC.  Last year, for the AMC 10 I had next to no sleep the night before and only got through about half of it, for about 13 or 14 right, don't remember exactly.  This year I did a little better with the AMC 12, but again, only got through about the first 16ish and then started skipping around, looking for problems I could solve quickly.  So I guess the question is are proofs easier or can you work extremely fast?  I just don't do as well with time limitations as I do without - more time to be stuck on a problem and eventually find a solution, instead of wasting valuable time.

Having said all that, USAMO is a proof-writing competition, so perhaps the USAMTS should convert to really good indices, ex by doubling score.  I don't know - I think its a different learning curve.

But back to my original question: does anyone know [u]for certain[/u] how USAMTS turns into a USAMO index?

H Potter is correct -- you get 100 points.

It's funny how people continue to ask question even after the correct response has been given.

h_s_potter2002 posted the correct answer. With USAMTS, you can qualify to AIME yet your score to count for USAMO is only 100, regardless of whatever your score. Now, the detailed rule about USAMTS and USAMO is as follows:

You may choose to use AMC score instead of USAMTS score if you want to. So, if you scored 80 on USAMTS to qualify for AIME and scored 9 on AIME, your index score is 100 + 10*9 = 190. Now, let's say that you also took AMC-12 and received 130. Your other index score that you could use is 130 + 10*9 = 220.

I'm [u]very certain[/u] that this is how they do the scoring. For one thing, recall that USAMTS's qualification to AIME is chosen AFTER the third round, not the final round, so it is not right to be considered for index as a whole anyway.

I believe the reasoning for this is that it's much easier to cheat on USAMTS than on AMC. They don't really have an issue with too many people taking AIME but USAMO takes a lot more per-student resources which they don't want to spend on someone who's there dishonestly.

[quote="worthawholebean"]I believe the reasoning for this is that it's much easier to cheat on USAMTS than on AMC.[/quote]

Also, maybe facis is right, but I actually think that it's easier to get, say, a 72 in three rounds of the USAMTS than a 144 on the AMC 12--the bottom line is that one has access to resources and basically unlimited time instead of a three-minutes-per-problem average.


There is a list of seven numbers. The average of the first four
numbers is 5, and the average of the last four numbers is 8. If the
average of all seven numbers is $ 6\frac{4}{7}$, then what is the number common to both sets of four numbers?

Multiplying an average by the number averaged will get you the sum of the numbers involved in the average:
So
Sum of all: $ \frac{46}{7}\cdot 7\equal{}46$
Sum of first four: $ 5\cdot 4\equal{}20$
Sum of last four $ 8\cdot 4\equal{}32$

Then the sum of the last three is $ 46\minus{}20\equal{}26$
But this means that the 4th number is $ 32\minus{}26\equal{}\boxed{6}$


modular arithmetic
number theory unsolved
number theory

Prove that for non-negative integer $k$ , the factor of

$4k^{2}+4k+3$

is always in the form of $1(\mod 8)$ or $3(\mod 8)$

If prime p is 5 or 7 modulo 8 then -2 is not a square modulo p (i.e., Legendre symbol $\left(\frac{-2}{p}\right)=-1$).
Therefore, $(2k+1)^{2}\equiv-2\pmod{p}$ cannot hold, implying that $p$ cannot divide $4k^{2}+4k+3.$
Hence, all prime factors of $4k^{2}+4k+3$ equal 1 or 3 modulo 8, and any product of such factors again equals 1 or 3 modulo 8. Therefore, all factors of $4k^{2}+4k+3$ equal 1 or 3 modulo 8.


Find all the function $f: \mathbb R^{\geq 0} \to \mathbb R^{\geq 0}$ such that for all $x, y \in \mathbb R^{\geq 0}$,
\[\sqrt[2]{f(\frac {x^2+y^2}{2})}=\frac {f(x)+f(y)}{2}.\]

[quote="ehsan2004"]Find all the function $ f: R^ \plus{} \cup {\{0}\}\longrightarrow R^ \plus{} \cup {\{0}\}$ such that:
$ \forall x,y\geq 0: \sqrt [2]{f(\frac {x^2 \plus{} y^2}{2})} \equal{} \frac {f(x) \plus{} f(y)}{2}$. :lol:[/quote]

Let $ g(x)\equal{}\sqrt{f(x)}$

We have $ g(\frac {x^2 \plus{} y^2}{2})\equal{} \frac {g^2(x) \plus{} g^2(y)}{2}$

And so, taking $ x\equal{}y\equal{}0$ : $ g(0)\equal{}g^2(0)$ and either $ g(0)\equal{}0$, either $ g(0)\equal{}1$

1) If $ g(0)\equal{}0$, taking $ y\equal{}0$, we have $ g(\frac {x^2}{2})\equal{} \frac {g^2(x)}{2}$ and so $ g(\frac {x^2 \plus{} y^2}{2})\equal{}g(\frac {x^2}{2})\plus{}g(\frac {y^2}{2})$
And so $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ (Cauchy equation) and, since $ g(x)\geq 0$, $ g(x)\equal{}ax$ (since any non continuous solution of Cauchy's equation is unbounded on any non empty open  interval).
Putting back this expression in the original equation, we find $ a\equal{}0$ or $ a\equal{}1$

And two solutions :
$ f(x)\equal{}0$
$ f(x)\equal{}x^2$

2) If $ g(0)\equal{}1$, taking $ y\equal{}0$, we have $ g(\frac {x^2}{2})\equal{} \frac {g^2(x)}{2}\plus{}\frac{1}{2}$ and so $ g(\frac {x^2 \plus{} y^2}{2})\equal{}g(\frac {x^2}{2})\plus{}g(\frac {y^2}{2})\minus{}1$
And so $ g(x\plus{}y)\minus{}1\equal{}(g(x)\minus{}1)\plus{}(g(y)\minus{}1)$ (Cauchy equation) and, since $ g(x)\geq 0$, $ g(x)\equal{}ax\plus{}1$ (since any non continuous solution of Cauchy's equation is unbounded on any non empty open  interval).
Putting back this expression in the original equation, we find $ a\equal{}0$

And a third solution :
$ f(x)\equal{}1$


[quote="navidj"]Let $ x_1 , x_2 ,... , x_n$ be positive intigers satisfying $ \sum \frac {1}{1 \plus{} x_i} \equal{} 1$ . Prove that :
$ \sum \sqrt {x_i} \geq (n \minus{} 1)\sum \frac {1}{\sqrt x_i}$[/quote]
A proof for reals $ x_{i}\geq\frac {1}{n \minus{} 1}$ see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60590


geometry
algebra
polynomial
function
calculus
derivative
real analysis

Let f(x,y) be an irreducible real polynomial. Consider the solution set V of f(x,y) = 0 in the real plane R2.

By Hilberts Nullstellensatz we have I(V) = (f), if I'm not mistaken. Let C be the smooth nonsingular curve part of V. Assume that C is not empty.

My question is this: Is it always possible to find a g in I(V) such that the gradient field of g restricted to C is the curvature field of C?

[quote="henrik"]By Hilberts Nullstellensatz we have I(V) = (f), if I'm not mistaken.[/quote]
$\mathbb{R}$ is not algebraically closed, so you can't use it. Indeed, $f(x,y)=x^{4}+y^{2}+1$ is a counterexample, $x^{2}+y^{2}+1$ is not a multiple of it.

All right, I realize there might be a problem there, but that's not essential to my question. Forget about I(V)=(f).

But your counter examples do not convince in the real plane because the varieties defined are both empty!

Hey, that's an interesting problem in its own right!

Let's call a real polynomial f(x,y) hilbert if it satisfies I(V(f)) = (f).

Here V(f) is all (x,y) in R2 (the real plane) such that f(x,y) = 0, i.e. the solution set of the equation. I(V) is the ideal in R[x,y] of all polynomials that vanish on V.

ZetaX points out that not all polynomials are hilbert.

f(x,y) = x is an example of a hilbert polynomial. Indeed, if g(x,y) vanishes on the y-axis, then g(0,y) is a polynomial in y that is 0 everywhere (as a function), so it must be 0 (as a polynomial). Therefore g(x,y) = x*h(x,y) for some h.

Another example of a hilbert polynomial is f(x,y) = x2 + y2 - 1 determining the circle. If g(x,y) vanishes on the circle then g(x,c) is a polynomial in x that has the same (real) roots as f(x,c) for all -1 < c < 1 (and possibly other roots as well). Therefore f(x,c) is a factor in g(x,c). Similarly f(c,y) is a factor in g(c,y).

Consider the function h(x,y) = g(x,y) / f(x,y). It is well defined and smooth everywhere in the open square <-1,1> x <-1,1> except possibly on the circle. But we have just seen that it extends to a smooth function in the whole square. All the partial derivatives of h of sufficiently high order vanish identically, so we are forced to conclude that h is a polynomial.

That proves that g(x,y) is a multiple of f(x,y), i.e. g is in (f). So f is hilbert.

I conjecture the following generalization: Let f(x,y) be an irreducible real polynomial. If the curve part C of V(f) is nonempty then f is hilbert.

In my first posting in this thread I assumed that C is nonempty, so f should be hilbert as stated, even if the Nullstellensatz cannot be used. Anyway, it's beside the point, but interesting!

Here's another problem:

Let C be a smooth curve in the real plane (possibly with selfintersections). Consider the set P(C) of all points $p \in \mathbb{R}^{2}$ with the projective property:

There is a unique point q on the curve with minimal distance to p, and qp is normal to C at q.

Everybody knows that if C is a straight line then $P(C) = \mathbb{R}^{2}$.

It should be clear (?) that for any smooth C, P(C) is a neighborhood around C. This is because the curve locally is approximated by a straight line (smoothness!). We may have to assume something about the embedding to arrive at this.

It is easy to give an example of a curve C for which $P(C) \neq \mathbb{R}^{2}$. C : $y = x^{2}$. On the positive y-axis sufficiently far away from the origin the points fail to have the projective property: There are TWO points on C with minimal distance.

An even simpler example is the circle C : $x^{2}+y^{2}= 1$. The origin is not projective.

But the exceptional sets are "small" in these examples and that suggests the following conjecture:

Let C be any smooth algebraic curve in the real plane $\mathbb{R}^{2}$. Then P(C) is dense in $\mathbb{R}^{2}$.

I believe it can be generalized to analytic curves.

The examples suggest that the exceptional set $\mathbb{R}^{2}\setminus P(C)$ is closely related to the symmetries of the curve.


geometry
parallelogram
trigonometry
Asymptote
circumcircle
geometric transformation
reflection

extend $ AP $ and $ AQ $ to meet $ BC $ at $ P' $ and $ Q' $, respectively.  since triangle  $ BMK $ is isosceles and $ AQ $ is parallel to $ MK $, it follows that triangle $ BAQ' $ is also isosceles, and thus that $ MA = KQ' $.  similarly, $ AL = P'K $.  since $ MA = AL $, it follows that $ KQ' = P'K $.  now, since triangles $ QKQ' $ and $ PP'K $ have their sides parallel to each other, they must then be congruent, and hence $ QQ' = PK $.  since $ QQ' $ and $ PK $ are also parallel, $ QQ'KP $ is thus a parallelogram, and hence $ PQ $ is parallel to $ KQ' $, and thus to $ BC $.  since $ PQ $ passes through the midpoint of cevian $ AK $ (as $ APKQ $ is a parallelogram), it then follows, finally, that $ PQ $ also passes through the midpoints of $ AB $ and $ AC $.

[quote="Mathx"]The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$, and the line through $A$ parallel to $MK$ meets $LK$ at $Q$. Show that the line $PQ$ bisects $AB$ and bisects $AC$.[/quote]

Another solution: As I showed in http://www.mathlinks.ro/Forum/viewtopic.php?t=23677 (in the Extended problem), if B' and C' are the midpoints of the sides CA and AB of triangle ABC, and if X is the orthogonal projection of the point A on the angle bisector of the angle BCA, then this point X lies on the lines B'C' and MK. Now, since the point X is the orthogonal projection of the point A the angle bisector of the angle BCA, the line AX is perpendicular to the angle bisector of the angle BCA; on the other hand, since the points L and K, being the points of tangency of the incircle of triangle ABC with the sides CA and BC, are symmetric to each other with respect to the angle bisector of the angle BCA, the line LK is also perpendicular to the angle bisector of the angle BCA. Hence, the line AX is parallel to the line LK. In other words, the point X lies on the parallel to the line LK through the point A. Since the point X also lies on the line MK, we thus can state that the point X is the point where the parallel to the line LK through the point A meets the line MK. On the other hand, we have defined the point P as the point where the parallel to the line LK through the point A meets the line MK. Hence, X = P. And since we know that the point X lies on the line B'C', we thus can conclude that the point P lies on the line B'C'. Similarly, the point Q lies on the line B'C'. Hence, the line PQ is the line B'C'. And, of course, this line must bisect the segments AB and CA (since it passes through the points C' and B' which are the midpoints of the segments AB and CA, respectively). Problem solved.

  Darij

[hide="Here is a short metrical solution for this nice and instructive problem."]
Denote the projection $D$ of the point $A$ on the line $BC$ and the
projections $R$, $S$ of the points $P$, $Q$ respectively on the line $AD$.
Prove easily that $m(\widehat {PAD})=\frac C2$, $m(\widehat {QAD})=\frac B2$.
From the theorem of Sinus in the triangles $AMP$ and $ALQ$ we obtain
$AP=\frac{(p-a)\cos \frac B2}{\cos \frac A2}$, $AQ=\frac{(p-a)\cos \frac C2}{\cos \frac A2}$ respectively.
Therefore, $AR=AS=\frac{(p-a)\cos \frac B2 \cos \frac C2}{\cos \frac A2}=\frac Sa=\frac{h_a}{2}$ q.e.d.[/hide]

[asy]
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[/asy]
Let $S = AK \cap PQ$. Since the incircle of $\triangle ABC$ is the circumcircle of $\triangle KLM$, $KS$ is the $K$-symmedian of $\triangle KLM$. Take any point $T$ on the internal angle bisector of $\angle LKM$. Reflect $\overline{KL}, \overline{KM}$ over $KT$ to form $\overline{KL'},\overline{KM'}$.  Let $S'$ be the midpoint of $\overline{L'M'}$. Since $KS$ was the $K$-symmedian of $\triangle KLM$, $KS$ is the $K$-median of $\triangle KL'M'$. Now since $L',M'$ lie on $KP,KQ$ respectively, and $KS$ is also the median of $\triangle KPQ$ (as $APKQ$ is a parallelogram), we have $PQ \parallel L'M' \implies \triangle KLM \sim \triangle KPQ$. So $\angle PQK = \angle LMK = \angle LKC$ (alternate segment theorem). So $PQ \parallel BC$. Since $S$ is the midpoint of $\overline{AK}$, and $PS \parallel BK$, so $PQ$ bisects $AB$ and similarly $AC$. $\blacksquare$

Dear Mathlinkers,
this problem is a little variation of a Lascases result :

http://jl.ayme.pagesperso-orange.fr/Docs/An%20unlikely%20concurrence.pdf  p. 3...

Sincerely
Jean-Louis

It is easy to find out that six points $A, P, M, I, L$ and $Q$ are on one common circle.
Here is the proof:
We know that: $\angle APK + \angle MKL = 180$ and since $\angle MKL = \angle ALM$ we can conclude that $\angle APM + \angle ALM = 180$.
So $APML$ is cyclic. In exactly the same manner, we can show that $AQLM$ is cyclic too. And though $APMLQ$ is cyclic.
Now note that $AMIL$ is cyclic too. So $APMILQ$ is cyclic.
Now that 6 points are on one circle, note that $\angle PMA = \angle PQA$. And since $\angle KPQ = \angle PQA$ we can conclude that $\angle KPQ = \angle PKB$ and though $PQ \parallel BC$.
Now let $F$ be the intersection of $PQ$ and $AB$. We may now show that $AF = FB = FQ$.
And to proceed that, cause $AF = FQ$, we may show that $\angle AQB = 90$. We know that $\angle IQA + \angle IMA = 180$. Thoght we have $\angle AQI = 90$.
So it is enough to show that $B, I$ and $Q$ are colinear. To proceed that, we show that $\angle PQI = \angle IBC = \angle B/2$.
Note that $\angle AQP + \angle PQI = 90$. And cause $\angle AQP = \angle BMK = 90 - \angle B/2$ we find out that: $\angle PQI = \angle B/2$.
Now we are done.

Let $AP \cap BC=X$ and $AQ \cap BC=Y$
$$ XK=CX-CK=AC-CL=AL=AM=AB-BM=BY-BK=KY \implies XK=KY \implies AP=PX \implies AQ=QY$$

Just notice that $AP \perp  bisector \angle C$ and combine [url=https://artofproblemsolving.com/community/c6h1834562p12296756]this[/url] and [url=https://artofproblemsolving.com/community/c6h1813322p12088802]this[/url].

This is not good :( [url=https://artofproblemsolving.com/community/c6h530567p3028101]All-Russia MO 1997[/url]

Let $AP$ and $AQ$ meet BC in $U$ and $T$ respectively . Let $PQ$ meet $BC$ in $X$. We see that $APKQ, PKTQ, PUKQ$ are parallelograms $\Longrightarrow \triangle KQT \cong \triangle PAQ \Longrightarrow Q$ is the midpoint of $AT$. Similarly, $\triangle PAQ \cong \triangle PUK \Longrightarrow P$ is the midpoint of $AU$. By midpoint theorem $PQ \parallel BC$. $PX \parallel BC \Longrightarrow \triangle APX \sim \triangle ABU \Longrightarrow X$ is the midpoint of $AB$.

[hide = This is a very easy problem.]

Using the tangent angle with chord properties as $AL$ is tangent to $(KLM)$ at $L$ and the fact that $AP \parallel LK$ ,we have $\measuredangle ALM = \measuredangle LKM =\measuredangle LKP = \measuredangle APK = \measuredangle APM$
$\implies APML$ is concyclic .

Similar argument gives that $AQML$ is concyclic

As both $(APML)$ and $(AQML)$ have three points $A,M,L$  in common , the two circles coincide .
$\implies A,P,M,L,Q$ is concyclic
$\implies PMLQ$ is concyclic

Using the above fact and  that $BK$ is tangent to $KLM$ at $K$, we again have a final angle chase killing the problem

$\measuredangle BKM =\measuredangle KLM =\measuredangle QLM =\measuredangle QPM =\measuredangle QPK$
$\implies PQ \parallel BK \subset BC$

As $APKQ$ is a parallelogram , $PQ$ bisects $AK$ , and the problem is solved.[\hide]

A simple solution with the Iran Lemma...

Let $I$ be the incenter of the triangle and let $M_B, M_C$ be the midpoints of $AC, AB$ respectively. Let $P'$ be the point lying on $(AI)$ such that $AP' || LK$. Since $AP' || LK$, we get $\angle P'AM = \frac{B-A}{2}$ which means $\angle P'FD = \angle P'IA = 90 - \angle P'AI = 90 - \frac{B}{2}$, but since $\angle BFD = 90 - \frac{B}{2}$, $D, F, P'$ are collinear which means $P' = P$ and so $P$ lies on $(AI)$, similarly, we can get that $Q$ lies on $(AI)$ as well. Also, since $\angle CID = 90 - \frac{C}{2}$, we get that $C, I, P$ are collinear

But now, by Iran Lemma, we know that since $P$ lies on $(AC)$ and on $DF$, it lies on $M_BM_C$. Similarly, so does $Q$. Therefore, $PQ$ bisects sides $AB, AC$

PS : This Problem was INMO 2000/1

[quote=grobber]We have $\angle AML=\angle MKL=\pi-\angle AQL$, meaning that $AMLQ$ is cyclic. In the same way we show that $ALMP$ is cyclic, so $PMLQ$ is cyclic.

Consider the inversion of pole $K$ and power $KL\cdot KQ$. It turns the line $PQ$ into the incircle of $ABC$, and it fixes the line $BC$. Since inversion preserves angles, and the angle between the incircle of $ABC$ and $BC$ is $0$ (i.e. they are tangent), it means that the angle between $PQ,BC$ is $0$, i.e. they are parallel. Call this $(*)$.

Since $APKQ$ is a parallelogram, it means that $PQ$ passes through the midpoint of $AK$. This, together with $(*)$, gives the desired conclusion.[/quote]

No need of inversion

@above please don't write absolutely useless things (at least in HSO) :)

IRAN lemma kills it. If BI cuts LK at Q' then AQB=90, BQ is perpendicular to MK => Q'=Q then by IRAN lemma Q is on the bisector of AB and AC. It is true for P as well.

[quote=Mathx]The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$, and the line through $A$ parallel to $MK$ meets $LK$ at $Q$. Show that the line $PQ$ bisects $AB$ and bisects $AC$.[/quote]

well activing this thread after a year , here's my solution:-
well i may have posted a LaTeX solution but i dont know how to draw diagrams through it  :maybe:


from $\left(x-\frac{1}{2}\right)^2\geq 0$ we have $x^2-x+\frac{1}{4}\geq 0$ .

$\Longleftrightarrow x+y\leq \frac{1}{4}$  ($\because x^2=-y$)

$\implies a^{x+y}\geq a^{\frac{1}{4}}$   ($\because 0<a<1$)

By AM-GM , $\frac{a^x+a^y}{2}\geq \sqrt{a^{x+y}}\geq \sqrt{a^{\frac{1}{4}}}=a^{\frac{1}{8}}$

$\Longleftrightarrow \log_a\left(a^x+a^y\right)\leq \log_a2+\frac{1}{8}$

 ;)

Nice question anyway  :)


inequalities
limit
real analysis
real analysis unsolved

Prove, that for all natural numbers $ n, k$ there exist $ M_{k}$ such as:

$ n^{k}\leqslant 2^{n}M_{k}$

For each $ k\in \mathbb{N}$ $ \lim_{n\to \infty}\frac {n^k}{2^n} = 0$ so, in particular, there exists an $ N_k\in \mathbb{N}$ so that for all $ n > N_k$ we have $ n^k < 2^n$.  The set $ \{\frac {n^k}{2^n}: 1\leq n \leq N_k\}$ is finite and therefore has a maximum; let $ M_k = \max  \{\max  \{\frac {n^k}{2^n}: 1\leq n \leq N_k \},1\}$, then by construction $ n^k\leq M_k 2^n$ for all $ n\in \mathbb{N}$.


听说用五笔字型比用拼音打字要快一些，怎么用五笔字型呀？需要背好多编码吗？

[quote="manchu_princess"]听说用五笔字型比用拼音打字要快一些，怎么用五笔字型呀？需要背好多编码吗？[/quote]
打繁體字可選擇用倉頡, 原理跟五筆應該差不多.


Let $ A\equal{}\{z\in \mathbb{C}\ |\ |z\minus{}1|\le 1\ ,\ |z\minus{}\epsilon |\le 1\ ,\ |z\minus{}\epsilon^2|\le 1\ ,\ \epsilon^3\equal{}1\ ,\ \epsilon \in \mathbb{C}\backslash \mathbb{R}\}$ . If $ z\in A$ prove that $ |z|\le 1$ .

It rather seems to me that $ A \equal{} \{0\}$, so $ z \in A$ implies $ z \equal{} 0$. Geometrically it's quite clear.

ANSWER to next post: Yes.

Can you prove algebracally that that set is $ \{0\}$

Algebraic proof. Start with following

[b]Lemma.[/b] [color=blue]For complex numbers, $ |z \minus{} \omega| \leq |\omega|$ is equivalent to $ |z|^2 \leq z\overline{\omega} \plus{} \overline{z}\omega$.[/color]

[i]Proof.[/i] $ |\omega|^2 \geq |z \minus{} \omega|^2 \equal{} (z \minus{} \omega)\overline{(z \minus{} \omega)} \equal{} (z \minus{} \omega)(\overline{z} \minus{} \overline{\omega}) \equal{} |z|^2 \minus{} (z\overline{\omega} \plus{} \overline{z}\omega) \plus{} |\omega|^2$, hence $ |z|^2 \leq z\overline{\omega} \plus{} \overline{z}\omega$. $ \quad \square$

Now use $ |\epsilon| \equal{} |\epsilon^2| \equal{} 1$, $ \overline{\epsilon} \equal{} \epsilon^2$, $ \overline{\epsilon^2} \equal{} \epsilon$, and $ 1 \plus{} \epsilon \plus{} \epsilon^2 \equal{} 0$.

By adding the three relations obtained by applying the Lemma, we get $ 3|z|^2 \leq (z \plus{} \overline{z})(1 \plus{} \epsilon \plus{} \epsilon^2) \equal{} 0$, hence $ |z| \leq 0$.
Therefore $ |z| \equal{} 0$ and so $ z \equal{} 0$.


1.

If (x,12,13) is Pythagorean triple, then x =____

2.

Write the conersion of 30 square yards to square inches unit miltipliers.

3.

Factor 80 into a product of prime factors.

1.) $5$

2.) I can't understand the question due to my limited english skills :)

3.) $2^4 * 5$

So change/convert 30 sqrae yards to square units for 2.

These problems are too easy for Getting Started.  (Moderator please move.)  Anyway,

[hide="Number 2"] 30 square yards = 270 square feet = 38880 square inches[/hide]

Agree with all answers posted.... the trickiest one is two but this is not a challenge....  :rotfl:
[hide]#2 \[ 30 sq. yds * (3 ft/yd)^2 * (12 in/ft)^2 = 38880 \] Simplicity be good :P [/hide]


[quote="GameBot"]What is the probability that Fatima gets fewer heads than tails if she flips 10 coins?[/quote]

The probability of getting more heads than tails is the same as less heads. So our probability is $ \frac{1}{2}(1\minus{}P(5))$, where P(5) is the probability of 5 heads.

Thus, our probability is $ \frac{1}{2}(1\minus{}\frac{\binom{10}{5}}{2^{10}}$, or $ \frac{1}{2}(1\minus{}\frac{63}{256})$.

Our final answer is thus $ \frac{193}{512}$.

Alternatively, we could use casework to find the probability to have 0, 1, 2, 3, and 4 heads:

0 heads: $\binom{10}{0}\left(\frac12\right)^{10}$
1 head: $\binom{10}{1}\left(\frac12\right)^{10}$
2 heads: $\binom{10}{2}\left(\frac12\right)^{10}$
3 heads: $\binom{10}{3}\left(\frac12\right)^{10}$
4 heads: ${\binom{10}4}\left(\frac12\right)^{10}$

Since $\binom{10}{n}=\binom{10}{10-n}$ for integers $n$ and $\binom{10}{0}+\binom{10}{1}+\cdots+\binom{10}{10}=2^{10}$, we see that $\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}=\frac{2^{10}-\binom{10}{5}}{2}=386$, so the probability of getting less heads than tails is $P(\text{1 head})+P(\text{2 head})+P(\text{3 head})+P(\text{4 head})=386\cdot \frac{1}{2^{10}}=\boxed{\frac{193}{512}}$.

Our probability has to be the same for heads and tails, so we find the probability of $5$ and $5$.  This is $\dfrac{\binom{10}{5}}{2^{10}}=\dfrac{63}{256}$.  We then take $1-\dfrac{63}{256}=\dfrac{193}{256}$ and divide by $2$ to get $\dfrac{193}{512}$

I’d think this would go a lot more in Binomial Probability than here

what's the probability of it landing on its edge

[quote=Skupp3]to @above. don't bump year(s) old posts. I know you are new to AoPS. Just a heads up.[/quote]

don't look at post number look at profile


How many ways can you make 45 cents with pennies, nickels, dimes, and quarters?  50 cents?  55?  60?

You can use casework:

First find the number of ways you can do this with zero quarters, then with one quarter.

0 quarters:
How many ways can you do this?  Test with 0 dimes, 1, 2, 3, 4 dimes.

1 quarter:
How many wyas? Test with 0,1, or 2 dimes.

This should be enough.

Same method for 50, 55, 60 cents.  As far as I know, there is no explicit formula that gives an answer to this.

This problem is equivalent to:

p+5n+10d+25q=45

How many solution this equation has, if p,n,d,q - non negative integers.

For $ 11 \minus{} 49$, the formula is $ 3 \plus{} \frac {\left\lfloor \frac n5\right\rfloor\left(\left\lfloor \frac n5\right\rfloor \minus{} 1\right)}2$.

I think its $ 3\plus{}\frac{\lfloor \frac{n}{5}\rfloor(\lfloor \frac{n}{5}\rfloor\minus{}1)}{2}$.

I typoed. Sorry!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231086

[quote="007math"][quote="math154"]For $ 11 \minus{} 49$, the formula is $ 3 \plus{} \frac {\lfloor n\rfloor(\lfloor n\rfloor \minus{} 1)}2$.[/quote]

Hmm...this formula looks like it doesn't work.  If we use 25, then we get 303.=, and obviously the answer is 13.  What do you mean by 11-49?  Does it mean 11-49 cents?[/quote]

$ \lfloor{n}\rfloor$ - the maximum number of nickels.

For example 25 cents: the maximum number of nickles - 5

3+ 5*4/2=3+20/2=3+10=13

For $ 50\plus{}$, just break it down into cases. If you have $ 1\plus{}$ quarters, you can take it away from $ n$ first and then apply the formula. But if you don't have any quarters, the casework is only with dimes, nickels, and pennies, which is a lot easier.

[quote="math154"]For $ 50 \plus{}$, just break it down into cases. If you have $ 1 \plus{}$ quarters, you can take it away from $ n$ first and then apply the formula. But if you don't have any quarters, the casework is only with dimes, nickels, and pennies, which is a lot easier.[/quote]

Okay....I do not get what you are saying....can you do that "take it away form n first" thing by showing how many possible ways for 50 or 55?

[quote="AIME15"]Sorry, the formula I gave is incorrect.  I forgot that quarters were allowed...[/quote]

The formula you gave is the formula that I meant to post which is the formula that I proved in a FTW Problems Discussion thread (a proof in the FTW fora?!!! :wink: ), so it's correct. The restrictions are $ 11\leq n\leq49$.

[quote="AIME15"]Doesn't $ \lfloor n \rfloor$ denote the floor value of $ n$, Bachukas?

Oh whoops math154, I took too many stupid pills this morning.[/quote]

What's the "floor"value?

No one has answered my other post...

[quote="math154"]For all real $ n$, $ \lfloor n\rfloor$ denotes the greatest integer less than or equal to $ n$.

And I don't understand your question.[/quote]

I was trying to say, can you explain your method after we pass 50 cents?  Thanks for the floors.

So say we have $ 73$ cents.
Then we know that we can either have no quarters or at least one quarter.
If we have at least $ 1$ quarter, then we have $ 73 \minus{} 25 \equal{} 48$ cents left to use. Since this is exactly what the formula can do, we plug it in: $ 3 \plus{} \frac {\left\lfloor \frac {48}5\right\rfloor\left(\left\lfloor \frac {48}5\right\rfloor \minus{} 1\right)}2 \equal{} 3 \plus{} \frac {9\cdot8}2 \equal{} 39$.
If we have no quarters, then this is a listing problem. Since the number of pennies can be calculated given the number of quarters (0 in this case), dimes, and nickels, we don't bother listing the pennies.
D N
7 0--1
6 (2,1,0)--3
5 (4-0)--5
4 (6-0)--7
3 (8-0)--9
2 (10-0)--11
1 (12-0)--13
0 (14-0)--15
For this case, there are $ 8^2 \equal{} 64$ ways. Adding up, we have $ 39 \plus{} 64 \equal{} \boxed{103}$.

So isn't there a general formula for past 50 cents? And don't forget half dollars.

Why does everyone insist on having a formula for everything....?

Method>Formula. Just because I found one formula doesn't mean I have to find 5 billion extensions. Find them yourself!


what is the maximum area of a quadrilatral with sides$1,4,7,8$?

if you remember the equation for area of a quadrilateral...

$A^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\frac{A+C}{2}$
then clearly to maximize the area, $\cos^2\frac{A+C}{2}=0$, so substituting, we get $A=18$

when the Cos is 0, you have a Cyclic quad

Yup, you can prove the max of any quadrilateral given the sides is cyclic

[quote="chess64"][hide]The maximum possible area of a quadrilateral is when it is cyclic, so by Brahmagupta's formula we have $K=\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{(10-1)(10-4)(10-7)(10-8)}=\sqrt{(3^2) \cdot (2 \cdot 3) \cdot (3) \cdot (2)}=3 \cdot 2 \cdot 3 = \boxed {18}.$[/hide][/quote]
[hide="A solution without the formula"]
Notice that the quadrilateral (let's name ABCD) is cyclic when $AB=1$, $BC=8$, $CD=4$, $DA=7$.
This is because $AB^2+BC^2=CD^2+DA^2=65$, which means $\angle ABC=\angle ADC= 90^\circ$.

Therefore, $\text{Area of ABCD}=\triangle ABC+\triangle ADC=\frac{1}{2}\cdot 1\cdot 8+\frac{1}{2}\cdot 4\cdot 7=\boxed{18}$ [/hide]


AwesomeMath
summer program
MathPath
AMC 8

I'm a 7th grader who got 23 out of 25 for AMC 8, and I made Nationals this year for mathcounts.

Do you think I'm good enough to go to AwesomeMath?
Well I was considering either MathPath or AwesomeMath.... any suggestions?

People who [i]didn't[/i] get to Nationals are considering AwesomeMath.

But the Admission Test is all about proofs. Did you take the Admission Test yet?

hey hey hey, making nationals isn't everything, especially if you live in a state like california. D:

to the OP: I haven't gone to mathpath but I've gone to amp so I can't really recommend one over the other, but I will say that amp is a LOT of fun and you get a lot of stuff done (but only if you are willing to work on your own time, because no one will tell you to do work sessions or anything). I think you're qualified enough to get in even if you don't have too much proof experience.

Well I am working on the admission test right now... I did the test A and it was pretty easy, then I realized that the early admission for AwesomeMath was already done, so I'm working on test B now.
For me, I find the admission test for MathPath a little more challenging, but i know that AwesomeMath and MathPath are both great camps and I'll learn a lot.... so thanks for everyone replying.

Umm.....I don't think the B test is even out yet....
Doesn't it come out the 28th?


the answer is not the square root of two

[quote="skatekld"]can anyone help me figure out how to do problems where say x^x^x=2
the answer is the square root of 2 and I don't understand how to do it.[/quote]
You have to make certain you are putting the formula correctly so that your question can be expressed more clearly.

You said x^x^x=2, to me it means $x^{x^{x}}=2$. This equation is not easy to solve. On the other hand, if your equation is

\[ x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=2 \]

with [b][size=150]3[/size][/b] extra dots on top of last x to indicate "and so on", then it is easier to solve.

The idea is this: you know
 \[ x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=2 \]
so
\[ x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=x^{\left(x^{x^{\cdot^{\cdot^{\cdot}}}}\right)}=x^2=2 \]
x has to be a positive real number, so $x=\sqrt2$

well, not the answer is the square rooot of two. it wasn't the way skatekld stated it.

If you mean $(x^x)^x$, then you use the rule $(a^b)^c = a^{bc}$.  In this case, $(x^x)^x = x^{x^2} = 2$.  Plugging in $x = \sqrt{2}$ gives $\sqrt{2} ^ 2 = 2$, which is true, so in that case $\sqrt{2}$ is a valid answer, although I am not certain how you are supposed to do this without already knowing the answer.

Like 10000th User said, if your question involves an infinite tower of $x^x$, then it's doable by a bit of reasoning.

I don't see any algebra way to can solve $x^{x^x}=2$. An approximate value of x is $x\approx 1.476684337$ by means of a graphing calculator

[quote="skatekld"]could you guys show me the way that JEngel thinks I mean:
x^x^x=2
Only three times x is used.
Thanks[/quote]

JEngel thinks you mean $x^{x^x} \neq (x^x)^x$.  To demonstrate why they are not the same thing, plug in $x = 3$, which gives you $x^{x^x} = 3^{27} = 7.62559... \times 10^{12}$, whereas $(x^x)^x = 27^3 = 19683$.

As a general rule of thumb, whenever you have a finite number of terms in a "exponent tower" evaluate it from the top to the bottom.

Well, x^x^x really means $x^{x^x}$, which has no solution in terms of elementary functions (that I know of, and I'm pretty sure about that one).

Solution to Ph-An's problem:

[hide="Solution"]
$f_{n+1}(x) = x^{f_n(x)}$

$\ln(f_{n+1}(x)) = f_n(x)\ln(x)$

Thus $\ln(L_{n+1}) = \lim_{x\rightarrow 0}{f_n(x)} \cdot \lim_{x\rightarrow 0}{\ln(x)}$.

If  $\lim_{x\rightarrow 0}{f_n(x)} = 0$, then $\ln(L_{n+1}) = 0 \Rightarrow L_{n+1} = 1$.

If $\lim_{x\rightarrow 0}{f_n(x)} \neq 0$, then $\ln(L_{n+1}) = -\infty \Rightarrow L_{n+1} = 0$.

We also have that $L_0$ = 1, so we have an oscilating sequence:

$L_0 = 1, L_1 = 0, L_2 = 1, L_3 = 0, \ldots$.

This can be expressed more succintly as $L_{2n} = 1, L_{2n+1} = 0$ for any $n\in\mathbb{N}$.

QED[/hide]

[quote="t0rajir0u"]

JEngel thinks you mean $x^{x^x} \neq (x^x)^x$.  To demonstrate why they are not the same thing[/quote]
They are the same thing if x = -1, 1 or 2  :P


linear algebra
matrix
superior algebra
superior algebra solved

let p be a prime number and A,B be nxn integer valued matrices so that AB=p(A+B). Prove that det (A-pI) - det (B-pI) is divisible with p^2-1 or p.

source: [i]Ferenc Kasco[/i]

[quote="Valentin Vornicu"]... AB=p(A+B)...[/quote]
Just easy observations:

(A-pI)(B-pI)=p^2I => (B-pI)(A-pI)=p^2I
after expansion and simplification AB=BA

det(B-pI)=p^(2n)1/det(A-pI)

Well, it shouldn't be that hard now, right? det(A-pI) and det(B-pI) are integers, so we have 2 cases:

(1) both of them are divisible by p => p | det(A-pI)-det(B-pI);

(2) Let's assume WLOG that p | det(A-pI). Then det(A-pI)=p2n => det(B-pI)=1 or det(A-pI)=-p2n => det(B-pI)=-1. It's easy to see that in this case p2 -1 | det(A-pI)-det(B-pI) because p2 -1 | p2n -1.

(A-pI)(B-pI)=p^2I => det(A-pI)det(B-pI)=p^(2n)
 but det(A-pI), det(B-pI) are integers each divides p^(2n),
p prime number.

det(A-pI)= \pm p^k, det(B-pI)= \pm p^(2n-k) where 0 \leq k \leq 2n

det(A-pI) - det(B-pI)= \pm  (p^k -  p^(2n-k))

If k=0 or 2n , p^k -  p^(2n-k)= \pm (p^(2n)-1) wich is divisible by p^2-1

If 1 \leq k \leq 2n-1, we can factor p in p^k -  p^(2n-k)


1)   Aflati cate cifre are numarul 2^2006.
  2)   Demonstrati ca 121 nu divide (n^2)-n-8, n natural.

Here's my solution for the second problem.Let's take a=(n^2)-n-8=>a=(n+5)(n-6)+22.Because (n+5)-(n-6)=11,
we have 11 devides n+5 or n-6.If 121 would devide a, then 11 devides (n+5)(n-6)+22=>11 devides (n+5)(n-6)=>
11/n+5 or 11/n-6.Because 11/(n+5)-(n-6)=>11/n+5 and 11/n-6=>121/(n+5)(n-6), but 121/a, then 121/22-false.

Oh, I'm sorry.I didn't see the first problem and I think is not good your solution.The number 2^2006 has at least
500 digits, because we have 2^2006>10^500(I think so).

era o mica greseala de calcul. am corectat-o.

Ciudat!Esti roman?Pai si atunci de ce vorbesti in engleza pe sectiunea asta?Oricum, asa fac si eu pe sectiunea francezilor...