Find the sum of the infinite geometric series $ 100\plus{}60\plus{}18....$

[hide="My Attempt"]
Seeing the common ratio for 100 to 60 is $ \frac{3}{5}$. But trying that once more, gets us 36 and not 18.
Can someone explain this?[/hide]

Where did you get this exercise? That sequence is not a geometric sequence.

Chris (ckck), I took that same test, MAO was it?
Just use the ratio of the first two if you want the given answer, but it's def. flawed.

[quote="Temperal"]Eh, the answer is just 250 if you take the first two terms as correct.[/quote]

[hide="click for solution"]Let $ S \equal{} 100 \plus{} 60 \plus{} 36 \plus{} \cdots$.

Thus, $ \frac {3S}{5} \equal{} 60 \plus{} 36 \plus{} \frac {108}5 \plus{} \cdots$

Subtracting the two, we have:

$ \frac {2S}{5} \equal{} 100$

$ S \equal{} 250$[/hide]

[quote="ZhangPeijin"]Are you sure its geometric?[/quote]

They're going under the assumtion that the third term is flawed and the first two are correct.

Assuming the sequence given is actually correct, it's not geometric and more terms would be needed to determine how the series is defined

[quote="b-flat"][quote="Temperal"]Eh, the answer is just 250 if you take the first two terms as correct.[/quote]

What an incredibly helpful post.

[hide="click for solution"]Let $ S \equal{} 100 \plus{} 60 \plus{} 36 \plus{} \cdots$.

Thus, $ \frac {3S}{5} \equal{} 60 \plus{} 36 \plus{} \frac {108}5 \plus{} \cdots$

Subtracting the two, we have:

$ \frac {2S}{5} \equal{} 100$

$ S \equal{} 250$[/hide][/quote]

Alternately, one could use the geometric series formula (which is the same thing as the subtraction, really, just rearranged, but...) which is what I did. Since it's just plugging into a formula, I didn't see a need for a solution.

Sarcasm is always incredibly helpful and will never offend anyone, by the way.

I'm not solving it but this might help. Try this sequence:

$ n\plus{}1\equal{}N(\frac{3}{5n})$

I'm not sure if this is the right way to write it, but it's saying to find a value for the n+1 term, use the following formula.

N=value of nth term.

Wait nvm.
Eh, perhaps it's not a typo?
And he just didn't post all the terms because he assumed we would get the pattern.

I don't think its geometric however a pattern that works is:

$ 100*\displaystyle\frac{3}{5}\equal{}60$

Then double the denominator:

$ 60*\displaystyle\frac{3}{10}\equal{}18$

Nothing new is being added / discussed, so this topic is locked.


Prove: n is composite if and only if $ \delta (n) > n \plus{} \sqrt{n}$.

Care to explain what $ \delta(n)$ is? The only $ \delta$ coming to mind are Kronecker's delta and the Dirac delta function, and I don't think you're talking about either one of those.

Perhaps the sum of the divisors?  (That plops this result into the category "true but not obviously interesting.")


A father divide 20 candys for his children is Tom,Marry,Max such that none of them has 20 that candys.After that ,each minute, one in three person gives  each  other person 1 candy. After some minute, the number candys of Tom,Marry,Max is $ a,b,c$ respectively.Find the possible value of $ (a,b,c)$.


IMO Shortlist
Functional inequality
algebra

Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m \plus{} n) \geq f(m) \plus{} f(f(n)) \minus{} 1
\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

[i]Author: Nikolai Nikolov, Bulgaria[/i]

Hints:
(1) The function $ f$ is nondecreasing.
(2) Let $ a: \equal{} f(1)$.  Then, show that $ f(n) \geq \left(f(a) \minus{} 1\right)n \plus{} 1$ and use this to prove that $ f(a) \geq a^{k} \minus{} a^{k \minus{} 1} \plus{} 1$ for all natural number $ k$.  Thus, $ f(1) \equal{} 1$.
(3) Let $ t$ be the maximum integer such that $ f(t) \equal{} 1$.  Show that $ f(n) \leq n \plus{} t$ for all $ n$.
(4) Suppose there exists $ b$ such that $ f(b) > b$, say, $ f(b) \equal{} b \plus{} d$.  Show that $ f(kb) \geq kb \plus{} k(d \minus{} 1) \plus{} 1$ for all $ k \in \mathbb{N}$.   Use this to justify that $ d \equal{} 1$.
(5) From (4), $ f(n) \leq n \plus{} 1$ for all $ n$.  In fact, for a given $ N$, $ f(N)$ can be anything from $ 1$, $ 2$, ..., $ N \plus{} 1$.
   (5.1) Let $ A < N$.  Define $ f(n) \equal{} 1$ if $ n \leq A$ and $ f(n) \equal{} A$ if $ n > A$.  Then, $ f$ satisfies the condition and $ f(N) \equal{} A$.
   (5.2) If $ f$ is the identity function, then $ f(N) \equal{} N$.
   (5.2) Any function in the form $ f(n) \equal{} c\left\lfloor\frac {n}{c}\right\rfloor \plus{} 1$, where $ c > 1$ is an integer, satisfies the condition.  When $ c \equal{} N$, we have $ f(N) \equal{} N \plus{} 1$.

(1) $ 1 \leq f(2007) \leq 2007$

if $ f(2007) \equal{} 2007 \minus{} k$, let $ f(x) \equal{} x \minus{} k (x \geq k \plus{} 1)$ and $ f(x) \equal{} 1 (x \leq k)$.

(2) $ f(2007) \equal{} 2008$

let $ f(2m) \equal{} 2m, f(2m \minus{} 1) \equal{} 2m$

(3) if $ f(n) \equal{} n \plus{} k ( k \geq 2)$

because $ f(m \plus{} n) \geq f(m) \plus{} f(f(n)) \minus{} 1 \geq f(m)$, $ f$ is increasing function.

Let $ m$ is greatest integer satisfy $ f(m) \equal{} 1$ (i)

If $ f(\alpha n) \geq \alpha (n \plus{} k \minus{} 1)$ then
$ f((\alpha \plus{} 1)n) \geq f(n) \plus{} f(f(\alpha n)) \minus{} 1 \geq n \plus{} k \plus{} f(\alpha (n \plus{} k \minus{} 1)) \minus{} 1$
$ \geq n \plus{} k \plus{} f(\alpha n) \minus{} 1 \geq n \plus{} k \plus{} \alpha(n \plus{} k \minus{} 1) \minus{} 1 \equal{} (\alpha \plus{} 1)(n \plus{} k \minus{} 1)$

Therefore, for all a, $ f(an) \geq a(n \plus{} k \minus{} 1)$

$ f(an \plus{} a(k \minus{} 1)) \geq f(a(k \minus{} 1)) \plus{} f(f(an)) \minus{} 1 \geq f(a(k \minus{} 1)) \plus{} f(a(n \plus{} k \minus{} 1)) \minus{} 1$
$ f(a(k \minus{} 1)) \leq 1 \Rightarrow f(a(k \minus{} 1)) \equal{} 1$ for all a.
Contradiction to (i)

$ \therefore f(2007)$'s possible value is 1 to 2008

[quote="orl"]Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m \plus{} n) \geq f(m) \plus{} f(f(n)) \minus{} 1\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

[i]Author: unknown author, Bulgaria[/i][/quote]
The function $ f(n) \equal{} n$ when $ 2007 \nmid n$ and $ f(n) \equal{} n \plus{} 1$ when $ 2007 \mid n$ satisfies the equation.
The function $ f(n) \equal{} 1$ when $ n \le k$ and $ f(n) \equal{} n \minus{} k \plus{} 1$ when $ n > k$ satisfies the equation for all $ k \in \mathbb{N}$.
So it is possible to archeive $ f(2007) \in \{1,2,3,...,2008\}$.

First of all $ f(n \plus{} 1) \ge f(n)$, so $ f$ is non-decreasing. Assume that $ f(n) > n \plus{} 1 \iff f(n) \equal{} n \plus{} c, c \ge 2$. Then $ f(2n) \ge f(n) \plus{} f(f(n)) \minus{} 1 \equal{} n \plus{} c \minus{} 1 \plus{} f(n \plus{} c) \ge n \plus{} 2(c \minus{} 1) \plus{} 1$, and by induction we get $ f(2^rn) \ge 2^rn \plus{} 2^r(c \minus{} 1) \plus{} 1$. Therefore:
$ f(2^rn \plus{} 1) \ge f(f(2^rn)) \ge f(2^rn \plus{} 2^r(c \minus{} 1) \plus{} 1)$, so $ f(2^rn \plus{} 1) \equal{} f(2^rn \plus{} 1) \equal{} \cdots \equal{} f(2^rn \plus{} 2^r(c \minus{} 1) \plus{} 1)$
For all $ m \in \mathbb{N}$ choose $ r$ such that $ 2^r(c \minus{} 1) > m$. Then:
$ f(2^rn \plus{} 1 \plus{} m) \ge f(2^rn \plus{} 1) \plus{} f(f(m)) \minus{} 1$. But $ f(2^rn \plus{} 1 \plus{} m) \equal{} f(2^rn \plus{} 1)$, so we obtain $ f(f(m)) \le 1$ or $ f(f(m)) \equal{} 1 \forall m \in \mathbb{N}$. Hence $ 1 \equal{} f(f(n)) \equal{} f(n \plus{} c) \ge f(n)$ giving $ f(n) \equal{} 1$, and we obtain a contradiction. Hence $ f(n) \le n \plus{} 1 \forall n \in \mathbb{N}$, and the only possible values are from the set $ \{1,2,3,...,2008\}$ which can all be reached.

[quote="Dumel"][quote="Mathias_DK"]by induction we get $ f(2^rn) \ge 2^rn \plus{} 2^r(c \minus{} 1) \plus{} 1$.[/quote]it's false for n=1  :wink:[/quote]
What do you mean by that?

[quote="Dumel"][quote="Mathias_DK"]Then $ f(2n) \ge f(n) \plus{} f(f(n)) \minus{} 1 \equal{} n \plus{} c \minus{} 1 \plus{} f(n \plus{} c) \ge n \plus{} 2(c \minus{} 1) \plus{} 1$ and by induction we get $ f(2^rn) \ge 2^rn \plus{} 2^r(c \minus{} 1) \plus{} 1$. [/quote]the basis of induction is false, because for r=1 we get
$ f(2n) \ge 2n \plus{} 2(c \minus{} 1) \plus{} 1$
not
$ f(2n) \ge n \plus{} 2(c \minus{} 1) \plus{} 1$
as you wrote above[/quote]
Ok. You mean $ r \equal{} 1$. That is just a typo. I forgot the $ 2$ in front of the $ n$. So the proof is ok - I can't edit my old post though.

Let $P(m,n)$ be the assertion that $f(m+n) \ge f(m) + f(f(n))-1$. We claim that $f(2007) \in \{ 1,2,\dots,2008 \}$. For all $k \in \{1,2,\dots, 2007 \}$ a function such that $f(2007)=k$ can be given by $f(n)=1$ for $1 \le n \le 2006$, $f(2007)=k$ and $f(n)=n$ for $2008 \le n$. A function such that $f(2007)=2008$ can be given by $f(n)=n$ for $2007 \not | n$ and $f(n)=n+1$ for $2007 |n$. Now we will prove that $f(n) \le n+1$. Assume for contradiction that there exists $k$ such that $f(k) = k+c$ where $c>1$. By $P(n,m-n)$ where $m>n$, it follows that $f(m) \ge f(n) + f(f(m-n))-1 \ge f(n)$. Hence $f$ is nondecreasing. Now we will prove by induction that $f(ik) \ge ik+i(c-1)+1$ for all $i \in \mathbb{N}$. Note that we have shown the case where $i=1$. Now assume that the claim holds for $i=1,2,\dots, n$. By $P(nk, k)$, it follows that $f((n+1)k) \ge f(nk)+f(f(k))-1=f(nk) + f(k+c)-1$ which since $f$ is nondecreasing satisfies that $f(nk) + f(k+c)-1 \ge f(nk)+k+c-1=(n+1)k+(n+1)(c-1)+1$. This implies that for any $m$, there exists $p \in \mathbb{N}$ such that $f(p)-p \ge m$. Now let $q$ be such that $f(q)-q \ge k$. By $P(f(q)-q, q)$, it follows that $f(f(q) \ge f(f(q)-q)+f(f(q))-1$ which implies that $f(f(q)-q) \le 1$. However, since $f$ is nondecreasing $f(f(q)-q) \ge f(k) = k+c > c > 1$ which is a contradiction. Hence $f(n) \le n+1$, which implies that $f(2007) \in \{ 1,2,\dots,2008 \}$.

thank you very much, NIKOLAI  NIKOLOV  for wonderful problems.

Someone  know official solution of the problem? [which  in shortlist or by autor!]

[hide=Solution] Let $P(m,n)$ be the assertion $f(m,n) \geq f(m) + f(f(n)) - 1$. Now, note that
1.$P(m,1) \Rightarrow f(m+1) \geq f(m)$
 and if for some $m$, $f(m) > m$, then
2. $P(f(m)-m, m) \Rightarrow f(f(m)-m) = 1$
3. $P(m,m) \Rightarrow f(2m) \geq f(m)+f(f(m))-1$
4. $1 \Rightarrow x \geq y \Rightarrow f(x) \geq y$

[b]Claim[/b]: $2008 \geq f(2007) \geq f(1)$.
[hide = Proof] Let $2007 = m$, and assume for the sake of contradiction $f(m) \geq m+2$. Then, from #3, we get $$f(2m) \geq f(m)+f(f(m)) -1 \geq f(m)+f(m)-1 = 2f(m)-1 \geq 2(m+2)-1 = 2m+3$$. By induction, we  have $$f(2^k m) \geq 2^km + 2^k - 1$$. Now, plugging $k=12$, we  get $$ f(2^{12} m) - m \geq 2^12-1 = 2047 \Rightarrow 1 = f( f(2^{12} m) - m) \geq (2^{12}-1 ) \geq f(2047) $$ (by #2) $$\Rightarrow 1 = f( f(2^{12} m) - m) \geq f(2047) \geq f(2017)  \geq 2019 $$, a contradiction. [/hide]

We now show that $f(2017)$ can attain all the values between $1, \cdots, 2018$ in the possibly wrong construction. [/hide]

Please check my construction :help:
[hide = Possibly wrong construction] (Okay, sorry I don't know how to display those curly bracket definations) Let $2007 \geq f(2007) = j$. Then define $f(x) = \text{1, if x} < \text{ 2017; j, if x = 2017 ; x otherwise}$.

For $f(2007) = 2008$, this construction should work:
$f(x) = \text{1 if x}< \text{2017, x+1 otherwise}$ [/hide]

We claim that for any such function $f,$ we have $f(2007)\in \{1,2,\dots, 2008\}.$ We first show everything is achieveable:

- In particular, $f(2007)=2008$ is witnessed by taking $f(2007k+m)=2007k+1,\ (0\leq m < 2007),$ and
- $f(2007)=k\leq 2007$ is achieved by taking $f(1)=\dots=f(2008-k)=1,f(2009-k)=2,f(2010-k)=3,\dots$

Now we show that we cannot have $f(n_0)\geq n_0+2$ for any $n_0,$ which will imply that these are the only possible values.

We first claim that $f$ is nondecreasing. Indeed, $P(n, m-n)$ for $m>n$ shows that $f(m)\geq f(n)+f(f(m-n))-1\geq f(n).$

Now assume FTSOC that we have $f(n_0) = n_0+d, d\geq 2.$ We can show with induction that $f(kn_0)\geq kn_0+(k-1)d-(k-1):$
\[f((k+1)n_0)\geq f(kn_0)+f(f(n_0))-1= f(kn_0)+f(n_0+d)-1\geq f(kn_0)+f(n_0)-1\geq (k+1)n_0+kd-k.\]
In particular, this shows that $f(x)-x$ can be arbitrarily large. Then
\[P(f(x)-x,x)\implies f(f(x))\geq f(f(x)-x)+f(f(x))-1\implies f(\text{big})=f(f(x)-x)=1,\]
whence from the nondecreasing property we conclude that $f$ must be constant at $1.$ This contradicts $f(n_0)\geq n_0+2,$ and we conclude with the solutions stated at the beginning.

Here's a solution sketch:

You get these series of things using an inductive-like telscoping procedure:

$\bullet f(1) = 1$
$\bullet$ For $r \neq 0$, if you have f(a) = qa+r, then
- $q = 1 \rightarrow f(r) = 1$
- $q \geq 2 \rightarrow f(r) = 1 \text{ and } f(qa+r) = 1$
$\bullet$ For $r = 0$, f(a) = 1 for $q \geq 2$.
$\bullet$ Otherwise, $f(a) \leq a$.

So let N be the largest integer for which $f(N) = 1$. This is a thing because not how $\forall n \leq N, f(n) = 1$. Hence, if N was arbitrarily large, all values under N would have a value of 1, which implies that N is the constant function. So we assume otherwise.

It follows that $f(n) \leq n+r_N \leq n+N$ for $0 \leq r_N \leq N$ and all $n \in \mathbb{N}$ from the above.

Then you use fact that f is nondecreasing and induction to get:
If $f(b) = b+d$, then $f(kb) \geq kb + k(d-1)+1$.
But for large enough k if $d \neq 1$, it breaks down from my thing. So d = 1. So $f(a) \leq a+1$.

For construction, look at Generic_Username ^.

So answer is f(2007) = {$1, 2, \dots 2008$}

Can anyone verify whether my solution is correct? :help:

We claim that $f(2007) \in \{1,2,\dots , 2008\}$. Firstly, assume that $f$ is non-constant, otherwise $f\equiv 1$. Denote by $P(m,n)$ the given assertion:
$$f(m+n) \ge f(m) + f(f(n))-1$$
Now, $P(1,1) \implies f(2) \ge f(1)+f(f(1))-1 \ge f(1)$. Also, $P(2,1) \implies f(3) \ge f(2) + f(f(1))-1 \ge f(2)$. Inductively we see that $$f(m+1) \ge f(m) \ \forall m  \in \mathbb{N}.$$
This shows that $f$ is a [i]non-decreasing[/i] function.
Also, $f(1) = \text{min} \{f(1),f(2),\dots\}$. Now, we proceed with the following claim:
[b]Claim:[/b] $f(1)=1$.
$\rightarrow$ Let $f(1)=k$, for some $k>1$. Then, $P(1,1) \implies f(2) \ge (k-1) + f(k) > f(k)$. Since $f$ is non-decreasing, it follows that $2>k$, contradiction. Hence, $f(1)=1$.

Now, $P(1,n) \implies f(1+n) \ge f(1) +f(f(n)) -1 =f(f(n))$. Again, non-decreasing nature of $f$ implies $n+1 \ge f(n) \forall n \in \mathbb{N}$.
Thus, $f(n) \le n+1$. Now, we give a construction which shows that all values up to $n+1$ are achievable (for $n=2007$).
[hide=construction]
Define $$f(N) = \begin{cases} f(N)=1 &, \forall N \le 2006\\
f(N)=k &, \text{for}\ N = 2007, k \in \{1,2,\dots , 2008\}\\
f(N)=N &, \forall N \ge 2008 \end{cases}$$

[/hide]

Suppose there existed some $x$ such that $f(x) \ge x+2$.

[color=#0f0][b]Claim 1:[/b][/color] $f$ is nondecreasing.
[color=#0f0][i]Proof:[/i][/color] Since $m+n>m$ and $f(m+n)\ge f(m)$, we get $f$ is nondecreasing. $\square$

[color=#0f0][b]Claim 2:[/b] [/color]$f(k)-k$ grows arbitrarily large as $k$ increases.
[color=#0f0][i]Proof:[/i][/color] Since $f(x)>x$, we know $f(f(x)) > x$. Then
\[ f(2x) \ge f(x)+f(f(x))-1 \ge 2f(x)-1 \ge 2x+3,\]
and similarly,
\[f(3x) \ge f(2x)+f(f(x))-1 \ge 2x+3+f(x)-1 \ge 3x+4, \]
and so on. In general, we can prove by induction that $f(nx) \ge nx + (n+1)$. In particular, $f(nx)-nx = n+1$ for any $n$, which means $f(k)-k$ can grow arbitrarily large over all $k$. $\square$

[color=#0f0][b]Claim 3:[/b][/color] $f$ is identically 1.
[color=#0f0][i]Proof:[/i][/color] $P(f(x)-x,x)$ gives $f(f(x)) \ge f(f(x)-x) + f(f(x))-1$, so $f(f(x)-x) \le 1$, which means $f(f(x)-x)=1$. Since $f(x)-x$ can grow arbitrarily large by claim 2, we conclude $f(N)=1$ for arbitrarily large $N$. But since $f$ is nondecreasing, $f\equiv 1$. $\square$

But now $1=f(x)\ge x+2$, contradiction. Hence, we cannot have an $x$ such that $f(x)\ge x+2$, and therefore $f(2007)\le 2008$. We can actually get $f(2007)=k$ for any $1\le k \le 2008$. If $k\le 2007$, just take $f(1)=f(2)=\cdots=f(2006)=1$, $f(2007)=k$, and $f(2008)=2008, f(2009)=2009, \ldots$. If $k=2008$, then let $f(x)$ be the greatest number at most $x$ that is 1 mod 2007.

Answer. $f(2007)\in\{1,2,\ldots,2008\}$.

Let $P(m,n)$ be the assertion.

$P(m,1)\implies f(m+1)\geq f(m)+f(f(1))-1$. By induction, $f(m)\geq (m-1)f(f(1))+f(1)-(m-1)$ for all $m$.
Thus, $f(f(m))\geq (f(m)-1)f(f(1))+f(1)-(f(m)-1)$. Taking here $m=1$, we get
$$f(f(1))\geq (f(1)-1)f(f(1))+f(1)-(f(1)-1)\implies 0\geq 1+(f(1)-2)f(f(1))\implies f(1)=1.$$
Thus, $f(m+1)\geq f(m)\forall m\in\mathbb{N}$. Hence we have $f(n)\geq f(m)$ for all $n\geq m$.

$P(1,n)\implies f(n+1)\geq f(f(n))\implies n+1\geq f(n)$. Thus, $2008\geq f(2007)$. Now, we claim that every such value is achievable.

For $f(2007)=1$, we can just use function $f(m)=1\forall m\in\mathbb{N}$.
For $2\leq f(2007)=a\leq 2007$, we can just use function $f(m)=1\forall m\leq 2007-a$ and $f(m)=m-(2007-a)\forall m\geq 2008-a$.
For $f(2007)=2008$, we can use $$  f(m) = \begin{cases} m & \text{if } m\not\equiv 0\pmod{2007}\\ m+1 &\text{if } m\equiv 0\pmod{2007} \end{cases} .$$
We are done.

The answer is $1\le n\le 2008$.
The construction for $n<2007$ is $f(x)=1$ for $1\le x\le n$ and $f(x)=n$ otherwise.
The construction for $n=2007$ is the identity.
The construction for $n=2008$ is $f(2007m)=2007m+1$ and $f(x)=x$ otherwise.
It is easy to verify that the above functions work.

We will proceed to prove that $n>2008$ is impossible.
Note that the function is nondecreasing by $f(1,x)$. Now, suppose that $f(2007)=n>2008$; then, $f(m,2007)$ gives that $f(m+2007)\ge f(m)+f(n)-1\ge f(m)+2008$. Let $N=1991104216969$; we know that $f(2007N)\ge 2008N+1$, then $f(1,2007N)$ gives that $f(2007N+1)\ge f(f(2007N))\ge f(2008N+1)>f(2007N+1)$.

Lowkey finding construction is hard

We claim that $f(m)\le m+1$ for all $m\in \mathbb{N}.$

Note that $f(m+1)\ge f(m)+f(f(1))-1\ge f(m)$ so the function is non-decreasing.

Assume for the sake of contradiction that $f(m)\ge m+2.$ Then substitute $m=f(m)-m$ and $n=m$ We get $f(m+f(m)-m)\ge f(f(m)-m)+f(f(m))-1,$ so $f(f(m)-m)=1.$ We claim that there must exist an integer $k$ such that $f(k)\ge k+c$ for any finite positive integer $c.$ Our proof is as follows, using induction:

[b]Base Case:[/b] Our assumption proves $c=2$ and note that $f(2m)\ge f(m)+f(m+2)-1\ge m+2+f(m)-1\ge 2m+3$ so $c=3$ also is true.

[b]Induction Hypothesis:[/b] Suppose $f(k)\ge k+d$ for some $k,d.$

[b]Induction Step:[/b] Note that $f(2k)\ge f(k)+f(f(k))-1\ge k+d+f(k+d)-1\ge 2k+2d-1,$ so if the statement is true for $c=d$ then it is true for all $c\le 2d-1.$

Taking this induction step enough times we can find that there must exist an integer $k$ such that $f(k)\ge k+c$ for any finite positive integer $c.$

Now, if $f(k)-k\ge c$ then $f(f(k)-k)\ge f(c)$ so $1\ge f(c)$ which implies $f(c)=1$ ... for all $c$ which is clearly a contradiction.

Therefore, $f(m)\le m+1$ which proves $f(2007)\le 2008.$ We provide the construction $f(1)=f(2)=\dots=f(2006)=1,$ which satisfies the equation, and then $f(2007)=k,$ $f(2008)=k+1,$ $f(2009)=k+2$ and so on $f(2007+c)=k+c.$ If both $m,n\le 2006$ obviously this works, and if $m\ge 2007>n$ then $f(m+n)=m+n+k-2007,$ $f(m)=m+k-2007,$ and $f(f(n))=1$ so the function works. If $n\ge 2007>m$ then $f(m+n)=m+n+k-2007$ and $f(f(n))=n+2k-4014$ while $f(m)-1=0$ so this is satisfied as long as $m\ge k-2007.$ If both are $\ge 2007$ then we have $f(m+n)=m+n+k-2007,f(m)+f(f(n))-1=m+n+3k-6022$ which is satisfied as long as $2k\le 4015$ or $k\le 2007.$ Then $m\ge k-2007$ is made true anyway.

Finally, we need to find a construction for $f(2007)=2008,$ so we provide $f(m)=2\lceil\frac{m}{2}\rceil$ which is easy to test that it works.

Note that $f$ is non decreasing. The identical 1 function obviously works. Hence assume $f$ is not identically 1. Let $M$ be the maximum such that $f(x)=x+M.$ It follows that $2x+M\geq 2f(2x)\geq f(x)+f(f(x))-1=f(x+M)+x+M-1\geq f(x)+x+M-1=2x+2M-1.$ Thus $M\leq 1\implies f(x)\leq x+1.$ So $f(2007)\in [1,2008].$

We can indeed construct such a function. For any $k\in [1,2008)$ let $f(x)=\max\{\lfloor kx/2007 \rfloor, 1\}$ and for $k=2008$ let $f(x)=x+1$ if $2007\nmid x$ or $f(x)=x,$ otherwise. Clearly these work.

[quote=mathleticguyyy]Let $N=1991104216969$; [/quote]

Any motivation behind this number?

[quote=TheCosineLaw] Now, $P(1,n) \implies f(1+n) \ge f(1) +f(f(n)) -1 =f(f(n))$. Again, non-decreasing nature of $f$ implies $n+1 \ge f(n) \forall n \in \mathbb{N}$. [/quote]
[quote=rafaello]Hence we have $f(n)\geq f(m)$ for all $n\geq m$.[/quote]

Am I being silly or non decreasing means $f(x)\geq f(y)~\forall x>y$ not $x\geq y.$ :)

The answer is all integers from $1$ to $2008$, inclusive. For $f(2007)=2008$ take $f(x)=2007\left\lfloor \frac{x}{2007}\right\rfloor + 1$ and for $f(2007)=k\le 2007$ take the function $f$ such that $f(2007-k+c)=c$ for all $c\ge 1$ and $f(n)=1$ for all $1\le n\le 2007-k$.

Now we show that $f(2007)<2009$ and in general that $f(n)<n+2$. Suppose FTSOC that $f(n)\ge n+2$. Note that
\[f(2n)\ge f(n)+f(f(n))-1\ge 2n+3\]
and continuing inductively we find that
\[f(2^kn)\ge f(2^{k-1}n)+f(f(2^{k-1}n))-1\ge 2^kn+2^k+1\]
where we make use of the fact that $f(n+1)-f(n)\ge f(f(1))-1\ge 0$ so that $f$ is nondecreasing. Then note that if $f(x)=x+k$ then we obtain $f(x+k)\ge f(k)+f(f(x))-1$ so that $f(k)=1$, where $k\ge 1$. Using this, we see that $f(2^kn)\ge 2^kn+2^k+1$ implies that $f(2^k+1)=1$ for arbitrarily large $k$, a contradiction. $\blacksquare$

[hide=proof that f(2007)<2009] We will prove that $f(n) \leq n+1$ for all $n$. Suppose that there exists $c$ such that $f(c) \geq c+2$.

Notice that for any $x$ such that $f(x)>x$, we have $f(f(x)) \geq f(f(x)-x)+f(f(x))-1$, which means $f(f(x)-x)=1$ for all such $x$. Take the largest $k$, such that $f(k)=1$ (it exists because $f$ can't be the constant $1$). Thus if $f(x)-x>0$, it follows that $f(x)-x \leq k$.

We now prove that $f(x)-x$ is unbounded using the existence of $c$. Obviously $f$ is non-decreasing, so $f(f(c))\geq f(c+2) \geq f(c) \geq c+2$. Plug $m=n=c$ to see that $f(2c) \geq f(c)+f(f(c))-1 \geq 2c+3$, so by induction using $m=(p-1)c, n=c$ we see that $f(pc)\geq pc+p+1$ for all positive integers $p\geq 2$. Thus $f(x)-x$ is unbounded, contradiction.

Doing this problem assuming N is positive integers not nonnegative, because someone clarified for me. Also can someone check my solution? After perusing the others it seems like mine is way too short and simple.. Honestly the hardest part was construction which is just intuition lol

We claim the answer is all integers from 1 to 2008.
First notice f is nondecreasing. $$1,n\implies f(1+n)\ge f(1)+f(f(n))-1\ge f(f(n))\implies 1+n\ge f(n).$$ It remains to show that these values all work. For f(2007)=1 take f as constant 1.
For $2\le f(2007)\le 2007$ we're motivated to let the inequality hold easily by just making the f() 1s, since they'll cancel out. Indeed, let $f(n)=1\forall n\le 2007-f(2007)$, which evidently has no "bad" impact on the condition, while $$f(n)=n+f(2007)-2007\le n\forall n\ge 2008-f(2007)\implies f(m+n)=m+n+f(2007)-2007\ge f(m)+f(f(n))-1=1\forall m+n\ge 2008-f(2007),m,n\in\mathbb{N},$$ while the inequality has equality case for $m+n\le 2007-f(2007)$ since it's just ones on both sides.
Finally, for $f(2007)=2008$ take $f(n)=n+1\forall 2007\nmid n$ and $f(n)=n\forall 2007|n$. We can easily check that this works by simply caseworking on whether or not m,n divisible by 2007 in the given condition to see that the ineq holds. $\blacksquare$


geometry
rectangle
analytic geometry
AMC

Rectangle $ ABCD$ has $ AB\equal{}8$ and $ BC\equal{}6$.  Point $ M$ is the midpoint of diagonal $ \overline{AC}$, and E is on $ \overline{AB}$ with $ \overline{ME}\perp\overline{AC}$.  What is the area of $ \triangle AME$?

$ \textbf{(A)}\ \frac{65}{8} \qquad
\textbf{(B)}\ \frac{25}{3} \qquad
\textbf{(C)}\ 9 \qquad
\textbf{(D)}\ \frac{75}{8} \qquad
\textbf{(E)}\ \frac{85}{8}$

It can be shown that $ \triangle AME \sim \triangle ABC$, because they both have a right angle and they share an angle in common. Thus, the scale factor of $ \triangle AME$ to $ \triangle ABC$ is $ \frac 58$. Therefore, $ ME\equal{}6\cdot\frac 58\equal{}\frac{15}{4}$, and $ AM\equal{}5$, so the area is
\[ AM \cdot ME \equal{} \frac 12 \cdot 5 \cdot \frac{15}{4}\equal{}\boxed{\frac{75}{8}\textbf{ (D)}}\]

A way to finish it faster is to observe that since the scale factor is $ \frac {5}{8}$, then $ MEC$ is $ (\frac {5}{8})^{2}$ or $ \frac {25}{64}$ of ABC. Since the area of $ ABC$ is $ 24$, $ \frac {25}{64}*24 \equal{} \frac {75}{8}$

Coordinate bashing is another way (I did it that way) but this is probably the more efficient way.


solve:
$ \frac {x(3 \minus{} x)}{x^{2} \minus{} 2} \equal{} \sqrt {5 \minus{} x}$


search
number theory solved
number theory

a and b are natural numbers so that (2^n)a+b is perfect square for each positeve integer of n.prove that a=0.

This one appeared at least three times on ML. Please try the search button...

Pierre.


Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle $I$ to that of circle $II$, then $r$ equals:

$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt 3 \qquad \text{(D)} \ 2\sqrt 2 \qquad \text{(E)} \ 2\sqrt 3$

[hide]Let $l$ be the side of the square.

Then $R_{I} = l\frac{\sqrt{2}}{2}$ and $R_{II} = \frac{l}{2}$.

So the ratio is $2$.[/hide]


Let $n$ be a positive integer with $n \geq 2$, and $0<a_{1}, a_{2},...,a_{n}< 1$. Find the maximum value of the sum
$\sum_{i=1}^{n}(a_{i}(1-a_{i+1}))^{\frac{1}{6}}$
where $a_{n+1}=a_{1}$


induction
real analysis
real analysis unsolved

Routine induction, all algebra. This has nothing to do with calculus.

Let $P_n=\prod_{k=1}^na_k$ and $S_n=\sum_{k=1}^n\frac{a_k}k$. We obviously have $S_1=P_1$ and $S_2=a+\frac{a}{2a-1}=\frac{2a^2}{2a-1}=P_2$.
In order for us to have $S_{n+1}=P_{n+1}$ given that $S_n=P_n$, we must have $P_n+\frac{a_{n+1}}{n+1}=a_{n+1}P_n$, which is equivalent to $P_n=\frac{a_{n+1}}{(n+1)(a_{n+1}-1)}$.
We prove this by induction. If $P_{n-1}=\frac{a_n}{n(a_n-1)}$, we have $P_n=a_nP_{n-1}=\frac{a_n^2}{n(a_n-1)}$.
Now, $\frac{a_{n+1}}{(n+1)(a_{n+1}-1)}= \frac{\frac{(n+1)a_n^2}{(n+1)a_n^2-na_n+n}}{(n+1)\cdot\frac{na_n-n}{(n+1)a_n^2-na_n+n}}$ $=\frac{a_n^2}{n(a_n-1)}=P_n$ and the induction step is complete. Our calculation of $S_n$ and $P_n$ for $n=1,2$ is a sufficient base case, and we are done.


calculus
ARML
algorithm
algebra
binomial theorem

Let x=1000^(1/3)-999^(1/3). Which integer is closest to 1/x? Can you generalize this result in a few ways?

Are you looking for like a differentials answer here?

This not a calculus problem.  Think factoring.

but it can be solved simply with calculus :\

Syntax Error wrote:[hide]i know this is cheating, but just to check the answer on the calculator. i got about 333.[/hide]

[hide]I got 299.8999:approx:300...[/hide]

ah. well i stand corrected then. sorry mathfanatic.

I believe the general statement you are looking for:

[hide]

x^(1/n) - (x-1)^(1/n)  :approx:  1 / (n*10^(n-1))

where x = 10^n.

*** EDIT ***

Actually, x doesnt have to be 10^n but the approximation is closest when x = 10^n

*** Edit ****

[/hide]

yes?

works for me.

P.S.
if you click the "view more emoticons" button, youll be able to find the approximate sign

Ok all this looks right. Do you remember the ARML problem (#6 2003)

Out of all the integers n, 1/n is closest to the fractional part of sqrt(123456789) for which value of n. See if you can use the same technique. (I know this problem has been asked before on the forum, but post it here too if you can figure it out.)

Complex, where do you or where are you planning on going to school?

I will be off to University of California at Santa Barbara, College of Creative Studies next month.

The expression is = 10-(1000-1)^(1/3).  So by using the binomial theorem, we have:

= 10 - [[C(1/3,0)*1000^(1/3)] +[C(1/3,1)*1000^(-2/3)*-1^1]....]
~10-10-1/3*1/100
~1/300

so 1/x=300.

ComplexZeta (or anyone else who did ARML) : For the ARML question, how did you solve it during the competition?  I used square root algorithm, but i also read other methods.

I realized that the fractional part of sqrt(123456789) is equal to sqrt(123456789)-sqrt(123454321) and then used the same method as for this problem.


let there be a race of $ 64$ horses. each time only $ 8$ horses can race and we can only know the order in which they finish. can we order the horses from the slowest to the fastest in $ 50$ races?
(each horse in every race has the same performance, that is its finishing time never changes)

Acording to me it's imposible, and we need 65 races

Let's call the horses by its finish place (the horse that will finish at the first place is "one")

We have the initial 8 races in which every horse race in an arbitrary order to get 8 horses at the 1st, 8 at the 2nd..... and 8 at the 8th place, but we know that one will finish at the first place, and two can finishi at the first place or at the second place, but in the match in which one played, three will finish at the 1st place, at the 2nd place (followed by one or two), or at the 3rd place followed by one and two and so on.

Then to get the one we do a match with all first places, to get two we do a race with the rest of the last match and with the horse that won the second place after one (because it has the greter posibilities of the second places to win in the firts place this match), and we repeat this operation 56 times to have the last 8 horses, then the final match will arrange those horses in their respectively places

If I'm wrong, let me know, because I got confuse like seven times with this problem

do we have a better result instead of $50$ here?


I have no idea how to solve 3., could you show me?
The others are not so hard, but I think you meant 2-hours?!


Worship the moderators, especially me :P .

Don't spam, as it's worthless. Also try to spoiler post ([hide]Stuff goes here[/hi*e] - just add the d in the second hide bracket instead of the *) whenever you're giving solutions.

0h ya dnt use chtsp33k cuz itz hard 2 unerstan n peepZ tink u dum if u use it

And talking intelligibly can help others help you, and it can help you help yourself.

 :D

Just try to have fun in the forums!

We don't like using internet kind of language,

like saying ,

y do u want 2 do something like that??

this meaning

why do you want to do something like that??

Treethings, you can just do this:

[hide]Stuff goes here[/hide]

Which is also what shinwoo has done to his signature.

[quote="EUCLA"][quote="Treething"]
0h ya dnt use chtsp33k cuz itz hard 2 unerstan n peepZ tink u dum if u use it[/quote]

what do you mean by that?
i can't understand...look at my flag :lol:[/quote]

He meant: Oh yeah don't use chat speak because its hard to understand and peeps think you dumb if you use it

thanks a lot...it could seems easy for you but not for me :lol:

can you explain your signature? just curious.

yes
this is a quote from Platone(I love him :blush: )...I don't know how to say his name in English..

it says" the guide Teeteto(his name), which help u to cross the river, said: the water itself will show you how deep it is "

can u understand?...if you speek Greek, i'll send you this quote and then you'll translate it. :lol:


trigonometry
function
calculus
calculus computations

How do you check that $ x(t)\equal{}\arcsin{\left(\frac{1}2 e^{\frac{1}2t^{2}}\right)}$ solves the
differential equation $ x'(t)\equal{}t\tan{x}$. Differentiating and substituting in the DE forces one to
recognize an identity i think, could someone help, please?

:oops: forgot the times $ t$ factor from the first equality on. After correction, e.g., the final equality will be your target function $ t\,\tan x(t)$


[quote="miyomiyo"][hide="Solution"]
Christmas is the $25{}^t{}^h$ of December.
Halloween is the $31{}^t{}^h$ of October.
We want to know why 25 DEC = 31 OCT
DEC refers to base ten
OCT refers to base eight
25 in base ten is 31 in base eight.
[/hide][/quote]

this was nice...


calculus
derivative
function
calculus computations

Find the derivative of the function at the point p in the direction of a.

f(x, y) = 4x^2 - xy + 5y^2 , p = (-3, -5), a = i - 2j

The formula is $ D_{\textbf{u}}f(\textbf{x})\equal{}\nabla f(\textbf{x})\cdot \textbf{u}$.  You can do the rest.  Remember, $ \textbf{u}$ is a unit vector.

I got the answer is 75, is that the correct one, the book shows different answer

$ \nabla f\equal{}(8x\minus{}y,\minus{}x\plus{}10y)$  We must normalize $ \vec{a}$: $ \textbf{u}\equal{}\frac{\vec{a}}{\parallel{}\vec{a}\parallel{}}\equal{}\frac{(1,\minus{}2)}{\sqrt{5}}$.  Then $ \nabla f(\minus{}3,\minus{}5)\equal{}(\minus{}19,\minus{}47)$ and $ (\minus{}19,\minus{}47)\cdot\textbf{u}\equal{}\minus{}\frac{19}{\sqrt{5}}\plus{}\frac{94}{\sqrt{5}}\equal{}\frac{75}{\sqrt{5}}\equal{}15\sqrt{5}$

You obviously ignored me when I said that $ \textbf{u}$ has to be a unit vector.


[hide=" Hint"]

If when multiplied by 3, the result is even, that means the numbers we're looking for must have all even digits. From here it's just casework.[/hide]

[quote="isabella2296"]that means the numbers we're looking for must have all even digits.[/quote]

Not necessarily; look at his example.

[hide="Hint"]You have to find multiples of 6.[/hide]

[hide="Another Hint"]Build a solution from the bottom-up, by constructing all numbers with $ d$ digits with just even digits, and sort them modulo a certain number, and then constructing numbers with $ d\plus{}1$ digits using a recurrence.

[/hide]

Well none of the odd integers would have all even digits when multiplied by 3, so we can narrow down our search to 1004 integers...
Then from there heavy casework...
bleh don't listen to me, im dumb

maybe ask nike?

[quote="AdithyaGanesh"][quote="mathboy101"]@ernie: 1334*3=4002, and 1334 isn't a mult. of 6...[/quote]

1334 may not be, but 4002 is.[/quote]

Exactly. ;)

[quote="mathboy101"]How many positive integers from 1-2009 have the property that, when tripled, they give a result having all even digits? Example: 216*3=648[/quote]

SarahAhmed - I don't know how to say this nicely, but try not to post if you have nothing to say.   :wink:

[hide="Another hint"]Basically, as ernie said, you are looking for multiples of 6.  Your highest number to consider is $ 2009/cdot3\equal{}6027$.

Therefore, we know that we only need to consider the following numbers:
$ 6, 12, 18, 24 \cdots 6024.$

Let's see if we can find a recurrence -

$ 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 \cdots$

$ 6$ works, $ 12$ doesn't, $ 18$ doesn't, and so on.

Let $ Y$ denote numbers that fulfill the conditions and $ N$ denote numbers that do not.

Now we have the sequence:
$ Y, N, N, Y, N, N, Y, Y, N, Y, N, N, Y, N, N$

We can easily eliminate the multiples with an odd hundreds or thousands digit.

You can easily continue from here I presume?  There might be an easier way though.[/hide]

was wanting to get an explanation by somebody who would be willing to if they saw me as lost instead of having nothing to say....... :(

If that is the case, express what you don't understand and someone will try to help you.

In any case, I'm sure the AoPS community is willing to help you understand.

[hide="Similar Solution"]Let your number be $ abcd$, and let $ bcd$ be an arbitrary three-digit number consisting of solely even digits. We limit ourselves to $ a \le 4$. We note that $ 0$, $ 2$, and $ 4$ are pairwise different modulo $ 3$, so exactly one of them can be appended to the $ bcd$ such that $ abcd$ is divisible by $ 3$ (since $ a\plus{}b\plus{}c\plus{}d\equiv 0$ mod $ 3$ in that case). There are $ 5^3$ possibilities for $ bcd$, minus one since $ 0$ is not a valid number, so we have $ 124$ valid numbers less than $ 6000$.

We then bruteforce check the rest of the numbers between $ 6000$ and $ 6027$, of which there are $ 3$, $ 6000$, $ 6006$, and $ 6024$. Thus, there are $ \boxed{127}$ solutions.

[/hide]


limit
induction
function
real analysis
real analysis unsolved

Find all positive real numbers like $\alpha$, so that if $a_{n}=({\alpha}^{\frac{1}{\alpha}})^{a_{n-1}},a_{1}=1$, then $\lim a_{n}=\alpha$.

If $\alpha=1$ then $a_{n}$ is constant and equal to 1. So $\lim a_{n}=1=\alpha$ and then $\alpha=1$ is a solution to the problem.

If we suppose now that $\alpha>1$ -> $\frac{ln(\alpha)}{\alpha}>0$ ->
$\alpha^{\frac{1}{\alpha}}>1$. And by induction I´ll prove that $1<a_{n}<\alpha$. For $n=2$ this is true and so we can write $\alpha^{\frac{1}{\alpha}}<(\alpha^{\frac{1}{\alpha}})^{a_{n-1}}<(\alpha^{\frac{1}{\alpha}})^{\alpha}$ -> $1<a_{n}<\alpha$.

We can notice that $ln(a_{n})=a_{n-1}\frac{ln(\alpha)}{\alpha}$ ->
$\frac{ln(a_{n})-ln(a_{n-1})}{a_{n-1}}=\frac{ln\alpha}{\alpha}-\frac{lna_{n-1}}{a_{n-1}}$.

And if $e \geq \alpha>1$, we will have $e \geq \alpha>a_{n-1}$. But the
function $f$ defined by $f(x)=\frac{ln(x)}{x}$ is increasing on $(1,e]$
and so we have $\frac{ln(\alpha)}{\alpha}-\frac{ln(a_{n-1})}{a_{n-1}}>0$ -> $ln(a_{n})-ln(a_{n-1})>0$ -> $a_{n}>a_{n-1}$. At this point we
can view that $a_{n}$ is monotonic and bounded and so it´s
convergente. Hence, exist $\lim a_{n}=L$ and so we´ll have $\frac{ln\alpha}{\alpha}-\frac{ln(L)}{L}=0$ if and only if $L=\alpha$.

If we assume that $\alpha>e$ we´ll have by induction again that $a_{n}<e$ for all natural $n$. For $n=1$ the result is true. Assum that $a_{n-1}<e$ -> $ln(a_{n})<e\frac{ln(\alpha)}{\alpha}$. Mas se $\alpha>e$ ->
$\frac{ln(\alpha)}{\alpha}<\frac{1}{e}$ -> $a_{n}<e$. So if this limit
exist it´ll never be equals to $\alpha$ and so we can´t have $\alpha>e$.

For $0 < \alpha < 1$ we can prove that the answer will be $[\frac{1}{e},e)$

Clearly the sequence isn´t definied for $\alpha \leq 0$.

[quote="mffmartinelli"]By a simliar way we can prove taht for $0<\alpha<1$ the limit will be equals to $\alpha$.[/quote]
I think the correct answer is the interval $[\frac{1}{e},e]$, not  $(0,e]$. For a hint take a look at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=89869[/url].

I didn´t take care with $0 < \alpha < 1$ because I thought that this will be analogue. and the answer is really $[\frac{1}{e},e]$. I´ll edit my solution.


Hi, does anyone know how to do the following questions?

In each of the following cases, find all roots of f(x) in F and factor f(x) as the product of irreducible polynomials in F[x].

(a) F=R, f(x) = (x^6)-1 in R[x]
(b) F=Z3, f(x) = (x^5)+x+1 in Z3[x]
(c) F=Z5, f(x) = (x^3)+2x+3 in Z5[x]

Thanks very much guys....


[color=green][b]hameye adade avalle p va q ra biaBd ke baraye anha hamneheshTe [/b][/color]

[color=olive]$ a^{3pq}\equiv{a}$  (be peymaneye $ 3pq$)[/color]

[color=green][b]baraye har adade sahihe a bargharar bashad!! [/b][/color]

 :wink:

em,nemidoonam dombale rahe halli,ya inke be nazaret soale ghashangi boodeo gozashti ke baghieam estefade konan,dar har soorat age kasi khas hallesho bebine mitoone bere [url=http://www.mathlinks.ro/viewtopic.php?p=335786#335786]inja[/url]

har soali ye juri ghashange :roll:

man in topic o Ddam!!amma  mikhastam rahaye Bshtari ham peyda she :wink:


Finitely many convex subsets of $\mathbb R^3$ are given, such that every three have non-empty intersection. Prove that there exists a line in $\mathbb R^3$ that intersects all of these subsets.

Seems pretty similar to Helly's theorem...

It is. Consider any plane $\pi$. The projection of any convex set $K\subseteq \mathbb{R}^3$ on $\pi$ is a plane convex set. Therefore  the family of the projections have the property of every three having non-empty intersection; thus, by Helly's theorem in the plane, the projections have a common point in $\pi$. The normal at that point to $\pi$ intersects all the original sets. Clearly, this may be generalized to any dimension $n$ where any $n$ of the convex sets have non-empty intersection. So we proved even more, namely that for any given direction there exists a stabbing line, parallel to that direction (take $\pi$ normal to the direction).

For the plane, this was Problem 23 in [Hadwiger & Debrunner - Combinatorial Geometry in the Plane], with a different proof, that showed that such a line exists through any point in the plane. My method above shows such a line exists, parallel to any given direction in the plane. The book result follows from it via a continuity argument.


We have to count the number of ways to put $ 10$ balls into $ 4$ urns such that each urn contains at least $ 1$ ball. For that first we put $ 1$ ball into each of the urns. Then it remains to count the number of ways to put the remaining $ 6$ balls into the $ 4$ urns which, by the balls in urns formula, is $ \binom{6 \plus{} 4 \minus{} 1}{6}$.

I've already thought of that. But teens are different  :)  :roll:

I mean they are distinguished  :D (new word for me hoho)

[quote="bibobeo"]Ten teens go and watch 4 films. How many cases that each film has at least 1 teen watches?[/quote]
I think using Exclusion principle will work in this case! Your problem is classical! :)
Hint:How many cases that each film doesn't have any teens?
PS: How are you doing? For a long time I haven't had any chance to talk to you privately! :lol:

[quote="bibobeo"]I've also thought of that, ghik hoho.
And, we think we'll have a lot cases to deal with, since it can be 1 film with no audience or 2 films, ...
Moreover, these films are distinguished too.

I think there should be some formula for it.
What I have in my mind is
<<How many surjectives from 10-element set to a 4-element one?>>

Sorry for my bad English.

@ghik: ý tui là số toàn ánh từ tập 10 sang tập 4. Nhưng tui ko biết tiếng Anh, ông bíêt viết lại dùm, thx.[/quote]
OK! I think I can "touch" your problem! It must be a special case of Euler problem! This problem can be proposed like the following:
Find positive integers for this equation: x_1+x_2+...+x_10=4
Using Exclusin will work when you examine everycases that have at least one teen not seen any films! :)
PS:I haven't known that words yet! :blush:

[b][EDIT: I made a mistake in this post. See my next post after this.][/b]

I may have misunderstood the original question, but isn’t the answer $ 4^{10} \minus{} 4\cdot3^{10} \minus{} 6\cdot2^{10} \minus{} 4$?

$ 4^{10} \equal{}$ number of ways they can watch what films, without restriction

$ 4\cdot3^{10} \equal{}$ number of ways they can watch the films such that exactly one film is not watched

$ 6\cdot2^{10} \equal{}$ number of ways they can watch the films such that exactly two films are not watched

$ 4 \equal{}$ when all of them watch the same film

[quote="Jane_Bennet"]I may have misunderstood the original question, but isn’t the answer $ 4^{10} \minus{} 4\cdot3^{10} \minus{} 6\cdot2^{10} \minus{} 4$?

$ 4^{10} \equal{}$ number of ways they can watch what films, without restriction

$ 4\cdot3^{10} \equal{}$ number of ways they can watch the films such that exactly one film is not watched

$ 6\cdot2^{10} \equal{}$ number of ways they can watch the films such that exactly two films are not watched

$ 4 \equal{}$ when all of them watch the same film[/quote]

There's something wrong in your solution, Jane.
For example, let say 4 films are A,B,C,D.
Then assume that film A is not watched. So we have B,C,D to watch.
Each teens have 3 choices, but they can all choose 1 film to watch.
 :?: Is that a mistake? Or I was so silly?
Hope that you can understand me.

[quote="Jane_Bennet"]I’m sorry, I got the whole thing the wrong way round. :oops:

This should be what I mean. My answer should be $ 10^{4} \minus{} 4\cdot10^3 \minus{} 6\cdot10^2 \minus{} 4$:

$ 10^{4} \equal{}$ number of ways they can watch what films, without restriction

$ 4\cdot10^3 \equal{}$ number of ways they can watch the films such that exactly one film is not watched

$ 6\cdot10^2 \equal{}$ number of ways they can watch the films such that exactly two films are not watched

$ 4 \equal{}$ when all of them watch the same film[/quote]

Again, I just don't get it  :oops:
Consider 4 films A,B,C,D and D is not watched.
Pick film A, there are 10 cases that teens watch it.
But 1 teen just watch 1 film, how do you know that how many teens watch that film???
Your second solution seems less persuasive than the first one.

In my 4th post, (also this) I think (just think, I have no answer yet) that we can count the number of surjectives from a 10-element set to a 4-element one?
What do you think? This confuses me a lot.(It's just a sudden question from my friend.)

@ghik: I don't think Euler problem works here! Reason? Teens are distinguished.

This constraint, I mean teens are distinguished just make this equation have less solutions than usual but it doesn't make the way to find those solutions satisfying x1<x2<...<x9<x10! You should consider the solution of Euler's problem again,bibobi! :)

I think that it should be x1+x2+x3+x4=10. Since your equation cannot have positive integer roots.
But I don't know what is the next step (for same reason).

The answer I get until now is
$ (4C0)*4^{10} \minus{} (4C1)*3^{10} \plus{} (4C2)*2^{10} \minus{} (4C3)*1^{10}$

This is definitely right.  I solved a similar problem with the same method.  The problem asked the number of ways of assigning 8 teachers to 4 schools given that each school must get at least one teacher.  So,

$ 4^8 \minus{} {4 \choose 1} \cdot 3^8 \plus{} {4 \choose 2} \cdot 2^8 \minus{} {4 \choose 3} \cdot 1^8 \equal{} 40824$

My friend wrote a java program and got the same answer.

here is the java program:

public class Main {
    public static void main(String[] args) {
        int count=0;
        int teacher[]=new int[8];
        for( teacher[0]=0;teacher[0]<4;teacher[0]++)
        for( teacher[1]=0;teacher[1]<4;teacher[1]++)
        for( teacher[2]=0;teacher[2]<4;teacher[2]++)
        for( teacher[3]=0;teacher[3]<4;teacher[3]++)
        for( teacher[4]=0;teacher[4]<4;teacher[4]++)
        for( teacher[5]=0;teacher[5]<4;teacher[5]++)
        for( teacher[6]=0;teacher[6]<4;teacher[6]++)
        for( teacher[7]=0;teacher[7]<4;teacher[7]++)//4^8
        {
            int school[]=new int[4];
            for(int i=0;i<teacher.length;i++)//counted num in ea school
                for(int j=0;j<4;j++)
                    if(teacher[i]==j)
                        school[j]++;
            if(school[0]!=0&&school[1]!=0&&school[2]!=0&&school[3]!=0)//restrictions
            {
//                for(int i=0;i<teacher.length;i++)
//                    System.out.print(teacher[i]+" ");
//                System.out.print(" ");
//                for(int i=0;i<school.length;i++)
//                    System.out.print(school[i]+" ");
//                System.out.println();
                count++;
            }
        }
        System.out.println(count);
    }
}

Can you explain your solution in Math, andrsncpr?
I have problems in explaining due to my poor English.
Thank you.

Well the math part is just the Principle of Inclusion and Exclusion.  I can't explain the java code: my friend wrote it...

[quote="madness"]This method is also called the Inclusion Exclusion Priciple or the Sieve Priciple
See http://en.wikipedia.org/wiki/Inclusion_exclusion_principle[/quote]

Just read that link that madness posted.

I’m sorry, my two posts above are wrong. Please ignore my two posts above. :wallbash:

I’ll start again. First, the total number of ways in which the 10 teens can watch the 4 films is $ 4^{10}$. (Each teen has four choices of fims, and there are 10 teens.)

Now we want to find the number of ways they can watch so that at least one film is not watched. This will then be subtracted from $ 4^{10}$ to get the required answer.

If exactly three films are not watched, then all of them have gone to see the same film. Hence there are 4 ways this can happen.

Now for if exactly two films are not watched. Each teen now has 2 choices of films, so they can watch them in $ 2^{10}$ ways. But this includes the 2 possibilities that they all watch the same film, so we must subtract 2 from that. And you can choose 2 films from 4 in $ ^4\mathrm{C}_2\equal{}6$ ways. Hence the number of ways the teens can watch such that exactly 2 films are not watched is $ 6\cdot(2^{10}\minus{}2)$.

And now for if exactly one film is not watched. Each teen has 3 choices of films, so they can watch them in $ 3^{10}$ ways. But this includes cases when they all watch the same film (3 possibilities) and when exactly one of the three films is not watched ($ 3\cdot2^{10}$ possibilities). So the number of ways of watching such that exactly one film is not watched is $ 4\cdot(3^{10}\minus{}3\cdot2^{10}\minus{}3)$.

Hence my final answer (total number of ways of watching such that no film is unwatched) is:

$ \color{white}.\hspace{25mm}\color{black}\boxed{4^{10}\minus{}4\minus{}6\cdot(2^{10}\minus{}2)\minus{}4\cdot(3^{10}\minus{}3\cdot2^{10}\minus{}3)}$

What if I increase the number of films? Will you consider every single little case?
I think there must be some formula for this kind of problem.  :)

Here's a solution that uses PIE.  Jane_Bennet's solution works, but it takes more work and doesn't generalize as well.  If you don't want to read the link about PIE, I'll explain it now.  PIE is a process you can use to count the number of ways for at least one of *something*.  Here we're counting the number of ways for there to be at least one unwatched film.  We count these possibilities by designating a film to be unwatched and then counting the possibilities with that film unwatched, and then we repeat for the three other films (we can just multiply by 4 since there's nothing special about any one film).  This clearly overcounts all cases where more than one film is unwatched, so we correct our overcounting by designating pairs of films to be unwatched and subtracting these cases the correct number of times.  Then  we correct our counting for the cases where more than two films are unwatched, and so on.  Here's the solution for this problem.

1 unwatched film: The 10 teens choose between 3 films.  There are $ 3^{10}$ possibilities here.  We must add this 4 times since there are 4 films.

2 unwatched films: For each pair of films, we counted the cases where both are unwatched twice above.  We only want to count these once, so we must subtract each possibility once.  Here the 10 teens choose between 2 films, so there are $ 2^{10}$ possibilities.  We must subtract this 6 times since there are $ \binom{4}{2}\equal{}6$ pairs of films.

3 unwatched films: For each group of 3 films, we counted the cases where all three are unwatched 3 times in the first step and subtracted these cases 3 times in the second step.  Therefore we must now add these cases once.  All the teens must go to the remaining movie, so there is only one possibility.  We must subtract this 4 times since there are $ \binom{4}{3}\equal{}4$ groups of 3 films.

4 unwatched films: This is impossible, so we don't have to correct for any more overcounting.

The answer is $ 4^{10}\minus{}(4\cdot 3^{10}\minus{}6\cdot 2^{10}\plus{}4)$.

[quote="matt276eagles"]

2 unwatched films: For each pair of films, we counted the cases where both are unwatched twice above.  We only want to count these once, so we must subtract each possibility once.  Here the 10 teens choose between 2 films, so there are $ 2^{10}$ possibilities.  We must subtract this 6 times since there are $ \binom{4}{2} \equal{} 6$ pairs of films.

[/quote]

hm when the 10 teens choose between 2 films, what if they all choose the same film? This would lead to 3 unwatched films, so wouldn't the answer for this case actually be $ 2^{10} \minus{} 2$? This would also mean for the case of 1 unwatched film, there would be $ 3^{10}\minus{}3$ possibilities.

Yes, but when we consider this case we're really looking at cases with 2 or more unwatched films.  After this case we've counted the correct number of possibilities for exactly one or two unwatched films, but we still need to correct our counting for 3 unwatched films - which we do in the next step.

This is the difference between a pure casework approach and a PIE (Principle of Inclusion-Exclusion) approach.

oh ok that makes sense. good solution btw  :)

[quote="Jane_Bennet"]
Hence my final answer (total number of ways of watching such that no film is unwatched) is:

$ \color{white}.\hspace{25mm}\color{black}\boxed{4^{10} \minus{} 4 \minus{} 6\cdot(2^{10} \minus{} 2) \minus{} 4\cdot(3^{10} \minus{} 3\cdot2^{10} \minus{} 3)}$[/quote]

Well you havent understood PIE correctly. You need not uncount the cases as you are using PIE. i.e for example you needn't do $ 6.(2^{10} \minus{} 2)$ you only need to do $ 6.(2^{10})$

This can be solved with recursion as well, but beware, it isn't pretty.


calculus
derivative
limit
calculus computations

need to solve this using the four steps or the long method i did my best but i am really no math person and ill fail the subject if i wont be able to pass this qualifying assignment soon

1. y=(7-2x)^3

2. y=5x-3x^2+4x^4

3. y = (x^2-1)/(x^2+1)

please help

I have no idea what you mean by "using the four steps or the long method." That's not standard calculus terminology. I do have the awful suspicion that it involves the raw definition of the derivative: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$ or its equivalent in different notation. Any situation in which the chain rule, the product rule, or the quotient rule could be usefully applied would make for a tedious (and pointless) exercise if you tried to use the raw definition.

If you just want to know these derivatives, then you [b]must[/b] have complete mastery of the chain rule, the product rule, and the quotient rule - especially the chain rule.


The eyes of a basketball player are 5 feet above the floor. The player is at the free-throw line, which is 15 feet away from the center of the basket rim. What is the angle of elevation from the player's eyes to the center of the rim?

NOTE: The rim is 10 feet above the ground.

[hide="Hint"]Try making a right triangle with the player's eyes, the center of the rim, and the point that is 5 feet below the center of the rim. Then just use basic trigonometry.[/hide]

You're basically just posting nearly identical problems over and over, but w/different #s. What exactly about these types of problems don't you understand?


[quote="munchman"]How many rays can be formed from 4 collinear points?[/quote]
If by ray you mean a half-line bounded by one point and extending to infinity in the other direction, then there's infinitely many rays from a given pt, so answer is infiniteky many...
What's wrong? :maybe:

He said How many rays can be formed from [color=red]4 collinear points[/color]?

So what's wrong!?

Look at one of the points: $A$. You can draw infinitely many rays from the point so that none are on the line on which the 4 pts lie...


LaTeX
integration
trigonometry
real analysis
real analysis unsolved

here is some good problem in asymptotic. try to solve

Why don't you use LaTeX to actually write your questions into the forum?  I think many users are dis-inclined to download files, and .doc is an unquestionably inferior product for writing mathematics.

You might want to look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]here[/url]. Math mode is enabled on the forum; the rest of $\LaTeX$ is not, and you don't need it.

$\frac{1}{2}-\frac 1{23}+\int^{1}_{0}f^{2}(x_{1}) dx = \cos x+\sin \frac{1}{x}$

thanks very much.
Only today In classroom we begin study program Latex

this/two/problems/solved easy, because inj first: $x=u+\alpha$ and then we put it in our equation and by tayler find all other


When you order old AMC tests at
http://www.unl.edu/amc/d-publication/publication.html

, and on the order form it says "10 sets of 1 contest for $5," does that mean I can get 10 different AMC 12 tests for 5 dollars or do I get 10 copies of the same test for 5 dollars? Thanks!

10 copies of one test.

The use of that is for school since it's illegal to copy them in anyway.

Hope this helps!

Alright thanks!

Also, is it useful to order AIME/Alternate AIME's even though they are on the Kalva page???

Kalva are X-brand. If you want the real ones, buy them.

There's no practical difference. It's just the answer packet has better explanation.

I didn't buy them though. :D

I just ordered by mail. To those who have been ordering, how long will it take for me to receive it?


Let $A,B,C,D$ be four distinct points on a circle such that the lines $(AC)$ and $(BD)$ intersect at $E$, the lines $(AD)$ and $(BC)$ intersect at $F$ and such that $(AB)$ and $(CD)$ are not parallel.

Prove that $C,D,E,F$ are on the same circle if, and only if, $(EF)\bot(AB)$.

Nice, but simple for problem 3.

"$\Rightarrow$" $CEDF$ is cyclic , so $\angle{ACB}=\angle{BDF}=\angle{BDA}$, and now because $A,D,F$ are collinear we have that $\angle{BDA}=\angle{BDF}=90$, , therefore $E$ is the ortocenter of triangle $\triangle{FAB}$, thus $EF \perp AB$.

"$\Leftarrow$" Denote $P \in AB \cap CD$, thus $EF$ is the polar of $P$ w.r.t to the circle, so $PO \perp EF$, where $O$ is the center of the circle.
But $EF \perp AB$, and $P \in AB$, so $P,A,O,B$ are collinear, thus $\angle{ADB}=\angle{ACB}=\angle{BDF}=\angle{ACF}=90$, so $CEDF$ is cyclic.

$ \Rightarrow$ $ CEDF$ is cyclic, then $ \angle{ABD}\equal{}\angle{ACD}\equal{}\angle{EFD}\equal{}\alpha$, and $ \angle{BAC}\equal{}\angle{BDC}\equal{}\angle{EFC}\equal{}\beta$. Because $ ABCD$ is cyclic we also have that $ \angle{DBC}\equal{}\angle{CAD}\equal{}\gamma$. Since $ F$ is the intersection of $ AD$ and $ BC$, we have that $ \angle{BAF}\plus{}\angle{ABF}\plus{}\angle{AFB}\equal{}2\alpha\plus{}2\beta\plus{}2\gamma\equal{}180$. Therefore $ \alpha\plus{}\beta\plus{}\gamma\equal{}90$. Finally, we notice that $ \angle{ABF}\plus{}\angle{BFE}\equal{}\alpha\plus{}\gamma\plus{}\beta\equal{}90$. We conclude that $ EF\perp AB$.

$ \Rightarrow EF\perp AB$. Since $ AB$ and $ CD$ are not parallel,  we define $ G\equal{}AB\cap CD$. Since $ ABCD$ is cyclic, and  $ G\equal{}AB\cap CD,  E\equal{}AC\cap BD,  F\equal{}AD\cap CB$, by Brocard's Theorem we can observe that $ O$ (center of the circle that pases through $ A,B,C,D$) is the orthocenter of $ \triangle{EFG}$. Considering that $ EF\perp AB$, or  $ EF\perp GB$, we can observe that $ O$ has to be somewhere between $ A$ and $ B$ (because $ BG$ is one of the altitudes of $ \triangle{EFG}$, and naturally the orthocenter O must be on this line and on the circle containing $ A,B,C,D$, therefore $ O\in AB$). This leads to $ \angle{BCA}\equal{}\angle{BDA}\equal{}\angle{ECF}\equal{}\angle{EDF}\equal{}90$. Finally, we conclude that $ \angle{ECF}\plus{}\angle{EDF}\equal{}180$, and $ ECFD$ is cyclic.

I have to agree with pohoatza. I doesn't seem like a problem 3.


calculus
integration
function
Support
search
real analysis
real analysis unsolved

Let $ D \equal{} \{ (x_1, x_2, x_3) | x_1^2 \plus{} x_2^2 \plus{} x_3^2 < 1 \}$ and $ f \in C^{1}(\mathbb{R}^3)$. Show that if a function $ \tilde{u} \in C^2(D \cup \partial D)$ with $ u(x) \equal{} f(x)$ for all $ x \in \partial D$ satisfies

$ \int_D |\bigtriangledown \tilde{u}|^2 \,dV \equal{} \min \{ \int_D |\bigtriangledown u|^2 \,dV \hskip 0.05in \mid u \in C^2(D \cup \partial D), u(x) \equal{} f(x) \forall x \in \partial D \}$

then it solves the Laplace equation:

$ \minus{}\bigtriangledown \cdot \bigtriangledown \tilde{u}(x) \equal{} 0, \hskip 0.05in \tilde{u}(x) \equal{} f(x) \hskip 0.05in \forall x \in \partial D$

[hide="hint"]$ \int_D\left\vert \bigtriangledown u \right \vert ^2 dV \equal{} \int_{\partial D}f\bigtriangledown u \cdot \nu dS \minus{} \int_D u\bigtriangleup u dV$ where $ \nu$ is the outward unit normal.  Suppose $ \bigtriangleup u\neq 0$ near some $ x_0\in D$ and consider functions of the form $ v \equal{} u \plus{} \xi$ where $ \xi$ is a function with support in a small neighborhood of $ x_0$.

This is a classical result, and a proof can be found in most texts on PDEs.[/hide]

What is this identity known as? I want to read up on it but I don't know what to search for.

$ \int_D\left\vert \bigtriangledown u \right \vert ^2 dV \equal{} \int_{\partial D}f\bigtriangledown u \cdot \nu dS \minus{} \int_D u\bigtriangleup u dV$ is just [url=http://en.wikipedia.org/wiki/Integration_by_parts]integration by parts[/url].


logarithms
calculus
calculus computations

Let $c>0, q>1$. For every positive integer number $p$ , let $k(p)$ is least element of set $\{k\in\mathbb{N}|(k+c)^{p}\leq qk^{p}\}$. Find lim$\frac{k(p)}{p}$.

[quote="didilica"]The limit is $\frac{c}{\ln q}$.
Observe that

$k(p)=[\frac{c}{q^{1/p}-1}]$, hence we have that

$\frac{c}{p(q^{1/p}-1)}\leq \frac{k(p)}{p}\leq \frac{c}{p(q^{1/p}-1)}+\frac{1}{p}$.

Letting $p$ concerge to $\infty$ we get that the desired limit holds.[/quote]

me too!  :)


calculus
integration
function
geometry
calculus computations

$ 7/6 \equal{} \int^1_0 f(y) dy$  I need to solve for f(y).. I was thinking of differentiating both sides but that would give my 0 for 7/6 and that doesnt help.. ALSO, im pretty sure the function is not linear but has something like $ x^2$ instead

ok.. that works for the one case which i thought would apply to the rest of them but it doesn't. at any point (a,b) on the curve $ x^2$, $ \frac 12 a^2 \equal{} \int^b_0 f(y) \minus{} \sqrt(y) dy$ I've been trying to solve for the function $ f(y)$ for days now..

Could you please rewrite your last statement for better understanding.

the area between $ y \equal{} x^2$ (which can be rewritten as $ sqrt(y)$  and the unknown curve $ f(y)$ is equal to $ \frac 12 a^2$. Keep in mind that $ y \equal{} x^2$ is greater than $ f(y)$.  the point (a,b) is any point on the curve $ y \equal{} x^2$ or ($ sqrt(y)$). for example, (1,1), (2,4) etc.. i came up with the equation.. $ \frac 12 a^2 \equal{} \int^b_0 f(y) \minus{} \sqrt(y) dy$ and I simplified it to $ \frac 12 a^2 \minus{} \frac 23 b^(3/2) \equal{} \int^b_0 f(y)$ I just don't know where to go from here.. I need to find the $ f(y)$ I hope this is more clear

That equation is actually very easy to solve. Write $ (\sqrt {t},t)$ instead of $ (a,b)$:
$ \frac12t \equal{} \int_0^t f(y) \minus{} \sqrt {y}\,dy$
Differentiating,
$ \frac12 \equal{} f(t) \minus{} sqrt{t}$
$ f(t) \equal{} \frac12\plus{}\sqrt{t}$

I have no idea whether this is actually a solution to the problem you intended; you've mixed up $ x$ and $ y$ too much for anything to be clear.

[Edit- fixed a silly mistake]

how does that relate to $ f(y)$? is $ f(t) \equal{} f(y)$?

If $ (a,b)$ is any point of $ y(x)\equal{}x^2$ then it's obvious that $ b\equal{}a^2$ so your equation can be rewritten as:

$ \frac{b}{2}\equal{}\int^{b}_{0}\left(f(y)\minus{}\sqrt{y}\right)\,dy$

Since the integral operator is linear, we get:

$ \frac{b}{2}\equal{}\int^{b}_{0}f(y)\,dy\minus{}\int^{b}_{0}\sqrt{y}\,dy$

$ \frac{b}{2}\plus{}\frac{2}{3}b^{\frac{3}{2}}\equal{}\int^{b}_{0}f(y)\,dy$

Now, let us denote:

$ F^{'}(y)\equal{}f(y)$

Then:

$ \int^{b}_{0}f(y)\,dy\equal{}F(b)\minus{}F(0)$

which gives us:

$ \frac{b}{2}\plus{}\frac{2}{3}b^{\frac{3}{2}}\equal{}F(b)\minus{}F(0)$

Now we can derivate with respect to $ b$ and get:

$ \frac{1}{2}\plus{}\sqrt{b}\equal{}F^{'}(b)\equal{}f(b)$

and therefore:

$ f(y)\equal{}\sqrt{y}\plus{}\frac{1}{2}$

[quote="ryanww"]how does that relate to f(y)? is f(t) = f(y)[/quote]
$ f$ is a function; $ t$ and $ y$ have no independent meaning. Understanding the language of functions is fundamental to higher mathematics, and you should probably review it.

If I say $ f(t)\equal{}\frac12\plus{}\sqrt{t}$, or $ f(y)\equal{}\frac12\sqrt{y}$, or $ f(\triangle)\equal{}\frac12\plus{}\sqrt{\triangle}$, these all mean the same thing, and all refer to the same function $ f$.


Could someone help me with this?

So, let t be the system of sets containing the empty set and every subset of the closed unit interval [0,1] obtained by deleting a finite/countable number of points from X. Let T=(X,t) be the corresponding topological space, where X=[0,1].
Prove that T satisfies neither the second nor the first axiom of countability. Prove that T is a T1 space, but not Hausdorff.
Prove that the only convergent sequences in T are the "stationary sequences" - the sequences all of whose terms are the same starting from some index n.
Prove that the set M=(0,1] has the point 0 as a contact point, but contains no sequence of points converging to 0.

If the second axiom of countability held, then the intersection of all open subsets would be a non-void open set $ U_0$ - and minimal w.r.t. this property. Next simply drop one more point in $ U_0$ to get an open open set strictly contained in $ U_0$. Contradiction.


Let $A,B,C$ be angles of triangle $ABC$. Prove that: $\sin\frac{A}{2}\sin\frac{B}{2}\sqrt{\sin\frac{C}{2}}\leq\frac{\sqrt{3}}{9}$


Let $a_i \in R^+$, such that $\sum_{n=1}^{10000} \frac{1}{a_i} = 1$. Let the number of indexs $i$ such that $k(k-1) < a_i < k (k + 1)$ be $N_k$. Prove the inequality:
\[ \sum_{n=1}^{\infty} {\sqrt{\frac{N_k}{k(k+1)}}} \leq 20 \]


algorithm
vector
absolute value
algebra unsolved
algebra

Given that a_i belongs to {-1,1} for 1 <= i <= n and 0 <= b <= 2*pi. Find the maximum absolute value (magnitude) of the following complex number:

z = a_1*e^(i*b*1) + a_2*e^(i*b*2) + a_3*e^(i*b*3) + ... + a_n*e^(i*b*n)

That is, is there any easy way to find the coefficients a_i so that |z| is maximum?

An intuitive algorithm is probably to keep the vectors aligned together as much as possible...but I don't know if there is any real mathematical expression for those coefficients? Or one can just formulate an algorithm like I wrote?

Let $S(a_{1},...,a_{n})=\sum_{j}a_{j}exp(ibj)=S_{j}+a_{j}exp(ibj), \ S_{j}=\sum_{1\le k\le n, k\not =j}a_{k}exp(ibk).$
Obviosly, if $Re(S_{j}exp(-ibj))>0$  $max(S)=S_{a_{j}=1}$ else $a_{j}=-1$.
Therefore maximum value is $S=\sum_{j}exp(i(bj+k_{j}\pi )), k_{j}=0,1$.

Hm...but you haven't given an expression for the maximum absolute value or some algorithm on how to find the largest absolute value of that number? Or am I missing out something?

[quote="DK"]Given that $a_{i}\in \{-1,1\}$ for $1 \leq  i \leq  n$ and $0 \leq  b \leq  2\pi$. Find the maximum absolute value (magnitude) of the following complex number:

\[z = a_{1}e^{bi}+a_{2}e^{2bi}+a_{3}*e^{3bi}+\cdots+a_{n}e^{nbi}\]

That is, is there any easy way to find the coefficients $a_{i}$ so that $|z|$ is maximum?[/quote]
Let $e^{bi}=\epsilon$ which is a complex number on the unit circle.
By the greedy algorithm, we can attain an apparent maximum as follows (from step n->n+1):
determine $max(|z_{n}+\epsilon^{n+1}|,|z_{n}-\epsilon^{n+1}|)$
By computing the end result of this, we get a formula for $z$. Then it remains to prove $z$ is a minimum. I have a feeling that it really is the greedy alg. which gives the result here.

Another way to think about it: we have a set of $2n$ vectors, $\pm\epsilon^{i}$ and want to choose one of each pair to minimize sum. This might have a nice geometric approach.

I have already thought about the greedy, the first idea I got on this problem actually. But unfortunately, it didn't work. I draw a geometric picture that contradicted it and also wrote a program in Matlab that contradicted the greedy. Instead, I think that the solution is to take those vectors that are at most pi angles from each other; that is, if you take the leftmost vector and the rightmost vector, the difference in angle (if you measure positive angles) between those two is smaller or equal to pi. You also get a solution if you measure negative angles, so there is always an even number of sequences maximizing the magnitude. I haven't had time to prove it, but I guess (if it really is true) it could be proved in a simple way, because if there are two differing with more than pi in angle, then you are reducing the magnitude (kind of a greedy reasoning here).


how can i solve 3 equations .....

such as
c1 + c2   + c3  = 2
c1 + 2c2 + 3c3 =5
c1 + 4c2 +9c3  =15
---------------------

please show me an easy way tio soilve system of equation like that

thanx

It would be less confusing if you let c1 c2 and c3 to be a, b and c.

$(1) a+b+c=2$
$(2) a+2b+3c=5$
$(3) a+4b+9c=15$

$(2)-(1) = b+2c = 3$
$(3)-(1) = 3b+8c=13$

$3b+8c=13$
$-$
$b+2c = 3$
==========
$2c=4$
$c=2$

Then you plug in with other equations to find b and then a.

hello, all solutions are $ c_1\equal{}1,c_2\equal{}\minus{}1,c_3\equal{}2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, all solutions are $ c_1 \equal{} 1,c_2 \equal{} \minus{} 1,c_3 \equal{} 2$.
Sonnhard.[/quote]

You really have to make "sonnhard" your signature...

anyways, for a more general approach towards problems like these:

1. look for symmetry, as this often greatly simplifies a problem
2. I think cramer's rule works as well
3. gaussian elimination
4.Look for patterns, in this example the coefficients of c2 and c3 are increasing powers of 2 and 3 respectively.

[quote="RobRoobiks"][quote="batteredbutnotdefeated"][quote="Dr Sonnhard Graubner"]hello, all solutions are $ c_1 \equal{} 1,c_2 \equal{} \minus{} 1,c_3 \equal{} 2$.
Sonnhard.[/quote]

You really have to make "sonnhard" your signature...
[/quote]

I agree. What about 'Hello from Sonnhard. Period.'  :rotfl:

Otherwise, I agree with battered. But these problems are taught in Algebra II no?[/quote]
Yeah these problems are taught in algebra II.
What he really needs to do is post his own solution  :mad:

I think he is too lazy, and just used mathematic to solve it...

Hi;

I don't think that anyone who has put in the time and effort to learn a CAS could ever be called lazy. Forman S. Acton the guru of numerical analysis wrote:

[quote]After 6 months of intensive use of Mathematica, I still am having problems with the syntax. I still get mixed up on the brackets, full colons, etc.[/quote]

Hello all idiots insulting @DrSonhardGraubner.
There is something in this world called respect, and I think you should learn that before mathematics, or are you too high and mighty to treat others politely?

shouldn´t this be on high school basics ?


probability
calculus
integration
function
symmetry
conditional probability
probability and stats

An  insurance company estimates that  the the time  of death of Thomas is  uniformely distributed  inside the interval  [0,5] and the time of death  of Samuel  is also  uniformely distributed inside the  interval [0,10].  The insurance company assumes  that  both  times of death  are independent  of each  other.  Determine the probability  that  Thomas dies before  Samuel.

  I just  want to  confirm  my  answer  which is 1/4.  Also  does there exist  a solution using  a conditional  probability?  i  thought  i  had one but it  integrates  to  something  greater  than  one. My logic was this :

 Given  x, the time of death  of Thomas , the probability  that Samuel  dies after Thomas is  10-x/10 . So i  integrate with  limits  0 and 5 and this  should(why not?) give me  the probability  desired. the result of such  integration is 15/4>1 so somethings wrong.
 My correct (i hope) solution uses  the fact  that  both events are independent so  i  can calculate  the joint  density  function which  is 1/50.  this way  i  can calculate directly  the probability  of X<Y.

 Anyway  many thanks in advance.

Just as a spot check, can the answer $ \frac{1}{4}$ possibly be right?

Suppose the time of Thomas' death was uniformly distributed in $ [0,10]$, then by symmetry the probability that Thomas dies before Samuel would have to be $ \frac{1}{2}$.  This would seem to suggest that the answer to the original question must certainly be greater than $ \frac{1}{2}$.

[hide="Answer"]$ \frac34$[/hide]

What went wrong with your integral calculation? You need to weight the integral by the probability density for Thomas. That's a $ \frac15$ factor.

hmmm..  does this 1/5  factor come from  the  fact  that  both deaths are independent?  would it change  if both deaths were not independent?   By  the way the second approach actually works its just  that  i calculated my  integral  incorrectly.

  I  thank you  very  much  for your time.

Independence isn't what matters here. It's that we always need to account for the density of the variable we're integrating over. Here, we have the conditional probability $ f(x)$ that $ T<S$ given $ T\equal{}x$ ($ T$ is Thomas' death time and $ S$ is Samuel's). If $ g(x)$ is the density function for Thomas' death, the probability we get is $ \int_{\mathbb{R}} f(x)g(x)\,dx$. That works whether we have independence or not- all you need is a density and a conditional probability function.

If you don't have that density in there you get something with the wrong units, which isn't a probability.

I  understand now. Many  thanks for your help. :)


analytic geometry
graphing lines
slope
trigonometry
complex numbers

[quote="1234567890"]Draw an equilateral triangle in the coordinate plane with sides of slopes $ m_{1},m_{2},m_{3}$. None of the sides are vertical.

Find $ m_{1}m_{2}+m_{1}m_{3}+m_{2}+m_{2}m_{3}$.

[hide="Hint/Discussion"]The problem is really easy if you let $ m_{1}=\tan{\theta}, m_{2}=\tan{(\theta+120)}, m_{1}=\tan{(\theta-120)}$, and simplify.
Anyone have a non-trig way?[/hide][/quote]

[hide]I don't know, but I have a different way a little bit.

$ \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\sqrt{3}$

So, $ \frac{m_{1}-m_{2}}{\sqrt{3}}-1=m_{1}m_{2}$.

Analogously, $ \frac{m_{2}-m_{3}}{\sqrt{3}}-1=m{2}m_{3}$ and $ \frac{m_{3}-m_{1}}{sqrt{3}}-1=m_{3}m_{1}$

Adding up the three equations gives us the answer of $ -3$

I think maybe complex numbers could help, but I don't know.[/hide]


Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by 3.

It's very easy problem.
Let's replace our numbers on it's remainders modulo $3$.
If there is $\geq 3$ same numbers,  their sum is divisible by $3$.
Else by Dirichlet's principle there is $3$ different numbers: $0,1,2$. Their sum is divisible by $3$ too, q.e.d

Or just Erdös-Ginzburg-Ziv for $ n \equal{} 3$ ;)


function
parameterization
algebra unsolved
algebra

f(x,y) is a function from R*R to R.so that when x is fix then f equals a polymonial which its parameter is y. also the reverse is true.
is it true that f is polymonial which its parameters are x and y?
thank you. :?:

No idea?
I know that its answer is Yes.but the solution is not primary.so give a primary one. :!:

I dont understand your solution.I said a primary one.but any way thank you.but if you find an easier solution please send it.


I'm looking at problems 24 and 25: [url]http://www.pha.jhu.edu/~fair/physics_challenge.pdf[/url]. (I can't find solutions on the given website.)

My intuition tells me that the answer to both problems is $ \displaystyle\frac{I}{4}$, since, using the idea of parallel resistors, each of the four resistors adjacent to $ A$ should have an equal current flowing across it because they have equal resistances. However, I'm not sure if that's correct, primarily because I think that the grid being infinite could potentially cause problems. What is the correct way of approaching this problem, and what would the correct answer be?

In the first one you can think of it like, the current can't flow in the opposite direction, and the electric potential has to be the same across the whole row of resistors since the ground is infinitely far away. I think that was correct...

[quote="genericme"]You're right in the first case - the answer in the second case is $ I/2$, by superposition.[/quote]

When is superposition applicable?

Hmm, well, given the linear relationships you find when dealing with electric circuits, and the fact that differential equations in basic physics are almost always linear, I can't immediately think of a situation in which the principle of superposition doesn't hold.  It might not always be useful, though.  Of course, that depends on the problem.


Find all pairs $(m,n)$ of natural numbers
such that $4^m+5^n$ is a perfect square!
 :)

[hide="hint"]$5^n = (a+2^m)(a-2^m)$ both  factors must be   power of $5$ =>
 $2^{m+1} = 5^j - 5^k = 5^k(5^l-1) $  etc   ( assigning j=k+l etc)
==> only solution would be $4+5=9$
(There is some more work left  - some of it may be non-trivial[hide] (eg showing 5^k-1 can not be written as 2^m except for the obvious case etc...  but  it is not too hard -  :)[/hide] [/hide]


Elizabeth biked 504 miles on a 6-day trip. On each day $ n$ of the trip, she biked $ n$ times as many miles as she biked on the first day. (For example, on day $ 3$, she biked three times as many miles as she had on the first day.) How many miles did she cycle on the first day?

Suppose she bikes x miles on the first day.

Then (1+2+3+4+5+6)x=21x=504, x=24.


vector
algebra
polynomial
abstract algebra
induction
linear algebra
matrix

Let A \in M_2(C) proove that det(A.A*+I_2)  \geq 0
where A* is complex conjugate of A
Gheorghe Andrei
----------------------------------------------------------------------------
* in the following is the complex conjugate

As I said to Valentin the result is true also for A \in M_n(C)
det(I_n+A.A*) \geq 0
To prove that we need a

Lemma: Let E a C-vector space dimE=n >0, u:E->E a semi-linear map
wich mean
for any x,y\in E, for any k\in C, u(x+y)=u(x)+u(y), u(kx)=k*u(x)
If u^2=uou has an eigenvalue  \lambda <0 then  \lambda is a root

of the characteristic polynomial of u^2 of even order wich mean
det(xI-u^2)=(x- \lambda )^(2 \gamma )Q(x) with  \gamma \in N

Here the proof
u is semi-linear, u^2 is linear map E->E
u^2(x+y)= u^2(x)+ u^2(y), u^2(kx)=u(k*u(x))=ku(u(x))=ku^2(x)
Let (k,a)\in ]-oo;0[ x E\(0) such that u^2(a)=ka we have u^2(u(a))=ku(a)
And the vectors a,u(a) is independant if not u(a)=qa and ka=u(u(a))=u(qa)=q*u(a)=q*.qa => k=q*.q=|q|^2 \geq 0 contradiction since k<0.
ker(u^2-k.Id) contain the plane F=vect(a,u(a)) and u(F) is included in F
This prove that n \geq 2, if n=2 the lemma is proved, consider the case n>2.
Let G a supplementary of F in E, E= G(direct sum)F, and p is the projector on G with kernel F.

Taking a base of E obtain by a base of F,G we have
det(xI-u^2) = (x-k)^2.det(xI-t) where t:G->G linear define by
t(x)=p(u^2(x))
It is easy to see that w(x)=p(u(x)) is semi-linear G-->G
We will prove that t=w^2 indeed if x\in G t(x)=pu(u(x))=pu(w(x)+y) with y\in F , pu(y)=0 , t(x)=pu(w(x)) but w(x) \in G => pu(w(x))=w(w(x))
t=w^2
And by induction with det(xI-u^2) = (x-k)^2.det(xI-t) we have k is even order in the polynomial characteristic of u^2
This prove the lemma.

Now coming back to det(I+A.A*)
Let u:C^n-->C^n defined by
u(x_1,...,x_n)=(x*_1,..,x*_n) is semi-linear
A.A* is the matrix of u^2 in the canonical base
det(xI-A.A*)=det(xI-A*A)=(det(xI-A.A*))* => det(xI-A.A*) is real coefficient
det(xI-A.A*)= (x-a_1)^k_1...(x-a_j)^k_j.Q(x) with Q \in R[X] monic,
degQ even .
det(I+A.A*)=(-1)^ndet(-I-A.A*)=(-1)^n.(-1)^k.Q(-1)(1+a_1)^k_1...(1+a_j)^k_j
but Q(-1)>0,
if all a_i \geq 0 det(I+AA*) \geq 0
if there is a_i<0 with the lemma k_i is even

The matrix $ I \plus{} AA^*$ is hermitian and, for $ 0 \neq x \in \mathbb{C}^n$ we have $ x^*(I \plus{} AA^*)x \equal{} x^*x \plus{} x^*AA^*x > 0$. Therefore, $ I \plus{} AA^*$ is positive definite, which implies that all its eigenvalues are positive and so its determinant must be positive.

by considering $ AA*$, you can prove that all the eigenvalues of $ AA*$ is non-negative
and then consider $ I\plus{}AA*$ you can prove that all the eigenvalues are $ \geq 1$, the determinant is just the product of all eigenvalues, so must be positive( in fact $ \geq 1$)

I think that the result still true when we replace $A*$ by $\bar{A}$, and its more complicated to prove...


analytic geometry
geometry unsolved
geometry

Let G is the intersection point of the medians of ABC. (I) is the inscribed circle.
Prove that
[B,C,I,G are on the same circle] is equivalent to $ \ AB^2\plus{}BC^2\plus{}AC^2\equal{}3AB.AC$ :lol:

We use barycentric coordinates with respect to $ \triangle ABC.$ The circle passing through $ B \equiv (0: 1: 0),$ $C \equiv (0: 0: 1)$ and $G \equiv (1: 1: 1)$ has equation

$ (O') \equiv a^2yz + b^2xz + c^2xy - \frac {_1}{^3}(a^2 + b^2 + c^2)(x + y + z)x = 0$

$ B,C,I,G$ are concyclic $ \Longleftrightarrow$ $ I \equiv (a: b: c)$ satisfies the equation of $ (O').$ Then

$ a^2bc + b^2ac + c^2ab + \frac {_1}{^3}(a^2 + b^2 + c^2)(a + b + c)a = 0$

$ abc(a + b \plus{} c) - \frac {_1}{^3}(a^2 + b^2 + c^2)(a + b + c)a = 0 \Longrightarrow  3bc = a^2 + b^2 + c^2.$

I also use coordinate(with axis Ox, Oy) to solve this problem but Sorry, I wonder what barycentric coordinate is. And how can you have 3 variables x,y,z on a plane.


function
combinatorics proposed
combinatorics

Suppose i have n boolean variables (value is either 0 or 1) X1, ... Xn which are used in the following boolean function, f, in disjunctive normal form:

f(X1...Xn) = (Clause 1) OR (Clause 2) OR ... (Clause [n choose k])

where each clause is a [u]unique[/u] conjunction of the Xi with the following property:
- [u]exactly[/u] k of the Xi are not negated, and the rest are negated.

Here's an example to make things clearer:
suppose we have the four boolean variables:
[code]X1    X2    X3    X4[/code]

Let k=2, then we have the following [4 choose 2] = 6 clauses:
[code]
            X1       X2      X3      X4
clause 1   (1   AND  1   AND 0   AND 0 )
clause 2   (1   AND  0   AND 1   AND 0 )
clause 3   (1   AND  0   AND 0   AND 1 )
clause 4   (0   AND  1   AND 1   AND 0 )
clause 5   (0   AND  1   AND 0   AND 1 )
clause 6   (0   AND  0   AND 1   AND 1 )
[/code]
where a one represents the unnegated variable and a zero the negated variable.
For the example above the function f then is the function that is true (1) whenever exactly two of the Xi are true and false otherwise. If we had k=3, then f would be the function that is true whenever exactly three of the input variables are true... etc
If i wished to simplify f, so that it doesn't have [n choose k] clauses, and if we define the length of a boolean function as the number of boolean operators it contains, can i estimate the minimum size of f based on n and k? In short, is it possible to have the following theorem:
If f(X1... Xn) is a boolean function as described above, then the minimum length of f, |f| is given by g(n, k)
Does g(n, k) exist?

On a first look it seems:
g(n, k) = g(n-1, k-1)+g(n-1, k)

Sorry if this is a trvial question... :P

There's actually a very interesting question underlying the problem i posted above.
I think i should've posted this in the combinatorics forum, could it be moved?

[quote="-Job-"]I think i should've posted this in the combinatorics forum, could it be moved?[/quote]
Done.


9 congruent spheres are packed into a unit cube in such a way that one of them has its center at the center of the cube and each of the others is tangent to the center sphere and to 3 faces of the cube. What is the radius of each sphere.

how would you do this problem?

[hide="Solution"]Observe five spheres: the central one and the four which are tangent to one side. Their centers form a regular quadrilateral pyramid whose base edge is $1-2r$, altitude ${1\over 2}-r$ and lateral edge $2r$. Therefore

$\left((1-2r)\frac{\sqrt{2}}{2}\right)^{2}+\left({1\over 2}-r\right)^{2}=4r^{2}$

$\frac{1-4r+4r^{2}}{2}+{1\over 4}-r+r^{2}=4r^{2}$

${3\over 4}-3r+3r^{2}=4r^{2}$

$r^{2}+3r-{3\over 4}=0$

$r=\frac{-3+\sqrt{9+3}}{2}=\frac{2\sqrt{3}-3}{2}=\sqrt{3}-{3\over 2}$[/hide]


I understand the manual recommends you take the test from 9-12 noon but are the AMC people strict about following that.  It says if you fail to comply witht their rules you will be disqualified.  Will my school be able to take the test at 3:00 after school ends, or will the AMC disqualify us if we do that?

I've contacted AMC about our situation. Our testing center doesn't have space for us till noon in our time zone, and one test-taker is traveling in from two hours away. We will just be REALLY SURE not to go online or to communicate with any other math-liking people in any way until after the test-takers' answer forms are in the mail.

Good luck to everyone testing on Tuesday.

We are flexible about the AIME adminstration times, within the following guidelines:
(1) The AIME I must be taken on Tuesday, March 7, 2006 (and the AIME II on Wednesday, March 22, 2006).
(2) If the adminstration of the AIME falls outside of the recommended 9 am - 12 noon time period, we ask that the school contact us.  We will fax to the school an Academic Integrity Form for the AIME participants to sign stating that the participants did not discuss the AIME with other students.  The Forms are to be returned with the answer forms.

Steve Dunbar
AMC Director


calculus derivative algebra polynomial integration function trigonometry

I have this problemI am trying to set up the particularsolution for but don't know where to even start. It's unlike any other differential problem I've been doing.

(D-1)^3 (D^2-4)y=xe^x +e^2x +e^-2x

Should I just multiply the LHS out and solve for y? It doesn't even appear to be a differential eq as I only have y, not any of its derivatives on the left???

Thanks for any help you can give  :?:  :?:

This problem isn't as difficult as it appears at first glance.

$D$ is an operator that means $\frac{d}{d x}.$ It's used because it can be treated as a variable in a polynomial expansion. Your problem $\left(D-1\right)^{3}\left(D^{2}-4\right)y = xe^{x}+e^{2x}+e^{-2x}$ can be expanded as follows:

$\left(D-1\right)^{3}\left(D^{2}-4\right)y = \left(D^{3}-3D^{2}+3D-1\right)\left(D^{2}-4\right)y$

$= \left(D^{5}-3D^{4}-D^{3}+11D^{2}-12D+4\right)y$

The right hand side can be simplified as:

$xe^{x}+e^{2x}+e^{-2x}= xe^{x}+2\cosh 2x$

You wind up with a fifth order differential equation that looks extremely nasty but in fact is not. It can be solved easily using any convenient method. Personally I prefer the Laplace transform for this as the transform of the right hand side is simple:

$\frac{1}{\left(s-1\right)^{2}}+\frac{2 s}{s^{2}-4}$

On the left hand side you have

$\left(s^{5}-3s^{4}-s^{3}+11s^{2}-12s+4\right)Y\left(s\right)$ + some icky fourth order polynomial in $s$ whose coefficients are determined by the initial conditions on the first four derivatives of $y.$

Solve for $Y\left(s\right)$ then find the inverse transform using residues (a technique for evaluating integrals of complex functions that can be often performed by inspection).

Of course if you want to be a smart Alec about it, you can always give the answer that $y = L^{-1}\lbrace Y\left(s\right) \rbrace$ at that point.

Note that Laplace transforms can be somewhat awkward unless you're given initial conditions.

This is a constant-coefficient linear differential equation, nonhomogeneous, but with a right hand side that is a linear combination of exponentials and exponentials multiplied by polynominal. (For future reference, sines and cosines are also in this exponential-polynomial class of functions.)

If you want to see the differential equation in a more conventional-looking form, you can write it as

$y^{[5]}-3y^{[4]}-y'''+11y''-12y'+4y= xe^{x}+e^{2x}+e^{-2x}$

(But trust me, you're going to want that factorization.)

Part of what we need is the solution to the corresponding homogeneous equation:

$(D-1)^{3}(D-2)(D+2)y=0.$

Those solutions are $(c_{1}+c_{2}x+c_{3}x^{2})e^{x}+c_{4}e^{2x}+c_{5}e^{-2x}$

Now you need one "particular solution" to add to this homogenous solution.

Reading suggestion: find some mainstream undergraduate differential equations textbook and look up the words "undetermined coefficients."

Then there's an alternative approach - it's not really different, but it's a different organization, and is suggested by the notation here. Find the linear constant coefficient differential operator that annihilates the right hand side. In this instance that would be $(D-1)^{2}(D-2)(D+2).$ Apply this operator to both sides of the original equation. You now have

$(D-1)^{5}(D-2)^{2}(D+2)^{2}y=0.$

We know the form of the solutions of this 9th order differential equation. Our solutions will be contained in expressions of the following form:

$(a_{1}+a_{2}x+a_{3}x^{2}+a_{4}x^{3}+a_{5}x^{5})e^{x}+(a_{6}+a_{7}x)e^{2x}+(a_{8}+a_{9})e^{-2x}.$

Now, which functions of this form satisfy the original equation? That's just a big ugly linear algebra problem.

Frankly, this looks to me like the kind of textbook exercise in which the authors are more interested in knowing whether you can set up the form of the solution than in having you actually carry out all the steps.

Returning to Dr. No's preferred solution, the one with Laplace transforms: he's going to be doing partial fractions on an expression whose denominator is $(s-1)^{5}(s-2)^{2}(s+2)^{2}$. It's the same big ugly linear algebra mess as any other method.

(His preference for using $\sinh$ and $\cosh$ might lead him to bundle those last factors as $(s^{2}-4)^{2}.$ It's still a 9th degree polynmial denominator.)

[quote="Kent Merryfield"]Returning to Dr. No's preferred solution, the one with Laplace transforms: he's going to be doing partial fractions on an expression whose denominator is $(s-1)^{5}(s-2)^{2}(s+2)^{2}$. It's the same big ugly linear algebra mess as any other method.[/quote]
Partial fractions? ... Not necessarily. The Bromwich integral (the inverse Laplace Transform) may be evaluated directly as a sum of residues. Partial fractions work too, but why use a technique that usually requires more work.

[quote="Kent Merryfield"](His preference for using $\sinh$ and $\cosh$ might lead him to bundle those last factors as $(s^{2}-4)^{2}.$ It's still a 9th degree polynmial denominator.)[/quote]
My preference for using $\cosh$ is due only to the fact that I recognized the $e^{2x}+e^{-2x}$ term in the forcing function as a $\cosh$ function, a function for which the Laplace Transform is usually the most appropriate method of solution.

As with all other Laplace transforms, however, the characteristic function in the denominator must be completely factored in order to evaluate the Bromwich integral using residues. (Repeated poles, this problem has three, require successive differentiation to find the residue associated with them. However finding the fourth derivative of a compound rational function - the most complicated part of actually solving this problem - is in itself hideously tedious, which is why I stopped short of solving it.)

[quote="Dr. No"]- the most complicated part of actually solving this problem - is in itself hideously tedious, which is why I stopped short of solving it.)[/quote]
As I said above, I suspect this of being a textbook problem which asks for the form of the answer (possibly in terms of understanding the technique of undetermined coefficient) but does not ask for the answer itself.

Computer algebra systems (Maple, Mathematica, the version of Derive that's in TI-92's and 89's) have their uses.


Prove that a plane can't be covered by a finite number of parabolas.

I know that it was posted before,but I hope to see new ideas now. Besides that one of the solutions posted was wrong.
Who has a very nice idea?Please help and post new solutions. :( .
Thank you.

Why not bump up the old topic? Creating a new one is useless. I remember seeing the problem in a rather elementary problem book (one that was given out to all participants of Vojtech Jarnik 2008 - some of the problems were on advanced mathematics, but most of them were rather easy), but I have not thought about it and don't have it here with me. Thanks for reading this useless information. :)

  darij


inequalities
function
superior algebra
superior algebra unsolved

Let $ G$ finite group and $ \chi: G\rightarrow \mathbb{C}^*$ character of a one dimensional complex representation, for $ \chi$ we define F-S indicator, as \[ ind(\chi)\equal{}\frac{1}{|G|}\sum_{a\in G} \chi(g^2).\] Prove $ ind(\chi)\equal{}1$ if $ \chi(G)\subseteq \mathbb{R}$, $ ind(\chi)\equal{}0$ otherwise.

Since it's a homework problem (an active one?) I'll just drop a hint.

$ \chi$ here is really just a homomorphism.  That should help.  The typical technique for showing that some sum like this is zero is to find something to multiply by that keeps the sum fixed but isn't equal to $ 1$...

It's easier than that. This expression is just the inner product of $ \chi$ with $ \overline{\chi}$. If they are the same, this inner product is $ 1$, and if not, it is zero.

How is the result evident from the above statement? While I agree that when $ \chi(G) \subseteq \mathbb{R}$ then
\[ \dfrac{1}{|G|} \sum_{g \in G} \chi(g^2) \equal{} \dfrac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\chi(g)} \equal{} (\chi | \chi) \equal{} 1,
\]
where $ (\cdot|\cdot)$ is the inner product on the space of all functions from $ G$ to $ \mathbb{C}$, however I fail to see how the above statement could be applied in the case when $ \chi(G) \subseteq \mathbb{C}$, that is I fail to see how is the expression for $ \textrm{ind}(x)$ equal to the inner product of $ \chi$ with $ \overline{\chi}$ when $ \chi(G)\subseteq \mathbb{C}$?

Here's the definition of the inner product:

$ (\chi | \lambda) \equal{} \dfrac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\lambda(g)}$.

When we set $ \lambda \equal{} \overline{\chi}$, we get $ \dfrac{1}{|G|} \sum_{g \in G} \chi(g^2)$.  When $ \chi \equal{} \overline{\chi}$, this is $ 1$, and when $ \chi \neq \overline{\chi}$, this is $ 0$.

Edit: sorry for deleting my entire post, I missed the conjugation sign and wrote a whole bunch of nonsense :) Yes, that is obvious now, thank you. :)

Just to straight things out for myself: every one dimensional representation is irreducible and so if $ \pi$ and $ \rho$ are such representations, then $ (\chi_\pi | \chi_\rho)$ is $ 1$ if $ \pi \simeq \rho$ and $ 0$ otherwise. There is the fact that two representations are isomorphic if and only if their characters are equal - so in the real case $ \chi \equal{} \overline{\chi}$ and the appropriate inner product is $ 1$, and in the complex case $ \chi \neq \overline{\chi}$ so the appropriate inner product is $ 0$.

All true.  But to be clear, you don't need any general theory of representations here, just properties of homomorphisms from $ G$ to $ \mathbb{C}$.


linear algebra
matrix
probability and stats

Let us consider a Markov chain with $ \mathbb{Z}$ as the state space and with the following transition matrix

\[ p_{0,k} \equal{} \frac{1}{3} \qquad \textrm{for } k \equal{} \minus{}1, 0, 1,\]
\[ p_{k,k\minus{}1} \equal{} q, \ p_{k,k\plus{}1} \equal{} p \qquad \textrm{for } k \leq \minus{}1,\]
\[ p_{k,k\minus{}1} \equal{} p, \ p_{k,k\plus{}1} \equal{} q \qquad \textrm{for } k \geq 1,\]

where $ q\equal{}1\minus{}p$ and $ p \in (0,1)$. Obviously the chain is irreducible and aperiodic. Investigate for what $ p$ it is recurrent.

do you know the answer for the simple one: $ p(k,k \plus{} 1) \equal{} 1 \minus{} p(k,k \minus{} 1) \equal{} p$ ? If so, this is the same answer since the two Markov chains behave the same after leaving the origin. (the law of large number could help ..)

We know that random walk on $ \mathbb{Z}$ is recurrent if and only if $ p \equal{} \frac{1}{2}$. However, to my mind in the chain I've postes the answer is different. Namely, it is recurrent for $ p \leq \frac{1}{2}$ (this is only intuition). Therefore could you be more exact and show how to prove that these two Markov chains behave the same after leaving the origin using the law of large number?

oh.. sorry. I have read it too fast (was thinking that $ p(k,k\minus{}1)\equal{}q$ everywhere except at the origin). So as you said, the markov chain is recurrent for $ p \leq \frac12$.

What I was suggesting was the following:
define for example $ \tau_{x}(X)\equal{}\inf \{t > 0 : X_{\tau_{x}}\equal{}x \}$, and $ X_n$ your Markov chain, and $ Y_n$ the standard p-q Markov chain we were talking about above.
Then for example, if $ X$ is recurrent, then $ P[ \tau_1(X) < \infty | X_0\equal{}2] \equal{} 1$, and of course $ P[ \tau_1(X) < \infty | X_0\equal{}2]\equal{}P[ \tau_1(Y) < \infty | Y_0\equal{}2]$. But $ P[ \tau_1(Y) < \infty | Y_0\equal{}2]$ iff $ p \leq \frac{1}{2}$. Hence if $ X$ is recurrent, then $ p \leq \frac12$. The same argument shows that if $ X$ is not recurent, then $ p>\frac12$


Hello everyone,
Given a regular polyhedra of n vertices and n distinct objects. In how many distinct ways can these n objects be arranged on the vertices.Please be careful with "distinct" objects

 :)

You just divide the number of permutations of the objects by the number of symmetries of the polyhedron.

Hi
But how do we get the number of symmetries of the polygonal of n vertices!!
Basically I am novice in the topic kindly inform me about the related articles.
 :)

[quote="Albanian Eagle"]whats a regular polyhedron on 5 vertices?  :)[/quote]
I think there isnt such a thing.Thank You for pointing it out.
There is an other question:
When can a regular polyhedron have  n vertices? What is the nature of n?

you'd be surprised  :lol:

http://en.wikipedia.org/wiki/Regular_polyhedra