limit inequalities logarithms calculus integration function derivative

please show that
                       $\lim_{n\to \infty}\left(1+\frac1n\right)^{n}=e$

Isn't it the definition of $e$?
Of course, we can show that $\lim (1+\frac 1n)^n$ does exist.

Well, I think we can use "ln" :)
$ln(1+\frac{1}{n})^n = n ln(1+\frac{1}{n}) = \frac{ln(1+\frac{1}{n})}{\frac{1}{n}} = \frac{ln(1+\frac{1}{n})-ln(1)}{\frac{1}{n}}$
Indeed, ln(1) = 0.
And then, taking the limit when n -> +oo, I think we could re-find e.

But, yeah, this limit is the definition of e, so..

[quote="osmantelli"]please show that
                       lim(1+1/n)^n=e[/quote]
If you want to proove that $(1+\frac{1}{n})^n$ is bounded it can easily been done. in fact it can be shown that this limit is always bigger than two and smaller than three. because the proof is a bit long tell me to write it only if you want.

I know two ways of proving that $\left( 1+ \frac1n \right)^n$ converges: 1) by binomial theorem; 2) by showing $\left( 1+\frac1n \right)^{n+1}$ is convergent via Bernoulli

Are there any nicer other ways?

You can show Mollaw's (right spelling?) inequality,but I couldn't find at mathworld or Google, Something like this,

\[ \frac{e}{2n+2}<e-\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n<\frac{e}{2n+1} \]

I will post Japanese university entrance exam regarding to the latter part.

You can show the equivalent limit $\lim_{x\to 0}\frac{\ln(1+x)}{x} = 1$ using the inequality
$\frac{x-1}{x} < \ln x < x -1, x\ne1$. Of course, it all depends on in what order you introduce things. The way I was taught it, we defined $\ln$ through an integral (the inequality can be shown from this), and then $\exp$ as its inverse and $e = \exp(1)$.

[quote="oooooh ya"]mathmanman, wont ln(1) / (1 / n) become intederminant as n approaches infinity?[/quote]
No, it's zero. Why would it be indeterminant? $0 / n = 0$ no matter how small n is.

$a_n=\left( 1+\frac 1n\right)^n,\ b_n=\left(1+\frac 1n\right)^{n+1},\ c_n=\left( 1-\frac 1n\right)^n$.

$\blacksquare\ 1^{\circ}.\ \left(\forall\right) n\in N^*,\ a_n<a_{n+1}$, i.e. $a_n\nearrow a\in \overline R \equiv [-\infty ,\infty ].$
(Apply the A.M.-G.M. to $x_k=1+\frac 1n,\ k\in \overline {1,n}$ and $x_{n+1}=1$).

$\blacksquare\ 2^{\circ}.\ \left(\forall\right) n\in N,\ n>1\ c_n<c_{n+1}$, i.e. $c_n\nearrow c\in \overline R.$
(Apply the A.M.-G.M. to $y_k=1-\frac 1n,\ k\in \overline {1,n},\ y_{n+1}=1$).

$\blacksquare\ 3^{\circ}.\ \left(\forall\right) n\in N^*,\ 2\le a_n<a_{n+1}<b_{n+1}<b_n=\frac{1}{c_{n+1}}\le 4$.

[b]Weierstrass' theorem:[/b] $\Longrightarrow a_n\nearrow a\in R,\ b_n\searrow b\in R:$

$2=a_1<\ldots <a_n<\ldots <a\le b<\ldots <b_n<\ldots <b_5\simeq 2.98$.

$\left(\forall\right)n\in N^*,\ b_n=\left( 1+\frac 1n\right) \cdot a_n;\ n\rightarrow \infty \Longrightarrow \underline {a=b\equiv e}$
\[ \underline {\overline {\left| \ \lim_{n\to \infty} \left( 1+\frac 1n\right)^n=e\simeq 2.718281\ldots\in (2,3)\ \right| }} \].

[quote="oooooh ya"]its $\lim_{n \to \infty} \frac{ln(1)}{\frac{1}{n}}$

which is $\frac{0}{0}$, indeterminate[/quote]
It is not. "$\frac{0}{0}$" refers to where both numerator and denominator tend to 0. $\lim_{n\to0}\frac0n = 0$. No matter what the value of x is (as long as it is not zero of course), $0/x = 0$. I don't see how this is a problem at all. You realize that a limit is only concerned with the value of the function AROUND the point, not in the point, right?

What is actually the definition of $e$ ?
In class, we learned about logarithm function. the natural logarithm function is defined by
\[ \ln x = \int_{1}^x \frac{1}{t} \, dt \qquad x > 0  \]
And the number $e$ , is the uniqe number for which $\ln e =1$. And also we learned about proving the properties of logarithm function (i.e  $e^{\ln x} = x$).  And to estimate the number $e$. we often use that limit defenition (without that lim sign).
In numerical  analysis, $e$ often calculate by using series expansion with using computer help.
About limit definition:
\[ \ln \left(1+ \frac{1}{n} \right) = \int_{1}^{1 + \frac{1}{n}} \frac{1}{t} \, dt \].
And since $\frac{1}{t} \leq 1$ and $\frac{1}{t} \geq \frac{1}{1+ \frac{1}{n}}$ on $[1, 1+ \frac{1}{n}]$. we have inequality \[ \frac{1}{n+1} \leq \ln \left( 1+ \frac{1}{n} \right) \leq \frac{1}{n}  \]
so \[ \left( 1+ \frac{1}{n} \right)^n \leq e \leq \left( 1+ \frac{1}{n} \right)^{n+1} \].
if you try  $n=10$, $n=20$ .... , you'll get the numerical estimation of $e$. The advantage of using this inequality instead of using series expansion is that we can get the desire correct desimal place without considering remainder like in series expansion. for instance if we let $n=10000$ we have $2.718145927 < e < 2.718417741$. so confidently we say that the first four digits of $e$ is $e=2.718$.
We can take $n \rightarrow \infty$. then the limit is $e$, as we can verify that like mathmanman or someone else did.

$y = \lim_{x\to\infty} ( 1 + \frac{1}{x} )^x$

$ln(y) = \lim_{x\to\infty} x ln( 1 + \frac{1}{x} )$

let x = 1 / u

as 1 / u approaches infinity, u approaches 0+

$ln(y) = \lim_{u\to0^+}\frac{ln( 1 + u )}{u}$

indeterminate \to l'hopitals rule

$ln(y) = \lim_{u\to0^+} \frac{1 / ( 1 + u )}{1}$

$ln(y) = 1$

$y = e$

[hide="its indeterminiant"]

$\frac{ln(1)}{\frac{1}{n}}$ is indeterminant

ln(1) = 0

1/n approaches 0 as n approaches infinity

$f'(x) = \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$

let x + h = c
h = c - x
as h approaches zero, x approaches c

$f'(x) = \lim_{x\to c}\frac{f(c) - f(x)}{c - x}$

if the left and right side limit are the same, its reversible and thats the way i usually see it, so i'll reverse it:

$f'(x) = \lim_{x\to c}\frac{f(x) - f(c)}{x - c}$

l'hopitals rule:

$\frac{f'(x)}{g'(x)} = \frac{\lim_{x\to c}\frac{f(x) - f(c)}{x - c}}{\lim_{x\to c}\frac{g(x) - g(c)}{x - c} }$

$\frac{f'(x)}{g'(x)}=\lim_{x\to c}\frac{f(x) - f(c)}{g(x) - g(c)}$

if f(c) equals zero and g(c) equals zero, than:

$\frac{f'(x)}{g'(x)}=\lim_{x\to c}\frac{f(x)}{g(x)}$

$f(x) = ln(1)$
$g(x) = 1 / x$

f(x) = 0 for all of x
g(x) = 0 when x approachs infinity

so

$f(c) = 0$
$g(c) = 0$

and

$\frac{f'(x)}{g'(x)}=\lim_{x\to c}\frac{f(x)}{g(x)}$

oooooooooooooooooh ya

[/hide]

I'm sorry, but you're just completely wrong. I don't think you truly appreciate what a limit means. I don't know how to prove L'hopitals rule or how it really works, but you're just absolutely wrong. For ANY $n \ne 0$ the expression is equal to 0, and so its limit is 0. For this reason we might as well just consider the function to be zero in the point $n=0$ too. THERE IS NO PROBLEM. End of discussion. Stop trying to make some sort of psuedoproof that there is a problem. Look up the definition of a limit.

can we have a third party comment?

that is NOT a pseudo proof, that is the real proof!  there are different proofs for the different cases, but thats the proof for 0/0.  wikipedia has proofs for it, you can check mine against that one, i've heard a lot about math world, you can check my proof against math words

[hide="on the expression being indeterminant"]

[quote="Kalle"]THERE IS NO PROBLEM. End of discussion.[/quote]

i took another look at mathmanman's equation, which he equated to e:

$e = lim_{n\to\infty}\frac{ln(1 + \frac{1}{n}) + ln(1)}{\frac{1}{n}}$

note: $lim_{n\to\infty}\frac{1}{n} = 0$

so:

a$e = lim_{n\to\infty}\frac{ln(1 + 0) + ln(1)}{0}$

also: $ln(1) = 0$

$e = \frac{0 - 0}{0}$

$e = \frac{0}{0}$[/hide]

ARGH. Ok. Here is a real proof that $\lim_{n\to\ 0}\frac{\ln1}{(1/n)} = 0$. Let $\epsilon > 0$. Now, choose $\delta > 0$ (that's right. this is so obvious we can choose any delta). If $0 < |x - 0| < \delta$ then $\frac{\ln 1}{(1/n)} = 0$ so we have that $\left|\frac{\ln(1)}{(1/n)} - 0\right| < \epsilon$. QED.

As I've said all along: the function is equal to zero EVERYWHERE. A limit is concerned with the value AROUND THE POINT, NOT IN IT. What is it you are suggesting? That the limit wouldn't exist, so that it tends to infinity? If it tends to infinity give me a point where the function is larger than 1.

[quote="mathmanman"] And then, taking the limit when n -> +oo, I think we could re-find e.[/quote]

its as n approachs infinity not zero, "and that's so obivious" he just had to use +oo.  but aside from petty shots, what you've written is all correct as n approaches 0.  i hate it when things go to such length due to an oversight.

Oh, ohps. Right. The exact same proof works for infinity except let $x > \delta$. There is still absolutely no problem. What would the problem be, exactly? It is zero everywhere. It should be even more obvious. No matter how large n is, the function is zero. It is so annoying how you won't seem to accept this.

I gave a method for completing it. You need to show the standard limit $frac{ln(1+x)}{x} \to 1$ as $x\to 0$, which you can do using the inequality I posted.

I think we finish this topic. Already you run over stock and block ! You read the Perfect_radio and my previous messages. It is clearly.

I hope this clarifies things.  It is not hard to see that the limit exists (monotone bounded sequences converge).

I will use the definition $\exp x=\sum^{\infty}_{i=0} \frac{x^i}{i!}$.  This means that I can take $\log$s and whatnot without worrying about circular reasoning.

Okay.

Take natural logarithms.  By continuity of $\log$ around $x=1$, we conclude that

$\exp\left[\lim_{n\to\infty} \log\left(1+\frac 1n\right)^n\right]=\lim_{n\to\infty} \left(1+\frac 1n\right)^n$

So it's enough to evaluate

$\lim_{n\to\infty} \log\left(1+\frac 1n\right)^n=\lim_{n\to\infty}\frac{\log(1+n^{-1})-\log 1}{n^{-1}}$

By a change of variable $t=n^{-1}$, we see that the latter limit is just

$\frac{d}{dt} \log (1+t)\big|_{t=0}=1$

Hence the limit is $e$.

[quote="oooooh ya"]complete the proof started by mathmanman, otherwise i'm finished posting until someone else comments.[/quote]

Mathmanman's proof was correct (as you can see by my post above)

What he did was add a term that was always zero, e.g, $n\ln 1$, so that it would be more easy to recognize the limit as the value of the derivative of a certain function at a point.  He also could have directly applied l'hopital, but apparently he (and I) prefer the latter method.

Also, the limit $\lim_{n\to\infty} \frac{\ln 1}{n^{-1}}$ is not an indeterminate form, because it is actually zero for any (non infinite) choice of $n$ - if you look at the epsilon delta definition of a limit, this should be clear.  Remember, a limit is essentially determined by the behavior of a function, or sequence, around a point, not at it.

[quote="blahblahblah"]
Mathmanman's proof was correct (as you can see by my post above)[/quote] Thanks :)

[quote="blahblahblah"]He also could have directly applied l'hopital, but apparently he (and I) prefer the latter method.[/quote]
Well, to be honest, I did not know l'hospital method until last week :) (I'm just an 11th grader)

[quote="blahblahblah"]I will use the definition $\exp x=\sum^{\infty}_{i=0} \frac{x^i}{i!}$.  This means that I can take $\log$s and whatnot without worrying about circular reasoning.[/quote]

This raises some questions which I cannot answer by myself:

1) how do we prove that the series converges?

2) can we derive properties like $\exp (ab)=\exp a + \exp b$ from that definition?

i'm interested in this because i have never seen a proof of very important properties like $\exp$ is increasing and bijective (has an inverse, $\ln$). maybe your definition can help us more  :lol:

Convergence can be proved using the ratio test.

Search for Walter Rudin's real and complex analysis on amazon.com and browse inside the first couple of pages - there should be all you need there.

[quote="perfect_radio"]1) how do we prove that the series converges?[/quote]
The ratio test, which is to say by comparison to a convergent geometric sequence (with arbitrarily small positive ratio.) This shows that the series converges for all $x\in\mathbb{C}.$

[quote]2) can we derive properties like $\exp (ab)=\exp a + \exp b$ from that definition? [/quote]
Certainly! I know of at least two paths. Both require some serious analysis along the way.

A sketch of the first path: we prove a lot of general theorems about power series, including that a power series (inside its radius of convergence) is a differentiable function and its derivative is given term by term as if it were a polynomial.

Define $\exp(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}.$ We can now easily show that $\exp'(x)=\exp(x).$

Now we work on showing (using the multiplying factor method from ordinary differential equations) that $y(x)=\exp(x)$ is the only function satiysfying $y'(x)=y(x)$ for all $x$ and $y(0)=1.$ Now just start playing around with taking derivatives of various expression; the result will eventually fall out.

But I prefer the second method. The serious analysis required here involves doubly indexed sums:

Suppose we have $a_{jk}.$ Then we have the following equivalences:

$\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}a_{jk} = \sum_{k=0}^{\infty}\sum_{j=0}^{\infty}a_{jk} = \lim_{n\to\infty}\sum_{j=0}^n\sum_{k=0}^na_{jk} = \lim_{n\to\infty}\sum_{k=0}^n\left(\sum_{j=0}^ka_{(k-j)j}\right)$

provided one of two possible hypotheses is met: either $a_{jk}\ge 0$ for all $j$ and $k,$ or the whole double sum is absolutely convergent, meaning the sum of the absolute values is finite. We'll justify what we do below by absolute convergence.

Now:

$\exp(x)\exp(y)=\left(\sum_{j=0}^{\infty}\frac{x^j}{j}\right) \left(\sum_{k=0}^{\infty}\frac{y^k}{k}\right)$

$=\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{x^jy^k}{j!k!}$

$=\lim_{n\to\infty}\sum_{k=0}^n\left(\sum_{j=0}^k \frac{x^{k-j}y^j}{j!(k-j)!}\right)$

$=\sum_{k=0}^{\infty}\frac1{k!}\left(\sum_{j=0}^k \binom{k}{j} x^{k-j}y^j\right)$

$=\sum_{k=0}^{\infty}\frac1{k!}(x+y)^k=\exp(x+y).$

This is the proof of Rudin's that blahblahblah was referring you to. Note that it's valid for all complex $x$ and $y,$ thus allowing us to develop the complex exponential.

more on e:

$e^x = e^x$

$x ln(e) = ln(e^x)$

derivitive, chain rule
$1 = \frac{1}{e^x}\frac{d}{dx}[e^x]$

$e^x = \frac{d}{dx}[e^x]$

which means the power series offers:

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} ...$

$e = 2.71828182846...$

Rather than rely on shaky formal manipulations, you can make it rigorous that $\exp'(x)=\exp(x)$.

First, it is a convergent power series, so we can differentiate term by term to get the original series back.

Second, we can use the fact that

$\exp(t)=1+t+O(t^2)$ in any neighborhood of $t$.

Hence

\begin{eqnarray*} \exp'(t)&=&\lim_{h\to 0} \frac{\exp(t+h)-\exp(t)}{h} \\ &=&\exp(t)\frac{\exp(h)-1}{h} \\ &=&\lim_{h\to 0}\exp(t)(1+O(h)) \\ &=&\exp t \end{eqnarray*}

im still unclear on the limit not being in $\frac{0}{0}$ indeterminate form.  $ln(1) = 0$,  and as n approaches infinity $\frac{1}{n}$ gets infinitly close to zero.  i understand that $\frac{1}{n}$ no matter what n equals, it will never equal zero, but as n approaches infinity than it equals 0, infinity being an idea not a number.

If you want, you can look at it as being indeterminate and apply L'Hopital, but it isn't necessary.

Moreover, you run into circular reasoning pretty quickly.

The derivative of $f(x)=c$ is

$\lim_{t\to 0} \frac{f(x+t)-f(t)}{t}=\lim_{t\to 0}\frac{0}{t}$, which is an indeterminate form according to you.  What do you do from there?

When we say "the indeterminate case $\frac00,$" we do not mean literally zero divided by literally zero. We mean a limit case in which we're dividing an arbitrarily small number by an arbitrarily small number. The essence of any indeterminate case is conflicting tendencies. Dividing a very small number by something should yield something very small; dividing by a very small number should yield something very large. There's the confliciting tendencies; we need to balance the relative sizes to resolve the ambiguity. All of the other indeterminate cases have a similar tension between conflicting tendencies.

If you have something that's always zero divided by something that is never zero but merely tends to zero, then there's no ambiguity, no tension. It simply is zero. And without that tension, it's not an indeterminate case.

"Indeterminate case" is a functional or practical definition. All it really means is "we can't just read off an answer here; we must do more work to decide the case."

Of course the expression that led to the whole digression with ooooh ya is an indeterminate case:

$\lim_{n\to\infty}\frac{\ln\left(1+\frac1n\right)}{\frac1n}.$

Taking the completely harmless act of subtracting (literally!) zero from the numerator to make it

$\lim_{n\to\infty}\frac{\ln\left(1+\frac1n\right)-\ln1}{\frac1n}$

gives us a form we can recognize as (essentially) the definition of the derivative.

A number of the responders on this thread then asked the key question: how do we know what the derivative of the logarithm is? And the answer to that leads us to the question of how we defined the logarithm and/or the exponential in the first place. And as for using L'Hôpital's rule - that's a derivative; it raises the exact same questions as to how we know what that derivative is.

ooooh ya - I think you're too trusting of symbols, and of symbolic manipulations that you regard as familiar. (In particular, don't think of L'Hôpital's rule as being some kind of fountain of truth - if you study many of my posts over the last year or so, you'll see a steady campaign to wean people away from overreliance on L'Hôpital.) You need to look behind those symbols and ask what the expressions you write mean, or how you justify the manipulations you perform.

I am sorry, here is College or some beginners?

[quote="oooooh ya"]i knew how to do the proof, i was just pissed off no one would accept that it was in indeterminant form, which my proof relied on.  i do rely on symbolic manipulation, but i can prove what i do, like the l'hopital proof.  the power series proof is a can of worms to write out understandibly, but i can do it if i put in enough time.[/quote]

Read post 34.  You'll see that you can't even take the derivative of a constant function if you treat it as an indeterminate form for which l'hopital must be applied.

[quote="perfect_radio"][quote="blahblahblah"]Search for Walter Rudin's real and complex analysis on amazon.com and browse inside the first couple of pages - there should be all you need there.[/quote]

Thanks. The beginning of the book is very promising.

But I also have a question: How does Rudin define the logarithm?[/quote]

I quote (from another text of his)

Since $E$ (the exponential function) is strictly increasing and differentiable on $\mathbb{R}$, it has an inverse function $L$ which is also strictly increasing and differentiable on and whose domain is $E(\mathbb{R})$ (resp. $\exp(\mathbb{R})$), that is, the set of all positive numbers.  $L$ is defined by
\[ L(E(x))=x  \]
for all real $x$.

Differentiating, we get
\[ L'(E(x))\cdot E(x)=1  \]
writing $y=E(x)=1$, this gives us
\[ L'(y)=\frac{1}{y}  \]
Taking $x=0$, we see that $L(1)=0$.  Hence this implies
\[ L(y)=\int^y_1 \frac{dx}{x}  \]

I guess the complex logarithm is defined the usual way as all necessary functions have been defined.

And one last thing: if $a>0$ and $b$ is irrational, how do we define $a^b$?

Sorry for making this topic excesively long  :blush:

Well :
$\forall b \in \mathbb{R}$
$\forall a > 0$
We have :
$a^b = e^{b \cdot ln(a)}$
Indeed, those two functions have the same values in $\mathbb{Q}$ (dense in $\mathbb{R}$); $e^{b \cdot ln(a)}$ is defined on $\mathbb{R}$ (by continuity).

That's useless and wrong in several ways. The most important is that we cannot say that if $\lim_{x\to\infty} f'(x)=0$, $\lim_{x\to\infty}f(x)$ exists.


wouldnt it be nice if we could post more than 4 smileys? :huh:  :huh:

No, because people would do like this:  :stretcher:  :cleaning:  :cleaning:  :starwars:  :starwars:  :spider:  :trampoline:  :bruce:  :yankchain:  :pilot:  :pilot:  :police:  :ddr:  :ddr:  :ddr:  :ddr:  :diablo:  :nhl:  :clap:  :flex:  <<--and that is just obnoxious.

(I'm a mod, with moderation power comes responsibility to be sane, and while I'm not always sane, I don't post ridiculous numbers of smiles usually.)


linear algebra
matrix
algebra
polynomial
complex numbers

Let n be a positive integer and let A be an n×n matrix with real numbers
as entries. Suppose 4A^4+I = 0, where I denotes the identity matrix. Prove
that the trace of A (i.e. the sum of the entries on the main diagonal) is an
integer.

The trace is the sum of the complex eigenvalues. The eigenvalues must all be roots of $4z^{4}+1=0,$ meaning that they are $\pm\frac12\pm\frac{i}2.$ Furthermore, the eigenvalues are roots of the characteristic polynomial, which has real coefficients. Hence the eigenvalues must come in complex conjugate pairs.

In listing the eigenvalues, with multiplicities, we have $k$ copies of the pair $\left(-\frac12+\frac{i}2,-\frac12-\frac{i}2\right)$ and $m$ copies of the pair $\left(\frac12+\frac{i}2,\frac12-\frac{i}2\right)$ That's the same thing as saying that the characteristic polynomial factors as $(2x^{2}+2x+1)^{k}(2x^{2}-2x+1)^{m}.$

Then the trace is $m-k,$ which is an integer.

Since $4 A^{4}+I=0$, we must have that the minimal polynomial $p_{m}(x)$ for $A$ divides $4x^{4}+1=0$.  If $p_{m}(x)=4x^{4}+1$ then the sum of its roots is the coefficient of $x^{3}$, which is equal to $0$.  If that is not the minimal polynomial, then it must be of lower degree.  Now note that since all of the entries of the matrix are real, the roots of the corresponding characteristic polynomial will, when complex, come in conjugate pairs.  So the only two candidates for minimal polynomial are $x^{2}-\frac{i}{2}=0$ or $x^{2}+\frac{i}{2}=0$.  For either equation the roots add up to $0$.  So we have established that the sum of the roots of the minimal polynomial of $A$ must add up to $0$.

Additionally we know there are only $2$ or $4$ possible roots of the characteristic polynomial and they necessarily come in conjugate pairs.  Call them $r_{1},\bar{r_{1}},r_{2},\bar{r_{2}}$ (these are just the roots of $4x^{4}+1=0$).  The characteristic polynomial will be of the form $(x^{2}-|r_{1}|^{2})^{a}(x^{2}-|r_{2}|^{2})^{b}=0$ with $a$ or $b$ possibly zero but not both.  The sum of these extra roots regardless of the values of $a$ or $b$ will always be an integer.  So the result is true.

[quote="edoo"]So the only two candidates for minimal polynomial are $x^{2}-\frac{i}{2}=0$ or $x^{2}+\frac{i}{2}=0$[/quote]
Since $A$ is real, its minimal polynomial must be real. Those aren't the candidates. The possible minimal polynomials are implicit in my post: $4x^{4}+1,2x^{2}+2x+1,$ or $2x^{2}-2x+1.$ But since we're talking about the trace, I'd rather talk about the characteristic polynomial than the minimal polynomial. Note that we can have $4x^{4}+1$ as the minimal polynomial but at the same time have the trace of $A$ be something other than zero. Just take $k\ne m$ with both $k$ and $m$ positive in my post above.

Thanks Kent for keeping me honest!  Yes, I definitely messed up the candidates for the minimal polynomial  :oops:

A word on my approach though, I know the trace of the minimal polynomial and the characteristic polynomial can differ from each other and from zero.  I was trying to figure out the minimal polynomial, its trace, infer the possible factors of the characteristic polynomial, and add up the additional roots that make up the characteristic polynomial that have not already been accounted for in the minimal polynomial and add that value to the trace of the minimal polynomial and show that the total value will still be an integer.  This will work with the correct minimal polynomial candidates, but it may be more pedantic than your approach.


I was just wondering how long have you guys been doing this stuff and when did you guys first take programming class and when did you take the AP computer science.  Also do you have any suggestions for me to take the exam since I have no background relates to programming.

Thanks a lot.

I'm not sure which contest you're referring to, although USACO (http://www.usaco.org) appears to be the main one. Topcoder (http://www.topcoder.com/tc) is another one. I personally never took APCS as a class, but I took the AP exam as a freshman. In general, learning about data structures, big-oh notation, and graph theory are useful on both USACO and Topcoder.


linear algebra
matrix
linear algebra unsolved

I can't solve the following problem:
[img]http://photos.piry.net/images/19819cap1.JPG[/img]
Any ideas?
Thanks  :)

Show that $ AB$ is nilpotent- in particular, $ (AB)^{2007}\equal{}0$.


It isn't a TST for IMO, is it?  :maybe:

Also, how many days of TST are more to go? And when?

It is the TST for the Balkan MO and the IMO as well. There will be around 5 more besides the above one (there is one that just finished today - the second one, and 4 more I belive).

what about the romanian national olympiad? has it been posted? when was it held?


[quote="Algadin"]my own!
[img]http://i757.photobucket.com/albums/xx217/changvi/toan3.jpg[/img]

 :roll:[/quote]
to problem5
i think is $ a^3 \plus{} b^3 \plus{} c^3 \geq \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\plus{}\frac{3}{2}$

[quote="ye109"][quote="Algadin"]my own!
[img]http://i757.photobucket.com/albums/xx217/changvi/toan3.jpg[/img]

 :roll:[/quote]
problem5.
$ \iff\sum_{cyc}a^3 \geq \sum_{cyc}\frac {a}{b \plus{} c} \plus{} \frac {3}{2}$
$ \iff\sum_{cyc}a^3 \minus{} 3abc \geq \sum_{cyc}\frac {a}{b \plus{} c} \minus{} \frac {3}{2}$
$ \iff \frac {1}{2}\sum_{cyc}{(\sum_{cyc}a \minus{} \frac {(a \plus{} b)abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}})(a \minus{} b)^2 \geq 0$[/quote]

We have:
$ LHS \ge \frac {1}{2}\sum ab(a \plus{} b) \equal{} \frac {1}{2}\sum \frac {a \plus{} b}{c} \ge RHS$

for [b]6[/b]

let $ a\equal{}\frac{2x}{y\plus{}z},b\equal{}\frac{2y}{z\plus{}x},c\equal{}\frac{2z}{x\plus{}y}$

the inequality becomes $ (x\plus{}y\plus{}z)\sum x(x\plus{}y)^{2}(x\plus{}z)^{2} \ge 2(x\plus{}y)(y\plus{}z)(z\plus{}x)\sum xy(2x\plus{}2y\minus{}3z).$

I think this can be solved by AM-GM, but I haven't checked. :oops:

[quote="earldbest"]for [b]6[/b]

let $ a \equal{} \frac {2x}{y \plus{} z},b \equal{} \frac {2y}{z \plus{} x},c \equal{} \frac {2z}{x \plus{} y}$

the inequality becomes $ (x \plus{} y \plus{} z)\sum x(x \plus{} y)^{2}(x \plus{} z)^{2} \ge 2(x \plus{} y)(y \plus{} z)(z \plus{} x)\sum xy(2x \plus{} 2y \minus{} 3z).$

I think this can be solved by AM-GM, but I haven't checked. :oops:[/quote]

Problem 6;7;8 see here:http://www.maths.vn/forums/showthread.php?t=16721&page=2

[quote="Algadin"]my own!
[img]http://i757.photobucket.com/albums/xx217/changvi/toan3.jpg[/img]

 :roll:[/quote]
All problem you post is in maths.vn??And nothing(canhang2007),nhocnhoc(nguoivn) solve it.
Are you tranquoc?I want to know?:)


invariant
calculus
derivative
calculus computations

What does it mean that ordinary differential equation is singular at the point (for example at g=0) ?

What does it mean that ordinary differential equation is scale invariant with respect to some variable?

:)

[url=http://lmgtfy.com/?q=singular+differential+equation]Let me google that for you.[/url]

This is not very usefull :(

Please answer my question... :)

i know google :P

Scale invariance is too particular; don't look for that.

The words are most commonly used for linear differential equations, and within linear differential equations, most commonly used for equations that have coefficients that are analytic except at isolated points.

Suppose you have a linear ordinary differential equation. You can put it into a standard form in which the coefficient of the highest order derivative is 1. Then the set of singular points for the equation are the union of the sets of singular points for the other coefficients.

For instance, $ (1\minus{}x^2)y''\minus{}2xy'\plus{}2y\equal{}0$ has singular points at $ 1$ and $ \minus{}1.$

Equations with analytic coefficients should have power series solutions about any nonsingular point with radius of convergence at least as large as the distance to the nearest singular point (in the complex plane.)

There is a further distinction between regular singular points and irregular singular points. There, something like scale invariance (at least asymptotically) does start to matter. This distinction is tied to the method of Frobenius for finding solutions of the form $ \sum_{n\equal{}0}^{\infty}a_n(x\minus{}x_0)^{n\plus{}r}$ where $ x_0$ is a regular singular point.

In much of the study of ODE's, classifying words are closely tied to methods. If you can name it, you have some idea of what you can do with it.


suppose b(n)=p*a(n)+q*a(n+1). if Lim(b(n)) exist, a(n),p,q is positive.
prove Lim(a(n)) exist or give a counterexample[/img][/list]

OK,Thanks a lot!
In fact this question is a change for an exercise.The exercise is:suppose b(n)=p*a(n)+q*a(n+1),|p|<|q|,Lim(b(n)) exist,then prove  Lim(a(n)) .And I have solved this exercise while I thought about if p|<|q| does not hold whether there is condition that makes Lim(a(n)) still exist.And I exclude some other conditions while there still be a condition demonstrated above that I can not make a conclusion.So I ask about this question.Since Emis gives a counterexample while p=q=1,I am wondering if there still be a counterexample such that p>q>0 or a  positive proof for this condition.And this will makes this question solved perfectly.Still hope some help!


function
real analysis
real analysis unsolved

Hi could anyone help me with this question:

Let $ A \in R^{m\times n}$ and $ b \in R^m$, prove that
$ f(x) \equal{} \parallel{}Ax \minus{} b\parallel{}^2 \plus{} \sum_{k \equal{} 1}^{n}{|x_k|}$ is convex

I tried to prove that $ \parallel{}Ax \minus{} b\parallel{}^2$ is convex but was not successful.

Thank you.

If $ g(x)\equal{}\|Ax\minus{}b\|^2,$ then we can write

$ g((1\minus{}t)x\plus{}ty)\equal{}(1\minus{}t)g(x)\plus{}tg(y)\minus{}t(1\minus{}t)\|Ax\minus{}Ay\|^2.$

I did that the hard way, by writing $ g(x)\equal{}x^TA^TAx\minus{}x^TA^Tb\minus{}b^TAx\plus{}b^Tb$ and just multiplying everything out; I won't show the details because there's probably a slicker way and I'll let someone else show that.

Once you have that, you have $ g((1\minus{}t)x\plus{}ty)\le(1\minus{}t)g(x)\plus{}tg(y)$ for all $ x$ and $ y$ and for $ 0\le t\le 1,$ and that is convexity.

You're done if you verify the following: 1) If $ T$ is a linear map from$ R^n$ to $ R^m$ and $ b \in R^m$, then $ x\rightarrow \parallel{}Tx \plus{} b\parallel{}$ is convex on $ R^n$. 2) If $ f$ is convex and increasing on $ [0,\infty)$, and $ g : R^n\rightarrow [0,\infty)$ is convex, then $ f(g(x))$ is convex.


algebra proposed
algebra
inequalities
n-variable inequality

[quote="JellyWorld"][quote="Valentin Vornicu"]Let $a_1,a_2,\ldots,a_n$ be a permutation of the numbers $1,2,\ldots,n$, with $n\geq 2$. Determine the largest possible value of the sum \[ S(n)=|a_2-a_1|+ |a_3-a_2| + \cdots + |a_n-a_{n-1}| .  \]
[i]Romania[/i][/quote]
The answer should be $\frac{n(n-1)}{2}$  I'll think over the proof for a while.[/quote]

Unfortunatelly the answer isn't the one that you have written.
The answer is: $\lbrack\cfrac{n^2}{2}\rbrack -1$.

For an odd $n$ you can arange them in this order: $\frac{n-1}{2},n,1,n-1,2,n-2,3,...$
For an even $n$ the arangement is: $\frac{n}{2},n,1,n-1,2,...,\frac{n}{2}+2,\frac{n}{2}-1,\frac{n}{2}+1$

The main idea is dividing this numbers into two groups with equal number of elements:
$(n,n-1,n-2,...)$ and $(1,2,3,...)$. (Note: If n is odd then the number $\frac{n-1}{2}$ is not situated in any of this two groups).
If there are two adjesent element from the same group ($a_i,a_{i+1}$) then there should exist two adjesent elements from the other group ($a_j,a_{j+1}$). But we have that: $|a_i-a_{i+1}|+|a_j-a_{j+1}|<|a_i-a_{j+1}|+|a_j-a_{i+1}|=|a_{j+1}-a_{i+1}|+|a_j-a_{i}|$, so the elements from the groups should alternate in the permutation to maximize the sum. The answer was exposed already.

[quote=Valentin Vornicu]Let $a_1,a_2,\ldots,a_n$ be a permutation of the numbers $1,2,\ldots,n$, with $n\geq 2$. Determine the largest possible value of the sum \[ S(n)=|a_2-a_1|+ |a_3-a_2| + \cdots + |a_n-a_{n-1}| . \]
[i]Romania[/i][/quote]
[url=https://artofproblemsolving.com/community/c6h2331807p18721764]1999 Kazakhstan[/url]


geometry
3D geometry
Mandelbrot Competition
puzzles

I will irregularily post cryptograms. Try to solve them. :)

Since this is the first one, I will make this easy. I used monoalphabetic substitution.

[color=red]jb ucwcuf k mnyf cojb jhb bjgfl mnyfx, k tbnljg vbhfl bl co rfoflkz kod vbhfl hgkjfwfl cojb jhb vbhflx bt jgf xkif ufobicokjcbo kybwf jgf xfmbou cx civbxxcyzf, kou c gkwf kxxnlfuzd tbnou ko kuiclkyzf vlbbt bt jgcx, ynj jgf iklrco cx jbb okllbh jb mbojkco cj.[/color]

Good luck!

[quote="Myself"][hide]TO DIVIDE A CUBE INTO TWO OTHER CUBES, A FOURTH POWER OR IN GENERAL ANY POWER WHATEVER INTO TWO POWERS OF THE SAME DENOMINATION ABOVE THE SECOND IS IMPOSSIBLE, AND I HAVE ASSUREDLY FOUND AN ADMIRABLE PROOF OF THIS, BUT THE MARGIN IS TOO NARROW TO CONTAIN IT.[/hide][/quote]

Shoot, you neat me to it!  Crytograms are fun, hehehe

Hmm. I guess it was too easy. Let's try a shorter one. :D

[color=red]fdimufdiroj rj d sdfu cndwuv doogxvrqs ig ouxidrq jrfcnu xknuj trim fudqrqsnujj fdxaj gq cdcux.[/color]

Nice... I suggest you can also post Cryptograms in the [url=http://www.mathlinks.ro/Forum/index.php?f=337]Puzzles & Brainteasers[/url] forum :P ;) .

Now, some fact that can be usefull

[hide="Facts"]
[b]Letter Frequency in the English Language[/b]
e t a o i n s r h l d c u m f p g w y b v k x j q z
[b]
Word Frequency of the Top Twenty Most Used Words found in Written English[/b]
the of to in and a for was is that on at he with by be it an as his

[b]Two-Letter Words and the Frequency in the English Language[/b]
of to in it is be as at so we he by or on do if me my up an go no us am

[b]Three-Letter Words and the Frequency in the English Language[/b]
the and for are but not you all any can had her was one our out day get has him his how man new now old see two way who boy did its let put say she too use
[b]
Four-Letter Words and the Frequency in the English Language[/b]
that with have this will your from they know want been good much some time very when come here just like long make many more only over such take than them well were

[b]Most Common Words and Frequency in the English Language[/b]
the of and to in a is that be it by are for was as he with on his at which but from has this will one have not were or all their an i there been many more so when had may today who would time we about after dollars if my other some them being its no only over very you into most than they day even made out first great must these can days every found general her here last new now people public said since still such through under up war well where while years before between country debts good him interest large like make our take upon what
[/hide]

And also remember that there are only two one letter words...[b]a[/b] and [b]I[/b]

[quote="MathFiend"]could someone clue me in as to how to decode the first one??? plz? lol[/quote]
About 5% of normal English is "the" :D . Eg) First cryptogram has three "the".
If the cryptogram is long enough, you should try to look for "the".
If the cryptogram is too short, you cna try to find "e" first. "e" occurs 12.51% of the time in spoken/written English. Next most frequent letter is "t" with 9.25%.

Also, assuming that the guy who gives you the cipher (the Cryptogram) is using only simple substitution (he chooses a fixed permutation of the letters and apply it letter by letter) then you can easely find other letters. For example, there are few letters that can repite in an english word, and some of them are more common than others (ss, cc, tt, ee, oo, mm, etc) and some of them will never occur at the end of a word. There are a lot of peculiarities of each language that are quite useful for decipher ciphers :D . It is actually quite interesting, and there are ways too much challenging to do it.

Best,

[quote="MathFiend"]could someone clue me in as to how to decode the first one??? plz? lol[/quote]

These things go along faster if you're on the CIA...

[quote="Myself"]Solution to the second one:
[hide]MATHEMATICS IS A GAME PLAYED ACCORDING TO CERTAIN SIMPLE RULES WITH MEANINGLESS MARKS ON PAPER.[/hide][/quote]
You got it! This is a quote made by Hilbert.
Now for next cryptogram.

[color=red]u gupsigupchcua cw u guhscai btx pyxacaq htbbii capt psitxigw.[/color]

[quote="Myself"]Third one: I got this one quickly because of a random guess that was correct...
[hide]A MATHEMATICIAN IS A MACHINE FOR TURNING COFFEE INTO THEOREMS.
I guessed that the first word was "mathematics" or something like that.[/hide][/quote]
Wow! That was fast! This is a very famous quote by Erdos.
I won't make the same mistake on this cryptogram.

[color=red]cp cwa'p psup psir hua'p wii psi wtnypcta. cp cw psup psir hua'p wii psi mxtonig.[/color]

[hide="another quote?"]It isn't that they can't see the solution.  It is that they can't see the problem.[/hide]

[hide] It isn't that they can see the solution, it is that they can't see the problem. [/hide]

Aww, peter beat me to it

[quote="MithsApprentice"][quote="MathFiend"]could someone clue me in as to how to decode the first one??? plz? lol[/quote]These things go along faster if you're on the CIA...[/quote]Actually the modern encryption use to be a lot more complex and in Digital Format, not in Text mode. This kind of ciphers can be easily analysed if you have some side information, as at least the language. Any way, a friend did like five or six years ago a program to solve this kind of things, that actually proved 5 languages (English, Spanish, German, French and Italian), he got some dictionaries and some grammar basis online. And that program could analyze about 500 words per second (In a pentium II of 400MHz) and give with a pretty low statistical error the language, the permutation and the text if the text was large enough, and that was usually just about fifteen words I think.  (It just worked with simple replacement)

[quote="lightrhee"]I don't believe it! 10 minutes to solve this! I guess "ps" was too obviously "th". This is a quote by Chesterton.
Next problem:
[color=red]byakugiapun mxtqxiww suw pt kt fcps psi xicapixmxipupcta tb ouwch ckiuw.[/color]
Enjoy![/quote]

[hide]FUNDAMENTAL PROGRESS HAS TO DO WITH THE REINTERPRETATION OF BASIC IDEAS. [/hide]

[quote="lightrhee"]I don't believe it! 10 minutes to solve this! I guess "ps" was too obviously "th". This is a quote by Chesterton.
Next problem:
[color=red]byakugiapun mxtqxiww suw pt kt fcps psi xicapixmxipupcta tb ouwch ckiuw.[/color]
Enjoy![/quote]

This one was hard, I gave up after a while.  Post another one, hehehe.

This quote was by Whitehead.
This cryptogram is still coded with monoalphabetic substitution.

[color=red]qyg nossglgxq blpxtygf ks ploqyrgqot -- prboqokx, nofqlptqokx, ihzosotpqokx, pxn nglofokx.[/color]

[quote="MithsApprentice"][quote="djimenez"](It just worked with simple replacement)[/quote]

But the point is that most cryptology today is NOT based just on simple replacement.[/quote]Yes I know. Actually, since Julius Ceasar the simple replacement is obsolete. These days, most of the cryptography today is based on either Elliptic Curves or Finite Fields (or Finite Grups) that have either hard Diffie Helman decisional problem or hard Discrite Log problem (or at least that they are not known to be easy... the problem of having a "primitive" complexity theory :D ) Any way, it is still a lot of fun to solve these cryptograms by hand, without a lot of knowledge about it.

[quote="peter"][quote="lightrhee"][color=red]qyg nossglgxq blpxtygf ks ploqyrgqot -- prboqokx, nofqlptqokx, ihzosotpqokx, pxn nglofokx.[/color][/quote]

THE DIFFERENT BRANCHES OF ARITHMETIC -- AMBITION, DISTRACTION, UGLIFICATION, AND DERISION.[/quote]

Good job! This quote was made by Charles Dodgson,also known as Lewis Carroll.
Next problem:

[color=red]myxi gupsigupchw, gur cp aidix oi tb uar ywi pt uartai.[/color]

[quote="jmadsen"]You can use the letter frequency counter at this site: [url=http://http://www.10ticks.co.uk/s_codebreaker_letter.asp]http://www.10ticks.co.uk/s_codebreaker_letter.asp[/url][/quote]
thanks, it's still really hard though

[quote="lightrhee"]
[color=red]vpx vwmvp jn t vpxjwc ds du cjmw fduq, ujv du cjmw xcxs.[/color][/quote]

[hide] The truth of a theory is in your mind, not in your eyes. [/hide]

[quote="lightrhee"][quote="ambierona"][quote="lightrhee"]
[color=red]myxi gupsigupchw, gur cp aidix oi tb uar ywi pt uartai.[/color][/quote]

[hide]pure mathematics, may it never be of any use to anyone[/hide][/quote]
I made same mistake again(including the word 'mathematics')! I won't make same mistake on next problem.

[color=red]vpx vwmvp jn t vpxjwc ds du cjmw fduq, ujv du cjmw xcxs.[/color][/quote]

haha yeah that helped a lot.
gahh the next one's already solved and I just saw it!!

[quote="MithsApprentice"]You know...it's really a shame. Heck...I think we should go back to the days of replacement and let us not-so-handy-with-codes people feel smart. :-)[/quote]I feel a sarcastic tone on this. :P
I am just saying that, as a puzzle, it is nice to solve. I would not feel secure any way writting my credit card on an online purchase with just a replacement cryptography technic.

[quote="Myself"][hide][color=red]UPRTKUNOGKNHSRVQNSXBIFHRFTITQTQRUNOUYNHNNMQENTWUPNKRAHCKBEPTKAKWQEDCMTIAQTIANKSRPNEDMCUPRHNNMQYPBEPTKRTHNWUQRXRIPWIAKRATIAOBOUCGTFRQUNFRUPRKTKRONKQUWARIUQYPNTKRBIURKRQURABIVTUPRVTUBEQNKENVGRURBIVTUPRVTUBEQENVGRUBUBNIQXNSWVRNIRUPRHTQBEQBQTGGKNGKBTURONKQUWARIUQLWQUHRFBIIBIFBIVTUPENIURQUQYPBSRXNSWVRUYNTIAHRCNIABQONKVNKRTAXTIERAQUWARIUQHNUPHNNMQORTUWKRPWIAKRAQNOGKNHSRVQOKNVTEUWTSVTUPRVTUBEQENVGRUBUBNIQQWEPTQUPRTVRKBETIVTUPRVTUBEQENVGRUBUBNIQUPRWIBURAQUTURQNOTVRKBETVTUPRVTUBETSNSCVGBTAUPRVTIARSHKNUENVGRUBUBNIVWTSGPTUPRUTVTUPRVTUBEQTIABIONKVTUBEQJWTKURKSCVTFTDBIRTIAUPRBIURKITUBNITSVTUPRVTUBETSNSCVGBTATSUPNWFPUPRHNNMQABQEWQQVTICUNGBEQUPTUNEEWKOKRJWRIUSCBIVTUPRVTUBEQENIURQUQUPRCTKRINULWQUTENSSREUBNINOUKBEMQBIQURTAUPRHNNMQURTEPQUWARIUQUNSRTKITIAWIARKQUTIAVRUPNAQYPBEPTSSNYQUPRVUNQNSXRTYBARXTKBRUCNOGKNHSRVQHRCNIAUPRUCGRQGKRQRIURABIUPRHNNMQ[/color][/hide]
I slightly modified this text.  I enciphered it using [url=http://www.reed.edu/~mcphailb/applets/crypto/]this[/url] web site, using a cipher alphabet related the topic.  This has to do with math.  Note that this is very long and contains no spaces.  Still a substitution cipher.[/quote]
Non-spaced cryptogram was going to be my next problem, interesting. The web site seems to be better than my Maple program. Mine is little bit slower than this web site.

[quote="Myself"][hide][color=red]UPRTKUNOGKNHSRVQNSXBIFHRFTITQTQRUNOUYNHNNMQENTWUPNKRAHCKBEPTKAKWQED
CMTIAQTIANKSRPNEDMCUPRHNNMQYPBEPTKRTHNWUQRXRIPWIAKRATIAOBOUCGTFRQUNFRUPRKTKRONKQUWARIUQYPNTKR
BIURKRQURABIVTUPRVTUBEQNKENVGRURBIVTUPRVTUBEQENVGRUBUBNIQXNSWVRNIRUPRHTQBEQBQTGGKNGKBTURONKQU
WARIUQLWQUHRFBIIBIFBIVTUPENIURQUQYPBSRXNSWVRUYNTIAHRCNIABQONKVNKRTAXTIERAQUWARIUQHNUPHNNMQORTU
WKRPWIAKRAQNOGKNHSRVQOKNVTEUWTSVTUPRVTUBEQENVGRUBUBNIQQWEPTQUPRTVRKBETIVTUPRVTUBEQENVGRUBUBN
IQUPRWIBURAQUTURQNOTVRKBETVTUPRVTUBETSNSCVGBTAUPRVTIARSHKNUENVGRUBUBNIVWTSGPTUPRUTVTUPRVTUBEQ
TIABIONKVTUBEQJWTKURKSCVTFTDBIRTIAUPRBIURKITUBNITSVTUPRVTUBETSNSCVGBTATSUPNWFPUPRHNNMQABQEWQQV
TICUNGBEQUPTUNEEWKOKRJWRIUSCBIVTUPRVTUBEQENIURQUQUPRCTKRINULWQUTENSSREUBNINOUKBEMQBIQURTAUPRH
NNMQURTEPQUWARIUQUNSRTKITIAWIARKQUTIAVRUPNAQYPBEPTSSNYQUPRVUNQNSXRTYBARXTKBRUCNOGKNHSRVQHRCN
IAUPRUCGRQGKRQRIURABIUPRHNNMQ[/color][/hide]
I slightly modified this text.  I enciphered it using [url=http://www.reed.edu/~mcphailb/applets/crypto/]this[/url] web site, using a cipher alphabet related the topic.
  This has to do with math.  Note that this is very long and contains no spaces.  Still a substitution cipher.[/quote]
Anyone soived it?
Since no one posted nothing for days, I am going to post the solution.

[hide]The art of problem solving began as a set of two books coauthored by Richard Rusczyk and Sandor Lehoczky. The books,
which are about seven hundred and fifty pages together, are for students who are interested in mathematics or compete in
mathematics competitions. Volume one the basics is appropriate for students just beginning in math contests while volume
two and beyond is for more advanced students both books feature hundreds of problems from actual mathematics competitions
such as the American mathematics competitions, the united states of America mathematical Olympiad, the Mandelbrot competition,
mu alpha theta mathematics, and informatic squarterly magazine and the international mathematical Olympiad. Although the
books discuss many topics that occur frequently in mathematics contests, they are not just a collection of tricks. Instead, the
books teach students to learn and understand methods, which allows them to solve a wide variety of problems beyond the types
presented in the books.[/hide]
Wow! That was very long. I'm quite surprised that no one got it(since it was really long). Maybe no one worked on it.


limit
real analysis
real analysis theorems

Let $(a_n)_{n \geq 1} \subset \mathbb R^n$. If $|a_n| \leq r$, then there is a convergent subsequence.

The proof should the same as in the $\mathbb R$ case. There you have intervals contained in each other and you take $\sup$ or $\inf$. What do you do in this case?  :blush:

Lemma: A sequence in $\mathbb{R}^n$ converges iff each of its components does.

Using this, take a subsequence that converges in the first component, a subsequence of that that converges in the second component, etc.

OK, thanks for the proof.

Is it true that if we have a sequence of "curves" $(C_n)_{n \geq 0}$ in $\mathbb R^n$ (don't really know how you call them, neither in romanian, nor in english) s.t. $C_{n+1} \subsetneq C_n$ and $\lim_{n \to \infty} \textrm{diameter} \ C_n = 0$ then there is a point $P$ s.t. $P \in C_k$ for all $k$?


You know guys I like simple stated things, so here it is another...

John Doe selected five different integers and formed the sums of pairs, then wrote down the ten results, some of which were wrong, as: $2,4,5,7,7,8,10,11,12,13$.
Dick, who was checking John Doe's work, told John Doe that the first three and the last three sums were correct, but that some of the others were wrong.
Find the five integers!

[quote="dasherm"]Does this mean that the first three and last three sums are necessarily the three least and greatest sums?[/quote]

No, we don't know which sum is involved. ;)
Otherwise is quite straighforward!

I got:

[hide]
$-1,1,2,3,10$
which yields the sums of
$0,1,2,3,4,5,9,11,12,13$

This contains none of the middle sums.
[/hide]
Is that what you're looking for, because it seems like there are other answers?


Hi, can anyone link me to a site with the dimensions of adult human organs (in particular, the stomach)? I'm trying to make a model that is supposed to be to scale. Thanks!

Is that the bio assignment half the grade 11s are working on. I heard this from someone at Wednesday Math, collectively as a class kill  a hobo and steal his organs!

Do you want some figures of the human stomach? By entering "anatomy" and "stomach", you can find quite a lot of results.

y don't u try wikipedia--u wud get exact exact results

[quote="Farnak"]Thanks for the replies. I have found many diagrams of the stomach but I haven't found dimensions/measurements which is what I'm more interested in. Can anyone help me with that?[/quote]
Hi. I checked an anotomy book "The Human Body" by Adolf Faller, M.D., Michael Schünke, M.D., Ph.D. and Gabriele Schünke, M.Sc.. It says, on page 405, that:

[i]Its shape and position may show wide variations (with degree of filling). The volume of the stomach is about 1200−1600 ml.[/i]


function
trigonometry
calculus
algebra proposed
algebra

Let  $f: \mathbb{R}\to\mathbb{R}$ is a function such that $f( \cot x ) = \cos 2x+\sin 2x$ for all $0 < x < \pi$. Define $g(x) = f(x) f(1-x)$ for $-1 \leq x \leq 1$. Find the maximum and minimum values of $g$ on the closed interval $[-1, 1].$

[quote="N.T.TUAN"]Who can?
[hide="hint"]1)Find $f(t)$ for $t\in\mathbb{R}$.
2)Denote $y=x(1-x)$, when $g(x)=h(y)$.[/hide][/quote]

I found that $f(x)=\frac{x^{2}+2x-1}{x^{2}+1}$ and $g(x)=\frac{(x^{2}+2x-1)(x^{2}-4x+2)}{(x^{2}+1)(x^{2}-2x+2)}$.

How can I find $min$ and $max$ without calculus?

You need find $h(y)$, but I so can't no use calculus.

You are right  :oops: . Because $\max_{-1\leq x\leq 1}x(1-x)=\frac{1}{4}$ and $\min_{-1\leq x\leq 1}x(1-x)=-2$.


algebra
polynomial
quadratics
quadratic formula

How many real roots does the polynomial
$P(x) = 2x^3+x^2+3x-2$
have?

From Rational Roots Theorem, $x=\frac{1}{2}$ is a root. Then \[ P(x)=2x^3+x^2+3x-2=(2x-1)(x^2+x+2). \]From the quadratic formula, the roots of the latter polynomial has no real roots. Thus, $P(x)$ has only one real root.


Let $n\in N$ and $\frac 21+\frac {2^2}{2}+\frac {2^3}{3}+...+\frac {2^n}{n}=\frac{p_n}{q_n}$ such that $gcd(p_n,q_n)=1$.

a) Prove that $p_n$ is the even number.

b)Prove that if $n>3$ then $8\mid p_n$

c) (main part of the problem) Prove that for every $k\in N$ there exist a number $n$

such that the numbers of $p_n,p_{n+1},p_{n+2},... $,divisibles by $2^k$.

Well, the special major was (as so many times) Peter Schoelze.


[quote="jeter123"]2) The magnitude E of an electric field depends on the radial distance r according to E = A/r^4, where A is a constant with unit volt-cubic meter. As a multiple of A, what is the magnitude of the electric potential difference between r = 1.24 m and r = 2.37 m?
[/quote]
[color=blue]$\large E=\frac{A}{r^4}=\frac{dV}{dr}$

$\large dV=\frac{Adr}{r^4}$

$\large \Delta V=\int_{r_1}^{r_2}\frac{Adr}{r^4}=\frac{A}{3}(\frac{1}{r_1^3}-\frac{1}{r_2^3})$[/color]

[color=blue]Really? I've got no idea[/color]

since it is a sphere the charge seems to act at the center.

$U=\frac{q}{4{\pi}{\epsilon}_0r}$


Constant-nth-Differences Proof Project:
Let f(x) = ax^3 + bx^2 +cx +d.

a. Find algebraically four consecutive values of f(x) for which the
x-values are k units apart. That is, find f(x), f(x + k), f(x+2k), and f(x + 3k). Expand the powers.

b. Show that the third differences between the values in part a are
independent of x and are equal to 6ak^3.

My Work Done:

I've already done part a, but I can't figure out how to do part b. Can you check my answers to part a and show me how to complete part b please?

Answers to Part a:

f(x) = ax^3 + bx^2 +cx +d
f(x + k) = ax^3 + 3ax^2k + 3axk^2 + k^3 + bx^2 + 2bxk + bk^2 + cx + ck +d
= ax^3 + x^2(3ak +b) + x(3ak^2 + 2bk +c) +ak^3 + bk^2 +ck + d

   f(x+2k) = ax^3 + 6akx^2 + 12ak^2x + 8ak^3 + bx^2 + 4bkx +4bk^2+cx+2ck+d
= ax^3 + x^2(6ak + b) + x(12ak^2 + 4bk +c) + 8ak^3 + 4bk^2 +2ck+d

   f(x+3k) = ax^3 +9akx^2+27ak^2x+27ak^3+bx^2 + 6bkx + 9bk^2 + cx + 3ck +d
= ax^3+x^2(9ak + b)+ x(27ak^2 + 6bk +c) + 27ak^3 + 9bk^2 + 3ck +d

[hide="Without expansion"] [b]Lemma:[/b]  If $f(x)$ is a polynomial of degree $n$ with leading term $ax^{n}$, then the first difference

$f_{1}(x) = f(x+1)-f(x)$

Is a polynomial of degree $n-1$ with leading term $nax^{n-1}$.

[b]Corollary:[/b]  If $f(x)$ is a polynomial of degree $n$ then $f_{n}(x)$ is a constant.

We conclude that for any third-degree polynomial with leading term $ax^{3}$ the third difference polynomial $f_{3}(x) = 6a$.

Your desired conclusion is found by considering the third difference of the polynomial

$g(x) = f(kx) = ak^{3}x^{3}+...$ [/hide]


Let $ ABCD$ is a cyclic quadrilateral. Construct $ ABCD$ show that $ AB \equal{} a, BC \equal{} b, CD \equal{} c, DA \equal{} d$, with $ a,b,c,d$ are constant.

[quote="kabi"][hide="hint"]
1.As all the four sides r given u can easily find the radius of the circle by the advance version of Brahmagupta's formula I don't think after that there would be any problem in construction.
2.Otherwise we can also find the two diagonals which will make the task easier.[/hide]

Well I still trying to find  a much much much easier way,

Thank u.[/quote]

Yes! I'm waiting to get a easier way.

Thank you! :P

The Sturm's construction :

[img]http://img255.imageshack.us/img255/9692/stumsm4.jpg[/img]

Domi

[quote="mihai miculita"]Applying theorems of Ptolemy:
$ |AC|.|BD| \equal{} |AB|.|CD| \plus{} |AD|.|BC| \equal{} a.c \plus{} b.d$ and $ \frac {|AC|}{|BD|} \equal{} \frac {|AB|.|AD| \plus{} |CB|.|CD|}{|BA|.|BC| \plus{} |DA|.|DC|} \equal{} \frac {ad \plus{} bc}{ab \plus{} cd}\Rightarrow$
$ \Rightarrow |AC|^2 \equal{} |AC|.|BD|.\frac {|AC|}{|BD|} \equal{} \frac {(ac \plus{} bd).(ad \plus{} bc)}{ab \plus{} cd}.$
Construct now $ \triangle{ABC}$ and $ \triangle{ACD}.$[/quote]

Thank Mihai miculita and domi very much. :P


limit
function
logarithms
real analysis
real analysis unsolved

Using the limitated dezvoltations, calculate :

[code]\begin{eqnarray*}
\ \lim_{x\longrightarrow 0}\frac{x^x-(sinx)^sinx}{x^p} \\
\end{eqnarray*}[/code]

hello, do you mean
\begin{eqnarray*}
\ \lim_{x\longrightarrow 0}\frac{x^x-(sinx)^{sinx}}{x^p} \\
\end{eqnarray*}
what do we know about $ p$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean
\begin{eqnarray*} \ \lim_{x\longrightarrow 0}\frac {x^x - (sinx)^{sinx}}{x^p} \\
\end{eqnarray*}
what do we know about $ p$?
Sonnhard.[/quote]

Well let's do it for all $ p$ in $ R$. For simplicity, write $ sinx = s$. Then $ x^x-s^s=x^x(1-\frac{s^s}{x^x}) \leq x^x(1-\frac{s^x}{x^x})$. We have $ \frac{s^x}{x^x}=e^{xln(s/x)}$. Now use the Taylor approximations for $ s$, $ ln(1-u)$ and $ e^u$ for $ x$ and $ u$ near 0. This should give  $ x^x-s^s\leq x^3/6 +O(x^5)$. We should get a similar estimate from below. So I'm getting the limit in question equal to $ 1/6$ when $ p=3$, $ 0$ if $ p<3$ and $ \infty$ if $ p>3$. But of course this needs to be checked, and there may be an easier way.

Scratch the last post. For one thing it's clear that the limit, if it exists, can't be positive. That's because the function $ f(x) \equal{} x^x$ is decreasing on the interval $ (0,1/e)$, hence $ x^x \minus{} (sinx)^{sinx}$ is negative for $ x$ in this interval. (Incidentally, we are taking a one sided limit as $ x\rightarrow0^ \plus{}$ for the problem to be sensible.)

An easier approach is to use the mean value theorem: For any small  $ x > 0$, our expression equals $ \frac {f'(c_x)(x \minus{} sinx)}{x^p}$. Now $ f'(x) \equal{} x^x(1 \plus{} lnx)$ and $ x \minus{} sinx$ is like $ x^3/6$ for small $ x$. We know $ x^x\rightarrow1$ and because logarithms can never beat powers of $ x$, the limit equals $ 0$ if $ p < 3$, and equals $ \minus{} \infty$ if $ p\geq3$.

just an asmptotic expansion near 0

 $ x^x \minus{} (sinx)^{sinx}\equal{} (1/6\plus{}(1/6)*ln(x))*x^3\plus{}O(x^4)$

[quote="Moubinool"]just an asmptotic expansion near 0

 $ x^x \minus{} (sinx)^{sinx} \equal{} (1/6 \plus{} (1/6)*ln(x))*x^3 \plus{} O(x^4)$[/quote]

How derived?


What are http://alt1.artofproblemsolving.com/ and http://alt2.artofproblemsolving.com/ for? Are they alternate servers if the main one crashes?

They do look like backup subdomains.  I think LaTeXRender is installed on those.


i've seen this a few times, but i'm never sure whether i'm right or not.

[size=59]0[/size], [size=75]1[/size], [size=100]4[/size], [size=150]9[/size],[size=200]16[/size]
are the numbers getting bigger or greater?

i think its supposed to be greater cuz the numbers' size is getting bigger but the value is increasing.

[size=200]0[/size],[size=150]1[/size],[size=100]4[/size],[size=75]9[/size],[size=59]16[/size]
are these numbers getting greater/less (or bigger/smaller, based on the answer of the first problem)? :D

Based on $ >$ meaning "greater than", I would conclude that 1. is "greater" and 2. is less. However, this is assuming that you mean the value of the numbers, not the glyphs representing the numbers.

Can the answer be bigger and greater simultaneously? :maybe:
Then, the first one would be bigger and greater, and the second one would be smaller and greater.

I think both can be the answer . :maybe:

 :maybe:

Well, it depends on what you mean when you say "number": the numerical value that is represented by the Arabic symbol or the symbol itself. :wink:


advanced fields
advanced fields unsolved

if <x,y>=<x,z> show that y=z.
x,y,z belong to an inner product space

Please give titles to your posts that convey some useful information about their content.

This is obviously wrong (consider e.g. $ x \equal{} 0$).  Probably the question you meant to ask is to show that if $ \langle x, y\rangle \equal{} \langle x, z \rangle$ [i]for all $ x$[/i] then $ y \equal{} z$.  Yes?


Find all $x,y\in N$ such that $x^2(x^2+y^2) = y^{100}$

Obviously x divides y.  So put y = xa.

Then $x^{96}a^{100} = 1 + a^2$

Suppose x,a > 1.  Then $x^{96}a^{100} > 1 + a^2$ obviously, so no solution.

Suppose x = 1.  Then we cant find a solution.
Suppose a = 1.  Then we cant find a solution.

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

another method is to look at discriminant of the quadratic in x^2.

I think all we can say is $x$ divides $y^{50}$, which does not mean $x$ divides $y$.

I nearly did the same thing. Solutions do exist, so lets try again?

looking in mod 4, $y^2$ = 0 or 1, trying 1 mod 4, cannot be since $x^2$ must be 0 or 1 mod 4, but that does not work since $0(0+1) \equiv 1$ or $1(1+1) \equiv 1$
trying $y = 0 \mod 4$, $x = 1 \mod 4$, we get $1(1+0)\equiv 0$, trying $x = 0 \mod 4$, we get $0(0+0)\equiv0$
so x and y are $0 \mod 4$, substituting,
$(4n)^2((4n)^2+(4m)^2) = (4m)^{100}$
$n^2(n^2+m^2) = m^{100}*4^{96}$
and i think that we could extend the same argument to this as well, RHS 0 mod4, so LHS 0 mod 4, x = 0 mod 4, y = 0 mod 4, and repeat forever
so x and y infinitely divisible by 4, but that only works for (0,0), but we are limited to natural numbers so there are no solutions


Show that the roots of the equation z^5-(z-i)^5=0, where z is not equal to i, are 1/2(cot(360k/5)+i) where k=1,2,3,4. (i is sqrt-1)

[hide="Hint"]$ (z\minus{}i)^5 \equal{} z^5 \implies |z\minus{}i| \equal{} |z| \implies z \equal{} x \plus{} \frac{1}{2} i, \; x \in \mathbb{R}$

Then, we have:

$ \left(x \minus{} \frac{i}{2} \right)^5 \equal{} \left( x \plus{} \frac{i}{2} \right)^5$.

Now draw it and consider the two quantities as vectors in polar form. What happens if we exponentiate?

[/hide]

[hide="Alternately..."] Let $ f(z) \equal{} \frac{z}{z \minus{} i}$.  The roots of the equation correspond to the roots of $ f(z)^5 \equal{} 1$, so $ f(z)$ is a fifth root of unity.  Now find $ f^{\minus{}1}(z)$. [/hide]


[quote="Tarento"]Say A, B, and C are n x n matrices and are invertible. Prove that ABC is also invertible.[/quote]

Try starting with just A and B.

Show that if A and B are are invertible, then AB is invertible.

Given invertible matrices $A,B$, the inverse of $AB$ is $B^{-1}A^{-1}$.  See if you can generalize it one step more...

[quote="spix"]like....isn`t it obvious? :rotfl:[/quote]

Not really... :blush:

$A,B,C$ are invertible ... that means that $\det(A),\det(B),\det(C)\neq 0$ so that means that$\det(ABC)=\det(A)\det(B)\det(C)\neq 0$ which concludes that $ABC$ is invertible...


function
induction
limit
algebra
functional equation
algebra proposed

Find all continuous functions $f$ : $\mathbb{R} \to \mathbb{R}$ such that:

1- $f(0)=0$

2- $f(\frac{1}{x})+f(\frac{1}{y}) = \frac{f(x+y)}{f(x)f(y)}$ for all  $x$ and $y$ non zero in $\mathbb{R}$

Wow, I think I'm making this problem more complicated than it actually is.

From the definition, the only zero of the function must be  $f(0)=0$. If there was another zero $z$, the function wouldn't be defined for $\frac{1}{z}$.

Taking $x=-y=\frac{1}{t}$, for every $t\in\mathbb{R^*}$, $f(t)+f(-t)=0$, so $f$ is an odd function. From now on, we can only worry about $f(t)$, with $t>0$.

Taking $x=y=1$, we reach $f(2)=2f^3(1)$.
Taking $x=y=\frac{1}{2}$, we reach $f^2\left(\frac{1}{2}\right)=\frac{1}{4f^2(1)}$, so $f\left(\frac{1}{2}\right)=\pm\frac{1}{2f(1)}$.
By induction, for every $n\in\mathbb{Z}$,
a) $f(2^n)=\pm 2^n f^{2n+1}(1)=f(1)(2f^2(1))^n=\pm Ca^n$ (where $C=f(1)$ and $a=2f^2(1)$),
b) $sign(f(2^{n+1}))=sign(f(2^{-n}))$.

Then, taking $x=2^n$ and $y=2^m$, we reach
$f(2^n+2^m)=(\pm Ca^n)(\pm Ca^m)(\pm Ca^{-n} \pm Ca^{-m})=C^3(\pm a^m \pm a^n)$.
Because of the continuity, if $n\to -\infty$, then $f(2^n+2^m)=C^3(\pm a^m \pm a^n)\to f(2^m)=\pm Ca^m$. That implies the following:
a) $C^3=C$, so $C=f(1)=\pm 1$, and $a=2f^2(1)=2$
b) taking a look at where the signs $\pm$ come from, $sign(f(2^n))=sign(f(2^{-n}))$, which, together with the result of our induction, implies that $sign(f(2^n))=sign(f(2^{n+1}))$, the sign of $f(2^n)$ is constant, so it's always like $f(1)$.

To put it together, $f(2^n)=sign(f(1))2^n$ for every $n\in\mathbb{Z}$.

I think we could proceed and try to prove that $f(\sum_{i\in\mathbb{Z}}c_i2^i)=sign(f(1))\sum_{i\in\mathbb{Z}}c_i2^i$, and then $f(t)=sign(f(1))t$ for every $t>0$.

The only solutions would then be $f(x)=x$ and $f(x)=-x$.

[quote="lordWings"]Wow, I think I'm making this problem more complicated than it actually is.

I think we could proceed and try to prove that $f(\sum_{i\in\mathbb{Z}}c_i2^i)=sign(f(1))\sum_{i\in\mathbb{Z}}c_i2^i$, and then $f(t)=sign(f(1))t$ for every $t>0$.[/quote]
You did the tricky part  :)
$y=1$ then $f(\frac{1}{x})+f(1)=\frac{f(x+1)}{f(x)f(1)}$ then $\lim_{x\to\ 0^+}f(\frac{1}{x})+f(1)=\frac{f(1)}{f(0^+)f(1)}=+-\infty=f(+\infty)$
As $f$ is continuous and the only zero is $f(0)=0$ we have $f(\mathbb{R^+})=\mathbb{R^+}$ or $f(\mathbb{R^+})=\mathbb{R^-}$ then $\forall x>0$ we have $f(x)>0$  or $\forall x>0$ we have $f(x)<0$ . ;)

[quote="AYMANE"]As $f$ is continuous and the only zero is $f(0)=0$ ... then $\forall x>0$ we have $f(x)>0$ or $\forall x>0$ we have $f(x)<0$.[/quote]
You're right! I could have avoided the whole sign thing!
And if you see that $f(x)$ is a solution iff $f(-x)$ is a solution, we could look for solutions with $f(x)>0$ $\forall x>0$s, and only at the end take the solutions $f(-x)$ too.

[quote="AYMANE"]$y=1$ then $f(\frac{1}{x})+f(1)=\frac{f(x+1)}{f(x)f(1)}$ then $\lim_{x\to\ 0^+}f(\frac{1}{x})+f(1)=\frac{f(1)}{f(0^+)f(1)}=+-\infty=f(+\infty)$.
...
As $f$ is continuous and the only zero is $f(0)=0$ we have $f(\mathbb{R^+})=\mathbb{R^+}$ or $f(\mathbb{R^+})=\mathbb{R^-}$...[/quote]
It's true, but... what's the point of proving this?

With this point you have this:

[quote="lordWings"]

Then, taking $x=2^n$ and $y=2^m$, we reach
$f(2^n+2^m)=\pm C^3(a^m+a^n)$.
.[/quote]

Nothing really new, but - here's the full proof. I'm proud of it.
[hide="Proof"]
1) The only zero of $f$ is $f(0)=0$.
Proof: If there was any other $z\neq 0$ with $f(z)=0$, $f(\frac{1}{z})$ wouldn't be defined.

2) $f$ is an odd function.
Proof: $x=-y=\frac{1}{t}$ $\to$ $f(t)+f(-t)=0$.
From now on, we'll only study $f(x)$ with $x>0$.

3) $f(x)$ is a solution of the functional equation iff $f(-x)$ is.
Proof: Direct from the functional equation.

4) $f(x)$ for $x>0$ is either always positive or always negative.
Proof: The only zero of $f$ is $f(0)=0$, and the function is continuous.
Thanks to 3) and 4), from now on, we'll only study functions with $f(x)>0$ for all $x>0$.

5) $f(2^n)=2^n f(1)^{2n+1}$ for every $n\in\mathbb{Z}$.
Proof by induction: It's obviously valid for $n=0$, $f(1)=f(1)$.
We'll now assume that $f(2^n)=2^n f(1)^{2n+1}$ and $f(2^{-n})=2^{-n} f(1)^{-2n+1}$.
$x=y=2^n$ $\to$ $f(2^{n+1})=2^{n+1} f(1)^{2n+3}$.
$x=y=2^{-n-1}$ $\to$ $f(2^{-n-1})=2^{-n-1}f(1)^{-2n-1}$.
Induction complete.
For simplicity, we'll call $C=f(1)$ and $a=2f(1)^2$ and then $f(2^n)=Ca^n$ for every $n\in\mathbb{Z}$.

6) $f(2^n)=2^n$ for every $n\in\mathbb{Z}$.
Proof: From 5), we only have to prove that $f(1)=1$.
$x=2^n$, $y=2^m$ $\to$ $f(2^n+2^m)=C^3(a^m+a^n)$.
Because of the continuity, if $m\to -\infty$, then $f(2^n+2^m)=C^3(a^m+a^n)\to f(2^n)=Ca^n$.
Therefore, $C^3=C$ $\to$ $C=f(1)=1>0$.

7) $f(\sum_{i=1}^n {2^{m_i}})=\sum_{i=1}^n {2^{m_i}}$, for every $m_i\in\mathbb{Z}$, $i\in{1,2,\dots,n}$.
Proof by induction:  It's obviously valid for $n=1$, $f(2^{m_1})=2^{m_1}$.
We'll now assume that $f(\sum_{i=1}^n {2^{m_i}})=\sum_{i=1}^n {2^{m_i}}$.
$x=y=\sum_{i=1}^n {2^{m_i}}$ $\to$ $f(\frac{1}{\sum_{i=1}^n {2^{m_i}}})=\frac{1}{\sum_{i=1}^n {2^{m_i}}}$.
$x=\sum_{i=1}^n {2^{m_i}}$, $y=2^{m_{n+1}}$ $\to$ $f(\sum_{i=1}^{n+1} {2^{m_i}})=\sum_{i=1}^{n+1} {2^{m_i}}$.
Induction complete.

8) The only two solutions to the functional equation are:
$f(x)=x$ for every $x\in\mathbb{R}$,
$f(x)=-x$ for every $x\in\mathbb{R}$.
Proof: Taking the base-2 expansion $x=\sum_{i\in\mathbb{Z}} {x_i 2^i}$ and using continuity in 7) we get $f(x)=x$ for every $x>0$. Thanks to 2), we can say the same for every $x\in\mathbb{R}$. And 3) reminds us $f(x)=-x$ is a solution too.[/hide]

[quote="abdurashidjon"]Lordwings: Can you explain clearly part (3) and (4). And why in part (4) you said it is continuous. I could not understand
Abdurashid[/quote]
(3) Direct proof:
Assume $f$ is a solution to the functional equation, so $f(\frac{1}{x})+f(\frac{1}{y}) = \frac{f(x+y)}{f(x)f(y)}$ for all $x$, $y$.
Substitute $g(x)=f(-x)$, $x'=-x$, $y'=-y$. You'll see that $g(\frac{1}{x'})+g(\frac{1}{y'}) = \frac{g(x'+y')}{g(x')g(y')}$ for all $x'$, $y'$.
$g$ is a solution to the functional equation too.

(4) Proof by contradiction:
Assume we have two values $x_1>0$ and $x_2>0$, with $f(x_1)>0$ and $f(x_2)<0$.
We've been told $f$ is continuous, so we can use Weierstrass' intermediate value theorem (or just intuition) to say that there must be a $z\in(x_1,x_2)$ with $f(z)=0$. BUT that would contradict part (1), which has already been proved.
Our assumption must be wrong. For every $x_1>0$ and $x_2>0$, $f(x_1)$ and $f(x_2)$ must be both bositive or both negative.

Luckily, the problem says "find all continuous functions", so we can assume the fact that $f$ is continuous whenever we want. Parts (4), (6) and (8) would have never been possible without continuity.


[quote="mecrazywong"]I know it's a pretty childish question in the eye of you these physics genius...but what's the definition of wavefront?[/quote]
I cannot find it in my dictionary. Perhaps you can describe it?

[quote="mecrazywong"]I know it's a pretty childish question in the eye of you these physics genius...but what's the definition of wavefront?[/quote]

[quote="A Dictionary of Physics, Oxford University Press, 1996"][b]wavefront[/b] A line or surface within a two- or three- dimensional medium through which waves are passing, being the locus of all adjacent points at which the disturbances are in phase. ...[/quote]


Find all real, complex solutions $x$ to:
$x^{7}+a^{7}=(x+a)^{7}$.


Let $ a,b$ are positive real numbers and $ b\le 1$, n is a constant such that $ n\ge 1$. Prove that:
\[ (a^2\plus{}n)(n\plus{}b^2)^2\ge (1\plus{}n)^3(2ab\plus{}b^2)\]

$ a\equal{}b\equal{}n\equal{}1$ gives a false inequality...?

:D I have made mistake  :P
Let $ a,b,c$ are positive real numbers, $ n\ge 1$. Prove that:
\[ (a^2\plus{}n)(b^2\plus{}n)(c^2\plus{}n)\ge (n\plus{}1)^3(ab\plus{}bc\plus{}ca)\]

$ a\equal{}b\equal{}c\equal{}n\equal{}1$ The inequality is false :P
This is problem
Let $ a,b,c$ be  positive real numbers, $ n\geq1$.Prove that:
$ (a^2\plus{}n)(b^2\plus{}n)(c^2\plus{}n)\geq (n\plus{}1)^2(ab\plus{}bc\plus{}ca)$

[quote="toanIneq"]$ a \equal{} b \equal{} c \equal{} n \equal{} 1$ The inequality is false :P
This is problem
Let $ a,b,c$ be  positive real numbers, $ n\geq1$.Prove that:
$ (a^2 \plus{} n)(b^2 \plus{} n)(c^2 \plus{} n)\geq (n \plus{} 1)^2(ab \plus{} bc \plus{} ca)$[/quote]
$ a \equal{} b \equal{} c \equal{} n \equal{} 1$ The inequality is false :P

Oh my god  :D above inequality hold on $ ab\plus{}bc\plus{}ca\ge 3$  :(

[quote="zaizai-hoang"]Oh my god  :D above inequality hold on $ ab \plus{} bc \plus{} ca\ge 3$  :([/quote]
Do you mean to this one
[quote="zaizai-hoang"]:D I have made mistake  :P
Let $ a,b,c$ are positive real numbers, $ n\ge 1$. Prove that:
\[ (a^2 \plus{} n)(b^2 \plus{} n)(c^2 \plus{} n)\ge (n \plus{} 1)^3(ab \plus{} bc \plus{} ca)
\]
[/quote]
It is still wrong.$ a\equal{}b\equal{}c\equal{}n\equal{}1$. :wink:

no, it is not what I want to say  :D
\[ (a^2 \plus{} n)(b^2 \plus{} n)(c^2 \plus{} n)\ge (n \plus{} 1)^3
\]
with $ ab \plus{} bc \plus{} ca\ge 3,n\ge \frac{1}{3}$

[b]Problem:[/b]Given $ a\plus{}b\plus{}c \geq 3$.Determine all real numbers $ k$ satisfying
$ (a^2\plus{}k)(b^2\plus{}k)(c^2\plus{}k) \geq (1\plus{}k)^3$
We can find $ k \equal{}1,109926818.........$ :P


I remember doing a MOEMS practice book a long time ago and seeing a lot of sequence problems. Here are a few practice ones for you guys:

Find the sum of the 2000th terms in the following sequences:

1, 4, 7, 10, 13, ...

2, 6, 10, 14, 18, ...

3, 5, 7, 9, 11, ...

4, 10, 16, 22, 28, ...

5, 9, 13, 17, ...

[quote="Drunken_Math"]I remember doing a MOEMS practice book a long time ago and seeing a lot of sequence problems. Here are a few practice ones for you guys:

Find the sum of the 2000th terms in the following sequences:

1, 4, 7, 11, 15, ...

2, 6, 10, 14, 18, ...

3, 5, 7, 9, 11, ...

4, 10, 16, 22, 28, ...

5, 9, 13, 17, ...[/quote]

2-5
[hide]
rule

2)4n-2
3)2n+1
4)6n-2
5)4n+1
^^^^^^^^^^^
2)7998
3)4001
4)11998
5)8001

[/hide]

The fifth term is supposed to be 16, right?

[quote="pianoforte"][quote="Drunken_Math"][hide="Hint for Question 1"]
*cough*$3n+1$*cough*
[/hide][/quote]

I don't think that works :?[/quote]

Oops, I made a typo. Try to solve it NOW.  :)

[quote="pascal_1623"]I didn't get what pianoforte got  :? :
(In all of these, $n$ is the number in the sequence, like the first, etc.)
[hide="a"]The equation is $1+3\cdot(n-1)$, so the 2000th term is: $\framebox{2998}$[/hide]
[hide="b"]The equation is $2+4\cdot(n-1)$, so the 2000th term is: $\framebox{3998}$[/hide]
[hide="c"]The equation is $3+2\cdot(n-1)$, so the 2000th term is: $\framebox{2001}$[/hide]
[hide="d"]The equation is $4+6\cdot(n-1)$, so the 2000th term is: $\framebox{5998}$[/hide]
[hide="e"]The equation is $5+4\cdot(n-1)$, so the 2000th term is: $\framebox{4001}$[/hide][/quote]

Something went wrong in your calculations. :?

BTW

[hide]$1+3\cdot(n-1)=$
$1+3\cdot n-3=$
$3n-2$

which is what I got.

same goes for all the other equations[/hide]

lol, these are all simple arithmetic sequences......  even moems students can comprehend simple pattenrs like $2n + 1 $ and $\frac{n+1}{2}$


the measures of the sides of a triangle $ABC$ are $BC=a;CA=b;AB=c$ and the measures of its altitudes issued respectively from $A;B;C$ are $x;y;z$ such that:$a^{n}+x^{n}=b^{n}+y^{n}=c^{n}+z^{n}$.Prove that $ABC$ is an equilateral triangle

Let $a^{n}+x^{n}=b^{n}+y^{n}=c^{n}+z^{n}=k$
Since $ax=by=cz=2S$ where $S$ is area of the triangle.
We have all three pairs are the solutions of the equation $x^{2}-(2S)^{n}x+k=0$. From here the pairs coincide. And since x<b,c triangle is equilateral.

Here is my solution:
$ax=by=cz=2S$
$\Leftrightarrow \frac{a^{n}}{b^{n}}=\frac{y^{n}}{x^{n}}=\frac{a^{n}-y^{n}}{b^{n}-x^{n}}$
We have $a^{n}+x^{n}=b^{n}+y^{n}\Rightarrow a^{n}-y^{n}=b^{n}-x^{n}$
$\Leftrightarrow \frac{a^{n}-y^{n}}{b^{n}-x^{n}}=1$
$\Rightarrow \frac{a^{n}}{b^{n}}=1$
$\Rightarrow a=b$
Similar we have $b=c$
So $ABC$ is an equilateral triangle


geometry
3D geometry
combinatorics open
combinatorics

A binary word is a finite sequence of 0 and 1's. A binary word is called $ n$-periodic if it is the repetition of the same word $ n$ times. For example, the word 011011011011 is a 4-periodic word. Every binary word of length 4 contains a 2-periodic word. (proof : If a word of length 4 does not contain 00 nor 11 then it must be either 0101 or 1010). But the word 010 and 101 does not contain a 2-periodic word.

What is the largest number $ M$ such that there exists a number $ n(M)$ such that every binary word of length $ n(M)$ contains an $ M$-periodic word?

Are you sure that $ n(M)$ does exist (I mean 'is finite')?
The Morse sequence contains no cube :
http://en.wikipedia.org/wiki/Thue-Morse_sequence

Pierre.

That makes the answer $ M\equal{}2$. I knew $ n(4)$ does not exist but didn't know Morse sequence does not contain cubes.


1. Describe the minimum structural, metabolic, and genetic equipment of a postprotobiont that you would consider to be a true primitive cell.

2. Darwin suggested looking at close relatives of a species of intrest to gain insight into what its ancestors may have been like. How does his suggestion anticipate the use of outgroups in modern cladistic analysis?

3. How can competition between scientists slow progress in a certain scientific field?

1.  A primitive cell would have a membrane, some form of genetic material, some form of metabolism, and would be capable of self-replication.  An RNA replicator inside a liposome would suffice

2.  An outgroup is a taxa that diverged earlier than two more closely associated taxa; its used to mark evolutionary divergences in cladistic analysis.  Here, Darwin has suggested the basic assumption underlying outgroup analysis; that more closely related species have a more recent common ancestor.


articles
geometry
geometric transformation
reflection
search

Hello everyone, I need your help.

I wanna have some articles of the appliance of math in our life.

Please tell me something about it.

I have no words about your reply......

I wanna have essential things...Cuz' I have to write an article of the title'MATH IN OUR LIFE'...

I need your help...... :mad:

DO NOT SAY SOMTHING LIKE 'SEARCH ON THE INTERNET AND YOU WILL GOT THE ANSWER~~~~~~'

 :mad:  :mad:  :mad:

I went on Google and searched for you, and this link will probably be helpful: [url=http://www.learner.org/interactives/dailymath/]click here.[/url]

Sometimes, math is not really the most important thing in contests such as MathCounts.  What is most important is problem solving and trying to think more creatively to solve different problems, both in math and life.

Alternatively, (this is probably what your teacher is thinking of) stop and reflect on all the places you use numbers.

Your height and weight.
When you measure stuff.
When you count stuff.
etc.
(This is very generic)

Going from here, you can probably think of many more things.

you should ask a teacher or just search appliance of math in our life on google. and don;t quote what he said.

OK...I have found something on a secret webpage...It's about the transfer of the rubbish...It's close to our life, I think.

thank you for your help...THANKS~~~~ :D  :D

Well math was used to build bridges and buildings, and it is used to predict the weather, and it is also used to make technology, and there are many other careers as well.

[color=orange]By the way, this was my 250th post which just made me a Reimann's Hypothesis!!![/color]


group theory
abstract algebra
superior algebra
superior algebra unsolved

It is easy to see that the group defined by generators $ a,b$ and relations $ a^{8} \equal{} b^{2}a^{4} \equal{} ab^{\minus{}1}ab \equal{} e$ has order at most $ 16$. I guess that it has order exactly $ 16$ but with awful solution.....too many computations.....do any one has a happy idea for this?

The order should be 32.

In fact $ G \equal{} \langle a,b| a^8 \equal{} b^4 \equal{} 1,a^b \equal{} a^{ \minus{} 1}\rangle$, we have $ \langle a\rangle \lhd G$. Obviously $ \langle a\rangle\cap\langle b\rangle \equal{} 1$, and hence $ G \equal{} \langle a\rangle : \langle b\rangle$.  So  $ |G| \equal{} 32$.

$ \langle a\rangle\cap\langle b\rangle \equal{} 1$? I don't see why is it so obvious , I think that $ a^{4} \equal{} b^{2} \neq 1$ implies that your argument is not true. But what about this?
Rename $ \langle a\rangle \equal{} H$ and $ \langle b\rangle \equal{} K$, $ b^{2} \in \langle a\rangle\cap\langle b\rangle$, and it is easy to see that $ |\langle a\rangle\cap\langle b\rangle|$ = 2 and $ G \equal{} HK$, then $ |G| \equal{} |H\parallel{}K|/|\langle a\rangle\cap\langle b\rangle| \equal{} (8*4)/2 \equal{} 16$.

Yes, alejnaser is right.

I have a mistake in the conditions of generators and relations. In fact the group G of mine  is not equivalent to the group of alejnaser's. This group should  $ G\equal{}\langle a,b|a^8\equal{}b^4\equal{}1,a^4\equal{}b^2,a^b\equal{}a^{\minus{}1}\rangle$.  sorry.

Perhaps we can reason this way: $ G$ ought to be a quotient of the semidirect product $ Z_8 \rtimes_\phi Z_4$, where $ \phi(b)(a) \equal{} a^{\minus{}1}$, as this is clearly the group we get if we omit $ a^4 b^2$ from the presentation.  Since $ a^4$ and $ b^2$ lie in the center of this group, $ a^4 b^2$ generates a normal subgroup of order $ 2$, so the quotient will have order $ 16$.


topology
advanced fields
advanced fields unsolved

Let K $ \subseteq$ V $ \subseteq$ M, where K is compact and V an open on a metric space M. Show that there is r > 0 where 	$ \cup$ B (x,r) $ \subseteq$ V, with x $ \subseteq$ K.

Lemma :If $ A$ is compact and $ B$ is closed and $ d(A,B) \equal{} 0$ then $ A\cap B \neq\emptyset$

the proof is very simple.

solution of the problem :since K is compact and $ v^c$ is closed and

 $ K\cap V^c\equal{}\emptyset$   $ \Rightarrow$    $ r\equal{}d(K,V^c)\neq0$

It is very easy to verify that $ r$ satisfies the required conditions.


logarithms
integration
calculus
calculus computations

Determine whether the series $ \sum_{n\equal{}2}^{\infty}a_n$ abcolutely convergent,conditionally convergent or divergent if

$ a_n\equal{}\frac{1}{ln(n!)}$

Can we say that it's convergent,because $ \frac{1}{ln(n!)}<\frac{1}{n}$ for n>5

Smaller than $ \frac {1}{n}$ is uninformative. The "direct" comparison test is one sided: smaller than a convergent series is convergent; bigger than a divergent series is divergent. But smaller than a divergent series is uninformative.

We need Stirling's approximation, or at least to go part of the way toward deriving Stirling, in order to answer this. Start with the notion that the logarithm of a product is the sum of the logarithms.

$ \ln(n!) \equal{} \sum_{k \equal{} 1}^{n}\ln k$ (or we could write it as $ \sum_{k \equal{} 2}^{n}\ln k.$)

$ \int_1^n\ln x\,dx < \sum_{k \equal{} 2}^n\ln k < \int_1^{n \plus{} 1}\ln x\,dx$

$ n\ln n \minus{} n \plus{} 1 < \ln(n!) < (n \plus{} 1)\ln(n \plus{} 1) \minus{} n.$

Note that $ n\ln n$ is much larger than $ n$ as $ n\to\infty,$ so $ n\ln n$ is the dominant term here.

$ \sum_n\frac {1}{\ln(n!)}$ is limit comparable to $ \sum_n\frac1{n\ln n}.$

The latter series can be shown to diverge by using either the integral test or the condensation test. Hence the original series diverges.

Is $ \sum_{n}\frac{1}{nln}\equal{}\int \frac{dx}{xlnx}$ ?

Not equal to, but comparable to in the sense that either both integral and series converge, or both integral and series diverge.


Suppose that $M$ is a set of $2003$ numbers such that, for any distinct $a,b,c\in M$, the number $a^{2}+bc$ is rational. Prove that there is a natural number $n$ such that $a\sqrt{n}$ is rational for all $a\in M$.

$a^{2}+bc+b^{2}+ac+c^{2}+ab \in \mathbb{Q}\rightarrow (a+b)^{2}+(b+c)^{2}+(a+c)^{2}\in \mathbb{Q}$. Considering that expression for each $3$-element subset of an arbitrary set $\{ a_{1},a_{2},a_{3},a_{4},a_{5}\}\subset M$ we get that each of all $\binom{5}{3}=10$ linearly indipendent rational combinations of $\binom{5}{2}=10$ numbers $(a_{i}+a_{j})^{2}$ are rational, and so (weekest part in the proof) for each $i,j$ $(a_{i}+a_{j})^{2}\in \mathbb{Q}$ which can be extended, being the set arbitrary, to any $(a+b)^{2}$ with $a,b\in M$, implying that $(a+b)$ is of the form $q\sqrt{n}$ for some $q\in \mathbb{Q}$ and some squarefree $n\in \mathbb N$. Now $a^{2}+cd \in \mathbb{Q}\wedge b^{2}+cd \in \mathbb{Q}\Rightarrow (a-b)(a+b) \in \mathbb{Q}$, so,  $\frac{1}{2}[(a-b)(a+b)+(a+b)^{2}]=a(a+b) \in \mathbb{Q}$ which means $a$ is also of the form $a=p\sqrt{n}$ with $p\in \mathbb{Q}$ and $n$ the previous one. Since we could have chosen any other pair $a,c$, then $c,d$ etc. we have that actually there is a unique squarefree $n$ such that $\forall a \in M\;\exists q\in \mathbb{Q}| a=q\sqrt{n}$. From this we get that:
$\forall a \in M \;\;a\sqrt{n}=q\sqrt{n}^{2}=qn \in \mathbb{Q}$.
Sleeps has victory over me now, sorry.


I need a LOT of practices on finding the range and domain of functions. I googled it but they only yield elementary exercises. Do you know of any good websites with worksheets that cover all the things I should know?

Or perhaps, could someone please start off with a range and domain question of their own?

Thanks!

[img]http://img401.imageshack.us/img401/5079/mathaf0.jpg[/img]

Find the domain where g[x] is continuous.

sorry I forgot to clarify- not the piecewise ones. Just the regular mixture of fractions/polynomials/absolute values.

[quote="i_like_pie"]What does $ \prec$ mean?[/quote]

How is it that you know that LaTeX for it but not what it means?  :o

$ \prec$ means "precedes".

I looked it up in the list of symbols, but \prec doesn't really give any hints.

Where can I find a list of latex symbols?

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols

but this is off topic.

Someone should solve the original problem.

[hide="Hint"]Precedes is the same as "less than". Take the open intervals for each possibility of the function.[/hide]


I've improved so much that I have absolutely no idea.


Four bugs are situated at each vertex of a unit square. Suddenly, each bug begins to chase its counterclockwise neighbor with speed 1 unit per minute. How much, in degrees, does each bug turn before they crash into each other?

Uh... i think it would be $\infty$ since they are all going the same speed...

They spiral inwards towards the center of the square (try it with your friends!).

The curves form logarithmic spirals with the polar equation of the form $r=e^{kq}$, where $r,q$ are the polar coordinates.  A logarithmic spiral has total angular distance traveled equal to infinity. Note that the mice have constant speed, and so they are approaching the center of the square.

Now, consider the problem of the distance traveled by each of the $n$ bugs initially situated at the vertices of a regular $n-\text{gon}$ after they intersect.

Masoud Zargar

Is there any easier way you can demonstrate this? There should be an elementary solution.

Well, after half a minute they would have been at least [b]some[/b] turning, i.e. nonzero; and they are still arranged in a square similar to the starting position. Therefore the angle turned is the same for the next 15 seconds. And for the next 7.5 seconds, ad nauseum. An infinite copies of the same angle adds up to infinity!
[quote="boxedexe"]
Now, consider the problem of the distance traveled by each of the $n$ bugs initially situated at the vertices of a regular $n-\text{gon}$ after they intersect.

Masoud Zargar[/quote]
Well, the side length shrinks from 1 unit to nothing, so 1 unit, right?
EDIT: no, wait, that only applies to a square because the bug you chase is going sideways......so taking into account that they are also running away from you at an angle of $\frac{90^\circ}{n}(n-4)$, so I guess it's $\frac{1}{1-\sin(\frac{90^\circ}{n}(n-4))}$

It takes only one minute for the bugs to crash :wink:

consider the component of their velocity pointing inward


algorithm
number theory
Euclidean algorithm

If the Highest Common Divisor of $ 6432$ and $ 132$ is diminished by $ 8$, it will equal:

$ \textbf{(A)}\ \minus{}6 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ \minus{}2 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\  4$

[hide="Click for solution"]
Using the Euclidean Algorithm we find that the "Highest Common Divisor" of $ 132$ and $ 6432$ is $ 12$.  Then $ 12\minus{}8\equal{}4$, so the answer is $ \boxed{\textbf{(E)}}$.
[/hide]


I am currently enrolled in AoPS classes, and I want to save my class transcripts for future reference. I tried to copy and paste them to Microsoft Word, but I can't copy anything that's in Latex text. Is there any way I can permanently save the transcripts--- without printing them all out?

In the my classes tab(above) click on the class name, then under the "transcripts" section is a list of links to past transcripts. Copy and paste those to word. Unless your version is old, then you should be able to see the images

If that doesn't work, you can virtually print them to PDF - (that is, your computer [i]thinks[/i] it's sending them to a printer, but in reality it's printing them to a PDF).

The best virtual printer is (in my experience) Go2PDF; a link is available on the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]AoPS Downloads page[/url].  However, it's only a 14-day trial.  Alternatives include [url=http://www.cutepdf.com/]CutePDF[/url], which is completely free and has no ads, but is somewhat slow and inaccurate, or [url=http://www.pdf995.com/]PDF995[/url], which opens an ad window each time you use it.

I think Go2PDF is freeware (we previously had linked the trial page by mistake).

I think that it prints a line at the bottom of every page encouraging you to buy it, though

I'd be happy if AoPS would create PDF files that we could d/l if it wasn't much of an inconvenience.

[quote="white_horse_king88"]Actually, you can save the files, as long as you copy and paste later. It shouldn't be too much of an inconvenience - I mean the transcripts are released right after class.[/quote]

Occasionally, the transcripts and challenge problems have been released up to an hour after the class.

I personally liked Mr. Crawford's idea of binders best. When the transcripts come out, copy into word or something, copy the images into the document, edit a bit and then print...

I usually include everything from a class in the binder...transcripts, problems, forum, midterm and final challenges, solutions, handouts, etc...sure it can be like 500 pages per class (especially the geometry ones :-) ) ...but I think it's worth it...Makes for great reference when you're working problems (which, sadly, is almost never for me :-( )...

If you have a lot of free time, you can $\text{\TeX}$ all the transcripts and stuff. That's what I'd do. You'll also learn a lot while $\text{\TeX}$ing, about both $\text{\LaTeX}$ and math.

I $\text{\TeX}$ the board problems up.  It is really good practice like Simon said.  The transcript are mostly text so I don't bother $\text{\TeX}$ ing them up as it would take forever.

when printing out the transcrips from the virtual pdf printer several of the lessons
have been truncated at the end of the long lines.

intermediate algebra   lessons 1 and 9,  all others concatenate correctly

introduction to counting   lessons 4,5,7,8,9   all others ok

what am i doing wrong?  thank you

I'm not sure if somebody mentioned it yet, but you can directly save the web page on your computer. Just go to my classes, then get to the transcript.

Afterwards, click the file option in the top left hand corner of the web browser. There's a save option. :D

I'm not sure if this works outside of internet explorer though, cause that's the only browser I've used before.

[quote="cowpi"]Is it possible to retrieve previous class tests (midterm, final) and solutions?  The archive has previous foruum discussions but not transcripts or tests / solutions.[/quote]

No.  We only archive the message board discussions.

This is kinda late, I guess... but PDF995 is also a good pdf convertor.


Let be $ ABC$ a triangle such that $ \widehat{C}\equal{}90^{o}$ , and $ m_a,m_b$ the medians . Show that :

$ \frac{r^2}{m_a^2\plus{}m_b^2}\le \frac{3\minus{}\sqrt{8}}{5}$

[quote="Dr Sonnhard Graubner"]hello, is here $ r$ the radius of the incircle of your triangle?
Sonnhard.[/quote]

Yes  :) I didn't said that because $ r$ is the usual notation for the radius of the incircle .

hello, ok, we have $ m_a^2\equal{}b^2\plus{}\frac{a^2}{4}$, $ m_b^2\equal{}\frac{b^2}{4}\plus{}a^2$ and $ r\equal{}\frac{a\plus{}b\minus{}c}{2}$, inserting this in your inequality, we get
$ \frac{(a\plus{}b\minus{}c)^2}{5c^2}\le \frac{3\minus{}2\sqrt{2}}{5}$ and this is equivalent with
$ a\plus{}b\minus{}c\le c\sqrt{3\minus{}2\sqrt{2}}$
squaring both sides we get
$ 2ab\le (a^2\plus{}b^2)\left((\sqrt{3\minus{}2\sqrt{2}}\plus{}1)^2\minus{}1\right)$
$ \frac{a}{b}\plus{}\frac{b}{a}\geq\frac{2}{\left(\sqrt{3\minus{}2\sqrt{2}}\plus{}1\right)^2\minus{}1}\equal{}2$ and this is true.
A nice inequality!
Sonnhard.


vector
algebra
polynomial
geometry
group theory
abstract algebra
number theory

The finite field $\mathbb{F}_{p^r}$ can be regarded as a vector space over $\mathbb{F}_p$, and we can consider the Frobenius morphism $\varphi: \mathbb{F}_{p^r} \to \mathbb{F}_{p^r}: x \mapsto x^p$ as a linear transformation of this vector space. What is the minimal polynomial of $\varphi$? :)

I see no point in explaining how i got it when someone else has a complete solution, but :

it will certainly be a divisor of $x^n-1$ and the degree will be at least the greatest common divisor of $p-1$ and $n$

pf $p-1$ divides n the degree is n and thus $x^n-1$ is the answer? :blush:

Do you mean $r$ where you wrote $n$? :)

And I don't really understand the part about the greatest common divisor of $p - 1$ and $n$, could you explain?

first of all yes, I'm just so used at working with $p^n$ elements (take a course in projective geometry over finite fields and you'll know why :lol: )

well there are couple of basic things I know about minimal polynomials :

they divide any polynomial 'annulating' the morphism
the roots of the min poly are eigenvalues, and conversely every eigenvalue appears at least once as root (the multiplicity is annoying)

so when in this situation is z as eigenvalue?

it is when z is in $F_p$ and is a solution of $x^p=x z$

so it has to be a p-1 power, there are exactly $\frac{p^r-1}{p-1}$ of them
however it also has to be in $F_p$ of course

now the big field has a multiplicative cyclic group and a cyclic subgroup of p-1 elements that are in the small field

we are looking for the intersection in a cyclic group of a subgroup of   $\frac{p^r-1}{p-1}$ elements and one of $p-1$ elements

the order is the gcd of both
however, one easily checks that that is the same ad $gcd(r,p-1)$

that is the number of eigenvalues and thus as lower bound for the degree

not much of a solution right?

i like finite fields a lot, but never really considered these 'classic' morphisms as linear morphism, more as field morphisms therefore... :|

Hmz, my thoughts on this question:

It's pretty obvious that the minimal polynomial is a divisor of $X^r - 1$, since we have $a^{p^r} - a = 0$ for all $a \in \mathbb{F}_{p^r}$. However, the degree of the polynomial cannot be less than $r$. Indeed, if the minimal polynomial would be of the form $c_0 + c_1X + \cdots + c_dX^d$ where $d < r$ then each $a \in \mathbb{F}_{p^r}$ would be a solution of the polynomial equation $c_0Y + c_1Y^p + \cdots + c_d Y^{p^d} = 0$, contradiction since this equation can have at most $p^d$ distinct roots.
So the minimal polynomial must be $X^r - 1$ indeed.

I also think this is kind of a weird question, certainly to be posed on an exam. And I found it difficult too :)

So just plain restriction on number on roots

I have been looking too far :mad:

Indeed, during the exam I showed that it divides $X^r - 1$ and I was looking for an argument to show that the degree must be at least $r$. I started messing around with generators of $\mathbb{F}_{p^r}^{\times}$ and other "too complicated" things, and I only realized I should look at the number of roots five minutes after the exam. It's kind of a tricky question. Too bad, but I solved all other questions completely, so I guess it's still okay :)

Hey, in fact: if $P\left(X\right)$ is a polynomial, the degree of which is strictly smaller than $r$, and if $\psi = P\left(\varphi\right)$ (as a transformation), then any generator $u$ of $\mathbb{F}_{p^r}^{\times}$ cannot satisfy the equation $\psi\left(u\right) = 0$. Am I right?