calculus
derivative
trigonometry
integration
function
ratio
calculus computations

Hey,
[b]1. The problem statement, all variables and given/known data.[/b]
Prove the following,

[b]Lemma 2.4[/b] - Suppose $ {f}$ has a second derivative everywhere, and that,
$ {{f}+{f^{\prime\prime}}} = {0}$
$ {f(0)} = {0}$
${ {f^{\prime\prime}(0)}} = {0}$

Then,
$ {f(0)} = {0}$
[b]2. Relevant equations.[/b]
Calculus.
[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I'm curious to see how this is proved, how would this be proved?
[/hide]
Thanks,

-APSStudent

Well, take $ f(x)\equal{}\sin x$ for a counterexample. If the second condition was supposed to be $ f'(0)\equal{}0$, then the statement becomes true and is just the standard uniqueness theorem for differential equations.

Hey,
[quote="fedja"]
Well, take $ f(x) \equal{} \sin x$ for a counterexample. If the second condition was supposed to be $ f'(0) \equal{} 0$, then the statement becomes true and is just the standard uniqueness theorem for differential equations.
[/quote]
Thanks...but I was looking for a more substantial proof.
Below is the proof from the lecture notes,
[hide="Click to See Proof."]
[img]http://d.imagehost.org/0948/Lemma_2_4.jpg[/img]
[/hide]
What I don't get is how they go from,
$ {{{{f}^{\prime}}{{f}^{\prime\prime}}}\plus{}{{f}{{f}^{\prime}}}} \equal{} {0}$
to
$ {{\left({{\left({f}^{\prime}\right)}^{2}}{\plus{}}{{f}^{2}}\right)}^{\prime}} \equal{} {{2}{\left({{{f}^{\prime}}{{f}^{\prime\prime}}}{\plus{}}{{f}{{f}^{\prime}}}\right)}} \equal{} {0}$
How do they do that?
Thanks,

-APSStudent

I see you used my lecture notes-nice. :D   If you want a differential equations solution, here it is:

Take the characteristic equation, which is $ r^{2}\plus{}1\equal{}0$, which has solutions
$ r\equal{}{\pm i}$.  Then the general solution is $ y(t)\equal{}e^{\lambda t}(A\cos{\mu t}\plus{}B\sin{\mu t})$, where in this case, $ \lambda\equal{}0$ and $ \mu\equal{}1$.  Then $ y(t)\equal{}A\cos{t}\plus{}B\sin{t}$.  From the initial conditions, $ 0\equal{}A$ and $ 0\equal{}B$, so $ y(t)\equal{}0$.

You could also use power series but it's tedious.

The proof that the characteristic equations method is valid is far more difficult than this one.  The one given is straightforward and proves exactly the desired result.

Hey,
[quote="JRav"]
I see you used my lecture notes-nice. :D
[/quote]
Yup, Thanks again :).
[quote="t0rajir0u"]
The first equality is the product rule; the second one is the statement of the previous line.  Which step is confusing?
[/quote]
Thanks for the clear up, I see it now.

In addition, the following line is interesting,
$ {{\left({{\left({f}^{\prime}\right)}^{2}} \plus{} {{f}^{2}}\right)}^{\prime}} \equal{} {{2}{\left({{{f}^{\prime}}{{f}^{\prime\prime}}} \plus{} {{f}{{f}^{\prime}}}\right)}}$
I see what they're doing by showing how, on the RHS we have,
$ {{\left({{f}^{\prime}}\right)}{\left({{f} \plus{} {{f}^{\prime\prime}}}\right)}} \equal{} {0}$

However, I feel proving the Lemma in this manner seems unmotivated and roundabout. Rather can the Lemma be proved in the following way?
$ {{f} \plus{} {f^{\prime\prime}}} \equal{} {0}$

$ {{\left({{f}^{\prime}}\right)}{\left({{f} \plus{} {{f}^{\prime\prime}}}\right)}} \equal{} {\left({0}\right)}{\left({{f}^{\prime}}\right)}$

$ {{{{f}^{\prime}}{{f}^{\prime\prime}}} \plus{} {{f}{{f}^{\prime}}}} \equal{} {0}$

$ {{\int_{}^{}}{{f}^{\prime}}{\left({{f} \plus{} {{f}^{\prime\prime}}}\right)}{d{?}}} \equal{} {{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}} \plus{} {C?}}$
If I could prove it this way, what would my differentiable variable (d?) be? In addition, would I need a constant on the RHS, if not why?

Thanks,

-APSStudent

[quote="APSStudent"]However, I feel proving the Lemma in this manner seems unmotivated and roundabout.[/quote]
Not at all; the motivation for this entire situation is classical mechanics.  A particle $ f$ moving so that $ f \plus{} f'' \equal{} 0$ has the property that at all times its acceleration is proportional to the negative of its position, which is precisely the situation of a mass against a spring satisfying [url=http://en.wikipedia.org/wiki/Hooke%27s_law]Hooke's law[/url].  But anyone who has played with a spring should suspect that such a mass undergoes [url=http://en.wikipedia.org/wiki/Simple_harmonic_motion]simple harmonic motion[/url].

Moreover, in such a situation $ f^2$ is proportional to the potential energy of the spring and $ f'^2$ is proportional to the kinetic energy of the mass, and so the fact that the sum of the two is a constant is a consequence of [i]conservation of energy.[/i]

[quote="APSStudent"]Rather can the Lemma be proved in the following way?[/quote]
What you did is not fundamentally different; it's just in the other direction and messier.  You do need the constant; it's the constant value of $ f'^2 \plus{} f^2$.  The variable you're integrating with respect to is whatever variable $ f$ is a function of.

The problem with doing things in this direction is that you don't know that $ f' f'' \plus{} f f'$ is integrable.  It's much easier to claim that $ f'^2 \plus{} f^2$ is differentiable because we are given that $ f$ has second derivatives everywhere.  So to really justify this argument you end up having to do things in the direction already given anyway.

[quote="t0rajir0u"]Not at all; the motivation for this entire situation is classical mechanics. A particle  moving so that  has the property that at all times its acceleration is proportional to the negative of its position,...[/quote]
Assuming that there is no dampening effect.  Then we have
$ m\frac{d^{2}x}{dt^{2}}=-kx-\gamma\frac{dx}{dt}\Rightarrow \frac{d^{2}x}{dt}+\frac{\gamma}{m}\frac{dx}{dt}+\frac{k}{m}x=0$
where $ \gamma$ is come constant.  I'll use the characteristic equation argument to solve it since it's faster:
$ r^{2}+\frac{\gamma}{m}r+\frac{k}{m}=0\Rightarrow r=\frac{-\frac{\gamma}{m}\pm\sqrt{\left(\frac{\gamma}{m}\right)^{2}-\frac{4k}{m}}}{2}$.
In this situation, we usually just assign new variables to these ratios of constants and solve for different discriminants.

Hey,
[quote="t0rajir0u"]
Not at all; the motivation for this entire situation is classical mechanics.
...[...]...
Moreover, in such a situation $ f^2$ is proportional to the potential energy of the spring and $ f'^2$ is proportional to the kinetic energy of the mass, and so the fact that the sum of the two is a constant is a consequence of [i]conservation of energy.[/i]
[/quote]
Thanks for pointing that out I should have suspected that reasoning, particularly since physics is one of my majors. :blush:
[quote="t0rajir0u"]
You do need the constant; it's the constant value of $ f'^2 \plus{} f^2$.  The variable you're integrating with respect to is whatever variable $ f$ is a function of.
[/quote]
Ok, so my integral than is,
$ {{\int_{}^{}}{{f}^{\prime}}{\left({{f} \plus{} {{f}^{\prime\prime}}}\right)}{dx}} \equal{} {{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}} \plus{} {C}}$
Where,
$ {C} \equal{} {0}?$
I'm not sure if the above is right, because how I would determine what the constant value of $ f'^2 \plus{} f^2$ is?
[quote="t0rajir0u"]
The problem with doing things in this direction is that you don't know that $ f' f'' \plus{} f f'$ is integrable.  It's much easier to claim that $ f'^2 \plus{} f^2$ is differentiable because we are given that $ f$ has second derivatives everywhere.  So to really justify this argument you end up having to do things in the direction already given anyway.
[/quote]
Hmm, isn't there a way I can test for integrability though? In other words, can I determine whether $ f' f'' \plus{} f f'$ is integrable?

Thanks,

-APSStudent

[quote="JRav"] I'll use the characteristic equation argument to solve it since it's faster:[/quote]
My point is that when $ \gamma \equal{} 0$ one has a physical system in which physical reasoning alone suggests that $ \frac{1}{2} kx^2 \plus{} \frac{1}{2} mv^2$ is a constant.  This in fact what comes out of these equations, but [i]we don't need to be aware of the characteristic equation method to understand this[/i]; we don't need to know the closed form of $ f$ to answer this question.

You seem to spend a lot of time applying methods you've learned in class to problems you see on the forum rather than thinking about the problems.  Mathematics is more than the rote memorization of technique.

[quote="APSStudent"]Ok, so my integral than is,
$ {{\int_{}^{}}{{f}^{\prime}}{\left({{f} \plus{} {{f}^{\prime\prime}}}\right)}{dx}} \equal{} {{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}} \plus{} {C}}$
Where,
$ {C} \equal{} {0}?$
I'm not sure if the above is right, because how I would determine what the constant value of $ f'^2 \plus{} f^2$ is?[/quote]
$ C$ is arbitrary as long as the integral is indefinite - you have to be careful to distinguish between definite and indefinite integrals here.  The definite integral is

$ \int_{0}^{x} 2f'(t) (f(t) \plus{} f''(t)) \, dt \equal{} f(x)^2 \plus{} f'(x)^2 \minus{} f(0)^2 \minus{} f'(0)^2 \equal{} 0 \Leftrightarrow$
$ f(x)^2 \plus{} f'(x)^2 \equal{} f(0)^2 \plus{} f'(0)^2 \equal{} 0^2 \plus{} 0^2 \equal{} 0$.

This argument works, but again, it's more convenient just to talk about derivatives; the two arguments are essentially the same by the Fundamental Theorem of Calculus.

[quote="APSStudent"]Hmm, isn't their a way I can test for integrability though? In other words, can I determine whether $ f' f'' \plus{} f f'$ is integrable?[/quote]
$ f' f'' \plus{} f f'$ is integrable because $ f^2 \plus{} f'^2$ is differentiable.  (There may be some subtleties associated with this statement, but for now it's more or less true.)

Are you starting to understand what I mean when I say it is more convenient to talk about derivatives?  :)

Hey,
[quote="t0rajir0u"]
$ {C}$ is arbitrary as long as the integral is indefinite - you have to be careful to distinguish between definite and indefinite integrals here.
[/quote]
Ok. The way you setup the integral makes sense since you're able to end up with the desired result. In this case, the integral was setup as a definite integral from: $ {0}$ to $ {x}$; with respect to $ {t}$. Where $ {t}$ is a parametric (or is it a dummy) variable, is that right?
[quote="t0rajir0u"]
The definite integral is
$ \int_{0}^{x} 2f'(t) (f(t) \plus{} f''(t)) \, dt \equal{} f(x)^2 \plus{} f'(x)^2 \minus{} f(0)^2 \minus{} f'(0)^2 \equal{} 0 \Leftrightarrow$
$ f(x)^2 \plus{} f'(x)^2 \equal{} f(0)^2 \plus{} f'(0)^2 \equal{} 0^2 \plus{} 0^2 \equal{} 0$.
[/quote]
Thanks, but on a slightly separate subject how would you prove the following (without just differentiating the RHS)?
$ {{\int_{}^{}}{{f}^{\prime}}{\left({{f} \plus{} {{f}^{\prime\prime}}}\right)}{dx}} \equal{} {{\frac {1}{2}}{\left({{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}}}\right)} \plus{} {C}}$
[quote="t0rajir0u"]
$ f' f'' \plus{} f f'$ is integrable because $ f^2 \plus{} f'^2$ is differentiable.  (There may be some subtleties associated with this statement, but for now it's more or less true.)
[/quote]
Really?...Is this the only way I can determine whether $ f' f'' \plus{} f f'$ is integrable?
[quote="t0rajir0u"]
Are you starting to understand what I mean when I say it is more convenient to talk about derivatives?  :)[/quote]
Yes, I'm definitly starting to appreciate the convenience of talking about derivatives as opposed to integrals.

Thanks,

-APSStudent
EDIT: I answered my own question on how to prove,
$ {{\int_{}^{}}{{f}^{\prime}}{\left({{f}\plus{}{{f}^{\prime\prime}}}\right)}{dx}} \equal{} {{\frac{1}{2}}{\left({{{f}^{2}}\plus{}{{{f}^{\prime}}^{2}}}\right)}\plus{}{C}}$
[hide="My Solution."]
Let,
$ {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} \equal{} {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}$

Find,
$ {I} \equal{} {{\int_{}^{}}{{f}^{\prime}}{\left({{f}\plus{}{{f}^{\prime\prime}}}\right)}{dx}}$

$ {I} \equal{} {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}\plus{}{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}}$

Let,
$ {\textcolor{white}{.}}
{\textcolor{white}{.}}
{u} \equal{} {f}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{v} \equal{} {{f}^{\prime}}$
$ {du} \equal{} {{{f}^{\prime}}{dx}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{dv} \equal{} {{{f}^{\prime\prime}}{dx}}$

$ {I} \equal{} {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}\plus{}{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}}$

$ {I} \equal{} {{{\int_{}^{}}{u}{du}}\plus{}{{\int_{}^{}}{v}{dv}}}$

$ {I} \equal{} {{\left({{\frac{{u}^{2}}{2}}\plus{}{{C}_{1}}}\right)}{\plus{}}{\left({{\frac{{v}^{2}}{2}}\plus{}{{C}_{2}}}\right)}}$

Let,
$ {{{C}_{1}}\plus{}{{C}_{2}}} \equal{} {C}$

$ {I} \equal{} {{{\frac{1}{2}}{\left[{{{\left({{f}^{2}}\right)}^{2}}\plus{}{{\left({{f}^{\prime}}\right)}^{2}}}\right]}}\plus{}{\left({C}\right)}}$

$ {I} \equal{} {{{\frac{1}{2}}{\left({{{f}^{2}}\plus{}{{{f}^{\prime}}^{2}}}\right)}}\plus{}{C}}$
$ \therefore$
$ {{\int_{}^{}}{{f}^{\prime}}{\left({{f}\plus{}{{f}^{\prime\prime}}}\right)}{dx}} \equal{} {{\frac{1}{2}}{\left({{{f}^{2}}\plus{}{{{f}^{\prime}}^{2}}}\right)}\plus{}{C}}$
[/hide]


analytic geometry
graphing lines
slope
complex numbers
AMC

The complex number $z$ satisfies $z+|z| = 2+8i$. What is $|z|^2$? Note: If $z = a+bi$, then $|z| = \sqrt{a^2+b^2}.$

$\text{(A)} \ 68 \qquad \text{(B)} \ 100 \qquad \text{(C)} \ 169 \qquad \text{(D)} \ 208 \qquad \text{(E)} \ 289$

Also think about this problem geometrically!

Geometrically?

[hide]

Since $|z|$ is real, b=8

Then, from the first formula, we have:

$a+\sqrt{a^2+64}=2$
$a^2+64=4-4a+a^2$
$4a=-60$
$a=-15$.

So, $|z|^2=a^2+b^2=225+64=289$

I love complex numbers!

[/hide]

I suppose, geometrically, we want the square of the identical side lengths of an isosceles triangle with vertices at $A(0, 0), B(2, 8)$, and $C(2 - |z|, 8)$ such that $AC = BC = |z|$.  But I think it's easier to do algebraically.

How about... [hide]take the midpoint of AB, which is (1, 4). Then take a line hitting that at slope -1/4, and you get it intersecting the x-axis at x = 17. That wasn't too hard.[/hide]


if one duck costs 3 dollars
2 ducks cost 8 dollars
3 ducks cost 8 dollars
4 ducks cost 11 dollars
5 ducks cost 14 dollars
6 ducks cost 17 dollars
then how much do 10 ducks cost?

[hide="BIGGEST HINT EVER"]
[hide="only click if you are stuck"]
[hide="are you sure you wanna click?"]

The equation is 2(number of ducks) + (number of duck -1)
[/hide][/hide][/hide]

[hide]29. I plugged in 10 for the equation: 2(#of ducks)+(#of ducks-1)[/hide]


how do u find thegeometric mean of the numbers
2,40,16

6,12,81

and

5/6,13/12,7/24

Geometric mean of numbers $x,y,z$ is $\sqrt[3]{xyz}$


How do you solve polynomials in the form, $ ax^4\plus{}bx^2\plus{}c$

Like solve for $ x$ if $ x^4\minus{}32x^2\plus{}16\equal{}0$?

hello, by substituting $ x^2=t$ you will get a quadratic equation like this:
$ at^2+bt+c=0$. this equation can solved by the following formula:
$ t{_1_{_2}}=-\frac{b}{2a}\pm\frac{1}{2a}\sqrt{b^2-4ac}$
After calculation the roots you have to solve
$ x^2=t{_1_{_2}}$ with respect to $ x$.
Sonnhard.


Find the number of square units in the area of the region bounded by the graphs of the equations: \\

(1) $ y\equal{}8$ \\

(2) $ y\equal{}|x|\plus{}2$

Translate downward by $ 2$, since it preserves area, to get $ y\equal{}|x|$ and $ y\equal{}6$.

The region satisfying these two equations is the triangle with vertices at $ (0,0),(\pm6,6)$, which clearly has area

$ \frac12\cdot12\cdot6\equal{}\boxed{36}$.


blogs
advanced fields
advanced fields theorems

what is Spec (k[[ X ]]) ?

k[[ X ]] is the ring of formal power series whose coefficients belong to field k.
spec (A) is the set of all prime ideals o a ring A.

Every ideal is generated by $ x^n$ for some $ n$. Conclude.

The important observation here is that a power series with a constant term is invertible; in other words, writing any power series in the form $ x^n p(x)$ where $ p$ has a nonzero constant term, $ \left< x^n p(x) \right> \equal{} \left< x^n \right>$.  Which of these ideals are prime?

Tks both of you!

And tks alkso for your blog t0rajir0u
it is absolutely awsome!!!

really love it!

(Ihaven't visited yours yet ZeaX.)


Prove by induction that the sum to 2^n where i=1 of 1/i <= 1 + n/2 for all n >=0. I am not sure how to do this, mainly because I cannot figure out how to replace the summation with an appropriate formula (if that is even possible.) I couldn't figure out how to represent a summation well in text, so if you did not understand what I wrote, this is how it is supposed to look: 2^n is on top of the capital sigma, i=1 is on the bottom of the capital sigma, and 1/i is to the right of the capital sigma. Thanks.

Written in $ \text{\LaTeX}$ for clarity:

Prove by induction that $ \sum_{i=1}^{2^n}\frac{1}{i}\leq1+\frac{n}{2}$.

You actually mean $ \sum_{i=1}^{2^n} \frac{1}{i} \ge 1 + \frac{n}{2}$.

[hide="Demonstration"] For $ n = 1$ we have $ 1 + \frac{1}{2} = 1 + \frac{1}{2}$, so our base case is valid.  The inductive step:  suppose we have shown that

$ \sum_{i=1}^{2^k} \frac{1}{i} \ge 1 + \frac{n}{2}$

then

$ \begin{eqnarray*}
\sum_{i=1}^{2^{k+1}} \frac{1}{i} &=& \sum_{i=1}^{2^k} \frac{1}{i} + \sum_{i=2^k + 1}^{2^{k+1}} \frac{1}{i} \\
 &\ge& 1 + \frac{n}{2} + \sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} \\
 &\ge& 1 + \frac{n}{2} + \frac{1}{2} \\
 &\ge& 1 + \frac{n+1}{2}$

so our result is true by induction. [/hide]

First of all, listen to Bruno... Use LATEX.

Induction is merely finding the first few terms, then generalizing to some $ n$, and finally proving that the expression holds true for $ n \plus{} 1$.

I believe that what you wanted to say is $ \displaystyle\sum_{i\equal{}1}^{2^{n}}\frac {1} {i}\le 1 \plus{} \frac {n} {2}$ for all $ n \ge 0$.

EDIT: Sorry, didn't see that some people had already posted this stuff. And Torajirou, he actually did mean less than or equal to.
[hide=" Solution"]Let us proceed by first proving that our base case is true, simply when $ n \equal{} 0$.

$ \displaystyle\sum_{i \equal{} 1}^{1}\frac {1} {i} \equal{} 1 \le 1$.
Thus the base case is true.
You may also let $ n \equal{} 1$ to further support your solution.  In fact, it will probably be better if you do.

Let us assume that this expression is true up to any $ n$. Then, we wish to prove that it holds for $ (n \plus{} 1)$.

$ \displaystyle\sum_{i \equal{} 1}^{2^{n \plus{} 1}}\frac {1} {i} \equal{} \displaystyle\sum_{i \equal{} 1}^{2^{n}}\frac {1} {i} \plus{} \frac {1} {2^{n \plus{} 1}}$, which we wish to prove is $ \le 1 \plus{} \frac {n \plus{} 1} {2}$

This is relatively simple, given that we know $ \displaystyle\sum_{i \equal{} 1}^{2^{n}}\frac {1} {i} \le 1 \plus{} \frac {n} {2}$.  The only thing we need to prove now is that $ \frac {1} {2^{n \plus{} 1}} \le \frac {1} {2}$.
This inequality holds as long as $ n \ge 0$, and that was given.
Thus, our induction is complete.[/hide]

[quote="mathcrazed"]$ \displaystyle\sum_{i \equal{} 1}^{2^{n \plus{} 1}}\frac {1} {i} \equal{} \displaystyle\sum_{i \equal{} 1}^{2^{n}}\frac {1} {i} \plus{} \frac {1} {2^{n \plus{} 1}}$[/quote]

This is not true.  What t0r has written is correct.

Oh right. Whoops, I computed the sum incorrectly, and I thought that he had the same thing, but merely made a mistake. Sorry for that ~_-.

Thank you for your help. I will be sure to use LaTeX when posting in the future. And yes, I did mean greater than or equal to like t0r said, I made a mistake in my original post, I apologize.


calculus
integration
trigonometry
real analysis
real analysis unsolved

f(y) = Integral from 0 to Infinity sinx cos (xy/x) dx if y>=0
Prove f(y) = PI /2 IF 0<= y <1
          f(y) = 0 IF  y > 1
Calculate Integral from 0 to a f(y) dy

Is there anything wrong in the definition of $f(y)$?

why did you write cos (xy/x), not cos y?

Of course that should be $\int_0^\infty \frac{\sin x}{x}\cos xy\, dx$. That is part of the Fourier transform of "sinc", a well-known problem.

that correct integral sinx /x cos xy
Jmerry , can u show me how to solve it


Remember that when adding fractions, look for a least common denominator.

$ \sqrt{\frac{1}{9}\plus{}\frac{16}{25}}\equal{}\sqrt{\frac{1(25)\plus{}16(9)}{9(25)}}\equal{}\sqrt{\frac{25\plus{}144}{225}}\equal{}\sqrt{\frac{169}{225}}\equal{}\frac{\sqrt{169}}{\sqrt{225}}\equal{}\frac{\sqrt{169}}{15}\equal{}\boxed{\frac{13}{15}}$


If $ z_1$ is diffrent from $ z_2$, $ z_1 \in C$(complex numbers), $ \frac{z_1\plus{}z_2}{z_1\minus{}z_2}\equal{}u\plus{}vi$, $ u,v$ are real numbers. Calculate $ u$ and $ v$ related to $ z_1,z_2$

[quote="balan razvan"]If $ z_1$ is diffrent from $ z_2$, $ z_1 \in C$(complex numbers), $ \frac {z_1 \plus{} z_2}{z_1 \minus{} z_2} \equal{} u \plus{} vi$, $ u,v$ are real numbers. Calculate $ u$ and $ v$ related to $ z_1,z_2$[/quote]

$ u\equal{}\frac{\frac {z_1 \plus{} z_2}{z_1 \minus{} z_2}\plus{}\overline{\frac {z_1 \plus{} z_2}{z_1 \minus{} z_2}}}{2}$

$ v\equal{}\frac{\frac {z_1 \plus{} z_2}{z_1 \minus{} z_2}\minus{}\overline{\frac {z_1 \plus{} z_2}{z_1 \minus{} z_2}}}{2}$

Maybe it simplifies more, but not so much...


geometry
incenter
perpendicular bisector

[quote="nayel"]Rays $AB$ and $AC$ touch the circle $\omega$ at $B$ and $C$ respectively. Prove that the excenter of $\triangle ABC$ opposing $A$ lies on $\omega$.[/quote]

[b] Really easy![/b]

 If the bisector of angle $ABC$ meets the circle at $I$ , then an easy angle chasing shows that $I$ is the incenter of triangle $ABC$. If line $AI$ meets the circle again at $K$ , then $IK$ is obviously a diameter  in this circle( $AK$ is the perpendicular bisector of chord $BC$).But then $BK \perp BI$ and since $BI$ is bisector of angle $ABC$ , $BK$ will be the bisector of angle $supl.ABC$.
[u]
 babis[/u]


calculus
derivative
trigonometry
calculus computations

I missed 2 of the 3 recent classes where we were taught how to find derivatives of trig equations. Was wondering if anyone can help with these questions:

Find derivative:

y = sin^3(cos(^2)x)

f(x) = cos^4(sinx^3)

y = xcos(tanx)

y = xsin2x

Two sides of a triangle have lengths 15 m and 20 m. The angle between them is increasing at a rate of "pi"/90 rad/s. At what rate is the length of the third side changing when the angle between the other twos sides is "pi"/3?

I was looking for rules in our text book, but apparently it isn't covered in the one I have. Thanks for any help.

BBob[/list][/code]

here is a general rule
if $y=f(g(x)); y'=g'(x).f'(g(x))$
thuse these are the answers to the first four:
1
$-3\sin{(2x)}.\sin^2{\cos^2{x}}$
2
$4\cos^3{\sin{(x^3)}}.3x^2\cos{(x^3)}$
3
$\cos{(\tan{x})}+x(-\sin{(\tan{x})}.\sec^2{x})$
4
$\sin{(2x)}+2x\cos{(2x)}$


maybe what i say is stupid but p is prime..then in the fatctorial p-1 there is no number such as their product is equal to p...then p=2..is it right? :?

Correct me wrong but this is just application of Wilson's Theorem.

Wilson's Theorem states that if $p$ is a prime, then $(p-1)! \equiv -1 \mod p$. In this problem, it is given that $p$ is a prime number. So, $(p-1)! \equiv -1 \mod p$ and therefore, $(p-1)! + 1 \equiv -1+1 \equiv 0 \mod p$. So, $(p-1)!+1$ is divisible by $p$?

[quote="Nea"]$p$ is a prime number. Which value of $p$, makes $(p-1)!+1$ a power of $p$.[/quote]I agree with Silverfalcon's answer, and if that's the case, it is too easy. Perhaps we need to find those $p$ whose power [b]is greater than 1[/b].  :D

PS Edit: I'm sorry, not all $p$'s yield a power. For instance, let $p=7$. Then $(p-1)!+1=6!+1=721$ but there does not exist any integer $\alpha$ such that $721=7^\alpha$. In other words, I think she meant find those primes $p$ for which there exists a positive integer $\alpha$ such that $(p-1)!+1=p^\alpha$ is true. A prime number that works is $p=5$.

[quote="Silverfalcon"][quote="chess64"]$p|(p-1)!+1$ does not imply that $p^k=(p-1)!+1$.[/quote]

Well.. Read what 10000th User said. I was talking about $k = 1$ ONLY.[/quote]

Well.. $p|(p-1)!+1$ does not imply that $p=(p-1)!+1$ either.

This is what I have so far:

Let $p^a=(p-1)!+1$. Then $p^a-1=(p-1)!$. If $a=1$, $p-1=(p-1)!$ or $1=(p-2)!$ since 1 is not prime. This happens iff $0!=1!=1$ and so $p=2,3$ are two possible primes. $p=5$ is found by trial-and-error. I'm not sure if there are any more higher primes but I doubt it.

[hide="there are no higher primes"]Suppose $p$ is a prime greater than 5 and $(p-1)!+1=p^n$. Subtracting 1 and removing a common factor,
\[ (p-2)!=1+p+p^2+\cdots +p^{n-1} \]

Since $p>5$, $\frac{p-1}{2}>2$ but also $\frac{p-1}{2}<p-2$. So $2$ and $\frac{p-1}{2}$ are distinct positive integers less than $p-2\Rightarrow p-1|(p-2)!$.

The right side must also be divisible by $p-1$. Reducing the right side $\pmod{p-1}$ gives $\underbrace{1+1+1+\cdots +1}_n$. So $n$ must be divisible by $p-1\Rightarrow n\geq p-1$.

But this means $p^n\geq p^{p-1}>p^{p-2}+1>2\cdot 3\cdot 4\cdot \cdots \cdot (p-1)+1=(p-1)!+1=p^n$, a contradiction. So there are no other solutions.[/hide]


LaTeX
quadratics
inequalities
algebra
quadratic formula

Share your designs/fonts of LaTeX here!
I'll start:

My Design: $ \ddagger$

My font: $ \mathcal{MATHCAL}$

$ A(r^{t})\text{ }\phi{f}\text{ }P(r_{0})b\cdot le^{m}\text{ }S_{0}l\cdot vi^{ng}$

$ \left(\begin{array}{lcr}\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\\\text{AoPS} & \text{AoPS} &\text{AoPS}\end{array}\right)$

$ \binom{I}{u}\pi\binom{k}{ \minus{} s}$

I made this about a year ago to pun at those who can't seem to be able to carry their imagination beyond $ \binom{I}{u} Pikachu$

$ \text{\LaTeX}\ is\ A(r^{t})\text{ }\phi{f}\text{ }P(r_{0})b\cdot le^{m}\text{ }S_{0}l\cdot vi^{ng}$

$ \sum\prod\alpha\beta\gamma\pm\mp\ge\le < > \ne$

$ All^{shall^{bow^{to^{me^{or^{else^{die^{a^{very^{painful^{death^{mwahahaha^{bwahahaha^{haha^{*evil laugh*}}}}}}}}}}}}}}}$

[color=blue][u]If $ 0 < x < 1$, what is the maximum value of $ x \minus{} x^3$? Figure it out WITHOUT using calculus.[/u][/color]

[color=blue][u]I lost THE GAME[hide="."][/u][/color]
We will set $ b$ as the value of $ x$ that will make $ x \minus{} x^3$ as large as possible. Then we set $ a$ as our "guess" value. We have only 3 cases:
$ a < b$
$ a \equal{} b$
$ a > b$
If $ a < b$, then $ b \minus{} a > b^3 \minus{} a^3$. We can factor $ b^3 \minus{} a^3$ as $ (b \minus{} a)(b^2 \plus{} ab \plus{} a^2)$. By dividing each side by $ b \minus{} a$, we get $ 1 > b^2 \plus{} ab \plus{} a^2$. Now we have the quadratic inequality $ a^2 \plus{} (b)a \plus{} (b^2 \minus{} 1) < 0$. Using the quadratic formula, we get $ a < \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. If $ a \equal{} b$, then $ b \minus{} a \equal{} b^3 \minus{} a^3$. However, we cannot divide by $ b \minus{} a$ as we did in the first step because doing so would result in $ \frac {0}{0}$. We skip this case for now. If $ a > b$, then $ a \minus{} b < a^3 \minus{} b^3$. Dividing by $ a \minus{} b$, we get $ 1 < a^2 \plus{} ab \plus{} b^2$, and that leads to the quadratic inequality $ a^2 \plus{} (b)a \plus{} (b^2 \minus{} 1) > 0$. The quadratic formula yields $ a > \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. The first and third cases cover every value between $ 0$ and $ 1$ except $ b$. The solution to the first and third cases are less and more than $ \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$, respectively. They cover every solution between $ 0$ and $ 1$ except for $ a \equal{} \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. Therefore, the solution to $ a \equal{} b$ has to be $ a \equal{} \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. By setting these expressions equal to $ a$, we are setting them equal to each other. Now we have the equation $ b \equal{} \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. Multiplying by $ 2$ and adding $ b$ to each side, we have $ 3b \equal{} \sqrt {4 \minus{} 3b^2}$. Squaring each side, we get $ 9b^2 \equal{} 4 \minus{} 3b^2$. Adding $ 3b^2$ to each side and dividing by $ 12$, we get $ b^2 \equal{} \frac {1}{3}$, so $ b \equal{} \sqrt {\frac {1}{3}}$. Plugging $ b$ in for $ x$, we get $ \frac {\sqrt {3}}{3} \minus{} \frac {\sqrt {3}}{9}$, which simplifies to $ \frac {2\sqrt {3}}{9}$.

If $ 0 < x < 1$, then the maximum value of $ x \minus{} x^3$ is $ \frac {2\sqrt {3}}{9}$.
[/hide]

See my signature (the one that you see and has been there, unchanged (except for about 7 hours), since I created a signature)

Nice solution, mathelete

just kidding.

$ \text{\LaTeX}$ is $ awesome^{awesomest}^{\infty}$

\[ \color{green}\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathfrak{Glory to God in the Highest! }}}}}}}}}}}}\]

[color=red]$ \ Art^{Of^{Problem^{Solving}}}$[/color]

This took a REALLY long time to figure out. PLEASE NOTE: THE SQUIGGLIES DO NOT APPEAR IN THE CLASSROOM! TESTED ABOUT 400 TIMES! (Yes, this is ALL my creation!)

$ \color{green}\frac {\color{red}Hello\sim \color{cyan}all}{\color{cyan}fellow\sim \color{red}earthlings}!\cdot\frac {\color{blue}Hello\sim \color{yellow}all}{\color{yellow}alien\sim \color{blue}entities}!\cdot\frac {\color{red}How\sim \color{cyan}are}{\color{cyan}you\sim \color{red}today}?$

$ \color{green}\frac {\color{red}Thanks\sim \color{cyan}for}{\color{cyan}the\sim \color{red}lesson}!\cdot\frac {\color{blue}It\sim \color{yellow}was}{\color{yellow}very\sim \color{blue}neat}!\cdot\frac {\color{red}I\sim \color{cyan}learned}{\color{cyan}so\sim \color{red}much}!$

$ \color{green}\frac {\color{red}Goodbye\sim \color{cyan}all}{\color{cyan}fellow\sim \color{red}earthlings}!\cdot\frac {\color{blue}Goodbye\sim \color{yellow}all}{\color{yellow}alien\sim \color{blue}entities}!\cdot\frac {\color{red}See\sim \color{cyan}you}{\color{cyan}next\sim \color{red}week}!$

Use these in your online courses and help it spread!! :D And feel free to edit the words!

How do I do spaces in $ \text{\LaTeX}$ ?

$ \spadesuit \clubsuit \color{red} \heartsuit \diamondsuit$

$ \[ \color{cyan}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathfrak{\pi rules}}}}}}}}}}}$


modular arithmetic
induction
function
algebra
polynomial

The sequence $a_{n}$ is defined by $a_{0}= 2$, $a_{n+1}= 3a_{n}+1$. Find for which $n$, $7 \mid a_{n}$.

[hide]$7|a_{n}$ for $n=6x+1$ which $x$ is an integer $\geq 0$.  Notice that the sequence is cyclic in mod7:
$a_{0}\equiv 2\pmod 7$
$a_{1}\equiv 0\pmod 7$
$a_{2}\equiv 1\pmod 7$
$a_{3}\equiv 4\pmod 7$
$a_{4}\equiv 6\pmod 7$
$a_{5}\equiv 5\pmod 7$
$a_{6}\equiv 2\pmod 7$

So $7|a_{n}$ for n=1, 7, 13, 19,...

This can be proved fairly easily with induction, I think.
[/hide]

Yeah, that's right.  :)
As a practice of generating functions, try to find $a_{n}$, and then use congruencies.  :D

[quote="pkerichang"]...what does that have to do with generating function??? Do you mean finding close form for recursive functions??? (or characteristic polynomial)??[/quote]

One method for solving recursions is via generating function.  Let $f(x) = \sum_{i=0}^\infty a_{i}x^{i}$.  Then
$3xf(x)+\frac1{1-x}= \sum_{i=0}^\infty 3a_{i}x^{i+1}+1+\sum_{i=0}^\infty x^{i+1}= 1+\sum_{i=0}^\infty (3a_{i}+1)x^{i+1}= 1+\sum_{i = 1}^\infty a_{i}x^{i}= f(x)-1$

And so $f(x) = \frac{2-x}{(1-x)(1-3x)}$.  Then you can use that to find the explicit form for $a_{n}$.


I don't know where this is supposed to go, so if one of the moderators can help me, that would be appreciated. :blush:

A farmer has died, and his five sons are about to see his will. When they see it, they are overjoyed to learn that their father has entrusted all his land(a sizeable amount) to them. However, the farmer asks that the land split has to occur like this:

1. The five sons must divide the land into five pieces. However, the parcels of land can be any size and any shape.
2. Each son had to be a neighbor to all the other sons, and to be neighbors, the sides of the parcels must share one common edge, not just a point.
3. Each brother's land had to be in one piece.

That's the end of the puzzle, but I should add that the farmer had an infinite amount of land. So make the land parcels as big or as little as you want, it doesn't matter.
Have fun! :D

By the [url=http://en.wikipedia.org/wiki/Four_color_theorem]Four Color Theorem[/url], this is impossible, so the farmer's sons should not be so overjoyed.

Maybe the Farmer owned a Torus shaped planet. :idea:

Or they construct a bridge. Or share land.

Well, a bridge is the equivalent to a torus surface.

The farmer lives all alone on a torus-shaped planet. His sons are easily able to divide up the land....if they ever move to Earth, they will have a sizeable problem.....


Is mathpath worth it? How much would it help me in mathcounts? What exactly do you do/learn there?

Are you going? :coolspeak:

It's definitely worth it. Although most of what you learn there is at a bit higher of a level than Mathcounts.

I heard John Conway's going to be there.

See http://www.mathpath.org

MathPath is great. It's possible to do very well in MATHCOUNTS without going to MathPath at all, but you won't learn all the other stuff that there is to learn at MathPath, nor will you make all the cool friends you can make at MathPath. I highly recommend MathPath and am glad my son went last year.

Be prepared to work hard (if you want) though, even though it's a middle school camp, the higher level classes have some pretty serious olympiad level problems.

I'm still working on the majority of the problems from last year... plus that last qualifying quiz problem. :blush:


function
algebra
polynomial
analytic geometry
graphing lines
slope
real analysis

Ascertain the continue function $ f: \mathcal R\rightarrow\mathcal R$ so that $ (f\circ f)(x)\minus{}3f(x)\plus{}2x\equal{}0$ , $ (\forall )\ x\in \mathcal R$ .

[b]A correction[/b] : the author of this problem is [b]M. Tetiva[/b] and appeared in G.M.B., 5-6, C:1401

Hi Virgil,

Looks like $ f(x)\equal{}2x$ by inspection.  Surely I must be missing something.

Yeap. If I don't mistake, you are missing infinitely many more solutions and a proof that there are no others ;)

Among those missed are $ f(x) \equal{} x$ and $ f(x) \equal{} 2x \plus{} c$ for any constant $ c$.  It's easy to see that these are the only polynomial solutions (constant functions are right out and polynomials of degree $ n > 1$ give us on the left a nonzero polynomial of degree $ n^2$), but this isn't much more interesting than what shawtoos wrote.

Here is one more solution: $ f(x)\equal{}x$ for $ x\le 1$ and $ f(x)\equal{}2x\minus{}1$ for $ x\ge 1$ ;)

[quote="Virgil Nicula"] [color=darkred][b]Ascertain the continue function $ f: \mathcal R\rightarrow\mathcal R$

so that $ f(f(x)) - 3f(x) + 2x = 0$ , $ (\forall )\ x\in \mathcal R\ \ (*)$ .[/b][/color] [/quote]
 [color=darkblue][b][u]Remark.[/u] Here are a solution : $ f_{m,n} = \{\begin{array}{ccc} 2x - m & \mathrm {if} & x < m \\
 \\
x & \mathrm {if} & m\le x\le n \\
 \\
2x - n & \mathrm {if} & n < x\end{array}$ ,

where $ \{m,n\}\subset \overline {\mathcal R} = [ - \infty , \infty ]$ and $ m\le n$ . Therefore, in detail

$ \|\begin{array}{ccc} - \infty = m < n = \infty & \implies & f_{ - \infty ,\infty}(x) = x\ ,\ (\forall )x\in\mathcal R \\
 \\
- \infty < m = n < \infty & \implies & f_{m,m}(x) = 2x - m\ ,\ (\forall ) x\in\mathcal R \\
 \\
- \infty < m < n = \infty & \implies & f_{m,\infty }(x) = \{\begin{array}{ccc} 2x - m & , & x < m \\
 \\
x & , & x\ge m\end{array} \\
 \\
- \infty = m < n < \infty & \implies & f_{ - \infty ,n}(x) = \{\begin{array}{ccc} x & , & x\le n \\
 \\
2x - n & , & x > n\end{array} \\
 \\
- \infty < m < n < \infty & \implies & f_{m,n}(x) = \{\begin{array}{ccc} 2x - m & , & x < m \\
 \\
x & , & m\le x\le n \\
 \\
2x - n & , & n < x\end{array}\end{array}$ .[/b] [/color]

[size=117][color=darkred][b]Are these the $ \boxed {\mathrm {all}}$ continue functions which verify the relation $ (*)$ ?![/b][/color][/size]

Call $ g(x) \equal{} f(x) \minus{} 2x$.  Then, $ g(f(x)) \equal{} f(f(x)) \minus{} 2f(x) \equal{} f(x) \minus{} 2x \equal{} g(x)$, or $ g(2x \plus{} g(x)) \equal{} g(x)$.  The trivial solutions of this last equation are $ g(x) \equal{} k$ for any constant $ k$, yielding $ f(x) \equal{} 2x \plus{} k$, and $ 2x \plus{} g(x) \equal{} x$, yielding $ f(x) \equal{} x$.  Note that, since $ f$ is continuous, so is $ g$, or $ g$ may be a continuous, piecewise linear function defined over intervals such that $ g(x) \equal{} k$ in some intervals ($ k$ may then be different in distinct intervals) and by $ g(x) \equal{} \minus{} x$ in some other intervals.  Two disconnected intervals where $ g(x) \equal{} \minus{} x$ mean that the graph of the function would have two pieces of the same straight line with a slope $ \minus{} 1$.  These two pieces of the graph cannot be connected by horizontal lines.  Or there is at most one interval where $ f(x) \equal{} x$, yielding the solution proposed by Virgil Nicula.  Now, in order to show that there are no other solutions, it suffices to show "only" that $ g(2x \plus{} g(x)) \equal{} g(x)$ has no other continuous solutions appart from the trivial ones, or that any other solution would not satisfy that $ f$ is injective (which is trivially shown since, if $ f(x) \equal{} f(y)$ for some $ x\neq y$, then $ 2x \equal{} 3f(x) \minus{} f(f(x)) \equal{} 3f(y) \minus{} f(f(y)) \equal{} 2y$, absurd...

One other remark about $ f(x)$: since it is injective and continuous, then it is either strictly increasing, or strictly decreasing.  Maybe this helps to show that the given solutions for $ g$ are the only possible ones, maybe not, I haven't worked it out completely yet...

Why not just differentiate the equation and solve the resulting DE for the proof:

$ \frac {d}{dx}\left\{f\circ f \minus{} 3f \plus{} 2x \equal{} 0\right\}$

$ \left(f'\right)^2 \minus{} 3f' \plus{} 2 \equal{} 0$

$ p \equal{} f'$

$ p \equal{} \frac {3\pm 1}{2}$

$ f' \equal{} 2$

$ f' \equal{} 1$

Thus all solutions are of the form:

$ f(x) \equal{} 2x \plus{} c$ and $ f(x) \equal{} x \plus{} c$

[edit] never mind, I just noticed only continuity was assumed

[quote="shawtoos"]Why not just differentiate the equation and solve the resulting DE for the proof: $ \frac {d}{dx}\left\{f\circ f \minus{} 3f \plus{} 2x \equal{} 0\right\}$ $ \left(f'\right)^2 \minus{} 3f' \plus{} 2 \equal{} 0$[/quote]$ (f\circ f)(x)\equal{}f(f(x))$ and $ [f(f(x))]'\equal{}f'(f(x))\cdot f'(x)$ ...

[quote="Virgil Nicula"][quote="shawtoos"]Why not just differentiate the equation and solve the resulting DE for the proof: $ \frac {d}{dx}\left\{f\circ f \minus{} 3f \plus{} 2x \equal{} 0\right\}$ $ \left(f'\right)^2 \minus{} 3f' \plus{} 2 \equal{} 0$[/quote]$ (f\circ f)(x) \equal{} f(f(x))$ and $ [f(f(x))]' \equal{} f'(f(x))\cdot f'(x)$ ...[/quote]

Furthermore, nothing guarantees that $ f$ will be differentiable.  In fact, if you look at Virgil's solution (which you can readily see that is correct), it is in fact NOT differentiable...

[quote="daniel73"][quote="Virgil Nicula"][quote="shawtoos"]Why not just differentiate the equation and solve the resulting DE for the proof: $ \frac {d}{dx}\left\{f\circ f \minus{} 3f \plus{} 2x \equal{} 0\right\}$ $ \left(f'\right)^2 \minus{} 3f' \plus{} 2 \equal{} 0$[/quote]$ (f\circ f)(x) \equal{} f(f(x))$ and $ [f(f(x))]' \equal{} f'(f(x))\cdot f'(x)$ ...[/quote][/quote]

thanks . . . got an F on that one . . . :oops:

[quote="Virgil Nicula"]
[size=117][color=darkred][b]Are these the $ \boxed {\mathrm {all}}$ continue functions which verify the relation $ (*)$ ?![/b][/color][/size][/quote]
The answer is "yes". I do not want to spoil the fun by posting an entire solution so soon but here is a small [hide="hint"]It is not hard to prove that $ f$ is monotone and onto. Fix $ x$ and consider the two-sided sequence $ \dots,f^{\minus{}1}(f^{\minus{}1}(x)),f^{\minus{}1}(x),x,f(x),f(f(x)),\dots$. What can you say about this sequence?[/hide]

This is really a damn hard problem, even though in appearance it's so classical. It's pretty clear that $ f$ is bijective (obviously increasing and cannot be bounded in any direction because of the functional equation). Next, as Fedja said, let's look at the sequence $ x_n\equal{}f^{\minus{}n}(x)$. Then $ 2x_{n\plus{}2}\equal{}3x_{n\plus{}1}\minus{}x_n$ and so $ x_n$ tends towards $ 2g(x)\minus{}x$, where $ g\equal{}f^{\minus{}1}$. But the limit of this sequence is a fixed point of $ f$ (thus of $ g$). So, if we put $ h(x)\equal{}2g(x)\minus{}x$, then the image of $ h$ is precisely the set of fixed points of $ g$ (we saw one inclusion and the other is clear), so it is an interval. We have two obvious solutions of the equation corresponding to the cases when this interval is a point or the whole real line. The other cases follow from the possibilities $ [t,\infty)$ and so on for this interval. In each case, using monotony of $ g$ it is immediate to deduce that it has the desired form.


linear algebra
matrix
quadratics
algebra
system of equations

I might be wrong, but I think there's something wrong with this problem.

[hide]
a^4 - b^4 = p1 * p2 * p3.
a^2 + b^2 = p1 * p2.

Since a^4 - b^4 factors into (a^2 + b^2)(a^2 - b^2), by replacement p3 = a^2 - b^2. But a^2 - b^2 factors into (a-b)(a+b), meaning that it is not prime.

Am I missing something?[/hide]

EDIT- NVM, That actually helps my solution!
[hide]
...in order for it to be prime, then either a+b or a-b would have to be 1. a-b is it, so a = b+1 (or b = a-1).
...and that's all I know right now...
[/hide]

So what to do in order to yield a solution?  :)

[hide=""answer""]

Try to find the following solutions: $(a|b|p_1|p_2|p_3|p_4) = (7|6|5|17|13|3)$ and $(a|b|p_1|p_2|p_3|p_4) = (7|6|17|5|13|3)$

[/hide]

[quote]a^2 - b^2 factors into (a-b)(a+b), meaning that it is not prime. [/quote]
[hide]a=2,b=1 , a^2-b^2 does factor into  (3)(1) yet you will not call  3   a prime :)
[/hide]

So far I have not seen a correct solution.  :)

[quote="orl"]Let $a,b$ be integers and $p_1, p_2, p_3, p_4$ be primes. Find the numbers satisfying the following system of equations:

$
\begin{matrix*}

a^4 - b^4 = p_1 \cdot p_2 \cdot p_3 \\

a^2 + b^2 = p_1 \cdot p_2 \\

a + b = p_3 \\

a \cdot b = p_3 \cdot p_4 + 3

\end{matrix*}
$[/quote]

[hide]a-b=1
\[
\begin{gathered}
  a = \frac{{p_3  + 1}}
{2},b = \frac{{p_3  - 1}}
{2} \hfill \\
  ab = \frac{{p_3^2  - 1}}
{4} = p_3 p_4  + 3 \hfill \\
  4p_3 p_4  + 12 = p_3^2  - 1 \hfill \\
  p_3 (p_3  - 4p_4 ) = 13 \hfill \\
  p_3  = 13,p_4  = 3 \hfill \\
\end{gathered}
\]
then it's easy to find out other numbers.[/hide]

Nice, I have another solution

[hide]Like everyone said its easy to show that $\displaystyle\ a-b=1$  It is also true that $\displaystyle\ (a+b)^2-2ab=a^2+b^2$.  So we can write $\displaystyle\ p_3^2-2p_3p_4 -6=p_1p_2$.  We can also write $\displaystyle\ (a-b)^2 + 2ab=a^2+b^2$ so that $\displaystyle\ 2p_3p_4 +7=p_1p_2$ combining this with our first equation we get $\displaystyle\ p_3^2-4p_4p_3-13=0$, and since $\displaystyle\ p_3$ must be an integer this quadratic must factor so 3 must be the only possible value for $\ p_4$.  Factoring this equation to find $\ p_3$ and finding the other primes is trivial.[/hide]

I must've done something wrong but nothing seems to be wrong.

$\displaystyle
a^4 - b^4 = p_1 \cdot p_2 \cdot p_3 \\
a^2 + b^2 = p_1 \cdot p_2 \\
a + b = p_3 \\
a \cdot b = p_3 \cdot p_4 + 3$

We are going to substitue the second and third equation into the first equation.

$a^4 - b^4 = (a^2 + b^2) \cdot (a + b) \\
(a^2 + b^2) (a^2 - b^2) = (a+b) (a^2+b^2) \\
(a^2 - b^2) = a+b\\
(a-b)(a+b) = a+b\\
a = b$

From here on, we find that since $a = b$ and $p_3$ is prime, $p_3$ must be $2$ and $a = b = 1$.

But then it kind of not work after this because substituting into equation 1 will give a $0$ in either $p_1$ or $p_2$ but substituting in equation 2 will only result $p_1$ and $p_2$ to be a 2 or 1.

I'm not too sure what mistakes I might have made.

[quote="kool_dudy"]
$(a-b)(a+b) = a+b\\
a = b$
.[/quote]

This doesn't make sense. $(a-b)(a+b)=a+b \Rightarrow a-b=1$.

Hmm...I think what I did when I got to...

$\displaystyle (a-b)(a+b) = a+b\\
a = b$

was that I divided both sides by $(a + b)$

so that's why I got $a = b$

Everyone else got $a -b = 1$  :blush:

[hide]$a^4 + b^4 = (a^2+b^2)(a+b)(a-b)$
$p_1 \cdot p_2 \cdot p_3 = (p_1 \cdot p_2) \cdot (p_3) \cdot (a-b)$

thus $a-b =1$

$(a-b)^2 = (a+b)^2 - 4ab = p_3^{2} - 4p_3 p_4 - 12 = 1$
$p_3(p_3 - 4p_4) = 13$

since $p_3$ is a prime and 13 is also a prime, $p_3 = 13$
$p_3 - 4p_4 = 13 - 4p_4 = 1 %Error. "arrow" is a bad command.
p_4 = 3$

$a^2+b^2 = p_1 \cdot p_2 = 13^2 - 2 \cdot (13 \cdot 3 + 3) = 85$

$p_1 = 5, p_2 = 17$ or $p_1 = 17, p_2 = 5$

So the master answers are:
$p_1 = 5 or 17$
$p_2 = 17 or 5$
$p_3 = 13$
$p_4 = 3$ $QED$[/hide]

Sorry, for some reason, sometimes my BBcodes dont work. So I could not hide. :(

[i]not sure why it doesn't work...but i made it hidden for you. -joml88[/i]


am I missing something or it should be like this: $B'$ is the midpoint of $AOB$... :?:   :maybe:

Nice problem.
I think $B'$ is midpoint of $AOC$ and it is ok.

Let $X$ and $Y$ be the points on lines $OB$ and $OC$ respectively s.t. $OX=OY=OA$. Let $P$ an $Q$ be points s.t. $AXP \sim AOC$ and $AYQ \sim AOB$. Let $D$ be a symetric point to $A$ w.r.t. $XY$.
We have
$AC'B \sim AOX$ so $ABX \sim AC'O$ and $AXP \sim AOC$. Thus $AXPB \sim AOCC'$ and $OCC' \sim XPB$.
$O$ is circumcenter of $AXY$ so $\angle AXP=\angle AOC=2\angle AXY=\angle AXD$ so $P$ lies on $XD$.
We also have
$\frac{XP}{XD}=\frac{XP}{XA}=\frac{OC}{OA}=\frac{XB}{XO}$ so
$OCC' \sim XPB \sim XDO$ and $OBB' \sim YDO$. Thus
$\angle BB'O+\angle CC'O=\angle DOY+\angle DOX=\angle BOC$ so cicumcircles $COC'$ and $BOB'$ are tangent. :)

[quote="Omid Hatami"]$O$ is a point inside triangle $ABC$ such that $OA=OB+OC$. Suppose $B',C'$ be midpoints of arcs $\overarc{AOC}$ and $AOB$. Prove that circumcircles $COC'$ and $BOB'$ are tangent to each other.[/quote]
use inversion around $O$ the problem will be:

[color=blue][b]Problem[/b].in triangle $ABC$ let $O$ be a point that $\frac{1}{OA}=\frac{1}{OB}+\frac 1{OC}$ let ex-anglebisector of $\angle AOB$ intersect $AB$ at $C'$ and $B'$ be intersection point of ex-anglebisector of $\angle AOC$ with $AC$.prove that $CC'\parallel BB'$[/color]

[i]proof[/i].as problem said we have $OB.OC=OA.OC+OA.OB\Longrightarrow \frac{OA}{OB}=\frac{OC-OA}{OC}$ also we have $\frac{OA}{OB}=\frac{AC'}{BC'}\ ,\ \ \frac{OA}{OC}=\frac{AB'}{CB'}\Longrightarrow \frac{OC-OA}{OC}=\frac{AC}{CB'}$ hence $\frac{AC'}{BC'}=\frac{AC}{CB'}$ hence $CC'\parallel BB'$.

[quote="Amir.S"]
use inversion around $O$ the problem will be:
[color=blue][b]Problem[/b].in triangle $ABC$ let $O$ be a point that $\frac{1}{OA}=\frac{1}{OB}+\frac 1{OC}$ let ex-anglebisector of $\angle AOB$ intersect $AB$ at $C'$ and $B'$ be intersection point of ex-anglebisector of $\angle AOC$ with $AC$.prove that $CC'\parallel BB'$[/color]
[/quote]
I can't understand how this happens(it doesn't seem obvious(at least for me))?

Uh sorry i disunderstood problem from @Babak Chalebi's post, thinking that B' is the midpoint of AC.Sorry

[b]Lemma.[/b] Given triangle $ABC$, bisectors of angle $B$ and $C$ meet $AC$ and $AB$ at $E,F$, respectively. $P$ is an arbitrary point on $EF$. Let $XYZ$ be pedal triangle of $P$ wrt $\triangle ABC$. Then $PX=PY+PZ.$

This lemma is well-known so I leave to the readers.

[b]Back to our problem.[/b]
Let $O_a, O_b, O_c$ be the circumcenters of triangles $BOC, COA, AOB$, $I_b, I_c$ be the circumcenters of triangles $BOY, COZ.$, $M,N,P$ be the midpoints of  $OA, OB, OC,$ respectively.
Since $I_bO_b\perp OY, O_bO_c\perp OA$ then $\angle I_bO_bO_c=\angle YOM.$
Similarly, $\angle I_bO_bO_a=180^\circ-\angle YOC.$
$Y$ is the midpoint of arc $AOC$ then $OY$ is the external bisector of $\angle AOC.$ We get $\angle I_bO_bO_c=\angle I_bO_bO_a$ or $I_b$ is the intersection of the bisector of angle $O_b$ and $O_aO_c.$
Similarly, $I_c$ is the intersection of the bisector of angle $O_c$ and $O_aO_b.$
But $OA=OB+OC$ hence $OM=ON+OP$, this means the distance from $O$ to $O_bO_c$ is equal to the sum of distances from $O$ to $O_aO_b$ and $O_aO_c$. According to lemma above, $O$ lies on $I_bI_c$. In other words, $(I_b)$ is tangent to $(I_c)$ at $O$.

Sorry to revive, but I'm not familiar with the above lemma about pedal triangles. Can someone share a proof or some references concerning the subject?


trigonometry
inequalities
inequalities unsolved

hello, i have found $\frac{1}{4}(4+3\sqrt{3})$
Sonnhard.

[quote="tranthanhnam"]find the value maximum  of expression :
 $\frac{a}{a+bc} +\frac{b}{b+ca} +\frac{\sqrt{abc}}{c+ab}$
 where $a+b+c=1$ and$a,b,c >0$[/quote]
Let $a+b=z$, $a+c=y$ and $b+c=z$.
Hence, $x+y+z=2$ and $\frac{a}{a+bc} +\frac{b}{b+ca} +\frac{\sqrt{abc}}{c+ab}=\frac{y+z-x}{2yz}+\frac{x+z-y}{2xz}+\frac{\sqrt{\frac{\prod\limits_{cyc}(x+y-z)}{8}}}{xy}=$
$=\frac{(y+z)^2-x^2}{4yz}+\frac{(x+z)^2-y^2}{4xz}+\frac{\sqrt{\frac{(x+y+z)\prod\limits_{cyc}(x+y-z)}{16}}}{xy}=$
$=1+\frac{1}{2}\left(\cos\alpha+\cos\beta+\sin\gamma\right)$.
If $\alpha$ or $\beta$ is a not acute angle, so $1+\frac{1}{2}\left(\cos\alpha+\cos\beta+\sin\gamma\right)<1+\frac{1}{2}(1+1)=2$.
Let $\alpha$ and $\beta$ be acute angles. Hence, by Jensen we obtain:
$1+\frac{1}{2}\left(\cos\alpha+\cos\beta+\sin\gamma\right)=1+\frac{1}{2}\left(\sin\left(90^{\circ}+\alpha\right)+\sin\left(90^{\circ}+\beta\right)+\sin\gamma\right)\leq$
$\leq1+\frac{3}{2}\sin\frac{\alpha+\beta+\gamma+180^{\circ}}{3}=1+\frac{3\sqrt3}{4}$.
Since $2<1+\frac{3\sqrt3}{4}$, we see that the answer is indeed $1+\frac{3\sqrt3}{4}$.


advanced fields
advanced fields unsolved

If n, x, and y are positive integers, prove that

http://www.artofproblemsolving.com/Forum/latexrender/pictures/e58a7d527155e74b2dea1b6617569176.gif

cannot be an integer for all n greater than 2.

um...

Fermat's last Theorem is:

show that there are no whole number solutions to the equation

x^n + y^n = z^n for n greater than 2.

We have $(1+s)^{t}= \sum_{k=0}^\infty s^{k}\binom{t}{k}$.
With $t=\frac{1}{n}$ and $s=\frac{y^{n}}{x^{n}}$ this gives $x \left( 1+\left(\frac{y}{x}\right)^{n}\right)^\frac{1}{n}= \sum_{m=0}^\infty \frac{y^{mn}}{x^{mn-1}}\binom{\frac{1}{n}}{k}$.

Lets write $z$ for the right hand side, then $x,y,z$ shall be integers with $x \left( 1+\left(\frac{y}{x}\right)^{n}\right)^\frac{1}{n}=z \iff x^{n}\left( 1+\left(\frac{y}{x}\right)^{n}\right) = z^{n}\iff x^{n}+y^{n}=z^{n}$.
Thus its indeed Fermat's theorem.


Solve the follow equation $ abc\equal{}a\plus{}b\plus{}c\plus{}2$ where $ a,b,c\in\mathbb{N}$ and $ a<b<c$.

[quote="Mîndrilă Claudiu"]Solve the follow equation $ abc \equal{} a \plus{} b \plus{} c \plus{} 2$ where $ a,b,c\in\mathbb{N}$ and $ a < b < c$.[/quote]

Here is my solution

take first case $ b\geq3$. Then $ a \equal{} \dfrac{b \plus{} c \plus{} 2}{bc \minus{} 1}$. So, because $ b \plus{} c \plus{} 2 > 2$ and $ a\in\mathbb{N}$, we get $ b \plus{} c \plus{} 2\geq bc \minus{} 1$. The last inequality is equivalent to $ b\leq\dfrac{c \plus{} 3}{c \minus{} 1}$. So $ 3\leq\dfrac{c \plus{} 3}{c \minus{} 1}$ or $ 3c \minus{} 3\leq c \plus{} 3$ or $ c\leq3$ but $ c > b\geq3$. So we get contradiction. $ b$ may be only $ 1,2$

If $ a \equal{} 0$ then $ 0 \equal{} b \plus{} c \plus{} 2$. So $ a\neq0$

So $ b \equal{} 2\Rightarrow 2c \equal{} c \plus{} 5\Rightarrow c \equal{} 5$

The only solutions is:$ a \equal{} 1, b \equal{} 2, c \equal{} 5$

$ abc\equal{}a\plus{}b\plus{}c\plus{}2<3c$, so $ ab \le 2$ , $ a,b\equal{}1,2$ gives $ (1,2,5)$ as the only solution.

[quote="enndb0x"][quote="stephencheng"]$ abc \equal{} a \plus{} b \plus{} c \plus{} 2 < 3c$, so $ ab \le 2$ , $ a,b \equal{} 1,2$ gives $ (1,2,5)$ as the only solution.[/quote]

why $ ab \le 2$[/quote]
Stephencheng proved that:
$ abc\equal{}a\plus{}b\plus{}c\plus{}2<3c\Longrightarrow ab<3$
As $ a, b\in \mathbb{N}$ we have $ ab\le2$ :roll:
Isn't it so?

hope you understand now.  :)


Someone is using a swastika as their avatar.   (Yes, I know the symbol had other meanings before the Nazis used it...)

I don't want to name names in public, but if someone in authority agrees that this is inappropriate, I will be happy to pm them the details.

i saw the same one today.....i think his username was
[hide]anithros, or something like that[/hide]

[quote="mathmom"]Are there any rules regarding what may be used as an avatar?  (I just saw one that I find offensive.)[/quote]

We suggest PMing or emailing an administrator with details.  The rules are roughly the same as for determining what's permissible in posts - some things are obviously bad, and others we have to handle on a case-by-case basis.


[quote="Ignite168"]If ${\frac{x^{2}}{c}+\frac{x}{a}+\frac{x^{3}}{b}=1^{2}+2^{2}+3^{2}+\ldots x^{2}}$ then find $ac+b$.[/quote]
[hide]just plug in the formula to get 15[/hide]

[quote="nutz_for2.718281828"]@bpms: What formula? Oh wait... I don't think you need to memorize it....................

@Ignite168: Do you need the formula for this problem?[/quote]
If for this problem you don't know the formula, the you have to derive it or you won't get the problem right (other than guessing). The formula is easy to derive since the problem tells you that it is a cubic trinomial.

[quote="13375P34K43V312"]By the way...

$\sum_{k=1}^{n}k^{2}=\frac{n(n-1)(2n-1)}{6}$.[/hide][/quote]

[color=red]EDIT: I see that the typo has been fixed[/color]

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/4dd28f5f3f523f9ca49336fba5d2f40f.gif[/img]

[b][color=Blue][hide="Proof by Mathematical Induction"]In order to understand how this formula works, I might as well prove it by Induction. (I had been challenged by bpms anyway before to prove this formula by induction sometime around the summer so I will do it now)

I. Show for ${k=1}$:

$\begin{cases}1^{2}=\frac{(1)(1+1)(2+1)}{6}=1\\ \\ 1^{2}=1 \end{cases}$

Therefore, the equation stands true for $k=1$.

II. Assume true for $k=n$:

$1^{2}+2^{2}+\dots+n^{2}=\frac{n(n+1)(2n+1)}{6}$

III. Show true for $k=n+1$:

$1^{2}+2^{2}+\dots+n^{2}+(n+1)^{2}=\frac{[(n+1)][(n+1)+1][2(n+1)+1]}{6}$

$1^{2}+2^{2}+\dots+n^{2}+(n+1)^{2}=\frac{(n+1)(n+2)(2n+3)}{6}$

$\frac{n(n+1)(2n+1)}{6}+(n+1)^{2}=\frac{(n+1)(n+2)(2n+3)}{6}$

$\frac{6n(2n+1)(n+1)^{3}}{6}=\frac{(n+1)(n+2)(2n+3)}{6}$

Then, expand the LHS, and then resimplify it, since you are technically not allowed to modify the form of the RHS. Therefore, the equation becomes true, as the RHS also becomes:

$\frac{(n+1)(n+2)(2n+3)}{6}$

So,
$\frac{(n+1)(n+2)(2n+3)}{6}=\frac{(n+1)(n+2)(2n+3)}{6}$

And, thus, the formula has been proved.

$\mathbb{Q.E.D.}$

[/hide][/color][/b]

OK, but how would you go about deriving it?

Well...i just proved it right there, so I don't get what you mean :maybe:

[quote="humblemastermind"]Proving is a little different from deriving. By "derive" I mean figuring it out from scratch. Your proof just starts by assuming that it's true, without giving any hint of how you would arrive at that formula in the first place.[/quote]
I know a couple ways of deriving. One is hockey stick, but the mathcounts level method is that since you know it is a cubic polynomial, you know it is in the form $ax^{3}+bx^{2}+cx+d$ you can plug in four values of x and get four equations then solve for a, b, c, and d.

Heh, whoops anirudh caught my typo. I'm editing it.

[quote="jhredsox"]should i memorize something like this for possible sucess at mc states?

-jorian[/quote]
Yes, in the format leet has edited his post to be in.
To anirudh, I don't recall challenging you to prove this with induction. I do recall challenging you to prove herons with induction though.

[quote="bpms"][quote="jhredsox"]I do recall challenging you to prove herons with induction though.[/quote][/quote]

No. You didn't. You said that you:

'would like to see me prove the egyptian formula by induction'

and that egyptian formula was the one I just proved. :wink:

It's possible to prove heron's formula by induction? ON which variable?

[quote="anirudh"][quote="bpms"][quote="jhredsox"]I do recall challenging you to prove herons with induction though.[/quote]

No. You didn't. You said that you:

'would like to see me prove the egyptian formula by induction'

and that egyptian formula was the one I just proved. :wink:[/quote][/quote]
Link to that and I'll eat what I said.

[quote="bpms"][quote="mathgeniuse^ln(x)"][quote="Ubemaya"]How do you do it using hockeystick?[/quote]

I think he meant for 1+2+3+...

I don't know how to do it for squares.[/quote]
Oh there's a way.
[hide="hint"]$\binom{n+1}{2}+\binom{n}{2}=n^{2}$[/hide][/quote]

Oh, of course, I forgot about hat.

Yeah

[hide]$\binom{2}{2}+\binom{1}{2}=1^{2}$
$\binom{3}{2}+\binom{2}{2}=2^{2}$
...
$\binom{n+1}{2}+\binom{n}{2}=n^{2}$

So adding up both sides

$\binom{n+2}{3}+\binom{n+1}{3}=n^{2}$

And once again

$\frac{n(n+1)(2n+1)}{6}=1^{2}+2^{2}+...n^{2}$[/hide]


For an article, how can I make the page number on the top right corner of the page and type something in front of it (like homework, 5)?

Use the fancyhdr package - see [[LaTeX:Packages#fancyhdr]]. For the effect that you want you need:
\rhead{homework \thepage}
\cfoot{}
The \cfoot command stops the page number appearing in its usual place, the centre of the footer.


search
number theory
relatively prime
number theory proposed

Let [i]a>b>1[/i] be relatively prime positive integers.
Define the weight of an integer [i]c[/i], denoted by [i]w(c)[/i] to be the minimal possible value of |[i]x[/i]|+|[i]y[/i]| taken over all pairs of integers [i]x[/i] and [i]y[/i] such that [i]ax[/i] + [i]by[/i] = [i]c[/i].
An integer [i]c[/i] is called a [i]local champion[/i] if [i]w(c)[/i]>=[i]w(c+a)[/i], [i]w(c)[/i]>= [i]w(c-a)[/i], [i]w(c)[/i]>=[i]w(c+b)[/i] and [i]w(c)[/i]>=[i]w(c-b)[/i].

Find all local champions and determine their number.

Next time try searching
http://www.mathlinks.ro/viewtopic.php?search_id=1994848479&t=214621


Hi,

I was working on LaTeX and I wanted to go to next page before finishing the previous page. In another word, is there any way you can go to the new page 2 when you're half-way in page 1?

I tried to use \vfill but it only goes to like 2/3 of the page 1. Any idea?

Thanks, this is the exact kind of code I was looking for.

I know \newpage, which does exactly what Silverfalcon wants and has no side-effects. Maybe this is the technically better soluzion?

I haven't tried \newpage yet but if \clearpage doesn't work (it works now but in case it doesn't) I'll use \newpage.

Thanks for telling.

what's against the normal way? (= \pagebreak )

\clearpage flushes all the floats (figures and tables)


Supposed we have a regular hexagon with the length of side is 5 cm. Can we draw a triangle with side 6 cm, 8 cm and 10 cm inside the hexagon without cutting the triangle or the regular hexagon. God bless you all.

The triangle is inside the hexagon so the 10cm-side must be the segment which connect 2 vertices of the hexagon and go through the center of the hexagon's circumcircle.
Since the triangle with sides 6, 8, 10 is a right triangle, the other vertex must lie on the circumcircle. Also, the triangle must be inside the hexagon so the triangle's vertex is one of the other vertices of the hexagon.
This triangle does not have sides 6, 8, 10 so we can't draw the mentioned triangle.


[quote="heaven_us"]Hello every body, i'm new member....every body help me this problem, thanks
Find all continuous functions f,g,h on R such that f(x+y)=g(x)+h(y)[/quote]
$ f(x)\equal{}g(x)\plus{}C_1$ and $ f(y)\equal{}h(y)\plus{}C_2$
Hence, $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}C_1\plus{}C_2$
This is function equation Cauchy's :
Let $ F(x)\equal{}f(x)\plus{}C_1\plus{}C_2$ then $ F(x\plus{}y)\equal{}F(x)\plus{}F(y) \to F(x)\equal{}Ax$
Thus,$ f(x)\equal{}Ax\minus{}C_1\minus{}C_2$,$ g(x)\equal{}Ax\minus{}C_2$,$ h(x)\equal{}Ax\minus{}C_1$

[quote="mathangel"]Nice solution, Hong Quy!

You just need to add $ C_1 \equal{} h(0)$ and $ C_2 \equal{} g(0)$ for clarification.[/quote]
yes, i knew that. with ur comnent, now  the solution's right 100% :))

[quote="heaven_us"]Hello every body, i'm new member....every body help me this problem, thanks
Find all continuous functions f,g,h on R such that f(x+y)=g(x)+h(y)[/quote]

Sorry all but this is a problem in Maths And Youth Magazine this issue. Please delete this topic.

See also http://www.mathlinks.ro/viewtopic.php?p=1352325#1352325.

Please more carefully.

Heaven_us, so sad for you.


Did anyone by any chance take the 30 question 90 minute, level one of the Massachusetts olympiad yesterday. If so, would you like to compare answers?

yea i took it

what school do you go to?

i go to boston Latin

I too took the olympiad, and I personally thought it was damn hard! I only answered 16 total, 1 of which I know is wrong already.

I dont remember most of the problems, but I do remember answering the one with the sum of the logs and $f_i$'s, I think it was #25 or something.

I answered 22, but I know I got a fair amount (3-4) wrong. Does anyone know their answers?

Hopefully we're allowed to post answers since everyone took it already...

I got 29 right, 1 blank, 0 wrong --> 146

1.E
2.A

Hopefully we're allowed to post answers since everyone took it already...

I got 29 right, 1 blank, 0 wrong --> 146

[hide]1.E
2.A
3.D
4.E
5.A
6.B
7.D
8.E
9.B
10.D
11.E
12.E
13.B
14.A
15.E
16.C
17.D
18.B
19.E
20.B
21.C
22.A
23.B
24.C
25.D
26.E
27.B
28.B
29.C
30.A

[/hide]

23. For Each side of the octagon, use it to be the base of an equalateral triangle. Whats remaining is 8 sector with radius 1 and they are each 15 degrees. So the area of the octagon is $2+ 2\sqrt{2}$, each equilateral triangle is $\frac{\sqrt{3}}{4}$, each sector is $\frac{\pi}{24}$. Thus the area of the shaded region is \[ 2+2\sqrt{2} - 8 \huge( \frac{\sqrt{3}}{4} \huge) - 8(\frac{\pi}{24})  \]
\[ \frac{6+6\sqrt{2} - 6\sqrt{3} -\pi}{3}  \]

29. Baiscally what I did was experiment with smaller cases, like 1 in a row, 2 in a row... and found that it was a fibonacci pattern. I still can't figure out why but it's like adding steps to a ladder thing....

How did other people do on MAML lvl 1??

Plus what schools are you guys from?

I'm from BLS(Boston Latin School)... we suck at tech

I posted the list of finalists and semifinalists at

http://wocomal.yanco.com/statistics/olympiad/200506/20051025o_level1.html

(The highlighted rows are for students at WOCOMAL schools.)

Enjoy.

Cool, I'm a finalist! (barely; I'm tied for 89th  :) )

Meh, I'm a finalist, but I had a pretty bad day, as evident.

is there a pdf online with the questions?

I could put it (and a few recent years) online, but I'm waiting for http://www.maml.org to be put back up.

Although it's not this year's contest, you can get the [i]2002[/i] level 1 at
http://www.maml.org/wwwroot/contestinfo/Level1_Exam.pdf

and its solutions at
http://www.maml.org/wwwroot/contestinfo/Level1_Solutions.pdf

If you poke around in the maml.org directories, you might be able to find other exams, but I can't find them.  A few years ago, the 2000 and 2001 level 1 exams were online, and the corresponding (2001 and 2002) level 2 exams were online, all with solutions.  Not online through the website, but hidden in those directories.


Γειά σας και πάλι. Γράφω 2 ακόμα ανισότητες, ελπίζω να τις βρείτε διασκεδαστικές.

1)   Αν a, b, c θετικοί πραγματικοί να αποδείξετε ότι:

     $ b \sqrt{\frac{a}{c}} \plus{} c \sqrt{\frac{b}{a}} \plus{} a \sqrt{\frac{c}{b}} \geq 3\sqrt[3]{abc}$

2)  Αν a, b, c θετικοί πραγματικοί να αποδείξετε ότι:

     $ \frac{2a}{b} \plus{} \frac{3b}{c} \plus{} \frac{6c}{a} \geq (\sqrt{\frac{a}{b}} \plus{} \sqrt{\frac{b}{c}} \plus{} \sqrt{\frac{c}{a}})^{2}$

Για  την 1) έχουμε απο $ AM \minus{} GM$ ότι $ b\sqrt {\frac {a}{c}} \plus{} c\sqrt {\frac {b}{a}} \plus{} a\sqrt {\frac {c}{b}}\geq 3\cdot \sqrt [3]{abc\cdot \sqrt {\frac {abc}{abc}}} \equal{} 3\cdot \sqrt [3]{abc}$.

Για την 2) θέτουμε $ \frac {a}{b} \equal{} x^{2},\frac {b}{c} \equal{} y^{2},\frac {c}{a} \equal{} z^{2}$ και παίρνουμε ότι

$ 2x^{2} \plus{} 3y^{2} \plus{} 6z^{2}\geq \left(x \plus{} y \plus{} z\right)^{2} \equal{} \sum x^{2} \plus{} 2\sum xy\Longleftrightarrow x^{2} \plus{} 2y^{2} \plus{} 5z^{2} \minus{} 2\sum xy\geq 0\Longleftrightarrow \sum (x \minus{} y)^{2} \minus{} x^{2} \plus{} 4z^{2}\geq 0\Longleftrightarrow 4z^{2} \plus{} (x \minus{} y)^{2} \plus{} (y \minus{} z)^{2} \plus{} (x \minus{} z)^{2}\geq x^{2}$.

Όμως απο Cauchy παριρνουμε $ 4z^{2} \plus{} (x \minus{} y)^{2} \plus{} (y \minus{} z)^{2} \plus{} (x \minus{} z)^{2}\geq \frac {\left(x \minus{} y \plus{} y \minus{} z \plus{} x \minus{} z \plus{} 2z\right)^{2}}{4} \equal{} \frac {(2x)^{2}}{4} \equal{} x^{2}$ και τελειώσαμε :)

1) Am-Gm όπως είναι
2) Πολλαπλασιάζουμε το δεξί μέλος με $ \frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{6}\equal{}1$ και μετά B.C.S

[quote="KostasGT"]Γιώργο, ήθελα να σου πώ ότι έκανες θαυμάσια σκέψη. Την ανισότητα την κατασκεύασα με Andreescu (η οποία φυσικά συνδέεται άμεσα με την B.C.S. που αναφέρει ο bilot4), αλλά η λύση σου είναι πολύ ωραία. Μπράβο για την εξαιρετική σκέψη.[/quote]

Σας ευχαριστώ κ. Κώστα για τα καλά σας λόγια :) .

P.S Σας περίμενα για κανένα συνταξιούχο μαθηματικό που βρήκε επιτέλους χρόνο να ασχοληθεί με τις ολυμπιάδες, αλλά τελικά είστε νέος! Καλή συνέχεια στο forum και καλό κουράγιο(if you know what i mean) :wink: .

Έγινε Γιώργο. Καλή επιτυχία στις Πανελλήνιες, αν φέτος εξετάζεσαι, και καλή επιτυχία στους διαγωνισμούς.


It looks like Dr. Andreescu started sending out the admissions letters.

I got in. Yay!

Now I have to figure out what to bring. I have a rubik's cube, eleven sealed decks of Bicycle cards, maybe I'll get a Go board and a Scrabble set (even though I'm mediocre at both) ... are we allowed to bring textbooks?

I got in too!

 i'm rather scared...first time going to a math camp eeeek x[
been to bunch of science programs..
ahahahaha

well yeahh :]
congrats to everyone who got in =)

Yay I'm so excited I also got in.  :lol:

I think everyone got in... How many problems did you guys do? I did 9 -_-... I didn't have any more days left to do number 8 :D... Bring a Go Board and we can play!!

I also got in!

Hmm... Maybe people are interested in speedknotting? (Gordian's Knot)

I also did everything except 8.

Can't do geometry. :(

I got in! Yay!
PS. geometry is my favorite part of math :)

I got in quite easily,
I only worked on the test for 2 hours

[quote="sooozyy"]hmm i thought number 8 was rather easy..
i solved that one in like 3 minutes..
but took forever on number six..since i went around and around with that problem lol[/quote]

Same, except reverse problem numbers.

I did 8 problems. Didn't do the geometry and forgot how to do the system of equations. But still got in.


Let $ m,n$ be two natural numbers with $ m > 1$ and $ 2^{2m \plus{} 1} \minus{} n^2\geq 0$. Prove that:
\[ 2^{2m \plus{} 1} \minus{} n^2\geq 7 .\]

[hide]We prove the claim by induction.  The base case is true: $ 2^5\minus{}5^2\ge7$. Suppose it is true that $ 2^{2m\plus{}1}\minus{}n^2\equal{}x\ge7$ where we take $ n$ to be maximal.  Thus we have $ 2n\plus{}1>x$  or else $ 2^{2m\plus{}1}\minus{}(n\plus{}1)^2\equal{}x\minus{}2n\minus{}1\ge0$.  Multiply the given equation by $ 4$ to get $ 2^{2m\plus{}3}\minus{}(2n)^2\equal{}4x$.  If $ 2n$ is maximal in this case then it is obvious that $ 4x\ge7$ and we're done.  Suppose that $ 2n$ is not maximal.  Then, since $ 8n\plus{}4>4x\Rightarrow 2^{2m\plus{}3}\minus{}(2n\plus{}2)^2\equal{}4x\minus{}8n\minus{}4<0$, the only possible maximal value of the square is $ 2n\plus{}1$.  For this to be maximal, we must have $ 2^{2m\plus{}3}\minus{}(2n\plus{}1)^2\equal{}4x\minus{}4n\minus{}1\ge0$.  That is, we have $ 4x\ge4n\plus{}1\Rightarrow x\ge n\plus{}\frac{1}{4}$.  Since $ x$ is integral, we have $ x\ge n\plus{}1$.  We split into two cases:

$ x\equal{}n\plus{}1$
If $ x\equal{}n\plus{}1$ we have $ 2^{2m\plus{}1}\minus{}n^2\equal{}n\plus{}1\Rightarrow2^{2m\plus{}1}\equal{}n^2\plus{}n\plus{}1$.  But, for all integral $ n$, $ n^2\plus{}n\plus{}1$ is odd, while $ 2^{2m\plus{}1}$ is even for all $ m>1$.  Thus $ x\ne n\plus{}1$.

$ x\ge n\plus{}2$
This case yields $ 2^{2m\plus{}3}\minus{}(2n)^2\ge4n\plus{}8\Rightarrow2^{2m\plus{}3}\minus{}(2n\plus{}1)^2\ge7$ and we're done.[/hide]

If you use modulo 3,4, and 8, you can easily prove that the expression cannot equal $ 0,1,2,\hdots ,6$.  Thus, $ 2^{2m\plus{}1}\minus{}n^2\geq 7$.


are there infinitely many primes p such that 2 is a primitive root in mod p???

   and what about prime q in mod p???

[quote="Lagrangia"]http://www.mathlinks.ro/phpBB/viewtopic.php?t=44[/quote]
Are you sure the link is working right? I got the following error page:
[quote]Object not found!
The requested URL was not found on this server. The link on the referring page seems to be wrong or outdated. Please inform the author of that page about the error.
If you think this is a server error, please contact the webmaster
Error 404
http://www.mathlinks.ro
Thu Mar 4 17:39:36 2004
Apache/2.0.40 (Red Hat Linux) [/quote]

that's because he also put the old /phpBB/ extension.

here's the correct link :)
http://www.mathlinks.ro/viewtopic.php?t=44

E. Artins conjeture asserts that giving any  integer a that is not an square and not -1, there are infinitely many primes p for which a is a primitive root.  however it's Known that this conjeture is an implication of  of the extension that Riemann Hipothesis holds in algebraic number fields.

2

I got something that may be of some help: if p is a prime of the form 8t+3, and (p-1)/2 is also a prime, then 2 is a primitive root mod p, i think that to prove this the key is to notice that 2 is not a square mod p and phi(p)=2q where q is a prime......

I guess that in the conditions you have stated 2 is indeed a primitive root mod p, but how does this help? Should we prove that there are infinitely many primes that satisfy those conditions? I'm sure that problem us unsolved (are there or are there not infinitely many primes p s.t. (p-1)/2 is also prime).

[size=59]ya halil olum
h be bunu da yapamıyorsan
herneyse bu soruya bir bakcam
yapınca bildiririm.
hadi yarın grrz.[/size]
[color=red][Admin Edit: Please use English Language][/color]
"""<<<<<there is not one potato there are many of them>>>>>"""


limit
integration
trigonometry
calculus
function
real analysis
real analysis unsolved

If $ c>0$, then find
\[ \lim_{n\to\infty}\int_{0}^{\infty}e^{\minus{}cx}\cot x\sin (2n\plus{}1)x\; dx\]

Let $ f(x) = e^{-c|x|}\cos x$. Then

\begin{eqnarray*}\lim_{n\to\infty} \int_{0}^{\infty} e^{-cx}\cot x\sin (2n+1)x \, dx
& = & \lim_{n\to\infty} \frac{1}{2} \int_{\mathbb{R}} f(x) \frac{\sin (2n+1)x}{\sin x} \, dx \\
& = & \lim_{n\to\infty} \frac{1}{2} \sum_{k=-n}^{n} \int_{\mathbb{R}} f(x) e^{-2 k i x} \, dx \\
& = & \frac{1}{2} \sum_{k=-\infty}^{\infty} \hat{f}\left( \frac{k}{\pi} \right).
\end{eqnarray*}

Since $ \hat{f}(w) = 2 \int_{0}^{\infty} e^{-cx}\cos x \cos(2 \pi w x) \, dx = O(w^{-2})$, Poisson summation formula is applicable and

$ \lim_{n\to\infty} \int_{0}^{\infty} e^{-cx}\cot x\sin (2n+1)x \, dx = \frac{\pi}{2} \sum_{k=-\infty}^{\infty} f(\pi k) = \frac{\pi}{2} + \frac{\pi}{1+e^{c \pi}}$

Or we can proceed as follows:

We are going to need:

1. $ \boxed{1 \plus{} 2\sum_{k \equal{} 1}^{n} \cos 2kx \equal{} \frac {\sin \left(2n \plus{} 1 \right)x}{\sin x}}$

2. $ \boxed{\cos 2kx \cdot \cos x \equal{} \frac {1}{2}\left[\cos \left(2k \minus{} 1\right)x \plus{} \cos \left(2k \plus{} 1\right)x\right]}$

3. $ \boxed{\int_{0}^{\infty} e^{ \minus{} st} \cos (at) dt \equal{} \frac {s}{s^2 \plus{} a^2}}$

4. $ \boxed{\frac {\pi}{4}\tanh \left(\frac {\pi c}{2}\right) \equal{} \sum_{k \equal{} 1}^{\infty} \frac {c}{c^2 \plus{} (2k \minus{} 1)^2}}$

Now, we have:

$ I(c) \equal{} \lim_{n \to \infty} \int_{0}^{\infty} e^{ \minus{} cx} \cot x \sin (2n \plus{} 1)x dx$

$ \equal{} \lim_{n \to \infty} \int_{0}^{\infty} e^{ \minus{} cx} \cos x \cdot \frac {\sin (2n \plus{} 1)x}{\sin x} dx$

$ \equal{} \lim_{n \to \infty} \int_{0}^{\infty} e^{ \minus{} cx} \cos x \cdot \left[1 \plus{} 2\sum_{k \equal{} 1}^{n} \cos 2kx \right]dx$

$ \equal{} \int_{0}^{\infty} e^{ \minus{} cx} \cos x dx \plus{} 2\lim_{n \to \infty} \int_{0}^{\infty} e^{ \minus{} cx} \cos x \cdot \sum_{k \equal{} 1}^{n} \cos 2kx dx$

$ \equal{} \int_{0}^{\infty} e^{ \minus{} cx} \cos x dx \plus{} 2\lim_{n\to \infty} \sum_{k \equal{} 1}^{n}\int_{0}^{\infty} e^{ \minus{} cx} \cos x\cos 2kx dx$

$ \equal{} \int_{0}^{\infty} e^{ \minus{} cx} \cos x dx \plus{} 2\sum_{k \equal{} 1}^{\infty}\int_{0}^{\infty} e^{ \minus{} cx} \left(\frac {1}{2}\left[\cos \left(2k \minus{} 1\right)x \plus{} \cos \left(2k \plus{} 1\right)x \right] \right)dx$

$ \equal{} \frac {c}{c^2 \plus{} 1} \plus{} \sum_{k \equal{} 1}^{\infty} \frac {c}{c^2 \plus{} (2k \minus{} 1)^2} \plus{} \sum_{k \equal{} 1}^{\infty} \frac {c}{c^2 \plus{} (2k \plus{} 1)^2}$

Now use (4) to get,

$ I(c) \equal{} \frac {c}{c^2 \plus{} 1} \plus{} \frac {\pi}{4}\tanh \left(\frac {\pi c}{2}\right) \plus{} \left(\frac {\pi}{4}\tanh \left(\frac {\pi c}{2}\right) \minus{} \frac {c}{c^2 \plus{} 1}\right)$

$ \equal{} \frac {\pi}{2} \tanh \left(\frac {\pi c}{2}\right) \equal{} \frac {\pi}{2} \minus{} \frac {\pi}{e^{c\pi } \plus{} 1}$

Edited after Kent's remarks.

Please give me the proof of these, thanks
[quote="akech"]
We are going to need:
3. $ \boxed{\int_{0}^{\infty} e^{ \minus{} st} \cos (at) dt \equal{} \frac {s}{s^2 \plus{} a^2}}$
4. $ \boxed{\frac {\pi}{4}\tanh \pi c \equal{} \sum_{k \equal{} 1}^{\infty} \frac {c}{c^2 \plus{} (2k \minus{} 1)^2}}$
[/quote]
and please give me the proof link of Poisson summation formula, thanks.

Everything I said [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298286]in this thread[/url] applies here as well - it's just a different $ g.$ The methods that akech, sos440, and I used are all linked and interlaced - Fourier series, Fourier transforms on the line, and the Poisson summation formula connect these approaches to each other. (Actually akech seems to mostly be using Laplace transforms - but that's another member of the family.)

Ad, chuong01:
Apply integration by parts twice to $ I \equal{} \int_{0}^{\infty} e^{\minus{}st}\cos at dt$ and you should be able to get the formula I gave in my thread.

Knowing Laplace Transforms of some elementary functions saves time because then one does not have to do integration by parts repeatedly to derive a formula such as the one in question.

As for the other derivation, one considers the series expansion of tanh x [some books on infinite series will contain the derivation] and then replace x with $ \pi x$.

In the thread I linked to, I derived the formula $ \frac{\pi}2(g(0)\plus{}2g(\pi)\plus{}2g(2\pi)\plus{}2g(3\pi)\plus{}\cdots)$ for this.

In this case, $ g(x)\equal{}e^{\minus{}cx}\cos x,$ so $ g(k\pi)\equal{}(\minus{}1)^ke^{\minus{}c\pi k}.$

Hence, I get that the answer should be

$ \frac{\pi}2\left(1\plus{}2\sum_{k\equal{}1}^{\infty}(\minus{}1)^ke^{\minus{}c\pi k}\right)$

The geometric sum is

$ \frac{\pi}2\left(1\minus{}\frac{2e^{\minus{}c\pi}}{1\plus{}e^{\minus{}c\pi}}\right)\equal{}\frac{\pi}2\left(\frac{1\minus{}e^{\minus{}c\pi}}{1\plus{}e^{\minus{}c\pi}}\right).$

At this point, I don't agree with either of the answers above. I would agree with sos440 with one sign change - if he were to say $ \frac{\pi}2\minus{}\frac{\pi}{1\plus{}e^{c\pi}}$ instead of having a + sign there.

As for akech's answer - where are those rational functions of $ c$ coming from, anyway? Check that you've appropriately accounted for the beginning of the sum. And shouldn't that be $ \tanh\left(\frac{\pi c}2\right)?$


email
number theory
relatively prime
number theory solved

Here is a mail I received from Antoine Verroken. His email address is [b]antoine.verroken%pandora.be[/b] where the [b]%[/b] is to be replaced by [b]@[/b] . (Protection against spam harvesters.)

Because of my very poor knowledge in number theory, I cannot manage the solution :?

[quote="Antoine Verroken"]Dear M.Grinberg,

I have a problem in number theory that I cannot solve :

  determine all groups (m,k) for  ( m – 1 )! = m^k – 1.

  I tried the following :

1.       because (m-1)! and m^k must have minimum 1 common factor, m is a prime.

2.       a.  k=1  -> (m-1)!= m-1 -> m=2 and 3

b.  k=2   there is a solution for m=5 ; for m>5 we can prove by induction that (m-1)!> m^2 – 1

c.  k=3   there is no solution up to m=6,and for m>6 we prove by induction that (m-1)!>m^3-1; thus there is no solution for k=3.

d.       k>3  I think,because m is prime and because there are no solutions for k=3,that there are no solutions for k>3.

I tried the following :

     -because m^3 divides m^4 ; if (m-1)! is congruent -1 mod.m^4 then (m-1)! is congruent -1 mod.m^3.

     -because (m-1)! is not equal m^3-1,then (m-1)! is not congruent -1mod.m^3

     -if (m-1)! is not congruent -1 mod.m^3 -> (m-1)! is not congruent -1mod.m^4

     -thus (m-1)! is not equal m^4 – 1.

e.if d. is correct we can do  the same for all k>3.

    The problem,I think, is to prove that (m-1)! is not congruent -1mod.m^3,because m is prime and (m-1)! is not equal m^3-1.I think this is true but up to now I cannot prove it.

    I would be very greatfull if you could help me with this problem.

    Thanking in advance,

    Antoine Verroken[/quote]

I already solved this problem ojn the forum, but I do not know where. So, take m-1=n. Then n!=(n+1)^k-1. Suppose k>1 and also that n>5. Then, if n+1 is not a prime, it isn't relatively prime with n!, false. Thus, n+1 is a prime. Also, we have (n-1)!= n^(k-1)+kn^(k-2)+...+k(k-1)/2n+k. Suppose n is not a prime. Then n|(n-1)! and thus n|k. It follows that k>=n and thus n!>=(n+1)^n-1, false. This means that n is also a prime and n+1 is also a prime, false. THus, we must have n<=5 and now it's simple.

Dear Harazi,

Thanks a lot! But one little question: You assume n > 5. Wouldn't n > 4 be enough?

  Darij

It is enough, I hurried a little bit and I said it would be better to take it reasonably large so not to have problems.

Aha, clear, thanks.

Some background: Two days ago I had to write an exam - one of the final TST exams intended to choose the 6 German IMO participants from the 16 candidates. And by a strange coincidence, one of the problems was the above (*) !!! Moreover, to that time you had already posted the solution, but I had no internet access at Bad Homburg and I couldn't read it, so I hadn't succeeded to solve the problem.

BTW, one of the contestants, Mihail Shkolnikov, solved it in the same way as you. Another one, Peter Scholze, used a slightly different way with powers of 2. (I don't remember exactly how this went.)

(*) The exam problem had a slightly different form:

Find all pairs of positive integers (n; k) such that $\left( n+1\right) ^{k}-1=n!$.

[b]EDIT:[/b] I found the topic where Harazi solved this problem before: http://www.mathlinks.ro/Forum/viewtopic.php?t=3452 .

  Darij

@harazi

There are a few things I don't understand:
With Wilson's Theorem I can proof that m has to be a prime in the equation $ \left(m - 1 \right) ! + 1 = m^k $.
However, I don't understand the following argumentation. Why you use these assumptions for m and k? What do you mean with Newtons Formula?
And please could you explain the two following contradictions?

Thanks a lot

Valentin


Pls help me solve the problem

Find all (n,x) such that \[ x(1\plus{}x)(1\plus{}x^2)\equal{}4n(n\minus{}1)\]

This may be quite an easy prob but I have got stuck...

Your equation equavalent to $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}(2n\minus{}1)^2$ or
$ (16n\minus{}8)^2\equal{}(8x^2\plus{}4x\plus{}3)^2\plus{}40x\plus{}75$. If $ x>3$, then $ 40x\plus{}75<2(8x^2\plus{}4x\plus{}3)\plus{}1$. If $ x<\minus{}2$, then $ \minus{}40x\minus{}75<2(8x^\plus{}4x\plus{}3)\minus{}1.$ Chek by mod 8 give $ 8|x$ or $ 4|x\plus{}1$. Therefore solutions are $ x\equal{}0,n\equal{}0$, $ x\equal{}0,n\equal{}1$, $ x\equal{}3,n\equal{}6$, $ x\equal{}3,n\equal{}\minus{}5$.


vector
linear algebra
matrix
linear algebra theorems

I have some very fundamental questions that I just can't seem to comprehend, which really irritate me. Help please...

Let T be the linear operator on $ R^2$ defined by:

$ T\dbinom{a}{b} \equal{} \dbinom{3a \minus{} b}{a \plus{} 3b}$,

and let $ \beta \equal{} \{(1,1),(1, \minus{} 1)\}$ and $ \beta' \equal{} \{(2,4),(3,1)\}$ be ordered bases.

What does $ [(2,4)]_{\beta}$ mean? Does it represent $ 2(1,1) \plus{} 4(1, \minus{} 1)$?
Does this mean that $ [(2,4)]_{\beta} \equal{} [(6, \minus{} 2)]_{SB}$?

Using the above knowledge, it is somehow supposed to be clear that $ [T]_{\beta}$ is $ \left( \begin{array}{ccc} 3 & 1 \\
\minus{} 1 & 3 \end{array} \right)$.

First, what does that mean? Is  $ [T]_{\beta}$  the transformation acting on R2, but only if every vector in R2 is expressed not in terms of $ \{(1,0),(0,1)\}$ but instead in terms of $ \beta$? Shouldn't T just be $ \left(\begin{array}{ccc}3 & \minus{} 1 \\
1 & 3\end{array}\right)$? Or is that the representation of T in the standard basis, since $ \left(\begin{array}{ccc}3 & \minus{} 1 \\
1 & 3\end{array}\right)\dbinom{a}{b} \equal{} \dbinom{3a \minus{} b}{a \plus{} 3b}$? How do we know that $ [T]_{\beta}$ is $ \left( \begin{array}{ccc} 3 & 1 \\
\minus{} 1 & 3 \end{array} \right)$ ?

Then what does $ [T]_{\beta}^{\beta'}$ mean? And is it nonsensical to say $ [(2,4)]_{\beta}^{\beta'}$?

Also, completely unrelated, but $ \beta\equal{}\{v_1,v_2,...,v_n\}$ and $ \gamma\equal{}\{w_1,w_2,...,2_m\}$ then $ T(v_{j})\equal{}\displaystyle\sum_{i\equal{}1}^{m}a_{ij}w_i$ for $ 1\le j\le n$. The book says that if $ A_{ij}\equal{}a_{ij}$ then A is the matrix representation of T, and that the jth column of A is simply $ [T(v_j)]_{\gamma}$. What does this mean?

I agree, this notation can be annoying and confusing.  Let me give you some examples and then you can attempt to draw the analogues and fill in the gaps:

First, let's define $ V\equal{}P_{2}(\mathbb{R})$, and let $ \beta\equal{}\{1,x,x^{2}\}$, which is the standard ordered basis.  Define our linear transformation $ f$ by $ f(x)\equal{}4\plus{}6x\minus{}7x^{2}$.  Then $ \left[f\right]_{\beta}\equal{}(4,6,\minus{}7)^{T}$ (I would normally write it vertically but I don't want to take up all the extra space and throw in extra syntax).

Now suppose $ V$ and $ W$ are finite-dimensional vector spaces with ordered bases $ \beta\equal{}\{v_{1},v_{2},\ldots,v_{n}\}$ and $ \gamma\equal{}\{w_{1},w_{2},\ldots,w_{n}\}$.  Let $ T: V\rightarrow W$ be linear.  Then for each$ j$, $ 1\leq j\leq n$, there exist unique scalars $ a_{ij}\in F$, a field, $ 1\leq \leq m$ such that $ T(v_{j})\equal{}\sum_{i\equal{}1}^{m}a_{ij}w_{i}$ for $ 1\leq j\leq n$.  Using this notation, the $ m\times n$ matrix $ A$ defined by $ A_{ij}\equal{}a_{ij}$ is called the matrix representation of $ T$ in the ordered bases $ \beta$ and $ \gamma$.  We write $ A\equal{}\left[T\right]_{\beta}^{\gamma}$ and if $ V\equal{}W$ and $ \beta\equal{}\gamma$, we can leave out the $ \gamma$.


OK, after help from here, I tried 4 more questions and thought i got the hang of this. Now i have another dilemma. :S

By expressing: tan(alpha + beta) = sin(alpha + beta) / cos(alpha + beta)

Prove that: tan(alpha + beta) = tan alpha + tan beta / 1 - tan alpha tan beta

Here are my first steps:
= sin(alpha + beta) / cos(alpha + beta)
= sin alpha cos beta + cos alpha sin beta / cos alpha cos beta - sin alpha sin beta

now im stuck,
should i cancel out the cos and sin from top and bottom?
i tried that, but that didn't work. after i thought, since i was lookin for tan in my answer, i would sin / cos = tan
that also did not work out.... (or maybe i did something wrong)
please guide me.

[hide="Previous Knowledge"]
$\boxed{\sin (\alpha+\beta ) = \sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta }$

$\boxed{\cos (\alpha+\beta ) = \cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta }$[/hide]

If $\tan (\alpha+\beta ) = \frac{{\sin (\alpha+\beta )}}{{\cos (\alpha+\beta )}}$, prove that $\tan (\alpha+\beta ) = \frac{{\tan \alpha+\tan \beta }}{{1-\tan \alpha \cdot \tan \beta }}$

[u]Proof[/u]:

\[\begin{equation*}\begin{aligned}\tan (\alpha+\beta ) &= \frac{{\sin (\alpha+\beta )}}{{\cos (\alpha+\beta )}}\hfill \\ &= \frac{{\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta }}{{\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \cos \beta }}\hfill \\ &= \frac{{\frac{{\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta }}{{\cos \alpha \cdot \cos \beta }}}}{{\frac{{\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \cos \beta }}{{\cos \alpha \cdot \cos \beta }}}}\hfill \\ &= \frac{{\frac{{\sin \alpha \cdot \cos \beta }}{{\cos \alpha \cdot \cos \beta }}+\frac{{\cos \alpha \cdot \sin \beta }}{{\cos \alpha \cdot \cos \beta }}}}{{\frac{{\cos \alpha \cdot \cos \beta }}{{\cos \alpha \cdot \cos \beta }}-\frac{{\sin \alpha \cdot \cos \beta }}{{\cos \alpha \cdot \cos \beta }}}}\hfill \\ &= \frac{{\tan \alpha+\sin \beta }}{{1-\tan \alpha \cdot \tan \beta }}\hfill \\ \end{aligned}\end{equation*}\]

$\emph{Quod Erat Demonstrandum}$


Find all points having a y-coordinate of -3 whose distance from the point (1, 2) = 13.

[quote="ckck"]Like I said in the other post do the distance formula backwards to get the point. You already have both points, just set it into an equation. The equation would be $13=\sqrt{(1-x)^{2}+25}$. I'm sure you can do the rest.[/quote]

[hide]$169 = (1-x)^{2}+25$

$144 = (1-x)^{2}$

$\pm 12 = 1-x$

$11,-13 =-x$

$x=-11, 13$

Are the answers $(-11,-3)$, $(13,-3)$?[/hide]

Good answers.

I did not know that the distance formula could be done backwards.

I also did not know that this question involved the distance formula at all.

Thanks.


algebra polynomial calculus integration trigonometry function domain

$P$ real polynomial $deg(P) \geq 1$

Prove that the system of equations

$\int_{0}^{x}P(t)\sin t dt=0$
$\int_{0}^{x}P(t)\cos t dt=0$

has a finite number of solutions

Rephrase this slightly. Define $F:\mathbb{R}\mapsto\mathbb{C}$ by

$F(x)=\int_0^xP(t)e^{it}dt.$

We're asked to show that $F(x)$ = 0 has at most finitely many solutions.

But by repeated integration by parts, we can write $F(x)=Q(x)e^{ix}+C$ for $Q$ a complex-coefficient polynomial of the same degree as $P.$

It is clear that the zeros of $F$ are confined to a bounded domain, since $\lim_{|x|\to\infty}|F(x)|=\infty.$ Since $P(x)\cos x$ and $P(x)\sin x$ each have only finitely many zeros in any bounded domain, it is clear by Rolle's Theorem that the total number of zeros of $F$ must be finite.


What year will it be 3700 days from December 1, 1995?

qball, could you plz post the solution as well the answer next time?
thanks

anyway, 3700 days is a little over 10 years, so the year would be 2005, however, there are enough days left over to bring us over to the next year

therefore, our answer is 2006

(i know it's not the most detailed i coulda given, but i'm listening to my ipod, and i'm a little distracted)

More detailed version of vallon's solution (and way more than qball's):
There are 10 full years in 3700 days, so we add 10 years to 1995 to get the year to be 2005. We have 3700-3653=47 days left over. Since the month is December, we have enough days to jump to the next year, so we subtract 30 from 47 to get 17 days left over on January 1, 2006. Then we add 17 in to get the exact date even though thats not what it asks, so the exact date is Jan 18, 2006.

Notice Dragon's good eye for detail while including the 3 leapdays that got him the 3653 days in those particular 10 years.

However 30 days after Dec 1 is Dec 31, so the date would end up being Jan 17, 2006.


linear algebra
matrix
calculus
trigonometry
trig identities

The problem says "Use the matrix of cofactors to calculate the inverse of $ A$".
$ A\equal{}\begin{pmatrix} \cos(t) & 0&\sin(t) \\ 0&1&0 \\ \sin(t)&0&\cos(t)     \end{pmatrix}$.
My attempt : I'm not sure about what they call "the matrix of cofactors".  I guess it's a $ 2\times 2$ matrix, but I don't see anything further than this.  And how could it help to find $ A^{\minus{}1}$?

You probably want to learn what the [url=http://en.wikipedia.org/wiki/Cofactor]matrix of cofactors[/url] actually is.

Yes, thank you for the link.
I'm not sure I did the calculus well (there was plenty of it), so I got that the cofactor matrix of $ A$, $ C$ is worth $ \begin{pmatrix} \cos(t)&0&\minus{}\sin(t) \\ 0& 1\minus{}2\sin^2(t)&0 \\ \minus{}\sin(t)&0&\cos(t) \end{pmatrix}$.
Furthermore on the page you gave, I've read that $ A^{\minus{}1}\equal{}\frac{1}{\det(A)}\cdot adj(A)$ where $ adj(A)$ is the adjugate matrix which is in fact the cofactor's transposed matrix.  In my case the cofactor matrix is equal to its transposed matrix thus it's equal to the adjugate.
I've calculated $ \det(A)$ to be $ 1\minus{}2\sin^2(t)$.
So finally $ A^{\minus{}1}\equal{}\frac{1}{1\minus{}2\sin^2(t)}\cdot \begin{pmatrix} \cos(t)&0&\minus{}\sin(t) \\ 0& 1\minus{}2\sin^2(t)&0 \\ \minus{}\sin(t)&0&\cos(t) \end{pmatrix}$.  It remains just to multiply the entries of the matrix by $ \frac{1}{1\minus{}2\sin^2(t)}$.
Did I make an error?

Small adjustment: this is the specific forum that is best for this problem, and those like it.

[quote]I'm not sure I did the calculus well (there was plenty of it)[/quote]
That wasn't calculus - that was algebra. Trig identities, to be specific.

As for wondering whether you got it right ... this is one thing that's a lot easier to check than it was to find. You have $ A$ given, and you have a proposed $ A^{\minus{}1},$ if your computations are correct. So just do the matrix multiplication needed to compute $ AA^{\minus{}1}$ (or $ A^{1}A$ - whichever one is more convenient or whichever one you think of first, it doesn't really matter.). If you get $ I$, then either you're correct or you somehow coincidentally made matching errors in both the computation or the check, with sounds pretty unlikely. If you don't get $ I,$ then something is wrong and you should stay with the problem until you root out the error.

(It does check, this time.)

[quote]That wasn't calculus - that was algebra. Trig identities, to be specific.[/quote] Sorry, my error.  English is not my mother tongue and it would fit well in French or in Spanish.   :roll:
And thank you very much for the explanations, I could do $ AA^{\minus{}1}$ and it gave me the identity.  Is that what you said by "(It does check, this time.)"?
And ok for the forum.


induction
combinatorics unsolved
combinatorics

We arrange  n boys and n girls to 1 rows ,then find how to split it to 2 parts such that in each part, number of boys are same as girls.Let A be number of cases which can't split.Let B be number of cases which have only one way to split.Prove B=2.A

[quote="JBL"]This is not your personal help-line, so be patient.

It may be helpful to introduce a value $ C_n$ equal to the number of ways to line up $ n$ girls and $ n$ boys such that the number of girls among the first $ k$ people is at least as large as the number of boys, for any $ 0 \leq k \leq 2n$.  Then it's easy to show that $ A \equal{} 2C_{n \minus{} 1}$ and $ B \equal{} \sum_{k \equal{} 1}^{n \minus{} 1} (2C_{k \minus{} 1}) \cdot (2 C_{n \minus{} k \minus{} 1})$.  Now the stated claim is reduced to a claim about the numbers $ C_n$, which is actually well-known (and is quite similar to the equalities I've outlined above).

I'm quite certain I've seen this problem on the forum before, but I wasn't able to find it with a search.  Some key words are Dyck path, ballot sequence, and Catalan number.[/quote]

Oh,i 'll calm down.But I think we can't solve by induction or difficult to solve above equation.
Can you do any things else?

[quote="math man"]But I think we can't solve by induction or difficult to solve above equation.[/quote]  I don't know what this means.  (Because you quoted my entire post, I don't know what part of it you're referring to.)  I've sketched an approach that reduces the given problem to a problem involving the numbers $ C_n$.  And I've noted that to solve the problem involving the $ C_n$ uses exactly the same sorts of ideas that are used in the reduction.  So, I've outlined (but not provided) a viable solution method.  Which part is unclear to you?


em là thành viên mới của mathlinks.Em dốt tiếng Anh lắm,vì vậy nhờ các anh chỉ bảo giùm em gõ công thức toán học ở đâu đc ko ạ?(nhờ các anh cho sẵn link luôn)Có thể vẽ đc hình ko ạ?

http://www.mathlinks.ro/LaTeX/AoPS_L_About.php . Trước đây nó có hỗ trợ vẽ hình nhưng bây giờ không thì phải.  :roll:


linear algebra
matrix
function
vector
calculus
linear algebra unsolved

Let A be a nxn matrix with elements $a_{ij}$ such that
$a_{ij}+a_{ji}=0$ if $i \ne j$ and 1 if i=j. Let functions
$y_{1}(t),y_{2}(t),...,y_{n}(t)$ be the solutions of differential
equation system $\left( \begin{array}{c}y'_{1}\\ \vdots \\ y'_{n}\\ \end{array}\right) = \begin{pmatrix}a_{11}& \cdots & a_{1n}\\ \vdots && \vdots \\ a_{n1}& \cdots & a_{nn}\end{pmatrix}\left(\begin{array}{c}y_{1}\\ \vdots \\ y_{n}\\ \end{array}\right)$

Find $\sum_{i=1}^{n}y_{i}^{2}(t)$ if $y_{i}(0)$ are given. (2points)

I assume that the matrix and the vectors are real. It's simpler to think about that way.

Write $y=y(t)$ as the column vector. It satisfies the differential equation $y'=Ay.$ The given condition on $A$ is that $A^{T}+A=I.$

We're asked to compute $\sum_{i=1}^{n}y_{i}=y^{T}y.$

$(y^{T}y)'=(y')^{T}y+y^{T}y'$

$=(Ay)^{T}y+y^{T}(Ay)=y^{T}A^{T}y+y^{T}Ay=y^{T}(A^{T}+A)y$

$=y^{T}Iy=y^{T}y.$

Thus if $u=y^{T}y,$ then $u'=u,$ hence $u(t)=u(0)e^{t}.$ That's our answer:

$\sum_{i=1}^{n}y_{i}(t)^{2}=e^{t}\sum_{i=1}^{n}y_{i}(0)^{2}.$

In fact the first-order ODE $y'=Ay$ has unique solution $y(t)=e^{tA}y_{0}$. where $y(0)=y_{0}$ so $\parallel y(t)\parallel ^{2}=\parallel e^{tA}\parallel ^{2}\cdot\parallel y_{0}\parallel ^{2}=e^{t}\cdot \parallel y_{0}\parallel ^{2}$ (because $A$ is anti-symetry, $|e^{tA}|=|e^{tA^{T}}|$ and $e^{tI}=e^{t}I$)

I started with goldhp's argument, before I switched to the more elementary argument that I posted, and there's a reason I abandoned it. I question one step: why can you say that $\|e^{tA}y_{0}\|^{2}=\|e^{tA}\|^{2}\cdot \|y_{0}\|^{2}\,?$ I don't see that. In fact,

$\|y(t)\|^{2}=\|e^{tA}y_{0}\|^{2}=y_{0}^{T}e^{tA^{T}}e^{tA}y_{0}$

I got that far and paused - do we know that $A$ commutes with $A^{T}$ ? No, we don't. And without that, we can't claim that $e^{tA^{T}}e^{tA}=e^{t(A^{T}+A)}.$

Is there a way to save this argument? I'm not seeing it.

[color=green][This problem involves both calculus and matrices, so it could plausibly be posted in more than one place. At this point, I'm seeing the difficulties with it lying primarily with the linear algebra side of the problem, hence I'm moving it to that forum.][/color]

Kent, but $A$ commutes with $I-A$ and since the traspose of $A$ is $I-A$...

Ah, yes - that does it. Then $e^{tA^{T}}e^{tA}=e^{t(A^{T}+A)}=e^{t}I$ and goldhp's argument works.

Really, I used the fact which harazi showed :D


$ a,b,c\in\mathbb{R}^{\plus{}}$,and $ abc\equal{}1$.
Try to prove that $ (\frac{a\plus{}b\plus{}c}{3})^2\ge \frac{(1\plus{}\frac{1}{a})(1\plus{}\frac{1}{b})(1\plus{}\frac{1}{c})}{8}$.

[quote="skywalkerJ.L."]$ a,b,c\in\mathbb{R}^{ \plus{} }$,and $ abc \equal{} 1$.
Try to prove that $ (\frac {a \plus{} b \plus{} c}{3})^2\ge \frac {(1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{b})(1 \plus{} \frac {1}{c})}{8}$.[/quote]

$ \frac {(a \plus{} b \plus{} c)^2}{9} \ge \frac {a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca \plus{} 2}{8} \Leftrightarrow 8(a^2 \plus{} b^2 \plus{} c^2) \plus{} 16(ab \plus{} bc \plus{} ca) \ge 9(a \plus{} b \plus{} c) \plus{} 9(ab \plus{} bc \plus{} ca) \plus{} 18 \Leftrightarrow 8(a^2 \plus{} b^2 \plus{} c^2) \plus{} 7(ab \plus{} bc \plus{} ca) \ge 9(a \plus{} b \plus{} c) \plus{} 18 \Leftrightarrow 3(a \plus{} b \plus{} c)^2 \plus{} 5(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca \ge 9(a \plus{} b \plus{} c) \plus{} 18$

But $ 3(a \plus{} b \plus{} c)^2 \ge 9(a \plus{} b \plus{} c)$ because $ a \plus{} b \plus{} c \ge 3\sqrt [3]{abc} \equal{} 3$
And $ 5(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca \ge 5 \cdot 3 \sqrt [3]{a^2b^2c^2} \plus{} 3 \sqrt [3]{a^2b^2c^2} \equal{} 18$ :wink: