Challenging Math Problems for the Curious Mind

[benorin]

Would someone please post a good problem (or at least an interesting one) for me to work on, say calculus, real/complex analysis, or some generating function stuff, or some problem anybody can understand but will scratch their head at? I really bored. Thanks for cherring me up,

-Ben

 

[0rthodontist]

OK-

If P is a polygon, prove that it cannot be the union of disjoint convex quadrilaterals, each of which has exactly one face that is also a face of P.

 

[Robokapp]

A particle is moving on the line y=x^3 in the first quadrant starting from Origin at a speed dy/dx=1 unit per...time unit. ignore units. :p

Question is...find a formula for the angle formed by the x-axis and a line created by the Origin and the particle's position.

In all honesty I saw it before but I never took the time to solve it. I doubt i can however ... :(

 

[Hurkyl]

Find the sum:

<br /> \sum_{i = 0}^{+\infty}<br /> \tan^{-1} \frac{1}{1 + x + x^2}<br />

(I think I have that right)

 

[WigneRacah]

Find the sum:

<br /> \sum_{i = 0}^{+\infty}<br /> \tan^{-1} \frac{1}{1 + x + x^2}<br />

(I think I have that right)

Quite easy: it is a diverging series ...

 

[benorin]

Hurkyl, do you mean \sum_{x = 0}^{+\infty}\tan^{-1} \frac{1}{1 + x + x^2} ?

 

[Hurkyl]

Yes, that one looks better!

 

[benorin]

Hurkyl,

\sum_{x = 0}^{+\infty}\tan^{-1} \frac{1}{1 + x + x^2} = \sum_{x = 0}^{+\infty}\tan^{-1} \frac{(x+1)-x}{1 + (x+1)x} = \sum_{x = 0}^{+\infty}\left( \tan^{-1}(x+1)-\tan^{-1}x\right) = \lim_{M\rightarrow\infty} \tan^{-1}(M+1)-\tan^{-1}(0)= \frac{\pi}{2}

Thanks,
-Ben

 

[WigneRacah]

Try with this.

Find the product

\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right).

 

[DavidBektas]

Well, I have a problem which currently bugs me (although I think I already solved it). Find the solution to the differential equation:

dv/dt = a*v+b*v^2.

It represents the movement of a particle with a velocity dependent friction force (which is proportional to v for small v and proportional to v^2 for large v). So the solution must be equal to the stokes case for small values of v and equal to the Newton case for large values of v, that´s how you can check if your answer is correct.

I made the problem up myself, so it might not be physically correct. But at least it´s a solvable DE which you can make into a linear first order ODE by substitution. Sorry for my english.

 

[benorin]

Try with this.

Find the product

\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right).

Since \frac{\sin \pi x}{\pi x}=\prod_{k=1}^{\infty} \left( 1-\frac{x^2}{k^2}\right) = \prod_{k=1}^{\infty} \frac{k^2-x^2}{k^2} it follows that

\frac{y}{x}\frac{\sin \pi x}{\sin \pi y}= \prod_{k=1}^{\infty} \frac{k^2-x^2}{k^2-y^2}

\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right) = \prod_{k=1}^{\infty} \frac{k^2+9}{k^2 + 7} = \frac{\sqrt{7}}{3}\frac{\sin 3\pi i}{\sin \sqrt{7}\pi i}​

recall that \sin iz = i\mbox{sinh}z to get the final value, namely

\boxed{\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right) = \frac{\sqrt{7}}{3}\frac{\mbox{sinh} 3\pi}{\mbox{sinh} \sqrt{7}\pi } = \frac{\sqrt{7}}{3}\frac{e^{3\pi}-e^{-3\pi} }{e^{\sqrt{7}\pi}-e^{-\sqrt{7}\pi} }}​

 

[lo2]

Wow! Are you studying math at the uni? Or how come you can solve all that?

 

[matt grime]

1. Prove from first principles that exp(x) is indeed (1+x/s)^s as s tends to infinity.

2. If M is a matrix over the complex numbers and Tr(M^r)=0 for all r show that all eigenvalues of M are zero.

3. If f is a function from C to C and the integral of f round any triangle is zero show that f is analytic/holomorphic.

4. What is the genus of the Riemann surface corresponding to w=sqrt((z-1)(z-2)(z-3)..(z-n))

5. Prove, using homology groups, the fundamental theorem of algebra. Hint, this is actually a fixed point theorem.

(Note: one of these is tricky, one of these probably requires more material than you've yet learnt.)

 

[benorin]

1. Prove from first principles that exp(x) is indeed (1+x/s)^s as s tends to infinity.

I'll do the easy one (it was homework in grad real analysis with Papa Rudin).

I assume that we have the definition e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}.


Let f_{s}(x)=\left(1+\frac{x}{s}\right) ^s. Define (a)_{k}:=a(a-1)\cdots (a-k+1) and (a)_{0}=1 (that is put (a)_{k}= \frac{\Gamma (a+1)}{\Gamma (a-k+1)} for k\in\mathbb{N}. Now notice that

f_{s}^{(k)}(x)=\frac{(s)_{k}}{s^k}\left(1+\frac{x}{s}\right) ^{s-k}, for k\in\mathbb{N}​

hence

f_{s}^{(k)}(0)=\frac{(s)_{k}}{s^k} for k\in\mathbb{N}​

so that we have the MacClaurin Series for f_s(x) as being

f_{s}(x)=\sum_{k=0}^{\infty}\frac{(s)_{k}}{k!}\left( \frac{x}{s}\right) ^{k}​

Consider the quantity

\left|f_{s}(x)-e^x\right| = \left|\sum_{k=0}^{\infty}\frac{(s)_{k}}{k!}\left( \frac{x}{s}\right) ^{k}-\sum_{k=0}^{\infty}\frac{x^k}{k!} \right|= \left|\sum_{k=0}^{\infty} \frac{x^k}{k!}\left(\frac{(s)_{k}}{s^k}-1\right)\right|
\leq \sum_{k=0}^{\infty} \frac{|x| ^k}{k!}\left|\frac{(s)_{k}}{s^k}-1\right|​

and note that (s)_{k}=s(s-1)\cdots (s-k+1) \sim s^k\mbox{ as }s\rightarrow\infty so that we have \left|f_{s}(x)-e^x\right|\rightarrow 0, \mbox{ as }s\rightarrow\infty.

 

[teclo]

derive a variation for the nambu-goto action!

 

[benorin]

BTW, Hurkyl, this problem was PUTNAM 1986/A-3.

Hurkyl,

\sum_{x = 0}^{+\infty}\tan^{-1} \frac{1}{1 + x + x^2} = \sum_{x = 0}^{+\infty}\tan^{-1} \frac{(x+1)-x}{1 + (x+1)x} = \sum_{x = 0}^{+\infty}\left( \tan^{-1}(x+1)-\tan^{-1}x\right) = \lim_{M\rightarrow\infty} \tan^{-1}(M+1)-\tan^{-1}(0)= \frac{\pi}{2}

Thanks,
-Ben

 

[Hurkyl]

Oh, I didn't know that!

 

[Noesis]

How much wood could a woodchuck chuck if a woodchuck could chuck wood?

Answer in terms of the variables x, y, and photons.

Then the cubed root of it.

This one's been bothering me personally.

Thanks.

 

[Noesis]

Ah, double posted my question.

 

[tehno]

Would someone please post a good problem (or at least an interesting one) for me to work on, say calculus, real/complex analysis, or some generating function stuff, or some problem anybody can understand but will scratch their head at? I really bored. Thanks for cherring me up,

-Ben

How about this one:

https://www.physicsforums.com/showthread.php?t=150136



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