Challenging Math Problems for the Curious Mind

  • Thread starter benorin
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In summary: The maximum amount of wood a woodchuck could chuck is impossible to determine, as woodchucks do not have the ability to chuck wood. The phrase "how much wood could a woodchuck chuck" is simply a tongue-twister and does not have a mathematical answer. Therefore, it is not possible to express it in terms of variables or find the cubed root of it.
  • #1
benorin
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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
 
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  • #2
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.
 
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  • #3
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 ... :(
 
  • #4
Find the sum:

[tex]
\sum_{i = 0}^{+\infty}
\tan^{-1} \frac{1}{1 + x + x^2}
[/tex]

(I think I have that right)
 
  • #5
Hurkyl said:
Find the sum:

[tex]
\sum_{i = 0}^{+\infty}
\tan^{-1} \frac{1}{1 + x + x^2}
[/tex]

(I think I have that right)

Quite easy: it is a diverging series ...
:rolleyes:
 
  • #6
Hurkyl, do you mean [tex]\sum_{x = 0}^{+\infty}\tan^{-1} \frac{1}{1 + x + x^2}[/tex] ?
 
  • #7
Yes, that one looks better!
 
  • #8
Hurkyl,

[tex]\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}[/tex]

Thanks,
-Ben
 
  • #9
Try with this.

Find the product

[tex]\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right)[/tex].
 
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  • #10
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.
 
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  • #11
WigneRacah said:
Try with this.

Find the product

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

Since [tex]\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}[/tex] it follows that

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

[tex]\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}[/tex]​

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

[tex]\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} }}[/tex]​
 
  • #12
Wow! Are you studying math at the uni? Or how come you can solve all that?
 
  • #13
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.)
 
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  • #14
matt grime said:
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 [tex]e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/tex].


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

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

hence

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

so that we have the MacClaurin Series for [tex]f_s(x)[/tex] as being

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

Consider the quantity

[tex]\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| [/tex]
[tex] \leq \sum_{k=0}^{\infty} \frac{|x| ^k}{k!}\left|\frac{(s)_{k}}{s^k}-1\right|[/tex]​

and note that [tex](s)_{k}=s(s-1)\cdots (s-k+1) \sim s^k\mbox{ as }s\rightarrow\infty[/tex] so that we have [tex]\left|f_{s}(x)-e^x\right|\rightarrow 0, \mbox{ as }s\rightarrow\infty[/tex].
 
  • #15
derive a variation for the nambu-goto action!
 
  • #16
BTW, Hurkyl, this problem was PUTNAM 1986/A-3.

benorin said:
Hurkyl,

[tex]\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}[/tex]

Thanks,
-Ben
 
  • #17
Oh, I didn't know that!
 
  • #18
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.
 
  • #19
Ah, double posted my question.
 
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  • #20
benorin said:
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


:smile:
?
 

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