# Master's Challenge

1. Dec 24, 2009

### FeDeX_LaTeX

Prove that this is the equation of a straight line;

$$-\frac{\zeta(x)\zeta(2)y\Gamma(x)}{\int_0^\infty \frac{u^{x - 1}}{e^u - 1} du \int_0^1 \frac{\ln(1 - x)}{x} dx} - \nabla \cdot (\nabla \times \vec{A}) =$$
$$\frac{L^{-1}[\frac{F(t)}{t}] \int_{k = 0}^1 \int_{\theta = 0}^{\pi /2} \frac{d\theta dk}{\sqrt{1 - k^2 \sin^2 \theta}} \prod_{n = 0}^\infty \left(m + \frac{4mx^2}{[\pi(2n + 1)]^2}\right) T_n \cos \theta \sum_{n = 1}^\infty \frac{1}{n^n}}{\int_0^1 \frac{dx}{x^x} \cosh x \frac{\cos n\theta}{2\pi ix} \int_{c - i\infty}^{c + i\infty} e^{st}\int_s^\infty f(u) du ds \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n + 1)^2}}$$
$$+ \frac{\oint_C \frac{dz}{z - a} \frac{\pi}{x \sinh \pi x} \sum_{n = 1}^\infty \frac{2^{n/2} \sin(n\pi/4) x^n}{n!}}{|\Gamma(ix)|^2 \frac{2\pi i}{b} e^x \sin x} + 2 + \sum_{i,j} [\sigma_0^m, \sigma_i^{m - 1}] \cdot [\sigma_i^{m - 1}, \sigma_j^{m - 2}] + \int_{\partial S} \omega - \int_S d\omega - \int_0^1 \ln x \ln(1 + x) dx - \eta(2) - 2\eta(1)$$,

where ζ is Riemann's zeta function, Γ(x) is the gamma function, ∇ is the del operator, L-1 denotes the inverse Laplace transform, Tn is the nth Chebyshev polynomial of the first kind, C is a simple closed curve bounding a region having z = a as an interior point, σ0m is a simplex of an oriented simplicial complex and [σm, σm - 1] is an incidence number, S is a compact, orientable, differentiable k-dimensional manifold with boundary in En and ω is a (k - 1)-form in En, defined, and C1 at all points of S, and η(x) is Dirichlet's eta function.

[ Credit to Zhylliolom. ]

EASIER CHALLENGE:

$$\frac{y(e^{\pi i})}{(acos^2(\int_{}^{}e^{x^2})+asin^2(\int_{}^{}e^{x^2}))b} = \frac{x(z+1)(z-1)}{b(z^2-1)} + \frac{gcd(p, p-1)}{a}$$

Where p is a prime number.

---

Last edited: Dec 24, 2009
2. Dec 24, 2009

### Mentallic

um... wow?

I always knew straight lines were a tricky subject...

3. Dec 24, 2009

### Char. Limit

I'll take one (and only one) step... subtract 2 from both sides!

4. Dec 24, 2009

### Mentallic

Don't you want to at least set the equation to zero?!? :tongue2:
But you have a point there... that little 2 seems kind of out of place.

Oh and just wondering, but is this a real problem? Even though this maths is way out of my league, I usually can sense when something isn't worth solving.
Seriously... all that to prove it's a line? Would it even be a "omg, that's really cool" moment if it were solved?

5. Dec 24, 2009

### FeDeX_LaTeX

Prove that this equation is identical to y = mx + c

It's actually surprisingly easy once you solve the first part - you'll notice that it isn't that complex at all.

Try the easier challenge at the bottom if you're having trouble.

6. Dec 24, 2009

### Char. Limit

OK, I'm pretty sure that even I can solve the easier challenge.

$$\frac{y(e^{\pi i})}{(acos^2(\int_{}^{}e^{x^2})+asin^2(\int_{}^{}e ^{x^2}))b} = \frac{x(z+1)(z-1)}{b(z^2-1)} + \frac{gcd(p, p-1)}{a}$$

First, let's work on the left side. Since $$e^{\pi i}=-1$$, the equation simplifies to:

$$-\frac{y}{(acos^2(\int_{}^{}e^{x^2})+asin^2(\int_{}^{}e ^{x^2}))b} = \frac{x(z+1)(z-1)}{b(z^2-1)} + \frac{gcd(p, p-1)}{a}$$

Then, factoring out the a and realizing that the cosine squared plus the sine squared of any number or function is 1 (I assume it is for that integral, at least)... we get this.

$$-\frac{y}{a b} = \frac{x(z+1)(z-1)}{b(z^2-1)} + \frac{gcd(p, p-1)}{a}$$

OK, the left side is probably as simple as we can get it. Now to the right side. First, I note that $$(z+1)(z-1)=(z^2-1)$$, thus changing the equation to:

$$-\frac{y}{a b} = \frac{x}{b} + \frac{gcd(p, p-1)}{a}$$

Next, if I remember right, the gcd of any prime number with any other number is 1, so...

$$-\frac{y}{a b} = \frac{x}{b} + \frac{1}{a}$$

Now, let's multiply x/b by a, and 1/a by b.

$$-\frac{y}{a b} = \frac{a x}{a b} + \frac{b}{a b}$$

Now, finally, we multiply all sides by a*b to get...

$$-y=ax+b$$

Which is the equation of a line.

Did I do good?

7. Dec 24, 2009

### jgens

This isn't quite right. For example, consider $\gcd{(7,14)}$ which, while $7$ is prime, is clearly equal to $7$. Perhaps what you meant is that $\gcd{(p,q)} = 1$ whenever $p$ is prime and $q/p \notin \mathbb{N}$. Either way, the result in this particular instance is the same.

8. Dec 24, 2009

### Mentallic

But the problem says $gcd(p,p-1)$ so if you had 7 as your prime, the other should be 6, not 14.
Char.limit just mixed up his words a bit, but he's still right.

And yes, you did good

9. Dec 24, 2009

### jgens

I understand what the problem says; my point was that Char. Limit's statement was incorrect. The $\gcd$ of two numbers where one of them is prime is not necesarrily $1$ as he/she stated during the proof. While it does not matter in this particular problem, it is of crucial importance of other problems.

10. Dec 24, 2009

### Mentallic

Which is why I believe he just wasn't thinking when he said that, considering how wrong his statement obviously is.

Anyway, who will take a shot at the "master's challenge"?

11. Dec 25, 2009

### Char. Limit

Sorry about that, I meant to write "the other number", but I was thinking about possibilities in the gcd and wrote "any" instead.

"The other number" is a vague thing anyway, and better to be obviously wrong than vague.

I'd gladly take up the challenge if I knew how to do anything to that other than subtract a 2.

12. Dec 25, 2009

### Mentallic

Same here. I can't even tell where the x and y is... if there's meant to be an x and y that is.

13. Dec 25, 2009

### Mute

The y is on the LHS, sandwiched between zeta(2) and Gamma(x). The "challenge" doesn't seem to require any mastery of anything other than a recognition of (or ability to look up) various identities which have been entered into the formula either as disguised 1's or 0's. For instance, the LHS of the equation reduces to y after noting that the divergence of a curl is zero, and

$$\Gamma(x)\zeta(x) = \int_0^\infty du \frac{u^{x-1}}{e^u - 1};$$

the $-\zeta(2)$ is cancelled by the integral in the denominator using the above form and a change of variables.

The right hand side presumably falls to a similar sort of identity application; I would guess the last two terms are zero and the first two fractions probably almost entirely cancel. I don't see a clear independent variable on the RHS, though. Anywho, I'll let the interested figure the rest of it out.

Last edited: Dec 25, 2009
14. Dec 25, 2009

### fourier jr

i noticed that too. there are lots of definite integrals that don't even have to be recognised as anything, and can be lumped together into one big constant... i guess the problem is to find a linear x term

15. Dec 25, 2009

### ice109

dumbest problem i've ever seen

16. Dec 25, 2009

### Mentallic

Agreed.

This question doesn't mean anything and it's merely there to make things unnecessarily complicated. It is but a simplication problem, for those that already know how to simplify...