Master's Challenge

**FeDeX_LaTeX** · Dec24-09, 02:23 AM

Prove that this is the equation of a straight line;

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

where ζ is Riemann's zeta function, Γ(x) is the gamma function, ∇ is the del operator, L^-1 denotes the inverse Laplace transform, T_n 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, σ₀^m is a simplex of an oriented simplicial complex and [σ^m, σ^{m - 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:

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

Where p is a prime number.

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**Mentallic** · Dec24-09, 07:00 AM

um... wow?

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

**Char. Limit** · Dec24-09, 07:14 AM

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

**Mentallic** · Dec24-09, 07:27 AM

Don't you want to at least set the equation to zero?!?

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?

**FeDeX_LaTeX** · Dec24-09, 08:45 AM

Sorry, worded badly.

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.

**Char. Limit** · Dec24-09, 05:29 PM

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

Start with the equation:

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

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

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

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.

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

OK, the left side is probably as simple as we can get it. Now to the right side. First, I note that

, thus changing the equation to:

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

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

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

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

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

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

Which is the equation of a line.

Did I do good?

**jgens** · Dec24-09, 08:54 PM

Originally Posted by Char. Limit

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

This isn't quite right. For example, consider $LaTeX Code: \\gcd{(7,14)}$ which, while

is prime, is clearly equal to

. Perhaps what you meant is that $LaTeX Code: \\gcd{(p,q)} = 1$ whenever

is prime and $LaTeX Code: q/p \\notin \\mathbb{N}$ . Either way, the result in this particular instance is the same.

**Mentallic** · Dec24-09, 09:05 PM

Originally Posted by jgens

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

But the problem says

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

**jgens** · Dec24-09, 09:31 PM

Originally Posted by Mentallic

But the problem says

I understand what the problem says; my point was that Char. Limit's statement was incorrect. The $LaTeX Code: \\gcd$ of two numbers where one of them is prime is not necesarrily

as he/she stated during the proof. While it does not matter in this particular problem, it is of crucial importance of other problems.

**Mentallic** · Dec24-09, 09:54 PM

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"?

**Char. Limit** · Dec25-09, 12:12 AM

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.

**Mentallic** · Dec25-09, 12:39 AM

Originally Posted by Char. Limit

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

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.

**Mute** · Dec25-09, 01:19 AM

Originally Posted by 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.

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

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

the $LaTeX Code: -\\zeta(2)$ 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.

**fourier jr** · Dec25-09, 01:38 PM

Originally Posted by 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.

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

~~ice109~~ · Dec25-09, 06:53 PM

dumbest problem i've ever seen

**Mentallic** · Dec25-09, 08:02 PM

Originally Posted by ice109

dumbest problem i've ever seen

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...