# Solving Master's Challenge: Prove Equation of a Straight Line

• FeDeX_LaTeX
In summary: I'll see what I can do.In summary, the conversation discusses an equation that is claimed to be the equation of a straight line. The conversation then moves on to an easier equation, where the left side is simplified to -y = ax + b, which is the equation of a line. The conversation then moves on to a "master's challenge" where the equation is more complex, involving prime numbers and various functions. The challenge is eventually solved to prove that the original equation is identical to y = mx + c.
FeDeX_LaTeX
Gold Member
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:
um... wow?

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

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

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?

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.

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?

Char. Limit said:
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 $\gcd{(7,14)}$ 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.

jgens said:
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.

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

Mentallic said:
But the problem says $gcd(p,p-1)$

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.

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"?

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.

Char. Limit said:
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.

Mentallic said:
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

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

the $-\zeta(2)$ is canceled 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:
Mute said:
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

dumbest problem I've ever seen

ice109 said:
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...

## 1. What is the equation of a straight line?

The equation of a straight line is typically written in the form y = mx + b, where m represents the slope of the line and b represents the y-intercept, or the point where the line intersects the y-axis.

## 2. How do you prove the equation of a straight line?

To prove the equation of a straight line, you must have at least two points on the line. Then, you can use the point-slope formula, y - y1 = m(x - x1), where m is the slope and (x1, y1) is one of the given points, to find the equation of the line.

## 3. What is the point-slope formula?

The point-slope formula is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line. This formula is used to find the equation of a line when given the slope and a point on the line.

## 4. What is the slope of a straight line?

The slope of a straight line is a measure of its steepness and is calculated by taking the change in the y-coordinates divided by the change in the x-coordinates between two points on the line. It is represented by the symbol m in the equation y = mx + b.

## 5. How is the y-intercept of a line represented in an equation?

The y-intercept of a line is represented by the constant b in the equation y = mx + b. It is the value of y when x is equal to 0, or the point where the line crosses the y-axis.

• General Math
Replies
4
Views
1K
• General Math
Replies
2
Views
836
• General Math
Replies
3
Views
1K
• General Math
Replies
1
Views
717
• General Math
Replies
1
Views
985
• General Math
Replies
2
Views
1K
• General Math
Replies
14
Views
1K
• General Math
Replies
4
Views
949
• General Math
Replies
2
Views
1K
• General Math
Replies
4
Views
1K