/sci/ - Science & Math

[Anonymous No.5298284]

After the first challenge got ruined by some ass holes, I decided to make another.
Prove that "<span class="math">\displaystyle\cot(x)=\frac{1}{x}-\sum\limits_{k=1}^\infty\left(\frac{2x}{(\pi \cdot k)^2-x^2)}\right)[/spoiler]" is true.

The problem requires very basic Math but is very challenging nonetheless. I proved the formula by accident but doing on purpose would be much harder.

Since it is hard, I will keep on adding hints and they will all be as a reply to the OP.

1st Hint: You'll need Euler's infinite product for the Sine function.

[Anonymous No.5298309]

why not use the McLaurin series?

[Anonymous No.5298320]

>>5298309
Use it. It won't work though. The Taylor Series of cot(x) at any point will never give that formula but you can always try to convince yourself.

[Anonymous No.5298358]

Hint 2 : Natural logarithm.

[Anonymous No.5298388]

Hey OP. I just looked at your last thread. How do you know those two don't differ by a constant? It looks like you only showed that their derivatives are equal. Still clever though.

[Anonymous No.5298420]

>>5298388
I thought about that and the reason is that there are no constant terms. The sum starts with 1/(1*x^1). If it was to start with 1 for example, then it would be a different story.

[Anonymous No.5298426]

>>5298420
Why does that matter?

[Anonymous No.5298431]

>>5298426
It matters because there are no constant terms that got canceled by the derivative because there are no constant terms.

[Anonymous No.5298458]

>>5298431
That doesn't mean anything. You're dealing with an infinite series on one side, and a named function on the other side. They aren't simple polynomials. If I were to define <span class="math">g(x) = \ln(x)+1<span class="math"> and replace your ln with g on the right side, your argument would prove that it was equal to g(x/(x-1)) as well.[/spoiler][/spoiler]

[Anonymous No.5298461]

>>5298431
That doesn't mean anything. You're dealing with an infinite series on one side, and a named function on the other side. They aren't simple polynomials. If I were to define g(x) = \ln(x)+1 and replace your ln with g on the right side, your argument would prove that it was equal to g(x/(x-1)) as well.

[Anonymous No.5298466]

>>5298431
That doesn't mean anything. You're dealing with an infinite series on one side, and a named function on the other side. They aren't simple polynomials. If I were to define <span class="math">g(x) = \ln(x)+1[/spoiler] and replace your ln with g on the right side, your argument would prove that it was equal to <span class="math">g(x/(x-1))[/spoiler] as well.

[Anonymous No.5298469]

<span class="quote deadlink">>>5298461[/spoiler]
I don't understand what you don't understand.
Is it the assumption that
<span class="math">\displaystyle Sum=\int \frac{d}{dx}\left(Sum\ right) dx[/spoiler]

[Anonymous No.5298473]

>>5298466

I don't understand what you don't understand.
Is it the assumption that
<span class="math">\displaystyle Sum=\int \frac{d}{dx}\left(Sum \right) dx[/spoiler]

[Anonymous No.5298478]

>>5298473
So first you need to prove that not having a constant term means that that holds for not just Taylor series, but infinite Laurent series in general, which may be true but is certainly non-trivial (if you know this is true, please provide a link)
But also remember that
<div class="math">\int \frac{1}{x} {dx} = \ln(x)</div> is wrong. You only have
<div class="math">\int \frac{1}{x} {dx} = \ln(x) + C</div>

[Anonymous No.5298488]

>>5298478
Actually, I guess the constant cancels out because you integrate it twice. So it's just the first part then.

[Anonymous No.5298505]

>>5298478
What you're missing is that the original sum has no constant term.
To show it to you in another way, we have
<span class="math">\displaystyle\sum\limits_{k=1}^{\infty}\left(\frac{1}{k \cdot x^k}\right)=\ln\left(\frac{x}{x-1}\right)+C[/spoiler]
We take the limits of both sides as x approaches infinity and we get.
<span class="math">\displaystyle\lim_{x\to\infty}\left(\sum\limits_{k=1}^{\infty}\left(\frac{1}{k \cdot x^k}\right)\right)=\lim_{x\to\infty}\left(\ln\left(\frac{x}{x-1}\right)\right)+C[/spoiler]

We evaluate the limits to get
<span class="math">0=0+C[/spoiler]

Or

<span class="math">C=0[/spoiler]

And that's why I didn't include any constant.

[Anonymous No.5298517]

>>5298505
>What you're missing is that the original sum has no constant term.
I know. You're just restating the same thing over again without improving your argument. Your other argument works with a tiny bit of elaboration though, so I'll accept that one (specifically, noting that the left hand is less than a geometric series so you can apply the squeeze theorem)