*solves your equation with ease*
*solves your equation with ease*
Xavier Adams
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Chase Allen
* Invents free Energy *
Nicholas Murphy
Prove [math]\sum_{n\ \geqslant\ 1} \frac{\sin\left(2\, n\right)}{1 \,+\, \left(\cos\,n\right)^4}[/math] is convergent.
Brayden Parker
*never accomplishes anything past 19*
Luke Phillips
what's the motivation for this problem?
Mason Diaz
Making the difference between legit geniuses and autists spoonfed by their parents.
Michael Richardson
Int test [math]\sum\limits_{n\geq 1}\frac{\sin(2x)}{1+\cos(x)^4}\Rightarrow \int\limits_{0}^{\infty}\frac{\sin(2x)\,\mathrm{d}x}{1+\cos(x)^4}=\lim\limits_{t\to\infty} \int\limits_{0}^{t}\frac{2\sin(x)\cos(x)\,\mathrm{d}x}{1+\cos(x)^4}\quad \mathscr{Q.E.D.}[/math]
Ian Murphy
What a qt
Daniel Sullivan
>[math]\int[/math]
Aiden Rivera
Joshua Cox
its pretty fucking obvious that this series doesn't converge
for a series to converge, the absolute value of the term must approach 0 as n approaches infinity, but this obviously doesn't happen.
2*n arbitrarily approximates pi/2 + 2*pi*m for sufficiently large pairs of n and m. Take the expression 2*n - 2*pi*m. This expression can approximate pi/2 by applying the steps of the Euclidean algorithm, except, instead of using the standard termination condition, you terminate when the deviation from pi/2 is smaller than epsilon. The algorithm constructs the appropriate n and m when the termination condition is met, and the algorithm will always terminate for any arbitrarily small epsilon. So, we can always find an integer value of n such that 2*n is arbitrarily close to pi/2 + 2*pi*m, so sin(2*n) can be made arbitrarily close to 1 for infinitely many n. The denominator is bounded between 1 and 2, so, the general term does not approach 0 as n approaches infinity, so the series cannot converge.
Liam Sanders
...which is exactly what he proved
Blake Cooper
>...which is exactly what he proved
>Prove the series is convergent
>...which is exactly what he proved
I mean, I know the facts are pretty obvious, but this seems like a question, and not an answer
Brayden Lewis
The integral test doesn't work because the u substitution is completely fucked.
taking the cosine of t doesn't work because of its periodicity.
Andrew Gonzalez
What is Jacob Barnett doing nowadays?
Gabriel Nelson
You also have sin(2*n) arbitrarily close to -1 for infinitely many n.
Clearly the terms don't go to zero, but it seems it may be possible to do a grouping of terms and have the sequence converge that way.
Brandon Turner
Dropped out of UNI and got into Red Devils. He's is trying to make it as a rapper now.
Thomas Brown
shitposting on /pol probably
Tyler Wood
Logan Ross
fuck off kurisu
Joshua Martin
time travel isn't possible
Colton Ortiz
you literally proved yourself wrong within two weeks brainlet
Alexander Carter
he's cute and the same age as me. how can i meet up with him and do lewd things, Veeky Forums?
Adam Sanders