Brainlet filter

>every time you take a test that uses the phrase "closed form"

never took a test that used that phrase, because i don't take brainlet classes

Not OP but holy shit, you know what he fucking means.

By closed form he means "remove the sum sign". It is a nice problem and so far I don't see a trivial answer.

>every time you take a test that uses the phrase "closed form"
Yeah, the formula in the OP is an example of a closed form equation similar to what you'd see on a test. What are you even trying to say? Retard

k=0

sin(k) = sin(0) = 0

fuck, then I don't find it possible here

no bully pls, I'm a hexagon specialist

>remove the sum sign
[eqn] \sin(0)+\sin(1)+...+\sin(N-1)+\sin(N) [/eqn]

good lad

answer is here OP

>Problem #1: Evaluate the integral of cos(x) from 0 to pi
>Answer: It's already in closed form XD

It's just not as clever as you think and it gets tiring seeing it over and over when people would rather be pedants than just solve the actual problem

do you want an interpolation?

>I am not afraid of derailing a good thread and showing I have the worst type of autism.

Consider suicide.

>integrals are evaluated in a finite number of operations

Consider suicide

>Consider suicide

Consider suicide.

that is not possible, op
good job, now that nobody can keep posting on Veeky Forums, it's dead

Yes they are? At least some of them

for a sum up to N
[eqn]\frac{\sin N - \cot\frac{1}{2}\cos N+\cot\frac{1}{2}}{2}[/eqn]

brainlets i swear...

>that is not possible, op

At least you can be confirmed for a fucking brainlet. Leave Veeky Forums immediately.

Answer is here: wolframalpha.com/input/?i=sum from k=0 to n of sin(k)

Now, let us grown men remember series and find how wolfram reached the solution.

I am sure r/science will take you in.

Proof or it didn't happen.

Anyone here can type ask wolfram to compute it.

Get out of my sight, brainlet
math.stackexchange.com/questions/1035231/properties-of-partial-sums-of-series-sum-sinn

For all the brainlets here: The hint is that its actually a sum of two geometric sums

Not by the real definition of the integral

>>Consider suicide
>Consider suicide.

Consider suicide

>I simply looked up the solution elsewhere

Killjoy. Consider suicide.

OP, next time find a more obscure problem.

lol manlet

I couldn't be bothered to use Veeky Forums built-in LaTeX.

>I'm a hexagon specialist
sophomores out

nice work, that's my solution as well

S-same

yes i arrived at the same conclusion

Now, what happens as N -> infinity?

Obviously, the series does not converge, but is it O(N) (this is my suspicion)?

It's the partial sum which is what the original problem entails.
just did the wolframalpha search.
It checks out.
This answer's correct too, but it's not detailing the partial sum. lol.

Here we go.

Those are completely different sorts of problems. You can't "solve" an expression, but you can evaluate a function at a point.

...

[math]\sin(k)[/math]

wat

it's a stupid joke about removing the sum sign

>All these retards who can't solve it
You literally just sum [math]\sum_{k \,=\, 0}^N \mathrm e^{\mathrm i\,k}[/math], use the half-arc technique to make nifty trigonometry functions appear and conclude by taking the imaginary part of that.

watching mythbusters is enuff

It's O(1), you turd.

It's literally 0, solved it in 1.35 seconds

I wish to stop posting but every path I take seems to bring me right back. I walk the path of peace and least resistance, so I only know to return here.

>>>Consider suicide
>>Consider suicide.

>Consider suicide

Consider suicide.

I'm a junior math major and I've never even heard this phrase before. How are we supposed to know what the fuck you mean if you're not using the accepted definition?

My thoughts exactly, but mine is regarding a different problem entirely.

>>>>Consider suicide
>>>Consider suicide.

>>Consider suicide

>Consider suicide.
Considering suicide.

>Considering suicide.

Good for ya.

Use method of differences

Using cosine angle formulas and MOD, the answer requires no complex exponentials, although it is more tedious to derive it this way

same

phew, glad i didn't have to type it all out!

Are we doing this with Maclaurin series or what?

The maximum

GEENIUS

I'd say the complex exponential is more elegant than any arcane cosine formula.

This sum is more fun goys, have a go:

[math]\text{Find} \ \ \sum_{n=1}^\infty \frac{\sin n}{n}[/math]

this problem was going to appear in a STEP paper (brainlet filter exam for cambridge maths), but they decided against it, so it's definitely possible

[eqn] \sum_{n = 1}^\infty \frac{\sin n}{n} = \frac{1}{2} (\pi - 1) [/eqn]

Prove the infinite series [math] \sum_{n = 0}^{\infty} \frac{ \sin 2n }{ 1 + \cos^4 n} [/math] is convergent.

yep, proof?

Just use a Barnettian integral convergence test

Isn't this too advanced for an entrance exam?

well I was being slightly misleading in my post, in reality this question would be part of a multi-part one, in which this the conclusion of multiple other results (like all other STEP questions), so for instance I'd imagine the question would go:

a) prove this sum of cosines formula
b) prove this integral representation for this harmonic sine sum
c) prove this upper and lower bound on the integral
d) hence prove the result

(which is how the proof would go in an elementary fashion)

what this guy said
I can't be bothered to write it properly, so here is a shitty photo of my shitty handwriting
maybe there is a mistake in there

You write really fucking ugly for a Frenchman

How did you guess that I'm French?
Genuinely interested

>op(N) = sum(sin(k) for k in 0:N)
did i do good?

The paper you are using is only used in France basically. Not the guy btw.

So it's just that
I was thinking something too complex about how i write the equations
I just used some draft paper that was lying about on my desk

I know that it's not zero because we're using integer whole numbers which are not going to cancel out.

Even as N approaches infinity you cannot say there will be an equal probability distribution.

Use the well known formula for Dirichlet kernel, shift the parameter and use symmetry.

How do wanting to post on sci and being able to solve something that isn't a problem? And why is it do? Is logic not essential in science?

I walk 2 paths at the same time. They go in opposite directions.

[eqn]f(\alpha)=\sum\limits_{n=1}^{\infty}\dfrac{\sin{\alpha n}}{n} = \sum\limits_{n=1}^{\infty}\dfrac{\sin{\alpha n}}{n}[/eqn]
[eqn]f'(\alpha) = \sum\limits_{n=1}^{\infty}{\cos{\alpha n}}=-1+\sum\limits_{n=0}^{\infty}{\cos{\alpha n}}=-1+\dfrac{1}{2}\left(\dfrac{1}{1-e^{i\alpha}}+\dfrac{1}{1-e^{-i\alpha}}\right)= -1
+\dfrac{1-\cos{\alpha}}{2\cos{\alpha}}
[/eqn]
[eqn]f(\alpha)=C-\alpha+\int \dfrac{1-\cos{\alpha}}{2\cos{\alpha}}d\alpha [/eqn]

Sorry I'm to lazy to integrate that shit

Oh yeah and it should be [math]f(0)=0[\math] to determinate C

[math]f(0)=0[/math]

wrong slash

Can you cross the street without staring at your cell-phone? Can you blow your nose, which most cell-phone addicts can't do? These should be on an intelligence test!

ITT people who don't know basic calculus

So using Mathematica, I even find that if you pass from sin(n) to sin(k·n), the sum equals

(1/2)·(pi - k)

But it returns this as

(i/2)·log(-e^(i·k))

Does anybody know how to get from the latter to the former?

I didn't expect anyone to get the reference, but here we are.

just use log(-x) = ipi + log(x)

>It takes me an infinite number of operations to evaluate any integral
Integrals are defined as the sum of infinite number of infinitesimally small elements but evaluating an integral does not take an infinite number of operations. Go take a calc 1 course you brainlet

>Integrals are defined as the sum of infinite number of infinitesimally small elements
no they aren't
go take an analysis course moran

-1/12

>this is the type of person who browses Veeky Forums

The real brainlet filter: reformulate sum i=1 to n sin((2*i + 1)*x) into a form that can easily be calculated with a table of values of sin(x) (of course, if x>2pi, you can just repeatedly subtract values of pi until you get in the range provided in the table).

What identify was used here?

I'm sorry I'm literally retarded

>you should be able to understand
>the meaning of the word "solve"
FTFY

>when retards use 'closed form' they're typically >not using a rigorous mathematical notion
FTFY

>I am not afraid of derailing a shit thread
FTFY

slowclap.gif

wolframalpha.com/input/?i=Tanx+1^(0.cosy-1)÷0.1cosy^(-0.cosx+1)×−1.cosy×(0.cosx+1÷(0.1cosy^(0.cos-1x)÷tany-1

Something like this?

XXDDDDDDDDD

i was a mistake

that [math]\exp{ix} = \cos x+ i\sin x[/math]

then that the sum of a geometric progression with multiplicative term [math]r[/math] up to [math]N[/math] is [math]\frac{1-r^{N+1}}{1-r}[/math]

Then a lot of fraction simplification