brainlet filter

>>8651760
solve what? It's an expression, not a problem.

>>8651765
simplify to a reasonable closed form formula. don't play dumb

>>8651769
https://en.wikipedia.org/wiki/Closed-form_expression

>In mathematics, a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations.

seems like it is already in "closed form"

>>8651769
what does N stand for?

>>8651785
/thread

>>8651788
Holy shit

>>8651794
no, I'm serious, what does it mean in this particular expression

>>8651785
wow you must be the first person to ever figure out that when people use 'closed form' they're typically not using a rigorous mathematical notion. Bravo to you, you're so fucking clever. I bet you link to that wikipedia article every time you take a test that uses the phrase "closed form"

>>8651800
He wants an expression in terms of N. For example, [math] \sum_{i=1}^N i = \frac{N^2-N}{2} [/math]

>>8651806
>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

>>8651785
>>8651794

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.

>>8651806
>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

>>8651808
fuck, then I don't find it possible here

no bully pls, I'm a hexagon specialist

>>8651811
>remove the sum sign
[eqn] \sin(0)+\sin(1)+...+\sin(N-1)+\sin(N) [/eqn]

>>8651818
good lad

answer is here OP

>>8651812
>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

>>8651760
do you want an interpolation?

>>8651818
>I am not afraid of derailing a good thread and showing I have the worst type of autism.

Consider suicide.

>>8651821
>integrals are evaluated in a finite number of operations

Consider suicide

>>8651824
>>8651828
>Consider suicide

Consider suicide.

>>8651760
that is not possible, op
good job, now that nobody can keep posting on /sci/, it's dead

>>8651828
Yes they are? At least some of them

>>8651760
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...

>>8651831
>that is not possible, op

At least you can be confirmed for a fucking brainlet. Leave /sci/ immediately.

Answer is here: http://www.wolframalpha.com/input/?i=sum+from+k%3D0+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.

>>8651838
Proof or it didn't happen.

Anyone here can type ask wolfram to compute it.

>>8651844
Get out of my sight, brainlet
http://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

>>8651834
Not by the real definition of the integral

>>8651831
>>Consider suicide
>Consider suicide.

Consider suicide

>>8651846
>I simply looked up the solution elsewhere

Killjoy. Consider suicide.

OP, next time find a more obscure problem.

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>>8651857
lol manlet

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I couldn't be bothered to use /sci/ built-in LaTeX.

>>8651817
>I'm a hexagon specialist
sophomores out

>>8651889
nice work, that's my solution as well

>>8651889
S-same

>>8651889
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)?

>>8651838
It's the partial sum which is what the original problem entails.
just did the wolframalpha search.
It checks out.
>>8651818
This answer's correct too, but it's not detailing the partial sum. lol.

Here we go.

>>8651821
Those are completely different sorts of problems. You can't "solve" an expression, but you can evaluate a function at a point.

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>>8651899
>>8651906
>>8651894

>>8651811
[math]\sin(k)[/math]

>>8652139
wat

>>8652313
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

>>8651923
It's O(1), you turd.

It's literally 0, solved it in 1.35 seconds

>>8651760
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.

>>8651854
>>>Consider suicide
>>Consider suicide.

>Consider suicide

Consider suicide.

>>8651806
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?

>>8652741
My thoughts exactly, but mine is regarding a different problem entirely.

>>8652700
>>>>Consider suicide
>>>Consider suicide.

>>Consider suicide

>Consider suicide.
Considering suicide.

>>8652820
>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

>>8651889
same

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

>>8651760
Are we doing this with Maclaurin series or what?

>>8651788
The maximum

>>8651818
GEENIUS

>>8652883
I'd say the complex exponential is more elegant than any arcane cosine formula.

>>8651760
This sum is more fun goys, have a go:

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

>>8653127
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

>>8653127
[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.

>>8653144
yep, proof?

>>8653155
Just use a Barnettian integral convergence test

>>8653132

Isn't this too advanced for an entrance exam?

>>8653187
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)

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>>8652429
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

>>8653199
You write really fucking ugly for a Frenchman

>>8653234
How did you guess that I'm French?
Genuinely interested

>>8651760
>op(N) = sum(sin(k) for k in 0:N)
did i do good?

>>8653243
The paper you are using is only used in France basically. Not the guy btw.

>>8653255
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.

>>8651760
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?

>>8652459
I walk 2 paths at the same time. They go in opposite directions.

>>8653127
[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

>>8653562
Oh yeah and it should be [math]f(0)=0[\math] to determinate C

>>8653568
[math]f(0)=0[/math]

wrong slash

>>8651760
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

>>8653144
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?

>>8653165
I didn't expect anyone to get the reference, but here we are.

>>8653759
just use log(-x) = ipi + log(x)

>>8651828
>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

>>8653951
>Integrals are defined as the sum of infinite number of infinitesimally small elements
no they aren't
go take an analysis course moran

>>8651760
-1/12

>>8653951
>this is the type of person who browses /sci/

>>8651760
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).

>>8651889
>>8653199
What identify was used here?

I'm sorry I'm literally retarded

>>8651760
>you should be able to understand
>the meaning of the word "solve"
FTFY

>>8651806
>when retards use 'closed form' they're typically >not using a rigorous mathematical notion
FTFY

>>8651824
>I am not afraid of derailing a shit thread
FTFY

>>8651818
slowclap.gif

>>8651760

http://www.wolframalpha.com/input/?i=Tanx%2B1%5E(0.cosy-1)%C3%B70.1cosy%5E(-0.cosx%2B1)%C3%97%E2%88%921.cosy%C3%97(0.cosx%2B1%C3%B7(0.1cosy%5E(0.cos-1x)%C3%B7tany-1

Something like this?

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XXDDDDDDDDD

>>8651889
i was a mistake

>>8655039
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

Depends on what N is. Give an actual limit and the answer is easy.

>Hurr no it's csc(1/2)sin(n/2+1/2)sin(n/2)
That doesn't solve it you theoretical maths fuck. You've just rewritten the problem. That's why nobody respects math majors.

>>8656617
wew lad

>>8656617
Top kek

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>>8656617

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>>8656617
lit af senpai

>>8656617
even a CS brainlet could see that the formula you quote is computationally more efficient than just straight up summing all the terms and thus that proving this identity is worthwhile

>>8651760
if you're summing over real numbers $k$ then
$$
\begin{align*}
\sum_{k=1}^{N} \sin(k) &= \int_0^{N \mod 2\pi} \sin(k) dk \\
& = -\cos(N \mod 2\pi) + \cos(0) \\
\end{align*}
$$

>>8657728
if you're summing over real numbers [math] k [/math] then
[math]
\begin{align*}
\sum_{k=1}^{N} \sin(k) &= \int_0^{N \mod 2\pi} \sin(k) dk \\
& = -\cos(N \mod 2\pi) + \cos(0) \\
\end{align*}
[/math]