Help please, sum of infinite series of sin(nx)/n

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Help please, sum of infinite series of sin(nx)/n

Postby ydy59585 » Fri Jan 29, 2010 3:18 am UTC

I am trying to sum up the infinite series sin(nx)/n.
I konw that the result depends on x, so i am interested in the range of 0<x<pi.
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Re: Help please, sum of infinite series of sin(nx)/n

Postby t0rajir0u » Fri Jan 29, 2010 3:33 am UTC

Think about taking the imaginary part of the power series of the logarithm.
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Re: Help please, sum of infinite series of sin(nx)/n

Postby ydy59585 » Fri Jan 29, 2010 4:38 am UTC

let me try this, create the cos series cos(nx)/n add to the sin series.
so the series become sum of cos(nx)/n + i*sin(nx)/n from 1 to infinite.
exp(ix)+exp(i2x)/2+exp(i3x/3)+....exp(inx)/n+... = exp(ix)+exp(ix)^2/2+exp(ix)^3/3+..+exp(ix)^n/n+...
how to find the sum of this series?
I dont know much about the complex number.
Thanks!
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Re: Help please, sum of infinite series of sin(nx)/n

Postby phlip » Fri Jan 29, 2010 4:53 am UTC

Do you recognise the power series z + \frac{z^2}2 + \frac{z^3}3 + \cdots? It's a reasonably common one...
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Re: Help please, sum of infinite series of sin(nx)/n

Postby letterX » Fri Jan 29, 2010 5:04 am UTC

Temporarily forget that it's a complex number. Let y= e^{ix}, and you have the infinite sum
y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots


What does this look like?

-- Drat, ninja'd.

Anyways, if you're feeling uneasy about the substitution here, it's ok because e^{ix} is 'within the radius of convergence' of the above series. Hopefully, if you're doing power series and infinite sums, you'll cover how complex numbers fit into them, but if not, you should check it out.

Also, I don't think I've seen this problem before, or if I have, I've forgotten it. It's a super slick answer! Is there some nifty geometrical intuition behind this?
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Re: Help please, sum of infinite series of sin(nx)/n

Postby ydy59585 » Fri Jan 29, 2010 6:19 am UTC

actually i try to solve a streamfunction in the fluid mechanics and i find that my math is not sufficient...
still cannot quite understand why y=exp(ix)^n is convergented, try to understand that...
but if the y is convergence then
y' = 1 +y +y^2 +y^3 +... so now i can solve it
result will be -ln(1-y) = -ln(1-exp(ix))
question again...
if 0<x<pi, how to get the image part? sounds like a stupid question but i really dont know :?
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Re: Help please, sum of infinite series of sin(nx)/n

Postby phlip » Fri Jan 29, 2010 7:15 am UTC

The function to get the imaginary part of a complex number is called Im, so your final answer is:
\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \mathrm{Im} \left( -\log \left( 1 - e^{ix} \right) \right)

However, we can simplify that a little... details behind the spoiler for those who want to work it out themselves, but if you're not familiar with complex numbers, you might not follow it...
Spoiler:
First:
\begin{align} e^{x + iy} &= e^x e^{iy} \\ &= e^x \angle y \end{align}
So:
\begin{align} \log(r \angle \theta) &= \log(r) + i\theta \\ \mathrm{Im}(log(r \angle \theta)) &= \theta \\ \mathrm{Im}(log(z)) &= arg(z) \end{align}


So we can simplify this to:
\sum_{n=1}^{\infty} \frac{\sin nx}{n} = -\arg \left(1 - e^{ix} \right)


However, that's not all... now looking at what this means geometrically... let z = 1 - e^{ix} and \theta = -\arg(z), and plot these on an Argand diagram:
argand.png
In case it's not clear, z is a point, x and θ are their respective angles. The circle is the possible locations of z, which is centred on 1 and has radius 1.
argand.png (4.79 KiB) Viewed 3321 times
Now we can use the "angle at a centre is double the angle at the circumference" rule to find that:
\theta = \begin{cases} \frac {\pi-x}2 & 0 < x \le \pi \\ \frac{-\pi-x}2 & -\pi \le x < 0 \end{cases}
The cases are necessary as that's where the path of z crosses the origin - the argument is discontinuous there.

To complete the cases, we can evaluate the original power series at x=0 to plug the gap we got when we went from having just sin to exp (the power series using sin converges at zero, but the power series using cos diverges at zero, so when we added the cos, we lost the value at x=0), and get our final result:
\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \begin{cases} \frac {\pi-x}2 & 0 < x \le \pi \\ 0 & x = 0 \\ \frac{-\pi-x}2 & -\pi \le x < 0 \end{cases}

So we get a stepped linear thing. Alpha agrees (don't try to get alpha to find the formula though, as you get some convoluted mess... because it takes x as complex, not real, which is necessary for pretty much all the simplifications we've done in the thread).

However, I'd be curious to hear if there's a geometric reason why \sum_{n=1}^{\infty} \frac{e^{inx}}{n} gives that circle around 1...
Last edited by phlip on Mon Feb 01, 2010 8:11 am UTC, edited 1 time in total.
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Re: Help please, sum of infinite series of sin(nx)/n

Postby ydy59585 » Fri Jan 29, 2010 7:36 am UTC

wa...i am impressed, may be i need a textbook to learn the complex number again...
Thanks so much!!!
the answer (pi-x)/2 is quite clean and neat~~~~ :D
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