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.

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

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

Think about taking the imaginary part of the power series of the logarithm.

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

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!

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!

- phlip
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### Re: Help please, sum of infinite series of sin(nx)/n

Do you recognise the power series [imath]z + \frac{z^2}2 + \frac{z^3}3 + \cdots[/imath]? It's a reasonably common one...

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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

Temporarily forget that it's a complex number. Let [imath]y= e^{ix}[/imath], and you have the infinite sum

[math]y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots[/math]

What does this look like?

-- Drat, ninja'd.

Anyways, if you're feeling uneasy about the substitution here, it's ok because [imath]e^{ix}[/imath] 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?

[math]y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots[/math]

What does this look like?

-- Drat, ninja'd.

Anyways, if you're feeling uneasy about the substitution here, it's ok because [imath]e^{ix}[/imath] 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?

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

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

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

- phlip
- Restorer of Worlds
**Posts:**7458**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
**Contact:**

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

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

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...

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:**

Last edited by phlip on Mon Feb 01, 2010 8:11 am UTC, edited 1 time in total.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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

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~~~~

Thanks so much!!!

the answer (pi-x)/2 is quite clean and neat~~~~

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