Fourier Series exercise

[wolly]

How does the first line equal the second line?This is a Fourier Series!
Where do 0 and $$\frac{1}{2n+1}$$ appear from?

 

[blamocur]

What are the values of $(-1)^{n+1}+1$ for odd and even $n$'s ?

 

[wolly]

What are the values of $(-1)^{n+1}+1$ for odd and even $n$'s ?

For $$n=0=>0$$$$n=1=>2$$$$n=2=>0$$$$n=3=>2$$So the values are 0 and 2!

 

[logistic_guy]

View attachment 39343
How does the first line equal the second line?This is a Fourier Series!
Where do 0 and $$\frac{1}{2n+1}$$ appear from?

The idea is that we want to sum only when $\displaystyle n$ is odd

Then,

$\displaystyle \frac{(-1)^{n+1} + 1}{n} = \frac{2}{2n + 1}$

We start in the first with $\displaystyle n = 1$ while we start in the second with $\displaystyle n = 0$.

But your book is wrong!

$\displaystyle \sum_{n=1}\frac{(-1)^{n+1} + 1}{n}\sin nx \neq \sum_{n=0}\frac{2}{2n + 1}\sin nx$

It should be:

$\displaystyle \sum_{n=1}\frac{(-1)^{n+1} + 1}{n}\sin nx = \sum_{n=0}\frac{2}{2n + 1}\sin (2n + 1)x$


Do you see the difference and why?

 

[wolly]

The idea is that we want to sum only when $\displaystyle n$ is odd

Then,

$\displaystyle \frac{(-1)^{n+1} + 1}{n} = \frac{2}{2n + 1}$

We start in the first with $\displaystyle n = 1$ while we start in the second with $\displaystyle n = 0$.

But your book is wrong!

$\displaystyle \sum_{n=1}\frac{(-1)^{n+1} + 1}{n}\sin nx \neq \sum_{n=0}\frac{2}{2n + 1}\sin nx$

It should be:

$\displaystyle \sum_{n=1}\frac{(-1)^{n+1} + 1}{n}\sin nx = \sum_{n=0}\frac{2}{2n + 1}\sin (2n + 1)x$


Do you see the difference and why?

Can you show me how you got the $$\frac{2}{2n+1}*sin(2n+1)x$$?

 

[logistic_guy]

Can you show me how you got the $$\frac{2}{2n+1}*sin(2n+1)x$$?

In your post #3 you have shown that only the odd index of summation matters, ie $\displaystyle n = 1,3,5,7....$

The trick is that if we want only even indices we change $\displaystyle n$ to $\displaystyle 2n$

And

If we want only odd indices we change $\displaystyle n$ to $\displaystyle 2n + 1$

Why?

Because $\displaystyle 2n + 1$ allows us to get the odd indices again. Here are some calculations:

$\displaystyle 2(0) + 1 = 1$
$\displaystyle 2(1) + 1 = 3$
$\displaystyle 2(2) + 1 = 5$
$\displaystyle 2(3) + 1 = 7$

Aren't these the same as $\displaystyle n = 1,3,5,7....$?

The second part is that if we change one $\displaystyle n$ to $\displaystyle 2n + 1$, we have to change all other $\displaystyle n$'s to $\displaystyle 2n + 1$.

In the original summation, we have:

$\displaystyle \sum_{n=1}\frac{(-1)^{n+1} + 1}{n}\sin nx$

It means that we have two $\displaystyle n$'s. So we have to change them both. And we get:

$\displaystyle \sum_{n=0}\frac{2}{2n + 1}\sin (2n + 1)x$

And because we care only for the odd $\displaystyle n$, $\displaystyle (-1)^{n+1} + 1$ is always $\displaystyle = 2$

 

[logistic_guy]

temperature in a hemisphere

The steady-state temperature in a hemisphere of radius r = c is determined from \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0 0 < r < c, \ \ \ \...

www.freemathhelp.com

See post #6. The idea of the odd index $\displaystyle 2n + 1$ was used there.