1. ## Fourier Series Question

Hi
The following two question i am having trouble finding the fourier series.
Can someone tell me what i have done wrong in question 1?

In Question 2 i am stuck after subbing t = 0 and 1 into the intergral. How should this be simplified?

P.S

2. Originally Posted by Paymemoney
Hi
The following two question i am having trouble finding the fourier series.
Can someone tell me what i have done wrong in question 1?

In Question 2 i am stuck after subbing t = 0 and 1 into the intergral. How should this be simplified?

P.S
Cos of an odd multiple of $\displaystyle \pi$ is $\displaystyle $$-1 and cos of \displaystyle$$0$ is $\displaystyle $$1 CB 3. Originally Posted by CaptainBlack Cos of an odd multiple of \displaystyle \pi is \displaystyle$$-1$ and cos of $\displaystyle $$0 is \displaystyle$$1$

CB
is that reply to question 1 or 2?

4. Originally Posted by Paymemoney
Hi
The following two question i am having trouble finding the fourier series.
Can someone tell me what i have done wrong in question 1?

In Question 2 i am stuck after subbing t = 0 and 1 into the intergral. How should this be simplified?

P.S

Why do you do what you do with $\displaystyle a_0$?? Do you hate yourself? Much easier to split the integral into a sum:

$\displaystyle \displaystyle{\int\limits^1_{-1}(1-|t|)dt=\int\limits^0_{-1} (1+t)dt+\int\limits^1_0(1-t)dt=1}$ , and then do the same for $\displaystyle a_n\,,\,b_n$

If this exercise is what I think it is then it's a very nice and beautiful one. You'll get both

the sums of $\displaystyle \displaystyle{\sum\limits^\infty_{n=1}\frac{1}{(2n-1)^2}\,\,and\,\,\sum\limits^\infty_{n=1}\frac{1}{n ^2}}$

Tonio

5. Originally Posted by Paymemoney
is that reply to question 1 or 2?
2

cb

6. quick question how would i simplify the following line:

$\displaystyle a_n = \frac{1}{\pi}[\frac{-cos(1-n)\pi}{2(1-n)} - \frac{cos(1+n)\pi}{2(1+n)} - 0]$

7. Originally Posted by Paymemoney
quick question how would i simplify the following line:

$\displaystyle a_n = \frac{1}{\pi}[\frac{-cos(1-n)\pi}{2(1-n)} - \frac{cos(1+n)\pi}{2(1+n)} - 0]$

Since $\displaystyle \cos m\pi =(-1)^m$ we have that

$\displaystyle \displaystyle{a_n=\frac{1}{\pi}\left[\frac{(-1)^n}{2(1-n)}+\frac{(-1)^n}{2(1+n)}\right]=\frac{(-1)^n}{\pi}\cdot\frac{2}{1-n^2}}$

Tonio

8. in this case would i need to determine when n-> even and n-> odd. Because i checked the answers and has it like $\displaystyle \frac{2}{\pi}\sum\limits^\infty_{n=1} \frac{1}{4n^2-1}$

9. if$\displaystyle cos(m\pi) = (-1)^{m}$

then what would

$\displaystyle sin(m\pi)$ = ???

10. Originally Posted by Paymemoney
if$\displaystyle cos(m\pi) = (-1)^{m}$

then what would

$\displaystyle sin(m\pi)$ = ???
Sketch a sine curve and observe its behaviour, if you actually try it is not difficult to answer your own question.

Also if you do not know the value of the sine and cosine at interger multiples of pi you are not ready to study Fourier series and need to repeat trig.

CB

11. Originally Posted by Paymemoney
if$\displaystyle cos(m\pi) = (-1)^{m}$

then what would

$\displaystyle sin(m\pi)$ = ???

It is a little weird, imho, to make this kind of trivial questions in very basic trigonometry

if you're dealing with Fourier series...even if you don't remember. and this is already weird,

why don't you check your high school books or, at least, google it?

The sine of any multiple of pi is zero...

Tonio

12. i think my question was not clear, my question was based on what does $\displaystyle sin(1-n)$ equal to.

We know that if $\displaystyle cos(1-n)\pi = (-1)^{1-n}$ so would this apply to $\displaystyle sin(1-n)\pi$ as well.

13. Originally Posted by Paymemoney
i think my question was not clear, my question was based on what does $\displaystyle sin(1-n)$ equal to.
No such term appears in this problem or its solution.

CB

14. Originally Posted by Paymemoney
i think my question was not clear, my question was based on what does $\displaystyle sin(1-n)$ equal to.

We know that if $\displaystyle cos(1-n)\pi = (-1)^{1-n}$ so would this apply to $\displaystyle sin(1-n)\pi$ as well.

This continues to be weird: first, as CB already pointed out, $\displaystyle \sin (1-n)$ has nothing to do with do with your problem,

and if you meant $\displaystyle \sin (1-n)\pi$ this again is the sine of an integer multiple of pi

and thus equals zero...

Tonio