Fourier series

Any integral of the form
whenever m not equal to n and m,n are integers.

for all values of m,n where m,n are integers.

whereas

(If you notice there is a summation from n = 1 to infinity.)
Thus in the summation whenever n doesn't equal m we get zero. And in only one case n equals m. In that case the integral becomes pi. And this integral is multiplied by a[sub]n[/sub] , but since n = m hence
a[sub]n[/sub] = a[sub]m[/sub].

Note -


integrate both the cos terms in the integral you will get zero.

when m = n ,


=

Integration results become zero when m&n are non negative(positive) integers.

Thanks for the explanations. I got it now. Now I have another question here.

And I don't get this part of the solution:

Isn't bn supposed to be 2/pi ???

Bn is 2/pi only when n=1.

Value of bn changes when n changes.

ReyaBn is 2/pi only when n=1.

Value of bn changes when n changes.

It a bit complicated at first.Now I get what you mean 😀

cos (n pi) = 1 when n is even and it is -1 when n is odd. Thus it is convenient to write cos (n pi) = (-1)[sup]n[/sup].
Thus b[sub]n[/sub] equals 2/(nπ) when (n-1) is odd and 0 when (n - 1) is even.

AruwinBut even if n is not equals to 1, i got both n as the numerator and denominator so doesn't that make it equals to 1?And hence,times it with 2/pi I get Bn=2/pi

There isn't an n in numerator see the equation of bn again.

Arnav Joshicos (n pi) = 1 when n is even and it is -1 when n is odd. Thus it is convenient to write cos (n pi) = (-1)[sup]n[/sup].
Thus b[sub]n[/sub] equals 2/(nπ) when (n-1) is odd and 0 when (n - 1) is even.



There isn't an n in numerator.

And what about f(x)??I don't understand how the answer turned out like that 😔

AruwinAnd what about f(x)??I don't understand how the answer turned out like that 😔

That is just the fourier expansion of f(x) in terms of bn.
The fourier series expression is a summation in terms of bn.

That summation is expanded.
All the a[sub]n[/sub] terms will be zero.

How come?

If you give even values for n, bn=0.

ReyaHow come?

If you give even values for n, bn=0.

bn has a term 1 - (-1)^n in numerator. When n is even you get 1 - 1. When n is odd you get 1 - (- 1). Odd Powers of -1 will give you -1 whereas even powers will give you 1.

Er..I asked how come to aruwin. Sorry for not quoting or mentioning his name 😛

ReyaEr..I asked how come to aruwin. Sorry for not quoting or mentioning his name 😛

The posts were coming in quick succession in this thread so I also got confused who was asking what. You could say that I had lost the 'thread'.👍

Arnav JoshiThat is just the fourier expansion of f(x) in terms of bn.
The fourier series expression is a summation in terms of bn.

That summation is expanded.
All the a[sub]n[/sub] terms will be zero.

In the final part, the summation becomes 1/2 + 2/pi[(sinx)/1 + (sin3x/1)....]

Can you explain why it becomes 2/pi instead of 1/pi and why only sin(odd number)x are the only ones shown??what happened when n is an even number?? Is sin nx = 0 whenever n is an even number???

AruwinIn the final part, the summation becomes 1/2 + 2/pi[(sinx)/1 + (sin3x/1)....]

Can you explain why it becomes 2/pi instead of 1/pi and why only sin(odd number)x are the only ones shown??what happened when n is an even number?? Is sin nx = 0 whenever n is an even number???

No , sin nx won't be zero when n is even number. But instead bn will be zero when n is even number.

First aim is to calculate bn using f(x). As you can see from image you posted first :-

bn comes out to be
1 -(-1)[sup]n[/sup] /(n*pi). When n is even -1[sup]n[/sup] becomes 1 hence b[sub]n[/sub] becomes (1-1)/n*pi. Thus it becomes zero when n is even. And when n is odd it becomes (1 - (-1)) / (n pi) = 2/(n pi).

Now our job is to put these values of bn in the formula for f(x).


(a[sub]n[/sub] terms are all zero. Other than a[sub]0[/sub]/2 which is 1/2)

Because b[sub]n[/sub] is zero for even n , even terms won't be present in the summation for b[sub]n[/sub]. And all odd terms will be of form 2/(n pi). Hence you will see b[sub]n[/sub]sin (nx) becomes 2sin(3x)/(3 pi) when n =3 . When n = 5 b[sub]n[/sub]sin (nx) becomes 2sin(5x)/(5 pi). And so on for all odd values of n. 2/pi is taken out of bracket as it is common in all sin (nx) terms.

Arnav JoshiNo , sin nx won't be zero when n is even number. But instead bn will be zero when n is even number.

First aim is to calculate bn using f(x). As you can see from image you posted first :-

Ok,I got the drift now but there is still one more thing that is bugging me. Now that the equation for Bn can be simplified as (1 - (-1)^n)/n*pi.
But at the very end, the summation becomes different. I don't understand why it becomes
(1+(-1)^n-1).

AruwinOk,I got the drift now but there is still one more thing that is bugging me. Now that the equation for Bn can be simplified as (1 - (-1)^n)/n*pi.
But at the very end, the summation becomes different. I don't understand why it becomes
(1+(-1)^n-1).

That is because of (-1). (-1)[sup]n[/sup] means -1 has been multiplied by itself n times , that is what 'raised to n' means ,right ? Now all we are doing is taking one '-1' from these n '-1's and are multiplying it with that ' - 'sign , hence that - becomes + and we get 1 + (-1)[sup]n-1[/sup] , as we are now left with (n-1) '-1's.

Arnav JoshiThat is because of (-1). (-1)[sup]n[/sup] means -1 has been multiplied by itself n times , that is what 'raised to n' means ,right ? Now all we are doing is taking one '-1' from these n '-1's and are multiplying it with that ' - 'sign , hence that - becomes + and we get 1 + (-1)[sup]n-1[/sup] , as we are now left with (n-1) '-1's.

AAAaaa...now I realize that.Thank you so much! But is it ok if I just leave the answer as it is in the first one? I mean, is it ok if I don't take out one of the (-1) out??The value is gonna be the same after all,isn't it?