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Fourier Series Help!

Hi all

I've got a question on Fourier series, I know the answer i should be getting, however i'm getting the opposite sign. I've checked my work and just cant see where i have gone wrong, please help. Ill put my working below.
If someone could inform me why im getting 12/(pi)(n) instead of -12/(pi)(n), i would be very grateful

Thanks in advance

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Reply 1

joostan

Original post by Ramjam

I'm a little puzzled as to where you got your expression for

b_n

. . .
Should it not be

b_n=\dfrac{1}{\pi} \displaystyle\int_{-\pi}^{\pi} f(t) \sin(nt) \ dt

?
With this expression the result you want follows.

Edit: Also what is

\omega

here? You've written it in some places, and not others. . .
An image of the question itself might be useful.

(edited 8 years ago)

Reply 2

Ramjam

Original post by joostan

I'm a little puzzled as to where you got your expression for

b_n

. . .
Should it not be

b_n=\dfrac{1}{\pi} \displaystyle\int_{-\pi}^{\pi} f(t) \sin(nt) \ dt

?
With this expression the result you want follows.

Edit: Also what is

\omega

here? You've written it in some places, and not others. . .
An image of the question itself might be useful.

It came from 'Modern Engineering Mathematics by Glyn James', it states that i can use that formula specifically for odd function. However I will work through using the original formula for bn and see what answer I achieve.

The question is on the working, in black pen, it is to find the fourier series of f (t) = 3 (-pi < t < 0) -3 (0 < t < pi

(edited 8 years ago)

Reply 3

joostan

Original post by Ramjam

I can't say I've seen that formula before, so I can't comment as to what's gone on there.
All I can see is that for an odd function

f

[-\pi,\pi]

one can rewrite:

b_n=\dfrac{2}{\pi} \displaystyle\int_0^{\pi} f(t) \sin(nt) \ dt

.

Yeah, I figured that was the question, but then I don't see what

\omega

has to do with anything.

Reply 4

Ramjam

Original post by joostan

I can't say I've seen that formula before, so I can't comment as to what's gone on there.
All I can see is that for an odd function

f

[-\pi,\pi]

one can rewrite:

b_n=\dfrac{2}{\pi} \displaystyle\int_0^{\pi} f(t) \sin(nt) \ dt

.

Yeah, I figured that was the question, but then I don't see what

\omega

has to do with anything.

The

\omega

issue is just me making mistakes again (sorry)

I've reworked the question and reached a point in which i'm unsure on where to go from here. Whats do i have to do next? The point at which you assign odd and even values of n, although sounds simple always confuses me.

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Reply 5

joostan

Original post by Ramjam

The

\omega

There seems to be a mistake prior to that:

f(t)

changes sign in the interval, so you'll want to write:

\displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt = \displaystyle\int_{-\pi}^{0} f(t)\sin(nt) \ dt + \displaystyle\int_{0}^{\pi} f(t)\sin(nt) \ dt

Reply 6

Ramjam

Original post by joostan

There seems to be a mistake prior to that:

f(t)

changes sign in the interval, so you'll want to write:

\displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt = \displaystyle\int_{-\pi}^{0} f(t)\sin(nt) \ dt + \displaystyle\int_{0}^{\pi} f(t)\sin(nt) \ dt

Whats happens after that? do you integrate between limits separately for both of those?

Reply 7

joostan

Original post by Ramjam

Whats happens after that? do you integrate between limits separately for both of those?

You could do that, or you could send

t \mapsto -t

in either integral to obtain a single integral, it doesn't really matter.

Reply 8

Ramjam

Original post by joostan

You could do that, or you could send

t \mapsto -t

in either integral to obtain a single integral, it doesn't really matter.

Well i'm getting no where quick, i just cant seem to get my head around this question, i must be making mistakes somewhere as it keeps going completly wrong

Reply 9

joostan

Original post by Ramjam

Well i'm getting no where quick, i just cant seem to get my head around this question, i must be making mistakes somewhere as it keeps going completly wrong

Spoiler

(edited 8 years ago)

Reply 10

Ramjam

Original post by joostan

Spoiler

from here would you integrate them to get it in terms of cos?

Reply 11

joostan

Original post by Ramjam

from here would you integrate them to get it in terms of cos?

Well I actually used the fact that

\sin(nt)

is odd to write

\displaystyle\int_{-\pi}^{0} 3\sin(nt) \ dt=-\displaystyle\int_{0}^{\pi} 3\sin(nt) \ dt

.
But you could just as well integrate now.

Reply 12

poorform

Hint:

\displaystyle b_n=\frac{2}{\pi} \int_0^{\pi} f(t)\sin (nt)~dt=-\frac{6}{\pi} \int_0^{\pi} \sin (nt)~dt

(Since the integrand is an even function.)

Then note

\displaystyle \cos(n\pi)=(-1)^n

to tidy up your answer.

(edited 8 years ago)

Reply 13

joostan

Original post by poorform

Hint:

\displaystyle b_n=\frac{2}{\pi} \int_0^{\pi} f(t)\sin (nt)~dt=\frac{6}{\pi} \int_0^{\pi} \sin (nt)~dt

(Since the integrand is an even function.)

Then note

\displaystyle \cos(n\pi)=(-1)^n

to tidy up your answer.

You dropped a minus in your expression for

b_n

Reply 14

poorform

Original post by joostan

You dropped a minus in your expression for

b_n

Will update now thanks!

I (wrongly) assumed that f(t) was positive for positive t and negative for negative t.

Since I did a very similar question the other day that was defined that way haaaaa.