Fourier series

[final sail]

Need help with a fourier series question. The series I'm getting is nowhere near the expected form after substituting the given x. While on it, how does x = 4pi + a even fit in the domain of the fourier series substitution?

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[fallen arrow]

Can you show what you have so far?

[final sail]

okay so, i'm assuming period = 3pi here
L = 3pi/2
am getting these rn
a0 = (2/3pi)*(b-a)
an = (3/2n)*(sin(2bn/3) - sin (2an/3))
bn = (3/2n)*(cos(2an/3) - cos (2bn/3))

[fallen arrow]

Let me check, hold on.

[fallen arrow]

I'm getting the following:
l = 3π/2
a(0) = 2(b - a)/(3π)
a(n > 0) = (sin(2bn/3) - sin(2an/3))/(πn)
b(n) = (cos(2an/3) - cos(2bn/3))/(πn)

Here's the definition I'm using, by the way.

[final sail]

oh mb, i'm dumb, those x shouldn't be there in my answer

although it's still different, brb i'll verify it

[final sail]

okay mb i forgot multiplying 1/l, yeah i'm getting the same now

[fallen arrow]

Great! What about the next part?

[final sail]

hmm, that was the part i was thinking about tbh
i'm guessing we'll simplify sin - sin to 2 cos sin?

and similarly for cos - cos

[fallen arrow]

Hm... Hold on, let me think about that.

[final sail]

well also, before that, is 4pi + a even in the domain of the fourier series

a is supposed to be greater than -pi because that's given

[fallen arrow]

The domain is the whole axis. However, you need to account for the fact that the function will be periodic.

[final sail]

and the substitution 4pi + a will be greater than 3pi, which will fall outside the range the fourier series is defined, right?

[fallen arrow]

So, f(a + 4π) = f(a + π).

[final sail]

from my notes, we were given this. I thought the values of x were "allowed" to substitute were "in" the first cycle of the function?

the alpha < x < alpha + 2pi part (seeing period is 2pi here)

[fallen arrow]

No. The point here is that it shouldn't be written as "f(x) = a(0)/2 + ...", but rather as "f(x) ~ a(0)/2 + ...".

That is because f(x) does indeed equal to that in the interval we define it, but outside of it it is continued periodically.

Here is what our function looks like for a = 0, b = 1, for example (and n = 500).
The vertical lines show the initial segment.

[final sail]

oh, TIL that helps a lot
so if the function itself was periodic, it will be equal to the function perfectly for all x the original function was defined?

[fallen arrow]

As long as we define on on an interval with the same length as its period, then yes.

Oh, by the way: if the function has jump discontinuities at some points (including the boundary points), then the value of the Fourier series will be the average of one-sided limits.

[final sail]

that... is interesting

we never learnt the topic well enough to understand much more than "here's the formulae, solve" ;-;

[fallen arrow]

Now, let's work with the expression.
πn(a(n)cos(2nx/3) + b(n)sin(2nx/3)) = (sin(2bn/3) - sin(2an/3))cos(2nx/3) + (cos(2an/3) - cos(2bn/3))sin(2nx/3) = (sin(2bn/3)cos(2nx/3) - cos(2bn/3)sin(2nx/3)) - (sin(2an/3)cos(2nx/3) - cos(2an/3)sin(2nx/3)) = sin(2(b - x)n/3) - sin(2(a - x)n/3) = 2sin((b - a)n/3)cos((a + b - 2x)n/3)
So:
a(n)cos(2nx/3) + b(n)sin(2nx/3) = 2sin((b - a)n/3)cos((a + b - 2x)n/3)/(πn)
Though, I don't really know why they offer such a value. I think x = a will give a better result.

[fallen arrow]

Ah... That's a shame. Well, now you know!

Ok, yeah: I checked, you can get the result for x = a.

[fallen arrow]

You do need this property, though.

[final sail]

...man, i need to work on my trig
this makes so much sense, damn
thanks ;-;

[jolly lark]

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[final sail]

i literally got scared by the trig simplification

[fallen arrow]

You're welcome! I did use a couple of trig formulas, of course.

I can DM you my result where I substituted x = a if you want. It will be in a spoiler so you can still try it.

[final sail]

yeah it works for x = a, tysm ;-;

[final sail]

oh i'll hopefully be able to proceed from here

almost got the whole answer anw

[dense plover]

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[final sail]

+close