Find the coefficients in the Fourier series

[math_04]

Homework Statement

The odd periodic function f(x) is defined by

f(x) = -4-x for -4 <= x <= 0 and 4-x for 0 <= x <= 4

f(x + 8) = f(x)

Sketch f(x) for -12 <= x <= 8

Find the coefficients in the Fourier series for the function defined by equation (1) and write out the series, explicitly giving the first three non zero terms in the series

Homework Equations

bn = 1/L integral of f(x)sin (npix / L) dx with limits from L to -L

The Attempt at a Solution

So do I do like this

bn = 1/4* integral of (-4-x) sin (npix/4) with limits from 4 to -4 plus 1/4*integral of 4-xsin(npix/4) with limits from 4 to -4?

Then i use integration by parts to solve the integrals?

Or is there an easier way like for an even Fourier series if you have L as 2 and -2, u can have the limits of the integral as 2 and 0?

 

[HallsofIvy]

Knowing that the function is odd means that the coefficient of cos(nx) (an even function) in the series is 0. And, of course, if f(-x)= -f(x), then because sin(-x)= -sin(x),
$$\int_{-4}^4 f(x)sin(nx)dx= \int_{-4}^0 f(x)sin(nx)dx+ \int_0^4 f(x)sin(nx) dx$$
and, making the substitution u= -x in the first integral, since dx= -du,
$$-\int_{4}^{0}(-f(u))(-sin(nu)du+ \int_0^4 f(x)sin(nx)dx= \int_0^4 f(u)sin(u)du+ \int_0^4 f(x)sin(nx)dx= 2\int_0^4 f(x)sin(nx)dx$$

Or, more simply, now that I look at it, since sin(nx) and f(x) are both odd, their product is an even function:
[tex]\int_{-4}^4 f(x)sin(nx)dx= 2\int_0^4 f(x)sin(nx)dx[/itex][/tex]

 

[math_04]

For the second part, it got kinda confusing because

$$-\int_{4}^{0}(-f(u))(-sin(nu)du+ \int_0^4 f(x)sin(nx)dx= \int_0^4 f(u)sin(u)du+ \int_0^4 f(x)sin(nx)dx= 2\int_0^4 f(x)sin(nx)dx[/quote] <br /> <br /> this whole sentence is kinda confusing haha, could u write with the mathematical symbols instead of the computer lingo. sorry about that. cheers.<br /> <br /> thanks for your reply.$$

 

[Defennder]

Which part is confusing? You didn't quote properly.

 

[kreil]

For the second part, it got kinda confusing because

$$- \int_{4}^{0}(-f(u))(-sin(nu))du+ \int_0^4 f(x)sin(nx)dx= \int_0^4 f(u)sin(u)du+ \int_0^4 f(x)sin(nx)dx= 2\int_0^4 f(x)sin(nx)dx$$

this whole sentence is kinda confusing haha, could u write with the mathematical symbols instead of the computer lingo. sorry about that. cheers.

thanks for your reply.

he uses integral properties to obtain this, namely

$$- \int_b^a f(x)dx = + \int_a^bf(x)dx$$

 

[HallsofIvy]

Which part is confusing? You didn't quote properly.

I didn't have the LaTex delineated properly. I have edited the post.

 

[math_04]

Isnt f(x) even

f(-x) = -(-x) - 4 = x - 4 which is even right?

and i still don't understand, you only got one integral so what do we use for f(x) -x-4? or 4-x?

 

[HallsofIvy]

Isnt f(x) even

f(-x) = -(-x) - 4 = x - 4 which is even right?

and i still don't understand, you only got one integral so what do we use for f(x) -x-4? or 4-x?

f(x) is defined by "f(x) = -4-x for -4 <= x <= 0 and 4-x for 0 <= x <= 4". Which of those formulas you use depends on the interval of integration of course. If you are integrating from 0 to 4, then you use 4- x.