Determine the Fourier Series

  • Thread starter lycraa
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  • #1
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Homework Statement



determine the Fourier Series for f(x)=cos3x

Homework Equations



f(x)=ao/2+(sum) an cos(nx)+ (sum) bncos(nx)

ao= (integral) f(x)dx (from -[tex]\pi[/tex] to[tex]\pi[/tex])

an= (integral) f(x)cos(nx)dx (from -[tex]\pi[/tex] to[tex]\pi[/tex])

bn= (integral) f(x)sin(nx)dx (from -[tex]\pi[/tex] to[tex]\pi[/tex])

The Attempt at a Solution



i worked out that ao and an are both zero, which is fine. however when i go to work out bn i get answers that are divided by (n-1) and (n-3) which means that when i try and find b1 and b3 I'm dividing by zero. i don't know what to do now! can this function be made into a Fourier Series?
 

Answers and Replies

  • #2
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have i done the integration wrong here?I feel liek i must have sence i dont really know how i would go about integrating cos3x sin(nx) OR cos3x cos(nx)
 
  • #3
HallsofIvy
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I don't see how you could get that "a0 and an" are 0. Since \(\displaystyle cos^3(x)sin(x)\) is an odd function, it immediately follows that the integral from [itex]-\pi[/itex] to [itex]\pi[/itex] is 0- that is that [itex]b_n= 0[/itex] for all n.

And, it is easy to show that [itex]a_0= 0[/itex] but
[tex]a_1= \int_{-\pi}^{\pi}cos^4(x)dx[/tex]

Since [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex], [itex]cos^4(x)= (1/4)(1+ cos(2x))^2=[/itex][itex] 1/4+ (1/2)cos(2x)+ (1/4)cos^2(2x))[/itex]
[itex]= 1/4+ (1/2)cos(2x)+ (1/4)(1+ cos(4x))=[/itex][itex] (3/4)+ (1/2)cos(2x)+ (1/4)cos(4x)[/itex]

Now, since the integral of cosine is sine and sine is 0 at any integer multiple of [itex]\pi[/itex], those two cosine integrals will be 0 but the integral of 3/4, from [itex]-\pi[/itex] to [itex]\pi[/itex] will be [itex]3\pi/2[/itex], not 0.

By the way, I think you are missing the normalzing factor of [itex]1/(2\pi)[/itex] in front of the integrals. [itex]a_1= (3\pi/2)(1/2\pi)= 3/4[/itex], no 0.

For higher order terms, to integrate3 [itex]cos^3(x)cos(nx)[/itex], with n> 1, you will need to reduce either [itex]cos^3(x)[/itex] or [itex]cos(nx)[/itex].
 
  • #4
vela
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You can avoid doing any integrals by doing as HallsofIvy did for cos4 x and use trig identities to expand cos3 x.
 

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