Find the fourier series of (sinx)^3

In summary, the listener was confused about finding the Fourier series of (sinx)^3 and asked if there was an easier way to solve it. The speaker suggested using standard identities and expanding it as a trigonometric polynomial using sin(x) and cos(x). They also mentioned using complex definitions to simplify the process.
  • #1
fhidiort
2
0
I'm confused.

Trying to find the Fourier series of (sinx)^3. This is an odd function, so I try to find the Fourier sine coefficient, with integral of (sinx)^3*sinkx. However, my answer comes up with all sine terms. Of course all these terms go to zero when integrating between 0 and pi. Am I doing something wrong? Or is there some identity for sinx^3 that makes it easier to solve?
 
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  • #2
fhidiort said:
Of course all these terms go to zero when integrating between 0 and pi. Am I doing something wrong?

If you draw a graph of the function you are trying to integrate, you will see that you are doing something wrong, since the area under the curve isn't zero.

There are some standard identities that will help. Try expanding sin 3x = sin (2x + x), then sin 2x = sin(x+x) and cos2x = cos(x+x) and see what you get.
 
  • #3
So you're telling me that

∫_{0}^{π}sin kxsin³xdx =0

necessarily?

What if k=3 or k=1 ?

Daniel.
 
  • #4
fhidiort said:
I'm confused. Trying to find the Fourier series of (sinx)^3.
Actually, you don't even have to go through all that suffering. Whenever you want to find the Fourier series of a [tex]\sin^3(x)[/tex] or [tex]\cos^5(x)[/tex] or something, you just have to expand it out as a sum of first order sine and cosine terms--just like you would expand [tex]\sin(x)^[/tex] into [tex]\frac{1-cos(2x)}{2}[/tex].

In order to expand things like [tex]\sin^3(x)[/tex] or [tex]\cos^5(x)[/tex], just replace the sine and cosine term with its standard complex definition i.e.

[tex]\cos z=\frac{e^{iz}+e^{-iz}}2[/tex]

[tex]\sin z=\frac{e^{iz}-e^{-iz}}{2i}[/tex]

and then just simplify and convert everything back to rectangular form.
 
  • #5
Use trig identities to write it as a "trigonometric polynomial". [itex]sin^2(x)= \frac{1}{2}(1- cos(2x))[/itex] so [itex]sin^3(x)= \frac{1}{2}(sin(x)- sin(x)cos(2x))[/itex]. Since sin(a+b)= sin(a)cos(b)+ cos(a)sin(b) and sin(a-b)= sin(a)cos(b)- cos(a)sin(b), sin(a)cos(b)= (1/2)(sin(a+b)+ sin(a-b)) and, in particular, sin(x)cos(2x)= (1/2)(sin(3x)- sin(x)). That is, [itex]sin^3(x)= \frac{1}{2}sin(x)- \frac{1}{4}(sin(3x)- sin(x)= \frac{3}{4}sin(x)- \frac{1}{4}sin(3x)[/itex] (modulo arithmetic errors!).
 

1. What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sine and cosine functions. It is used to analyze and approximate periodic signals in various fields such as physics, engineering, and mathematics.

2. How do you find the Fourier series of a function?

To find the Fourier series of a function, you need to follow a series of steps. First, you need to determine the period of the function. Then, you need to find the coefficients of the sine and cosine functions by using the Fourier series formulas. Finally, you can express the function as a sum of these coefficients and sine and cosine functions.

3. What is the period of (sinx)^3?

The period of (sinx)^3 is 2π, which is the same as the period of the sine function. This is because raising a periodic function (such as sine) to a power does not change its period.

4. Can the Fourier series of (sinx)^3 be simplified?

Yes, the Fourier series of (sinx)^3 can be simplified by using trigonometric identities. For example, you can use the identity sin^3(x) = (3sin(x) - sin(3x))/4 to simplify the series.

5. What is the significance of finding the Fourier series of a function?

Finding the Fourier series of a function allows us to represent the function as a sum of simpler trigonometric functions. This can help in understanding the behavior of the function and making predictions about its future values. It also has practical applications in signal processing, control systems, and other fields.

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