Expanding the Function (x^2 sin x) in a Fourier Series

The concept of Fourier series is fundamental in the analysis and representation of periodic functions. This article explores the expansion of the function f(x) x^2 sin x over the interval [0, pi], focusing on its Fourier sine series representation. Understanding this process can provide valuable insights into the orthogonality of trigonometric functions and their application in various fields of mathematics and engineering.

Introduction to Fourier Series

A Fourier series is a way to express a periodic function as a sum of sine and cosine functions. However, when the function is defined on a semi-infinite interval, as in our case, the Fourier sine series is more appropriate. The Fourier sine series for a function f(x) on the interval [0, L] is given by:

f(x) ~ sum_{n1}^{infty} b_n sinleft(frac{npi x}{L}right)

The coefficients b_n are computed using the following integral formula:

b_n frac{2}{L} int_0^L f(x) sinleft(frac{npi x}{L}right) dx

Defining the Problem

In our case, the function is f(x) x^2 sin x and the interval is [0, pi]. Since the function has no cosine terms and is defined on a semi-infinite interval, we derive the Fourier sine series representation. Here, L pi, so the formula for the coefficients simplifies to:

b_n frac{2}{pi} int_0^{pi} x^2 sin x sin n x , dx

Calculating the Coefficients b_n

The first step is to simplify the integrand x^2 sin x sin n x. Using the product-to-sum identities, we have:

sin x sin n x frac{1}{2} left(cos (n-1)x - cos (n 1)xright)

Substituting this into the integral, we get:

b_n frac{2}{pi} int_0^{pi} x^2 cdot frac{1}{2} left(cos (n-1)x - cos (n 1)xright) dx

This simplifies to:

b_n frac{1}{pi} left(int_0^{pi} x^2 cos (n-1)x , dx - int_0^{pi} x^2 cos (n 1)x , dxright)

Evaluating the Integrals

To evaluate these integrals, we use integration by parts. Let's denote:

I_k int_0^{pi} x^2 cos kx , dx

Using integration by parts, we let:

u x^2 and dv cos kx , dxdu 2x , dx and v frac{sin kx}{k}

Applying integration by parts, we have:

I_k left[ x^2 cdot frac{sin kx}{k} right]_0^{pi} - int_0^{pi} frac{sin kx}{k} cdot 2x , dx

The boundary term evaluates to zero since sin kpi 0. Therefore:

I_k -frac{2}{k} int_0^{pi} x sin kx , dx

Again, using integration by parts on the integral of x sin kx:

u x and dv sin kx , dxdu dx and v -frac{cos kx}{k}

We get:

int_0^{pi} x sin kx , dx left[ -frac{x cos kx}{k} right]_0^{pi} frac{1}{k} int_0^{pi} cos kx , dx

The boundary term evaluates to zero and the integral of cos kx over [0, pi] is zero for non-zero k. Therefore:

int_0^{pi} x sin kx , dx 0

Substituting back into our expression for I_k, we find:

I_k -frac{2}{k} cdot 0 0

For both n-1 and n 1, we get b_n 0. Hence, all coefficients b_n for n geq 1 are zero, leading us to conclude:

x^2 sin x ~ 0

Conclusion

The Fourier sine series expansion of x^2 sin x over [0, pi] yields a series that converges to zero, indicating that x^2 sin x is orthogonal to the sine functions on this interval. This result is fascinating and demonstrates the orthogonality of trigonometric functions in a practical context. If you wish to explore different forms of expansion or need further assistance, feel free to ask!