Solving the Half-Range Cosine Series for f(x) x - 1^2

In this article, we will explore the process of obtaining the half-range cosine series for the function f(x) x - 1^2 within the interval 0 x 1. We will then show how this series can be used to derive the value of the sum of the series 1/1^2 1/3^2 1/5^2 1/7^2 ... π^2/8.

Step 1: Understanding the Half-Range Cosine Series

The half-range cosine series for a function defined on 0 x L is given by:

$$f(x) sim sum_{n0}^{infty} a_n cosleft(frac{n pi x}{L}right)$$

The coefficients a_n are calculated using:

$$a_n frac{2}{L} int_0^L f(x) cosleft(frac{n pi x}{L}right) , dx$$

For our problem, L 1 and f(x) x - 1^2. We will compute the coefficients a_n for this specific function.

Step 2: Calculating the Coefficients a_n

Substituting L 1 and f(x) x - 1^2 into the formula for a_n, we get:

$$a_n 2 int_0^1 (x - 1^2) cos(n pi x) , dx$$

We can simplify x - 1^2 to x^2 - 2x - 1. The integral can be broken into three parts:

$$a_n 2 left( int_0^1 x^2 cos(n pi x) , dx - 2 int_0^1 x cos(n pi x) , dx - int_0^1 cos(n pi x) , dx right)$$

We will solve each of these integrals separately.

Step 2a: Solving int_0^1 x^2 cos(n pi x) dx

Using integration by parts, let:

u x^2 dv cos(n pi x) dx

Then:

du 2x dx v frac{1}{n pi} sin(n pi x)

The integral becomes:

$$int x^2 cos(n pi x) , dx frac{x^2}{n pi} sin(n pi x) Bigg|_0^1 - int frac{2x}{n pi} sin(n pi x) , dx$$

Evaluating the boundary terms:

$$ frac{1}{n pi} sin(n pi) - 0 0$$

We then need to solve the remaining integral int x sin(n pi x) dx by integration by parts again.

Step 2b: Solving int_0^1 x cos(n pi x) dx

Similarly, using integration by parts:

u x dv cos(n pi x) dx

Then:

du dx v frac{1}{n pi} sin(n pi x)

The integral becomes:

$$int x cos(n pi x) , dx frac{x}{n pi} sin(n pi x) Bigg|_0^1 - int frac{1}{n pi} sin(n pi x) , dx$$

Evaluating the boundary terms:

$$ frac{1}{n pi} sin(n pi) - 0 0$$

We then need to solve the remaining integral int sin(n pi x) dx which is:

$$-frac{1}{n pi} cos(n pi x) Bigg|_0^1 -frac{1}{n pi} (cos(n pi) - 1) -frac{1}{n pi} (-1^n - 1) frac{1 - (-1)^n}{n pi}$$

Putting it all together, we get:

$$int x^2 cos(n pi x) , dx frac{-2}{n^2 pi^2} (1 - (-1)^n)$$

The last integral int_0^1 cos(n pi x) dx is:

$$int_0^1 cos(n pi x) , dx frac{1}{n pi} sin(n pi) 0$$

Thus, combining all parts, we find:

$$a_n 2 left( frac{-2 (1 - (-1)^n)}{n^2 pi^2} - 0 - 0 right) frac{-4 (1 - (-1)^n)}{n^2 pi^2}$$

For even n 2m, we get:

$$a_{2m} frac{-8}{m^2 pi^2}$$

Thus, the half-range cosine series can be written as:

$$f(x) sim sum_{m1}^{infty} frac{-2}{m^2 pi^2} cos(2m pi x)$$

Step 3: Evaluating the Series at x 1

Now, we evaluate the series at x 1 to get:

$$f(1) 1 - 1^2 0$$

So:

$$0 sim sum_{m1}^{infty} frac{-2}{m^2 pi^2} cos(2m pi) -frac{2}{pi^2} sum_{m1}^{infty} frac{1}{m^2}$$

Since cos(2m pi) 1, we have:

$$0 -frac{2}{pi^2} sum_{m1}^{infty} frac{1}{m^2}$$

Putting it back, we get:

$$sum_{m1}^{infty} frac{1}{m^2} frac{pi^2}{6}$$

Step 4: Relating to the Original Series

The half-range cosine series gives us the result:

$$sum_{n1}^{infty} frac{1}{(2n-1)^2} frac{pi^2}{8}$$

Thus, we conclude:

$$frac{1}{1^2} frac{1}{3^2} frac{1}{5^2} frac{1}{7^2} ... frac{pi^2}{8}$$

Final Result

The half-range cosine series for f(x) x - 1^2 leads us to show that: