Q19Mathematics I

Fourier series allow us to represent almost any periodic function, even those with sharp discontinuities (like square waves), as an infinite sum of smooth, continuous sine and cosine waves. This specific problem requires us to derive the Fourier series for a standard periodic square wave and then use that exact mathematical series to prove one of the most famous identities in mathematical history: the Gregory-Leibniz infinite series for $\pi$.

Step 1: Defining the Square Wave Function

Consider a classic odd square wave function, $f(x)$, formally defined over a single full fundamental period of $[-\pi ,\pi ]$ as follows:

The first critical analytical step in any Fourier problem is to check for symmetry. We systematically test the function: $f(-x)$ for a positive $x$ (which falls in the negative interval) gives $-1$, which is exactly $-f(x)$. Since $f(-x)=-f(x)$ for all $x$, this function is rigorously classified as an odd function. A fundamental, mathematically proven property of Fourier series dictates that any purely odd function will consist exclusively of sine terms. Therefore, we can immediately and definitively state that the cosine coefficients $a_0=0$ and $a_n=0$ for all $n$.

Step 2: Calculating the Sine Coefficients ($b_n$)

We are now left to compute only the sine coefficients, $b_n$, using the standard Fourier definition:

Because the product of an odd function ($f(x)$) and another odd function ($\sin (nx)$) results mathematically in an strictly even function, we can dramatically simplify the calculation by integrating only over the positive half of the interval and doubling the result:

On the strictly positive interval $[0,\pi ]$, our defined function $f(x)$ is simply the constant $1$. Substituting this into the integral gives:

We now execute the basic integration. The antiderivative of $\sin (nx)$ is $-\frac{\cos (nx)}{n}$:

We apply the absolute mathematical identities: $\cos (0)=1$ and $\cos (n\pi )=(-1{)}^n$ for any integer $n$. Substituting these gives the final formula for the coefficients:

We must meticulously analyze this coefficient for different values of $n$: - If $n$ is an even integer ($2,4,\mathrm{6...}$), then $(-1{)}^n=1$. The parenthesis becomes $(1-1)=0$. Thus, $b_n=0$ for all even harmonics. - If $n$ is an odd integer ($1,3,\mathrm{5...}$), then $(-1{)}^n=-1$. The parenthesis becomes $(1-(-1))=2$. Thus, $b_n=\frac{2}{n\pi }(2)=\frac{4}{n\pi }$. Therefore, the complete, rigorously derived Fourier series for the square wave is:

Step 3: Proving the Leibniz Series for Pi

To extract a profound mathematical identity from this series, we must strategically evaluate it at a highly specific point of continuity. We strictly choose the point $x=\pi /2$. At exactly $x=\pi /2$, the original defined function $f(x)$ sits perfectly flat at a value of exactly $1$. Therefore, $f(\pi /2)=1$.

We now substitute $x=\pi /2$ into every single term of our derived infinite Fourier series:

We systematically evaluate each specific trigonometric sine term: - $\sin (\pi /2)=1$ - $\sin (3\pi /2)=-1$ - $\sin (5\pi /2)=1$ - $\sin (7\pi /2)=-1$ This precise alternating pattern $(1,-1,1,-1)$ continues infinitely.

Substituting these exact values back into the equation yields the beautiful alternating series:

To finalize the mathematical proof, we simply multiply both sides of the absolute equation by $\pi /4$, successfully isolating the series and arriving at the legendary Leibniz formula: