Q. 2TF - Calculus

Question

Euler’s Formula: Use the results of the previous exercise and the Maclaurin series for the sine, the cosine, and the exponential function to derive Euler’s formula \(e^{ix}=cosx+isinx\).

Step-by-Step Solution

Answer

By using the Maclaurin series for the sine, the cosine, and the exponential function, we get Euler’s formula \(e^{ix}=cosx+isinx\). 

Step 1. Derive Euler's formula

To derive Euler's formula we will use the Maclaurin series for the sine, the cosine, and the exponential function. 

So, the Maclaurin series of \(sinx\) is

\(sinx=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n+1 \right )!}x^{2n+1}\)

\(sinx=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-... \)

And the Maclaurin series of \(cosx\) is 

\(cosx=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n \right )!}x^{2n}\)

\(cosx=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-... \)

And the Maclaurin series of the exponential function is

\(e^{z}=\sum_{n=0}^{\infty }\frac{z^{n}}{n!}\)

\(e^{z}=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+... \) ......(a)

Step 2. Derive Euler's formula

Let's substitute \(z=ix\) in the equation (a)

\(e^{ix}=1+ix+\frac{ix^{2}}{2!}+\frac{ix^{3}}{3!}+...\) 

\(e^{ix}=1+ix-\frac{x^{2}}{2!}-i\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+i\frac{x^{5}}{5!}\)  

\(e^{ix}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+...+i\left ( x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} \right )\) 

\(e^{ix}=cosx+isinx\)