পাওয়ার সিরিজের মাধ্যমে অয়লারের ফর্মুলা

To complete this proof, we need to be familiar with the power series for $e^x,\,\cos x,\,$and $\sin x$.

So, just as a refresher:

Great! Now that we have that out of the way, let's revisit $\cos x$ and $\sin x$: $$\cos x=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\cdots$$ $$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots$$

Looking at these series, one thing should be immediately apparent: they complement each other. $\cos$ has all the even powers of $x$ and $\sin$ has all the odd ones.

So the next question is: how can we combine the two?

Well, the main thing is to get rid of that pesky $(-1)^n$ in each of their sums, since they prevent us from getting a clean expression for each coefficient. So, we "absorb" the $-1$ into the power series:

$$\cos x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{i^{2n}x^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{(xi)^{2n}}{(2n)!}$$

Similarly,

$$\sin x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{i^{2n}x^{2n+1}}{(2n+1)!}=i^{-1}\sum_{n=0}^\infty \frac{(xi)^{2n+1}}{(2n+1)!}$$

Or, $i\sin x=\sum_{n=0}^\infty \frac{(xi)^{2n+1}}{(2n+1)!}$

Now, we get:

$$\cos x+i\sin x=1+(ix)+\frac{1}{2!}(ix)^2+\frac{1}{3!}(ix)^3+\frac{1}{4!}(ix)^4+\frac{1}{5!}(ix)^5+\cdots$$

But wait! That series looks familiar!

That's right, it's the power series of $e^{ix}$! As it turns out,

$$\cos x+i\sin x=\sum_{n=0}^\infty\frac{(ix)^n}{n!}=e^{ix}$$

This powerful result is known as Euler's Formula, and can be used to show results such as DeMoivre's Theorem and Trigonometric double-angle formulas.

Additionally, this gives rise to what is often referred to as "the most beautiful equation in math," Euler's identity: $$e^{i\pi}+1=0$$

However, some may argue that $e^{i\tau}=1$ is more aesthetically pleasing (see section 2.3: Euler's Identity).

Warning: The manipulations we did earlier weren't entirely justified. We never defined power series for complex numbers, and we didn't worry about rearranging terms in the series (actually, the latter problem is easy to fix by noting that the Taylor series in question are absolutely convergent). So the real issue is that we haven't defined what it means to have a complex power series or how to talk about convergence of a complex power series - these topics are part of complex analysis.

Euler's Formula is not one necessary for most basic algebra classes but is something that still fascinates many students. Throughout one's studies of mathematics, he or she will likely see the usage of $e$, $\pi$, and $i$ in many different contexts, but rarely in the same ones. This formula is a marriage of these three very different constants. The equation for Euler's Formula is

$e^{i\pi} + 1 = 0$

This allows for an interesting set of manipulations to be done with cool results. Consider the Maclaurin Series for $e^x$:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Now plug in $x = i\pi$:

$e^{i\pi} = 1 + i\pi + \frac{(i\pi)^2}{2!} + \frac{(i\pi)^3}{3!} + ...$

Considering that $i^2 = -1$, this allows the series to be rewritten as

$e^{i\pi} = 1 + i\pi - \frac{\pi^2}{2!} - \frac{i\pi^3}{3!} + \frac{\pi^4}{4!} + ...$

Substituting $e^{i\pi} = -1$ and moving all the constants to one side gives

$i\pi - \frac{\pi^2}{2!} - \frac{i\pi^3}{3!} + \frac{\pi^4}{4!} + ... = -2$

Euler's Formula

We've already seen a nice way to represent complex numbers: cis notation. However, there's a certain interesting thing about these. When we multiply them, we add the arguments. So, let's explore this deeper.

Suppose we defined the function $f(\theta) = \cos\theta + i\sin\theta$. By deMoivre's formula, this satisfies: $$f(a+b) = f(a)f(b)$$

So, we can try to express complex numbers in some way that involves exponents for the argument. Therefore, we express $\text{cis}\theta = \lambda^\theta$ for some $\lambda$. Now, to try to find $\lambda$, we look at the power series expansion of $\cos\theta + i\sin\theta$: $$\begin{align} \cos\theta + i\sin\theta &= \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots\right) + i\left(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots\right) \\ &= 1 + ix - \dfrac{x^2}{2!} - \dfrac{ix^3}{3!} + \dfrac{x^4}{4!} + \dfrac{ix^5}{5!} - \cdots \\ &= 1 + ix + \dfrac{(ix)^2}{2!} + \dfrac{(ix)^3}{3!} + \dfrac{(ix)^4}{4!} + \dfrac{(ix)^5}{5!} + \cdots \\ &= \boxed{e^{ix}} \end{align}$$ So, we have $\lambda = e^i$, and we've discovered (and proved) euler's formula! $$e^{\theta i} = \cos\theta + i\sin\theta$$

That's it for this explanation about Euler's Formula. I hope you've learned something new!