Maclaurin and Taylor series

Write a power series as:

$f(x)=\sum^\infty_0\frac{a_n}{n!}x^n$

$\text{i.e. }\quad f(x)=a_0+a_1x+\frac{a_2}{2}x^2+\frac{a_3}{3\times 2}x^3+\frac{a_4}{4\times 3\times 2}x^4+\dots$

When $x=0$ , every term is $0$ except the first, so $f(0)=a_0$ .
Differentiate the series:

$f'(x)&=a_1+a_2x+\frac{a_3}{2}x^2+\frac{a_4}{3\times 2}x^3+\frac{a_5}{4\times 3\times 2}x^4+\dots$

Now set $x=0$ and what remains is $f'(0)=a_1$ .
Differentiate again and get $f''(0)=a_2$ , and evidently $f^{(n)}(0)=a_n}$
So the series can be written in the form:

$\boxed{f(x)=\sum^\infty_0\frac{f^{(n)}(0)}{n!}x^n}$

This is the Maclaurin series.
If the derivatives are evaluated at $x=a$ instead, we get the Taylor series:

$\boxed{f(x)=\sum^\infty_0\frac{f^{(n)}(a)}{n!}(x-a)^n}$

series for $e$

For the function $y(x)=e^{ax}, \quad y'(x)=ae^{ax}, \quad y''(x)=a^2e^{ax}$ etc.
At $x=0$ , since $e^0=1$ , the $n$ th derivative is just $a^n$ .
So the series can be represented:

$e^{ax}=1+ax+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\frac{(ax)^4}{4!}+\dots$

When $x=1/a$ :

$e=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots$

series for $\sin$

When $y=\sin x,\quad y'=\cos x, \quad y''=-\sin x, \quad y'''=-\cos x$ etc.
The derivatives at $x=0$ are $0,1,0,-1\dots$ repeating endlessly.
So the series is:

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\dots$

series for $\cos$

When $y=\cos x,\quad y'=-\sin x, \quad y''=-\cos x,\quad y'''=\sin x,$ etc.
The derivatives at $x=0$ are $1,0,-1,0\dots$ repeating endlessly.
So the series is:

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\dots$

(or simply differentiating the $\sin$ series term by term gets the same series.)

Euler’s identity

Substitute $ax=i\theta$ into the series for $e$ to get:

$e^{i\theta}=1+(i\theta)+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}\dots$

Group the real and imaginary coefficients to get:

$\begin{align*} e^{i\theta}&=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots\right)+i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)\\ &=\cos\theta+i\sin\theta \end{align*}$

Plug in $\theta=\pi$ to get $e^{i\pi}=-1$ , or $e^{i\pi}+1=0$ , “the most famous formula in all mathematics”.
A poll of readers conducted by The Mathematical Intelligencer in 1990 named Euler’s identity as the “most beautiful theorem in mathematics”. In another poll of readers…by Physics World in 2004, Euler’s identity tied with Maxwell’s equations (of electromagnetism) as the “greatest equation ever”. – wikipedia
Great article here explaining its significance intuitively.

binomial expansion of $y=(1+x)^n$

Proceeding as above,

$\begin{align*} y(x)&=(1+x)^n & y(0)&=1 &\\ y'(x)&=n(1+x)^{n-1} & y'(0)&=n &\\ y''(x)&=n(n-1)(1+x)^{n-2} & y''(0)&=n(n-1) &\\ y'''(x)&=n(n-1)(n-2)(1+x)^{n-3} & y'''(0)&=n(n-1)(n-2) & \end{align*}$

$\begin{align*} y^{(k)}(x)&=n(n-1)(n-2)\dots (n-k+1)(1+x)^{n-k}\\ y^{(k)}(0)&=n(n-1)(n-2)\dots (n-k+1) \end{align*}$

So the series can be written:

$\begin{align*} (1+x)^n&=\sum^n_{k=0}\frac{y^{(k)}(0)}{k!}x^k\\ &=1+\frac{n}{1!}x+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3\\[8pt] &+\dots+\frac{n(n-1)(n-2)\dots (n-k+1)}{k!}x^k+\dots+\frac{n!}{n!}x^n \end{align*}$

The coefficient of $x^k$ is the binomial coefficient $C(n,k)$ , also written $\binom{n}{k}$ :

$\binom{n}{k}=\frac{n(n-1)(n-2)\dots (n-k+1)}{k!}=\frac{n!}{k!(n-k)!}$

Further reading

Hamming, Richard – Methods of Mathematics, ch20-21

adamponting.com

Maclaurin and Taylor series

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