Deriving Taylor series without applying Taylor's theorem.

Question

First, a neat little 'proof' of the Taylor series of $e^x$.

Start by proving with L'Hospital's rule or similar that

$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$

and then binomial expand into

$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-1)(n-2)(n-3)\dots(n-k+1))}{n^{k-1}}\frac{x^k}{k!}+\dots$$

Evaluating the limit, we are left with

$$e^x=1+x+\frac{x^2}2+\dots+\frac{x^k}{k!}+\dots$$

which is our well known Taylor series of $e^x$.

As dxiv mentions, we can exploit the geometric series:

$$\frac1{1-x}=1+x+x^2+\dots$$

$$\ln(1+x)=x-\frac12x^2+\frac13x^3-\dots$$

$$\arctan(x)=x-\frac13x^3+\frac15x^5-\dots$$

which are found by integrating the geometric series and variants of it.

I was wondering if other known Taylor series can be created without applying Taylor's theorem. Specifically, can we derive the expansions of $\sin$ or $\cos$?

Answer 1

The series for $\sin$ and $\cos$ can be derived from the expansion of $e^x$.

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040}+\cdots$$

Sub in $ix$ to get:

$$e^{ix}=1+ix+\frac{(ix)^2}{2}+\frac{(ix)^3}{6}+\frac{(ix)^4}{24}+\frac{(ix)^5}{120}+\frac{(ix)^6}{720}+\frac{(ix)^7}{5040}+\cdots$$

$$\cos x+i\sin x=1+ix-\frac{x^2}{2}-\frac{ix^3}{6}+\frac{x^4}{24}+\frac{ix^5}{120}-\frac{x^6}{720}-\frac{ix^7}{5040}+\cdots$$

Compare real and imaginary parts:

$$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots$$

$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots$$

EDIT:

Consider the function $f(x)=(\cos x+i\sin x)e^{-ix}$

$$f'(x)=(-\sin x+i\cos x)e^{-ix}-i(\cos x+i\sin x)e^{ix}$$

$$f'(x)=-e^{ix}\sin x+ie^{ix}\cos x-ie^{-x}\cos x+e^{ix}\sin x$$

$$f'(x)=0$$

Hence $f(x)=c$ and $f(0)=(\cos0+i\sin0)e^0=1$ so $f(x)=1$

Therefore $e^{ix}=\cos x+i\sin x$

SECOND EDIT:

Another way springs to mind as well:

$$f(x)=\cos x+i\sin x$$

$$f'(x)=-sin(x)+i\cos x$$

$$f'(x)=i(\cos x+i\sin x)$$

$$f'(x)=i\cdot f(x)$$

$$\frac{f'(x)}{f(x)}=i$$

$$\ln(f(x))=ix+c$$

$$f(x)=e^{ix+c}$$

$$f(0)=\cos 0+i\sin 0=1\implies c=0$$

$$\therefore f(x)=e^{ix}$$

Answer 2

Alan Turing, at a young age, derived the series expansion of $\arctan$ without using (and, purportedly without knowing) calculus whatsoever.

Using the identity

$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x},$$

he obtained

$$\tan(2 \arctan x) = \frac{2x}{1-x^2},$$

and

$$2 \arctan x = \arctan\left( \frac{2x}{1-x^2}\right).$$

Using the geometric series for $|x| < 1$,

$$\tag{1}2 \arctan x =\arctan [2x(1 + x^2 + x^4 + x^6 + \ldots)]$$

Assuming $\arctan x = a_0 + a_1x + a_2x^2 + \ldots$ and matching coefficients in the expansions of each side of (1), he obtained

$$\arctan x = a_1\left(x - \frac{1}{3} x^3 + \frac{1}{5}x^5 - \ldots \right).$$

Some basic trigonometry reveals

$$a_1 = \lim_{x \to 0} \frac{\arctan x}{x} = 1.$$

Answer 3

There is a neat trick which allows the Taylor series to "suggest itself". It is not a rigorous derivation whatsoever, but maybe it satisfies what you want.

Consider the following "chain", where each is the derivative of the previous

$$\sin(x),$$ $$\cos(x),$$ $$-\sin(x),$$ $$-\cos(x).$$

Since $\cos(0)=1$, let's write $\cos(x)$ as $1+\text{something}$. We get $$\sin(x)=?,$$ $$\cos(x)=1+?,$$ $$-\sin(x)=?,$$ $$-\cos(x)=-1+?.$$ Since the derivative of $\sin$ is $\cos$, it is a nice guess then that $\sin(x)$ is equal to $x+ \text{something}$. We get $$\sin(x)=x+?,$$ $$\cos(x)=1+,$$ $$-\sin(x)=-x+?,$$ $$-\cos(x)=-1+?.$$ Since the derivative of $\cos$ is $-\sin$, it is a nice guess that $\cos(x)$ has a factor of $-x^2/2$. We get $$\sin(x)=x+?,$$ $$\cos(x)=1-\frac{x^2}{2}+?,$$ $$-\sin(x)=-x+?,$$ $$-\cos(x)=-1+\frac{x^2}{2}+?.$$ Since the derivative of $\sin$ is $\cos$, it is a nice guess that $\sin(x)$ is equal then to $x-x^3/3\cdot 2 + \text{something}$. We get $$\sin(x)=x-\frac{x^3}{3\cdot 2}+?,$$ $$\cos(x)=1-\frac{x^2}{2}+?,$$ $$-\sin(x)=-x+ \frac{x^3}{3\cdot 2}+?,$$ $$-\cos(x)=-1+\frac{x^2}{2}+?.$$ Rinse and repeat.

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Addendum: Now that the series suggest themselves, let $c(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$ and $s(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$. Here we need some more theory to justify why those things are well-defined for all $x \in \mathbb{R}$ (and also to derivative termwise as will be used to the following).

It is clear that $s'=c$ and $c'=-s$. Also that $s(0)=0$ and $c(0)=1$.

Now, consider

$$h(x)=(s-\sin)^2+(c-\cos)^2.$$

Derivating, we get $$h'=2(s-\sin)(c-\cos)+2(c-\cos)(-s+\sin)\equiv 0$$ Hence, $h$ is constant. It is clear that $h(0)=0$. Hence, $h(x)=0$ for all $x$. But this is only possible if $s=\sin$ and $c=\cos$.

Answer 4

[A].From "101 Great Problems In Elementary Mathematics" by H. Dorrie : For $x\geq 0$ we have

(1).$\;\sin x \leq x\implies 1-\cos x=\int_0^x\sin y \;dy\leq \int_0^xy\;dy=x^2/2\implies \cos x\geq 1-x^2/2!.$

(2). From (1), $\;\sin x=\int_0^x\cos y\;dy\geq \int_0^y(1-y^2/2!)dy=x-x^3/3!.$

(3).From (2), $1-\cos x=\int_0^x\sin y dy\geq \int_0^x(y-y^3/3!)dy=y^2-y^4/4!\implies \cos x\leq 1-x^2/2!+x^4/4!.$

Et cetera.

(4). So in general for $n>0$ we have $$\sum_{j=1}^{2n}(-1)^{j-1}x^{2j-1}/(2j-1)!\leq \sin x \leq\sum_{j=1}^{2n-1}(-1)^{j-1}x^{2j-1}/(2j-1)!$$ and a similar set of inequalities for $\cos x$.

This gives us the power series for $\sin x$ and $\cos x$ for $x\geq 0.$ Since $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x,$ this gives us their power series for all real $x.$

[B]. For $x\geq 0:$

(1). $e^{-x}\leq 1\implies 1-e^{-x}=\int_0^xe^{-y}dy\leq \int_0^x1dy=x\implies e^{-x}\geq 1-x.$

(2).From (1), $$1-e^{-x}=\int_0^xe^{-y}dy\geq \int_0^x(1-y)dy=y-y^2/2!$$ $$\implies e^{-x}\leq 1-x+x^2/2!.$$

(3). From (2), $$1-e^{-x}=\int_0^xe^{-y}dy\leq \int_0^x(1-y+y^2/2!)dy=x-x^2/2!+x^3/3!$$ $$\implies e^{-x}\geq 1-x+x^2/2!-x^3/3!.$$ Et cetera.

This gives us the power series for $e^{-x}$ for $x\geq 0.$ And it is an elementary exercise to show that $\lim_{n\to \infty}(\sum_{j=0}^nx^{-j}/j!)(\sum_{j=0}^nx^j/j!)=1,$ so we obtain the series for $e^x$ for all real $x.$