# Deriving Taylor series without applying Taylor's theorem.

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 $\cos$?

• For example $\frac{1}{1+x}$ by the geometric progression sum formula, and $\ln(1+x)$ by integrating the former.
– dxiv
Commented Nov 8, 2016 at 23:19
• @dxiv Ah, well, you can get quite a lot from $\frac1{1-x}$. Definitely forgot about that. And from there, we get $\ln$ and $\arctan$... Hm, good ones. I guess I was more specifically looking at the trig functions. Commented Nov 8, 2016 at 23:20
• The Mandhavan school, as well as Newton and Leibniz, found the series expansions of sine and cosine before knowing anything about Taylor series in general by using what in modern terms would be called the binomial expansion of the arcsine's integral expression. Commented Nov 9, 2016 at 0:11
• An observation: Your "evaluation of the limit" to derive the series for $e^x$ is not properly justified. The sum has number of terms depending on $n$, you can't pass the limit "termwise" like that without some justification. Commented Nov 9, 2016 at 0:44
• @AloizioMacedo According to generalized binomial expansion, the expansion goes on forever, independent of $n$. It just happens to be that for $n\in\mathbb N$, we eventually reach $(n-1)(n-2)(n-3)\dots(n-n)$, which equals $0$, and so will all terms thereafter. Commented Nov 9, 2016 at 0:49

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

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

• @user254665 I love all of these, but I can't upvote them more than once nor can I mark them all with the awesome green checkmark. Commented Nov 9, 2016 at 1:04
• That's quite alright............ Commented Nov 9, 2016 at 1:06
• @user254665: and Simple Art. Thanks. Nice to see this sort of question every now and again.
– RRL
Commented Nov 9, 2016 at 1:52
• See See this question Commented Nov 9, 2016 at 8:45
• @RRL I try to post these types of questions every day. Commented Nov 9, 2016 at 14:34

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:

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

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

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

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

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

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

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

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

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

• Then how do I prove $e^{ix}=\cos(x)+i\sin(x)$ without the Taylor expansion? Commented Nov 9, 2016 at 0:28
• @SimpleArt Thought of an additional way. Commented Nov 9, 2016 at 0:55
• Hm, that's quite funny your proof. I basically found it on Wolfram the other day... Commented Nov 9, 2016 at 1:03
• I can't remember where I originally saw it. It might have been there. Commented Nov 9, 2016 at 1:08

[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 $x.$

• Basically equivalent to Aloizio Macedo's answer, but I can't complain. Commented Nov 9, 2016 at 1:00

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.

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 $c=\cos$.

• Nice. All it does is assume $\cos(x)=1+?$, with another catch that it also assumes the $?$ approaches $0$ as we expand. Requires a periodic derivative though. Commented Nov 9, 2016 at 0:47
• @SimpleArt As I said, the intention is not to be a proof, but to be a way for the series to suggest itself. I'll add a proof that the series indeed represent the functions to the answer. Commented Nov 9, 2016 at 0:50
• Oh no, that's not necessary for the question. :) If I could, I'd mark your answer as "answers my question" along with the other one. Commented Nov 9, 2016 at 0:52
• Derivating is a word? I do think you meant differentiating. Commented Nov 9, 2016 at 1:01