Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?

Question

Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?

Can someone point me to a proof, or explain if it's a simple answer?

What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series. A lot of the information I find when looking for that seems to point back to arctan.

Answer 1

The derivative of the arc tangent is $$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}.$$

From the formula for geometric series (see for example this answer for a proof) shows that $$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if }|y|\lt 1.$$ Plugging in $-x^2$ for $y$, we get that $$\begin{align*} \frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\ &= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\ &= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots \end{align*}$$ provided that $|-x^2| \lt 1$; that is, provided $|x|\lt 1$. All the computations below are done under this hypothesis (see comments at the end).

So we have that: $$\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1$$ Because this is a Taylor series, it can be integrated term by term. That is, up to a constant, we have: $$\begin{align*} \arctan(x) &= \int\left(\frac{d}{dx}\arctan (x)\right)\,dx \\ &= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right)\,dx\\ &= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right)\,dx\\ &= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n}\,dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n}\,dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1}}{2n+1}\right) + C\\ &= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} +\cdots\right). \end{align*}$$ Evaluating at $x=0$ gives $0 = \arctan(0) = C$, so we get $$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \cdots,\qquad\text{if }|x|\lt 1.$$ the equality you ask about.

Note however that this does not hold for all $x$: it certainly works if $|x|\lt 1$, by the general properties of Taylor series. But the arc tangent is defined for all real numbers. The series we have here is $$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}.$$ Using the Ratio Test, we have that $$\begin{align*} \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3}}{2n+3}\quad}{\frac{|x|^{2n+1}}{2n+1}}\\ &= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3}}{(2n+3)|x|^{2n+1}}\\ &= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\ &= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\ &= |x|^2. \end{align*}$$ By the Ratio Test, the series converges absolutely if $|x|^2\lt 1$ (that is, if $|x|\lt 1$) and diverges if $|x|\gt 1$. At $x=1$ and $x=-1$, the series is known to converge. So the radius of convergence is $1$, and the equality is valid for $x\in [-1,1]$ only (that is, if $|x|\leq 1$; we gained two points in the process).

However, the arc tangent has a nice property, namely that $$\arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan(x),$$ So, given a value of $x$ with $|x|\gt 1$, you can use this identity to compute $\arctan(x)$ by computing $\arctan(\frac{1}{x})$ instead, and for this argument the series is valid.

Answer 2

Well the usual way to get this series representation for the $\arctan$ is to use the geometric series

$$\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x}$$

and then substitute $-x^2$ in it to get

$$\sum_{n=0}^{\infty} (-1)^n x^{2n} = \frac{1}{1 + x^2}$$

Now the next step is to integrate both sides and then you get

$$\arctan{x} = \int \frac{1}{1 + x^2} \, dx = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n + 1} }{2n + 1} + C$$

and you can easily show that the constant $C = 0$. You can find this done in almost any calculus book, it's one of the classic series that most calculus students must know I guess.

Answer 3

Using $$\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x},$$

$$\tan^{-1}(x)=\int_0^x\frac{1}{1+t^2}dt=\int_0^x\left(\sum_{k=0}^n(-t^2)^k+\frac{(-t^2)^{n+1}}{1+t^2}\right)dt.$$ This implies that $$\tan^{-1}(x)=\sum_{k=0}^n(-1)^k\frac{x^{2k+1}}{2k+1}+R_n(x)$$ where $$R_n(x)=\int_0^x\frac{(-t^2)^{n+1}}{1+t^2}dt$$ Since $$|R_n(x)|\le\int_{\min(0,x)}^{\max(0,x)}\left|\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}\right|dt\le \int_{\min(0,x)}^{\max(0,x)}\frac{t^{2n+2}}{t^2}dt=\frac{|x|^{2n+1}}{2n+1}$$ So for $-1\le x\le 1$ it follows that $R_n(x)\to 0$ when $n\to\infty$. Consequently $$\tan^{-1}(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{2k+1}\quad {-1}\le x\le 1$$

Answer 4

Let $f(x)=\arctan(x)$. Then $f'(x)=\frac{1}{1+x^2}$ and $f(0)=0$, so by the fundamental theorem of calculus, $f(x)=\int_0^x\frac{dt}{1+t^2}$ for all $x$. When $|t|<1$, the integrand can be expressed as a geometric series $\frac{1}{1+t^2}=1-t^2+t^4-t^6+t^8-\cdots$. This series converges uniformly on compact subintervals of $(-1,1)$, so when $|x|<1$ we can integrate term by term to get $$f(x)=\int_0^x 1-t^2+t^4-t^6+\cdots dt= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$

(Adrián Barquero posted while I was writing.)

Here is a way to see that $f'(x)=\frac{1}{1+x^2}$. By definition of $\arctan$, $\tan(f(x))=x$ for all $x$. Taking the derivative of both sides, using the chain rule on the left-hand side, yields $\tan'(f(x))\cdot f'(x)=1$. Now $\tan'=\sec^2=1+\tan^2$, so $(1+\tan^2(f(x)))\cdot f'(x)=1\Rightarrow (1+x^2)f'(x)=1\Rightarrow f'(x)=\frac{1}{1+x^2}.$

A similar method gives the power series expansion for $g(x)=\arcsin(x)$. You have $g'(x)=(1-x^2)^{-1/2}$ and $g(0)=0$, so by the fundamental theorem of calculus, $g(x)=\int_0^x(1-t^2)^{-1/2}dt$ for all $x$ with $|x|<1$. The integrand can be expanded using the binomial theorem and integrated term by term to obtain the power series.

Answer 5

You can show that $\arctan'(x) =\frac1{1+x^2} $ from the functional equation $\arctan(x)-\arctan(y) =\arctan(\frac{x-y}{1+xy}) $ (gotten from $\tan(x+y) =\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} $).

$\begin{array}\\ \arctan(x+h)-\arctan(x) &=\arctan(\frac{(x+h)-x}{1+(x+h)x})\\ &=\arctan(\frac{h}{1+(x+h)x})\\ \end{array} $

From $\sin(x) \approx x$ and $\cos(x) \approx 1-x^2/2$ for small $x$, $\arctan(x) \approx x$ so, for small $h$,

$\arctan(\frac{h}{1+(x+h)x}) \approx \frac{h}{1+x^2} $, so $\frac{\arctan(x+h)-\arctan(x)}{h} \approx \frac{1}{1+x^2} $.

Note how the $x^2$ (in $1+x^2$) comes from the $\tan(x)\tan(y)$ in the tangent addition formula.

Answer 6

Many of the functions you encounter on a regular basis are analytic functions, meaning that they can be written as a Taylor series or Taylor expansion. A Taylor series of $f(x)$ is an infinite series of the form $\sum_{i=0}^\infty a_ix^i$ which satisfies $f(x) = \sum_{i=0}^\infty a_ix^i$ wherever the series converges. Trigonometric functions are examples of analytic functions, and the series you are asking about is the Taylor series of $\operatorname{arctan}(x)$ about $0$ (the meaning of this is explained in the link). You can read more about Taylor series here.

Answer 7

Marty Cohen's answer gives an explanation for the following fact: $$ \frac{d}{dx} \arctan(x) = \frac{1}{1+x^2} $$ Here is an alternative explanation.

Let $y = \arctan(x)$, and try to find $\frac{dy}{dx}$. We have $$ y = \arctan(x) $$ $$ \tan(y) = \tan(\arctan(x)) $$

We'd like to simplify the right-hand side. The definition of $\arctan(x)$ is: $$ \arctan(x) = \textrm{the angle between }-\frac{\pi}{2} \textrm{ and } \frac{\pi}{2}\textrm{ whose tangent is } x. $$ So whatever $\arctan(x)$ is, its tangent must be $x$. That is, $$ \tan(\arctan(x)) = x. $$ So we have $$ \tan(y) = x $$ Now differentiate both sides with respect to $x$ (this is called implicit differentiation): $$ \frac{d}{dx} \tan(y) = \frac{d}{dx} x $$ $$ \sec^2(y) \frac{dy}{dx} = 1 $$ $$ \frac{dy}{dx} = \cos^2(y) $$ $$ \frac{d}{dx}\arctan(x) = \cos^2(\arctan(x)) $$ But it turns out that $\cos^2(\arctan(x)) = \frac{1}{{1+x^2}}$. To see this, you can draw a right triangle with vertices at $(0,0), (1, 0)$, and $(1, x)$. The angle at the origin is $\arctan(x)$, and you can easily compute its cosine. (Try it!)

Or, you can try some algebra. Using the notation from above, we have $$ \tan(y) = x $$ $$ \frac{\sin(y)}{\cos(y)} = x $$ $$ \sin(y) = x\cos(y) $$ $$ \sin^2(y) = x^2\cos^2(y) $$ $$ 1-\cos^2(y) = x^2 \cos^2(y) $$ $$ 1 = \cos^2(y)(1 + x^2), $$ $$ \frac{1}{1 + x^2} = \cos^2(y), $$ as desired.

Answer 8

Differentiate $\arctan(x)$ and evaluate it at $x=0$. Repeat. Divide the $n$th term by $n!$. You should get the series. Uh, it might be useful to note that
$$\frac{1}{1+x^2} = 1-x^2 +x^4-x^6+x^8 ...$$

Answer 9

There might have been too many answers but I would like to add my favorite one:

Let $y = \arctan x$. We know that $y' = \frac{1}{x^2 + 1}$, hence $y'(x^2+1) = 1$ (*).

Now differentiate the equation (*) $n - 1$ times ($n > 1$), with Newton-Leibniz rule we have: $$ (x^2 + 1) y^{(n)} + 2 (n - 1) x y^{(n-1)} + (n - 1)(n - 2) y^{(n-2)}. $$

We set $x = 0$, and get: $$ y^{(n)} = - (n - 1)(n - 2) y^{(n-2)}, $$

Notice that $y(0) = 0$, $y'(0) = 1$, $$ y^{(2k + 1)} = (-1)^k (2k)! \qquad y^{(2k)} = 0. $$

Therefore we finally get the Taylor's series: $$ \arctan x = \sum_{k \in \mathbb N} (-1)^k \frac{x^k}{2k + 1}. $$

Answer 10

I'll give answer that doesn't utilize the derivative of $\tan^{-1}x$. Consider the result

$$\tanh^{-1}x=-i\tan^{-1}ix$$ $$\implies\frac12\ln\left(\frac{1+x}{1-x}\right)=-i\tan^{-1}ix$$

Now, we can use the Maclaurin expansion for $\ln(1\pm x)$ to compute that of $\tan^{-1}x$. The series obtained so will converge for $x\in(-1,1]\cap[-1,1)=(-1,1)$ since $\ln(1+x)$ converges for $x\in(-1,1]$. It may or may not converge at $x=\pm1$ since at these points as of now, both sides of the above identity become undefined. So, this case needs to be investigated separately.

$\textbf{Case 1, $x\in(-1,1)$:}$

$$-i\tan^{-1}ix=\frac{\ln(1+x)-\ln(1-x)}2$$ $$=\frac{\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5\cdots\right)-\left(-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4-\frac{x^5}5\cdots\right)}2$$ $$=x+\frac{x^3}3+\frac{x^5}5\cdots$$

Replacing $x$ with $-ix$,

$$\tan^{-1}x=i\left(-ix+\frac{ix^3}3-\frac{ix^5}5\cdots\right)$$ $$=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7\cdots$$

$\textbf{Case 2, $x=\pm1$:}$

Let the value of the above series at $x=1$ be $S$:

$$S=1-\frac13+\frac15-\frac17\cdots$$

By the alternating series test, $S$ converges. Similarly, the value of the series of $\tan^{-1}x$ also converges for $x=-1$. Alternatively, this can be explained by the fact that $\tan^{-1}x$ is an odd function.

Hence, the power series expansion for $\tan^{-1}x$ converges $\forall x\in[-1,1]$.

Answer 11

You can show that this series converges for $r = 1$ using the alternating series test. The sequence of the absolute values of the terms of this series is monotonically decreasing and approach zero. Hence the series converges for $r = 1$.

Also roughly speaking you need ten times as many terms in the partial sums for the approximation of pi to be one decimal place more accurate. i.e. If if the first 100 terms have about two decimal places of accuracy, then the first 1000 will give you about three decimal places of accuracy. Thus, the current record of 12 trillion digits would require a summation of about $10^{12000000000000}$ terms with this series.