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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.

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Based on the answers, perhaps what I'm really looking for is the proof that the derivative of arctan is $\frac{1}{1+x^2}$. –  gerber Mar 29 '11 at 3:39
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The derivative of $arctan x$ follows from: $$\begin{align}f \circ f^{-1} (x) = x &\Longrightarrow f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1 \\ &\Longrightarrow (f^{-1})'(x) = 1 /f'(f^{-1}(x)) \end{align}$$ (assuming all terms appearing exist of course. Now use $f(x) = \tan x$. –  JavaMan Mar 29 '11 at 3:48
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This video cleared it up for me. youtube.com/watch?v=tky25AUK7Io –  gerber Mar 29 '11 at 4:45
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5 Answers

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.

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Minor typo in the exponent of -1 =) –  Adrián Barquero Mar 29 '11 at 3:49
    
@Adrian: Thanks! –  Arturo Magidin Mar 29 '11 at 3:51
    
Alternating series test would be much simpler in this case. –  Superbus Apr 22 at 12:23
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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.

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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.

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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.

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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
$$\frac{1}{1+x^2} = 1-x^2 +x^4-x^6+x^8 ...$$

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