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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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  • $\begingroup$ Based on the answers, perhaps what I'm really looking for is the proof that the derivative of arctan is $\frac{1}{1+x^2}$. $\endgroup$ – gerber Mar 29 '11 at 3:39
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    $\begingroup$ 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 $f(x) = \tan x$. $\endgroup$ – JavaMan Mar 29 '11 at 3:48
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    $\begingroup$ This video cleared it up for me. youtube.com/watch?v=tky25AUK7Io $\endgroup$ – gerber Mar 29 '11 at 4:45
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    $\begingroup$ That $\arctan'(x)=1/(1+x^2)$ may be a definition. That depends on what you chose as a starting point for the construction of so-called "elementary functions". One such construction starts with the complex exponential, another starts from the definition of arctan and log as antiderivatives of respectively $1/(1+x^2)$ and $1/x$. There are still other ways to proceed. $\endgroup$ – Jean-Claude Arbaut Apr 13 '15 at 13:51
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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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    $\begingroup$ Alternating series test would be much simpler in this case. $\endgroup$ – Superbus Apr 22 '14 at 12:23
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    $\begingroup$ Indeed, the power series converges at x = 1. But why is this equal to arctan(1) since the expansion for 1/(1+x^2) does not hold at x = 1? $\endgroup$ – Zhanfeng Lim Apr 22 '16 at 16:30
  • $\begingroup$ But why is $arctan(x)$ at $x=0$ not $arctan(0)=0$? $\endgroup$ – mavavilj Sep 17 '16 at 10:30
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    $\begingroup$ Note: It follows from Abel's theorem that, for example, $\arctan(1)$ will equal the value of the power series evaluated at $x=1$. This is a subtlety that is glossed over in most calculus textbooks. $\endgroup$ – Morgan Sherman Jul 14 '17 at 17:35
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    $\begingroup$ @JohnDo: If you read what I say first, I say that because it is a Taylor series, you can integrate term by term. The fact that it is a Taylor series is what justifies the integration term by term, and that by itself also shows that the function is continuous: the Taylor series defines a continuous, infinitely differentiable function in its interval of convergence. So, no; I don't have to give any extra arguments and I don't have to show it is continuous, because all of that is already taken care of. $\endgroup$ – Arturo Magidin Sep 26 '17 at 22:42
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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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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.

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

$$\tan^{-1}(x)=\int_0^x\frac{1}{1+t^2}dt=\int_0^x\sum_{k=0}^n(-t^2)^k+\frac{(-t^2)^{n+1}}{1+t^2}dt$$ 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 $$\tan^{-1}(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{2k+1}\quad {-1}\le x\le 1$$

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  • $\begingroup$ I like this answer for including an elementary proof that the series converges (and to the correct value) even at $x = \pm1$. $\endgroup$ – Toby Bartels Feb 12 '18 at 10:16
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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.

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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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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.

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  • $\begingroup$ I assume that you mean $x = 1$, not $r = 1$ (and in fact your argument works whenever $0 \leq x \leq 1$). This doesn't answer the question (even for $x = 1$), since it doesn't explain why the sum is $\arctan(x)$. But you make a good point about the approximation provided for $\pi/4$. (You can get a much better result for $\pi$ using $x = 1/\sqrt{3}$ instead.) $\endgroup$ – Toby Bartels Feb 12 '18 at 10:42

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