How can I find the Maclaurin series up to degree 4 for:
$$\arctan(x)$$
Calculating the derivatives becomes complex very quickly. Is there a special expansion for $\arctan$ like there is for $\cos(x)$ and $\sin(x)$?
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I'm guessing that by $bgtan$ you mean $\tan^{-1}$? – ryagami Dec 27 '14 at 17:07
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Is “bgtan” the arctangent? – egreg Dec 27 '14 at 17:08
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@egreg Yes it is. – Stanko Dec 27 '14 at 17:09
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@ryagami Yes correct. – Stanko Dec 27 '14 at 17:10
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1@Michiel There is a nice solution to this in the first answer: http://math.stackexchange.com/questions/348711/maclaurin-series-for-arctanx-by-successive-differentiation – ryagami Dec 27 '14 at 17:13
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Googling suggests that $\operatorname{bgtan}$ is Dutch for $\arctan$. Is this so? Are there other languages where it is also used? – Daniel R Dec 27 '14 at 17:35
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At least my answer is the shortest one so far..... ${}\qquad{}$ – Michael Hardy Dec 27 '14 at 18:14
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@MichaelHardy It's still too long. The shortest answer is: "duplicate". – Dec 27 '14 at 18:18
3 Answers
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If by “bgtan” you mean the arctangent, implicitly defined by $$ \tan\arctan x=x $$ with values in $(-\pi/2,\pi/2)$, then the Taylor expansion at $0$ (also known as MacLaurin expansion) is the famous Gregory series: $$ \arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dotsb $$ which converges for $x\in(-1,1]$.
Computing it by derivatives is not really difficult. Consider $f(x)=\arctan x$; then $$ f'(x)=\frac{1}{1+x^2} $$ and so $$ f''(x)=-\frac{2x}{(1+x^2)^2} $$ The third derivative is $$ f'''(x)=-2\frac{(1+x^2)^2-4x(1+x^2)}{(1+x^2)^4}= -2\frac{1-4x+x^2}{(1+x^2)^3} $$ You don't need to compute the fourth derivative, because $f(x)=-f(-x)$, so all derivatives of even order are zero.
Now $$ f(0)=0,\quad f'(0)=1,\quad f''(0)=0,\quad f'''(0)=-2\quad f''''(0)=0 $$ so $$ \arctan x=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+ \frac{f''''(0)}{24}+o(x^4)=x-\frac{x^3}{3}+o(x^4) $$
A different approach would be to notice that $$ \frac{1}{1+x^2}=1-x^2+x^4+o(x^4) $$ because of geometric series. Integrating between $0$ and $x$ gives $$ \arctan x=\int_0^x (1-t^2+t^4+o(t^4))\,dt= x-\frac{x^3}{3}+\frac{x^5}{5}+o(x^5) $$
egreg
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Depending if bgtan is arctan or tan and not anything else, then you can find the series here: http://mathworld.wolfram.com/MaclaurinSeries.html along with some other maclaurin series.
user47475
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$$ \arctan x = \int_0^x \frac{dw}{1+w^2} = \int_0^x \left( 1-w^2+w^4 - w^6+\cdots \right)\,dw $$ Integrate term by term.
