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Find the Remainder term in the Maclaurian series of $\tan^{-1}(x)$ for $-1\le x \le 1$

$f'(x)= \left( {x}^{2}+1 \right) ^{-1}$

$f''(x)=-2\,{\frac {x}{ \left( {x}^{2}+1 \right) ^{2}}}$

$f'''(x)=8\,{\frac {{x}^{2}}{ \left( {x}^{2}+1 \right) ^{3}}}-2\, \left( {x}^{2}+1 \right) ^{-2} $

$f^{iv}=-48\,{\frac {{x}^{3}}{ \left( {x}^{2}+1 \right) ^{4}}}+24\,{\frac {x}{ \left( {x}^{2}+1 \right) ^{3}}}$

$f^v(x)=384\,{\frac {{x}^{4}}{ \left( {x}^{2}+1 \right) ^{5}}}-288\,{\frac {{x }^{2}}{ \left( {x}^{2}+1 \right) ^{4}}}+24\, \left( {x}^{2}+1 \right) ^{-3} $

then $f(0)=0=f''(0)=f^{iv}(0)=\cdots$, $f'(0)=1$,$f'(0)=1$, $f'''(0)=-2$ etc.

My problem is to check the convergency of the series. I am failed to show that the remainder term (Lagrange) $R_n=\frac{x^n}{n!} f^n(\theta x)\rightarrow 0$ as $n\rightarrow \infty$.

Edited:

$f^n(x)=\frac{(-1)^{n-1}(n-1)!}{2i}[\frac{1}{(x-i)^n}-\frac{1}{(x+i)^n}]=(-1)^n(n-1)!\sin^n{\theta}\sin{n\theta}$ where $x=\cot{\theta}$.

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    From your given $R_n$, $\displaystyle |R_n|=\bigg|\frac{x^n}{n!}(-1)^n(n-1)!\sin^n\theta\sin n\theta\bigg|\leq\frac{|x|^n}{n}\rightarrow 0$ as $n\rightarrow\infty$ since $-1\leq x\leq 1.$ It follows at $R_n\rightarrow 0.$ – 2017-01-30

3 Answers 3

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A simple way to open the series is the following.

$\displaystyle\tan^{-1}x=\int_0^x\frac{1}{1+t^2}dt=\sum_{n=0}^{\infty}\int_0^x(-1)^nt^{2n}dt=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}.$

Use the ratio test to check the convergence in the given interval.

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    I would be happy if I get the solution in general approch as I have mentioned above. – 2017-01-30
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    How you have checked the convergence by ratio test, a -ve sign in coming? – 2017-01-30
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    Let $\displaystyle a_n=(-1)^n\frac{x^{2n+1}}{2n+1}$, then for convergence $\displaystyle\lim_{n\rightarrow \infty}|\frac{a_n+1}{a_n}|<1$. The absolute value eliminate the negative sign. – 2017-01-30
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    Thank you very much. Would you please suggest where is the problem in my calculation above? – 2017-01-30
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    I am worried about your nth derivative. If n=2, $\displaystyle f''(x)=\frac{-1}{2i}\bigg[\frac{1}{(x+i)^2}+\frac{1}{(x-i)^2}\bigg]=\frac{-1}{i}\bigg[\frac{x^2-1}{(x^2+1)^2}\bigg]$ which is not correct – 2017-01-30
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    I have edited. Very sorry – 2017-01-30
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For $|x|<1$ $$\tan^{-1}x=\int_0^x\frac{1}{1+t^2}dt=\int_0^x1-t^2+t^4-t^6+\cdots=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

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    How to check its convergency? I need the the solution in general approch as I have mentioned above. – 2017-01-30
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    $t\leq x<1$ so the geometric series is converge, beside the convergence radius of a series and it's integral is the same. – 2017-01-30
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\begin{align*} \tan^{-1} x &= \int_{0}^{x} \frac{dt}{1+t^2} \\ &= \int_{0}^{x} \frac{1-(-1)^{n}t^{2n}}{1+t^2} \, dt+(-1)^{n} \int_{0}^{x} \frac{t^{2n}}{1+t^2} \, dt \\ &= \int_{0}^{x} \left[ 1-t^2+t^4-\ldots+(-1)^{n-1} t^{2n-2} \right] dt+(-1)^{n} \int_{0}^{x} \frac{t^{2n}}{1+t^2} \, dt \\ &= x-\frac{x^3}{3}+\frac{x^5}{5}-\ldots+(-1)^{n-1} \frac{x^{2n-1}}{2n-1}+ (-1)^{n} \int_{0}^{x} \frac{t^{2n}}{1+t^2} \, dt \end{align*}

Now for $x^2\le 1$, \begin{align*} R_{n} &= (-1)^{n} \int_{0}^{x} \frac{t^{2n}}{1+t^2} \, dt \\ |R_{n}| & \le \left| \int_{0}^{x} t^{2n} \, dt \right| \\ &= \left| \frac{x^{2n+1}}{2n+1} \right| \\ & \le \frac{1}{2n+1} \end{align*}