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Using power series representation find an approximation to $$\int_{0}^{0.1} \arctan(2x)dx$$

This is my solution:

$$\frac{d}{dx}\arctan(2x)=\ \frac{2}{1+4x^2} $$ $$\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$ $$\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$

$$= \int_{0}^{0.1}2\frac{1}{1-(-4x^2)} \, dx =\int_{0}^{0.1} \sum_{n=0}^\infty 2(-1)^n (4x^2)^n \, dx= \sum_{n=0}^{\infty}\frac{2(-1)^{n}4^{n}0.1^{2n+1}}{2n+1}$$

When I input the formula that I got in the calculator I get a value of 0.1973. But when I input the integral of arctangent I get 0.0099. I guess I missed something and can't find what. Where is my mistake?

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  • $\begingroup$ as a good OP, you should include the details that you need the series formula of integral. $\endgroup$ – Zack Ni Jul 27 '16 at 7:49
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    $\begingroup$ @ZackNi: They are included in both the title, and the first line. It's now up to the reader to read attentively. $\endgroup$ – Alex M. Jul 27 '16 at 8:10
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No, you are confusing things and you are not doing what you are required to do: you are not required to integrate the derivative of $\arctan$, but $\arctan$ itself.

Start by finding the Taylor series of $\arctan x$ around $0$:

$$(\arctan x)' = \frac 1 {1 + x^2} = \sum _{n = 0} ^\infty (-x^2)^n \implies \arctan x = \int \sum _{n = 0} ^\infty (-x^2)^n \ \Bbb d x = \sum _{n = 0} ^\infty (-1)^n \frac {x^{2n+1}} {2n+1} + C .$$

Since $\arctan 0 = 0$, we get that $C=0$.

Next, plugging $2x$ instead of $x$ in the above series leads to

$$\arctan 2x = \sum _{n = 0} ^\infty (-1)^n 2^{2n+1} \frac {x^{2n+1}} {2n+1} .$$

Using this series we may write

$$\int \limits _0 ^{0.1} \arctan 2x \ \Bbb d x = \int \limits _0 ^{0.1} \sum _{n = 0} ^\infty (-1)^n 2^{2n+1} \frac {x^{2n+1}} {2n+1} \ \Bbb d x = \sum _{n = 0} ^\infty \frac {(-1)^n 2^{2n+1}} {2n+1} \int \limits _0 ^{0.1} x^{2n+1} \ \Bbb d x = \\ \sum _{n = 0} ^\infty \frac {(-1)^n 2^{2n+1}} {2n+1} \frac {x^{2n+2}} {2n+2} \Bigg| _0 ^{0.1} = \sum _{n = 0} ^\infty \frac {(-1)^n 2^{2n+1}} {2n+1} \frac {(0.1)^{2n+2}} {2n+2} = \\ 0.02 \sum _{n = 0} ^\infty \frac {(-0.04)^n} {(2n+1)(2n+2)} .$$

Since the terms of this series decrease very quickly, it is enough to compute just the first 3 of it in order to obtain a satisfactory approximation:

$$0.02 \sum _{n = 0} ^2 \frac {(-0.04)^n} {(2n+1)(2n+2)} = 0.02 \left( \frac 1 2 - \frac {0.04} {12} + \frac {0.04^2} {30} \right) = 0.0099344 .$$

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  • $\begingroup$ You need to edit this because the limits you have applied are incorrect $\endgroup$ – David Quinn Jul 27 '16 at 7:59
  • $\begingroup$ @DavidQuinn: Done and thank you. $\endgroup$ – Alex M. Jul 27 '16 at 8:08
  • $\begingroup$ Your approximation is 0.0195008, but the integral of arctan is 0.0099 $\endgroup$ – ReeSSult Jul 27 '16 at 16:02
  • $\begingroup$ @ReeSSult: True, because due to some mental shortcircuit I wrote $4^{2n+1}$ everywhere instead of $2^{2n+1}$, and the error naturally propagated to the end result. Now it's correct. $\endgroup$ – Alex M. Jul 27 '16 at 16:23
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The Taylor series (for $|x|\leq 1$) is $$\tan^{-1}(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$ So $$\tan^{-1}(2x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}2^{2n+1}x^{2n+1}$$ Integrating $$\int \tan^{-1}(2x)\,dx=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)}2^{2n+1}x^{2n+2}$$ Using the bounds $$\int_0^{\frac 1{10}} \tan^{-1}(2x)\,dx=\frac{1}{100}-\frac{1}{15000}+\frac{1}{937500}-\frac{1}{43750000}+\cdots$$

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You need to obtain a power series for $\arctan 2x$ and integrate that. What you have done is to obtain a power series for its derivative and integrate that, which is clearly wrong.

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  • $\begingroup$ Your answer is correct but what's wrong with my answer? $\endgroup$ – Zack Ni Jul 27 '16 at 7:45
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    $\begingroup$ @ZackNi I didn't downvote your answer, but maybe suggesting evaluating the integral by parts is not what the question is looking for... $\endgroup$ – David Quinn Jul 27 '16 at 7:47
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We have:

$$ I=\int_{0}^{\frac{1}{10}}\arctan(2x)\,dx = \frac{1}{2}\int_{0}^{\frac{1}{5}}\arctan(x)\,dx \stackrel{IBP}{=} \frac{\arctan\frac{1}{5}}{10}-\frac{\log\left(1+\frac{1}{25}\right)}{4}\tag{1}$$ where, for any $x\in(0,1)$, $$ \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)},\qquad \log(1+x^2)=\sum_{n\geq 1}\frac{(-1)^{n+1} x^{2n}}{n}.\tag{2} $$ Due to Leibniz' test, $$ \left|\arctan\frac{1}{5}-\sum_{n=0}^{1}\frac{(-1)^n}{5^{2n+1}(2n+1)}\right|<\frac{1}{5^6},\qquad \left|\log\left(1+\frac{1}{25}\right)-\sum_{n=1}^{2}\frac{(-1)^{n+1}}{5^{2n}n}\right|<\frac{1}{3\cdot 5^6} $$ hence it follows that: $$ I \approx \frac{1}{10}\left(\frac{1}{5}-\frac{1}{3\cdot 5^3}\right)-\frac{1}{4}\left(\frac{1}{5^2}-\frac{1}{2\cdot 5^4}\right) = \color{red}{\frac{149}{15000}}=0.00993333\ldots \tag{3}$$ and the approximation error is less than $2\cdot 10^{-5}$.

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All the calculation is right but the answer is for $[arctan(2x)]_{0}^{0.1}$ not for $\int_{0}^{0.1} \arctan(2x)dx$ since $\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=[arctan(2x)]_{0}^{0.1}$ By fundamental theorem of calculus.

To solve the question, try to use integration by parts. And the answer is $x \arctan(2x) - \frac{1}{4} log(4x^2+1)$ which can be written as:

$$ \sum_{k =0}^\infty \frac{1}{1+2k}(-1)^k 2^{1+2k} x ^{2+2k} - \frac{1}{4} \frac{1}{k}(-4)^k (x ^{2})^k $$

Evaluating it on $x = 0.1$ and $x =0$.

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    $\begingroup$ I didn't, I up voted $\endgroup$ – ReeSSult Jul 27 '16 at 7:55

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