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Write down the Taylor series expansion about the origin in the region $|x| < 1$ for the function $$f(x)=x\tan^{-1}x-\frac{1}{2}\log(1+x^2)$$

I cant find the general formula for $f^{(n)}(x)$.an I get some help?

ghugni
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  • what is $f'(x)$?? what is $f''(x)$?? what is $f'''(x)$?? you see some pattern there?? It would be better if you can at least show us till what order you have calculated the derivative? –  Jan 21 '14 at 05:28

2 Answers2

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Hint:

  1. Write down the standard expansion of $\arctan x$. Multiply term by term by $x$.

  2. Write down the standard expansion of $\ln(1+t)$. Replace $t$ everywhere by $x^2$. Multiply term by term by $-\frac{1}{2}$.

  3. Add the results of (1) and (2), collecting the coefficients of like powers of $x$ together.

Remark: If you do not know the power series expansion of $\arctan x$, you can probably write down the power series expansion of $\frac{1}{1-u}$, and hence of $\frac{1}{1+t^2}$. Then use the fact that $$\arctan x=\int_0^x \frac{dt}{1+t^2}\,dt,$$ integrating the series for $\frac{1}{1+t^2}$ term by term.

If you do not know the power series expansion of $\ln(1+t)$, integrate the power series expansion of $\frac{1}{1+u}$ term by term from $0$ to $t$.

André Nicolas
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  • This seem to be natural but isn't this a cheating :D?? I really like this and would like to know if there are any conditions on $f(x)$ so that i can do this kind of trick.. –  Jan 21 '14 at 05:36
  • Under the right convergence assumptions, the power series for $x^kf(x)$ ($k$ a non-negative integer) is obtained by multiplying the power series of $f(x)$ term by term by $x^k$. If $c$ is the minimum of the radius of convergence of the power series for $f(x)$ and $g(x)$, then the power series of $f(x)+g(x)$ is obtained by addition, and we have convergence to $f(x)+g(x)$ when $|x|\lt c$. Similar considerations hold for integration and differentiation of functions given by power series. – André Nicolas Jan 21 '14 at 05:42
  • I do not totally understand what you have said... I need some time to understand this... Thank you :) –  Jan 21 '14 at 05:46
  • I tried something and i am stuck at the conclusion.. Please have a look at that and help me... –  Jan 21 '14 at 06:34
  • oh yes... I missed that $x^2$ thank you... –  Jan 21 '14 at 06:47
  • I am still in trouble... please have a look at the last equations... I am gettion all coefficients to be $0$ :O –  Jan 21 '14 at 07:01
  • Just algebra. In the first part, the coefficient of $x^{2n+2}$ is $(-1)^{n} \frac{1}{2n+1}$. In the second part, it is $-(-1)^n \frac{1}{2n+2}$. When you add, the coefficient is $(-1)^n \left[\frac{1}{2n+1}-\frac{1}{2n+2}\right]$, which is non-zero. – André Nicolas Jan 21 '14 at 07:15
  • yes yes.. I was dumb :D –  Jan 21 '14 at 07:17
  • Minor algebraic slips are a fact of life. – André Nicolas Jan 21 '14 at 07:17
  • May be you are correct... I hope now it is fine... :) Thank you so much for this wonderful hint and You made my day :) –  Jan 21 '14 at 07:23
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$f(u)=\frac{1}{1-u}\Rightarrow f'(x)=\dfrac{(-1)(-1)}{(1-u)^2}=\dfrac{1}{(1-u)^2}$

$f''(u)=\dfrac{1.(-2).(-1)}{(1-u)^3}=\dfrac{1.2}{(1-u)^3}$

$f'''(u)=\dfrac{1.2.(-3)(-1)}{(1-u)^4}=\dfrac{1.2.3}{(1-u)^4}$

More generally ; $f^{(n)}(u)=\dfrac{n!}{(1-u)^{n+1}}$

Taylor series for $f(u)$ is $$\sum_{n=0}^{\infty} f(u)=\sum_{n=0}^{\infty}u^n$$

For $u=-t^2$ we have $$\sum_{n=0}^{\infty} f(t)=\sum_{n=0}^{\infty}(-1)^n(t)^{2n}$$

$$\arctan x=\int_0^x \frac{dt}{1+t^2}$$ $$=\int_0^x \sum_{n=0}^{\infty}(-1)^n(t)^{2n}\,dt=\sum_{n=0}^{\infty}(-1)^n\int_0^x (t)^{2n}dt=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$

So, $$\arctan x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$

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

$f(u)=\dfrac{1}{1+u}\Rightarrow f'(u)=\dfrac{-1}{(1+u)^2}$

$f''(u)=\dfrac{(-1)(-2)}{(1+u)^3}=\frac{1.2}{(1+u)^3}$

$f'''(u)=\dfrac{1.2.(-3)}{(1+u)^4}=\frac{(-1).1.2.3}{(1+u)^4}$

More generally , $f^{(n)}(u)=\frac{(-1)^nn!}{(1+u)^{n+1}}$

Taylor series for $f(u)$ is $$\sum_{n=0}^{\infty} f(u)=\sum_{n=0}^{\infty}(-1)^n u^n$$

$$\log (1+x)=\int_0^x \frac{du}{1+u}$$

$$=\int_0^x \sum_{n=0}^{\infty}(-1)^n u^n du=\sum_{n=0}^{\infty}(-1)^n \frac{x^{n+1}}{n+1}$$

We have $$\log (1+x)=\sum_{n=0}^{\infty}(-1)^n \frac{x^{n+1}}{n+1}\Rightarrow \log (1+x^2)=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+2}}{n+1}$$ $$\Rightarrow \frac{1}{2}\log (1+x^2)=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+2}}{2(n+1)}$$

$$\Rightarrow -\frac{1}{2}\log (1+x^2)=\sum_{n=0}^{\infty}(-1)^{n+1} \frac{x^{2n+2}}{2(n+1)}$$

We have : $$x\arctan x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+2}}{2n+1}; -\frac{1}{2}\log (1+x^2)=\sum_{n=0}^{\infty}(-1)^{n+1} \frac{x^{2n+2}}{2(n+1)}$$

$$x\arctan x-\frac{1}{2}\log (1+x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+2}}{2n+1}+\sum_{n=0}^{\infty}(-1)^{n+1} \frac{x^{2n+2}}{2(n+1)}$$

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

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

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

$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n(2n-1)}x^{2n}$$