I hope that i could receive some help regarding the next problem:
Expand the function $f(x) = arctan(x)$ into a power series.
My solution:
I am using the following predication in solving this problem:
Function $f$ can be represented by a series $\sum\limits_{n = 0}^{+\infty} a_n(x - x_0)^n$ on the interval $I = (x_0 - r, x_0 + r)$, $r > 0$, if and only if:
1) $f$ is infinitely diferentiable on the $I$;
2) Remainder $R_n(x)$ in the Taylor series $f()x = \sum\limits_{k = 0}^{+\infty} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k + R_n(x)$ is converging to $0$, ($\forall x \in I$, $n \to +\infty$).
So, i understand that the next steps needs to be taken in solving this problem:
1) Find $I$ and $x_0$.
2) Prove that $f$ is infinitely diferentiable.
3) Prove that $\lim\limits_{n \to +\infty} R_n(x) = 0$.
4) Use the rest of Taylor series as the final solution.
I did next:
1) Because the domain of the $f$ is $\mathbb{R}$, i took $I = \mathbb{R}$ and $x_0 = 0$.
2) There is the tricky part. $$f' = \frac{1}{1 + x^2}$$ $$f" = \frac{-2x}{(1 + x^2)^2}$$ $$f^{(3)} = \frac{6x^2 - 2}{(1 + x^2)^3}$$ $$f^{(4)} = \frac{-24(x^2 - 1)}{(1 + x^2)^4}$$ $$f^{(5)} = \frac{24(5x^4 - 10x^2 + 1)}{(1 + x^2)^5}$$ $$f^{(6)} = \frac{-240x(3x^4 - 10x^2 + 3)}{(1 + x^2)^6}$$ I assume that $f$ has the infinite number of the derivatives, but i dont know how to prove it.
3) Because $x_0 = 0$, Taylor series becomes the Maclaurin series. Now i am using the Lagrange remainder for $R_n(x)$. It looks like this: $R_n(x) = \frac{x^{n + 1}}{(n + 1)!} f^{(n + 1)}(\theta x)$. At this point, i don't know how to find $\lim\limits_{n \to +\infty} R_n(x)$.
Please, could you tell me if there are any mistakes in the procedure and show me some hints about the limit?
Take the sequence of functions $f_n(x)=\dfrac{g_n(x)}{(1+x^2)^{2^n}}$ where $g_0(x)$ is continuous and differentiable in $\Bbb R$ and $g_{n+1}(x)=g_n'(x)(1+x^2)-2xg_n(x)$. It is obvious that $(1+x^2)^{2^n}$ has no roots and is continuous and differentiable and so is $g_n(x)$ because of continuity and differentiability of $g_0(x)$. Therefore $f_0(x)$ is continuous and differentiable in $\Bbb R$. By induction, assume $f_n(x)$ is continuous and differentiable everywhere. We know that: $$f_{n+1}(x)=\dfrac{g_{n+1}(x)}{(1+x^2)^{2^{n+1}}}=\dfrac{g_n'(x)(1+x^2)-2xg_n(x)}{((1+x^2)^{2^n})^2}=(\dfrac{g_n(x)}{(1+x^2)^{2^n}})'=f_n'(x)$$ which is continuous and differetiable and we have proven what we wanted.
Clearly:$$\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+...\qquad ,\qquad |x|<1\\\dfrac{1}{1+x^2}=\dfrac{1}{x^2}-\dfrac{1}{x^4}+\dfrac{1}{x^6}-\dfrac{1}{x^8}+...\qquad ,\qquad |x|>1$$Therefore:$$\tan^{-1}(x)=C_1+x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+...\qquad ,\qquad |x|<1\\\tan^{-1}(x)=C_2-\dfrac{1}{x}+\dfrac{1}{3x^3}-\dfrac{1}{5x^5}+\dfrac{1}{7x^7}+...\qquad ,\qquad |x|>1$$ where $C_1=0$ and $C_2=\dfrac{\pi}{2}$ for $x>1$ and $C_2=-\dfrac{\pi}{2}$ for $x<-1$ (why?) therefore:$$\tan^{-1}(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+...\qquad ,\qquad |x|<1\\\tan^{-1}(x)=\dfrac{\pi}{2}-\dfrac{1}{x}+\dfrac{1}{3x^3}-\dfrac{1}{5x^5}+\dfrac{1}{7x^7}-...\qquad ,\qquad x>1\\\tan^{-1}(x)=-\dfrac{\pi}{2}-\dfrac{1}{x}+\dfrac{1}{3x^3}-\dfrac{1}{5x^5}+\dfrac{1}{7x^7}-...\qquad ,\qquad x<-1$$