Answer to Question #339937 in Calculus for PEACEBOY

Question

Find the Maclaurin series for the function$f(x)=\sqrt{(x^2+2-x)^5}$and its radius of convergence. 

Answer 1

Maclaurin series is the Taylor series considered at point $x=0$. The Maclaurin series has the form: $f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+...$. Consider the function $g=(2-z)^{\frac{5}2}=2^{\frac52}(1-\frac{z}2)^{\frac52}$. Maclaurin series for the function $(1-x)^{\alpha}$ has the form: $(1-x)^{\alpha}=\sum_{n=0}^{+\infty}C_{\alpha}^n(-1)^nx^{n},$ where $C_{\alpha}^n=\frac{\alpha(\alpha-1)...(\alpha-n+1)}{n!}$. It is a known series and it converges for $|x|<1$. We get: $g=(2-z)^{\frac{5}2}=2^{\frac52}(1-\frac{z}2)^{\frac52}=\sum_{n=0}^{+\infty}C_{\alpha}^n(-1)^n\left(\frac12\right)^nz^{n}$. Its radius of convergence is: $|z|<2$. Now we set $z=x-x^2$. We receive: $\sum_{n=0}^{+\infty}C_{\alpha}^n(-1)^n\left(\frac12\right)^n(x-x^2)^{n}$.

It remains to expand it using the binomial formula $(x-x^2)^n=\sum_{k=0}^nC_n^kx^{k}(-1)^{n-k}x^{2n-2k}$,where $C_n^k=\frac{n!}{k!(n-k)!}$. We get: $\sum_{n=0}^{+\infty}C_{\alpha}^n(-1)^n\left(\frac12\right)^n\sum_{k=0}^nC_n^k(-1)^{n-k}x^{2n-k}$. The series converges for $|x^2-x|<2$. It is enough to solve the latter inequality to get the radius of convergence. Consider the expression: $x(x-1)$. It is positive for $x\in(-\infty,0)\cup(1,+\infty)$. For $x\in(-\infty,0)\cup(1,+\infty)$ we receive: $x^2-x-2<0$. The latter is equivalent to: $(x+1)(x-2)<0$. The solution is $(-1,0)\cup(1,2)$. For $x\in[0,1]$ we get: $x^2-x+2>0$. The latter is satisfied for $x\in[0,1].$ Thus, the Maclaurin series has the form: $\sum_{n=0}^{+\infty}\sum_{k=0}^nC_{\alpha}^nC_n^k(-1)^{2n-k}\left(\frac12\right)^nx^{2n-k}$ and it converges for $x\in(-1,2)$.

Answer: the Maclaurin series is: $\sum_{n=0}^{+\infty}\sum_{k=0}^nC_{\alpha}^nC_n^k(-1)^{2n-k}\left(\frac12\right)^nx^{2n-k}$. It converges for $x\in(-1,2).$