Maclaurin series of function $f(x)= x+\ln(9-x^2)^\frac{1}{3}$ Find the Macluarin series. I'm trying for hours to understand how should I solve it. Please explain it to me step by step. 
$f(x)=x+\ln {\sqrt[3]{9-x^2}}$
 A: $$y = x + \ln\sqrt[3]{9-x^2} = x + \frac 13(9-x^2)^{1/3}=x + \frac133^{2/3}(1-x^2/9)^{1/3}$$ now use the binomial expansion $$(1-x^2/9)^{1/3}=1 - \frac13(x^2/9) -\frac1{2!}\frac 13 \frac 23(x^2/9)^2 - \frac1{3!}\frac 13 \frac 23\frac 53(x^2/9)^2 -\cdots$$
A: The idea with a Maclaurin series is that you need to differentiate the function and at each step put in the value 0 for $x$ in each one. Then they must be combined in the standard Maclaurin formula,
$$f(x)=f(0)+f^{(1)}(0) {x\over 1!}+ f^{(2)}(0) {x^2\over 2!}+f^{(3)}(0){x^3\over 3!}+...+f^{(n)}(0){x^n\over n!}+...$$
where $f^{(n)}(0)$ means the $n$-th derivative of $f$ evaluated at $x=0$
A: $$f(x)=x+\ln {\sqrt[3]{9-x^2}}\\=f(x)=x+\frac13\ln (9-x^2)\\=x+\frac13\ln\left( \frac{(9-x^2)}{9}*9\right)\\=x+\frac13\ln((9)(1-x^2/9))\\=x+\frac13\ln((9)(1-(x/3)^2))=x+\frac13(\ln 9+\ln(1-(x/3)^2))\\=\frac23\ln 3+x-\left(\sum_{k=1}^{\infty}\frac {x^{2k}}{k\cdot9^k}\right)\\=\ldots$$
A: Ok. First of all notice that the function is not defined for $|x|\geq 3$, so you won’t be able to find a series which is valid for all the real numbers. Then you notice that we already know the series for $\ln(1-y)$:
$$ln(1-y)=-\sum_{n=1}^{+\infty}\frac{y^{n}}{n}$$
As you know, the MacLaurin series of a function is given by summing all the powers of its unknown, $y^{n}$, taken with the coefficients $a_{n}\ $ given by the $n$-th derivative of the function at point zero, divided by the factorial $n!$:
$$f(y)=\sum_{n=1}^{+\infty}\frac{f^{(n)}(0)}{n!}\,y^{n}$$
If you calculate the derivatives of $ln(1-y)$ in zero, you get that the coefficients are $a_{n}=-1/n\ $ for every $n$ except for $n=0$, for which $a_{0}=0$, so the series that we are given is correct. Then you work out the right expression for your $f(x)$, so as to be able to use the MacLaurin series of $ln(1-y)$. As
$$\ln(9-x^{2})^{\frac{1}{3}}=\frac{1}{3}\,\ln(9-x^{2})=\frac{1}{3}\,\ln[9(1-\frac{x^{2}}{9})]=\frac{1}{3}\,[\ln9+\ln(1-\frac{x^{2}}{9})]=\frac{2}{3}\,\ln3+\frac{1}{3}\ln(1-\frac{x^{2}}{9})$$
the expression you’re looking for is
$$f(x)=\frac{2}{3}\,\ln3+x+\frac{1}{3}\ln(1-\frac{x^{2}}{9})$$
Then, if
$$ln(1-y)=-\sum_{n=1}^{+\infty}\frac{y^{n}}{n}$$
substituting $y=\frac{x^{2}}{9}$ you get
$$f(x)=\frac{2}{3}\,\ln3+x-\frac{1}{3}\,\sum_{n=1}^{+\infty}\frac{x^{2n}}{9^{n}\,n}$$
This is the series that you’re looking for.
