Question

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.

$\sqrt{1-x}$

Short answer

Ans:The Maclaurin series of the function $f\left(x\right)=1-\frac{1}{2}x-\sum _{k=2}^{\infty }\frac{1·3·5\cdots (2k-3)}{2^{k}k!}{x}^k$

Step-by-step solution

Step 1. Given information:

$f\left(x\right)=\sqrt{1-x}$

Step 2. Finding the general form  Maclaurin series:  

Since for any function $f$ with derivatives of all orders at the point $x_{0}=0$, then the Maclurin series is

$f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}}\left(0\right)}{3!}{x}^3+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{4!}{x}^4+\dots$

Or, we can write the general form Maclurin series of the function $f$ is

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n$

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the function $f\left(x\right)=\sqrt{1-x}$

localid="1649437869098" $\begin{array}{|cccc|}\hline n& f^n\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& \sqrt{1-x}& 1& 1\\ 1& -\frac{1}{2\sqrt{1-x}}& -\frac{1}{2}& -\frac{1}{2}\\ 2& -\frac{1}{2^2}(1-x{)}^{-\frac{3}{2}}& -\frac{1}{2^2}& -\frac{1}{2^2}\\ 3& -\frac{1·3}{2^3}(1-x{)}^{-\frac{5}{2}}& -\frac{1·3}{2^3}& -\frac{1·3}{2^3}\\ 4& -\frac{1·3·5}{2^4}(1-x{)}^{-\frac{7}{2}}& -\frac{1·3·5}{2^4}& -\frac{1·3·5}{2^4}\\ ·& ·& ·& 4!\\ 5& -\frac{1·3·5\cdots (2k-3)}{2^k}(1-x{)}^{-\frac{2k-1}{2}}& -\frac{1·3·5\cdots (2k-3)}{2k}& -\frac{1·3·5\cdots (2k-3)}{2^k}\\ \hline\end{array}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f\left(x\right)=\sqrt{1-x}$ is

$1+-\frac{1}{2}·x+\frac{-\frac{1}{2^2}}{2!}{x}^2+\frac{-\frac{1·3}{2^3}}{3!}{x}^3+\frac{-\frac{1·3·5}{2^4}}{4!}{x}^4+\dots +\frac{-\frac{1·3·5\cdots (2k-3)}{2^k}}{k!}{x}^k$

Or, we can write as

$f\left(x\right)=1-\frac{1}{2}x-\sum _{k=2}^{\infty }\frac{1·3·5\cdots (2k-3)}{2^{k}k!}{x}^k$