Q. 58 - Calculus

Question

Show that the radius of convergence for the binomial series is $1$when $p$ is not a positive integer. What is the radius of convergence when $p$ is a positive integer? (Hint: Consider Exercise 57.)

Step-by-Step Solution

Answer

If $p$ is not a positive integer, then the radius of convergence of the binomial series is $1$ .

Step 1. Given Information

Radius of convergence for the binomial series is $1$ when $p$ is not a positive integer.

Step 2. Finding the radius of convergence when p is a positive integer

Considering $f\left(x\right)={(1+x)}^p$from exercise 57

Step 3. Defining the series

Since for any function $f$ with derivatives of all orders  at the point $x_0=0$, then the Maclaurin series is

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

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

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

Step 4. Constructing a table

A table of the Maclaurin series for the function $f\left(x\right)=f{(1+x)}^p$

$n$	$f^n\left(x\right)$	$f^n\left(0\right)$
$0$	${(1+x)}^p$	$1$
$1$	$p{(1+x)}^{p-1}$	$p$
2	$p{(p-1)(1+x)}^{p-2}$	$p(p-1)$
3	$p(p-1)(p-2){(1+x)}^{p-3}$	$p(p-1)(p-2)$
. . .	. . .	. . .
$k$	$p(p-1)(p-2)...(p-k+1){(1+x)}^{p-k}$

Step 5. Solving the table

So, the Maclaurin series for the function $f\left(x\right)={(1+x)}^p$ is

$1+px+\frac{p(p-1)}{2!}{x}^2+\frac{p(p-1)(p-2)}{3!}{x}^3+...+\frac{p(p-1)(p-2)...(p-k+1)}{3!}{x}^k+...$

Step 6. Solving with p as not a positive integer

If $p$ is not a positive integer, we'll take $p$as $-1$

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

Implies that

$1-x+x^2-x^3+x^4-...$

So if we take $x=1$then the sum of the series revolves around $[0,1]$

Hence, if $p$ is a positive integer, then the radius of convergence of the binomial series is $R$.