Need help understanding the binomial series

[stfz]

Homework Statement

My math textbook is currently on the Binomial Series now, after completing the Binomial Theorem (no problems with that one). I believe most of my trouble comes from the book's rather glancing explanation of it, only giving examples of the form $(1 +/- kx)$. Now have encountered this question:

Q
Obtain the first three terms, in ascending powers of x, of the expansion of $(8+3x)^\frac{2}{3}$, stating the set of values for which this expansion is valid.

Homework Equations

The binomial series is defined as :

$(1+x)^n = 1 + nx + \frac{n(n+1)}{1*2}x^2 + \frac{n(n+1)(n-2)}{1*2*3}x^3 + ...$, provided that |x| < 1.

Expansion of $(1+kx)^n$ is valid for $\frac{-1}{k} < x < \frac{1}{k}$

The Attempt at a Solution

I am able to find the expansion by following the binomial series definition, but multiplying everything by $8^n$, where n is 2/3, -1/3, -4/3, and so on.

$(8+3x)^\frac{2}{3} = 8^\frac{2}{3} + (8^\frac{-1}{3})(\frac{2}{3})(3x) + (8^\frac{-4}{3})\frac{(\frac{2}{3})(\frac{-1}{3})}{2}(3x^2)$

Which is simplified to :

$4 + x - \frac{x^2}{16}$ This is correct, according to the textbook answers.

However, I cannot find the range of x for which it is valid. The textbook answer is :
$\frac{-8}{3} < x < \frac{8}{3}$.

Why is that so? Could I have an explanation?

 

[jedishrfu]

If you pull the 8 out of the expression (8 + 3x) = 8*(1 + (3/8)x ) does that help?

 

[Ray Vickson]

Homework Statement

My math textbook is currently on the Binomial Series now, after completing the Binomial Theorem (no problems with that one). I believe most of my trouble comes from the book's rather glancing explanation of it, only giving examples of the form $(1 +/- kx)$. Now have encountered this question:

Q
Obtain the first three terms, in ascending powers of x, of the expansion of $(8+3x)^\frac{2}{3}$, stating the set of values for which this expansion is valid.

Homework Equations

The binomial series is defined as :

$(1+x)^n = 1 + nx + \frac{n(n+1)}{1*2}x^2 + \frac{n(n+1)(n-2)}{1*2*3}x^3 + ...$, provided that |x| < 1.

Expansion of $(1+kx)^n$ is valid for $\frac{-1}{k} < x < \frac{1}{k}$

The Attempt at a Solution

I am able to find the expansion by following the binomial series definition, but multiplying everything by $8^n$, where n is 2/3, -1/3, -4/3, and so on.

$(8+3x)^\frac{2}{3} = 8^\frac{2}{3} + (8^\frac{-1}{3})(\frac{2}{3})(3x) + (8^\frac{-4}{3})\frac{(\frac{2}{3})(\frac{-1}{3})}{2}(3x^2)$

Which is simplified to :

$4 + x - \frac{x^2}{16}$ This is correct, according to the textbook answers.

However, I cannot find the range of x for which it is valid. The textbook answer is :
$\frac{-8}{3} < x < \frac{8}{3}$.

Why is that so? Could I have an explanation?

If you write (8+3x)^n = 8^n \left(1 + \frac{3}{8}x \right)^n the reason should become clear.

Also: I hope you realize that the restrictions on $x$ apply only if $n$ is not a positive integer. When $n$ is a positive integer (n = 1 or 2 or 3 or ... ) the expansion is valid for all values of $x$.

 

[stfz]

D'oh. Thanks guys :P