Need help understanding the binomial series

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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?
 

Answers and Replies

  • #2
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If you pull the 8 out of the expression (8 + 3x) = 8*(1 + (3/8)x ) does that help?
 
  • #3
Ray Vickson
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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 [tex](8+3x)^n = 8^n \left(1 + \frac{3}{8}x \right)^n[/tex] 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##.
 
  • #4
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D'oh. Thanks guys :P
 

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