Show the first three terms in the expansion of #1/(1-(x/2))^3# in ascending powers of x?

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Bill K. Share
Dec 31, 2017

Answer:

It sounds like you are after the Taylor (or Maclaurin) series for this function centered at #x=0#, which is really a calculus topic, but it can be answered here using Newton's generalized binomial theorem (to get a so-called "binomial series ").

Explanation:

Newton's generalized binomial theorem says that, if #p# is any real number and #|y|<1#, then:

#(1+y)^{p}=1+py+(p(p-1))/(2!)y^{2}+(p(p-1)(p-2))/(3!)y^{3}+(p(p-1)(p-2)(p-3))/(4!)y^{4}+cdots#

(this truncates to a finite sum that converges for all #y# when #p# is a non-negative integer.)

Since #1/(1-(x/2))^{3}=(1+(-x/2))^{-3}#, we can apply this result with #p=-3# and #y=-x/2# to get:

#1/(1-(x/2))^{3}=1-3(-x/2)+((-3)(-4))/(2!)(-x/2)^{2}+((-3)(-4)(-5))/(3!)(-x/2)^{3}+((-3)(-4)(-5)(-6))/(4!)(-x/2)^{4}+cdots#

Simplifying this gives

#1/(1-(x/2))^{3}=1+3/2 x+3/2 x^{2}+ 5/4 x^[3}+ 15/16 x^{4}+cdots#.

This converges for #|-x/2| < 1#, which is equivalent to #|x| < 2# and #-2 < x < 2#.

Take the first three of these terms to get the answer that is requested.

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