How do you find the Taylor polynomial for #1/(2-x)#?

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Answer 1 · 2017-09-07T14:32:55

#1/(2-x) =sum_(n=0)^oo x^n/2^(n+1)#

for #x in (-2,2)#.

Note that the sum of a geometric series is:

#sum_(n=0)^oo q^n = 1/(1-q)#

for #q in (-1,1)#

Write then the function as:

#1/(2-x) = 1/2 1/(1-x/2)#

Posing: #q=x/2#, we have:

#1/(2-x) = 1/2 sum_(n=0)^oo (x/2)^n = sum_(n=0)^oo x^n/2^(n+1)#

for #x in (-2,2)#.