How do you find the taylor series for #(1/(1-x^2))# centered around #a = 3#?

Question

Scarlett Allison · Accepted Answer

#1/(1-x^2) equiv -1/2sum_{n=0}^oo (-1)^n/2^{n+1}(1-1/2^{n+1})(x-3)^n#

for # 1 < x < 5# Making now #y = x-3# #f(x)=1/(1-x^2) equiv 1/(1-(y+3)^2) = C_1/(1+y+3)+C_2/(1-y-3)# #=C_1/(y+4)-C_2/(y+2)# Solving for #C_1,C_2# the condition #C_1(y+2)-C_2(y+4)=1, forall y in RR# or equivalently #(C_1-C_2)y+2C_1-4C_2=1, forall y in RR# we obtain #C_1=C_2 = -1/2# so #1/(1-x^2) equiv 1/2(1/(y+2)-1/(y+4))=1/2(1/(2(y/2+1))-1/(4(y/4+1)))# Now considering that for #abs z < 1# #lim_{n->oo}(z^n-1)/(z-1) = -1/(z-1)=sum_{k = 0}^oo z^n# restraining #abs(y/2) < 1# then #1/(y/2+1) equiv -sum_{n=0}^oo (-y/2)^2# and #1/(y/4+1) equiv -sum_{n=0}^oo (-y/4)^2# Putting all together #1/2(1/(y+2)-1/(y+4)) equiv s(y) # #s(y)= -1/2(1/2 sum_{n=0}^oo (-y/2)^n-1/4sum_{n=0}^oo (-y/4)^n)# This serie in convergent for #abs y < 2# or equivalently # 1 < x < 5# Finally, for # 1 < x < 5# we have #f(x) equiv s(x-3)# #s(x-3) = -1/2sum_{n=0}^oo (-1)^n/2^{n+1}(1-1/2^{n+1})(x-3)^n#

Evelyn Agard · Answer

undefined To find the Taylor series for $ \frac{1}{1-x^2} $ centered around $ a = 3 $, you would first find the derivatives of the function. Then, evaluate these derivatives at $ x = 3 $ to find the coefficients of the Taylor series. The general formula for the Taylor series expansion centered at $ a $ is given by:

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n $$

For $ \frac{1}{1-x^2} $, start by finding the derivatives of $ f(x) $. Then, evaluate them at $ x = 3 $ to find the coefficients of the Taylor series.