How do you find the maclaurin series expansion of #1/sqrt(1-x^2)#?

1 Answer
George C.
Jan 1, 2016

Construct the Maclaurin series for #1/sqrt(1-t)# then substitute #t = x^2# to find:

#1/(sqrt(1-x^2)) = sum_(k=0)^oo (prod_(i=1)^k (2k-1))/(2^k k!) x^(2k)#

Explanation:

Let #f(t) = 1/sqrt(1-t)#

#f^((0))(t) = (1-t)^(-1/2)#

#f^((1))(t) = 1/2 (1-t)^(-3/2)#

#f^((2))(t) = 3/2^2 (1-t)^(-5/2)#

#f^((3))(t) = 15/2^3 (1-t)^(-7/2)#
...
#f^((k))(t) = (prod_(i=1)^k (2k-1))/2^k (1-t)^(-(2k+1)/2)#

So:

#f^((k))(0) = (prod_(i=1)^k (2k-1))/2^k#

Then the Maclaurin series for #f(t)# is:

#f(t) = sum_(k=0)^oo f^((k))(0)/(k!) t^k = sum_(k=0)^oo (prod_(i=1)^k (2k-1))/(2^k k!) t^k#

Substitute #t = x^2# to find:

#1/(sqrt(1-x^2)) = sum_(k=0)^oo (prod_(i=1)^k (2k-1))/(2^k k!) x^(2k)#

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