How do you find the Maclaurin Series for #arcsin(x)/x#?

We seek a Maclaurin series for the function:

# arcsin(x)/x #

We will start by gaining the Maclaurin series for #arcsinx#, and then simply divide all terms by #x# to get the series we seek:

We would generally form a general Maclaurin series using:

# f(x) = f(0) + f^(1)(0)/(1!) x + f^(2)(0)/(2!) x^2 + f^(3)(0)/(3!) x^3 + ...#

So we would take:

# f(x) = arcsinx#, say

and form the first derivative (from tables):

# f'(x) = d/dx (arcsinx) = 1/sqrt(1-x^2) # ..... [A]

Along with higher derivatives. But this method quickly gets cumbersome, and in fact there is a much faster way of generating the series we need, by using the result [A] whose Maclaurin series we can actually generate very quickly using the the binomial series:

# (1+x)^n = 1 + nx + (n(n-1))/(2!) x^2 + (n(n-1)(n-2))/(3!) x^3 + ... #

As follows:

# d/dx (arcsinx) = 1/sqrt(1-x^2) #
# " " = (1-x^2)^(-1/2) #

# " " = 1 + (-1/2)(-x^2) + ((-1/2)(-3/2))/(2!) (-x^2)^2 + ((-1/2)(-3/2)(-5/2))/(3!) (-x^2)^3 + ... #

# " " = 1 + 1/2 x^2 + (3/4)/(2) x^4 + (15/8)/(6) x^6 + ... #

# " " = 1 + 1/2 x^2 + 3/8 x^4 + 5/16 x^6 + 35/128x^8... #

Then, we can integrate term by term, to gain the power series for #arcsinx#

# arcsinx = int_0^x{1 + 1/2 t^2 +3/8 t^4 + 5/16 t^6 + 35/128t^8... } dt #
# " " = x + 1/6x^3 + 3/40 x^5 + 5/112 x^7 + 35/1162 x^9 + ... #

Then, we have:

# arcsinx/x = 1/x{x + 1/6x^3 + 3/40 x^5 + 5/112 x^7 +35/1162 x^9 + ... }#
# " " = 1 + 1/6x^2 + 3/40 x^4 + 5/112 x^6 + 35/1162 x^8 + ... #