How do you find the maclaurin series expansion of #f(x) = (x-sin x)/ x^3#?

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Answer 1 · 2017-01-07T14:39:44

#(x-sin x)/x^3 = 1/(3!)-x^2/(5!)-...+(-1)^(n-1)x^(2n-1)/((2n+1)+...#,
#x in (-oo, 0)# U #(0, oo)#.

I do not think that there is any need to find limits f(0), f'(0), f''(0), ...,

for the coefficients in the Maclaurin series for

f(x) = (x-sin x)/x^3 that has the indeterminate form #0/0# at x = 0.

Let g(x)= x -sinx that has the Maclaurin series

#x^3/(3!)-x^5/(51)-...+(_1)^(n-1)x^(2n+1)/((2n+1)!)+..., x in (-oo, oo)#.

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

So, sans x = 0,

#f(x) = g(x)/x^3=(x-sin x)/x^3#

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

After all, Maclaurin series is the power series for the function, about

x = 0.