Question

How do you derive the maclaurin series for `(sin(4x^2))/x^3`?

Answer 1

`sin(4x^2)/x^3 = 4/x - 32/3x^3 + 128/15x^7 - 1024/315x^11 + ...`

Explanation:

Whilst this could be generated from first principles using first, second, third ... derivatives it is easier to use the known series for `sinX`. i.e.

`sinX = X - X^3/(3!) + X^5/(5!) - X^7/(7!) + X^9/(9!) - ...`

So,
`sin(4x^2)/x^3 = {(4x^2) - (4x^2)^3/(3!) + (4x^2)^5/(5!) - (4x^2)^7/(7!) + ... }/x^3`
`:. sin(4x^2)/x^3 = {4x^2 - (64x^6)/6 + (1024x^10)/120 - (16384x^14)/5040 + ... }/x^3`
`:. sin(4x^2)/x^3 = {4x^2 - (32x^6)/3 + (128x^10)/15 - (1024x^14)/315 + ... }/x^3`
`:. sin(4x^2)/x^3 = 4/x - 32/3x^3 + 128/15x^7 - 1024/315x^11 + ...`