Question

What is the MacLaurin Series for `sin(2x)ln(1+x)`?

(portions of this question have been edited or deleted!)

Answer 1

The given series is incorrect.

The correct series is:

`sin(2x)ln(1+x) = 2x^2 - x^3 - 2/3x^4 +1/6x^5 + ...`

Explanation:

The given series is incorrect.

Let:

`S = sin(2x)ln(1+x)`

I won't derive the well known series for `sin` and `ln`, but just quote them:

So, we have the following well established series:

`ln(1+x) = x-1/2x^2+1/3x^3 -1/4x^4 + 1/5x^5 ...`
`sinx = x-x^3/(3!) + x^5/(5!) - x^7/(7!) + ...`

Thus the series for `sin2x` is:

`sin2x = (2x)-(2x)^3/(3!) + (2x)^5/(5!) - (2x)^7/(7!) + ...`
`" " = 2x-(8x^3)/6 + (32x^5)/(120) - (128x^7)/(5040) + ...`
`" " = 2x - 4/3x^3 + 4/15x^5 - 8/315x^7 + ...`

So if we take terms up to `O(x^4)` then:

`S = sin(2x)ln(1+x)`
`\ \ = {2x - 4/3x^3 + ...} * {x-1/2x^2+1/3x^3 -1/4x^4 + ...}`
`\ \ = (2x){x-1/2x^2+1/3x^3 -1/4x^4 + ...} -4/3x^3{x-1/2x^2+1/3x^3 -1/4x^4 + ...}`
`\ \ = 2x^2-x^3+2/3x^4 -1/2x^5 + ... -4/3x^4 + 2/3x^5...`

`\ \ = 2x^2 - x^3 - 2/3x^4 +1/6x^5 + ...`