Question

If x is very small such that the terms in `x^4` and higher can be neglected, how to find sin2xln(1+x) in first three terms by using maclaurin series?

Answer 1

`sin2xln(1+x) = 2 x^2 - x^3 - 2/3 x^4 +O(x^5)`

Explanation:

We seek the Maclaurin Series for:

`S = sin2xln(1+x)`

If we were to use the definition of the Maclaurin series and find the derivatives then there is no need to consider the restriction on `x` being small, thus suggesting we take an alternative approach, which is to use the well known series for the two seperate functions:

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

Thus we can write

`S = {(2x)-(2x)^3/(3!)+(2x)^5/(5!) - ... }{x-x^2/2+x^3/3-x^4/4+x^5/5 ... }`

By neglecting terms higher than `x^4` we can write a finite product:

`S = {2x-(8x^3)/(6)}{x-x^2/2+x^3/3-x^4/4} + O(x^5)`

`\ \ \ = {2x-(4x^3)/(3)}{x-x^2/2+x^3/3-x^4/4} + O(x^5)`

And now we expand the product:

`S = 2x{ x - 1/2 x^2 + 1/3 x^3 - 1/4 x^4}`
`\ \ \ \ \ \ \ -4/3 x^3{ x - 1/2 x^2 + 1/3 x^3 - 1/4 x^4} + O(x^5)`

`\ \ \ = 2{ x^2 - 1/2 x^3 + 1/3 x^4 - 1/4 x^5 + ... }`
`\ \ \ \ \ \ \ -4/3 { x^4 - 1/2 x^5 + 1/3 x^6 - ... } + O(x^5)`

Again, By neglecting terms higher than `x^4`:

`S = 2{ x^2 - 1/2 x^3 + 1/3 x^4 } -4/3 { x^4 } + O(x^5)`

`\ \ \ = {2}x^2 + {2(-1/2)}x^3 + {2(1/3)-4/3}x^4 +O(x^5)`

`\ \ \ = 2 x^2 - x^3 - 2/3 x^4 +O(x^5)`