Question

How do you find the maclaurin series expansion of `sin2x`?

Answer 1

`sin2x=sum_(n=0)^oo(2*4^n)/((2n+1)!)x^(2n+1)`.

Explanation:

Any general Maclauren series for a function `f(x)` may be expressed as

`f(x)=sum_(n=0)^oo(f^ (n) (x))/(n!)x^n`

So in this case, if we write the first few derivatives of the function out we get

`f=sin2x=>f(0)=0`
`f'=2cos2x=>f'(0)=2`
`f''=-4sin2x=>f''(0)=0`
`f'''=8cos2x=>f'''(0)=8`
`f^4=-16sin2x=>f^4(0)=0`
`f^5=32cos2x=>f^5(0)=32`

Continuing in this way, we eventually get that every even term is zero and hence vanishes from the Maclaren series, and every odd term has value `(2*4^n)`.

Therefore the Mclauen power series expansion for this function is :

`sin2x=sum_(n=0)^oo(2*4^n)/((2n+1)!)x^(2n+1)`.