Question

What would be the expansion of sin x in powers of x?

Answer 1

`sum_"n=0->oo" (-1)^nx^(2n+1)/((2n+1)!)`

Explanation:

Consider the derivatives of `f(x)=sinx`:

`f'(x)= cosx, f''(x)-sinx, f'''(x) = -cosx, f''''(x)=sinx....`

From the Taylor/Mclaurin series expansion we have:
`f(x) = f(0)+f'(0)x/(1!)+f''(0)x^2/(2!)+f'''(0)x^3/(3!)+ ........`

In this example we have `f(x) =sinx`

Note that since `sin0 = 0` all even powers of `x` will equal 0 in the series expansion.

Thus: `f(x) = sinx = cos(0)x/(1!)-cos(0)x^3/(3!)+ cos0x^5/(5!)-.......`

Now, since `cos0=1` the series reduces to:

`sinx = x/(1!)-x^3/(3!)+x^5/(5!)- .......`

The series is infinite in odd powers of x with alternating sign and thus can be written as:

`sum_"n=0->oo" (-1)^nx^(2n+1)/((2n+1)!)`