Question

How do you find the MacLaurin series for the function `f(x) = 5 cos pix`, and write it in sumation-notation?

Answer 1

`5cospix=5sum_(k=0)^oo ((-1)^k(pix)^(2k))/(2k!)`

Explanation:

The MacLaurin of a function is given by:

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

We therefore have to determine the expression of:

`(d^n)/(dx^n)(5cospix)`

for every `n`.

We start from:

`d/(dx)(5cospix) =-5pisinpix=5picos(pix+pi/2)`
`d^2/(dx^2)(5cospix) =-5pi^2cospix= (5pi)^2cos(pix+2pi/2)`

and we can see (and demonstrate using mathematical induction ) that in general:

`d^n/(dx^n)(5cospix) = 5pi^ncos(pix+npi/2)`

For `x=0` this becomes:

`f^((n))(0)=[d^n/(dx^n)(5cospix)]_(x=0) = 5pi^ncos(npi/2)= [(0 " for n odd"),((-1)^(n/2)5pi^n " for n even")]`

We can therefore substitute this expression for `f^((n))(0)` in `[1]` and keep only the terms for n even:

`5cospix= 5sum_(k=0)^oo ((-1)^kpi^(2k))/(2k!)x^(2k)=5sum_(k=0)^oo ((-1)^k(pix)^(2k))/(2k!)`