Question

Find 5th-orden Maclaurin polynomial P_5(x) for f ?

Given the function `f(x) = e^x` for all `x` `in` `RR`.

Find 5th-order Maclaurin polynomial `P_5` `(x)` for `f`.

Answer 1

`p_5(x)=1+x+x^2/2+x^3/6+x^4/24+x^5/120`

Explanation:

The Maclaurin Polynomial will be the Taylor Polynomial centered at `a=0`. If we want the `5`th order Maclaurin Polynomial, we will first need to determine the first `5` terms of the Maclaurin series for `e^x.`

Let us first take the first five derivatives, evaluated at `x=0`. We will adopt the convention that the `0`th derivative is just the function itself.

`f^(0)(x)=e^x -> f^(0)(0)=e^0=1`

`f'(x)=e^x -> f'(0)=e^0=1`

`f''(x)=e^x -> f''(0)=e^0=1`

`f'''(x)=e^x -> f'''(0)=e^(0)=1`

`f^(4)(x)=e^x -> f^(4)(0)=e^0=1`

`f^(5)(x)=e^x -> f^(5)(0)=e^0=1`

We see `f^((n))(0)=1` for all `n`

Furthermore, recall that for a function `f(x),` the `n`th order Maclaurin polynomial expansion is given by

`p_N(x)=sum_(n=0)^Nf^((n))(0)x^n/(n!) = f(0)+f'(0)x+(f''(0)x^2)/(2!)+(f'''(0)x^3)/(3!)+...+(f^((N))(0)x^N)/(N!)`

So, for `e^x,`

`p_5(x)=sum_(n=0)^5f^((n))(0)x^n/(n!)`

Knowing that `f^((n))(0)=1` for all `n` when dealing with `e^x,` and `f(0)=e^0=1`,

`p_5(x)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+x^5/(5!)`

Note that `0! =1, 1! =1`

Simplifying the factorials yields

`p_5(x)=1+x+x^2/2+x^3/6+x^4/24+x^5/120`