# How do you determine the third Taylor polynomial of the given function at x = 0 f(x)= e^(-x/2)?

Aug 9, 2017

e^(-x/2) = sum_(n = 0)^oo ((-1)^(n)(x^n))/((2^n)n!) = 1 - x/2 + x^2/(4(2!)) - x^3/(8(3!))...

#### Explanation:

We know the Maclaurin Series (taylor series centered at 0) for ${e}^{x}$.

e^x = 1 + x + x^2/(2!) + x^3/(3!) + ... + x^n/(n!)

If you substitute $- \frac{x}{2}$ for all of the x's in the series for ${e}^{x}$, you get:

e^(-x/2) = 1 - x/2 + (-x/2)^2/(2!) + (-x/2)^3/(3!) ...

e^(-x/2) = 1 - x/2 + x^2/(4(2!)) - x^3/(8(3!))...

This can be modelled by

sum_(n = 0)^oo ((-1)^(n)(x^n))/((2^n)n!)

Hopefully this helps!