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

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Answer 1 · 2017-08-09T15:38:31

#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!))...#

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 #-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!