My book states the following: Assume that f^(n+1)(x) exists and is continuous. L

Question

My book states the following:

Assume that f^(n+1)(x) exists and is continuous. Let K be a number such that abs(f^(n+1)(u)) is less than or equal to K for all u between a and x. Then
abs(Tn(x)-f(x)) is less than or equal to K*abs(x-a)^(n+1)/(n+1)!

I don't really understand how to find K.

The problem I am working on is:
Let Tn(x) be the nth Maclaurin polynomial for f(x) = e^x.
Determine a value of n so that |Tn(2) - e^2| < 10^(-4)

Being that f^(n)(0)=1 for all n, would I use K=1? Then, should I just plug in numbers for n until I get an error less than 10^(-4)?

Explanation / Answer

No you use the MAXimum of f^(n+1) in [x,a] which is [0,2] in your case. K(n+1) = max f^(n+1)(x) = e^2 for all n K abs(x-a)^(n+1)/(n+1)!

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