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)! < e^2 * 2^(n+1)/(n+1)! < 6.32e-5 when n=11.

So the answer is n=11.

--------------------------------------…
Numerical validation. The actual error |Tn - e^2| is given below. The table is generated by computer. When n=11, the error is about 1.01e-5, not too far away from the estimated error upper bound.


n = 1, error = -4.38906
n = 2, error = -2.38906
n = 3, error = -1.05572
n = 4, error = -0.389056
n = 5, error = -0.122389
n = 6, error = -0.0335005
n = 7, error = -0.00810372
n = 8, error = -0.00175451
n = 9, error = -0.000343577
n = 10, error = -6.13899e-005
n = 11, error = -1.00832e-005 *****
n = 12, error = -1.5321e-006
n = 13, error = -2.16541e-007
n = 14, error = -2.86047e-008

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