MATH 152 Chapter 10.9

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Error Analysis in Taylor Polynomials

We can find the Taylor series of any differentiable function f(x)=∑n=0∞cn(x−a)n, where cn=f(n)(a)n!.

We can approximate the function with a finite polynomial by taking the Nth degree Taylor Polynomial of f at x=a:

TN(x)=∑n=0Nf(n)(a)n!(x−a)n


Remainder

Everything else past the finite polynomial that we chopped off:

RN(x)=∑n=N+1∞f(n)(a)n!(x−a)n=f(x)−TN(x)

We can estimate RN; this shows how far off we are at any point on the given interval (of x) when we stop at the Nth degree Taylor Polynomial:

Alternating Series

|RN(x)|≤aN+1≤f(N+1)(a)(N+1)!|x−a|N+1

Taylor's Inequality

If |f(N+1)(x)|≤M on a given interval (in other words, if we can get an upper bound on the derivative), then

|RN(x)|≤M(N+1)!|x−a|N+1
Note: given; do not memorize

Example

Use a 3rd degree Taylor polynomial at a=1 to approximate f(x)=x on the interval [0.9, 1.1] and determine the accuracy of your results using the remainder theorem

n f(n)(x) f(n)(1) f(n)(1)n!(x−1)n
0 x 1 1
1 12x−1/2 12 1/21!(x−1)1
2 −14x−3/2 −14 −1/42!(x−1)2
3 38x−5/2 38 3/83!(x−1)3
T3(x)=1+12(x−1)−18(x−1)2+116(x−1)3


  • N=3
  • a=1
  • x∈[0.9,1.1] . . . Which value of x maximizes the error bound?
    • M=|f(4)(x)| (max. of next derivative on the interval of x)
    • Mmax occurs at x=0.9: 15/16(0.9)7/24!|0.9−1|4


|R3(x)|≤15/16(0.9)7/24!|0.9−1|4

Example

Determine the degree of the Taylor Polynomial needed to approximate sin⁡(0.1) to within 0.00001 accuracy.

sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!

|s−sN|≤|aN+!|<.00001(0.1)2N+3(2N+3)!<.00001use tech. to solveN=1