Taylor Polynomials- Lagrange remainder

[rambo5330]

So I'm studying for a final, and it just so happens my professor threw taylor polynomials at us in the last week.. I understand the concept of a taylor polynomial but i need some help fully understand the LaGrange remainder theorem

if we have a function that has n derivatives on the interval [a,b] and n+1 derivatives on (a,b). fix a point Xo [tex]\in[/tex] (a,b). then for any x [tex]\in[/tex] (a,b) there exists a number Z between Xo and x ...put z into the lagrange formula and it gives you the bounds on the error from what i understand?

The issue I am having is how to we pick the interval (a,b) ...

for example in the text the most basic question is e^x
if i wanted to estimate the value of e with a 3rd degree taylor polynomial then I calculate the 4 derivatives of e^x (which are all the same) put the first 3 derivatives in a taylor series.. and then for the remainder term use the fourth derivative and then ?

let Xo = 1 ... then the interval (a,b) needs to contain 1. so for example can i pick the interval (1/2, 6) or is it better to have a smaller interval?

 

[Office_Shredder]

Because you're picking z between x₀ and x, it doesn't matter what (a,b) is as long as it contains x₀ and x. If you want to use the Taylor series for e^x centered around 1 to estimate e², you could do two things:

1) Look at the interval (0,3). We apply the theorem to find a point in the interval [1,2] that we can plug into the fourth derivative to find the error.

2) Suppose we instead look at the interval (0,5). Then we would still find the same point in [1,2] that we would use to find the error. The extra part of the interval doesn't do anything here

 

[rambo5330]

so for the remainder term... if Xo = 1 ... the point its centered around I am assuming... and
x = 2 for estimating e² I should pick a value Z in [1,2] such that it creates the maximum amount of error i may get ?

 

[Office_Shredder]

I should pick a value Z in [1,2] such that it creates the maximum amount of error i may get ?

There exists a value Z in [1,2] that gives you the error. It's hard in general to figure out what it is, but we can take a look at what the remainder theorem to get a decent upper bound

The error is [tex] \leq \frac{2-1}{4!} f^{(4)}(Z)[/tex] where [tex]f(x)=e^x[/tex]. Z is some number that we don't know immediately (probably can't calculate it at all in fact), but we do know that [tex] f^{(4)}(x) = e^x[/tex] is an increasing function, so is largest at x=2 on [1,2]. Therefore the upper bound for our error is [tex] \frac{1}{4!} e^2=.31[/tex] (here f⁽⁴⁾ is the fourth derivative)

If we had a more complicated function it would be harder to find out what the maximum of its derivatives are