Taylor Polynomials- Lagrange remainder

rambo5330
Messages
83
Reaction score
0
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 \in (a,b). then for any x \in (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 ex
if i wanted to estimate the value of e with a 3rd degree taylor polynomial then I calculate the 4 derivatives of ex (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?
 
Physics news on Phys.org
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 ex 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
 
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 ?
 
rambo5330 said:
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 \leq \frac{2-1}{4!} f^{(4)}(Z) where f(x)=e^x. Z is some number that we don't know immediately (probably can't calculate it at all in fact), but we do know that f^{(4)}(x) = e^x is an increasing function, so is largest at x=2 on [1,2]. Therefore the upper bound for our error is \frac{1}{4!} e^2=.31 (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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top