# Lagrange Error Bound Question/Problem

• March 18th 2012, 01:03 PM
Xeritas
Lagrange Error Bound Question/Problem
My Calculus teacher gave me the following problem:

The function f has derivatives of all orders for all real numbers, and f(4) (x) = esin(x). If the third degree Taylor polynomial for f about x=0 is used to approximate f on the interval [0,1], what is the Lagrange error bound for the maximum error on the interval [0,1]?

My peers and I had a disagreement on the correct answer and I'm curious which one of us is right.
• March 18th 2012, 01:28 PM
skeeter
Re: Lagrange Error Bound Question/Problem
Quote:

Originally Posted by Xeritas
My Calculus teacher gave me the following problem:

The function f has derivatives of all orders for all real numbers, and f(4) (x) = esin(x). If the third degree Taylor polynomial for f about x=0 is used to approximate f on the interval [0,1], what is the Lagrange error bound for the maximum error on the interval [0,1]?

My peers and I had a disagreement on the correct answer and I'm curious which one of us is right.

so ... what say the both of you w/ regard to the maximum error?
• March 18th 2012, 02:57 PM
Xeritas
Re: Lagrange Error Bound Question/Problem
I'm saying the Lagrange error bound should be .113 while the other individual says it should be .097.<--period

We're using the following definition:

|Rn(x)|<= (M/((n+1)!))|x-a|n+1
• March 18th 2012, 03:15 PM
skeeter
Re: Lagrange Error Bound Question/Problem
for any $x \in [0,1]$ ...

$|f(x) - P_3(x)| \le \frac{e^{\sin(1)}}{4!} \cdot x^4$

note that the error depends on the x-value in [0,1] where the function is being approximated.
• March 18th 2012, 04:07 PM
Xeritas
Re: Lagrange Error Bound Question/Problem
Quote:

Originally Posted by skeeter
for any $x \in [0,1]$ ...

$|f(x) - P_3(x)| \le \frac{e^{\sin(1)}}{4!} \cdot x^4$

note that the error depends on the x-value in [0,1] where the function is being approximated.

Yes, but the maximum error should occur furthest from the origin, which is at x=1, correct?
And M is referring to the maximum value of the function, which would then make it:
$\frac{e^{\1}}{4!} \cdot 1^4$
Correct?
Or is the x value within M restricted to [0,1]?
• March 18th 2012, 04:19 PM
skeeter
Re: Lagrange Error Bound Question/Problem
Quote:

Originally Posted by Xeritas
Yes, but the maximum error should occur furthest from the origin, which is at x=1, correct?
And M is referring to the maximum value of the function, which would then make it:
$\frac{e^{\1}}{4!} \cdot 1^4$
Correct?
Or is the x value within M restricted to [0,1]?

M is the maximum value of the (n+1)st derivative on the interval [0,1], $M = e^{\sin(1)}$ in this case ... x is the specific value in [0,1] where the function is being approximated.
• March 18th 2012, 04:25 PM
Xeritas
Re: Lagrange Error Bound Question/Problem
Ok, I guess I heard wrong from our teacher, thanks again.