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) = e^{sin(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.

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) = e^{sin(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?

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:

|R_{n}(x)|<= (M/((n+1)!))|x-a|^{n+1}

Re: Lagrange Error Bound Question/Problem

for any $\displaystyle x \in [0,1]$ ...

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

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

Re: Lagrange Error Bound Question/Problem

Quote:

Originally Posted by

**skeeter** for any $\displaystyle x \in [0,1]$ ...

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

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:

$\displaystyle \frac{e^{\1}}{4!} \cdot 1^4$

Correct?

Or is the x value within M restricted to [0,1]?

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:

$\displaystyle \frac{e^{\1}}{4!} \cdot 1^4$

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], $\displaystyle M = e^{\sin(1)}$ in this case ... x is the **specific value** in [0,1] where the function is being approximated.

Re: Lagrange Error Bound Question/Problem

Ok, I guess I heard wrong from our teacher, thanks again.