Tangent Line Error Bound with Taylor Series

I have an equation, $e^x$, based at 0 (b=0).

I am supposed to us the Tangent Line Error Bound to bound the error $|f(x)-T1(x)|$ on the interval I=[-1,1].

(Aside: I have already computed the first Taylor Polynomial $T1(x)$ for $e^x$ at b=0 to be $1+x$)

To find the error, I have done the following:

• Found the second derivative of $f(x)$ to be $f''(x)=e^x$.
• Found $M$, the maximum of $f''(x)$, to be $e^1$ on the interval [-1,1].

Now, the notes I have on finding the error are slightly confusing in that they indicate the method to find the error is one of the following:

• $\frac{M}{(n+1)!} | x - b |^{n+1}$
• $\frac{M}{(n+1)!} | d - b |^{n+1}$

However, using either of these in the homework submission website indicate wrong answers. What am I doing wrong?

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2 Answers

The usual Lagrange estimate for the absolute value of the error at $x$ when we stop at the $k$-th derivative is $$\left|\frac{f^{(k+1)}(\xi)}{(k+1)!}(x-a)^{k+1}\right|,$$ where $\xi$ is between $a$ and $x$. In our case that would be $$\frac{e^\xi}{2}x^2.$$ As you wrote, $e^\xi\le e$, so $\frac{e}{2}x^2$ is an upper bound on the error at $x$.

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Thanks for the explanation and for confirming my finding. Now I just have to figure out why the homework website isn't accepting my solution... –  Andrew M Nov 29 '12 at 0:33
@AndrewM: Conceivably if I knew h exact wording of the question, and of your answer, I might be able to guess. But probably not, luckily we are not so "advanced." –  André Nicolas Nov 29 '12 at 1:35

Same exact problem here. You must be using webassign. It will work if you put in a good decimal approximation according to the Lagrange estimate explained in the other answer. I put in 1.359 and it finally marked my answer correct.

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