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I missed the lecture on this and was wondering if someone could explain the steps involved with this problem. I think that what I have to do is evaluate the polynomial up to the second derivative (degree $2$) and then I can choose $x=7$ and evaluate the polynomial at that point. Then with those values plug it into the error series? However I do not know the error series. Am I on the right track?

Finally, Part B I have absolutely NO idea what it is asking.

Thanks!

(a) Use Taylor's Theorem to estimate the error in using the Taylor Polynomial of $f(x)$=$\sqrt{x}$ of degree $2$ to approximate $\sqrt{8}$. (The answer should be something like $1/2 \times 8^{-7/2}$).

(b) Find a bound on the difference of $\sin(x)$ and $\,x-\frac{x^{3}}{6} + \frac{x^{5}}{120}$ for $x \in [0,1]$

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A not unreasonable thing to do, and probably what you are expected to do, is to expand $\sqrt{x}$ in powers of $x-9$.

The formula for the $n$-th degree Taylor polynomial for $f(x)$ in powers of $x-a$ is $$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a)^n.$$ In our case we have $f(x)=x^{1/2}$, $a=9$, and $n=2$. To finish getting the second degree polynomial, you need to calculate $f(9)$, $f'(9)$, and $f''(9)$.

We now turn to the error term. By the Lagrange form of the remainder, the absolute value of the error if we use the Taylor polynomial of degree $2$ is $$\frac{|f'''(\xi)|}{3!}|x-9|^3,$$ where $f(x)=x^{1/2}$ and $\xi$ is a number between $x$ and $9$. In our case, we have $x=8$.

Note that $f'''(x)=\frac{3}{8}x^{-5/2}$. Between $8$ and $9$, it is positive and bounded above by $\frac{3}{8}8^{-5/2}$. Thus, putting things together, and noting that $|8-9|=1$, we find that the absolute value of the error is less than $\frac{3}{(8)(3!)}8^{-5/2}$.

For the sine function, the best thing to do is to note that when $|x|\lt 1$ then the usual series for $\sin x$ is an alternating series. For such a series, the absolute value of the error is less than the absolute value of the first "neglected" term. In our case, that term is $-\frac{x^7}{7!}$, so in the interval $[0,1]$ the error has absolute value $\lt \frac{1}{7!}$.

Alternately, but less attractively, we can use the Lagrange formula for the remainder. It is useful to note that the given series up to $\frac{x^5}{5!}$ can be considered to be the expansion up to the term in $x^6$, which happens to be $0$. Thus the Lagrange formula gives absolute value of the error equal to $$\frac{|f^{(7)}(\xi)|}{7!}|x|^7,$$ where $f(x)=\sin x$. The $7$-th derivative of $f(x)$ is $-\cos x$, so the absolute value of the $7$-th derivative is $\lt 1$. That gives us that the absolute value of the error if $|x|\lt 1$ is $\lt \frac{1}{7!}$.

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  • $\begingroup$ unfortunately, we have not used Lagrange form for the remainder so a lot of this sounds very foreign to me. I am in a 2nd year real analysis class. $\endgroup$ – xc92 May 13 '14 at 3:15
  • $\begingroup$ What form of the remainder have you used? For information about the various forms of the remainder, you might look at Wikipedia, Taylor's Theorem/ $\endgroup$ – André Nicolas May 13 '14 at 3:19
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    $\begingroup$ Ah your edit made it a tad bit clearer, I didn't realize how close $3/(8)(3!)$ $8^{-5/2}$ was to the approximation they gave us! Thank you Andre! $\endgroup$ – xc92 May 13 '14 at 3:21

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