Krdytkyu

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 $x in [0,1]$

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 $lt frac{1}{7!}$.