Question

Ok let's look at this example.
(a) Approximate the function \[f(x)=\sqrt[3]{x}\] by a Taylor polynomial of degree 2 at a=8.
\[f(x)=\sqrt[3]{x}\]
\[f'(x)=\frac13 x^{-2/3}\]
\[f''(x)=-\frac{2}{9} x^{-5/3}\]
\[f'''(x)=\frac{10}{27} x^{-8/3}\]
f(8)=2
\[f'(8)=\frac{1}{12}\]
\[f''(8)=-\frac{1}{114}\]
Thus:
\[T_2(x)=f(8)+\frac{f'(8)}{1!}(x-8)+\frac{f''(8)}{2!}(x-8)^2\]
so...
\[\sqrt[3]{x}=T_2(x)=2+\frac{1}{12}(x-8)-\frac{1}{288}(x-8)^2\]
This part makes sense, but what about this:
(b) How accurate is this approximation when \[7\le x \le 9\]?
Taylor inequality Theorem states
\[f^{(n+

Answer 1

I have to complete writing the second part. Give me a second

Answer 2

The Taylor inequality Theorem States
if
\[\left|f^{(n+1)}(x)\right|\le M\]
then
\[\left|R_n(x) \right|\le \frac{M}{(n+1)!}{\left| x-1\right|}^{n+1}\]

Answer 3

if n=2 and a=8 then
\[\left|R_2(x) \right|\le \frac{M}{3!}{\left| x-8\right|}^{3}\]

Answer 4

where \[f'''(x)\le M\] Because \[x\le 7\], we have \[x^{8/3} \le 7^{8/3}\] and so

Answer 5

and so \[f'''(x)=\frac{10}{27}* \frac{1}{x^{8/3}} \le \frac{10}{27}*\frac{1}{7^{8/3}}<0.0021\]

Answer 6

I don't understand part (b) of this question. Please please help

Answer 7

Looks like you've already done it.

Answer 8

It's an example from the book. I don't understand the example....
I can't do my homework if I don't even understand the example. I posted my hw question yesterday and no one could help me.

Answer 9

It's example 1 in Chapter 11 section 11 of Stewart's Calculus.

Answer 10

Oh, okay. Which part of it confuses you? For one thing, I think you typed the original equation incorrectly.
\(\Large |R_n(x) |\le \frac{M}{(n+1)!}{\left| x-1\right|}^{n+1} \)
should be
\(\Large |R_n(x) |\le \frac{M}{(n+1)!}{\left| x-a\right|}^{n+1} \)
right?

Answer 11

Yes you are right. I guess I understand everything up to the 4th post but this part is still a mystery.
\[f'''(x)=\frac{10}{27}* \frac{1}{x^{8/3}} \le \frac{10}{27}*\frac{1}{7^{8/3}}<0.0021\]

Answer 12

hold on!!!!

Answer 13

sry don't mean to scream but I think something just clicked

Answer 14

so we replace M with f'''(x) in that inequality

Answer 15

Right. Exactly. M comes from looking at the derivative one degree higher, so we look at
f′′′(x)=(10/27)x^(−8/3)
On the one side, at 7, we get some number, and as x increases, that M gets smaller and smaller. So we take the largest value of M because f''' will always be less than or equal to that M over the interval.

Then we use M in the second equation.

Answer 16

I didn't really state that very well, but we look at the n+1 derivative, and
M is the greatest value of that derivative over the x interval.

Answer 17

Let's step back for a moment and look at the bigger picture. I think I'm losing track of what the purpose of this inequality is.

Answer 18

So we have this function f(x) and I'm trying to approximate that line by taking it's 2nd derivative. How am I doing so far?

Answer 19

Well, you're trying to approximate the line by creating a polynomial. The polynomial uses the first derivative and the second derivative in this case. To make it more accurate, you can use the third derivative or even more.

Answer 20

Understood. So when we say \[7 \le x \le9\]
are we zooming in on a value between 7 and 9 (graphically speaking)

Answer 21

Right right. We're interested in the accuracy of our approximation between those two x values. We're not worried about how accurate it is when x is 10 or 10000

Answer 22

ok. What is M? Just a value that is greater than next derivative up?

Answer 23

Right. So if we look at \(\large f^{n+1}(x)\) over the interval and see that it's always less than or equal to some number, then we call that number M. We could use a number that's dramatically larger, but it's better if we use something that's the max value of the deriv or close to it.