Question

Please help....Please help....Taylor series
f(x)=sinx a=1 n=3
\[0.8 \le x \le 1.2\]
\[sinx\approx T_3(x)=sin(1)+\frac{cos(1)}{1}(x-1)-\frac{sin(1)}{2}(x-1)^2-\frac{cos(1)}{6}(x-1)^3\]
(b) Use Taylor's Inequality to estimate the accuracy of the approximation \[f(x)\approx T_n(x)\] lies in the given interval.
\[\left|R_3(x)\right|\le \frac{M}{4!} {\left|x-1\right|}^4\]
I'm almost there.....

Answer 1

here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d

Answer 2

Which part is giving you trouble?

Answer 3

It looks like you made some mistakes in constructing the Taylor Series.

Answer 4

where?

Answer 5

i see it

Answer 6

Kind of a lot of mistakes. Let me see you try again.

Answer 7

\[\large T_3 = f(a) +\frac{f'(a)}{1!}*(x-a) + \frac{f''(a)}{2!}*(x-a)^2 + \frac{f'''(a)}{3!}*(x-a)^3\]

Answer 8

I fixed it

Answer 9

That's just by the definition of the Taylor series.

Answer 10

Much better =)

Answer 11

Okay, that's the first part. Now for the second part, we need to get M.

Answer 12

\[f^4(x)=sinx\le M\]

Answer 13

\[\left|R_3(x)\right|\le \frac{sinx}{4!} {\left|x-1\right|}^4\]
shoot me

Answer 14

Hold on. We need to get an actual value of M. Remember how we get that?

Answer 15

it's the absolute value of f^4 (x)

Answer 16

don't give up on me quite yet please.....
sin(1)
or
I have to do something with \[x \ge 0.8\]

Answer 17

Let's step back, slow down, and understand what it is that we're doing. Just some theory.

Answer 18

\[f^{(n+1)}\le f(lower limit)<\]

Answer 19

\[f^{(n+1)}\le f(lower limit)<M\]

Answer 20

Here'es the idea. We've made a taylor polynomial to approximate the function. Now we're trying to make a statement about how accurate the approximation is.
Taylor's inequality allows us to do that. Here's how.
We look at the derivative one higher than our approximation, and we look at it on the interval that we're interested in. If we can pick out a number, call it M, that the derivative is less than on the whole interval, then that allows us to use that M to limit the error.

Answer 21

The basic statement is:
\( \large f^{n+1} \le M\)
Therefore, \(R_n < \text{some thing dependent on M}\)

Answer 22

The lower the M I'm able to pick, the better. Why? Because it allows me to put a lower cieling on the possible error.

Answer 23

ceiling*

Answer 24

yep. So we pick the lowest limit that we're allowed 0.8 in this case

Answer 25

I don't think you're right.

Answer 26

Let me talk about the maximum of a function for a bit.

Answer 27

ok

Answer 28

Okay, I'm thinking just a few examples will help you to see what it is we're doing.