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.....

- anonymous

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- anonymous

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

- anonymous

Which part is giving you trouble?

- anonymous

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

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## More answers

- anonymous

where?

- anonymous

i see it

- anonymous

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

- anonymous

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

- anonymous

I fixed it

- anonymous

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

- anonymous

Much better =)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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.

- anonymous

ceiling*

- anonymous

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

- anonymous

I don't think you're right.

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

|dw:1342915184406:dw|

- anonymous

Text on right says max=10

- anonymous

|dw:1342915323520:dw|

- anonymous

|dw:1342915451038:dw|

- anonymous

Tell me that you understand. Because I can't make the words to explain this.

- anonymous

I'm really sorry, John. I'm really trying to understand this. Ok here is what I understand.
I see that you have drawn three unique functions, and each function was given an interval, and a maximum was chosen.
The maximum that you have chosen is the M.
The derivative should be less than M on the whole interval.

- anonymous

That M allows us to put a cap on the error

- anonymous

What we're trying to do with M is to put a cap on the error. That's why we need to find an actual value for M.

- anonymous

That value M is greater than the next derivative up.

- anonymous

hold on smooth

- anonymous

Well, let me plot the 4th derivative for us.

- anonymous

no hold on

- anonymous

http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5

- anonymous

x=1.2 will give me the ultimate cap.
\[f^4(x)=sinx\le sin(1.2)<1.932\]

- anonymous

ultimate cap for our interval

- anonymous

sin(1.2) is not 1.932.

- anonymous

0.932 :P

- anonymous

That's impossible, since sin oscillates between 1 and -1.

- anonymous

Good good =)

- anonymous

typo dude...chill out

- anonymous

and I didn't look at your wolfram ;P

- anonymous

Victory?

- anonymous

Almost there.

- anonymous

The example is what confused me. They picked 7 instead of 9....but 1/7 is greater than 1/9 that's why they picked 7

- anonymous

remember the interval \[7 \le x \le 9\]

- anonymous

I do remember. Please realize that the maximum is not always going to be at one of the endpoints.

- anonymous

I understand that now

- anonymous

You really have to look at what the function is. It's going to be different every time and you have to analyze the function to figure out where the upper bound is.

- anonymous

Yes sir.

- anonymous

Okay good =)

- anonymous

\[\left|R_3(x)\right|\le \frac{.932}{4!} {\left|1.2-1\right|}^4\]

- anonymous

yes?

- anonymous

\[\left|R_3(x)\right|\le 0.000054\]

- anonymous

Looks good, pal.

- anonymous

Thank ya! You're amazingly patient. Thank you soo much John. :)

- anonymous

My pleasure! =D

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