anonymous
  • anonymous
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.....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d
anonymous
  • anonymous
Which part is giving you trouble?
anonymous
  • anonymous
It looks like you made some mistakes in constructing the Taylor Series.

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anonymous
  • anonymous
where?
anonymous
  • anonymous
i see it
anonymous
  • anonymous
Kind of a lot of mistakes. Let me see you try again.
anonymous
  • 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
  • anonymous
I fixed it
anonymous
  • anonymous
That's just by the definition of the Taylor series.
anonymous
  • anonymous
Much better =)
anonymous
  • anonymous
Okay, that's the first part. Now for the second part, we need to get M.
anonymous
  • anonymous
\[f^4(x)=sinx\le M\]
anonymous
  • anonymous
\[\left|R_3(x)\right|\le \frac{sinx}{4!} {\left|x-1\right|}^4\] shoot me
anonymous
  • anonymous
Hold on. We need to get an actual value of M. Remember how we get that?
anonymous
  • anonymous
it's the absolute value of f^4 (x)
anonymous
  • anonymous
don't give up on me quite yet please..... sin(1) or I have to do something with \[x \ge 0.8\]
anonymous
  • anonymous
Let's step back, slow down, and understand what it is that we're doing. Just some theory.
anonymous
  • anonymous
\[f^{(n+1)}\le f(lower limit)<\]
anonymous
  • anonymous
\[f^{(n+1)}\le f(lower limit)
anonymous
  • 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
  • anonymous
The basic statement is: \( \large f^{n+1} \le M\) Therefore, \(R_n < \text{some thing dependent on M}\)
anonymous
  • 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
  • anonymous
ceiling*
anonymous
  • anonymous
yep. So we pick the lowest limit that we're allowed 0.8 in this case
anonymous
  • anonymous
I don't think you're right.
anonymous
  • anonymous
Let me talk about the maximum of a function for a bit.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Okay, I'm thinking just a few examples will help you to see what it is we're doing.
anonymous
  • anonymous
|dw:1342915184406:dw|
anonymous
  • anonymous
Text on right says max=10
anonymous
  • anonymous
|dw:1342915323520:dw|
anonymous
  • anonymous
|dw:1342915451038:dw|
anonymous
  • anonymous
Tell me that you understand. Because I can't make the words to explain this.
anonymous
  • 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
  • anonymous
That M allows us to put a cap on the error
anonymous
  • 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
  • anonymous
That value M is greater than the next derivative up.
anonymous
  • anonymous
hold on smooth
anonymous
  • anonymous
Well, let me plot the 4th derivative for us.
anonymous
  • anonymous
no hold on
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5
anonymous
  • anonymous
x=1.2 will give me the ultimate cap. \[f^4(x)=sinx\le sin(1.2)<1.932\]
anonymous
  • anonymous
ultimate cap for our interval
anonymous
  • anonymous
sin(1.2) is not 1.932.
anonymous
  • anonymous
0.932 :P
anonymous
  • anonymous
That's impossible, since sin oscillates between 1 and -1.
anonymous
  • anonymous
Good good =)
anonymous
  • anonymous
typo dude...chill out
anonymous
  • anonymous
and I didn't look at your wolfram ;P
anonymous
  • anonymous
Victory?
anonymous
  • anonymous
Almost there.
anonymous
  • 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
  • anonymous
remember the interval \[7 \le x \le 9\]
anonymous
  • anonymous
I do remember. Please realize that the maximum is not always going to be at one of the endpoints.
anonymous
  • anonymous
I understand that now
anonymous
  • 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
  • anonymous
Yes sir.
anonymous
  • anonymous
Okay good =)
anonymous
  • anonymous
\[\left|R_3(x)\right|\le \frac{.932}{4!} {\left|1.2-1\right|}^4\]
anonymous
  • anonymous
yes?
anonymous
  • anonymous
\[\left|R_3(x)\right|\le 0.000054\]
anonymous
  • anonymous
Looks good, pal.
anonymous
  • anonymous
Thank ya! You're amazingly patient. Thank you soo much John. :)
anonymous
  • anonymous
My pleasure! =D

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