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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
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Which part is giving you trouble?
It looks like you made some mistakes in constructing the Taylor Series.

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Other answers:

where?
i see it
Kind of a lot of mistakes. Let me see you try again.
\[\large T_3 = f(a) +\frac{f'(a)}{1!}*(x-a) + \frac{f''(a)}{2!}*(x-a)^2 + \frac{f'''(a)}{3!}*(x-a)^3\]
I fixed it
That's just by the definition of the Taylor series.
Much better =)
Okay, that's the first part. Now for the second part, we need to get M.
\[f^4(x)=sinx\le M\]
\[\left|R_3(x)\right|\le \frac{sinx}{4!} {\left|x-1\right|}^4\] shoot me
Hold on. We need to get an actual value of M. Remember how we get that?
it's the absolute value of f^4 (x)
don't give up on me quite yet please..... sin(1) or I have to do something with \[x \ge 0.8\]
Let's step back, slow down, and understand what it is that we're doing. Just some theory.
\[f^{(n+1)}\le f(lower limit)<\]
\[f^{(n+1)}\le f(lower limit)
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.
The basic statement is: \( \large f^{n+1} \le M\) Therefore, \(R_n < \text{some thing dependent on M}\)
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.
ceiling*
yep. So we pick the lowest limit that we're allowed 0.8 in this case
I don't think you're right.
Let me talk about the maximum of a function for a bit.
ok
Okay, I'm thinking just a few examples will help you to see what it is we're doing.
|dw:1342915184406:dw|
Text on right says max=10
|dw:1342915323520:dw|
|dw:1342915451038:dw|
Tell me that you understand. Because I can't make the words to explain this.
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.
That M allows us to put a cap on the error
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.
That value M is greater than the next derivative up.
hold on smooth
Well, let me plot the 4th derivative for us.
no hold on
http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5
x=1.2 will give me the ultimate cap. \[f^4(x)=sinx\le sin(1.2)<1.932\]
ultimate cap for our interval
sin(1.2) is not 1.932.
0.932 :P
That's impossible, since sin oscillates between 1 and -1.
Good good =)
typo dude...chill out
and I didn't look at your wolfram ;P
Victory?
Almost there.
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
remember the interval \[7 \le x \le 9\]
I do remember. Please realize that the maximum is not always going to be at one of the endpoints.
I understand that now
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.
Yes sir.
Okay good =)
\[\left|R_3(x)\right|\le \frac{.932}{4!} {\left|1.2-1\right|}^4\]
yes?
\[\left|R_3(x)\right|\le 0.000054\]
Looks good, pal.
Thank ya! You're amazingly patient. Thank you soo much John. :)
My pleasure! =D

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