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.....
here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d
Which part is giving you trouble?
It looks like you made some mistakes in constructing the Taylor Series.
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.
\[f^{(n+1)}\le f(lower limit)<\]
\[f^{(n+1)}\le f(lower limit)<M\]
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.
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