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anonymous
 4 years ago
Please help....Please help....Taylor series
f(x)=sinx a=1 n=3
\[0.8 \le x \le 1.2\]
\[sinx\approx T_3(x)=sin(1)+\frac{cos(1)}{1}(x1)\frac{sin(1)}{2}(x1)^2\frac{cos(1)}{6}(x1)^3\]
(b) Use Taylor's Inequality to estimate the accuracy of the approximation \[f(x)\approx T_n(x)\] lies in the given interval.
\[\leftR_3(x)\right\le \frac{M}{4!} {\leftx1\right}^4\]
I'm almost there.....
anonymous
 4 years ago
Please help....Please help....Taylor series f(x)=sinx a=1 n=3 \[0.8 \le x \le 1.2\] \[sinx\approx T_3(x)=sin(1)+\frac{cos(1)}{1}(x1)\frac{sin(1)}{2}(x1)^2\frac{cos(1)}{6}(x1)^3\] (b) Use Taylor's Inequality to estimate the accuracy of the approximation \[f(x)\approx T_n(x)\] lies in the given interval. \[\leftR_3(x)\right\le \frac{M}{4!} {\leftx1\right}^4\] I'm almost there.....

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Which part is giving you trouble?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It looks like you made some mistakes in constructing the Taylor Series.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Kind of a lot of mistakes. Let me see you try again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large T_3 = f(a) +\frac{f'(a)}{1!}*(xa) + \frac{f''(a)}{2!}*(xa)^2 + \frac{f'''(a)}{3!}*(xa)^3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's just by the definition of the Taylor series.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, that's the first part. Now for the second part, we need to get M.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\leftR_3(x)\right\le \frac{sinx}{4!} {\leftx1\right}^4\] shoot me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hold on. We need to get an actual value of M. Remember how we get that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's the absolute value of f^4 (x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0don't give up on me quite yet please..... sin(1) or I have to do something with \[x \ge 0.8\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let's step back, slow down, and understand what it is that we're doing. Just some theory.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f^{(n+1)}\le f(lower limit)<\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f^{(n+1)}\le f(lower limit)<M\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here'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
 4 years ago
Best ResponseYou've already chosen the best response.0The basic statement is: \( \large f^{n+1} \le M\) Therefore, \(R_n < \text{some thing dependent on M}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0yep. So we pick the lowest limit that we're allowed 0.8 in this case

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think you're right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let me talk about the maximum of a function for a bit.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, I'm thinking just a few examples will help you to see what it is we're doing.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342915184406:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Text on right says max=10

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342915323520:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342915451038:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Tell me that you understand. Because I can't make the words to explain this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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
 4 years ago
Best ResponseYou've already chosen the best response.0That M allows us to put a cap on the error

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What 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
 4 years ago
Best ResponseYou've already chosen the best response.0That value M is greater than the next derivative up.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, let me plot the 4th derivative for us.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x=1.2 will give me the ultimate cap. \[f^4(x)=sinx\le sin(1.2)<1.932\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ultimate cap for our interval

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sin(1.2) is not 1.932.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's impossible, since sin oscillates between 1 and 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0typo dude...chill out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and I didn't look at your wolfram ;P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0remember the interval \[7 \le x \le 9\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I do remember. Please realize that the maximum is not always going to be at one of the endpoints.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand that now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\leftR_3(x)\right\le \frac{.932}{4!} {\left1.21\right}^4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\leftR_3(x)\right\le 0.000054\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank ya! You're amazingly patient. Thank you soo much John. :)
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