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Please help....Please help....Taylor series
f(x)=sinx a=1 n=3
\[0.8 \le x \le 1.2\]
\[sinx\approx T_3(x)=sin(1)+\frac{cos(1)}{1}(x1)\frac{sin(1)}{2}(x1)^2\frac{cos(1)}{6}(x1)^3\]
(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.....
 one year ago
 one year 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.....
 one year ago
 one year ago

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MathSofiyaBest ResponseYou've already chosen the best response.1
here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Which part is giving you trouble?
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
It looks like you made some mistakes in constructing the Taylor Series.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Kind of a lot of mistakes. Let me see you try again.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
\[\large T_3 = f(a) +\frac{f'(a)}{1!}*(xa) + \frac{f''(a)}{2!}*(xa)^2 + \frac{f'''(a)}{3!}*(xa)^3\]
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
That's just by the definition of the Taylor series.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Okay, that's the first part. Now for the second part, we need to get M.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[f^4(x)=sinx\le M\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\leftR_3(x)\right\le \frac{sinx}{4!} {\leftx1\right}^4\] shoot me
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Hold on. We need to get an actual value of M. Remember how we get that?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
it's the absolute value of f^4 (x)
 one year ago

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

SmoothMathBest ResponseYou've already chosen the best response.1
Let's step back, slow down, and understand what it is that we're doing. Just some theory.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[f^{(n+1)}\le f(lower limit)<\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[f^{(n+1)}\le f(lower limit)<M\]
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

SmoothMathBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yep. So we pick the lowest limit that we're allowed 0.8 in this case
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
I don't think you're right.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Let me talk about the maximum of a function for a bit.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Okay, I'm thinking just a few examples will help you to see what it is we're doing.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
dw:1342915184406:dw
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Text on right says max=10
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
dw:1342915323520:dw
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
dw:1342915451038:dw
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Tell me that you understand. Because I can't make the words to explain this.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
That M allows us to put a cap on the error
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
That value M is greater than the next derivative up.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
Well, let me plot the 4th derivative for us.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
x=1.2 will give me the ultimate cap. \[f^4(x)=sinx\le sin(1.2)<1.932\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ultimate cap for our interval
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
sin(1.2) is not 1.932.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
That's impossible, since sin oscillates between 1 and 1.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
typo dude...chill out
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
and I didn't look at your wolfram ;P
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
remember the interval \[7 \le x \le 9\]
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
I do remember. Please realize that the maximum is not always going to be at one of the endpoints.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I understand that now
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\leftR_3(x)\right\le \frac{.932}{4!} {\left1.21\right}^4\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\leftR_3(x)\right\le 0.000054\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Thank ya! You're amazingly patient. Thank you soo much John. :)
 one year ago
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