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Error of a Taylor polynomial 
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#1
Feb513, 08:42 PM

P: 142

the error of a taylor series of order(I think that's the right word) n is given by
[tex] \frac{f^{n+1} (s)}{n!} (xa)^n [/tex] I think this is right. The error in a linear approximation would simply be [tex] \frac{f''(s)}{2} (xa)^2 [/tex] My question is what is s and how do I find it. Use linear approximation of √(47) if you need to use an example because I just did that question so it might be easier to explain. 


#2
Feb613, 12:51 AM

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$$\frac{f^{n+1} (s)}{(n+1)!} (xa)^{n+1}$$ I think this is right. The error in a linear approximation would simply be [tex] \frac{f''(s)}{2} (xa)^2 [/tex] $$f(x) = x^{1/2}$$ $$f'(x) = \frac{1}{2}x^{1/2}$$ $$f''(x) = \frac{1}{4}x^{3/2}$$ so $$\begin{align}f(a) + f'(a)(xa) + \frac{1}{2}f''(s)(xa)^2 &= (49)^{1/2} + \frac{1}{2}(49)^{1/2}(x49) + \frac{1}{8}s^{3/2}(x49)^2\\ &= 7 + \frac{1}{14}(x49)  \frac{1}{8}s^{3/2}(x49)^2\\ \end{align}$$ where ##s## is some number between ##a## and ##x##. Setting ##x=47##, we get $$\sqrt{47} = 7  \frac{1}{7}  \frac{1}{2}s^{3/2}$$ where ##s## is some number satisfying ##47 \leq s \leq 49##. Putting it another way, if we use only the linear approximation ##7  1/7##, then the approximation error is ##\frac{1}{2}s^{3/2}## for some ##47 \leq s \leq 49##. We can bound this error by taking the worst case ##s## (the value that maximizes the absolute approximation error), which in this case is obtained by choosing ##s=47##. Then the approximation error is guaranteed to satisfy $$\textrm{approximation error} \leq \frac{1}{2}(47)^{3/2} = 0.00155176...$$ Actually, we can also obtain a lower bound for the approximation error by choosing the bestcase value of ##s##, namely ##s = 49##. The error will be at least ##\frac{1}{2}(49)^{3/2} = 0.00145772...##. Thus we in fact have $$0.00145772... \leq \textrm{approximation error} \leq 0.00155176...$$ We can check this: the actual value of ##\sqrt{47}## is $$\sqrt{47} = 6.8556546...$$ and our approximation is $$7  1/7 = 6.85714285...$$ and the error is therefore ##6.85714285...  6.8556546... = 0.00148825...## which is indeed within the computed range. 


#3
Feb613, 07:44 PM

P: 142

Thanks that makes more sense than my book



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