Trouble with taylor's Inequality

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Homework Statement


Ok so I have been having trouble computing |R_{n}|
in problems.

Homework Equations


|R_{n}|\leq \frac{M}{(n+1)!}|x-a|^(n+1)
Edit:
M\geq f^(n+1) (h)

The Attempt at a Solution


Can some one walk me through calculatng some generic solution.
Like finding M and all that? I need more insight on what it actually does too.
Thanks
 
Last edited:
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if f is differentiable, you can use the methods of calculus to find the extremas of f and hence its absolute max. such a number is as M in your post.

in other cases, you can find such an M w/o calculus. If foor instance, f(x)=4sin(x). Well, |sin(x)|\leq1, so |f(x)|\leq4. So M=4 will do
 
If you use the nth Taylor polynomial for ex, about x= 0, the error is less than
\frac{M}{(n+1)!}|x|^{n+1}
where M is an upperbound on the n+1 derivative between 0 and x. Of course, every derivative of ex is ex itself so we just need an upperbound on that. For example if x= 1, We know the derivative is less than e so we can use that as an upper bound. The error is less than
\frac{e}{(n+1)!}[/itex]<br /> <br /> Calculating errors for the Taylor Polynomials for sine and cosine is even easier: all derivatives are either sine or cosine and 1 is always an upper bound.
 
Ok so suppose I have my function f(x)=\sqrt(x) and the interval [.5,1.5] using n=2 centered at 1. So:
|R_{2}|\leq \frac{M}{3!} |x-1|^3
f^3(x)= \frac{3}{8x^(5/2)}\leq M. My upperbound on f^3(x) is 1.5 so f^3(1.5)=\frac{\sqrt(3)}{72}
So: |R_{2}|\leq \frac{\sqrt(3)}{72*3!} |x-1|^3 X=1.5
So: So: |R_{2}|\leq \frac{\sqrt(3)}{72*3!} |.5|^3
How does that sound?
 
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Say f(x) = \sqrt{x+1} centered at 0, the reason why I am doing it this way is because you can get same result upon a basic substitution. On (-1,1).

Thus,
f(x) = (x+1)^{1/2} \implies f&#039;(0)=1
f&#039;(x) = \frac{1}{2}(x+1)^{-1/2} \implies f&#039;(0)=\frac{1}{2}
f&#039;&#039;(x) = - \frac{1}{4}(x+1)^{-3/2} \implies f&#039;&#039;(0) = -\frac{1}{4}

That means for some y between x\not =0 and 0 we have:
R_2(x) = f(x) - T_2(x) = \frac{f&#039;&#039;&#039;(y)}{3!}x^3

Now,
\left| \frac{f&#039;&#039;&#039;(y)}{3!}x^3 \right| = \left| \frac{3}{3!\cdot 8}\cdot \frac{(y+1)^{-5/2}}{1} \cdot x^3 \right| \leq \frac{3}{3!\cdot 8}
 
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