Trouble with taylor's Inequality

1. Aug 8, 2007

Winzer

1. The problem statement, all variables and given/known data
Ok so I have been having trouble computing $$|R_{n}|$$
in problems.

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

3. 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: Aug 9, 2007
2. Aug 9, 2007

quasar987

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)|$\leq$1, so |f(x)|$\leq$4. So M=4 will do

3. Aug 9, 2007

HallsofIvy

Staff Emeritus
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
$$\frac{e}{(n+1)!}[/itex] Calculating errors for the Taylor Polynomials for sine and cosine is even easier: all derivatives are either sine or cosine and 1 is always an upper bound. 4. Aug 9, 2007 Winzer Ok so suppose I have my function [tex]f(x)=\sqrt(x)$$ 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?

Last edited: Aug 9, 2007
5. Aug 9, 2007

Kummer

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,
$$f(x) = (x+1)^{1/2} \implies f'(0)=1$$
$$f'(x) = \frac{1}{2}(x+1)^{-1/2} \implies f'(0)=\frac{1}{2}$$
$$f''(x) = - \frac{1}{4}(x+1)^{-3/2} \implies f''(0) = -\frac{1}{4}$$

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

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