Numerical Analysis: Approximations — Discrete Average Value Theorem

I am asked to compute approximations to $f'(1)$ using $h=\frac{1}{16}$ for $f(x)=\sqrt{x+1}$ with the following formulas

$$f'(x)-\frac{f(x+h)-f(x)}{h}=-\frac{1}{2}hf''(\xi_{x,h})=\mathcal{O}(h)....(1)$$ and $$f'(x)=\frac{f(x+h)-f(x-h)}{2h}=-\frac{1}{6}h^2f'''(\xi_{x,h})......(2)$$

Using (1) we have

$$\frac{1}{2}(x+1)^{-\frac{1}{2}}-\frac{(x+h+1)^\frac{1}{2}-(x+1)^{\frac{1}{2}}}{h} = -(\frac{1}{6})(\frac{1}{16})^2(-\frac{1}{2})(\frac{1}{2})(\xi_{x,h}+1)^{-\frac{3}{2}}$$

We know that $f'(1)=\frac{1}{2}(1+1)^{-\frac{1}{2}}=\frac{1}{2\sqrt{2}}$

So using $h=\frac{1}{16}$ then

$$\frac{1}{2\sqrt{2}}-16(x+\frac{17}{16})^{\frac{1}{2}}-(x+1)^{\frac{1}{2}}$$

I suppose I could take $x=0$ and therefore solve for $\mathcal{O}(h)$, but what is $\xi_{x,h}$ and is that what I am solving for?

• You are missing $f'(x)$ to the right of the equals signs in both (1) and (2). – Conrad Turner Apr 22 '16 at 4:55
• Much obliged. Allow me some time for more calculations. – cryptomath Apr 22 '16 at 4:58

You are asked to compute the two appoximations at $x=1$ using $h=1/16$, which you can then compare with the exact value.
$$f'(1)\approx \frac{f(17/16)-f(1)}{1/16}=\frac{\sqrt{17/16+1}-sqrt(2)}{1/16}\approx 0.35083$$
which you can compare with the value of $1/(2\sqrt{2})\approx 0.35355$.