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Let $y(x)=\dfrac1x$.

$\quad$(a) Calculate the second finite difference $\Delta^2y$ using the $x$ values $x_0,x_0+\Delta x$ and $x_0+2\Delta x$.

$\quad$(b) Verify that in this case $$\frac{\Delta^2y}{\Delta x^2}\to y''(x_0)$$

$\qquad\quad$as $\Delta x\to 0$.

I tried to represent $Y_0$ ($Y$ initial) as $(1/X_0)$, but I have tried different computations, and all I am getting are huge variables that don’t match.

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2 Answers

up vote 2 down vote accepted

A useful tip: it’s often easier to do as much as you can before you substitute some possibly messy specific function into a general formula.

Your first differences are $y(x_0+\Delta x)-y(x_0)$ and $y(x_0+2\Delta x)-y(x_0+\Delta x)$, so your second difference is

$$\Big(y(x_0+2\Delta x)-y(x_0+\Delta x)\Big)-\Big(y(x_0+\Delta x)-y(x_0)\Big)\;.$$

This simplifies to

$$y(x_0+2\Delta x)-2y(x_0+\Delta x)+y(x_0)\;,$$

which becomes $$\frac1{x_0+2\Delta x}-\frac2{x_0+\Delta x}+\frac1{x_0}$$ when we plug in the actual function. Combining this over the least common denominator, we get $$\frac{x_0(x_0+\Delta x)-2x_0(x_0+2\Delta x)+(x_0+\Delta x)(x_0+2\Delta x)}{x_0(x_0+\Delta x)(x_0+2\Delta x)}\;.$$ The numerator expands to

$$x_0^2+x_0\Delta x-2x_0^2-4x_0\Delta x+x_0^2+3x_0\Delta x+2\Delta x^2\;,$$

and everything cancels out except the $2\Delta x^2$, so your second difference is just $$\frac{2\Delta x^2}{x_0(x_0+\Delta x)(x_0+2\Delta x)} ;,$$ so $$\frac{\Delta^2y}{\Delta x^2}=\frac2{x_0(x_0+\Delta x)(x_0+2\Delta x)}\;,$$ and from here you should be home free.

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thank you so much, I had uptil where you said "which becomes" at that line I thought I had something super whacky so I stopped trying afterwards! didnt know this question had so many steps. Thanks a lot! –  Raynos Jan 20 '12 at 15:37
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@Raynos: A tip: If you find an answer helpful, please upvote it. This not only motivates answerers to answer more of your questions, but also helps the future visitors tell apart helpful answers from the others. Regards, –  Srivatsan Jan 20 '12 at 20:29
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The answer below differs only notationally from the one posted by Brian M. Scott. We are interested in the behaviour of our function at three points in arithmetic progression. These points can be called $x_0$, $x_0+\Delta x$, and $x_0+2\Delta x$. That leads to expressions that may get in the way of seeing what's happening. Let's call the points $x$, $x+h$, and $x+2h$.

The function values are then $$\frac{1}{x},\quad \frac{1}{x+h},\quad\text{and}\quad \frac{1}{x+2h}.$$

The two first differences are $$\frac{1}{x+h}-\dfrac{1}{x}\quad\text{and}\quad \frac{1}{x+2h}-\dfrac{1}{x+h},$$ which simplify to $$\frac{-h}{x(x+h)}\quad\text{and}\quad \frac{-h}{(x+h)(x+2h)}.$$

The second difference is then $$\frac{h}{x+h}\left(\frac{1}{x}-\frac{1}{x+2h}\right),\qquad \text{that is,}\qquad \frac{2h^2}{x(x+h)(x+2h)}.$$

Divide by $h^2$, and let $h$ approach $0$.

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