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My physics professor made the following approximation: $$\frac{d^2y}{dx^2}\approx\frac{\Delta y}{\Delta x^2}$$

How can you fundament this? I get how you can do $\displaystyle\frac{dy}{dx}\approx\frac{\Delta y}{\Delta x}$, but it doesn't really click with the second order derivative.

Shiv Tavker
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Belen
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    What is $\Delta y$ here? If its $y(x+\Delta x) - y(x)$ then its very wrong. If its $\Delta y = y(x+\Delta x) + y(x-\Delta x) - 2y(x)$ then its correct. Some more details and context needed here. – Winther May 10 '20 at 20:47
  • He meant $\Delta y = y(x+\Delta x)-y(x)$. I thought so, I have no idea where he got that from. – Belen May 10 '20 at 21:14
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    Ok. I would doubt it was really intended that way (but all people make mistakes so its not impossible). In any case its best to ask this question to the person who actually said this and let him clear up what he really meant. Without any context on how it was used (which could tell what was really intended) the answer here is just "no its not correct". – Winther May 10 '20 at 22:27

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