Second derivatives using delta notation

[bodensee9]

Homework Statement

Can some one explain why the second derivative of f(x) is (f(x+h)-2f(x)+f(x-h))/h^2? If you take the intervals to be x = 0 to x = 1 and you divide up the segment into little “hs” so that each x=h, 2h, 3h, nh, and so on?

I see that first derivatives can be approximated using (f(x+h)-f(x-h))/2h, but if you were to try to measure the change in (f(x+h)-f(x-h))/2h, wouldn't you get (f(x+2h)-f(x)-f(x)-f(x-2h))/2h^2?

Thanks.

 

[AlephZero]

I see that first derivatives can be approximated using (f(x+h)-f(x-h))/2h, but if you were to try to measure the change in (f(x+h)-f(x-h))/2h, wouldn't you get (f(x+2h)-f(x)-f(x)-f(x-2h))/2h^2?

No, you would get (f(x+2h)-f(x)-f(x)-f(x-2h))/4h^2

And if you substitute g = 2h you get the standard formula.

Another way to derive it is to use f''(x) is approximately (f'(x+h/2)-f'(x-h/2))/h and f'(x+h/2) is approximately (f(x+h)-f(x))/h

 

[bodensee9]

Oh I see it now. Thanks!