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Show that, for a twice differentiable function $f$, $$\lim_{h\to 0} \frac{f(a+2h)-2f(a+h)+f(a)}{h^2} = f''(a)$$

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Unfortunately your recent string of questions has the look of you posting a problem set in a course. Could you slow down a bit, concentrate on one question at a time, and tell us (i) in what context these questions are arising and (ii) what thoughts you have had / work you have done on them already? –  Pete L. Clark May 16 '12 at 22:04
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1 Answer 1

f '(x) = lim h->0 [f(x + h)) - f(x)]/h

Now following the same formula for f''(a) except now instead of lim h->0 [f(x + h) - f(x)]/h, you use:

lim h->0 [f '(a + h) - f '(a)]/h

f "(a) = lim h->0 [f(a + 2h) - f(a + h)]/h - [f(a + h) - f(a)]/h

= lim h->0 [f(a + 2h) - 2*f(a + h) + f(a)]/h

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Given Pete Clark's comment above, you should know that giving an answer to this ill-formulated question is not helping the OP. –  M Turgeon May 29 '12 at 17:33
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gurl, check out meta.math.stackexchange.com/questions/107/… and en.wikipedia.org/wiki/… to find out how to typeset formulas nicely here –  Yrogirg May 30 '12 at 12:05
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