Show that, for a twice differentiable function $f$, $$\lim_{h\to 0} \frac{f(a+2h)-2f(a+h)+f(a)}{h^2} = f''(a)$$
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
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 |
|||||||||
|
