Why is $2C(x)−C(x+b)−C(x−b)≈−b^2\frac{∂^2C(x)}{∂x^2}?$ This is quoted from Feynman's lectures The Dependence of Amplitudes on Position
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In Chapter 13 we then proposed that the amplitudes $C(x_n)$ should vary with time in a way described by the Hamiltonian equation. In our new notation this equation is
  $$iℏ\frac{∂C(x_n)}{∂t}=E_0C(x_n)−AC(x_n+b)−AC(x_n−b).\tag{16.7}$$
  The last two terms on the right-hand side represent the process in which an electron at atom $(n+1)$ or at atom $(n−1)$ can feed into atom $n.$
  We found that Eq. $(16.7)$ has solutions corresponding to definite energy states, which we wrote as
  $$C(x_n)=e^{−iEt/ℏ}e^{ikx_n}.\tag{16.8}$$
  For the low-energy states the wavelengths are large ($k$ is small), and the energy is related to $k$ by
  $$E=(E_0−2A)+Ak^2b^2,\tag{16.9}$$
  or, choosing our zero of energy so that $(E_0−2A)=0$, the energy is given by Eq. $(16.1).$
  Let’s see what might happen if we were to let the lattice spacing $b$ go to zero, keeping the wave number $k$ fixed. If that is all that were to happen the last term in Eq. $(16.9)$ would just go to zero and there would be no physics. But suppose $A$ and $b$ are varied together so that as $b$ goes to zero the product $Ab^2$ is kept constant—using Eq. $(16.2)$ we will write $Ab^2$ as the constant $ℏ^2/2m_\text{eff}.$ Under these circumstances, Eq. $(16.9)$ would be unchanged, but what would happen to the differential equation $(16.7)$?
  First we will rewrite Eq. $(16.7)$ as
  $$iℏ\frac{∂C(x_n)}{∂t}=(E_0−2A)C(x_n)+A[2C(x_n)−C(x_n+b)−C(x_n−b)]. \tag{16.10}$$
  For our choice of $E_0,$ the first term drops out. Next, we can think of a continuous function $C(x)$ that goes smoothly through the proper values $C(x_n)$ at each $x_n.$ As the spacing $b$ goes to zero, the points $x_n$ get closer and closer together, and (if we keep the variation of $C(x)$ fairly smooth) the quantity in the brackets is just proportional to the second derivative of $C(x).$ We can write—as you can see by making a Taylor expansion of each term—the equality
  $$2C(x)−C(x+b)−C(x−b)≈−b^2\frac{∂^2C(x)}{∂x^2}.\tag{16.11}$$
  In the limit, then, as $b$ goes to zero, keeping $b^2A$ equal to $ℏ^2/2m_\text{eff},$ Eq. $(16.7)$ goes over into
  $$i\hbar\frac{∂C(x)}{∂t}=-\frac{\hbar^2}{2m_{\text{eff}}}\,
\frac{\partial^2C(x)}{\partial x^2}.$$

Can anyone tell me how Feynman equated the terms in the box $2C(x)−C(x+b)−C(x−b)$ with $\frac{\partial^2C(x)}{\partial x^2}?$ How could he tell '_the quantity in the brackets is just proportional to the second derivative of$C(x)$ _? How did he use the Taylor expansion here?
 A: HINT:
Taylor's Theorem with the Peano form of the remainder is
$$C(x\pm b)=C(x)\pm C'(x)b+\frac12 C''(x)b^2+h(x;b)b^2 \tag 1$$
where $\lim_{b\to 0}h(x;b)=0$.
Now, add the positve and negative terms in $(1)$ and see what happens.
A: Let's Taylor expand the $C$ terms to second order about $x$:
$$ C(x) = C(x) $$
So far so good...
$$ C(x+b) = C(x)+\frac{\partial C}{\partial x}(x) b + \frac{1}{2}\frac{\partial^2 C}{\partial x^2}(x)b^2 + o(b^2) $$
(I'm not a fan of this notation, but there's not really another way to make it consistent with Feynman's dodgy notation.) Doing the same thing for $C(x-b)$ gives you the same thing with $b \mapsto -b$:
$$ C(x-b) = C(x)+\frac{\partial C}{\partial x}(x) b - \frac{1}{2}\frac{\partial^2 C}{\partial x^2}(x)b^2 + o(b^2) $$
Then the combination Feynman takes looks like
$$ 2C(x)-C(x+b)-C(x-b) \\
= (2C(x)-C(x)-C(x)) + b \left( -\frac{\partial C}{\partial x}(x)+\frac{\partial C}{\partial x}(x) \right) - b^2 \left( \frac{1}{2}\frac{\partial^2 C}{\partial x^2}(x) + \frac{1}{2}\frac{\partial^2 C}{\partial x^2}(x) \right) + o(b^2) \\
= -b^2\frac{1}{2}\frac{\partial^2 C}{\partial x^2}(x) + o(b^2) $$
$o(b^2)$ just means a term that goes to zero faster than $b^2$ as $b \to 0$.
A: It can be easily proved via L'Hospital's Rule that $$\lim_{b \to 0}\frac{C(a + b) + C(a - b) - 2C(a)}{b^{2}} = C''(a)\tag{1}$$ provided that the second derivative $C''(a)$ exists. Hence we can write $$\frac{C(a + b) + C(a - b) - 2C(a)}{b^{2}} \approx C''(a)\tag{2}$$ when $b$ is small. This is same as writing $$2C(x) - C(x + b) - C(x - b) \approx -b^{2}C''(x)\tag{3}$$ where the variable $x$ has been used in place of $a$. To prove $(1)$ let's apply L'Hospital's Rule once (and only once) to get
\begin{align}
L &= \lim_{b \to 0}\frac{C(a + b) + C(a - b) - 2C(a)}{b^{2}}\notag\\
&= \lim_{b \to 0}\frac{C'(a + b) - C'(a - b)}{2b}\notag\\
&= \frac{1}{2}\lim_{b \to 0}\frac{C'(a + b) - C'(a)}{b} + \frac{C'(a - b) - C'(a)}{-b}\notag\\
&= \frac{1}{2}\{C''(a) + C''(a)\}\notag\\
&= C''(a)\notag
\end{align}
