Finite Difference Method Stability with diffusion equation The diffusion equation is:
$ \frac{\partial T}{\partial t}
   = \alpha \left( \frac{\partial^2 T}{\partial x^2} \right) $
An explicit finite difference approach can be used to solve this, forward in time and central
differences in space. Approximating the diffusion equation at a node i, yields,
$\frac{T_i^{n+1}-T_i^n}{\Delta t} = \alpha\frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{\Delta x^2}$
which gives 
$T_i^{n+1} = T_i^n(1-\omega)+\omega(0.5T_{i+1}^n+0.5T_{i-1}^n)$
where $\omega = \frac{2\alpha \Delta t}{\Delta x^2}$ 
The book states that for stability condition, the coefficients of the right-hand side terms must be positive which implies that
$\Delta t \lt \frac{\Delta x^2}{2\alpha}$
My question is why should the coefficients of the right-hand side terms be positive for stability?
 A: $$\frac{T_i^{n+1}-T_i^n}{\Delta t} = \alpha\frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{\Delta x^2}$$
Starting with the above and collect terms as you did
$$
T_i^{n+1} = T_i^n(1-\omega) + \omega(0.5T_{i+1}^n + 0.5T_{i-1}^n)
$$
we use a plane wave solution for the stability of $T_i^n = T_0\mathrm{e}^{at+ikx}$ leads to
$$
T_0\mathrm{e}^{a(t + \Delta t)+ikx} = T_0\mathrm{e}^{at+ikx} (1-\omega) +\omega(0.5T_0\mathrm{e}^{at+ik(x+\Delta x)} + 0.5T_0\mathrm{e}^{at+ik(x-\Delta x)})
$$
Dividing through by $T_i^n$ leads to
$$
$$
\mathrm{e}^{a\Delta t} = (1-\omega) + \omega(0.5\mathrm{e}^{ik\Delta x}+0.5\mathrm{e}^{-ik\Delta x}) = (1-\omega) +\omega\cos(k\Delta x) = 1 - \omega\left[2\cos^2\left(\frac{k\Delta x}{2}\right)\right]
$$
since we require the growth to be bounded i.e. $\vert \mathrm{e}^{a\Delta t} \vert < 1$
therefore 
$$
\vert 1 - \omega\left[2\cos^2\left(\frac{k\Delta x}{2}\right)\right]\vert < 1\implies 0< 2\omega < 2
$$
finally reaching 
$$
0 <  \frac{2\alpha \Delta t}{(\Delta x)^2} < 1
$$
and you finally reach your condition
$$
0 < \Delta t < \frac{(\Delta x)^2}{2\alpha}
$$
Hopefully that this derivation which you may already of done is not being overkill, but it highlights why the result is so.
