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At the end he concludes $V(x) = V''(x_0)(x-x_0)^2$. How does he get to know that the rest are $0$? How does he conclude $V''(x_0) = k$. Please try to explain in easy ways and tough vocabulary. I don't even known what a Hilbert space is.

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He chooses $V(x_0)=0$, so $x_0$ as our reference point. And note that $V^\prime (x_0)=0$ since the potential is at a minimum at $x_0$. This gives you to second order (ignoring higher order terms):

$V(x) \approx \frac{1}{2} k (x-x_0)^2$.

You can just call $k=V''(x_0)$ since it is a constant. After using the differential equation $F=-kx=m\ddot{x}$ we see that $\omega = \sqrt{\frac{k}{m}}$.

And this is not related to a Hilbert space directly? A Hilbert space is basically a vector space that contains of elements and their elements have finite norm. Mathematically it can be described more rigorously of course but you should look this kind of things up yourself.

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