Second order derivation optimization Recently I am thinking about a problem that might be easy to answer but for me is a big challenge.
Assume you have a function $f(x)$ that is second order derivative. So I am looking for a way to minimize this function just based on second derivation (not the first derivation). Then, my question is, "is that possible to minimize a function directly just by using second order derivation?". If yes, can you please explain me the method (with an example please) or give me a reference, or keywords to search on google,
Thanks.
 A: The second derivative does not, by itself tell you anything about the minimum of the function, since for any function with a minimum, you can create another function with a different minimum but the same second derivative.
Proof
Let $f(x)$ be a twice differentiable surjective function defined from the interval $[a,b]$ where $a<b$ to the interval $[c,d]$ where $c\le d$ and let $m$ be a value such that $f(m)=c$
Choose an $\xi\in[a,b]$ such that $\xi\ne m$
Let $\delta=m-\xi$
Let $\Delta=f(\xi)-f(m)$
Let $\gamma\in\mathbb R$ be a value such that $\Delta<\delta\cdot\gamma$
Note that such a value always exists since $\delta\ne0$
Now let the surjective function $g(x)$ be defined from $[a,b]$ as follows:
$$g(x)=f(x)+x\cdot\gamma$$
For this function $g(x)$ it is true that $g(\xi)<g(m)$ as shown below
$$\Delta<\delta\cdot\gamma$$
$$f(\xi)-f(m)<\gamma(m-\xi)$$
$$f(\xi)-f(m)<\gamma\cdot m-\gamma\cdot\xi$$
$$f(\xi)+\gamma\cdot\xi<f(m)+\gamma\cdot m$$
$$g(\xi)<g(m)$$
So $m$ is not the minimum of the function $g(x)$ now notice that
$$f(x)=f(x)$$
$$f'(x)=\frac{d}{dx}\left[f(x)+\gamma\right]$$
$$f''(x)=\frac{d}{dx}\frac{d}{dx}\left[f(x)+x\cdot\gamma\right]$$
$$f''(x)=g''(x)$$
Which was what we wanted
