I was reading Networn's Method of Optimization in the book "An Introduction to optimization" by Chong Zak. The text shows the Taylor polynomial expansion of $f(x)$ on $x_{k}$. Then it claims $q^{\prime\prime}(x^{(k)}) = f^{\prime\prime}(x^{(k)})$
We assume now that at each measurement point $x^{(k)}$ we can calculate $f(x^{(k)}), f^{\prime}(x^{(k)}),f^{\prime\prime}(x^{(k)})$. We can fit a quadratic function through $x^{(k)}$ that matches its first and second derivative with that of function $f$. This quadratic has the form $$g(x) = f(x^{(k)})+f^{\prime}(x^{(k)})(x-x^{(k)})+\frac{1}{2}f^{\prime\prime}(x^{(k)})(x-x^{(k)})^2$$
Note that $q(x^{(k)})=f(x^{(k)})$, $q^{\prime}(x^{(k)})=f^{\prime}(x^{(k)})$, $q^{\prime\prime}(x^{(k)})=f^{\prime\prime}(x^{(k)})$
I can verify the first two. But for the last one I am getting $q^{\prime\prime}(x^{(k)})=2f^{\prime\prime}(x^{(k)})$. I think I am wrong somewhere.
$$ q^{\prime}(x) = f^{\prime}(x^{(k)})+f^{\prime\prime}(x^{(k)})(x-x^{(k)})+f^{\prime\prime}(x^{(k)})(x-x^{(k)})+\frac{1}{2}f^{\prime\prime\prime}(x^{(k)})(x-x^{(k)})^{2}\\ = f^{\prime}(x^{(k)})+2f^{\prime\prime}(x^{(k)})(x-x^{(k)})+\frac{1}{2}f^{\prime\prime\prime}(x^{(k)})(x-x^{(k)})^{2}\\ q^{\prime\prime}(x) = 2f^{\prime\prime}(x^{(k)})+2(x-x^{(k)})f^{\prime\prime\prime}(x^{(k)}) + \dots $$
So $q^{\prime\prime}(x^{(k)}) = 2f^{\prime\prime}(x^{(k)})$ . I am not getting where I am missing.
I think something is wrong, unless we drop the 3rd term from $g(x)$ we cannot have the 3rd claim.
Here the variable is $x$, but $x^{(k)}$ is constant
therefore $g'(x)=f'(x^{(k)})+f''(x^{(k)})(x-x^{(k)})$ and $g'(x^{(k)})=f'(x^{(k)})$
$g''(x)=f''(x^{(k)})$ is constant