Find the Taylor expansion of $f(x-yh)$.
The exercise don't say if $h$ and $y$ are variables too. I tried to do a Taylor series at $yh$
$$f(x)=f(x-y)+\frac{f'(x-yh)}{1!}(x-yh)+\frac{f''(x-yh)}{2!}(x-yh)^2...$$
But the answer says that Taylor expansion of $f(x-y)$ is
$$f(x-yh)=f(x)+hyf'(x)+\frac{1}{2}h^2y^2f''(x)+o(h^2)$$
Anyone can help me understand it?
From the answer, it seems obvious that they expand $f(x-\epsilon)$ around $\epsilon=0$. So, $$f(x-\epsilon)=f(x)-\epsilon f'(x)+\frac{1}{2} \epsilon ^2 f''(x)+O\left(\epsilon ^3\right)$$ and replaced later $\epsilon$ by $hy$.
Not only the $x^n$ are missing (just as Simple Art commented) by it seems that there are also sign errors.
Also, what they give cannot be $O\left(h^2\right)$ but $O\left(h^3\right)$ instead.
We can bet that, once more, the textbook contains typo's.