The Taylor series of a function $f(x)$, (assuming differentiability) is:
$$\tag{1}f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+...$$
If I take $x=x+h$ I get:
$$\tag{2}f(x+h)=f(a)+f'(a)(x+h-a)+\frac{1}{2}f''(a)(x+h-a)^2+...$$
- From what I understand, $a$ in above statement can be equal to any value, whichever makes computations easy. But what If I take $a=x$? From eq.(1) I get: $$f(x)=f(x)+f'(x)(x-x)+\frac{1}{2}f''(x)(x-x)^2+....$$ $$\Rightarrow f(x)=f(x)$$ Seems valid...?
Now from eq.(1) and eq.(2) I can get: $$f(x+h)-f(x)=f(a)-f(a)+f'(a)(x+h-a-(x-a))+\frac{1}{2}f''(a)((x=h-a)^2-(x-a)^2)+...$$ $$\tag{3}\Rightarrow f(x+h)-f(x)=f'(a)h+\frac{1}{2}f''(a)(h^2+2h(x-a))+...$$
- Now in eq.(3) can I replace $a$ with $x$ like this: $$f(x+h)-f(x)=f'(x)h+\frac{1}{2}f''(x)(h^2+2h(x-x))+...$$ to get $$f(x+h)-f(x)=f'(x)h+\frac{1}{2}f''(x)(h^2)+...$$
Are the manipulations in 1. and 2. valid? What would be the underlying conditions be for them to be valid?
For the above function $f(x), f:R\to R$. I'm assuming $f(x)$ is differentiable $\forall x$, and $h$ is just a constant.