Are $f(x)$ and $f(x+ \delta x)$ the same after Taylor series expansion? According to  15.2.1 from https://www.rsmas.miami.edu/users/miskandarani/Courses/MSC321/lectfiniteDifference.pdf, the Taylor series of u(x) can be written as

However, according to wikipedia, the Taylor series is 

The difference is in $\delta x$. My question is are $f(x_i)$ and $f(x_i+\delta_x)$ the same?
 A: Both of them are the same. The second  Taylor expansion that you have written is the Taylor expansion of $f(x)$ about the point $x=a$.
So in the second Taylor expansion, put $x-a=\Delta x$. See what happens.
Hope this helps you.
A: In the second formula:
$f(x)_{about\space x=a}= f(a) + \dfrac{f'(a)}{1!}(x-a) + \dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3$
Replace $x \rightarrow x+\Delta x$ & $a \rightarrow x$ to get:
$f(x+\Delta x)_{about\space x=x}= f(x) + \dfrac{f'(x)}{1!}(\Delta x) + \dfrac{f''(x)}{2!}(\Delta x)^2+\dfrac{f'''(x)}{3!}(\Delta x)^3$
which is the first formula itself.
A: In most places I looked for the Taylor expansion, I found the 2nd equation (e.g. Wikipedia, Brilliant, other math websites). The 1st equation was more difficult to find, except in all the proofs of this specific problem I was looking at (Harris Corner Detector).
To create the first equation from the second, just substitute $x = a+\delta x$ into the 2nd equation.
$$
f(x) = f(a+\delta x) = f(a) + f^{'}(a)(a + \delta x-a) + \frac{f^{''}(a)(a + \delta x-a)^2}{2!} + \text{...}
$$
$$
= f(a) +f^{'}(a)\delta x+ \frac{f^{''}(a)}{2!}(\delta x)^2 + \text{...}
$$
You'll see the $a$'s cancel, and the $\delta x$ being in the right place. I think the intuition for whats happening is:
Both your equations 1 and 2 are taylor series that approximates a function at a. We're asking the function to tell us what the value is almost exactly at a, but with a small change ($\delta x$).
