I Taylor expansion of f(x+a)

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1. Nov 1, 2017

I recently found out the rule regarding the Taylor expansion of a translated function:
$f(x+h)=f(x)+f′(x)⋅h+\frac 1 2 h^ 2 \cdot f′′(x)+⋯+\frac 1 {n!}h^n \cdot f^n(x)+...$

But why exactly is this the case? The normal Taylor expansion tells us that
$f(x)=f(a)+f'(a)(x-a)+\frac 1 {2!}f''(a)(x-a)^2+...+\frac 1 {n!} f^n(x)(x-a)^n+...$

So how do you come from the original expansion to the second one? Simply substituting x with x+h doesn't do it, not to mention that the f's and the coefficients seem to have changed roles, with the f's now being functions instead of constants and the coefficients now being constants instead of functions.

An explanation would be appreciated.

2. Nov 1, 2017

hilbert2

You just have to write $f(x) = f(a + (x - a))$ and use the first formula with $a$ replacing x and $x-a$ replacing $h$ to get the second one.

3. Nov 1, 2017

PeroK

There's an important point here about what exactly is a "variable" and what is a "constant". You might ask (in the second equation) in what way is $a$ a constant and $x$ a variable? Both can be any real number. Likewise $h$ in the first expression. So, can't you interpret both equations as being an equation in two variables? $x$ and $h$ in the first equation and $x$ and $a$ in the second.

4. Nov 1, 2017

I think hilbert's practical explanation combined with perok's intuitive one answered my question.

I was under the impression that in both equation you fix $a$ and $h$ and leave x as a variable, but since the second equation applies for all real $a$ then it can also be treated as a variable. So if you take the original rule, replace $a$ with $x$ and $x$ with #x+h## you get the second formula.

Thanks for the help.