I was looking at proofs of the differentiation of exponentiation and I'm confused about the last step of this part of the proof:
$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$ = $\lim_{x \to x_0} \frac{x^n-x_0^n}{x-x_0}$ = $\lim_{x \to x_0}(x^{n-1}+x^{n-2} \cdot x_0+x^{n-3} \cdot x_0^2+...+x^2 \cdot x_0^{n-3}+x \cdot x_0^{n-2}+x_0^{n-1})$
The proof continues but I'm not sure what exactly the last step has been. If somebody could explain what has been done I would be really thankful.
Firstly we can write the expression with a sigma sign.
$$S_n=x^{n-1}+x^{n-2} \cdot x_0+x^{n-3} \cdot x_0^2+...+x^2 \cdot x_0^{n-3}+x \cdot x_0^{n-2}+x_0^{n-1})=\frac1{x_0}x^n\sum\limits_{i=1}^n \left(\frac{x_0}{x}\right)^i$$
We have $\sum\limits_{i=1}^n q^i$ is $q\cdot \frac{q^n-1}{q-1}$. This is the partial sum of a geometric series.
For $q=\frac{x_0}{x}$ we obtain
$$\sum\limits_{i=1}^n \left(\frac{x_0}{x}\right)^i=\frac{x_0}{x}\cdot \frac{\left(\frac{x_0}{x}\right)^n-1}{\frac{x_0}{x}-1}=x_0\cdot \frac{\left(\frac{x_0}{x}\right)^n-1}{x_0-x}$$
Expanding the fraction by $x^n$
$\large{x_0\cdot \frac{x_0^n-x^n}{x_0-x}\cdot \frac1{x^n}}$
Finally we have to regard the factor $\frac1{x_0}x^n$
$$S_n=x_0\cdot \frac{x_0^n-x^n}{x_0-x}\cdot \frac1{x^n}\cdot \frac1{x_0}x^n=\frac{x_0^n-x^n}{x_0-x}=\frac{x^n-x_0^n}{x-x_0}$$
If you want to do it the other way round just make a polynomial division. I started by calculating the first two summands.
However, I am still interested how the proof does it the other way?
It starts off with $\frac{x^n-x_0^n}{x-x_0}$ My problem seems to be how to calculate the division.
– NotSureHowToDoStuff Aug 02 '19 at 18:28
I guess you could say I had trouble calculating it more than understanding what was done, I was confusing in my initial question.
– NotSureHowToDoStuff Aug 02 '19 at 17:41