Derivation of Taylor expansion with the $a$ term

Question

If we have

$$f(x) = \sum_{n=0}^\infty a_n x^n$$

The $k$th derivative is

$$f^{(k)}(x) = \sum_{n=0}^{\infty} a_{n+k} \frac{(n+k)!}{n!} x^n$$

Which also means that

$$f^{(k)}(0) = k! a_k$$

Implying

$$a_k = \frac{f^{(k)}(0)}{k!}$$

So we can substitute this back into the original definition for $f(x)$:

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$

And this shows the Taylor series expansion.

However I've also noticed this other definition:

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$$

But I can't seem to use the previous logic to get the math to work out when I use $x=a$ instead of $x=0$, since:

$$f^{(k)}(a) = \sum_{n=0}^{\infty} a_{n+k} \frac{(n+k)!}{n!} a^n$$

But from here it doesn't exactly condense nicely and so I don't know how to substitute it to make it work out properly.

Answer 1

Same logic still applies

$f(x) = \sum_{n=0}^{\infty} a_{n} (x-a)^n\\ f^{(k)}(x) = \sum_{n=0}^{\infty} a_{n+k} \frac{(n+k)!}{n!} (x-a)^n\\ f^{(k)}(a) = a_{k} k!\\ a_k = \frac {f^{(k)}(a)}{k!}$

Answer 2

$f(x) = \sum_{k \geq 0} a_{k} (x-a)^{k}$

$f'(x) = \sum_{k \geq 1} k a_{k} (x-a)^{k-1}$, so $f'(a) = a_{1}$

$f''(x) = \sum_{k \geq 2} (k-1) k a_{k} (x-a)^{k-2}$, so $f''(a) = 2! \, a_{2}$

$f'''(x) = \sum_{k \geq 3} (k-2) (k-1) k a_{k} (x-a)^{k-2}$, so $f'''(a) = 3! \, a_{3}$.

So on.