Proof of Taylor expansion

Taylor's Theorem:

• The function $f(x)$ is infinitely differentiable
• The function $f(x)$ and all its derivatives exist at $x=\xi$
• The function $f(x)$ can be expressed as a series of the form: \[ f(x) = \sum_{k=0}^{\infty} a_k (x-\xi)^k \]

Differentiating term by term: \[\begin{align*} f(x) &= a_0 + a_1(x-\xi) + a_2(x-\xi)^2 + \cdots \\ \\ f'(x) &= a_1 + 2a_2(x-\xi) + 3a_3(x-\xi)^2 + \cdots \\ \\ f''(x) &= 2a_2 + 2\times 3 a_3(x-\xi) + 3 \times 4 a_4(x-\xi)^2 +\cdots \\ \\ f'''(x) &= 2 \times 3 a_3 + 2 \times 3 \times 4 a_4 (x-\xi) + 3 \times 4 \times 5 a_5(x-\xi)^2 + \cdots \end{align*} \]

Solve for $f(x)$ around the point $x=\xi$: \[\begin{align*} f(\xi) &= a_0 \\ f'(\xi) &= a_1 \\ f''(\xi) &= 2a_2 \\ f'''(\xi) &= 2 \times 3a_3 \end{align*}\] this can be expressed more generally as: \[ f^{(k)}(\xi) = a_k k! \] so: \[ a_k = \frac{f^{(k)}(\xi)}{k!} \] Substituting this into the original series: \[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(\xi)}{k!} (x-\xi)^k \] This is the Taylor expansion for a single variable.

A little more on the Taylor series:

• In general the Taylor series is only valid in a small region of x values. Often the series will diverge if $|x-\xi| > 1 $, so in general the series is only valid for $ \xi - 1 \le x \le \xi + 1$. However this is not the case for all functions.
• When the function $f(x)$ is not expanded completely there is some uncertainty in the expansion, this is denoted as: \[f(x) = S_n(x) + R_n (x)\] Where $S_n(x)$ is the sum of the first n terms and $R_n (x)$ is the sum of the remaining terms. As $n \to \infty $ : $S_n (x) \to f(x)$ and $R_n(x) \to 0$