Vector form for Taylor series

Question

What is the vector form of Taylor series for a vector valued function of a scalar variable $f:\mathbb R\to\mathbb R^n $? I presume it is exactly the same as the classical Taylor series but for confirmation I have been searching the internet to no avail. Can anyone also point out a reference for the proof? I think the proof should follow from applying the classical case to each coordinate.

Answer 1

If it's a function from $f:\mathbb{R} \rightarrow \mathbb{R}^n$, the Taylor expansion would be exactly similar expansion for $f:\mathbb{R} \rightarrow \mathbb{R}$.

Let the $f:\mathbb{R} \rightarrow \mathbb{R}^n$ be equal to $f\left(x\right)=(g_1\left(x\right),...,g_n\left(x\right))$

Consider the Taylor expansion for each of the functions $g_1\left(x\right),...,g_n\left(x\right)$ around point $a$

$g_1\left(x\right)={\displaystyle g_1(a)+{\frac {g_1'(a)}{1!}}(x-a)+{\frac {g_1''(a)}{2!}}(x-a)^{2}+{\frac {g_1'''(a)}{3!}}(x-a)^{3}+\cdots ,}$ and so on... $g_n\left(x\right)={\displaystyle g_n(a)+{\frac {g_n'(a)}{1!}}(x-a)+{\frac {g_n''(a)}{2!}}(x-a)^{2}+{\frac {g_n'''(a)}{3!}}(x-a)^{3}+\cdots ,}$

Now, $f\left(x\right)=(g_1\left(x\right),...,g_n\left(x\right))$ $=({\displaystyle g_1(a)+{\frac {g_1'(a)}{1!}}(x-a)+{\frac {g_1''(a)}{2!}}(x-a)^{2}+\cdots }$,...,${\displaystyle g_n(a)+{\frac {g_n'(a)}{1!}}(x-a)+{\frac {g_n''(a)}{2!}}(x-a)^{2}+\cdots })$ This is equal to: $=(g_1\left(a\right),...,g_n\left(a\right))+({\frac {g_1'(a)}{1!}},...,{\frac {g_n'(a)}{1!}})(x-a)+({\frac {g_1''(a)}{2!}},...,{\frac {g_n''(a)}{2!}})(x-a)^2$...

Clearly, this is: $f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^2...$, which is the Taylor expansion of $f(x)$ at the point $a$.