Taylor Expannsion

Introdction

日本語 ver

Today, I introduce Taylor Expansion.
I write not only One dimensional Taylor Expansion but also Multi dimensional Taylor Expansion.

One dimensional Taylor Expansion

f(X) is continuously differentiable for n-times on (a,b)
f(x) is expressed following.
\[\exists c ~~s.t~~ f(b) = \sum_{k=0}^{n-1} f^{(k)}(a)\frac{(b-a)^k}{k!} + f^{(n)}(c) \frac{(b-a)^n}{n!}, c \in (a,b)\]
This is called Maclaurin Expansion.
The last item is called Remainder term.

Multi dimensional Taylor Expansion

Multi dimensional Taylor Expansion is complex.
f is n-variable function.
f is continuously differentiable for m-times.
\(f(x_1+h_1,x_2+h_2,.....,x_n+h_n)\) is expressed following.
\[\exists \theta ~~s.t~~\]
\[f(x_1+h_1,x_2+h_2,...,x_n+h_n)=f(x_1,x_2,...,x_n) + \]
\[\sum_{m=0}^{n-1} \frac{1}{m-1} \sum_{k_1=1}^{n} \sum{k_2=1}^{n} ... \sum{k_{m-1}=1}^{n} \frac{\partial^{m-1} f}{\partial x_{k_1} \partial x_{k_2} .... \partial x_{k_{m-1}} }(x_1,x_2,..,x_n)h_{k_1}h_{k_2} ..... h_{k_m-1} \]
\[+ \frac{1}{m} \sum_{k_1=1}^{n} \sum_{k_2=1}^{n} ... \sum_{k_m=1}^{n} \frac{\partial^{m} f}{\partial x_{k_1} \partial x_{k_2} ... \partial x_{k_m} }(x_1 + \theta h_1, x_2 + \theta h_2,...., x_n + \theta h_n) h_k{k_1}h_{k_2}....h_{k_n}\]
Last item is Remainder term in Multi Taylor Expansion.

Proof

I prove only one dimensional Taylor Expansion.
This proof is used by Rolle’s theorem.
Rolle's Theorem is this page
To assume f(x) is continuously differentiable for n-times on (a,b).
This thorem is proved by founding A such that
\[f(b) = \sum_{k=}^{n-1} f^{(k)} (a) \frac{(b-a)^k}{k!} + A \frac{(b-a)^n}{n!}\]
Now, I define following function such that
\[g(x) = f(b) - \sum_{k=0}^{n-1} f^{(k)}(x) \frac{(b-a)^k}{k!} - A \frac{(b-x)^n}{n!}\]
This g(x) fulfill following.

• g(a) = 0
• g(b) = 0

Thus, by Rolle’s theorem
\[\exists c \in (a,b) ~~s.t~~ g'(c) = 0\]
\[\begin{eqnarray*} g'(x) &=& - \sum_{k=0} ^{n-1} f^{(k+1)} (x) \frac{(b-x)^k}{k!} + \sum_{k=1}^{n-1} f^{(k)} (x) \frac{(b-x)^{k-1}}{(k-1)!} + A \frac{(b-x)^{n-1}}{(n-1)!} \\ &=& -\sum_{k=1}^{n} f^{(k)} (x) \frac{(b-x)^{n-1}}{(k-1)!} + \sum_{k=1}^{n-1} f^{(k)} (x) \frac{(b-x)^{k-1}}{(k-1)!} + A \frac{(b-x)^{n-1}}{(n-1)!}\\ &=& -f^n (x) \frac{(b-x)^{n-1}}{(n-1)!} + A \frac{(b-x)^{n-1}}{(n-1)!} \end{eqnarray*}\]
I substitiute c for x on this form.
\[g'(c) = \frac{(b-x)^{n-1}}{(n-1)!} (A - f^{(n)}(x))\]
\[A = f^{(n)}(x)\]
Q.E.D

Reference
https://mathtrain.jp/taylortheorem
http://www.ne.jp/asahi/search-center/internationalrelation/mathWeb/Differentiation/TheoremsDffrntlNvarFnctn/TaylorTheorem.htm