Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I express the Taylor development of $$f(t+\Delta t, x+h)$$ When one of the variables is fixed, I can find it but with both of them varying, I have no idea what to do.

$$f(t+\Delta t, x)=f(t,x) + \Delta t \cfrac{\partial f}{\partial t}(t,x)+(\Delta t)^2 \cfrac{\partial^2 f}{\partial t^2}(t,x) + O((\Delta t)^3)$$ and $$f(t, x+h)=f(t,x) + h \cfrac{\partial f}{\partial x}(t,x)+h^2 \cfrac{\partial^2 f}{\partial x^2}(t,x) + O(h^3)$$

share|cite|improve this question
Intuitively, you may first Taylor-expand $f(t+\Delta t,x+\Delta x)$ in $t$, and then Taylor-expand the result in $x$. – Eckhard Dec 22 '12 at 11:05
Or, you can taylor expand $f(t + s \Delta t, x + s \Delta x)$ in $s$ – Hurkyl Dec 22 '12 at 11:09
@Hurkyl how will that work out? – user31280 Dec 22 '12 at 11:10

2 Answers 2

up vote 1 down vote accepted

$$f(t+\Delta{t},\, x+h)=\sum\limits_{n=0}^{\infty}{{\frac{1}{n!}}\;{\sum\limits_{k=0}^{n}{\binom{n}{k}}\frac{\partial^n{f}}{\partial{x}^k {}\partial{y}^{n-k}}(\Delta{t})^k h^{n-k}}} \tag{*}$$ Denote $u=(t,\,x),\;\; \Delta{u}=(\Delta{t},\,h), \;F(\lambda)=f(u+\lambda\Delta{u}) \;\;(\lambda\in \mathbb{R}).$ Expansion $(*)$ can be obtained from Taylor expansion $$F(\lambda)=\sum\limits_{n=0}^{\infty}{\frac{1}{n!}}F^{(n)}(0)\,\lambda^n$$ by differentiating $F(\lambda)$ by $\lambda$ and then putting $\lambda=1.$

share|cite|improve this answer

One approach to work out the multi-variable Taylor expansion of $f(\vec{v})$ about a point $\vec{v}_0$ is to reduce it to the single variable case by writing the function

$$ g(s) = f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) $$


$$ g(s) = \sum_{n=0}^{+\infty} \frac{s^n}{n!} g^{(n)}(0) $$

The repeated derivatives of $g$ are repeated directional derivatives of $f$: (if this isn't clear, try working out $g'(s)$ yourself by whatever means you like)

$$ g^{(n)}(s) = (\nabla_{\vec{v} - \vec{v}_0})^n f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) $$

If you've only seen directional derivatives defined for unit vectors, the generalization to all vectors is the obvious one: $\nabla_{\vec{w}} h = \vec{w} \cdot \nabla h$

So the Taylor series is

$$ f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) = \sum_{n=0}^{+\infty} \frac{s^n}{n!} \left( \nabla_{\vec{v} - \vec{v}_0}\right)^n f (\vec{v}_0) $$

which simplifies to, if it converges,

$$ \begin{align} f(\vec{v}) &= \sum_{n=0}^{+\infty} \frac{1}{n!} \left( \nabla_{\vec{v} - \vec{v}_0}\right)^n f (\vec{v}_0) \\ &= \sum_{n=0}^{+\infty} \frac{1}{n!} \left( (\vec{v} - \vec{v}_0) \cdot \nabla \right)^n f (\vec{v}_0) \end{align} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.