0
$\begingroup$

This is a probably very basic question about the Taylor series

$$\begin{align} X^\alpha\,(x + \delta x) &= X^\alpha (x)+\delta x^b\;\partial_bX^\alpha+\cdots\\[2ex] &=X^\alpha(x) + \delta x^b\;\frac{\partial X^\alpha}{\partial x^b}+\cdots\\[2ex] &= X^\alpha(x) + \delta X^\alpha + \cdots \end{align}$$

corresponding to a diagram such as

enter image description here

and introduced in the exposition of the covariant derivative at this point of a lecture on tensors by XylyXylyX .

So the very basic question is why is the expression above the Taylor expansion. I don't see the value of $X^\alpha(x + \delta x)$ as the first component, for example. And the first derivative doesn't seem to include $X^\alpha(x).$

$\endgroup$
6
  • 1
    $\begingroup$ What's the exact question? $\endgroup$ Oct 6 '17 at 15:39
  • 1
    $\begingroup$ I edited the OP. As I said, it's probably too basic for the forum... $\endgroup$ Oct 6 '17 at 15:46
  • $\begingroup$ Suppose we have a function $f(x)$. If x changes by a small amount $\delta x$, then $f(x)$ also changes by a tiny amount, given by $\delta f=\frac{df}{dx}\delta x$. Here $X^\alpha$ is a function of all the components $x_a$, and this is a Taylor expansion in multidimentional calculus. $\endgroup$ Oct 6 '17 at 15:59
  • $\begingroup$ @SayanMandal Thanks. Comparing to $f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}+\cdots$, the first term in the Taylor series is the value of the function evaluated at $a$, i.e. $f(a)$ in regular calculus, and here it would be evaluated at $x,$ not $x+ \delta x,$ right... And the expression $\delta x^b\frac{\partial X^\alpha}{\partial x^b}$ is also the first derivative evaluated at $a$ with $x-a=\delta x^b,$ and $\frac{f'(a)}{1!}= \frac{\partial X^\alpha}{\partial x^b} ,$ correct? $\endgroup$ Oct 6 '17 at 16:08
  • $\begingroup$ Right. In this case $x$ taken on the role of $a$ in your more familiar examples of Taylor Series. Also, I am sure you have realized that in the second term there is a sum over the index $b$. $\endgroup$ Oct 6 '17 at 16:10

This site is temporarily in read only mode and not accepting new answers.

Browse other questions tagged .