I Recovering some math notions: Variations

egozenovius
Messages
2
Reaction score
0
I have a paper and on that paper I only can read:

Let $$f:\mathbb{S^{1}} \to \mathbb{R^2}$$ be a function and $$f_{\epsilon}=f+\epsilon hn$$ and $$\mathbb{S^1}$$ is the unit circle.

$$\dot{f_\epsilon}=\dot{f}+\epsilon\dot{h}n+\epsilon h\dot{n}$$
$$\delta\dot{f}=\dot{h}n+h\dot{n}$$

can you tell me please, what does it mean that $$\delta$$, from where is it? Why $$\dot{f}$$ is not anymore in last equation?

also, I managed to recover the following formula:
$$r_{\epsilon}^{2}=\left(\dot{f}+\epsilon\dot{h}n+\epsilon h\dot{n}\right)^2$$
Can you help me to recover the $$\delta{r}$$?

What I tried:
$$r_{\epsilon}^2=(\dot{f})^2+(\epsilon \dot{h}n)^2+(\epsilon h\dot{n})^2+2(\dot{f}\epsilon\dot{h}n+\dot{f}\epsilon h\dot{n}+\epsilon^2\dot{h}nh\dot{n})$$
But now, I do not know how to compute $$\delta r$$.
 
Physics news on Phys.org
It looks like the ##\delta## is the same as ##\Delta## and so ##\delta f = f_{\epsilon} - f = \epsilon h n ##

The dots over ##\dot{f}## mean a time derivative I think so that's why there's an ##\dot{h}## and ##\dot{n}##
 
jedishrfu said:
It looks like the ##\delta## is the same as ##\Delta## and so ##\delta f = f_{\epsilon} - f = \epsilon h n ##

The dots over ##\dot{f}## mean a time derivative I think so that's why there's an ##\dot{h}## and ##\dot{n}##
Not quite. It is ##\delta f=\dfrac{f_\varepsilon - f}{\varepsilon}##, the differential operator. ##h,n## are functions, the rest is the Leibniz rule. To obtain ##\dot{r}## differentiate ##\delta \dot{r}^2=2 \cdot r \cdot \dot{r}##.
 
fresh_42 said:
Not quite. It is ##\delta f=\dfrac{f_\varepsilon - f}{\varepsilon}##, the differential operator. ##h,n## are functions, the rest is the Leibniz rule. To obtain ##\dot{r}## differentiate ##\delta \dot{r}^2=2 \cdot r \cdot \dot{r}##.

Thank for your answer. Please, if it is possible, can you recommend me some books/papers? OK, it seems to be Leibniz rule, but why it was used ##\delta r## instead of ##\dot{r}##... I have the feeling that something is missing me.
 
egozenovius said:
Thank for your answer. Please, if it is possible, can you recommend me some books/papers? OK, it seems to be Leibniz rule, but why it was used ##\delta r## instead of ##\dot{r}##... I have the feeling that something is missing me.
I don't think there is a difference, will say I think ##\delta r =\dot{r}##. But if we take what you wrote word by word, then ##\delta r = \delta_\varepsilon r= \dfrac{r_\varepsilon - r}{\varepsilon}## and ##\dot{r}=\lim_{\varepsilon \to 0}\delta_\varepsilon r##. However, chances are that it is meant to be the same and it's only a bit sloppy noted, i.e. the limit is skipped and replaced by ##\varepsilon## as something going to zero anyway.

I'm not sure what you mean by papers or books. The notation in the paper which you quoted is nowhere else defined. There is no universal truth how to write derivatives, so it's up to the author how they do it. Here is an article I wrote about derivatives in general, but I cannot promise that I have listed all possible notations.
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
 
Back
Top