Delta, derivative, differential etc.

[emma83]

I am still very confused about the differences between all the d's and delta's used to represent infinitesimal elements and/or derivatives and never know when and where to use what:
- du
- \partial u
- \delta u

For instance what can be simplified exactly in the chain rule and also what to use in an integral.

Thanks for your help!

 

[djeitnstine]

I think you are referring to \Delta x, dx and \partial x

\Delta x simply means some change in x

dx is an infinitesimal change in x

\partial x is also an infinitesimal change however most commonly used when referring to some other dependent variable.

for instance if y and t are both dependent variables in some function g(y,t). For example g(y,t) = t + yt^{2} then \frac{\partial g}{\partial t} = 1+ 2yt and \frac{\partial g}{\partial y} = t^{2}

Notice that one is held constant while the other is taken the derivative of

Also I am not sure wat you mean by "what can be simplified exactly in the chain rule" and as for the integral only dx and \partial x of course. It will do you good on finding out exactly what you are doing when you are integrating and deriving. Integrating is the summation of infinitesimally small sums for example.

 

[Fredrik]

If f:\mathbb R\rightarrow\mathbb R, then df:\mathbb R^2\rightarrow\mathbb R is defined by

df(x,h)=f'(x)h

Sometimes the h is written as "dx", and df(x,h) as "dy", so that the expression above becomes

dy=f'(x)dx

Note that h=dx doesn't have to be "small" for this to be well-defined. It does however have to be small for f'(x)dx to be a good approximation of f(x+dx)-f(x) since

f(x+h)=f(x)+hf'(x)+\frac 1 2 h^2f''(x)+\mathcal O(h^3)

when h\rightarrow 0.

\Delta x is the same as h=dx, i.e. it's just a number. \Delta y or \Delta f is the actual change of the value of the function f, i.e.

\Delta f=f(x+h)-f(x)

so \Delta x=dx, but \Delta y\approx dy with \Delta y-dy\rightarrow 0 when dx\rightarrow 0.

The lowercase delta is used in the calculus of variations. See e.g. the Wikipedia definition of a functional derivative. I'm not sure if it's used outside of that context.

I don't think I have ever seen the expression \partial x defined, or even used other than as a part of the expression for a partial derivative:

D_if(x)=f_{,i}(x)=\frac{\partial}{\partial x}f(x)=\frac{\partial f(x)}{\partial x}

The expression \partial_\mu is however used in differential geometry, as a short version of \partial/\partial x^\mu, a partial derivative operator constructed from some coordinate system x:U\rightarrow\mathbb R^n, where U is an open subset of the manifold.

\partial_\mu|_p f=\frac{\partial}{\partial x^\mu}\bigg|_p f=(f\circ x^{-1})_{,\mu}(x(p))

There are few things in the physics literature that I find as confusing as the use of the word "infinitesimal". It took me some time to realize that when the word "infinitesimal" is used (in a physics book), the relevant variables aren't infinitesimals at all, and do not even need to be "small" in any sense of the word. The only thing that the word "infinitesimal" represents in a physics book is that the equation that follows the word "infinitesimal" has been Taylor expanded around 0 and only the terms up to some finite order in the variables have been kept. Physicists apparently use the word "infinitesimal" only to avoid having to use the "big O" notation.

 

[Fredrik]

It should be clear from the definition of df above why the chain rule needs to be proved even though it looks so "obvious" when you use the notation

\frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}

to represent the more precise

(f\circ g)'(x)=f'(g(x))g'(x)

Let's try to express this using differentials (expressions like "df").

\frac{d(f\circ g)(x,h)}{h}=\frac{df(g(x),k)}{k}\frac{dg(x,l)}{l}

It looks a bit less obvious now, doesn't it? Now choose h=l=dx, and k=dg(x,dx) so that the denominator of the first factor on the right cancels the numerator of the second factor.

\frac{d(f\circ g)(x,dx)}{dx}=\frac{df(g(x),dg(x,dx))}{dg(x,dx)}\frac{dg(x,dx)}{dx}=\frac{df(g(x),dg(x,dx))}{dx}

d(f\circ g)(x,dx)=df(g(x),dg(x,dx))

We have now canceled the "dg", but what we ended up with isn't a trivial equality (which would have proved the chain rule). (The "df" on the left isn't the same thing as the "df" on the right). Instead we got an equality that isn't at all obvious. We can verify that it's equivalent to a statement of the chain rule by using the definition of the differentials again, but that just brings us back to the start.

d(f\circ g)(x,dx)=(f\circ g)'(x)dx

df(g(x),dg(x,dx))=f'(g(x))dg(x,dx)=f'(g(x))g'(x)dx