# Translation operator expansion

1. Oct 5, 2009

### valjok

Excuse me my lack of expertise, but it is very curious. Recently, I have https://www.physicsforums.com/showthread.php?t=54055", which I qualify 'absolutely amaizing' when I read it. It almost satisfies my curiosity on Laplace because I can almost understand it, except the shift operator. The author brings it from quantum physics remarking that its justification is purely mathematical (so, asking here, in 'mathematics of change and motion', I must be appropriate) and can be understood from the "series expansion of the function f(x+a) around x":
$$f(x + a) = \sum{ a^n f^{(n)}_x \over n!}$$

What kind of expansion is it? It lacks the member xn to complement d/dx for the Taylor expansion.

Last edited by a moderator: Apr 24, 2017
2. Oct 5, 2009

### mathman

You have a Taylor expansion where x is fixed and a is variable, so you get terms in an not xn.

Last edited by a moderator: Apr 24, 2017
3. Oct 6, 2009

### valjok

d(x+a)/dx = 1 = d(x+a)/da and, therefore, df/dx = df/da = df/d(x+a)?

4. Oct 6, 2009

### g_edgar

That seems nonsense to me.

Write it as
$$f(x + a) = \sum_{n=0}^\infty {f^{(n)}(x) \over n!}a^n$$

5. Oct 6, 2009

### valjok

Thanks. It became trivial since I have realized that df/dx = df/d(a+x) * d(a+x)/dx = df/d(a+x) * 1 = df/da. If this is what they do, the topic does not make any interest. Can we remove it?

6. Oct 6, 2009

### valjok

This is a Chain rule for a composite function. I need to prove that df/dx = df/da. Otherwise, I cannot write df/dx * a^n in Taylor's numerator.

This makes nonsense to me. That is, it is an ambigous sentence because (x) has two meanings. What you do is hiding the diverative variable. It might advantageous when we agree on the variable and omit to save the writing. But in this case, the derivative variable in the focus of question, if it is df/dx or df/da. Hiding does not solve a problem.

Last edited: Oct 6, 2009
7. Oct 6, 2009

### lurflurf

Mathematics does seem like nonsense at times.
There is no one way to write a taylor expansion.

Here is a nice one.
$$f(x+a)=exp\left (a \frac{d}{dx}\right) f(x)$$

8. Oct 6, 2009

### lurflurf

$$\frac{d}{dx}f(x+a)=\frac{d}{da}f(x+a)$$
If that is what you mean.
The expression df/dx = df/da was ambiguous.
The expansion was not ambiguous as differentiation was clearly with respect to x.
What two meanings do you think x has?

Last edited: Oct 6, 2009
9. Oct 7, 2009

### valjok

Why not with respect to a? Does f'(a) imply the differentiation over a? How do I know if f'(x) is fa'(x) or fx'(x) or fx'(a)? The (x) meanings are: 1) the point of taking value of a function and 2) specification of variable in partial derivative.

There is only one function, which has arguments x and a. A variable must be specified when taking a partial derivative. Especially, when somebody creates a topic to resolve the controversy with the differentiation variable. Now, how variable hiding resolves the ambiguity and what are other meanings of df/da and df/dx? Thanks.

10. Oct 7, 2009

### lurflurf

Okay.
So f here only has one variable as it is written f(x).
The function g(x,a)=f(x+a) has two variables, but partials with respect to each variable are equal.
You are correct that all of these notations have a slight amount of ambiguity, but they are not being used in an ambiguous manner. The letter used in the argument of a function is the value of the argument not the variable of integration.
f'(x) would be read the first derivative of f evaluated at x
f''(a) would be read the first derivative of f evaluated at a
The ambiguous thing about that is not what derivative is being taken as there is only one, but if x or a are known, unknown, or variable.
fa'(x) or fx'(x) or fx'(a)
These are all confusing.

Now let us consider a function of several variables.
h(x,y,z)
f(1,0,0)(x,y,z)
f(0,1,0)(x,y,z)
f(0,0,1)(x,y,z)
would be nice ways to denote differentiation with respect to each variable.
fx(x,y,z)
fy(x,y,z)
fz(x,y,z)
are also used especially when the variable have meaning
confusion can result though
f(0,1,0)(z,x,x)
is more clear than
fy(z,x,x)

The original
f(n)x
should be taken to read the nth derivative of the single variable function f evaluated at x.