Translation operator

matematikuvol

[tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...=\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\frac{d^n}{dx^n}[/tex]
Why this is translational operator?
##e^{\alpha\frac{d}{dx}}f(x)=f(x+\alpha)##

tiny-tim

taylor expansion?

G01

Consider alpha to be an infinitesimal translation. Expand [itex]f(x+\alpha )[/itex] for small [itex]\alpha[/itex] to first order.

Do the same for the LHS of the equation and you should see that the equality is true for infinitesimal translations. We say that the operator [itex]\frac{d}{dx}[/itex] (technically [itex]\frac{d}{idx}[/itex]) is the 'generator' of the translation.

EDIT: Beaten to the punch by TT!

matematikuvol

I have a problem with that. So
[tex]f(x+\alpha)=f(x)+\alpha f'(x)+...[/tex]
My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##.

Questionasker

I am not sure about your question. But that translation operator is a generic operator, which translate a function value at x to x+a.

LagrangeEuler

In Taylor series x is fixed, while in ##\frac{df}{dx}## ##x## isn't fixed. Well you suppose that is.

tiny-tim

yes, but x here is a constant, and only α is the variable

if you prefer, write f(x_o + α) = f(x_o) + α∂f/∂x|_xo + …

matematikuvol

Ok but that is equal to
[tex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}(\frac{df}{dx})_{x_0}[/tex]
and how to expand now
[tex]e^{\alpha\frac{d}{dx}}[/tex]

tiny-tim

no, it's [itex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}[/itex]

matematikuvol

I made a mistake. But I'm asking when you get that ##(\frac{d^n}{dx^n})_{x_0}##? Please answer my question if you know. In
[tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...[/tex]
you never have ##x_0##.

tiny-tim

[tex]\left(e^{\alpha\frac{d}{dx}}(f(x))\right)_{x_o}[/tex]
[tex]= \left(\left(1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+.. .\right)(f(x))\right)_{x_o}[/tex]
[tex]= \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}[/tex]