Understanding the Translational Operator and Its Applications

[matematikuvol]

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}
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 f(x+\alpha ) for small \alpha 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 \frac{d}{dx} (technically \frac{d}{idx}) is the 'generator' of the translation.

EDIT: Beaten to the punch by TT!

 

[matematikuvol]

I have a problem with that. So
f(x+\alpha)=f(x)+\alpha f'(x)+...
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 have a problem with that. So
f(x+\alpha)=f(x)+\alpha f'(x)+...
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 $(\frac{df}{dx})_{x_0}$.

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]

I have a problem with that. So
f(x+\alpha)=f(x)+\alpha f'(x)+...
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 $(\frac{df}{dx})_{x_0}$.

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]

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 + …

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

 

[tiny-tim]

no, it's \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x_0}

 

[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
e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...
you never have $x_0$.

 

[tiny-tim]

\left(e^{\alpha\frac{d}{dx}}(f(x))\right)_{x_o}
= \left(\left(1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...\right)(f(x))\right)_{x_o}
= \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x_0}