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]

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