What Should (x-a) Look Like in This Series Expansion?

Qubix
Messages
82
Reaction score
1

Homework Statement



Ok so I have to expand in a power series of ({\alpha} Z)^{2}, the equation

<br /> E_{nj}=mc^{2}\left\{ \left[1+\left(\frac{Z{\alpha}}{n-(j+1/2)+\sqrt{(j+1/2)^{2}-\alpha^{2}Z^{2}}}\right)^{2}\right]^{-\frac{1}{2}}-1\right\} <br />

Homework Equations



I know that a series expansion of a function f(x) around a point a is of the form

f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.
3. Question

In my above formula, if a is represented by ({\alpha} Z)^{2} , who is x ? What does E depend on? In other words , what should my (x-a) look like?

Thanks!
 
Last edited:
Physics news on Phys.org
Actually in your formula a = 0 and x = (\alpha Z)^2. Though I am wondering where your Z is in the formula for E. I guess you set it to 1. Can you do the expansion now?
 
hy, sorry about that, I modified now. So I should take x = (\alpha Z)^2 and a = 0.
My last question is, the variable of my equation, from what you're saying is x = (\alpha Z)^2 , so in my series expansion I must take the derivative with respect to this x ?
 
yes, absolutely.
 
worked, thanks :)
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top