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

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Homework Help Overview

The discussion revolves around expanding a complex equation in a power series, specifically focusing on the variable representation within the context of the series expansion. The subject area includes series expansions and potentially quantum mechanics, given the notation used.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to clarify the relationship between variables in their equation, specifically identifying what 'x' should be in the context of the series expansion. Some participants question the role of the variable Z in the equation for E and whether it has been set to a specific value. Others inquire about the implications of taking derivatives with respect to the identified variable.

Discussion Status

The discussion has progressed with participants confirming the identification of variables and the steps needed for the series expansion. There is acknowledgment of the adjustments made by the original poster, and a sense of collaborative exploration is evident.

Contextual Notes

There are indications of potential confusion regarding the variable definitions and their roles in the series expansion, as well as the implications of setting certain variables to specific values. The original poster's modifications suggest an evolving understanding of the problem context.

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Homework Statement



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

[tex] 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\} [/tex]

Homework Equations



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

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

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

Thanks!
 
Last edited:
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Actually in your formula a = 0 and x = [tex](\alpha Z)^2[/tex]. 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 = [tex](\alpha Z)^2[/tex] and a = 0.
My last question is, the variable of my equation, from what you're saying is x = [tex](\alpha Z)^2[/tex] , so in my series expansion I must take the derivative with respect to this x ?
 
yes, absolutely.
 
worked, thanks :)
 

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