I'm trying to figure out the theory behind how to obtain an equation, the whole thing is very long so I couldn't post the entire thing here, but part of it requires a Taylor expansion in certain terms and I've completely forgotten how to do so.

The Taylor expansion is of $\frac{(u + v)}2$ and $\frac{(u - v)}2$ where $$u = \frac{2{\pi}x}{\lambda_{Na}+\frac{\Delta\lambda}2}$$ and $$v = \frac{2{\pi}x}{\lambda_{Na}-\frac{\Delta\lambda}2}$$ but each expansion is supposed to be in powers of $\Delta\lambda$ since $\Delta\lambda << \lambda_{Na}$. Also, I'm told to only retain one term (the first non-vanishing term) in each series. I'm not sure what to do for this and am at a loss. Help would be very greatly appreciated.

Let me know if there's anything I left out that could be useful and I'll see what I can do.

  • 1
    $\begingroup$ Write $u=\frac{2\pi x}{\lambda_{Na}} (1-\tau+\tau^2-\tau^3\cdots)$, where $\tau = \Delta\lambda/(2\lambda_{Na})$, and so on. $\endgroup$ – kimchi lover Sep 23 '17 at 20:54
  • 1
    $\begingroup$ Take $\lambda_{Na}$ common in the denominator, and then expand $(1 + \frac{\Delta \lambda}{2 \lambda_{Na}})^{-1}$. $\endgroup$ – sbp Sep 23 '17 at 20:54
  • 1
    $\begingroup$ I wrote "and so on". Once you know how to expand $u$ you should know how to expand $v$, too, and combine the results accordingly. $\endgroup$ – kimchi lover Sep 23 '17 at 21:18
  • 1
    $\begingroup$ Don't forget to use the $1/(a+b)\pm 1/(a-b) = ((a-b)\pm(a+b))/(a^2-b^2)$ trick, too. $\endgroup$ – kimchi lover Sep 23 '17 at 21:21
  • 1
    $\begingroup$ Once you know how to expand one, you can carry over this to the other terms. $\endgroup$ – sbp Sep 23 '17 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.