Question

Show that only odd value of l = 66(l = (j - 1)) the solution you can use the following relation $$ \frac{(1-a^2/r^2)}{(1+a^2/r^2)^{1/2}} = (\frac{(1-a^2/r^2)}{(1+a^2/r^2)})^{-1/2}$$ to prove the following $$ \phi(z=r) = \frac{v}{\sqrt{\pi}} \sum_{d=1}^{\infty} (-1)^{j-1} \cdot (2j-1) \frac{\Gamma(j-\nu_2) \cdot (a|r)^{2j}}{j!} $$

Name: show that only odd value of l 66l j 1 the solution you can use the following relation frac1 a2r21a2r212 frac1 a2r21a2r2 12 to prove the following phizr fracvsqrtpi sumd1infty 1j 1 cdot 2j 1 51087
Uploaded: 2021-10-22T13:00:42-08:00
Duration: 6 min 11 s
Channel: Khoobchandra Agrawal
Description: show that only odd value of l 66l j 1 the solution you can use the following relation frac1 a2r21a2r212 frac1 a2r21a2r2 12 to prove the following phizr fracvsqrtpi sumd1infty 1j 1 cdot 2j 1 51087

          Show that only odd value of 
l = 66(l = (j - 1)) the solution
you can use the following
relation
$$ \frac{(1-a^2/r^2)}{(1+a^2/r^2)^{1/2}} = (\frac{(1-a^2/r^2)}{(1+a^2/r^2)})^{-1/2}$$
to prove the following
$$ \phi(z=r) = \frac{v}{\sqrt{\pi}} \sum_{d=1}^{\infty} (-1)^{j-1} \cdot (2j-1) \frac{\Gamma(j-\nu_2) \cdot (a|r)^{2j}}{j!} $$

$show that only odd value of l 66l j 1 the solution you can use the following relation frac1 a2r21a2r212 frac1 a2r21a2r2 12 to prove the following phizr fracvsqrtpi sumd1infty 1j 1 cdot 2j 1 51087$

Added by Peter H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Khoobchandra Agrawal

10/22/2021

Step 1

Step 1: The given relation is: $$ \frac{(1-a^2/r^2)}{(1+a^2/r^2)^{1/2}} = (\frac{(1-a^2/r^2)}{(1+a^2/r^2)})^{-1/2}$$ This relation can be simplified as: $$ \frac{(1-a^2/r^2)}{(1+a^2/r^2)^{1/2}} = \frac{(1+a^2/r^2)^{1/2}}{(1-a^2/r^2)}$$ This relation is true for all Show more…

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Show that only odd value of l = 66(l = (j - 1)) the solution you can use the following relation $$ \frac{(1-a^2/r^2)}{(1+a^2/r^2)^{1/2}} = (\frac{(1-a^2/r^2)}{(1+a^2/r^2)})^{-1/2}$$ to prove the following $$ \phi(z=r) = \frac{v}{\sqrt{\pi}} \sum_{d=1}^{\infty} (-1)^{j-1} \cdot (2j-1) \frac{\Gamma(j-\nu_2) \cdot (a|r)^{2j}}{j!} $$