Hey all. I have posted a thread regarding this question a while back. I did get an answer and everything. (Here is my old post along with the original question if you are interested: https://www.physicsforums.com/showthread.php?t=592885).(adsbygoogle = window.adsbygoogle || []).push({});

So i tried doing that problem again like this:

Given:

L = Hbar * √(l*(l+1))

Binomial expansion approximation: (1+z)^n = 1 + nz

My try:

L = Hbar * √l * (1 + l/2)

If you opened the link of my previous post you will see that I must obtain L = Hbar * (l + 1/2). As you can see my just calculated expression is not the same as the previous expression, however plugging in a certain value of l, one obtains the same value for either expression which leads me to believe the two expressions are equivalent. So my question is: How would I play with my current equation to obtain the equation from the previous post. Yes, this is completely an algebra question. Thank you.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Binomial expansion.

Loading...

Similar Threads - Binomial expansion | Date |
---|---|

I Normalisation constant expansion of spinor field | Jan 14, 2018 |

A Field Expansion in QFT | Jan 12, 2018 |

I Eigenvalues, eigenvectors and the expansion theorem | Sep 17, 2016 |

I Why do we require conditions for the Poisson Distribution? | Sep 3, 2016 |

Binomial expansion of massive spin 0 propagator | Feb 4, 2011 |

**Physics Forums - The Fusion of Science and Community**