# Binomial expansion.

1. May 2, 2012

### perplexabot

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).

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.

Last edited: May 3, 2012
2. May 3, 2012

### The_Duck

I'm going to call your lower-case l capital L because l looks like 1.

Hbar * √(L*(L+1)) is not equal to Hbar * √L * (1 + L/2), if that is what you are claiming? To see this, just plug in L = 1. Then the first expression becomes hbar*sqrt(2) while the second becomes hbar * (3/2).

Nor is Hbar * √L * (1 + L/2) equal to Hbar * (L + 1/2), if that is what you are claiming. Just plug in L = 2; then the first expression is hbar * sqrt(2) * 2 while the second is hbar * 5/2.

3. May 3, 2012

### perplexabot

Oh, i made a stupid mistake i guess. Thank you for your correction. Can you explain why the two equations are not equivalent? I got both of them from the same equation but they are not equal.