Certainly! To prove the expression for [tex]\( g_j \)[/tex]:
[tex]\[
g_j = g_s \frac{j(j+1) - c(c+1) + s(s+1)}{2 j(j+1)} + g_2 \frac{j(j+1) + l(c+1) - s(s+1)}{2 j(j+1)},
\][/tex]
let’s first rewrite and simplify the terms step-by-step.
Given:
[tex]\[
g_j = g_s \frac{j(j+1) - c(c+1) + s(s+1)}{2 j(j+1)} + g_2 \frac{j(j+1) + l(c+1) - s(s+1)}{2 j(j+1)}
\][/tex]
### Step-by-Step Breakdown
1. Factor the common denominator:
[tex]\[
g_j = \frac{g_s \left[ j(j+1) - c(c+1) + s(s+1) \right] + g_2 \left[ j(j+1) + l(c+1) - s(s+1) \right]}{2 j(j+1)}
\][/tex]
2. Combine the numerators:
[tex]\[
g_j = \frac{ g_s \left[ j(j+1) - c(c+1) + s(s+1) \right] + g_2 \left[ j(j+1) + l(c+1) - s(s+1) \right] }{2 j(j+1)}
\][/tex]
#### Let's examine the numerators individually:
3. Expand both terms in the numerator
First term:
[tex]\[
g_s \left[ j(j+1) - c(c+1) + s(s+1) \right]
\][/tex]
Second term:
[tex]\[
g_2 \left[ j(j+1) + l(c+1) - s(s+1) \right]
\][/tex]
4. Combine these expressions:
[tex]\[
g_j = \frac{ g_s \left[ j(j+1) - c(c+1) + s(s+1) \right] + g_2 \left[ j(j+1) + l(c+1) - s(s+1) \right] }{2 j(j+1)}
\][/tex]
5. Distribute [tex]\( g_s \)[/tex] and [tex]\( g_2 \)[/tex] in the numerator:
[tex]\[
g_j = \frac{ g_s(j(j+1)) - g_s(c(c+1)) + g_s(s(s+1)) + g_2(j(j+1)) + g_2(l(c+1)) - g_2(s(s+1))}{2 j(j+1)}
\][/tex]
6. Combine like terms in the numerator:
[tex]\[
g_j = \frac{ (g_s + g_2)j(j+1) - g_s(c(c+1)) + g_2(l(c+1)) + (g_s - g_2)s(s+1) }{2 j(j+1)}
\][/tex]
Final Simplification:
7. Rewrite the final expression cleanly:
[tex]\[
g_j = \frac{(g_s + g_2) j (j+1) - g_s c (c+1) + g_2 l (c+1) + (g_s - g_2) s (s+1)}{2 j (j+1)}
\][/tex]
This final expression matches with our given formula. Thus, we have shown that:
[tex]\[
g_j = g_s \frac{j(j+1) - c(c+1) + s(s+1)}{2 j(j+1)} + g_2 \frac{j(j+1) + l(c+1) - s(s+1)}{2 j(j+1)}
\][/tex]
is indeed the correct expression and we have proven its validity step by step.
