Are these two equations equivalent

Question

Can I say that this equation

is equivalent to

$$\frac{1}{S}\frac{1}{U} \sum_{p=1}^{S} \sum_{u=1}^{U} PL . SF (|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 )$$

Thank you.

Answer 1

The two expressions are equal.

We have \begin{align*} \frac{1}{S}\sum_{p=1}^{S} &\left(PL \cdot SF \cdot \frac{1}{U}\sum_{u=1}^{U} \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\right)\\ &=\frac{1}{S}\sum_{p=1}^{S} PL \cdot SF \cdot \frac{1}{U}\sum_{u=1}^{U} \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\tag{1}\\ &=\frac{1}{S} \frac{1}{U}\sum_{p=1}^{S} PL \cdot SF \sum_{u=1}^{U} \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\tag{2}\\ &=\frac{1}{S}\frac{1}{U} \sum_{p=1}^{S} \sum_{u=1}^{U} PL \cdot SF \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\tag{3} \end{align*}

Comment:

• In (1) we skip the outer parentheses. They do not have any effect since multiplication of terms has a higher precedence level than addition of terms.

• In (2) we factor out $\frac{1}{U}$ using the distributive law: $a(b+c)=ab+ac$.

• In (3) we multiply the terms of the third-left inner sum with $PL \cdot SF$ again using the distributive law.