How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$ My understanding would lead me to believe that:
$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$
But it actually equals: $\sqrt{1 + e^4}$
Can you explain why?
 A: You have
$$
\frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{4}\sqrt{1 + e^4} = \frac{1}{2}2\sqrt{1+e^4} = \sqrt{1+e^4}
$$
It looks like you where thinking that 
$$
\sqrt{a + b} = \sqrt{a} + \sqrt{b}.
$$
But that is not true (try to check this with $a=b=2$). And even if you did that it looks like you  forgot that $\sqrt{4e^4}= 2 e^2$ (as pointed out by TMM in the comment below.)
A: Also you can use squaring both sides:
$$
\sqrt{4+4e^4}=2\sqrt{1+e^4}\\
4+4e^4=4(1+e^4)
$$
since  $\sqrt{x^2}= \pm x$ and the expressions you start with is one of these two solutions (positive)
A: Your mistake is to say that $\sqrt{4(a+b)}=2(a+b)$, which is not true. For instance $\sqrt{4(3^2+4^2)}=10\not=2(3^2+4^2)$ 
The correct equalities you need is
$$\sqrt{(c^2a+c^2b)^2}=\sqrt{c^2(a+b)}=\sqrt{c^2}\sqrt{a+b}=c\sqrt{a+b}$$
Now $\frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{2^2(1+e^4)} = \frac{1}{2}\sqrt{2^2}\sqrt{1 + e^4} = \frac{1}{2}2\sqrt{1+e^4} = \sqrt{1+e^4}$
