Einstein summation confusion I was trying to evaluate the following expression:
$$\frac{1}{\sqrt{u_iu_i}}.$$
I know the author used this as short notation for 
$$\frac{1}{\sqrt{u_1^2+u_2^2+u_3^2}}$$
and I always thought of it as this expression. Now, I can rewrite 
$$u_i^{-1/2}u_i^{-1/2}.$$
Using the summation convention (https://en.wikipedia.org/wiki/Einstein_notation) 
$\sum_{i=1}^{3}c_ix^i=c_ix^i$
this would result in:
$u_i^{-1/2}u_i^{-1/2}=\sum_{i=1}^{n}u_i^{-1/2}u_i^{-1/2}=u_1^{-1/2}u_1^{-1/2}+u_2^{-1/2}u_2^{-1/2}+u_3^{-1/2}u_3^{-1/2}.$
Which is clearly not what was intended by the author. 

My question: Which intrepretation is right and which is wrong? I would be glad > if someone could give a trustable source.

 A: Think of it like this(*)
$$
\frac{1}{\sqrt{u_i u_i}} = \frac{1}{\sqrt{\sum u_i u_i}} = \frac{1}{\sqrt{u1^2 + u_2^2 + u_3^2}}
$$
(*)Important: There is however an abuse of notation here: $x_ix_i$ does not expand to $x_1^2 + x_2^2 + x_3^2$ using Einstein's convention, because it violates one rule: one upper index sums with one and only one lower index. In this expression there're only lower indices! You can fix this by introducing a Kronecker delta
$$
\frac{1}{\sqrt{u_i\delta^{ij}u_j}} = \frac{1}{\sqrt{u1^2 + u_2^2 + u_3^2}}
$$
A: $u^iu_i$ is shortened for $\sum_i u_iu_i$ ; which is $u_1^2+u_2^2+u_3^2$ when we implicitly have $i\in\{1,2,3\}$ 
Thus $(u^iu_i)^{-1/2} = (u_1^2+u_2^2+u_3^2)^{-1/2}$ because we resolve the Einstein notation before applying the exponent to the series.
This is quite distinct from $u^iv_i$ in which the exponent is the bound variable of the series.

PS: Also note that when using Einstein notation $(u^iv_i)^2 \neq (u^i)^2(v_i)^2$ as the square of a series is not equal to a series of squares.
$$(u^iv_i)^2=  (u_1v_1+u_2v_2+u_3v_3)^2~\neq~ (u_1+u_2+u_3)^2(u_1+u_2+u_3)^2=(u^i)^2(v_i)^2$$
Similarly $(u^iu_i)^{-1/2}\neq (u^i)^{-1/2}(u_i)^{-1/2}$.   Exponents do not distribute over series.
