Rewrite $\sin \left(2 \tan ^{-1} u\right)$ as an algebraic expression in $u$. \[ \sin \left(2 \tan ^{-1} u\right)= \]

Step 1 ：We are given the expression \(\sin \left(2 \tan ^{-1} u\right)\) and we are asked to rewrite it as an algebraic expression in terms of \(u\).

Step 2 ：We know that \(\sin(2x) = 2\sin(x)\cos(x)\). We can use this identity to rewrite the expression.

Step 3 ：We also know that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), so we can express \(\sin(x)\) and \(\cos(x)\) in terms of \(u\) using the right triangle definition of sine and cosine.

Step 4 ：In a right triangle, if \(\tan(x) = u\), then we can think of \(u\) as the ratio of the opposite side to the adjacent side. We can then use the Pythagorean theorem to find the hypotenuse, which will allow us to find \(\sin(x)\) and \(\cos(x)\).

Step 5 ：Let's denote the opposite side as \(u\), the adjacent side as 1, and the hypotenuse as \(\sqrt{u^2 + 1}\).

Step 6 ：Then, \(\sin(x) = \frac{u}{\sqrt{u^2 + 1}}\) and \(\cos(x) = \frac{1}{\sqrt{u^2 + 1}}\).

Step 7 ：Substituting these values into the identity \(\sin(2x) = 2\sin(x)\cos(x)\), we get \(\sin(2x) = \frac{2u}{u^2 + 1}\).

Step 8 ：Finally, substituting \(x = \tan^{-1}(u)\) back into the equation, we get \(\sin \left(2 \tan ^{-1} u\right)= \frac{2u}{u^2 + 1}\).

Step 9 ：\(\boxed{\sin \left(2 \tan ^{-1} u\right)= \frac{2u}{u^2 + 1}}\) is the final answer.