Quadratic equation form

Question

I have the relation $u=\sqrt{(a_1+b_1t)^2+(a_2+b_2t)^2+(a_3+b_3t)^2} \tag 1$

I need to write $t$ as a function of $u$ ($t=f(u)$). How will I get that ?

NB: $a_1,a_2,a_3,b_1,b_2,b_3$ are constants

Answer 1

If you square both sides, you get $$ u^{2}=\left(a_{1}+b_{1}t\right)^{2}+\left(a_{2}+b_{2}t\right)^{2}+\left(a_{3}+b_{3}t\right){}^{2}. $$ Then, expanding, \begin{align*} u^{2} & =\sum_{n=1}^{3}a_{n}^{2}+2a_{n}b_{n}t+b_{n}^{2}t^{2}\\ & =\left(\sum a_{n}^2\right)+\left(2\sum a_{n}b_{n}\right)t+\left(\sum b_{n}^{2}\right)t^{2} \end{align*} and now we have a quadratic equation for $u^{2}$ in terms of $t$. $$ u=\pm\sqrt{\frac{-2\sum\left(a_{n}b_{n}\right)\pm\sqrt{4\left(\sum a_{n}b_{n}\right)^{2}-4\left(\sum a_{n}^2\right)\left(\sum b_{n}^{2}\right)}}{2\sum a_{n}^2}.} $$

Simplify this further to get $$ u=\pm\sqrt{\frac{\sum\left(a_{n}b_{n}\right)\pm\sqrt{\left(\sum a_{n}b_{n}\right)^{2}-\left(\sum a_{n}^{2}\right)\left(\sum b_{n}^{2}\right)}}{\sum a_{n}^{2}}} $$ Thanks to Claude Leibovici for pointing out my mistake. Now, due to the Cauchy-Schwarz inequality: $$ \left(\sum a_{n}b_{n}\right)^{2}\leq\sum a_{n}^{2}\sum b_{n}^{2} $$ and hence $$ u=\pm\sqrt{\frac{\sum\left(a_{n}b_{n}\right)\pm i\sqrt{\left(\sum a_{n}^{2}\right)\left(\sum b_{n}^{2}\right)-\left(\sum a_{n}b_{n}\right)^{2}}}{\sum a_{n}^{2}}} $$ with the quantity in the innermost square root being nonnegative.