Let $K$ and $\bar K$ be two cartesian co-orditate systems in $\mathbb{R}^3$. The element: $$s^2=(\Delta x^1)^2+(\Delta x^2)^2+(\Delta x^3)^2$$ is an invariant in all co-ordinate system. I want prove that the relations between de co-ordinates of $K$ and $K$ must be linear.
In the book: "The meaning of relativity, by A. Einstein" we can find the following reasoning:
Suppose that $\bar x^\nu=\bar x^\nu(x^1,x^2,x^3)$ and note that: $$\left\{\begin{array}{lll} (\Delta x^1)^2+(\Delta x^2)^2+(\Delta x^3)^2= \text{const.}&(1)\\ (\Delta \bar x^1)^2+(\Delta \bar x^2)^2+(\Delta \bar x^3)^2 = \text{const.}&(2) \end{array}\right.$$ Also by Taylor's theorem: $$\Delta \bar x^\nu = \sum_\alpha \frac{\partial \bar x^\nu}{\partial x^\alpha}\Delta x^\alpha+\frac{1}{2}\sum_{\alpha,\beta} \frac{\partial^2\bar x^\nu}{\partial x^\alpha\partial x^\beta}\Delta x^\alpha\Delta x^\beta+\ldots\qquad (3)$$ Substituting $\Delta x^\nu$ given by $(2)$ to $(3)$ and compare this with $(*)$ then we deduce that $\bar x^\nu=\bar x^\nu(x^1,x^2,x^3,x^4)$ must be linear.
Can anybody help me to take $\Delta x^\nu$ given by $(2)$ and put into $(3)$ to deduce that the transformation must be linear?
