Are my results new? [closed]

Solution 1:

About the determinant:

Such a matrix is said to be of the Vandermonde form.

Its determinant is obviously a polynomial in the $x_i$. It must cancel whenever two $x_i$ are equal (making two equal rows), so that it must be a multiple of every $(x_i-x_j)$. There are $\dfrac{n(n+1)}2$ such factors. On another hand, the degree of the polynomial must be the the sum of the degrees along a row, i.e. the $n^{th}$ triangular number, $\dfrac{n(n+1)}2$. This is enough to say that

$$\det M=m\prod_{i<j}(x_i-x_j)$$ for some nonzero constant $m$.

About the infinite radical:

You must be careful about this notation because you don't say what is "at the end of the dots". You can express the nested radicals as a recurrence

$$r_{n+1}=\sqrt{r_n+2},$$ but some $r_0$ must be specified. If you take $r_0=2$, then for all $n$, $r_n=2$ and… $\pi=0$ !

You probably obtained your formula from the perimeter of the circle, by successive doublings of the number of sides, using the angle halving formula

$$2\cos\frac x2=\sqrt{2\cos x + 2}$$ and $x=\dfrac\pi{2^m}$ for some $m$. Then starting from $m=1$, $r_0=0$, and your estimates of $\pi$ are

$$2^n\sin\frac\pi{2^n}.$$

Congratulations, you rejoined Archimedes' findings !