Are my results new?

Question

I'm eighteen and sometimes I like doing math on my own when I'm inspired. I would like to know if some of my "discoveries" are new (I don't think so :) ). These are some of the results I found in the last 3 years:

Infinite radical converging to pi $$\pi= \lim_{x\to\infty} 2^x\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+...}}}} \tag{1}$$ Where in the radical there are $x$ numbers $2$

Determinant of a matrix for the interpolation of polynomials

Let $p(x)=[a_n,a_{n-1},...,a_1,a_0]$ be a polynomial passing through $n+1$ points $P_i(x_i,y_i)$ . This polynomial is intrinsically connected with the matrix: $$\begin{bmatrix} 1 & x_1 & x_1^2 & ... & x_1^{n-1} & x_1^{n} \\ 1 & x_2 & x_2^2 & ... & x_2^{n-1} & x_2^{n} \\ . & & . & & . & \\ . & & & . & . & \\ 1 & x_{n-1} & x_{n-1}^2 & ... & x_{n-1}^{n-1} & x_{n-1}^{n} \\ 1 & x_n & x_n^2 & ... & x_n^{n-1} & x_n^{n} \\ \end{bmatrix} \tag{2}$$

Moreover: $$det\begin{bmatrix} 1 & x_1 & x_1^2 & ... & x_1^{n-1} & x_1^{n} \\ 1 & x_2 & x_2^2 & ... & x_2^{n-1} & x_2^{n} \\ . & & . & & . & \\ . & & & . & . & \\ 1 & x_{n-1} & x_{n-1}^2 & ... & x_{n-1}^{n-1} & x_{n-1}^{n} \\ 1 & x_n & x_n^2 & ... & x_n^{n-1} & x_n^{n} \\ \end{bmatrix}=\prod _{k>j} (x_k-x_j) \tag{3}$$

Bounds for the sum of prime factors

Let $\ \Pi(x)\ $ be the function that gives in output the sum of the primes factors of $x$ for example $\ \Pi(8)=6,\ $ then: $$\log_3(x^3)\leq \Pi(x) \leq x \tag{4}$$

An alternative formula for the divisor function

$\sigma_0(x)$ is defined as the function that gives in output the number of divisors of $x$, then: $$\sigma_0(x)=\sum_{i=1}^x \log_x\left ( i^{\left \lfloor \frac{x}{i} \right \rfloor-\left \lfloor \frac{x-1}{i} \right \rfloor}\frac{\left \lfloor \frac{x}{i} \right \rfloor!}{\left \lfloor \frac{x-1}{i} \right \rfloor!} \right ) \tag{5}$$

Conjecture about the n-nacci period

Let $\ T(n)\ $ be the function the gives in output the $n$-nacci period, i.e. the periodicity of the units digit in a $n$-nacci sequence with all of the starting terms that are equal to $1$ ($T(2)=60$ is the classic Fibonacci sequence). Then my conjecture is that:

$$T(2n+1)=\frac{5^{2n+1}-1}{4} \tag{6}$$

Clearly, $n$ is integer. I almost forgot $2n+1$ must not terminate with digit $1$ (in that case the problem is trivial).

I hope you can tell me whether this results are trivial or not and if they are eventually wrong. Thank you for the time :) .

Answer 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 $m$.

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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 !