Dtjytyk

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
$$pi= lim_{xtoinfty} 2^xsqrt{2-sqrt{2+sqrt{2+sqrt{2+...}}}} tag{1}$$
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:
$$detbegin{bmatrix}
1 & x_1 & x_1^2 & ... & x_1^{n-1} & x_1^{n} \
1 & x_2 & x_2^2 & ... & x_2^{n-1} & x_2^{n} \
. & & . & & . & \
. & & & . & . & \
1 & x_{n-1} & x_{n-1}^2 & ... & x_{n-1}^{n-1} & x_{n-1}^{n} \
1 & x_n & x_n^2 & ... & x_n^{n-1} & x_n^{n} \
end{bmatrix}=prod _{k>j} (x_k-x_j) tag{3}$$

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:
$$sigma_0(x)=sum_{i=1}^x log_xleft ( i^{left lfloor frac{x}{i} right rfloor-left lfloor frac{x-1}{i} right rfloor}frac{left lfloor frac{x}{i} right rfloor!}{left lfloor frac{x-1}{i} right rfloor!} right ) tag{5}$$

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 :) .

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

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

About the infinite radical:

You must be careful about this notation because you don't say what is "a 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

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

$$2^nsinfracpi{2^n}.$$

Congratulations, you rejoined Archimedes' findings !