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In the chatroom we discussed about the asymptotic of $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$, and if we think of the inverse tangent integral, it's easy to see that $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2...
Let $n\in\mathbb{N}^*,A=(a_1,\dots,a_n)\in\mathbb{K}[X]^n$ all different numbers and $B=(b_1,...,b_n)in\mathbb{K}[X]^n$ all different numbers.
Let $L_{A,B}$ be the polynomial verifying $\forall i\in[|1,n|],L_{A,B}(a_i)=b_i$.
We know that this is a Lagrange interpolation polynomial and can be wr...18
In the chatroom we discussed about the asymptotic of $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$, and if we think of the inverse tangent integral, it's easy to see that $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2...
By Complementarity Law,
$$P + \overline P = 1 \space\text{ and }\space P \cdot \overline P = 0$$
(Note: I shall only be using $P + \overline P = 1$ as its dual is automatically true)
First Law::
DeMorgan's $1^{\text{st}}$ law states $\overline{X+Y} = \overline X \cdot \overline Y$
It is suffi...
