Your calculation shows that, if a^2=2b^2 then, with x=a-b, you also have (a-2x)^2=2x^2, i.e. (2b-a)^2=2(a-b)^2. This last equation could be checked more simply by expanding both sides, ...

One way would be to use a quadratic equation to change every value in your dataset from x_i to y_i= a \cdot x_i^2+b \cdot x_i+ c . The equation for the new minimum would then be 0.3 = a \cdot (-0.8)^2+b \cdot (-0.8) + c\tag{1} ...

No. For instance you can assume that your data comes from a parameterised family of distributions and try to learn the parameters, e. g. with maximum likelihood. Or you can use non parametric ...

Is this correct? No. i) This isn't a confidence interval you're calculating (since those are for parameters or functions of them), nor is it really a prediction interval, a tolerance interval, or ...

As you write, the statement in case c_{m,n} = 1 is immediate because clearly \Phi_m(x) \in \mathbb{Z}[x] = \mathbb{Z}[x] + (\Phi_n(x)) = \sqrt{\mathbb{Z}[x] + (\Phi_n(x))}. Since this is true for ...

\sum_{n=1}^\infty a_n = \sum_{k=0}^\infty \frac1{2^k} ..is clearly a convergent sum (\lt\infty). n a_n = \begin{cases} 1 , &\text{if }n = 2^k,\quad k= 0,1,2,,,,\\ 0 ,&\text{otherwise}\end{cases} ...