## Available data

Available to me is a set of points which can be represented as shown in image 1: Also available to me is a non-continuous path derived from this data. It is not important how this non-continuous path is obtained. It is however important, that it roughly represents a curve I intent to approximate. This non-continuous path is shown in image 2: ## Goal

I want to approximate this data using a polynomial of either second or third degree. Examples of these approximations are shown in images 3, 4:  ## Problem / question

Now I am looking for a way to obtain the red curve. Some details confuse me, where my knowledge is likely to be simply lacking. For example, how to fit a polynomial, when technically what I require is not a function, at least not in this coordinate system, because there will be situations where an x value is being assigned two y values.

I thought of possibly using the ends of my path to define a new x-axis, but I consider this approach faulty. Another consideration are Splines.

How should I go about obtaining this red curve from the non-continuous path (preferred) or from original data? What sources should I look into?

Apologies if this is an already answered question which I suspect it might be. However I have been issuing search queries for this without success, hence my question.

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## Would $\Sigma_i^P \neq \Pi_i^P$ imply that polynomial hierachy cannot collapse to the $i$-th level?

If $$\Sigma_i^P = \Pi_i^P$$, then it follows that the polynomial hierarchy collapses to the $$i$$-th level.

What about the case $$\Sigma_i^P \neq \Pi_i^P$$? For example, consider the case of $$NP \neq coNP$$. As far as I understand, this would imply the polynomial hierarchy cannot collapse to the first level, since if $$PH =NP$$, then in particular, $$coNP \subseteq NP$$, which means $$NP = coNP$$. Can we expand this idea to proof the general case: $$\Sigma_i^P \neq \Pi_i^P$$ implies $$PH$$ cannot collapse to $$i$$-th level?

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## Useful conditions for proving super polynomial time for some kind of recurrences

Given a recurrence of the form $$\forall n,m.\ \ T(n,m)=\begin{cases}1,&,m=1\\sum_i{T(n_i,m_i)}&,\text{else}\end{cases}$$

Note: both $$n_i$$ and $$m_i$$ are dependent on $$n,m$$ so they should have been written above as $$n_{i,n,m}$$ and $$m_{i,n,m}$$, but they are written above as $$n_i$$ and $$m_i$$ in a way of abbreviation for more readablity.

I am looking for a useful, necessary and sufficient condition so that $$\exists k. T(n,m)=O\bigg(\Big(\frac{\log(m)}{n}\Big)^k\bigg)$$

The importance of such conditions stems from their applicabilty in concluding super-polynomial lower bounds on complex recurrences of the above-mentioned form.

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## Why does NTM need to derive certificate to prove “If a language is in in NP iff it is decidable by some nondeterministic polynomial time TM”?   (Sipser’s Chapter 7: Time COmplexity, Pgs 294-295)

If we have to prove the forward direction, then we must have the certificate along with the verifier. I don’t get why we are “guessing the certificate”–essentially constructing the verifier– if we already know one exists.

On the other hand, if you don’t have the certificate, you can use a NTM to derive it and check w against c. So you don’t need to pass any input into V. So V is useless.

I don’t really understand this proof.

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## Polynomial – using Newton’s method, or not?

I have one problem that is expressed as polynomial of 2…n degree. In order to solve it I can use general Newton’s method for all degrees, or only for 5th+, and for 2,3 and 4 I can use algebraic equations (quadratic, cubic and quartic equations).

Since algebraic equations need to use sqrt, acos and similar functions – does it make sense at all to use algebraic equations – or is it better to use Newton’s method for all solutions? I guess it will actually be faster?

UPDATE

This is my polynomial:

$$\sum_{i=0}^nf^i\binom{\{b_1,…,b_n\}}{i}(-k+n-i), n>=2$$

And I need to write code to calculate $$f$$ for different values of $$n$$.

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## Traversing a Graph polynomial time

Given a directed graph $$G = (V, E)$$ and a starting vertex $$v_1$$.

The graphs edges is this

$$\{(v_1, v_2), (v_2, v_3), (v_3, v_4)\}$$

basically below

$$(v_1) -> (v_2) -> (v_3) -> (v_4)$$

Can we traverse this graph and record the path starting at $$v_1$$ in $$O(|V|)$$ time?

For example Making a list like $$[(v_1, v_2), (v_2, v_3), (v_3, v_4)]$$

I’m confusing whether this takes $$O(|V|^2)$$ worst case time because the edges are in a set

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## Simple interepretation problem regarding Polynomial Hierarchy?

So $$NP$$ stands for problems where we have small verifiable witnesses for $$YES$$ instances and $$coNP$$ for small verifiable witnesses for $$NO$$ instances. How does this work for

1. $$P^{NP}$$

2. $$NP^{NP}$$

3. $$coNP^{NP}$$

4. and so on?

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## Coefficient of power $p$ in polynomial expansion : mathematica doesn’t answer me

I would like to access the coefficient in front of $$n^p$$ for a polynomial expansion.

I wrote the following code:

f[n_, q1_, q2_] := (n^q1 + (n + 1)^q1)* (n^q2 + (n + 1)^q2)  SeriesCoefficient[f[n, q1, q2], {n, 0, p},   Assumptions -> {Element[q1, Integers] && q1 >= 0,     Element[q2, Integers] && q2 >= 0 }] 

However, mathematica doesn’t compute anything. Why ?

I would expect an expression that would make appear the binomial coefficient (I can do it by hand it is not very complicated thus I wonder why mathematica cannot).

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## Polynomial size Boolean circuit for counting number of bits

Given a natural number $$n \geq 1$$, I am looking for a Boolean circuit over $$2n$$ variables, $$\varphi(x_1, y_1, \dots, x_n, y_n)$$, such that the output is true if and only if the assignment that makes it true verifies

$$\sum_{i = 1}^{i = n} (x_i + y_i) \not\equiv n \bmod 3$$

I should specify that this I am looking for a Boolean circuit, not necessarily a Boolean formula as it is usually written in Conjunctive Normal Form (CNF). This is because when written in CNF, a formula like the one before has a trivial representation where the number of clauses is approximately $$\frac{4^n}{3}$$, as it contains a clause for every assignment $$(x_1, y_1, \dots, x_n, y_n)$$ whose bits sum to a value which is congruent with $$n \bmod 3$$. Constructing such a formula would therefore take exponential time.

I have been told that a Boolean circuit can be found for this formula that accepts a representation of size polynomial in $$n$$. However, so far I have been unable to find it. I would use some help; thanks.

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## Is the certificate for primality testing polynomial in the length of the input?

If we were to assume that primality testing was in NP.

What would the certificate be, so that a polynomial time verifier can check the number X is indeed prime?

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