Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [lagrange-interpolation]

A method of generating a polynomial that crosses through a set of data. The degree of this polynomial is equal to the size of the data.

Let $(x_j)_{j=0}^{m-1},(y_j)_{j=0}^{m-1}$ be real numbers such that no $2$ $x_j$s are the same. The Lagrange interpolating polynomial is given by $$ l(x) = \sum_{j=0}^{m-1} y_j \prod_{j \neq k \in [0..m-1]} \frac{x-x_k}{x_j - x_k} $$

387 questions
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For what $n$ can we find a degree $\leq n-2$ polynomial such that $P(i) \in \{0 , 1 \}$ for $i \in [n]$, but not all identical.

For what $n$ is the following statement true: There exists a choice of $ a_1, a_2, \ldots a_n \in \{ 0, 1 \}$, not all identical, such that there is a polynomial $F(x) \in \mathbb{R}[x]$ of degree at most $n-2$ such that $F(i ) = a_i$ for $i = 1 $…
Calvin Lin
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Newton form vs. Lagrange form for interpolating polynomials

I'm just wondering, what are the advantages of using either the Newton form of polynomial interpolation or the Lagrange form over the other? It seems to me, that the computational cost of the two are equal, and seeing as the interpolated polynomial…
Phillip Bredahl
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Analytic "Lagrange" interpolation for a countably infinite set of points?

Suppose I have a finite set of points on the real plane, and I want to find the univariate polynomial interpolating all of them. Lagrange interpolation gives me the least-degree polynomial going through all of those. Is there an analogous construct…
Mike Battaglia
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Are multivariate polynomials determined by their values on a lattice?

In one variable, a polynomial (of any degree) is determined by its values on a finite set of points. More specifically if $p$ is a polynomial of degree $k$, and $x_0 , \dots x_{k}$ are points for which we know the values $\{p(x_{i})\}_{i=0}^{k}$,…
Dan1618
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Is there a specialized formula for Lagrangian interpolation on equispaced points?

If we know $f(0),f(1),f(2),\cdots f(n)$, is there a specialized version of the Lagrangian interpolation formula and a shortcut to compute the coefficients ? (Stability is not a concern.)
user65203
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Constructing a cubic given four points

Question: Is there an easier way to solve this problem? Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$. My Attempt: I started off with the general cubic…
Frank
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Polynomial $P(x)$ such that $P(3k)=2$, $P(3k+1)=1$, $P(3k+2)=0$ for $k=0,1,2,\ldots,n-1$, $P(3n)=2$, and $P(3n+1)=730$

Let $n$ be a positive integer such that there exists a polynomial $P(x)$ over $\mathbb{Q}$ of degree $3n$ satisfying the conditions below: $$P(0) = P(3) = \ldots = P(3n) = 2\,,$$ $$P(1) = P(4) = \ldots= P(3n - 2) = 1\,,$$ $$P(2) = P(5) = \ldots =…
Niharika
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Existence of real analytic diffeomorphisms with prescribed values on a finite set

My question is the following: given a finite set $ F = \{ x_1, \dots, x_k \} \subset \mathbb{R} $ such that $x_i < x_{i+1}$ for all $i = 1, \dots, k-1$ and numbers $a_1, \dots, a_k \in \mathbb{R}$ such that $ a_i < a_{i+1}$ for $i=1,\dots,k-1 $, is…
Mauro
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Lagrange interpolation of multivariate polynomials

Given a univariate polynomial of degree $n$ with coefficients from a field, the polynomial is uniquely defined by $n+1$ evaluation points. That is, given $n+1$ points $(i, f(i))$, one can uniquely determine $f$. Is there a similar statement for…
Cryptonaut
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Compute $\sum \limits_{k=0}^{n}(-1)^{k}k^{m}\binom{n}{k}$ using Lagrange interpolation.

Using Lagrange interpolation (I think identity $\sum \limits_{k=0}^{n}k^{m}\prod \limits_{\substack{i=0\\i\neq k}}^{n}\frac{x-i}{k-i}=x^m$) shows that $$\sum \limits_{k=0}^{n}(-1)^{k}k^{m}\binom{n}{k}=0 \text{ if }\ 0≤m
Shingle
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Interpolating polynomial with integer coefficients

Let $n_1 < \dotsb < n_k$ and $m_1, \dotsc, m_k$ be integers. Can we find conditions for the existence polynomial $p$ with integer coefficients such that $p(n_i) = m_i$ for every $i$? This isn't always possible: for example, since $p(i) \equiv p(0)…
Oliver Miller
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$\ \forall x_1,x_2,...,x_n \in \mathbb{R} (x_i\not=x_j)$ in the range of $[-1,1]$ prove:$\sum_{i=1}^{n}\frac{1}{\Pi_{k\not=i}|x_k-x_i|}\ge2^{n-2}$

$\ \forall$ $x_1,x_2,...,x_n$ $\in \mathbb{R}$ $(x_i\not=x_j)$ in the range of $[-1,1]$ prove : $$\sum_{i=1}^{n}\frac{1}{\Pi_{k\not=i}|x_k-x_i|}\ge2^{n-2}$$ my attempt : $$p(x) =…
farnood gholampoor
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Prove that $\sum_{1\leq i\leq n}\prod_{j\neq i} \frac{1-x_ix_j}{x_i-x_j} = 0$ if $n$ is even and $1$ otherwise

Let $x_1,\cdots, x_n$ be different real numbers. Prove that $$\sum_{1\leq i\leq n} \prod_{j\neq i} \frac{1-x_ix_j}{x_i-x_j} = \begin{cases} 0,&\text{if }n\text{ is even}\\ 1,&\text{if }n\text{ is odd}\end{cases}.$$ Let $f(t) = \prod_{i=1}^n (1-x_i…
Gord452
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What is the difference between nodes and knots in interpolation?

I have been reading literature about polynomial interpolation (Lagrange) where the principles are described around nodes. The literature I have read about spline interpolation, however, talks only about knots. From this, it is not entirely clear to…
Thomas Arildsen
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Estimating the double derivative of lagrange basis polynomial for specific evaluation nodes

Consider the set of points $x_j=j/t,\ j\in\{0,1,\dots t\}$ (so we have equally spaced points on the unit interval). The lagrange basis polynomials are $$L_j(x)=\prod_{0\le m\le t,\ m\ne j}\frac{x-x_m}{x_j-x_m}=\prod_{0\le m\le t,\ m\ne…
HackR
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