### luosw's blog

By luosw, history, 11 months ago,

## Introduction

### Summary

How to determine the analytical formula? The original undetermined coefficient method not only has small scalability, but also has a large amount of calculation, and it performs poorly in computer science design operations. The Lagrangian interpolation method can solve this problem. This article first introduces the Lagrangian interpolation method, and explores the process of using the Lagrange interpolation method to solve the problem. It discusses the application of Lagrangian interpolation in the four aspects of basic function properties, factorization, proof of algebraic identities, and finding the relationship between the serial numbers and numerical values ​​of several ordered and regular numbers.

### Symbol Convention

1. $\prod$ means accumulative multiplication operation, generally its subscript will be used as the condition of accumulative multiplication;

2. $\sum\limits^n_{i=1}i$ means $1+2+3+\cdots+n$;

3. $f(k)$ represents the value of the function of $x$ when $x=k$;

4. $y=f(x)$ means that $y$ is a function of $x$.

## Source of the problem

In class, the teacher talked about using the method of undetermined coefficients to find the analytical formula of a function. If it is a linear function, set it to the form of $f(x)=y=kx+b$, and the result can be obtained after substituting two sets of values. For higher power functions, the undetermined coefficient method is also feasible, but the increase in the number of elements in the solution equations makes the undetermined coefficient method lose its original simplicity. Therefore, we will discuss how to determine the analytical expression of a function in combination with polynomials. Let's first study how many points on the graph of a given function, we can determine a unique function. The functions explored here are all polynomial functions, that is, the functions are polynomials.

### Lemma

$n+1$ points determine a polynomial function of degree $n$.

### Proof

We consider the most primitive method of undetermined coefficients. Suppose a polynomial function of degree $n$ $y=f(x)=\sum\limits_{i=0}^na_ix^i$ is the required function, and we finally have the sum in the equation system composed of the undetermined coefficient method A system of equations with the same number of a\$. Therefore, it takes $n+1$ equations to solve all $a$.

Then we will give $n$ points below and ask to determine a polynomial function of degree $n-1$.

## Questions raised

Given $n$ points on the function graph $P_i(x_i,y_i)(i=1,2,\cdots,n)$, at most $n-1$ degree polynomials passing through these $n$ points The function is set to $y=f(x)=\sum\limits_{i=0}^{n-1}a_ix^i$, and find: (1) When $x=k$, the value of $y$; (2) Find the function expression of $y$ and $x$.

## Problem solving

The undetermined coefficient method is indeed a method, but it is too cumbersome. Even if it is solved in a computer, the asymptotic function of the time complexity is $\Theta(n^3)$, which is not efficient. So, is there a more efficient method? Below we introduce Lagrangian interpolation and its application.

Suppose $y$ is the required function of $x$ (its image is represented by a black line), and we use $P_i(x_i,y_i)$ to make the perpendicular line of the $x$ axis intersect the $x$ axis at $H_i(x_i, 0)$, for each $i$, construct the following function $g_i(x)$ (the image is represented by a colored line) so that its image passes through all the points $H$ except $H_i$ and passes Click $P_i$.

If we consider describing such a function, we can find that the value of the function $g_i(x)$ at $x_i$ is $g_i(x_i)=y_i$, and the other values ​​at $x_j(i\neq j)$ $g_i(x_j)=0$, so we can define the function $g_i(x)$ like this:

$g_i(x)=\begin{cases}y_i \qquad x=x_i,\\0\qquad x=x_j(i\neq j).\end{cases}$

It is easy to construct its analytical formula

$g_i(x)=y_i\ell_i(x).^1$

among them

\begin{aligned} \ell _i(x) & =\prod _{i\neq j}\dfrac{x-x_j}{x_i-x_j}, \\ & =\dfrac{(x-x_1)(x-x_2)\cdots(x-x_{i-1})(x-x_{i+1})\cdots(x-x_n)}{(x_i-x_1 )(x_i-x_2)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)} \end{aligned}

It is called the interpolation basis function, which satisfies $\ell_i(x_i)=1$, $\ell_i(x_j)(i\neq j)=0$.

### Theorem

The final function $y=f(x)=\sum\limits_{i=0}^{n-1}a_ix^i$ is the sum of all $g_i(x)$. Formally,

$y=f(x)=\sum\limits_{i=1}^n y_i\prod _{i\neq j}\dfrac{x-x_j}{x_i-x_j}.\qquad (*)$

### Proof

Since the image of $g_i(x)$ passes $P_i$, and the others pass $H_j(i\neq j)$, the function image of the sum of all $z_i$ must pass all $P_i$, and the final function is determined .

In the process of solving the above problems, we also got the Lagrangian formula, that is, the Lagrangian interpolation method is used to express the form of the function. This will be very common in the following applications. It is also a common method to prove algebraic identities by comparing the coefficients of the same term in different representations of functions (using Lagrange formula here).

### Lagrange formula

That is the $(*)$ formula.

### Description

All of the above $i,j=1,2,\cdots,n$ and $i\neq j$.

## Lagrangian interpolation and the application of Lagrangian formula

We can extend Lagrangian interpolation to other mathematical applications, so that it is not only used to find functional analytical expressions. For example, the following factorization problem:

### Example

Decompose the factor $a^2(m-b)(m-c)(c-b)+b^2(m-c)(m-a)(a-c)+c^2(m-a)(m-b)(b-a)$.

### Analysis

Observe the form of factorization and find that it is a bit like the form on the right side of the Lagrangian formula of $f(x)=x^2$. Consider using the Lagrangian formula.

Note that by the Lagrangian interpolation method, let $a, b, c$ be the abscissa of a given point $x_1, x_2, x_3$, then the ordinate of the given point $y_1, y_2, y_3$ is $f( a),f(b),f(c)$, $y=f(x)=x^2$ are the functions on the left side of the Lagrange formula, then

$f(x)=f(a)\dfrac{x-b}{a-b}\cdot\dfrac{x-c}{a-c}+f(b)\dfrac{x-a}{b-a}\cdot\dfrac{x-c}{b-c}+f(c)\dfrac{x-a}{c-a}\cdot\dfrac{x-b}{c-b}.$

So there is

$f(x)(a-b)(a-c)(b-c)=f(a)(x-b)(x-c)(b-c)+f(b)(x-a)(x-c)(c-a)+f(c)(x-a)(x-b)(a-b).$

In the above formula, let $x=m$ to get

$\texttt{Original}=m^2(a-b)(a-c)(b-c).$

The following is an example of using the Lagrange formula to prove the effect of algebraic identities by comparing the coefficients of a certain term.

### Example

(Olympic classics and algebra problems in Olympic mathematics) $^2$ Let $a,b,c$ be the three sides of non-isosceles $\triangle ABC$, and $S$ is the area, verify: $\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)}>2\times \sqrt{\sqrt{3}^3}\sqrt{S}$.

### Analysis

First deal with the denominators of the fractions on the left side of the inequality, and observe that it is in the form of an interpolation basis function, so we can construct a function so that the elements on the right side of the Lagrangian formula are the content on the left side of the inequality sign, so we consider constructing a quadratic function . Furthermore, it can be found that the coefficient of $x^2$ in the Lagrangian interpolation representation of the quadratic function $y=f(x)=x^3-(x-a)(x-b)(x-c)$ is exactly on the left side of the inequality sign Formula, which is equal to $a+b+c$, so the Helen formula is considered. The result can be obtained by using Helen's formula.

### Proof

Consider the following function $y=f(x)=x^3-(x-a)(x-b)(x-c)$, consider setting $a,b,c$ as the abscissa of a given point $x_1,x_2,x_3$ , The ordinate $y_1, y_2, y_3$ of the given point are respectively $a^3, b^3, c^3$. Consider the Lagrangian representation of the function, that is, the sum of all the above $g_i(x)$, we can get

\begin{aligned} y=f(x) & =x^3-(x-a)(x-b)(x-c), \\ & =f(a)\dfrac{x-b}{a-b}\cdot\dfrac{x-c}{a-c}+f(b)\dfrac{x-a}{b-a}\cdot\dfrac{x-c}{b-c}+f(c)\dfrac{x-b}{c-b}\cdot\dfrac{x-a}{c-a}, \\ & =a^3\dfrac{x-b}{a-b}\cdot\dfrac{x-c}{a-c}+b^3\dfrac{x-a}{b-a}\cdot\dfrac{x-c}{b-c}+c^3\dfrac{x-b}{c-b}\cdot\dfrac{x-a}{c-a}. \end{aligned}

Compare the coefficient of $x^2$ to get

$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)}=a+b+c.$

So the original question is transformed into proof $a+b+c<2\times \sqrt{\sqrt{3}^3}\sqrt{S}$. The following is a geometric problem, the proof is omitted.

The following is an example of using the Lagrangian formula to find the formula of $a_i$ for any $n$ given $a_i$.

### Example

Given that $a_1=14,a_2=58,a_3=166,a_4=368$, for $n=1,2,3,4$, try the polynomial with the highest degree of 3 containing $n$ to express $a_n$ .

Consider setting the general term formula as $a_n=f(n)$, then $f(1)=14, f(2)=58, f(3)=166, f(4)=368$.

The following regards $1,2,3,4$ as the abscissa of a given point $x_1,\dots,x_4$, and $14,58,166,368$ as the ordinate of a given point $y_1,\cdots,y_4$, then

\begin{aligned} a_n & = 14\times\dfrac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)}+\cdots+368\times \dfrac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}, \\ & = 5x^3+2x^2+3x+4. \end{aligned}

The following example is about discussing the range of a certain point value of a polynomial.

### Example

Given the function $f(x)=x^3+px^2+qx+r$, where $p,q,r$ are real numbers. Proof: One of the $|f(1)|,|f(2)|,|f(3)|,|f(4)|$ must be no less than $\dfrac{3}{4}$.

### Analysis

Consider the Lagrangian formula of the polynomial function, take $x_1, x_2, x_3, x_4$ as $1,2,3,4$ respectively; take $y_1, y_2, y_3, y_4$ respectively to the value of the corresponding function, and finally Then consider the coefficient of $x^3$.

### Proof

Take $x_1,x_2,x_3,x_4$ as $1,2,3,4$ respectively; take $y_1,y_2,y_3,y_4$ respectively as the value of the corresponding function, which is obtained by Lagrangian formula

\begin{aligned} f(x) & =f(1) \dfrac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)}+\cdots+ f(3) \dfrac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}+f(4)\dfrac{(x -1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}, \\ & =\dfrac{-1}{6} f(1)\left(x^{3}-9 x^{2} + 26 x-24 \right) + \dfrac{1}{2} f(2 )\left(x^{3}-8 x^{2} + 19 x-12 \right)-\dfrac{1}{2} f(3)\left(x^{3}-7 x^{ 2} + 14 x-8 \right) + \cdots, \\\ & =\left(-\dfrac{f(1)}{6}+\dfrac{f(2)}{2}-\dfrac{f(3)}{2}+\dfrac{f(4)} {6}\right)x^3+\cdots. \\ \end{aligned}

Comparing the coefficients around $x^3$ in the above formula, we can get

\begin{aligned} 1 & =-\dfrac{f(1)}{6}+\dfrac{f(2)}{2}-\dfrac{f(3)}{2}+\dfrac{f(4)}{6 }, \\ & \leq \dfrac{1}{6}|f(1)|+\dfrac{1}{2}|f(2)|+\dfrac{1}{2}|f(3)|+\dfrac {1}{6}|f(4)|, \\ & \leq \left(\dfrac{1}{6}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{6}\right)\max\left\{ |f(1)|,|f(2)|,|f(3)|,|f(4)|\right\}, \\ & =\dfrac{4}{3}\max\left\{|f(1)|,|f(2)|,|f(3)|,|f(4)|\right\}. \end{aligned}

So

$\max\left\{|f(1)|,|f(2)|,|f(3)|,|f(4)|\right\}\ge \dfrac{3}{4}.$

The following example is still about the range of a point value of the polynomial.

### Example

The known function $f(x)=ax^3+bx^2+cx+d$, if $p\leq f(-1)\leq q,p\leq f(1)\leq r,q\leq f(2)\leq s,t\leq f(3)\leq p$, use $p,q,r,s,t$ to denote the value range of $f(4)$.

### Analysis

Considering that for the cubic polynomial function, $4$ corresponding value ranges are given, and the conditions required for Lagrangian interpolation are met, you can take $x_1=-1,x_2=1,x_3=2,x_4=3$ , So we can use Lagrangian interpolation to express $f(4)$ (see the question), and then according to the value range of the four function values, we can get that when $x=4$, The value range of $y$.

Take $x_i=-1,1,2,3$, which can be obtained by Lagrangian formula

\begin{aligned} f(4) & =\dfrac{(4-1)(4-2)(4-3)}{(-1-1)(-1-2)(-1-3)}f(-1)+\dfrac{(4+1)(4-2)(4-3)}{(1+1)(1-2)(1-3)}f(1) \\ & +\dfrac{(4+1)(4-1)(4-3)}{(2+1)(2-1)(2-3)}f(2)+\dfrac{(4+1)(4-1)(4-2)}{(3+1)(3-1)(3-2)}f(3), \\ & =-\dfrac{1}{4}f(-1)+\dfrac{5}{2}f(1)-5f(2)+\dfrac{5}{4}f(3). \end{aligned}

So $\dfrac{9}{4}p-5q+\dfrac{5}{4}t\leq f(4)\leq -\dfrac{1}{4}q+\dfrac{5}{2}r- 5s+\dfrac{5}{4}p$.

## References

[1] OI Wiki. Lagrangian interpolation [EB/OL].

[2] Shen Wenxuan, Zhang Yao, Leng Gangsong. Olympiad Classics · Algebraic Problems in Olympic Mathematics [M]. Changsha: Hunan Normal University Press, 2009.5: 322.

If you have some more applications of the Lagrange Interpolation in Informatics or Mathematics, please comment here and make this post better.

Thanks to ghoshsai5000 for the two sample questions!

I would like to share two CodeForces problems which use Interpolation in their solutions

• +150

 » 11 months ago, # |   +1 I never expected that I'll see lagrangian interpolation anywhere other than my numerical methods course.
•  » » 11 months ago, # ^ |   0 What? I've seen a couple of problems where lagrangian interpolation was the intended solution in codeforces.
•  » » » 11 months ago, # ^ |   +12 XD..... you're literally comparing my experience with yours
•  » » » » 11 months ago, # ^ |   +9 Are you telling me you don't browse through random editorials? https://codeforces.com/problemset/problem/622/F I saw this problem when I wasn't even div1. AlsoIf you can't tell this is half joking.
 » 11 months ago, # |   0 Auto comment: topic has been updated by luosw (previous revision, new revision, compare).
 » 11 months ago, # | ← Rev. 2 →   +11 Good editorial, added to favorite <3But how the hell can this blog is #basic math :<
 » 11 months ago, # | ← Rev. 2 →   0 I would like to share two CodeForces problems which use Interpolation in their solutions
 » 11 months ago, # |   0 Auto comment: topic has been updated by luosw (previous revision, new revision, compare).
 » 8 months ago, # |   +2 Can you tell any real-life application of Lagrange interpolation formula?
•  » » 8 months ago, # ^ |   +24 Who cares about real life nowadays
•  » » » 8 months ago, # ^ |   -18 actually I am an undergraduate student from NIT Raipur , India and I have to write a term paper on application of Lagrange Interpolation. But I am not getting anything to write on this topic. Can anyone help me ?
 » 8 months ago, # |   -9 bores me a lot.