Lagrange polynomial, Free Linear space and Dual basis

Give a sequence like $a_1=1,a_2=5,a_3=1145,a_4=31415$$a_1=1,a_2=5,a_3=1145,a_4=31415$

How could we find a formula $f(i)=a_i$$f(i)=a_i$?

Maybe some of the readers here have Lagrange polynomials before; this method can solve this problem directly.

But what is the background of the Lagrange polynomial? How could we understand it?

Let me introduce sth to you.

We need some knowledge about Free Linear Space.

What is ''Free''? Free is like, give you a set, and you Free it a structure, just like touching a stone and turning it to gold.

For free linear space, it gives you a set, for example, $S=\{1,2,3,4\}$$S=\{1,2,3,4\}$

And you can consider all the functions from $S$$S$ to $\mathbb{F}$$\mathbb F$; for instance, $\mathbb{F}=\mathbb{R}$$\mathbb F=\mathbb R$, you get a linear space!

$(f+g)(x)=f(x)+g(x),(\lambda f)(x)=\lambda f(x)$$(f+g)(x)=f(x)+g(x),(\lambda f)(x)=\lambda f(x)$

You can check that it is a linear space $F(S)$$F(S)$; by the way, putting an $a\in S$$a\in S$ in to function gives a Linear map!

$a(f+g):=(f+g)(a)=f(a)+g(a)=a(f)+a(g),a(\lambda f):=(\lambda f)(a)=\lambda f(a)=\lambda a(f)$$a(f+g):=(f+g)(a)=f(a)+g(a)=a(f)+a(g),a(\lambda f):=(\lambda f)(a)=\lambda f(a)=\lambda a(f)$

And , what is the basis of this linear space?

Actually you can consider those functions like $\begin{array}{c}{f}_i(j)=\{\begin{array}{l}1,i=j\\ 0,i\ne j\end{array}\end{array}$$f_i(j)=\begin{cases}1,i=j\\0,i\ne j\end{cases}$

And a function $f:S\to \mathbb{F},f(1)=a,f(2)=b,f(3)=c,f(d)=4$$f:S\to\mathbb F,f(1)=a,f(2)=b,f(3)=c,f(d)=4$

It can be written as $a{f}_1+b{f}_2+c{f}_3+d{f}_4$$a f_1+bf_2+cf_3+d f_4$

Thus $f_i$$f_i$ is a basis for this linear space! Therefore $\dim F(S)=4$$\dim F(S)=4$

And in general, we have $\dim F(S)=|S|$$\dim F(S)=|S|$, $|S|$$|S|$ is the cardinality of $S$$S$

And you can try to consider the isomorphism between $F(S)\cong {\mathbb{F}}^{|S|}$$F(S)\cong\mathbb F^{|S|}$ by $f_i\mapsto e_i$$f_i\mapsto e_i$

And now the question is, how to construct the $f_i$$f_i$ ? I mean, give it a formula.

In the first step, we can consider $\begin{array}{c}{F}_i(j)=\{\begin{array}{l}\ne 0,i=j\\ 0,i=j\end{array}\end{array}$$F_i(j)=\begin{cases}\ne 0,i=j\\0,i=j\end{cases}$

And a natural way is to consider $\begin{array}{c}\prod _{j\ne i}(x-j)\end{array}$$\prod_{j\ne i}(x-j)\\$

The second step is, to let $f_i=\lambda F_i$$f_i=\lambda F_i$ and let $F_i(j)=1,i=j$$F_i(j)=1,i=j$

We know that $\begin{array}{c}{F}_i(i)=\prod _{j\ne i}(i-j)\end{array}$$F_i(i)=\prod_{j\ne i}(i-j)\\$

So $\begin{array}{c}{f}_i(j)=\frac{F_i(j)}{F_i(i)}=\frac{\prod _{j\ne i}(x-j)}{\prod _{j\ne i}(i-j)}\end{array}$$f_i(j)= \frac{F_i(j)}{F_i(i)}=\frac{\prod_{j\ne i}(x-j)}{\prod_{j\ne i}(i-j)}\\$

And maybe some readers who have read Math Essays: Dual basis and Taylor series (wuyulanliulongblog.blogspot.com)

Will think $\begin{array}{c}{f}_i(j)=\{\begin{array}{l}1,i=j\\ 0,i\ne j\end{array}\end{array}$$f_i(j)=\begin{cases}1,i=j\\0,i\ne j\end{cases}$ is just like dual basis,

YES!

Review that putting an $a\in S$$a\in S$ in to function gives a Linear map!

$a(f+g):=(f+g)(a)=f(a)+g(a)=a(f)+a(g),a(\lambda f):=(\lambda f)(a)=\lambda f(a)=\lambda a(f)$$a(f+g):=(f+g)(a)=f(a)+g(a)=a(f)+a(g),a(\lambda f):=(\lambda f)(a)=\lambda f(a)=\lambda a(f)$

And actually, it is linear functional, so it is the element of dual space of $F(S)$$F(S)$.

Give an function $g(x)\in F(S)$$g(x)\in F(S)$ and consider $g(i)$$g(i)$ will give the coordinate under the basis $f_i(x)$$f_i(x)$

So for any function $g\in F(S)$$g\in F(S)$ , it always can be written as $\begin{array}{c}g(x)=\sum _{i=1}^{n}g(i)f_i(x)\end{array}$$g(x)=\sum_{i=1}^n g(i) f_i(x)\\$