1

I have found an interesting property of polynomial function

When I have multiple of 2nd degree polynomial functions. I could use method of 3 points to solve for $a,b,c$

$$a_nx^2 + b_nx + c_n = p_{nx}$$ $$a_n + b_n + c_n = p_{n1}$$ $$c_n = p_{n0}$$

But then. If I carefully select $n$ number of $p_{nx},p_{n1},p_{n0}$ so that

$$\sum_{i=0}^np_{nx} = \sum_{i=0}^np_{n0} = \sum_{i=0}^np_{n1} = 1$$

It seem like all the point of these polynomial could be sum to 1 for every $x$. Also I think it don't have to be 1, it could be any number just that if the 3 group of points will sum to the same number

For example. I have made these 3 lines. Red Blue and Purple

At $x=0$ is $0,0.4,0.6$

At $x=1$ is all $\frac13$

At $x=\frac12$ is $0,\frac13,\frac23$

The green line is the sum of these function. Which become just linear 1

https://www.desmos.com/calculator/95fpjobdht

enter image description here

I wonder if these property is also true for all $n$ all $p$ and all degree of polynomial with the same construct?

And if it is true why is that? Can we prove it? Are there any name of this property?

Thaina
  • 672

1 Answers1

3

The quadratic polynomials form a three dimensional vector space. $\{1,x,x^2\}$ forms a basis of that space. $1$ is a fine quadratic polynomial that is $0x^2+0x+1$ as an element of the vector space. If you fit a quadratic through three points such that the value of the quadratic is $1$ at all three points you will get the constant $1$ polynomial. Because $1,x,x^2$ are linearly independent, no other quadratic will go through those three points.

Ross Millikan
  • 374,822