# Essays/PolynomialsIntersection

## Find the intersection of polynomials

### Two polynomials

Given two polynomials, for example:

 $f(x)= 3 + x$ (a) $f(x)= 1 + 2x + x^2$ (b)

how can we find their intersection?

The coefficients of the above polynomials from lowest to highest order are:

```a=: 3 1
b=: 1 2 1
```

The dyadic form of the J primitive p. (Polynomial) evaluates a polynomial of order #x with coefficients x for the argument y. So $f(-2)$ for polynomial b is

```   1 2 1 p. _2
1
```
 plot _3 3;'a p. y ` b p. y'

We can plot these polynomials as follows:

```   load 'plot'
plot _3 3;'3 1 p. y ` 1 2 1 p. y'
NB. or
plot _3 3;'a p. y ` b p. y'
```

From the plot we can see that the two polynomials intersect when $x$ is _2 or 1. How can we get that result using J?

This is equivalent to finding the roots (the values for $x$ where a polynomial is zero) of a polynomial formed by subtracting polynomial a from polynomial b.

Firstly subtract one polynomial from another:

```   a (-/@,:) b
2 _1 _1
```

i.e.    $f(x)= 2 - x - x^2$      (c)

Then find the roots of the resulting polynomial c using the monadic form of the J primitive p. (Roots):

```   p. 2 _1 _1
┌──┬────┐
│_1│_2 1│
└──┴────┘
```

The roots are contained in the 2nd box (the first contains the multiplier) so we can put these ideas all together to give a vector of the $x$ values where two polynomials intersect:

```   findIntersect=: 1 {:: [: p. -/@,:

a findIntersect b
_2 1
```

Where the roots of the polynomial formed by subtraction are complex, findIntersect will return the complex roots.

```   6 2 1 findIntersect 1 2
0j2.236068 0j_2.236068
```

If we only wished to return real roots we could extend findIntersect as follows:

```   6 2 1 (#~ (= +))@findIntersect 1 2
```

### Two or more polynomials

The examples given above are probably the best way to find the intersection of 2 polynomials. If you want the common intersection of several, here is a less-succinct method.

You can test whether $f_0, f_1, ...$ are all equal by forming the sum of $(f_i - f_j)^2$ and setting it to zero.

```ppr   =: +//.@(*/)  NB. polynomial product
pdiff =: -/@,:      NB. polynomial difference
pps   =: ppr~      NB. polynomial square

comb=: 4 : 0
k=. i.>:d=.y-x
z=. (d\$<i.0 0),<i.1 0
for. i.x do. z=. k ,.&.> ,&.>/\. >:&.> z end.
; z
)

NB. findIntersectM <list of boxes of coefficients>
NB. returns x-coordinates of common intersection points
findIntersectM=:3 : 0
a=. >y
c=. 2 comb #a
~. (#~ (= +)) 1{:: p. +/ pps@pdiff /"2  c { a
)
```

For example to find the intersection of the polynomials:

 $f(x)= x^2$ (d) $f(x)= 2 - x^2$ (e) $f(x)= 1$ (f)
```d=: 0 0 1
e=: 2 0 _1
f=: 1
```
 plot _3 3;'d p. y`e p. y`f p. y'

We can plot these as follows.

```   require 'plot'
plot _3 3;'d p. y ` e p. y ` f p. y'
```

By inspection these polynomials intersect when $x$ is 1 or _1.

Given a list of boxed coefficients for each polynomial we can find the $x$ coordinates where they all intersect.

```   ]r=: findIntersectM d;e;f
1 _1
```

We can evaluate each polynomial at those $x$ values to show that they all have the same $f(x)$.

```   d p. r
1 1
e p. r
1 1
f p. r
1 1
```

## Contributions

This essay was compiled by Ric Sherlock from posts in this forum thread by John Randall, Raul Miller and Henry Rich.