Locally invert a Polynomial

Question

Challenge

Given a polynomial \$p\$ with real coefficients of order \1ドル\$ and degree \$n\$, find another polynomial \$q\$ of degree at most \$n\$ such that \$(p∘q)(X) = p(q(X)) \equiv X \mod X^{n+1}\$, or in other words such that \$p(q(X)) = X + h(X)\$ where \$h\$ is an arbitrary polynomial with \$ord(h) \geqslant n+1\$. The polynomial \$q\$ is uniquely determined by \$p\$.

For a polynomial \$p(X) = a_nX^n + a_{n+1}X^{n+1} + ... + a_m X^m\$ where \$n \leqslant m\$ and \$a_n ≠ 0\$,\$a_m ≠ 0\$, we say \$n\$ is the order of \$p\$ and \$m\$ is the degree of \$p\$.

Simplification: You can assume that \$p\$ has integer coefficients, and \$a_1 = 1\$ (so \$p(X) = X + \text{[some integral polynomial of order 2]}\$). In this case \$q\$ has integral coefficients too.

The purpose of this simplification is to avoid the issues with floating point numbers. There is however a non-integral example for illustration purposes.

Examples

• Consider the Taylor series of \$\exp(x)-1 = x + x^2/2 + x^3/6 + x^4/24 + ...\$ and \$\ln(x+1) = x - x^2/2 + x^3/3 - x^4/4 + ... \$ then obviously \$\ln(\exp(x)-1+1)= x\$. If we just consider the Taylor polynomials of degree 4 of those two functions we get with the notation from below (see testcases) \$p = [-1/4,1/3,-1/2,1,0]\$ and \$q = [1/24, 1/6, 1/2, 1,0]\$ and \$(p∘q)(X) \equiv X \mod X^5\$

• Consider the polynomial \$p(X) = X + X^2 + X^3 + X^4\$. Then for \$q(X) = X - X^2 + X^3 - X^4\$ we get

$$(p \circ q)(X) = p(q(X)) = X - 2X^5 + 3X^6 - 10X^7 +...+ X^{16} \equiv X \mod X^5$$

Testcases

Here the input and output polynomials are written as lists of coefficients (with the coefficient of the highest degree monomial first, the constant term last):

p = [4,3,2,0]; q=[0.3125,-.375,0.5,0]

Integral Testcases:

p = [1,0]; q = [1,0]
p = [9,8,7,6,5,4,3,2,1,0]; q = [4862,-1430,429,-132,42,-14,5,-2,1,0]
p = [-1,3,-3,1,0]; q = [91,15,3,1,0]

Answer 1

Python 2 + sympy, 128 bytes

We locally invert the polynomial by assuming that q(x) = x, composing it with p, checking the coefficient for x², and subtracting it from q. Let's say the coefficient was 4, then the new polynomial becomes q(x) = x - 4x². We then again compose this with p, but look up the coefficient for x³. Etc...

from sympy import*
i=input()
p=Poly(i,var('x'));q=p*0+x
n=2
for _ in i[2:]:q-=compose(p,q).nth(n)*x**n;n+=1
print q.all_coeffs()

Answer 2

Mathematica, 45 bytes

Normal@InverseSeries[#+O@x^(#~Exponent~x+1)]&

Yeah, Mathematica has a builtin for that....

Unnamed function taking as input a polynomial in the variable x, such as -x^4+3x^3-3x^2+x for the last test case, and returning a polynomial with similar syntax, such as x+3x^2+15x^3+91x^4 for the last test case.

#+O@x^(#~Exponent~x+1) turns the input # into a power series object, truncated at the degree of #; InverseSeries does what it says; and Normal turns the resulting truncated power series back into a polynomial. (We could save those initial 7 bytes if an answer in the form x+3x^2+15x^3+91x^4+O[x]^5 were acceptable. Indeed, if that were an acceptable format for both input and output, then InverseSeries alone would be a 13-byte solution.)

Answer 3

JavaScript (ES6), 138 bytes

a=>a.reduce((r,_,i)=>[...r,i<2?i:a.map(l=>c=p.map((m,j)=>(r.map((n,k)=>p[k+=j]=m*n+(p[k]||0)),m*l+(c[j]||0)),p=[]),c=[],p=[1])&&-c[i]],[])

Port of @orlp's answer. I/O is in the form of arrays of coefficients in reverse order i.e. the first two coefficients are always 0 and 1.

Answer 4

Pari/GP, 29 bytes

p->Pol(serreverse(p+O(x^#p)))

Try it online!