# Summing roots of polynomials¶

Let’s suppose we are given a univariate polynomial $$f(z)$$ and a univariate rational function $$g(z)$$, and we wish to compute:

$g(r_1) + g(r_2) + \ldots + g(r_n)$

where $$r_i$$ for $$i = 1 \ldots n$$ are the roots of $$f$$ (i.e. $$f(r_i) = 0$$).

In theory this is a very simple task. We just have to compute roots of $$f$$, using the roots() function, substitute those roots for $$z$$ in $$g$$ and add resulting values together.

Let’s consider the following polynomial and rational function:

>>> f = z**5 + z + 3
>>> f
5
z  + z + 3

>>> g = 1/z
>>> g
1
─
z


Following the trivial approach, let’s compute the roots of $$f$$:

>>> roots(f)
{}


We got a very unfortunate result: no roots! By the fundamental theorem of algebra we should get five, possibly complex, roots, including multiplicities. Unfortunately, there is no way to express roots in terms of radicals of some polynomials of degree five and higher. For certain instances of polynomials of this kind it may be possible to compute their roots (e.g. roots() recognizes cyclotomic polynomials of high degree), but in general we will most likely be unlucky.

Instead, we could switch to numerical root finding algorithms and compute approximations of roots of $$f$$ and proceed with summation of roots. This can be done by using nroots():

>>> R = nroots(f)

>>> for ri, r in zip(numbered_symbols('r'), R):
...     pprint(Eq(ri, r))
...
r₀ = -1.13299756588507
r₁ = -0.47538075666955 - 1.12970172509541⋅ⅈ
r₂ = -0.47538075666955 + 1.12970172509541⋅ⅈ
r₃ = 1.04187953961208 - 0.822870338109958⋅ⅈ
r₄ = 1.04187953961208 + 0.822870338109958⋅ⅈ


We can substitute those roots for $$z$$ in $$g$$ and add together:

>>> sum([ g.subs(z, r) for r in R ]).evalf(chop=True)
-0.333333333333332


It was necessary to evaluate this sum with evalf(), because otherwise we would get an unsimplified result. The additional parameter chop=True was necessary to remove a tiny and insignificant imaginary part. Next we can use nsimplify() to get an exact result from numerical approximation:

>>> nsimplify(_)
-1/3


Is this result correct? The best way is to figure out a purely symbolic method that doesn’t require computing roots of $$f$$. In SymPy it possible to represent a root of a univariate polynomial with rational coefficients using RootOf:

>>> RootOf(f, 0)
⎛ 5           ⎞
RootOf⎝z  + z + 3, 0⎠

>>> _.evalf()
-1.13299756588507


We can obtain all roots using list comprehensions:

>>> R = [ RootOf(f, i) for i in xrange(degree(f)) ]

>>> for r in R:
...     pprint(r)
...
⎛ 5           ⎞
RootOf⎝z  + z + 3, 0⎠
⎛ 5           ⎞
RootOf⎝z  + z + 3, 1⎠
⎛ 5           ⎞
RootOf⎝z  + z + 3, 2⎠
⎛ 5           ⎞
RootOf⎝z  + z + 3, 3⎠
⎛ 5           ⎞
RootOf⎝z  + z + 3, 4⎠


Alternatively we can use Poly(f).all_roots() which gives the same result, but is much faster when $$f$$ is a composite polynomial, because the preprocessing step in RootOf is executed only once.

Unfortunately we can’t get anywhere from here, because SymPy is not yet capable of simplifying expressions with RootOf:

>>> G = sum([ g.subs(z, r) for r in R ])
True

>>> _ = simplify(G)
True


We can, however, evaluate sums of RootOf‘s using evalf():

>>> G.evalf()
-0.333333333333333

>>> nsimplify(_)
-1/3


which gave us the same result as before. The difference is that now numerical approximations of roots of $$f$$ were computed using a hybrid symbolic–numeric method, where first disjoint isolating intervals (rectangles) where computed for all roots of $$f$$ and then a numerical root finding algorithm was used in each interval.

Let’s approach this problem differently, using a purely symbolic approach. We know that a polynomial of degree $$n$$ has exactly $$n$$ complex roots, counting multiplicities. In our case $$f$$ has five roots:

>>> R = var('r:5')
>>> R
(r₀, r₁, r₂, r₃, r₄)


Let’s now substitute those “roots” for $$z$$ in $$g$$:

>>> [ g.subs(z, r) for r in R ]
⎡1   1   1   1   1 ⎤
⎢──, ──, ──, ──, ──⎥
⎣r₀  r₁  r₂  r₃  r₄⎦


>>> sum(_)
1    1    1    1    1
── + ── + ── + ── + ──
r₄   r₃   r₂   r₁   r₀


We got a sum of simple rational functions. The next step is to put those rational functions over a common denominator:

>>> G = together(_)
>>> G
r₀⋅r₁⋅r₂⋅r₃ + r₀⋅r₁⋅r₂⋅r₄ + r₀⋅r₁⋅r₃⋅r₄ + r₀⋅r₂⋅r₃⋅r₄ + r₁⋅r₂⋅r₃⋅r₄
───────────────────────────────────────────────────────────────────
r₀⋅r₁⋅r₂⋅r₃⋅r₄


We got very peculiar numerator and denominator, which are very specific functions of roots of $$f$$ (symmetric polynomials). Polynomials of this kind can be generated using viete():

>>> V = viete(f, R, z)

>>> for lhs, rhs in V:
....     pprint(Eq(lhs, rhs))
....
r₀ + r₁ + r₂ + r₃ + r₄ = 0
r₀⋅r₁ + r₀⋅r₂ + r₀⋅r₃ + r₀⋅r₄ + r₁⋅r₂ + r₁⋅r₃ + r₁⋅r₄ + r₂⋅r₃ + r₂⋅r₄ + r₃⋅r₄ = 0
r₀⋅r₁⋅r₂ + r₀⋅r₁⋅r₃ + r₀⋅r₁⋅r₄ + r₀⋅r₂⋅r₃ + r₀⋅r₂⋅r₄ + r₀⋅r₃⋅r₄ + r₁⋅r₂⋅r₃ + r₁⋅r₂⋅r₄ + r₁⋅r₃⋅r₄ + r₂⋅r₃⋅r₄ = 0
r₀⋅r₁⋅r₂⋅r₃ + r₀⋅r₁⋅r₂⋅r₄ + r₀⋅r₁⋅r₃⋅r₄ + r₀⋅r₂⋅r₃⋅r₄ + r₁⋅r₂⋅r₃⋅r₄ = 1
r₀⋅r₁⋅r₂⋅r₃⋅r₄ = -3


Viete formulas show the relationship between roots of a polynomial and its coefficients:

$V_{i-1} = (-1)^i \frac{a_{n-i}}{a_n}$

where $$f(z)=a_nz^n + a_{n-1}z^{n-1} + \ldots + a_1z + a_0$$ and $$i = 1 \ldots n$$. To obtain the final result it sufficient to take $$V_3$$ and $$V_4$$ and substitute in $$G$$:

>>> numer(G).subs(*V[3])/denom(G).subs(*V[4])
-1/3


Or we could simply use G.subs(V), but due to a bug in SymPy (#2552) this doesn’t work as expected, leaving the denominator unchanged.

We obtained the same result as before, just this time using purely symbolic techniques. This simple procedure can be extended to form an algorithm for solving the root summation problem in the general setup. SymPy implements this algorithm in RootSum:

>>> RootSum(f, Lambda(z, g))
-1/3


The choice of $$g$$ allowed us to recognize Viete formulas very easily in $$G$$, but is this the case also for more complicated rational functions? Let’s modify $$g$$ a little:

>>> g = 1/(z + 2)
1
─────
z + 2


Now let’s repeat the procedure for the new $$g$$:

>>> G = together(sum([ g.subs(z, r) for r in R ]))

>>> p = expand(numer(G))
>>> q = expand(denom(G))

>>> p
r₀⋅r₁⋅r₂⋅r₃ + r₀⋅r₁⋅r₂⋅r₄ + 4⋅r₀⋅r₁⋅r₂ + r₀⋅r₁⋅r₃⋅r₄ + 4⋅r₀⋅r₁⋅r₃ + 4⋅r₀⋅r₁⋅r₄ + 12⋅r₀⋅r₁ + r₀⋅r₂⋅r₃⋅r₄ + \
4⋅r₀⋅r₂⋅r₃ + 4⋅r₀⋅r₂⋅r₄ + 12⋅r₀⋅r₂ + 4⋅r₀⋅r₃⋅r₄ + 12⋅r₀⋅r₃ + 12⋅r₀⋅r₄ + 32⋅r₀ + r₁⋅r₂⋅r₃⋅r₄ + 4⋅r₁⋅r₂⋅r₃ + \
4⋅r₁⋅r₂⋅r₄ + 12⋅r₁⋅r₂ + 4⋅r₁⋅r₃⋅r₄ + 12⋅r₁⋅r₃ + 12⋅r₁⋅r₄ + 32⋅r₁ + 4⋅r₂⋅r₃⋅r₄ + 12⋅r₂⋅r₃ + 12⋅r₂⋅r₄ + 32⋅r₂ + \
12⋅r₃⋅r₄ + 32⋅r₃ + 32⋅r₄ + 80

>>> q
r₀⋅r₁⋅r₂⋅r₃⋅r₄ + 2⋅r₀⋅r₁⋅r₂⋅r₃ + 2⋅r₀⋅r₁⋅r₂⋅r₄ + 4⋅r₀⋅r₁⋅r₂ + 2⋅r₀⋅r₁⋅r₃⋅r₄ + 4⋅r₀⋅r₁⋅r₃ + 4⋅r₀⋅r₁⋅r₄ + \
8⋅r₀⋅r₁ + 2⋅r₀⋅r₂⋅r₃⋅r₄ + 4⋅r₀⋅r₂⋅r₃ + 4⋅r₀⋅r₂⋅r₄ + 8⋅r₀⋅r₂ + 4⋅r₀⋅r₃⋅r₄ + 8⋅r₀⋅r₃ + 8⋅r₀⋅r₄ + 16⋅r₀ + \
2⋅r₁⋅r₂⋅r₃⋅r₄ + 4⋅r₁⋅r₂⋅r₃ + 4⋅r₁⋅r₂⋅r₄ + 8⋅r₁⋅r₂ + 4⋅r₁⋅r₃⋅r₄ + 8⋅r₁⋅r₃ + 8⋅r₁⋅r₄ + 16⋅r₁ + 4⋅r₂⋅r₃⋅r₄ + \
8⋅r₂⋅r₃ + 8⋅r₂⋅r₄ + 16⋅r₂ + 8⋅r₃⋅r₄ + 16⋅r₃ + 16⋅r₄ + 32


This doesn’t look that familiar anymore. Let’s try to apply Viete formulas to the numerator and denominator:

>>> p.subs(V).has(*R)
True
>>> q.subs(V).has(*R)
True


We weren’t able to get rid of the symbolic roots of $$f$$. We can, however, try to rewrite $$p$$ and $$q$$ as polynomials in elementary symmetric polynomials. This procedure is called symmetric reduction, and an algorithm for this is implemented in symmetrize():

>>> (P, Q), mapping = symmetrize((p, q), R, formal=True)

>>> P
(32⋅s₁ + 12⋅s₂ + 4⋅s₃ + s₄ + 80, 0)
>>> Q
(16⋅s₁ + 8⋅s₂ + 4⋅s₃ + 2⋅s₄ + s₅ + 32, 0)

>>> for s, poly in mapping:
...     pprint(Eq(s, poly))
...
s₁ = r₀ + r₁ + r₂ + r₃ + r₄
s₂ = r₀⋅r₁ + r₀⋅r₂ + r₀⋅r₃ + r₀⋅r₄ + r₁⋅r₂ + r₁⋅r₃ + r₁⋅r₄ + r₂⋅r₃ + r₂⋅r₄ + r₃⋅r₄
s₃ = r₀⋅r₁⋅r₂ + r₀⋅r₁⋅r₃ + r₀⋅r₁⋅r₄ + r₀⋅r₂⋅r₃ + r₀⋅r₂⋅r₄ + r₀⋅r₃⋅r₄ + r₁⋅r₂⋅r₃ + r₁⋅r₂⋅r₄ + r₁⋅r₃⋅r₄ + r₂⋅r₃⋅r₄
s₄ = r₀⋅r₁⋅r₂⋅r₃ + r₀⋅r₁⋅r₂⋅r₄ + r₀⋅r₁⋅r₃⋅r₄ + r₀⋅r₂⋅r₃⋅r₄ + r₁⋅r₂⋅r₃⋅r₄
s₅ = r₀⋅r₁⋅r₂⋅r₃⋅r₄


Here we performed the formal simultaneous symmetric reduction of the polynomials $$p$$ and $$q$$, obtaining their representation in terms of elementary symmetric polynomials, non-symmetric remainders, and elementary symmetric polynomials. Remainders are always zero for symmetric inputs.

We can zip this mapping and Viete formulas together, obtaining:

>>> [ (s, c) for (s, _), (_, c) in zip(mapping, V) ]
[(s₁, 0), (s₂, 0), (s₃, 0), (s₄, 1), (s₅, -3)]


Now we can take head of P and Q and perform substitution:

>>> P[0].subs(_)/Q[0].subs(_)
81
──
31


Let’s verify this result using RootSum:

>>> RootSum(f, Lambda(z, g))
81
──
31


The numerical approach also works in this case:

>>> sum([ g.subs(z, r) for r in Poly(f).all_roots() ]).evalf()
2.61290322580645

>>> nsimplify(_)
81
──
31


1. Repeat this procedure for:
• $$f = z^5 + z + a$$ and $$g = \frac{1}{z + 1}$$
• $$f = z^5 + z + a$$ and $$g = \frac{1}{z + b}$$

(solution)

1. Can this or a similar procedure be used with other classes of expressions than rational functions? If so, what kind of expressions can be allowed?