Computing all the roots of a pseudopolynomial with one indeterminate exponent Abstract We compute all countably infinitely many complex roots \[ z \] of a pseudo-trionmial \[ z^\alpha + w \cdot z + 1 \] for arbitrary complex numbers \[ \alpha \] and \[ w \] . We combine the use of Mishkov’s formula and Egorychev’s generalization of the Lagrange Inversion Formula with formulae for all \[ \left| n \right| \] roots of the polynomial \[ z^n + w \cdot z + 1 \] for a given integer \[ n \] . We point to ongoing efforts to generalize Mishkov’s formula in order to compute all the roots of pseudopolynomials \[ \sum\limits_{i = 1}^n {w_i \cdot z^{\alpha _i } } \] or any expression involving a finite number of parameters \[ \{ \alpha _i \} _{i = 1}^n \] . Introduction Given a complex-valued function \[ w = g(z) \] of a complex variable \[ z \] with \[ g(0) = 0 \] , which is analytic around 0, the Lagrange Inversion Formula gives an expression for a Maclaurin power series \[ z = f(w) \] with \[ f(0) = 0 \] which is the inverse of a \[ g \] . However, Lagrange does not give us the other (in general) infinitely many inverses \[ \{ f_k (w)\} _{k \in \mathbb{N}} \] of \[ g \] . Let \[ \alpha \] be a constant indeterminate over a number field of characteristic zero. For any complex number \[ z \] the symbol \[ z^\alpha \] means the germ of \[ \exp (\alpha \cdot \ln (z)) = \sum\limits_{n \in \mathbb{N}} {\frac{1} {{n!}}\alpha ^n (\ln (z))^n } \] . Define \[ G(t,\alpha ) \equiv t^\alpha + t + 1 \] . Suppose \[ G(z,\alpha ) = 0 \] . Mishkov Theorem 2.1 p481 If \[ G \] and \[ t \] are scalars, \[ (x_1 (t),...,x_r (t)) \] is an \[ r \] -vector and \[ G(x_1 (t),...,x_r (t)) \] is a composite function, for which all necessary derivatives are defined, \[ D \] is the differential operator with respect to \[ t \] , then \[ D^n G(x_1 (t),...,x_r (t)) = \sum\limits_0 {\sum\limits_1 { \cdots \sum\limits_n {\frac{{n!}} {{\left( {\prod\limits_{i = 1}^n {(i!)^{k_i } } } \right) \cdot \left( {\prod\limits_{i = 1}^n {\prod\limits_{j = 1}^r {q_{i,j} !} } } \right)}} \cdot \frac{{\partial ^k G}} {{\prod\limits_{j = 1}^r {\partial x_j^{p_j } } }} \cdot \Upsilon _q } } } \] where \[ \Upsilon _q = \prod\limits_{i = 1}^n {\prod\limits_{j = 1}^r {\left( {\frac{{d^i x_j }} {{dt^i }}} \right)^{q_{i,j} } } } \] where the respective sum \[ \sum\limits_i {} \] for a given \[ i \in [n] \] is over all nonnegative integer solutions of the Diophantine equation \[ k_i = \sum\limits_{j = 1}^r {q_{i,j} } \] and \[ \sum\limits_0 {} \] is over all nonnegative integer solutions of \[ \sum\limits_{i = 1}^n {i \cdot k_i } = n \] . The order \[ p_j \] of the partial derivative of \[ G \] with respect to \[ x_j \] satisfies \[ p_j = \sum\limits_{i = 1}^n {q_{i,j} } \] (Mishkov (2.3) p483). The total order \[ k \] of the derivative of \[ G \] with respect to all \[ x_j \] s satisfies \[ k = \sum\limits_{j = 1}^r {p_j } = \sum\limits_{i = 1}^n {k_i } \] . In ongoing research we need to extend the domain slightly of \[ q \] to include. Mishkov’s Theorem can be expressed without specifying the \[ k_i \] s nor the \[ p_j \] s. In other words, Corollary 1. Mishkov’s Theorem may be rewritten more succinctly as \[ D^n G(x_1 (t),...,x_r (t)) = \sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 1}^n {(i!)^{k_i } } } \right) \cdot \left( {\prod\limits_{i = 1}^n {\prod\limits_{j = 1}^r {q_{i,j} !} } } \right)}} \cdot \frac{{\partial ^k G}} {{\prod\limits_{j = 1}^r {\partial x_j^{p_j } } }} \cdot \prod\limits_{i = 1}^n {\prod\limits_{j = 1}^r {\left( {\frac{{d^i x_j }} {{dt^i }}} \right)^{q_{i,j} } } } } \] where the sum is over all maps \[ q:[n] \times [r] \to [n]_0 \] which satisfy \[ \sum\limits_{i = 1}^n {i \cdot \sum\limits_{j = 1}^r {q_{i,j} } } = n \] . Corollary 2. Let \[ r = 2 \] , \[ x_1 = \alpha \] , \[ x_2 = z \] in Mishkov’s theorem. Then \[ D^n G(\alpha (t),z(t)) = \] \[ \sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 1}^n {(i!)^{k_i } } } \right) \cdot \left( {\prod\limits_{i = 1}^n {(q_{i,1} !q_{i,2} !)} } \right)}} \cdot \frac{{\partial ^{\sum\limits_{i = 1}^n {q(i,1)} + \sum\limits_{i = 1}^n {q(i,2)} } G}} {{\partial \alpha ^{\sum\limits_{i = 1}^n {q(i,1)} } \partial z^{\sum\limits_{i = 1}^n {q(i,2)} } }} \cdot \prod\limits_{i = 1}^n {\left( {\left( {\frac{{d^i \alpha }} {{dt^i }}} \right)^{q(i,1)} \left( {\frac{{d^i z}} {{dt^i }}} \right)^{q(i,2)} } \right)} } \] where the sum is over all maps \[ q:[n] \times [2] \to [n]_0 \] which satisfy \[ \sum\limits_{i = 1}^n {i \cdot (q_{i,1} + q_{i,2} )} = n \] . Commentary: the other Italian authors Lemma 3. \[ \frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z(\alpha )) = \] \[ \sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 2}^n {(i!)^{q(i)} } } \right) \cdot \left( {(n - \sum\limits_{i = 1}^n {i \cdot q_i } )!} \right)\left( {\prod\limits_{i = 1}^n {(q_i !)} } \right)}} \cdot \frac{{\partial ^{n + \sum\limits_{i = 1}^n {(1 - i) \cdot q(i)} } G}} {{\partial \alpha ^{n - \sum\limits_{i = 1}^n {i \cdot q(i)} } \partial z^{\sum\limits_{i = 1}^n {q(i)} } }} \cdot \prod\limits_{i = 1}^n {\left( {\left( {\frac{{d^i z}} {{d\alpha ^i }}} \right)^{q(i)} } \right)} } \] where we sum over all maps \[ q:[n] \to [n]_0 \] . Proof. Specialize \[ t \to \alpha \] in Corollary 2. Then \[ \frac{{d^i \alpha }} {{dt^i }} \to \frac{{d^i \alpha }} {{d\alpha ^i }} \] and \[ \frac{{d^i z}} {{dt^i }} \to \frac{{d^i z}} {{d\alpha ^i }} \] . So \[ \frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z(\alpha )) = \] \[ \sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 1}^n {(i!)^{k_i } } } \right) \cdot \left( {\prod\limits_{i = 1}^n {(q_{i,1} !q_{i,2} !)} } \right)}} \cdot \frac{{\partial ^{\sum\limits_{i = 1}^n {q(i,1)} + \sum\limits_{i = 1}^n {q(i,2)} } G}} {{\partial \alpha ^{\sum\limits_{i = 1}^n {q(i,1)} } \partial z^{\sum\limits_{i = 1}^n {q(i,2)} } }} \cdot \prod\limits_{i = 1}^n {\left( {\left( {\frac{{d^i \alpha }} {{d\alpha ^i }}} \right)^{q(i,1)} \left( {\frac{{d^i z}} {{d\alpha ^i }}} \right)^{q(i,2)} } \right)} } \] Since \[ i > 1 \Rightarrow \frac{{d^i \alpha }} {{d\alpha ^i }} = 0 \] and since \[ q(i,1) > 0 \Rightarrow 0^{q(i,1)} = 0 \] , we may ignore all maps \[ q \] such that \[ i > 1 \Rightarrow q(i,1) > 0 \] . So the condition \[ \sum\limits_{i = 1}^n {i \cdot (q_{i,1} + q_{i,2} )} = n \] is reduced to \[ q_{1,1} + \sum\limits_{i = 1}^n {i \cdot q_{i,2} } = n \] . Since \[ i = 1 \Rightarrow \frac{{d^i \alpha }} {{d\alpha ^i }} = 1 \] and \[ 1^{q(1,1)} = 1 \] for any value of \[ q(1,1) \] we may rewrite \[ q \] as a map with just one argument, so \[ q:[n] \to [n]_0 \] . Since \[ q(1,1) \] can range over all \[ [n]_0 \] , our condition \[ \sum\limits_{i = 1}^n {i \cdot q_{i,2} } = n - q_{1,1} \] is further simplified to \[ \sum\limits_{i = 1}^n {i \cdot q_i } \] ranging over all \[ [n]_0 \] . So \[ \frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z(\alpha )) = \] \[ \sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 2}^n {(i!)^{q(i)} } } \right) \cdot \left( {(n - \sum\limits_{i = 1}^n {i \cdot q_i } )!} \right)\left( {\prod\limits_{i = 1}^n {(q_i !)} } \right)}} \cdot \frac{{\partial ^{n + \sum\limits_{i = 1}^n {(1 - i) \cdot q(i)} } G}} {{\partial \alpha ^{n - \sum\limits_{i = 1}^n {i \cdot q(i)} } \partial z^{\sum\limits_{i = 1}^n {q(i)} } }} \cdot \prod\limits_{i = 1}^n {\left( {\left( {\frac{{d^i z}} {{d\alpha ^i }}} \right)^{q(i)} } \right)} } \] where we sum over all maps \[ q:[n] \to [n]_0 \] . Q.E.D. Lemma 4. If \[ G(\alpha ,z) = 0 \] , then for each \[ n \in \mathbb{N} \] we have \[ 0 = \sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 2}^n {(i!)^{q(i)} } } \right) \cdot \left( {(n - \sum\limits_{i = 1}^n {i \cdot q_i } )!} \right)\left( {\prod\limits_{i = 1}^n {(q_i !)} } \right)}} \cdot \frac{{\partial ^{n + \sum\limits_{i = 1}^n {(1 - i) \cdot q(i)} } G}} {{\partial \alpha ^{n - \sum\limits_{i = 1}^n {i \cdot q(i)} } \partial z^{\sum\limits_{i = 1}^n {q(i)} } }} \cdot \prod\limits_{i = 1}^n {\left( {\left( {\frac{{d^i z}} {{d\alpha ^i }}} \right)^{q(i)} } \right)} } \] where we sum over all maps \[ q:[n] \to [n]_0 \] . Proof. All total derivatives of \[ G(\alpha ,z) = 0 \] with respect to \[ \alpha \] are zero. The result follows trivially from Lemma 3. Q.E.D. Definition 5. Define the multivariable polynomial \[ H_n (\alpha ,u_1 ,...,u_n ) \equiv \] . \[ \frac{1} {{G_z }}\sum\limits_q {\frac{{n!}} {{\left( {\prod\limits_{i = 2}^n {(i!)^{q(i)} } } \right) \cdot \left( {(n - \sum\limits_{i = 1}^n {i \cdot q_i } )!} \right)\left( {\prod\limits_{i = 1}^n {(q_i !)} } \right)}} \cdot \frac{{\partial ^{n + \sum\limits_{i = 1}^n {(1 - i) \cdot q(i)} } G}} {{\partial \alpha ^{n - \sum\limits_{i = 1}^n {i \cdot q(i)} } \partial z^{\sum\limits_{i = 1}^n {q(i)} } }} \cdot \prod\limits_{i = 1}^n {\left( {u_i^{q(i)} } \right)} } \] where we leave the partial derivative of \[ G \] with respect to \[ \alpha \] and \[ z \] as an indeterminate function of \[ \alpha \] and \[ z \] . So, strictly, Definition 5 is not polynomial in \[ \alpha \] . Since \[ G(\alpha ,z) = 0 \Rightarrow \forall n \in \mathbb{N}:\frac{1} {{G_z }}\frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z) = 0 \Rightarrow \forall n \in \mathbb{N}:H_n (\alpha ,\frac{{dz}} {{d\alpha }},...,\frac{{d^n z}} {{d\alpha ^n }}) = 0 \] , in other words \[ \forall n \in \mathbb{N}:H_n = 0 \] after specialize \[ \forall i \in [n]:u_i \to \frac{{d^i z}} {{d\alpha ^i }} \] . Definition 6. A singularity of \[ G(z,\alpha ) \] is a value of \[ \alpha \] such that at least one of the roots \[ z \] of \[ G(z,\alpha ) = 0 \] is infinity, or two or more of the roots \[ z \] of \[ G(z,\alpha ) = 0 \] are equal. In other words, one of the roots \[ z \] of \[ G(z,\alpha ) = 0 \] is also a root of \[ G_z \equiv \frac{{\partial G(z,\alpha )}} {{\partial z}} = 0 \] . Theorem 7. Egorychev Lagrange Inversion \[ \frac{{\partial ^{m + n} }} {{\partial \alpha ^m \partial u^n }}\left[ {u_j \cdot \prod\limits_{k = 1}^n {\left( {H_k (\alpha ,u) - u_k } \right)^{\beta (k)} } } \right] \] Proof. The highest order of the derivative of \[ z \] with respect to \[ \alpha \] in \[ \frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z(\alpha )) \] is \[ n \] , and the highest power of \[ \frac{{d^n z}} {{d\alpha ^n }} \] in the summation in Lemma 4 is 1, which occurs when \[ i < n \Rightarrow q(i) = 0 \] and \[ q(n) = 1 \] and \[ \sum\limits_{i = 1}^n {i \cdot q_i } = n \cdot q_n = n \cdot 1 = n \] , the coefficient of \[ \frac{{d^n z}} {{d\alpha ^n }} \] in \[ \frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z(\alpha )) \] is \[ \frac{{n!}} {{(n!)^1 \cdot 1!}} \cdot \frac{{\partial G}} {{\partial z}} = G_z \] . So, the coefficient of \[ \frac{{d^n z}} {{d\alpha ^n }} \] in \[ H_n \] is 1. First check that for each \[ n \in \mathbb{N} \] and each \[ k \in [n] \] that \[ H_k (\alpha ,u_1 ,...,u_n ) \] satisfies \[ \left. {\frac{\partial } {{\partial u_j }}H_k (\alpha ,u_1 ,...,u_n )} \right|_{u(1) = \cdots = u(n) = 0} = \delta _{j,k} \] . So \[ G(\alpha ,z) = 0 \Rightarrow \forall n \in \mathbb{N}:\frac{1} {{G_z }}\frac{{d^n }} {{d\alpha ^n }}G(\alpha ,z) = 0 \Rightarrow \forall n \in \mathbb{N}:H_n (\alpha ,z) = 0 \] . We must compute \[ \begin{gathered} \frac{{\partial ^{m + \sum\limits_i {\beta (i)} } }} {{\partial \alpha ^m \prod\limits_{i = 1}^n {\partial u_i^{\beta (i)} } }} \hfill \\ \left[ {u_n \cdot \prod\limits_{k = 1}^n {\left( {\frac{1} {{G_z }}\sum\limits_q {\frac{{k!}} {{\left( {\prod\limits_{i = 1}^k {(i!)^{q(i)} } } \right) \cdot \left( {(k - \sum\limits_{i = 1}^k {i \cdot q_i } )!} \right)\left( {\prod\limits_{i = 1}^k {(q_i !)} } \right)}} \cdot \frac{{\partial ^{k + \sum\limits_{i = 1}^k {(1 - i) \cdot q(i)} } G}} {{\partial \alpha ^{k - \sum\limits_{i = 1}^k {i \cdot q(i)} } \partial z^{\sum\limits_{i = 1}^k {q(i)} } }} \cdot \prod\limits_{i = 1}^k {\left( {u_i^{q(i)} } \right)} } - u_k } \right)^{\beta (k)} } } \right] \hfill \\ \end{gathered} \] and then evaluate this partial derivative at \[ u_1 = \cdots = 0 = \cdots = u_n \] . Commentary Theorem 5. Taylor series Any root \[ z \] of a given function \[ G(z,\alpha ) \] of \[ z \] and a single complex-valued parameter \[ \alpha \] is given by \[ z = \sum\limits_{n = 0}^\infty {\left. {\frac{1} {{n!}}\frac{{d^n z}} {{d\alpha ^n }}} \right|_{\alpha = m} \cdot (\alpha - m)^n } \] whose radius of convergence is the distance from \[ \alpha \] to the nearest singularity of \[ G(z,\alpha ) \] . Corollary. For \[ G(z,\alpha ,w) \equiv z^\alpha + w \cdot z - 1 \] , Commentary Theorem 5. Belidarnelli formula for resolvent of a trinomial Mishkov, Rumen L. Generalization of the Formula of Faa Di Bruno for a Composite Function with a Vector Argument Internat. J. Math. & Math. Sci. Vol 24, No. 7 (2000) 481-491.