Find unique solution to certain $N$ and related $N+1$ order polynomial (one solution, numerically known, looking for proof)

Question

Basically this question consists of two questions that are intertwined:

1) I would like to find the (unique) solution $x_{\mathrm{solution}}$ to the N-th polynomial in $x$ for $x>\frac{-1}{max(a)}$, namely

$\left(\sum\limits_{i=1}^{N}a_i\cdot \left( \prod\limits_{j=1,j \neq i}^{N} (1+a_j \cdot x ) \right) \cdot (1-x)\right) - C \cdot \prod\limits_{i=1}^{N} (1+a_i \cdot x) =0$

for general $a_i >0$ and for any $C \in \Bbb{N}$. As you can see all summands are of order N.

It is known, that there is only one real solution for $x>\frac{-1}{max(a)}$ (even though the polynomial can be of high order), and from numerical studies I know $x_{\mathrm{solution}}$ also solves the related polynomial

$\left(\sum\limits_{i=1}^{N}\left( \prod\limits_{j=1,j \neq i}^{N} (1+a_j \cdot x ) \right) \cdot (1-x)\right) + C \cdot \prod\limits_{i=1}^{N} (1+a_i \cdot x) =(N+C)\cdot \left(\prod\limits_{i=1}^{N} (1+a_i \cdot x)\right)\cdot (1-x)$

2) Is there any way to prove that both these polynomials are solved by the same real $x$? I know it is true numerically, I just dont know how to prove it.

For $N=1$, for example, the first equation is linear, the second quadratic, and one can easily check that $x_{\mathrm{solution}}=\frac{a_1-C}{a_1 \cdot (C+1)}$.

Maybe one can somehow work by induction to higher $N$?

Answer 1

Let $f(x) =(1-x)\sum\limits_{i=1}^{n}a_i\prod\limits_{j \neq i}^{n} (1+a_j x ) $, $g(x) =\prod\limits_{i=1}^{n} (1+a_i x) $, and $h(x) =(1-x)\sum\limits_{i=1}^{n}\prod\limits_{j \neq i}^{n} (1+a_j x ) $.

Your claim is that $u(x) =f(x)-cg(x)$ and $v(x) =h(x)+cg(x)-(n+c)(1-x)g(x) $ have a common root for some $c$.

I will show that if $0 < c \lt \sum_{i=1}^n a_i $, there is a common root $x$ such that $c =(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x } $.

$\begin{array}\\ f(x) &=(1-x)\sum\limits_{i=1}^{n}a_i\prod\limits_{j \neq i}^{n} (1+a_j x )\\ &=(1-x)\sum\limits_{i=1}^{n}a_i\dfrac{\prod\limits_{j=1}^{n} (1+a_j x )}{1+a_i x }\\ &=(1-x)g(x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x }\\ &=(1-x)g(x)f_1(x)\\ \end{array} $

$\begin{array}\\ h(x) &=(1-x)\sum\limits_{i=1}^{n}\prod\limits_{j \neq i}^{n} (1+a_j x )\\ &=(1-x)\sum\limits_{i=1}^{n}\dfrac{\prod\limits_{j=1}^{n} (1+a_j x )}{1+a_ix}\\ &=(1-x)g(x)\sum\limits_{i=1}^{n}\dfrac{1}{1+a_ix}\\ &=(1-x)g(x)h_1(x)\\ \end{array} $

Therefore

$\begin{array}\\ u(x) &=f(x)-cg(x)\\ &=(1-x)g(x)f_1(x)-cg(x)\\ &=g(x)((1-x)f_1(x)-c)\\ \end{array} $

and

$\begin{array}\\ v(x) &=h(x)+cg(x)-(n+c)(1-x)g(x)\\ &=(1-x)g(x)h_1(x)+cg(x)-(n+c)(1-x)g(x)\\ &=g(x)((1-x)h_1(x)+c-(n+c)(1-x))\\ &=g(x)((1-x)(h_1(x)-(n+c))+c)\\ \end{array} $

The positive root(s) of $u(x)$ satisfy $(1-x)f_1(x)=c $ or $x = 1-\dfrac{c}{f_1(x)} $.

The positive root(s) of $v(x)$ satisfy $(1-x)(n+c-h_1(x))=c $ or $x = 1-\dfrac{c}{n+c-h_1(x)} $.

These are equal if $f_1(x) =n+c-h_1(x) $ or $\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x } =n+c-\sum\limits_{i=1}^{n}\dfrac{1}{1+a_i x } $ or $n+c =\sum\limits_{i=1}^{n}\dfrac{1+a_i}{1+a_i x } $.

If all $a_i = a$, this is $n+c =n\dfrac{1+a}{1+ax} $ or $(n+c)(1+ax) =n(1+a) $ or $(n+c)ax =n(1+a)-(n+c) =na-c $ or $x =\dfrac{na-c}{a(n+c)} $.

In the general case,

$\begin{array}\\ \sum\limits_{i=1}^{n}\dfrac{1+a_i}{1+a_i x } &=\sum\limits_{i=1}^{n}\dfrac{1+a_ix-a_ix+a_i}{1+a_i x }\\ &=n+\sum\limits_{i=1}^{n}\dfrac{a_i(1-x)}{1+a_i x }\\ &=n+(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x }\\ \end{array} $

so

$n+c =\sum\limits_{i=1}^{n}\dfrac{1+a_i}{1+a_i x } $ when $c =(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x } $.

If $d(x) =(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x } $, then $d(0) =\sum_{i=1}^n a_i $ and $d(1) = 0$, so if $0 < c \lt \sum_{i=1}^n a_i $, there is an $x$ such that $d(x) = c$.

For this $x$, $f_1(x) = d(x)$ and

$\begin{array}\\ \dfrac{v(x)}{g(x)} &=(1-x)(h_1(x)-(n+c))+c\\ &=(1-x)(h_1(x)-(n+d(x)))+d(x)\\ &=(1-x)(h_1(x)-(n+(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x }))+(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x }\\ &=(1-x)(\sum\limits_{i=1}^{n}\dfrac{1}{1+a_ix}-(n+(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x }))+(1-x)\sum\limits_{i=1}^{n}\dfrac{a_i}{1+a_i x }\\ &=(1-x)(\sum\limits_{i=1}^{n}\left(\dfrac{1}{1+a_ix}-\dfrac{(1-x)a_i}{1+a_i x }+\dfrac{a_i}{1+a_i x }\right)-n)\\ &=(1-x)(\sum\limits_{i=1}^{n}\left(\dfrac{1+a_i-a_i+xa_i}{1+a_ix}\right)-n)\\ &=(1-x)(\sum\limits_{i=1}^{n}\left(\dfrac{1+xa_i}{1+a_ix}\right)-n)\\ &=0\\ \end{array} $

Answer 2

Are you sure that the upper limit in $\prod\limits_{j \neq i}^{n-1} (1+a_j x ) $ is not $n$? This is of degree $n-2$ except when $i=n$.

I will make this modification below.

Let $f(x) =(1-x)\sum\limits_{i=1}^{n}a_i\prod\limits_{j \neq i}^{n} (1+a_j x ) $, $g(x) =\prod\limits_{i=1}^{n} (1+a_i x) $, and $h(x) =(1-x)\sum\limits_{i=1}^{n}\prod\limits_{j \neq i}^{n} (1+a_j x ) $.

Your claim is that $u(x) =f(x)-cg(x)$ and $v(x) =h(x)+cg(x)-(n+c)(1-x)g(x) $ have a common root for some $c$.

I will show that this is true if all the $a_i$ are equal to $a$ and the root is $x = \dfrac{an-c}{a(n+c)} $.

If all the $a_i = 1$, $f(x) = h(x)$ so this becomes $f(x)-cg(x)$ and $f(x)+cg(x)-(n+c)(1-x)g(x)$ have a common root. For this root, $-cg(x) = cg(x)-(n+c)(1-x)g(x) $ or $0 =g(x)(2c-(1-x)(n+c)) $.

If $x > 0$ then $g(x)\ne 0$, so $2c =(1-x)(n+c) $ or $x(n+c) =n-c $ or $x = \dfrac{n-c}{n+c} $.

Also, in this case, $f(x) = h(x) =n(1-x)(1+x)^{n-1} $ and $g(x) =(1+x)^n $, so

$\begin{array}\\ u(x) &=n(1-x)(1+x)^{n-1}-c(1+x)^n\\ &=(1+x)^{n-1}(n(1-x)-c(1+x))\\ \end{array} $

with root $n(1-x)=c(1+x)$ or $x = \dfrac{n-c}{n+c} $;

$\begin{array}\\ v(x) &=h(x)+cg(x)-(n+c)(1-x)g(x)\\ &=n(1-x)(1+x)^{n-1}+(1+x)^n(c-(n+c)(1-x))\\ &=(1+x)^{n-1}(n(1-x)+(1+x)(c-(n+c)(1-x))\\ &=(1+x)^{n-1}(n(1-x)+c(1+x)-(n+c)(1-x^2))\\ &=(1+x)^{n-1}((n+c)x^2+x(c-n)+n+c-(n+c))\\ &=x(1+x)^{n-1}((n+c)x+(c-n))\\ \end{array} $

Again, this has root $x = \dfrac{n-c}{n+c} $.

If all the $a_i = a$, $f(x) = ah(x)$ and $h(x) =n(1-x)(1+ax)^{n-1} $ and $g(x) =(1+ax)^n $, so

$\begin{array}\\ u(x) &=an(1-x)(1+ax)^{n-1}-c(1+ax)^n\\ &=(1+ax)^{n-1}(an(1-x)-c(1+ax))\\ \end{array} $

with root $an(1-x)=c(1+ax)$ or $an-c = ax(n+c) $ or $x = \dfrac{an-c}{a(n+c)} $;

$\begin{array}\\ v(x) &=h(x)+cg(x)-(n+c)(1-x)g(x)\\ &=n(1-x)(1+ax)^{n-1}+(1+ax)^n(c-(n+c)(1-x))\\ &=(1+ax)^{n-1}(n(1-x)+(1+ax)(c-(n+c)(1-x)))\\ &=(1+x)^{n-1}(n(1-x)+c(1+ax)-(n+c)(1+(a-1)x-ax^2))\\ &=(1+x)^{n-1}(a(n+c)x^2+x(ac-n-(a-1)(n+c))+n+c-(n+c))\\ &=(1+x)^{n-1}(a(n+c)x^2+x(c-an))\\ &=x(1+x)^{n-1}(a(n+c)x+(c-an))\\ \end{array} $

Again, this has root $x = \dfrac{an-c}{a(n+c)} $.