Determine the maximum number of inflection points on $y=(1+x^{u_1})(1+x^{u_2})\dots(1+x^{u_n})$ where $u_i\in\mathbb{Z^+}$.

Question

Determine the maximum number of inflection points on $y=(1+x^{u_1})(1+x^{u_2})\dots(1+x^{u_n})$ where $u_i$ are chosen among the positive integers (not necessarily distinct).

Through experimentation, it seems that the maximum is five, for example $y=(1+x)(1+x^3)(1+x^4)(1+x^6)(1+x^7)$. (Here is the expansion.) Its graph is shown in red, with $y''$ in green.

Another example is $y=(1+x)(1+x^3)(1+x^4)(1+x^{13})$, provided by @Srini in the comments.

For a general function $y=(1+x^{u_1})(1+x^{u_2})\dots(1+x^{u_n})$, I can set $y''$ equal to $0$, but I do not know how to maximize the number of roots.

Not sure if this helps, but if we change all the plus signs to minus signs, i.e. $y=(1-x^{u_1})(1-x^{u_2})\dots(1-x^{u_n})$, it seems that the maximum number of inflection points is six, for example $y=(1-x)(1-x^3)(1-x^4)(1-x^6)(1-x^7)(1-x^8)(1-x^9)$. EDIT: This is not true; for example, $(1-x)(1-x^3)(1-x^4)(1-x^9)(1-x^{11})(1-x^{20})(1-x^{28})$ has eight inflection points (example provided by @Srini in chat).

Possibly related: this and this.

Context

I thought of this question when I noticed that $y=\prod\limits_{k=1}^\infty(1+x^k)$ seems to have exactly two inflection points, which I found peculiar.

Answer 1

This is not an answer.

As @jazzblaster wrote, for $$f_n(x)=\prod\limits_{k=1}^n (1+x^k)$$ excluding the multiple $x=-1$, there are only two inflection points.

I give below some of them $$\left( \begin{array}{ccc} n & x_1 & x_2 \\ 5 & -0.22589119064517197427 & -0.74028242166391524886 \\ 6 & -0.23484859262502136880 & -0.80000556779789281900 \\ 7 & -0.23159441913388658895 & -0.75894677414044432188 \\ 8 & -0.23256896711783560914 & -0.79246270029727908255 \\ 9 & -0.23227033506118166376 & -0.77117077732778199650 \\ 10 & -0.23235711575038931877 & -0.78971292565570820724 \\ 11 & -0.23233228909199430843 & -0.77926768123948581860 \\ 12 & -0.23233923645768187701 & -0.78846418191419616220 \\ 13 & -0.23233732152041012169 & -0.78453750049360149301 \\ 14 & -0.23233784213784544104 & -0.78785712011131879201 \\ 15 & -0.23233770220235508450 & -0.78781772572715844166 \\ 16 & -0.23233773944296335950 & -0.78763429876554399320 \\ 17 & -0.23233772961756267343 & -0.78970940998646838787 \\ 18 & -0.23233773219018299666 & -0.78766858325306359065 \\ 19 & -0.23233773152111734581 & -0.79066987381256456493 \\ 20 & -0.23233773169407638571 & -0.78786085603955366628 \\ 21 & -0.23233773164960654356 & -0.79104641260051896762 \\ 22 & -0.23233773166098447188 & -0.78813172393748620480 \\ 23 & -0.23233773165808624700 & -0.79108985860251648744 \\ 24 & -0.23233773165882150928 & -0.78842455903880744350 \\ 25 & -0.23233773165863566796 & -0.79096779242682615756 \\ \end{array} \right)$$

What is interesting is to notice that the roots follow two different lines depending on the parity of $n$.

Moreover, they are not monotonic. For example, if $n$ is odd, $x_2$ is minimum for $n=21$.

Using $$f_\infty (x)=\prod\limits_{k=1}^\infty (1+x^k)=\frac{1}{2}(-1;x)_{\infty }$$ where appears the q-Pochhammer symbol, the solutions are $$x_1=-0.23233773165867308048797254771753539814240447692346\cdots$$ $$x_2=-0.78983485691011228544265146474693127184443405841832\cdots$$

Edit

It is interesting to notice that, if we consider the case where the same powers appear multiple times such as with $$g_n(x)=\Bigg(\prod\limits_{k=1}^\infty (1+x^k)\Bigg)^n$$ the inflection points seem to disappear as soon as $n>1$.

Answer 2

Earlier I had searched for distinct $u_i$ and found nothing that had $\ge 6$ inflection points. But, OP says $u_i$ need not be distinct. So I expanded my search in Wolfram Cloud. Since I don't have a subscription, my search is limited. But I found several candidates that have $6$ inflection points. At least one that I verified in both Wolfram Cloud and Desmos is the following:

$$p_0(x) = (1+x)(1+x^3)(1+x^4)^\color{blue}{2}(1+x^{14})^\color{blue}{2}$$

Wolfram cloud lists the inflection points as $x=-0.957277,\color{red}{-0.796716,-0.787841},-0.589495,-0.228351,0$

Notice that the red highlighted roots are close to each other and may appear to be double root in Desmos graph non-distinct (so, zoom in more :) )

A few more candidates from my Wolfram search below (note: unverified in Desmos if they are true inflection points)

$$p_1(x) = (1+x)(1+x^3)(1+x^4)^\color{blue}{2}(1+x^{\color{red}{m}})^\color{blue}{2}$$ ($m=16,18,20,22,24,26$)

Edit 1: Using the same inelegant brute force search as in my comments reply to jazzblaster, the following polynomial was found with $\color{red}{6 \text{ distinct}}$ inflection points, while $u_i$ are distinct. Now, I am confident that $5$ is not a magic number at all.

$$p_2(x) = (1+x)(1+x^3)(1+x^4)(1+x^6)(1+x^{18})(1+x^{24})$$. Desmos graph distinct

Inflection points are at $x= -0.966089,-0.908615,-0.755146,-0.720714,-0.329131,0$

Edit 2: $\color{red}{7 \text{ Distinct}}$

$$p_3(x) = (1+x)(1+x^3)(1+x^4)(1+x^6)(1+x^{20})(1+x^{41})$$

Inflection points are at $x = -1, -0.913252, -0.883765, -0.844574, -0.68945, -0.329132, 0$

Answer 3

(I have some partial results that are getting difficult to fit into a comment.Here goes.)

Partial Result: Critical Points

Let

$$f(x)=\Pi_{j=1}^N (1+x^{u_j})$$

where $u_j$ are positive integers, not necessarily distinct.

We can try to convert this product to a sum as follows:

$$\ln{f(x)}=\Sigma_{j=1}^N\ln{(1+x^{u_j})}$$

(Note that this works only if $f(x)$ is greater than zero. This is assured on the interval $(-1,\infty]$. The interval $[-1, -\infty)$ might still be analyzed if the number of terms with odd $u_j$ is even, or if all $u_j$ even.)

Differentiating both sides gives the following.

$$\frac{f'(x)}{f(x)}=\Sigma_{j=1}^N \frac{u_jx^{u_j - 1}}{1+x^{u_j}}$$

It follows that

$$\textbf{ (Eq.1) } f'(x)=f(x) \cdot \Sigma_{j=1}^N \frac{u_jx^{u_j - 1}}{1+x^{u_j}}$$

(If $f(x)$ doesn't have a zero at -1, this is well defined. If it does, we can extend our domain to include -1 by taking limits.)

Define

$$h(x)=\Sigma_{j=1}^N \frac{u_jx^{u_j - 1}}{1+x^{u_j}}$$

If we now assume that $u_j$ are all even or odd and sketch the graph of $\frac{u_jx^{u_j - 1}}{1+x^{u_j}}$ by using standard calculus techniques, we can see that

$$|h(x)| > 0$$

for all x≠0. This puts the maximum number of critical points at 2.

In the cases with mixed $u_j$, the terms of $h(x)$ can end up cancelling out in between -1 and 0. For the other intervals, we should be able to resort to some inequalities to show that there can be no critical point.

The upper bound given by the Mean Value Theorem would be $\Sigma u_j -1$. There's a significant reduction on the upper bound in these two cases.

Inflection Points:

The strategy found above gives us hope for the second derivative, so we proceed as follows:

$$f''(x)=f(x) \cdot (h(x)^2+h'(x))$$

where we have used the chain rule and Eq. 1.

The derivative is given by

$$h'(x)=\Sigma_{j=1}^N \frac{-u_jx^{u_j - 2}(x^{u_j}-u_j+1)}{(1+x^{u_j})^2}$$

(This is admittedly not very useful.)

Answer 4

We can exploit that $m_i$ does not need to be unique to find a solution that has any number of inflection points. All this inflection points are rational numbers and are also extremal points of the polynomial.

Let $$ P(x)= \prod_{i=1}^N (1+x^{m_i})^{A_i} $$ where $m_i$ have altenating parity. Let $c_j$ $j=1,\cdots r$ rational numbers belonging to $(-\varepsilon,0)$, where $\varepsilon$ is a small enough number and assume $N\ge 2(r+1)$ is even.

We want to find $A_i$ such that each of the $c_j$ is an inflection point of $P(x)$

Let $F(x)=\log P(x)$. We have $P'(x)=P(x) F'(x)$ and $P^{''}(x)=P(x) \left(F^{''}(x)+(F^{'}(x))^2\right)$. So a sufficient condition for $P^{''}(c_j)$ to be zero is having both $F'(c_j)=0= F^{''}(c_j)$.

Let $$f_{m_i}(x)= \frac{m_i x^{m_i-1}}{1+x^m} \quad g_{m_i}(x)= \frac{d}{dx} f_{m_i}(x)$$

The equations became $$ \begin{align} &\sum_{i=1}^N A_i f_{m_i}(c_j)=0\\ & \sum_{i=1}^N A_i g_{m_i}(c_j)=0 \end{align} $$ for each $c_j$. This is an homogeneous linear system over $\mathbb{Q}^{N}$ and it admits a not trivial solution.

If we can prove that all the $A_i$ are not negative, as the system is homogeneous we can multiplicate for the common denominator to obtain non negative integers $A_i$

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Proof that the system admits a not negative solution

near $0^{-}$ we have that $f_{m_i}(x) \approx m_i x^{m_i-1}, g_{m_i}(x) \approx m_i(m_i-1) x^{m_i-2}$ In particular $\operatorname{sign} (f_{m_i}(c_j))=-\operatorname{sign}(g_{m_i}(c_j))$ for each $j$

By Stiemke's theorem the existence of a positive solution is equivalent to the statement that does not exist $p \in \mathbb{R}^{2r}$ such that $A^T p \ge 0$.

Let call $C_i=(f_{m_i}(c_1) \cdots f_{m_i}(c_r),g_{m_i}(c_1) \cdots g_{m_i}(c_r))$ a column of $A$. It has $r$ positive coordinates and $r$ negative coordinates. Then the interior of $C:=\operatorname{cone}((C_i)_{i=1,\cdots N})$ ( the conic hull of the columns) contains the origin.

But the interior of the cone consists of the points that have strictly positive inner product with each non zero element of its dual cone, so $C^*:= \{ y : A^T y \ge 0 \}= \{0\}$, that is what we want to prove.

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I have found (using python) a polynomial with $10$ inflection points stricly less than $-1$ by choosing $$ \prod_{i=1}^{22} (1+x^i)^{A_i} $$ and finding $A_i$ such that the points $-\frac{1}{2^{j}}$ ($j=1,\cdots 10$) are inflection points. The code seems to work, but each $A_i$ has thousands of digits, so for obvious reasons I will not post them