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Let $H(z)$ be a continuous solution of the problem $$ H(z)=\frac1{1-z}\int_z^1 \frac{2\zeta}{1+\zeta} H(\zeta^2)\,d\zeta,\ \ \ z\in[0,1);\ \ \ H(1)=1. $$ Is it true that $H(0)=1-\ln2$? The question is directly related to this one https://math.stackexchange.com/questions/4748129/asymptotics-of-sequence-of-rational-numbers .

Small update: I did not check all the details, but, it seems that the existence of the solution of this equation can be proved by applying the standard arguments - it is a contraction mapping acting on continuous functions satisfying $H(1)=1$. Hence, the corresponding fixed point (the solution $H(z)$) exists. I already calculated this by using the Picard-Lindelöf scheme $$ H_{n+1}(z)=\frac1{1-z}\int_z^1 \frac{2\zeta}{1+\zeta} H_n(\zeta^2) \, d\zeta,\ \ \ H_0(z)\equiv1. $$ In fact, the iterations work well even for discontinuous functions, but the continuity at $z=1$ is required. Results of the computation

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    $\begingroup$ I think it is worth mentioning that there can be only one such function (and it can be obtained by the iterative procedure starting with $H(z) = 1$). $\endgroup$ Nov 22, 2023 at 15:36
  • $\begingroup$ OK, done. I added a few comments. $\endgroup$
    – AAK
    Nov 22, 2023 at 15:56

3 Answers 3

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That is a rather tough puzzle (took me two full days) with a rather short solution.

The first step is the differential equation Fred already mentioned: $$ (1-z^2)H'(z)-(1+z)H(z)+2zH(z^2)=0\,. $$ Now define $F(\alpha)=\int_0^1 z^\alpha H(z)\,dz$, $\alpha>-1$. Then, by the condition $H(1)=1$, we have $F(\alpha)\approx 1/\alpha$ for large $\alpha$.

Integration of the differential equation against $z^\alpha$, $\alpha>0$ yields $$ -\alpha F(\alpha-1)+(\alpha+2)F(\alpha+1)-F(\alpha)-F(\alpha+1)+F(\alpha/2)=0. $$ Summing over odd $\alpha$ from $1$ to $A$, we get $$ -\sum_{0\le \alpha\le A+1}F(\alpha)+(A+2)F(A+1)+\sum_{1\le \alpha\le A, \alpha\text{ odd}}F(\alpha/2)=0 $$ where the first sum is over all integer $\alpha$ in the range (each positive integer is either odd, or odd plus $1$), or, equivalently, $$\sum_{0\le\alpha\le A}(-1)^\alpha F(\alpha/2)+\sum_{\frac A2<\alpha\le A+1}F(\alpha)=(A+2)F(A+1)\,.$$

Now notice that $H(0)=\int_0^1\frac 1{1+\sqrt z}H(z)\,dz=\sum_{\alpha\ge 0}(-1)^{\alpha}F(\alpha/2)$ (I leave it to you to justify the convergence). Thus, passing to the limit as $A\to+\infty$ in the last displayed equation and using the large $\alpha$ asymptotics for $F(\alpha)$ with $\alpha>A/2$, we get $$ H(0)+\log 2=1\,, $$ as desired.

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  • $\begingroup$ Super, this solution is quite nice! $\endgroup$
    – AAK
    Nov 24, 2023 at 2:45
  • $\begingroup$ @AAK Thanks! I'm glad I could be of some help :-) $\endgroup$
    – fedja
    Nov 24, 2023 at 2:56
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Alternative simple proof - integration by parts: $$ \int_0^{1-a}\frac{H(z)}{1-z}dz=\int_0^{1-a}\frac1{(1-z)^2}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta= $$ $$ \frac1{1-z}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta\bigg\rvert_{z=0}^{z=1-a}+\int_0^{1-a}\frac{2\zeta}{1-\zeta^2}H(\zeta^2)d\zeta, $$ which leads to $$ \int_{1-2a+a^2}^{1-a}\frac{H(z)}{1-z}dz= H(1-a)-H(0) $$ Since $H(1)=1$ and $$ \int_{1-2a}^{1-a}\frac{dz}{1-z}\to\ln2\ \ \ for\ \ \ a\to0, $$ we obtain the result.

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  • $\begingroup$ I cannot believe that I have not seen this before, very strange in my case. $\endgroup$
    – AAK
    Nov 25, 2023 at 11:19
  • $\begingroup$ Beautiful! It is curious that the integration against $\frac 1{1-z}$ up to $a\approx 1$ you used and adding the moments up to $A\approx\infty$ (i.e., integrating against $1+z+\dots+z^A$) I used are quite close to each other :-) $\endgroup$
    – fedja
    Nov 25, 2023 at 19:57
  • $\begingroup$ Yes, thanks! And another goal was to obtain this new integral functional equation useful for other expansions. $\endgroup$
    – AAK
    Nov 25, 2023 at 21:13
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This is an incomplete answer, the last step is missing (yet).

We can differentiate the OPs equation to get \begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)-(z+1) H(z)+2 z H(z^2)=0. \end{align} The series expansion of $H(z)$ around $z=0$, \begin{align}\tag{2}\label{eq:2} H(z)=\sum_{n=0}^\infty h_n z^n, \end{align} fulfills the recurrence relation (assuming $H(0)=1$) \begin{align} h_0&=h_1=1\\ h_n&=\frac 1 n\left[ (n-1)h_{n-2}+h_{n-1} + ((-1)^n+1)h_{n/2-1}\right], \tag{3}\label{eq:3} \end{align} such that \begin{align}\tag{4}\label{eq:4} H(z)=1+z+\frac{2 z^3}{3}-\frac{z^4}{3}+\frac{7 z^5}{15}-\frac{z^6}{5}+\frac{13 z^7}{35}-\frac{31 z^8}{105}+\frac{281 z^9}{945}+O\left(z^{10}\right) \end{align} Some Mathematica code:

Clear[hn]; hn[0]=1; hn[1]=1;
hn[n_Integer]:=hn[n] = ((n-1) hn[n-2] + hn[n-1] - If[EvenQ[n], 2 hn[n/2-1], 0])/n
Hs[z_]=Sum[hn[n] z^n, {n, 0, 10}]
(1-z^2) Hs'[z] + 2 z Hs[z^2] - (z+1) Hs[z] + O[z]^10

So we have to show that \begin{align}\tag{5}\label{eq:5} H(1)=\sum_{n=0}^\infty h_n = \frac 1 {1-\ln 2}. \end{align} Numerically, this is the case, the partial sums of \eqref{eq:5} converge with $O(1/n)$.

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  • $\begingroup$ Yes, and there is a little bit more simple sequence (not a series) that is equivalent to the problem, see the original question in math.stackexchange.com. In both, the difficult component $h_{n/2}$ appears. We have a few equivalent formulations of this problem - discrete and continuous, but I do not know which one is simpler to solve. $\endgroup$
    – AAK
    Nov 23, 2023 at 9:42

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