I have come upon with the next expression:

\begin{equation} P_r=\prod_{k=0}^r \left(1+\frac{1}{2}\left(\frac{\frac{1}{2}+k+1}{\frac{1}{2}+k} -\frac{\frac{1}{2}+k}{\frac{1}{2}+k+1}\right)\right) \end{equation}

And simulations have shown that its value can be approximated quite accurately to a linear function in the form: \begin{equation} P_r \approx O+M \cdot \left(\frac{1}{2}+r \right) \end{equation}

Using such knowledge, I have tried to obtain values for $O$ and $M$ using the first two values of $P_r$, that is, for $r=0$ and $r=1$. This approach is only reasonably accurate for small values of $r$, but tends to diverge from the actual value for higher values of $r$.

Could anyone help finding out the theoretical derivation to obtain a good linear approximation, and therefore the most accurate values for $O$ and $M$?

Some Matlab code:

x                   = [5];
x0                  = 5;
x_r                 = zeros(200,1);
x_r(1)              = x0/2;
P_r                 = zeros(201,1);
P_r(1)              = 1;

x_r(2)              = x(randi(length(x)));
zeta_r(2)           = sum(x_r(1:2));
P_r(1)              = (1+1/2*(sum(x_r(1:2))/sum(x_r(1:1))-sum(x_r(1:1))/sum(x_r(1:2))));

for r=3:201
    x_r(r)              = x(randi(length(x)));
    zeta_r(r)           = sum(x_r(1:r));
    P_r(r-1)            = P_r(r-2)*(1+1/2*(sum(x_r(1:r))/sum(x_r(1:r-1))-sum(x_r(1:r-1))/sum(x_r(1:r))));
 % P_r         = P_r(2:201);
 hold on;

 %Using two first points
 M   = (P_r(2)-P_r(1))/x_r(2);
 O   = P_r(1)-M*x0/2;

 %Using first and third points
 M   = (P_r(3)-P_r(1))/(x_r(2)+x_r(3));
 O   = P_r(1)-M*x0/2;

 %Using first and some other point
 M   = (P_r(10)-P_r(1))/(sum(x_r(2:10)));
 O   = P_r(1)-M*x0/2;

 h   = fit(cumsum(x_r(1:200)),P_r(1:200),'poly1');
  • $\begingroup$ It gives me the impression I might try to relate it to some function series expansion... exponential maybe? $\endgroup$ – Christian Mar 3 '16 at 13:18

Hint. One may first observe that $$ 1+\frac{1}{2}\left(\frac{1/2+k+1}{1/2+k}-\frac{1/2+k}{1/2+k+1}\right)=\frac{\left(k+\frac{3+\sqrt{2}}2\right) \left(k+\frac{3-\sqrt{2}}2\right)}{\left(k+\frac12\right)\left(k+\frac32\right)} \tag1 $$ then, using the $\Gamma$ function, we get $$ \prod_{k=0}^{r}(k+a)=\frac{\Gamma(r+a+1)}{\Gamma(a)} \tag2 $$ applying it to $(1)$ gives the following closed form of the initial product $$ \begin{align} &\prod_{k=0}^{r}\left(1+\frac{1}{2}\left(\frac{1/2+k+1}{1/2+k}-\frac{1/2+k}{1/2+k+1}\right)\right) \\\\&=\frac{\Gamma\left(r+\frac{5+\sqrt{2}}2\right) \Gamma\left(r+\frac{5-\sqrt{2}}2\right)\Gamma\left(\frac12\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac{3+\sqrt{2}}2\right)\Gamma\left(\frac{3-\sqrt{2}}2\right)\Gamma\left(r+\frac32\right)\Gamma\left(r+\frac52\right)}, \quad r\geq0.\tag3 \end{align} $$

or, with some simplifications,

$$ \begin{align} &\prod_{k=0}^{r}\left(1+\frac{1}{2}\left(\frac{1/2+k+1}{1/2+k}-\frac{1/2+k}{1/2+k+1}\right)\right) \\\\&=-2\cos(\pi/\sqrt{2})\frac{\Gamma\left(r+\frac{5+\sqrt{2}}2\right) \Gamma\left(r+\frac{5-\sqrt{2}}2\right)}{\Gamma\left(r+\frac32\right)\Gamma\left(r+\frac52\right)}, \quad r\geq0.\tag4 \end{align} $$

The function $r \mapsto f(r)$ on the right hand side of $(4)$ is $C^{\infty}$ over $r \in [0,\infty)$. It then admits a linear approximation on the form $ f(r)=f(r_0)+f'(r_0)(r-r_0)$ near each $r_0 \in [0,\infty)$.

For example, as $r$ is small, one finds that

$$ f(r) = \frac73+1.2635768599897183678498379747135790662125\cdots \times r+O(r^2)\tag5 $$

and, using the asymptotic expansion of the $\Gamma $ function, as the argument $r$ is great, one gets

$$ f(r) = -4\cos(\pi/\sqrt{2})-2\cos(\pi/\sqrt{2})\times r+O\left(\frac1{r}\right)\tag6 $$ with $$ \begin{align} -4\cos(\pi/\sqrt{2})&=2.422799468315253715217745425000817763852155123944\cdots \\\\-2\cos(\pi/\sqrt{2})&=1.2113997341576268576088727125004088819260775619722\cdots. \end{align} $$


Note that $$\frac{\frac12+k+1}{\frac12+k}- \frac{\frac12+k}{\frac12+k+1}=\frac{(\frac12+k+1)^2-(\frac12+k)^2}{(\frac12+k)(\frac12+k+1)}=\frac{2k+2}{(k+1)^2-\frac14}$$ hence we have $$\frac{P_r}{P_{r-1}}=1+\frac{r+1}{(r+1)^2-\frac14}\approx 1+\frac1{r+1}=\frac{r+2}{r+1} $$ which matches well with $P_r\sim r+2$ or at any rate witm $P_r\approx ar+b$. Numerically, I find $a\approx 1.211399734$ (by plugging in $r=10^7$ and $r=10^7-1$) and $b\approx 2.42$ (which seems to be less stable than the value for $a$), but surprises regarding the limit might still be possible.

  • $\begingroup$ Thank you Hagen. It is nice approximation I might end up using. $\endgroup$ – Christian Mar 3 '16 at 13:16

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