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Dalzell's integral $$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$

is case $n=2$ of the generalization

$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx = \frac{p_n}{q_n}-\pi$$

Such an integral gives rational approximations to $\pi$ from above: $4$, $\dfrac{19}{6}$, $\dfrac{22}{7}$, $\dfrac{377}{120}$,...

The qualities M of the latter three fractions are $2.057, 3.429, 1.98669,$ according to the definition

$$\Bigg \|{\pi-\frac{p_n}{q_n}}\Bigg \|=\frac{1}{q^{M_n}}$$

How is the asymptotic quality of this approximation sequence $\displaystyle\lim_{n \to \infty} M_n$ computed?

Antonio Vargas
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Jaume Oliver Lafont
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  • I saw your previous posts, I think you are in love with $\displaystyle\pi$! :-p (I didn't mean $\pi$ factorial :D ) – Jaideep Khare Apr 11 '17 at 21:26
  • Is there a connection with the continued fraction of $\pi$? – Jean Marie Apr 11 '17 at 21:38
  • @JeanMarie I don't know whether all fractions are semiconvergents, those displayed here are. The convergent $\frac{333}{106}$ is $\frac{377-2·22}{120-2·7}$, see http://math.stackexchange.com/a/2220563/134791 – Jaume Oliver Lafont Apr 11 '17 at 21:43
  • @JaideepKhare Thanks a lot for your help in improving format! – Jaume Oliver Lafont Apr 11 '17 at 21:45
  • Thanks for your answer. Sorry not to be able to say more. I knew the first integral, but not the generalization. – Jean Marie Apr 11 '17 at 21:45
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    Probably relevant notion: "irrationality measure" –  Apr 11 '17 at 21:47
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    @JaumeOliverLafont Welcome :)! Also, I recommend you to use \dfrac instead of \frac as it makes the visibility of fractions more clear; and use \displaystyle while writing any integral, sum etc. in mid of a line, to trigger the displaymode. – Jaideep Khare Apr 11 '17 at 21:49
  • Yes, @Hurkyl, I took the definition for quality from section "Irrationality measures" in Beukers' paper http://www.nieuwarchief.nl/serie5/pdf/naw5-2000-01-4-372.pdf, but I am failing to see how to actually compute those asymptotic qualities... – Jaume Oliver Lafont Apr 11 '17 at 22:06
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    I'm not sure if it's useful, but applying the Laplace method to the integral in your question yields the asymptotic $$\frac{p_n}{q_n} - \pi \sim \frac{4}{15} \left(\frac{2}{27}\right)^n \sqrt{\frac{\pi}{3n}}.$$ – Antonio Vargas Apr 12 '17 at 04:28
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    @AntonioVargas Tanks a lot! I think an expression for $q_n$ would also be needed, am I right? $$M=-\dfrac{log{\Bigg |\dfrac{p_n}{q_n}-\pi\Bigg |}}{log(q_n)}$$

    Would you recommend me a resource to try and learn how to compute your result?

     – Jaume Oliver Lafont Apr 12 '17 at 05:07
  • https://en.wikipedia.org/wiki/Laplace%27s_method – Jaume Oliver Lafont Apr 12 '17 at 06:58
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    The partial fraction expansion of the integrand is some polynomial of degree $3n$ with integer coefficients divided by $2^{n-2}$, minus $4/(1+x^2)$ (whose integral over this interval is $\pi$). The integral of that polynomial from $0$ to $1$ will be an integer linear combination of rational numbers with denominators between $1$ and $3n+1$, all divided by $2^{n-2}$. So one can take $q_n$ to be $2^{n-2} \mathop{\rm lcm}[1,\dots,3n+1]$, which is roughly $(2e^3)^n$ by the prime number theorem. – Greg Martin Apr 12 '17 at 22:01
  • Thank you very much, @GregMartin! I think this completes the answer. I understand that the quality of these approximations is low, despite the initial good values. – Jaume Oliver Lafont Apr 13 '17 at 06:49
  • So the $-2$ in $2^{n-2}$ or the $+1$ in $3n+1$ do not change the asymptotics, the same as the $\left(\dfrac{4}{15}\right)\sqrt{\dfrac{\pi}{3n}}$ in the numerator... – Jaume Oliver Lafont Apr 13 '17 at 09:26

1 Answers1

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This answer combines @AntonioVargas and @GregMartin results.

Let us start from the approximation in Laplace's method (https://en.wikipedia.org/wiki/Laplace%27s_method)

$$\int_a^b h(x)e^{Mg(x)}dx \approx \sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)}$$

and rewrite the general integral in the question as

$$\int_0^1 \frac{4x^2}{1+x^2}e^{n\log\left(\dfrac{x(1-x)^2}{2}\right)} dx$$

so we may identify

$$h(x)=\dfrac{4x^2}{1+x^2}$$

$$g(x)=\log\left(\dfrac{x(1-x)^2}{2}\right)$$

$$g'(x)=\dfrac{1-3x}{x(1-x)}$$

$$g''(x)=-\dfrac{1-2x+3x^2}{x^2(1-x)^2}$$

The position of the unique global maximum of $g(x)$ is obtained from $g'(x_0)=0$

$$x_0=\frac{1}{3}$$

Substituting into $h(x)$, $g(x)$ and $g''(x)$, we obtain

$$h(x_0)=h\left(\dfrac{1}{3}\right)=\frac{4\left(\dfrac{1}{3}\right)^2}{1+\left(\dfrac{1}{3}\right)^2}=\frac{2}{5}$$

$$g(x_0)=log\left(\dfrac{x_0(1-x_0)^2}{2}\right) = log\left(\frac{2}{27}\right)$$

$$g''(x_0)=g''\left(\dfrac{1}{3}\right)=-\frac{27}{2}$$

Finally,

$$\sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)} = \sqrt{\dfrac{2\pi}{n\frac{27}{2}}}\frac{2}{5}e^{n\log\left(\dfrac{2}{27}\right)} = \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n,$$

so

$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx \sim \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n$$

for large $n$.

The asymptotic quality is therefore

$$\theta = \lim_{n \to \infty} -\frac{\log\left(\dfrac{p_n}{q_n}-\pi\right)}{\log(q_n)} = \lim_{n \to \infty} -\frac{\log\left( \dfrac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n \right)}{\log( (2e^3)^n)} = \frac{3log(3)-log(2)}{3+log(2)} \approx 0.7 < 1,$$

which is low.

Jaume Oliver Lafont
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