MHB Unsolved analysis and number theory from other sites....

[chisigma]

Posted the 12 31 2013 on www.matematicamente.it by the user N56VZ [original in Italian...] and not yet solved...

... I searched on Internet how to solve this ODE but I didn't find anything...

$\displaystyle u^{\ '} = u^{2}\ \sin u,\ u(0)= - \frac{3}{2}\ \pi$

In the site this problem has been proposed this ODE has been classified as 'Riccati equation' and the conclusion has been 'unsolvable'. With a little more precise evaluation it is easy to see that the ODE belongs to the most 'comfortable' first order ODE : the separable variables type. The solving procedure is well known...

$\displaystyle \int \frac{d u}{u^{2}\ \sin u} = t + c\ (1)$

Unfortunately the integral in (1) 'cannot be expressed in term of elementary functions' so that an alternative solution has to be tried. We can use the Laurent expansion...

$\displaystyle \frac{1}{u^{2}\ \sin u} = \frac{1}{u^{3}} + \frac{1}{6\ u} + \frac{7}{360}\ u + \frac {31}{15120}\ u^{3} + \mathcal {O}\ (u^{5})\ (2)$

... and integrate (2) 'term by term' but now we meet a new criticity: the (2) converges for $\displaystyle 0 < |u| < \pi$, so that it's impossible to compute the initial condition $t(0) = u (- \frac{3}{2}\ \pi)$. Fortunately we can overcome all that searching as solution a function t (u) that is analytical in $\displaystyle u = - \frac{3}{2}\ \pi$. Setting $\displaystyle v= u + \frac{3}{2}\ \pi$ the ODE becomes...$\displaystyle \frac{d t}{d v} = - \frac{1} {(v - \frac{3}{2}\ \pi)^{2}\ \cos v},\ t(0)=0\ (3)$

... where...

$\displaystyle t(v) = \sum_{n=1}^{\infty} a_{n}\ v^{n}\ (4)$

The coefficients of (4) can be found as follows...

$\displaystyle t^{\ '} = - \frac{1} {(v - \frac{3}{2}\ \pi)^{2}\ \cos v} \implies t^{\ '} (0) = - (\frac{2}{3\ \pi})^{2} = a_{1}$

$\displaystyle t^{(2)} = - \frac{(v - \frac{3}{2}\ \pi)\ \tan v + 2}{(v - \frac{3}{2}\ \pi)^{3}} \implies t^{(2)} (0) = 2\ (\frac{2}{3\ \pi})^{3} \implies a_{2} = (\frac{2}{3\ \pi})^{3}$

$\displaystyle t^{(3)} = \frac{\{(\frac{3}{2}\ \pi)^{2} - 3\ \pi\ v + v^{2} + 6\}\ \cos 2 v - 3\ \{(\frac{3}{2}\ \pi)^{2} - 3\ \pi\ v + v^{2} -2\}}{ \cos^{3} v\ (\frac{3}{2}\ \pi - v)^{4}} \implies t^{(3)} (0) = \frac{6 - (\frac{3}{2}\ \pi)^{2}}{(\frac{3}{2}\ \pi)^{4}} \implies a_{3}= \frac{1 - \frac{1}{6}\ (\frac{3}{2}\ \pi)^{2}}{(\frac{3}{2}\ \pi)^{4}}$

Of course, if necessary [... with a little of patience (Wasntme)...] other terms can be computed...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 03 2014 on www.artofproblemsolving.com by the user Pirkuliyev Rovsen and not yet solved... ... calculate $f^{(n)} (0)$ of $f(x)= (\sin^{-1} x)^{2}$... Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 01 10 2014 on www.artofproblemsolving.com by the user carlover1 and not yet solved...

... find the sum...

$\displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k\ (n+1)^{k}}$

... where n is a natural number greater than 1...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 10 2014 on www.artofproblemsolving.com by the user carlover1 and not yet solved...

... find the sum...

$\displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k\ (n+1)^{k}}$

... where n is a natural number greater than 1...

For |x|< 1 is...

$\displaystyle \sum_{k=1}^{\infty} \frac{x^{k}}{k} = \ln \frac{1}{1 - x}\ (1)$

... so that setting $\displaystyle x = \frac{1}{n+1}$ we have... $\displaystyle S(n) = \ln \frac{1}{1 - \frac{1}{n+1}} = \ln (1 + \frac{1}{n})\ (2)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 14 2014 on www.mathhelpforum.com by the user Latsabb and not yet solved... ... we have just started working on the Taylor series, and one of our problems is to find the MacLaurin series for $\displaystyle \frac{\cos (x^{3})}{x}$...Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 01 14 2014 on www.mathhelpforum.com by the user Latsabb and not yet solved... ... we have just started working on the Taylor series, and one of our problems is to find the MacLaurin series for $\displaystyle \frac{\cos (x^{3})}{x}$...

It is quite surprising that none of the experts of the site have considered that the function $\displaystyle f(x)= \frac{\cos (x^{3})}{x}$ in x=0 has a singularity and that means that the McLaurin series of f(x) [i.e. the Taylor series in x=0...] doesn't exist... the Laurent series exists and it is easily found as follows...

$\displaystyle \cos x = 1 - \frac{x^{2}}{2} + ... + (-1)^{n}\ \frac{x^{2\ n}}{(2 n)!} + ... $

$\displaystyle \cos (x^{3}) = 1 - \frac{x^{6}}{2} + ... + (-1)^{n}\ \frac{x^{6\ n}}{(2 n)!} + ... $

$\displaystyle \frac{\cos (x^{3})}{x} = \frac{1}{x} - \frac{x^{5}}{2} + ... + (-1)^{n}\ \frac{x^{6\ n- 1}}{(2 n)!} + ... \ (1)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 21 2014 on www.artofproblemsolving.com by the user aziiri and noy yet solved...

For $\displaystyle n \ge 1$ we define a sequence... $\displaystyle a_{n} = \int_{0}^{\infty} \frac{d x} {(1 + x^{2})^{n}}$

1) Prove that $\displaystyle a_{n} \rightarrow 0$

2) Find $\alpha$,$\beta$ such that... $\displaystyle a_{n} - \sqrt{\frac{\pi}{4\ n}} \sim \alpha\ n^{\beta}$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 21 2014 on www.artofproblemsolving.com by the user aziiri and noy yet solved...

For $\displaystyle n \ge 1$ we define a sequence... $\displaystyle a_{n} = \int_{0}^{\infty} \frac{d x} {(1 + x^{2})^{n}}$

1) Prove that $\displaystyle a_{n} \rightarrow 0$

2) Find $\alpha$,$\beta$ such that... $\displaystyle a_{n} - \sqrt{\frac{\pi}{4\ n}} \sim \alpha\ n^{\beta}$

Using the Cauchy integral theorem on the upper 'big half circle' in the complex plane we obtain directly for $n \ge 2$ ...$\displaystyle a_{n}= \int_{0}^{\infty} \frac{d x}{(1 + x^{2})^{n}} = \frac{2\ \pi\ i}{(n-1)!} \lim_{z \rightarrow i} \frac{d^{n-1}}{d z^{n-1}} \frac{(z-i)^{n}}{(1 + z^{2})^{n}} = \frac{\pi}{2^{n}}\ \frac{(2 n - 3)!}{(n-1)!}\ (1)$

... but that isn't what is requested and we will proceed from (1) on in next post... Kind regards $\chi$ $\sigma$

 

[alyafey22]

Another way is looking at the beta integral

$$a_n = \int^\infty_0 \frac{1}{(1+x^2)^n}\, dx =\frac{1}{2} \int^\infty_0 \frac{dx}{\sqrt{x}(1+x)^{n}}=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(n-\frac{1}{2} \right)}{\Gamma(n)}$$

Then the denominator goes faster using the Sitrling approximation.

 

[chisigma]

Posted the 01 21 2014 on www.artofproblemsolving.com by the user aziiri and noy yet solved...

For $\displaystyle n \ge 1$ we define a sequence... $\displaystyle a_{n} = \int_{0}^{\infty} \frac{d x} {(1 + x^{2})^{n}}$

1) Prove that $\displaystyle a_{n} \rightarrow 0$

2) Find $\alpha$,$\beta$ such that... $\displaystyle a_{n} - \sqrt{\frac{\pi}{4\ n}} \sim \alpha\ n^{\beta}$

Using the Cauchy integral theorem on the upper 'big half circle' in the complex plane we obtain directly for $n \ge 2$ ...$\displaystyle a_{n}= \int_{0}^{\infty} \frac{d x}{(1 + x^{2})^{n}} = \frac{2\ \pi\ i}{(n-1)!} \lim_{z \rightarrow i} \frac{d^{n-1}}{d z^{n-1}} \frac{(z-i)^{n}}{(1 + z^{2})^{n}} = \frac{\pi}{2^{n}}\ \frac{(2 n - 3)!}{(n-1)!}\ (1)$

... but that isn't what is requested and we will proceed from (1) on in next post...

If we write the (1) in a 'more rational form' we obtain... $\displaystyle a_{n}= \int_{0}^{\infty} \frac{d x}{(1 + x^{2})^{n}} = \frac{\pi}{2}\ \prod_{k=1}^{n-1} (1 - \frac{1}{2\ k})\ (1)$... and, because the series $\displaystyle - \sum_{k =1}^{\infty} \frac{1}{2\ k}$ is a divergent all negative terms series, the infinite product $\displaystyle \prod_{k=1}^{\infty} (1 - \frac{1}{2\ k})$ diverges to 0.

It is easy to see that the sequence $a_{n}$ is the solution of the difference equation...

$\displaystyle a_{n+1} = a_{n} (1 - \frac{1}{2\ n}),\ a_{1} = \frac{\pi}{2}\ (2)$

... so that is $\displaystyle \Delta_{n} = a_{n+1} - a_{n} = - \frac{a_{n}}{2\ n}$ and that leads to approximate (2) with the ODE...

$\displaystyle y^{\ '} = - \frac{y}{2\ x}\ (3)$

... the solution of which is $\displaystyle y= \frac{c}{\sqrt{x}}$ so that is $\displaystyle a_{n} \sim \frac{c}{\sqrt{n}}$... Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 02 27 2014 on www.artofproblemsolving.com by the user AndrewTom and not yet solved...

Prove that...

$\displaystyle \int_{0}^{1} \frac{x^{n} - 1}{\ln x}\ d x = \ln\ (n+1)$



Kind regards

$\chi$ $\sigma$

 

[Saitama]

Posted the 02 27 2014 on www.artofproblemsolving.com by the user AndrewTom and not yet solved...

Prove that...

$\displaystyle \int_{0}^{1} \frac{x^{n} - 1}{\ln x}\ d x = \ln\ (n+1)$



Kind regards

$\chi$ $\sigma$

The exact same integral is evaluated in Zaid's thread: http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233.html

 

[Saitama]

Posted the 01 03 2014 on www.artofproblemsolving.com by the user Pirkuliyev Rovsen and not yet solved... ... calculate $f^{(n)} (0)$ of $f(x)= (\sin^{-1} x)^{2}$... Kind regards $\chi$ $\sigma$

Hi chisigma!

Did you find a solution to this? I am confused by the notation. Does $f^2(x)$ means $f(f(x))$ or the second derivative $f''(x)$?

Thanks!

 

[chisigma]

Hi chisigma!

Did you find a solution to this? I am confused by the notation. Does $f^2(x)$ means $f(f(x))$ or the second derivative $f''(x)$?

Thanks!

The notation $f^n (x)$ means $f(x)$ eleved to the n-th power, so that $f^{2} (x)$ means $f(x)*f(x)$. The notation $f^{(n)} (x)$ means $\frac{d^{n}}{d x^{n}} f(x)$ so that $f^{(2)} (x)$ means $\frac{d^{2}}{d x^{2}} f(x)$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 03 02 2014 on www.artofproblemsolving.com by the user bobthesmartypants and not yet solved...

A sequence $\{a_{n}\}$ satisfies $\displaystyle a_{n+3} = \frac{a_{n+2} + a_{n+1} + a_{n}}{3}$. Given $a_{0},a_{1},a_{2}$ find $\displaystyle \lim_{n \rightarrow \infty} a_{n}$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 03 02 2014 on www.artofproblemsolving.com by the user bobthesmartypants and not yet solved...

A sequence $\{a_{n}\}$ satisfies $\displaystyle a_{n+3} = \frac{a_{n+2} + a_{n+1} + a_{n}}{3}$. Given $a_{0},a_{1},a_{2}$ find $\displaystyle \lim_{n \rightarrow \infty} a_{n}$...

The characteristic equation related to the difference equation is...

$\displaystyle x^{3} - \frac{x^{2}}{3} - \frac{x}{3} - \frac{1}{3} = 0\ (1)$

... the solutions of which are $x_{1} = 1,\ x_{2,3} = - \frac{1}{3} \pm i\ \frac{\sqrt{2}}{3}$, so that the general solution is... $\displaystyle a_{n} = c_{1} + (-\frac{1}{3})^{n}\ \{c_{2}\ \cos (n\ \frac{\sqrt{2}}{3}) + c_{3}\ \sin (n\ \frac{\sqrt{2}}{3})\}\ (2)$

... and...

$\displaystyle \lim_{n \rightarrow \infty} a_{n} = c_{1}\ (3)$

The constant $c_{1}$ can be derived from the initial conditions solving the following linear system...

$\displaystyle c_{1} + c_{2} = a_{0}$

$\displaystyle c_{1} - \frac{1}{3}\ \cos \frac{\sqrt{2}}{3}\ c_{2} - \frac{1}{3}\ \sin \frac{\sqrt{2}}{3}\ c_{3} = a_{1}$

$\displaystyle c_{1} + \frac{1}{9}\ \cos \frac{2\ \sqrt{2}}{3}\ c_{2} + \frac{1}{9}\ \sin \frac{2\ \sqrt{2}}{3}\ c_{3} = a_{2}\ (4)$

Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 03 19 2014 on www.artofproblemsolving by the user james digol and not jet solved...

Solve the differential equation...

$\displaystyle y^{\ ''} + \frac{1}{x}\ y^{\ '} = c\ (1)$

... where c is a constant...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 03 19 2014 on www.artofproblemsolving by the user james digol and not jet solved...

Solve the differential equation...

$\displaystyle y^{\ ''} + \frac{1}{x}\ y^{\ '} = c\ (1)$

... where c is a constant...

With the substitution y ' = z the DE becomes...

$\displaystyle z^{\ '} + \frac{z}{x} = c\ (1)$

... and with standard procedure we arrive to...

$\displaystyle z + e^{- \int \frac{d x}{x}}\ (\int c\ e^{\int \frac{d x}{x}}\ d x + c_{1} ) = c\ \frac{x}{2} + \frac{c_{1}}{x}\ (2)$

... and with another integration we arrive at...

$\displaystyle y = c\ \frac{x^{2}}{2} + c_{1}\ \ln |x| + c_{2}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 05 17 2014 on www.artofproblemsolving.com by the user uniquesaylor and not yet solved...

Evaluate PV of the following improper integral...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{\cos (\alpha\ t)}{1 + t^{3}}\ dt\ (1)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 08 16 2014 on www.mathhelpforum.com by the user superzhangmch and not yet solved...

... prove...

$\displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n} = 0\ (1)$...or show it is not true...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 08 16 2014 on www.mathhelpforum.com by the user superzhangmch and not yet solved...

... prove...

$\displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n} = 0\ (1)$...or show it is not true...

From the expression of the function sin x as infinite product ...

$\displaystyle \sin x = x\ \prod_{k=1}^{\infty} (1 - \frac{x^{2}}{k^{2}\ \pi^{2}})\ (1)$

… we derive... $\displaystyle a_{n} = \frac{\ln |\sin n|}{n} = \frac{\ln n}{n} +\sum_{k=1}^{\infty} \frac{\ln |1 - \frac{n^{2}}{k^{2}\ \pi^{2}}|}{n}\ (2)$

The proposed question is not trivial since to show that $\displaystyle \lim_{n \rightarrow \infty} a_{n} = 0$ it is necessary to show that each term of the series (2) tends to zero as n tends to infinity. This can be critical when it is $\displaystyle \frac{n}{k} \sim \pi$, that is when $\displaystyle \frac{n}{k}$ is a 'good approximation' of $\pi$ since the logarithm can take negative values even higher. The workload needed for this investigation is not light but fortunately with a short research it has found a German text of the late nineteenth century, where are example values ​​$\displaystyle \frac{n}{k}$ 'good approximations' of $\pi$...

Archimedes,Huygens, Lambert, Legendre.

Vier Abhandlungen über die Kreismessung. Deutsh hrsg. und mit einerÜbersicht über die
Geschichte des Problemes von der Quadratur des Zirkels

Published 1892 by B.G. Teubner in Leipzig .

Written in German.


pages 146-147 has a table of rational approximations of pi...1:3
7:22
106:333
113 : 355
33102: 103993
33215 : 104348
66317: 208341
99532 : 312689
265381: 833719
364913 : 1146408
1360120: 4272943
1725033 : 5419351
25510582: 80143857
52746197 : 165707065
78256779: 245850922
131002976 :411557987
340262713 :1068966896
811528438 : 2549491779
1963319607 : 6167950454
4738167652: 14885392687
6701487259 : 21053343141
567663097408 : 1783366216531
1142027682075 : 3587785776203
1709690779483 : 5371151992734
2851718461558 : 8958937768937
107223273857129 : 336851849443403
324521540032945 : 1019514486099146

The next job in the next posted ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

From the expression of the function sin x as infinite product ...

$\displaystyle \sin x = x\ \prod_{k=1}^{\infty} (1 - \frac{x^{2}}{k^{2}\ \pi^{2}})\ (1)$

… we derive... $\displaystyle a_{n} = \frac{\ln |\sin n|}{n} = \frac{\ln n}{n} +\sum_{k=1}^{\infty} \frac{\ln |1 - \frac{n^{2}}{k^{2}\ \pi^{2}}|}{n}\ (2)$

The proposed question is not trivial since to show that $\displaystyle \lim_{n \rightarrow \infty} a_{n} = 0$ it is necessary to show that each term of the series (2) tends to zero as n tends to infinity. This can be critical when it is $\displaystyle \frac{n}{k} \sim \pi$, that is when $\displaystyle \frac{n}{k}$ is a 'good approximation' of $\pi$ since the logarithm can take negative values even higher. The workload needed for this investigation is not light but fortunately with a short research it has found a German text of the late nineteenth century, where are example values ​​$\displaystyle \frac{n}{k}$ 'good approximations' of $\pi$...

Archimedes,Huygens, Lambert, Legendre.

Vier Abhandlungen über die Kreismessung. Deutsh hrsg. und mit einerÜbersicht über die
Geschichte des Problemes von der Quadratur des Zirkels

Published 1892 by B.G. Teubner in Leipzig .

Written in German.


pages 146-147 has a table of rational approximations of pi...1:3
7:22
106:333
113 : 355
33102: 103993
33215 : 104348
66317: 208341
99532 : 312689
265381: 833719
364913 : 1146408
1360120: 4272943
1725033 : 5419351
25510582: 80143857
52746197 : 165707065
78256779: 245850922
131002976 :411557987
340262713 :1068966896
811528438 : 2549491779
1963319607 : 6167950454
4738167652: 14885392687
6701487259 : 21053343141
567663097408 : 1783366216531
1142027682075 : 3587785776203
1709690779483 : 5371151992734
2851718461558 : 8958937768937
107223273857129 : 336851849443403
324521540032945 : 1019514486099146

The next job in the next posted ...

Using the calculator of windows and the table of 'best approximations' of $\pi$ found in the German text of the late nineteenth century, we have calculated the values ​​of ...

$\displaystyle p(\frac{k}{n}) = \frac{\ln |1 - \frac{n^{2}}{k^{2}\ \pi^{2}}|}{n}\ (1)$

The results are as follows ...

$\frac{k}{n}= \frac{1}{3} \implies p(\frac{k}{n}) = -2.429...$

$\frac{k}{n} = \frac{7}{22} \implies p(\frac{k}{n}) = -.3238...$

$\frac{k}{n}= \frac{106}{333} \implies p(\frac{k}{n}) = - .2956...$

$\frac{k}{n}= \frac{113}{355} \implies p(\frac{k}{n}) = -.0439...$

$\frac{k}{n}= \frac{33102}{103993} \implies p(\frac{k}{n}) = - 2.0889...\ 10^{-4}$

$\frac{k}{n}= \frac{33215}{104348} \implies p(\frac{k}{n}) = -2.135...\ 10^{-4}$

$\frac{k}{n}= \frac{66317}{208341} \implies p(\frac{k}{n}) = -1.117...\ 10^{-4}$

$\frac{k}{n}= \frac{99532}{312689} \implies p(\frac{k}{n}) = -7.902...\ 10^{-5}$

$\frac{k}{n}= \frac{265381}{833719} \implies p(\frac{k}{n}) = -3.108...\ 10^{-5}$

$\frac{k}{n}= \frac{364913}{1146408} \implies p(\frac{k}{n}) = -2.408...\ 10^{-5}$

$\frac{k}{n}= \frac{1360120}{4272943} \implies p(\frac{k}{n}) = -6.784...\ 10^{-6}$

$\frac{k}{n}= \frac{1725033}{5419351} \implies p(\frac{k}{n}) = -5.8849...\ 10^{-6}$

$\frac{k}{n}= \frac{25510582}{80143857} \implies p(\frac{k}{n}) = -4.4341...\ 10^{-7}$

$\frac{k}{n}= \frac{52746197}{165707065} \implies p(\frac{k}{n}) = -2.22...\ 10^{-7}$

$\frac{k}{n}= \frac{78256779}{245850922} \implies p(\frac{k}{n}) = -1.5269...\ 10^{-7}$

$\frac{k}{n}= \frac{31002976}{411557987} \implies p(\frac{k}{n}) = -9.4603...\ 10^{– 8}$

$\frac{k}{n}= \frac{340262713}{1068966896} \implies p(\frac{k}{n}) = -1.502...\ 10^{-8}$

$\frac{k}{n}= \frac{811528438}{2549491779} \implies p(\frac{k}{n}) = -1.6667...\ 10^{-8}$

$\frac{k}{n}= \frac{1963319607}{6167950454} \implies p(\frac{k}{n}) = -7.21...\ 10^{-9}$

$\frac{k}{n}= \frac{4738167652}{14885392687} \implies p(\frac{k}{n}) = -3.04758...\ 10^{-9}$

$\frac{k}{n}= \frac{6701487259}{21053343141} \implies p(\frac{k}{n}) = -2.38...\ 10^{-9}$

$\frac{k}{n}= \frac{567663097408}{1783366216531} \implies p(\frac{k}{n}) = -3.112...\ 10^{-11}$

$\frac{k}{n}= \frac{1142027682075}{3587785776203} \implies p(\frac{k}{n}) = -1.585...\ 10^{-11}$

$\frac{k}{n}= \frac{1709690779483}{5371151992734} \implies p(\frac{k}{n}) = -1.067...\ 10^{-11}$

$\frac{k}{n}= \frac{2851718461558}{8958937768937} \implies p(\frac{k}{n}) = -6.76...\ 10^{-12}$

$\frac{k}{n}= \frac{107223273857129}{336851849443403} \implies p(\frac{k}{n}) = -1.81...\ 10^{-13}$


At this point I have to stop because The subsequent step provides a value $\frac{n}{k} = \frac{1019514486099146}{324521540032945}$ that coincides to the twenty-four decimal approximation of $\pi$ of the calculator I used. It is clear that even with a more powerful calculation tool is difficult, following this way, to give a definite answer to the proposed question and a different way must be tried ... always that what has been already done by someone in the past ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 8 18 2014 on www.artofproblemsolving.com by the user Kid_Dynamite and not jet solved...

How to prove that $\displaystyle \int_{0}^{x} \frac{dt} {\ln t} = \mathcal{O}\ \{\frac{x}{\ln x}\}$ ?...

Kind regards

$\chi$ $\sigma$

 

[Euge]

Posted the 8 18 2014 on www.artofproblemsolving.com by the user Kid_Dynamite and not jet solved...

How to prove that $\displaystyle \int_{0}^{x} \frac{dt} {\ln t} = \mathcal{O}\ \{\frac{x}{\ln x}\}$ ?...

Kind regards

$\chi$ $\sigma$

Does Kid_Dynamite want this to be proved for $x \to 0$ or $x \to \infty$?

 

[chisigma]

Does Kid_Dynamite want this to be proved for $x \to 0$ or $x \to \infty$?

I stronghly suppose $\displaystyle x \rightarrow \infty$...

Kind regards

$\chi$ $\sigma$

 

[Euge]

I stronghly suppose $\displaystyle x \rightarrow \infty$...

Kind regards

$\chi$ $\sigma$

By L'hospital's rule,

$$\lim_{x\to \infty} \dfrac{\int_0^x \frac{dt}{\ln t}}{\frac{x}{\ln x}} = \lim_{x\to \infty} \dfrac{\frac{1}{\ln x}}{\frac{\ln x - 1}{\ln^2 x}} = \lim_{x \to \infty} \frac{\ln x}{\ln x - 1} = 1$$.

So for all sufficiently large $x$,

$$ |\int_0^x \frac{dt}{\ln t}| < \frac{3}{2} |\frac{x}{\ln x}|$$

Consequently,

$$ \int_0^x \frac{dt}{\ln t} = \mathcal{O}\left(\frac{x}{\ln x}\right)$$ as $$x\to \infty$$.

 

[mathbalarka]

Euge, can you elaborate on how you're getting the inequality at the second step from the first.

Also, I might as well mention that

$$\int_0^x \frac{dt}{\log(t)}$$

Doesn't quite make sense, as the usual integral blows up at $t = 1$. One usually makes an indentation around the contour of the integral around the pole, to have

$$\int_0^{1-\epsilon} \frac{dt}{\log(t)} + \int_{1+\epsilon}^x \frac{dt}{\log(t)}$$

 

[Euge]

Euge, can you elaborate on how you're getting the inequality at the second step from the first.

Also, I might as well mention that

$$\int_0^x \frac{dt}{\log(t)}$$

Doesn't quite make sense, as the usual integral blows up at $t = 1$. One usually makes an indentation around the contour of the integral around the pole, to have

$$\int_0^{1-\epsilon} \frac{dt}{\log(t)} + \int_{1+\epsilon}^x \frac{dt}{\log(t)}$$

This logarithmic integral is a special function, usually denoted $\text{li}(x)$. It is defined for all positive $x$ different from 1, and when $x > 1$, the integral is to interpreted as the Cauchy principal value. In any case, I argued thinking of $\text{Li}(x)$ (where 2 is the lower limit) instead of $\text{li}(x)$, so I must approach this differently. I'll come back later.

Edit: For $x \ge 2$,

$\displaystyle \mathrm{li}(x) = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \varepsilon}^x \frac{dt}{2}\right) = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \epsilon}^2 \frac{dt}{\ln t} + \int_2^x \frac{dt}{\ln t}\right)$
$\displaystyle = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \epsilon}^2 \frac{dt}{\ln t}\right) + \mathrm{Li}(x) =\mathrm{li}(2) + \mathrm{Li}(x).$

In the argument I gave earlier, the lower limit of $0$ is to be replaced with $2$. Then the argment is valid and $\mathrm{Li}(x) = \mathcal{O}(x/\ln x)$ as $x \to \infty$. Since

$\displaystyle \lim_{x\to \infty} \dfrac{\mathrm{li}(2)}{\frac{x}{\ln x}} = \lim_{x\to \infty} \mathrm{li}(2) \frac{\ln x}{x} = 0$,

it follows that $\mathrm{li}(2) = \mathcal{O}(x/\ln x)$ and thus

$\displaystyle \mathrm{li}(x) = \mathcal{O}\left(\frac{x}{\ln x}\right)$ as $\displaystyle x \to \infty$.

 

[mathbalarka]

$\displaystyle \lim_{x\to \infty} \dfrac{\mathrm{li}(2)}{\frac{x}{\ln x}} = \lim_{x\to \infty} \mathrm{li}(2) \frac{\ln x}{x} = 0$

Typo 1 : You mean $1$ instead of $0$. We evidently don't want our estimates to be little-o!

it follows that $\mathrm{li}(2) = \mathcal{O}(x/\ln x)$ and thus ...

Typo-2 : You meant $\mathrm{li}(2) = O(1)$.

OK, now I'll just zip the lip (Lipssealed)

 

[Euge]

Typo 1 : You mean $1$ instead of $0$. We evidently don't want our estimates to be little-o!

No, it's 0 by L'hospital's rule. Also, a function that is $o(g)$ is also $O(g)$ (although the converse does not hold).

Typo-2 : You meant $\mathrm{li}(2) = O(1)$.

No. Since $\mathrm{li}(2)/(x/\ln x) \to 0$ as $x\to \infty$, there exists a positive integer $x_0$ such that $|\mathrm{li}(2)| < |x/\ln x|$ for all $x \ge x_0$. Thus $\mathrm{li}(2) = \mathcal{O}(x/\ln x)$.