Asymptotics of the solution of a differential equation

Let $f$ be twice continuously differentiable function defined on $\mathbb{R}$, such that $f(x)f''(x)=1$ for all $x\ge 0$, $f(0)=1$ and $f'(0)=0$.

Find $$\lim_{x\to\infty} \frac{f(x)}{x\sqrt{\ln(x)}}.$$

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I.

The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.

 Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds 

$$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$

Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$

$\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence

$$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$

and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_{n\to\infty}x_n=0.$

$\textbf{2)}$ Now $1/x_n\to +\infty$  increasingly and we can use Stolz theorem as follows:

$$\lim_{n\to\infty}nx_n =\lim_{n\to\infty}\frac{n}{\frac{1}{x_n}}=\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}$$

Now consider the defining equation. It can be rewritten as $(1+x_{n+1})^n=1+n x_n$ hence $ x_n=x_{n+1}+S/n$ where $\displaystyle S=\sum_{k=2}^n{n\choose k}x_{n+1}^k.$

Thus 

$$\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}=\lim_{n\to\infty}\frac{x_{n+1}\left(x_{n+1}+\frac{S}{n}\right)}{\frac{S}{n}}=\lim_{n\to\infty}\frac{nx_{n+1}^2}{S},$$

where we have used that $x_{n+1}\to 0$. Now this can be rewritten as follows

$$\lim_{n\to\infty}\frac{nx_{n+1}^2}{S}=\lim_{n\to\infty}\frac{n}{{n\choose 2}+S'},$$

 where $\displaystyle S'=\sum_{k=3}^n{n\choose k}x_{n+1}^{k-2}>0.$ This shows that the last limit is $0$.

$\textbf{3)}$  Apply the Stolz theorem in the very same way as above -

$$\lim_{n\to\infty}n^2x_n =\lim_{n\to\infty}\frac{n^2}{\frac{1}{x_n}}=\frac{(n+1)^2-n^2}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}(2n+1)\frac{x_{n+1}x_n}{x_n-x_{n+1}}=2\lim_{n\to\infty}\frac{n^2x_{n+1}^2}{S}.$$

For convenience we work with the reciprocal limit 

$$\lim_{n\to\infty}\frac{S}{n^2x_{n+1}^2}=\lim_{n\to\infty}\frac{\sum_{k=2}^n{n\choose k}x_{n+1}^k}{n^2x_{n+1}^2}=\frac{1}{2}+\lim_{n\to\infty}\sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}.$$

 We would be done if we prove, that the last limit is $0$.  Denote $A_n=\displaystyle \sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}$. Clearly $A_n>0$. Fix $\varepsilon>0$. According to $2)$, for large enough $n$ holds $\displaystyle x_{n+1}<\frac{\varepsilon}{n}.$

Thus (after completing to the binomial formula) $$A_n<\sum_{k=3}^n{n\choose k}\frac{\varepsilon^{k-2}}{n^k}=\frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2},$$

hence $$\limsup_{n\to\infty}A_n\le \lim_{n\to\infty} \frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2}=\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}.$$

This is true for arbitrary $\varepsilon>0$. Letting $\varepsilon\to 0$ in the last bound we obtain (using Taylor expansion of the exponent near $0$) that  

$$\lim_{\varepsilon\to 0}\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}=0,$$

whence  $\displaystyle\limsup_{n\to\infty}A_n=0,$ which finishes the proof.

IMO shortlist, 1998

G7. $ABC$ is a triangle with $\angle ACB = 2 \angle ABC$. $D$ is a point on the side $BC$ such that $DC = 2 BD$. $E$ is a point on the line $AD$ such that $D$ is the midpoint of $AE$. Show that $\angle ECB + 180 = 2 ∠EBC$.

 

Solution.  Put $D$ in the center. We can rewrite everything in terms of $a$ - the number corresponding to $A$ and $b$ - the number corresponding to $B$. Thus $C$ is $-2b$, E is $-a$.
The relation between the angles at $B$ and $C$ is easily reduced to the following equation:
$$\frac{(a-b)^2}{|a-b|^2}\frac{a+2b}{|a+2b|}=\frac{b^3}{|b|^3}.$$
Squaring and representing modules with conjugates, we obtain
$$\frac{(a-b)^2}{(\bar a-\bar b)^2}\frac{a+2b}{\bar a+2\bar b}=\frac{b^3}{\bar b^3}.$$
Clearing the denominator and setting everything to the left we obtain
$$a^3\bar b^3-3a\bar b^3 b^3-\bar a^3b^3+3\bar a\bar b^2 b^3=0\ (*)$$
Similarly, the equality we want to proof is equivalent to
$$\left(\frac{\frac{-b}{|-b|}}{\frac{-a-b}{|-a-b|}}\right)^2=(-1)
\left(\frac{\frac{-a+2b}{|-a+2b|}}{\frac{b}{|b|}}\right)$$
Both of the arguments of the complex numbers on both sides are between 180 and 360, so squaring is equivalent transformation here. As above we obtain
$$\frac{(a+b)^2}{(\bar a+\bar b)^2}\frac{a-2b}{\bar a-2\bar b}=\frac{b^3}{\bar b^3}.$$
Everything on the left and expand - we obtain
$$-a^3\bar b^3+3a\bar b^3 b^3+\bar a^3b^3-3\bar a\bar b^2 b^3=0.$$
But this is just the same as $(*)$.

A problem told by Zhivko Petrov

 The following integral was proposed for homework to Applied Math, by Zhivko Petrov. 

$\textbf{Problem}.$ Evaluate

$$\int_{0}^1\frac{\ln(1-x+x^2)}{x^2-x}\text{d} x.$$

Later I communicated the problem to prof. Gadjev and he proposed a neat solution (to be presented later).

Now we elaborate a solution, on an idea proposed by prof. Babev. I would like to thank the afformentioned people, as well as David Petrov for pointing me to that idea.

$\textbf{Solution}.$ Introduce $$I(y)=\int_0^1\frac{\ln(1-y(x-x^2))}{x^2-x}\text{d}x.$$

We need to find $I(1)$. Clearly $I(0)=0$. Using differentiation under the integral sign we obtain 

$$I'(y)=\int_0^1\frac{1}{1-y(x-x^2)}\text{d}x.$$

Assuming that $y\in [0,1]$, one can easily integrate the last to obtain 

$$I'(y)=\frac{4 \arcsin\left(\frac{\sqrt{y}}{2}\right)}{\sqrt{y(4-y) }}.$$

The  latter is very easy to integrate (for example making the change $y=4t^2$) in order to obtain

$$I'(y)=\left(4 \arcsin\left(\frac{\sqrt{y}}{2}\right)^2\right)'.$$

Thus 

$$I(1)=\int_0^1I'(y)\text{d}y+I(0)= 4 \arcsin\left(\frac{\sqrt{1}}{2}\right)^2-4 \arcsin\left(\frac{\sqrt{0}}{2}\right)^2+0=4 \arcsin\left(\frac{1}{2}\right)^2=\frac{\pi^2}{9}.$$

A problem proposed by prof. Gadjev

Evaluate 

$$\int_0^{\infty}\frac{\ln x}{x^2+2x+5}\, \text{d}x$$

$\textbf{Solution.}$

The indefinite integral is not expressible in elementary functions. Denote the integral by $I$. First make the change of variables $\displaystyle x\to \frac{1}{x}$ to obtain 

$$I=-\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x\ \ $$

In the original integral make the change of variables $x\to 5x$ to obtain 

$$ I=\int_0^{\infty}\frac{\ln (5x)}{25x^2+10x+5}5\, \text{d}x=\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=$$

$$-I+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x$$

whence 

$$I=\frac{1}{2} \int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=\frac{1}{8}\arctan(2)\ln(5)$$

Geometric-analytic problem

Let $f:\mathbb{R}^2\to\mathbb{R}$ be strictly positive Lipschitz function with constant $1/2$. Let $A$ be a nonempty subset of $\mathbb{R}^2$, such that if $x\in A$ and $y\in\mathbb{R}^2$ with $\|x-y\|=f(x)$, then $y\in A$. Prove that $A=\mathbb{R}^2$.

The problem was proposed to Bulgarian TST, 2009. Do you know earlier source or some context of the problem?

A problem proposed by prof. Babev

Consider the sequence
$$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$
It could be proven that
$$\lim_{n\to \infty}n a_n=\frac{1}{2}$$
 Now we are going to prove that
$$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$
and show how to derive all such limits.
$\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as
$$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$
$$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$
Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to
$$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$
Since for $s\in[0,1]$, $\displaystyle \frac{n}{n+s}$ is bounded between $1$ and $ \displaystyle \frac{n}{n+1}$, the last limit is bounded between the limits
$$ \lim_{n\to \infty}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s\ \ \mbox{and}\ \ \lim_{n\to \infty}\frac{n}{n+1}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s$$
which both are equal to $\displaystyle  \int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s=-\frac{1}{4}$, hence the result follows.

$\textbf{Comment.}$ In a similar vein it could be proven that
$$\lim_{n\to\infty}n\left(n\left(na_n-\frac{1}{2}\right)+\frac{1}{4}\right)=\int_0^1\sin^2(\pi s)s^2\text{d}s$$
and we could extend this further.
This could be viewed as a "Taylor series of the sequence at infinity", namely
$$a_n=\sum_{j=1}^\infty b_j\left(\frac{1}{n}\right)^j$$
where $$b_j= \int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\text{d}s$$
Having this perspective leads to general and more insightful solution. Consider the series
$$\sum_{j=1}^\infty b^j \left(\frac{1}{n}\right)^j=\sum_{j=1}^\infty\int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\left(\frac{1}{n}\right)^j\text{d}s=\int_{0}^{1}\sin^2(\pi s)\frac{1}{n+s}\text{d}s.$$
(where we have exchanged summation and integration and then calculated the sum of the geometric progression)
After changing the variables $s\to t-n$ we arrive at the integral
$$\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t$$
which is exactly $a_n$.