The following is a problem proposed in Pólya and Szegő's book "Problems and Theorems in Analysis"

Assume that $0<f(x)<x$ and $$f(x)=x-ax^k+bx^\ell+x^\ell \varepsilon(x),\,\;\;\;\lim_{x\to 0}\varepsilon(x)=0$$ for $0<x<x_0$, where $1<k<\ell$ and $a,b$ positive. Put $$v_0=x,\;\; v_1=f(v_0),\;\;v_2=f(v_1),\ldots,\;\;\;v_n=f(v_{n-1}),\ldots$$ Then we have for $n\to\infty$$$n^{\frac{1}{k-1}}v_n\to[(k-1)a]^{\frac{-1}{k-1}}$$

PROOF I will prove the better looking$$nv_n^{k-1}\to \frac{1}{a(k-1)}$$

To this end, first note that $v_n\to 0$, since $v_n$ decreases because of $0<f(x)<x$ and because of continuity at the origin, along with $f(0)=0$. Also note $f(x)/x\to 1$. Now, we have after some algebraic meddling that $$\frac{1}{{{x^{k - 1}}}} - \frac{1}{{f{{\left( x \right)}^{k - 1}}}} = \frac{{f{{\left( x \right)}^{k - 1}} - {x^{k - 1}}}}{{{x^{k - 1}}f{{\left( x \right)}^{k - 1}}}} = \frac{{f\left( x \right) - x}}{{{x^k}}}\frac{{f\left( x \right)}}{x}{\sum\limits_{m = 0}^{k - 2} {\left( {\frac{x}{{f\left( x \right)}}} \right)} ^m}$$

It follows that $$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{f{{\left( x \right)}^{k - 1}}}} - \frac{1}{{{x^{k - 1}}}}} \right) = a\left( {k - 1} \right)$$

Thus, since $v_n\to 0$, we must have $$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{v_{n + 1}^{k - 1}}} - \frac{1}{{v_n^{k - 1}}}} \right) = a\left( {k - 1} \right)$$

Appealing to Cesàro's theorem, it follows that $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\sum\limits_{m = 0}^{n - 1} {\frac{1}{{v_{m + 1}^{k - 1}}} - \frac{1}{{v_m^{k - 1}}}} } \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{1}{{v_n^{k - 1}}} - \frac{1}{{v_0^{k - 1}}}} \right) = a\left( {k - 1} \right)$$

Which is what we wanted to prove. $\blacktriangle$

DSC The above is inspired in this particular proof when $f(x)=\sin x$.

Do you know any other proofs?

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    $\begingroup$ I guess you are aware that Pólya and Szegő provide full solutions to almost all of the exercises. $\endgroup$
    – Martin
    Jun 23, 2013 at 1:20
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    $\begingroup$ @Peter : Did you understand it or did you just not like it? Sometimes I don't like proofs because I feel like the ideas are not natural ; once I understand why they are natural in another context I like them better. $\endgroup$ Jun 23, 2013 at 2:14
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    $\begingroup$ @PatrickDaSilva Now? =P $\endgroup$
    – Pedro
    Jun 23, 2013 at 19:39
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    $\begingroup$ @PeterTamaroff: are you sure that $v_n\to0$? If $f$ is not continuous, then $$ f(x)=\left\{\begin{array}{}\frac32-\frac2{2x+1}&\text{for }x\gt\frac12\\\frac{x}2&\text{for }x\le\frac12\end{array}\right. $$ satisfies the conditions, but if $v_0=1$, then $v_n\to\frac12$. $\endgroup$
    – robjohn
    Jun 25, 2013 at 1:09
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    $\begingroup$ What is $x_0$? You said neither $x_0\le1$ nor $v_0\lt x_0$. In any case, just set $v_0=\frac56$ (previously known as $v_1$) and you get the same sequence tending to $v_k\to\frac12$. And we can change the definition for $x\le\frac12$ to $x\sqrt{1-x}$ to give it the behavior you want near $x=0$. I think you need to add that $f$ is continuous. $\endgroup$
    – robjohn
    Jul 9, 2013 at 19:31

1 Answer 1


I'll try a short answer, then I can make it longer and go into more details. Let $x_n$ be a sequence such that $x_{n+1}=f(x_n)$ with $f(x) = x-ax^{k}+o(x^k)$.

I'll provide you another proof, or say, a manner of thinking it :

We have $x_{n+1}-x_n = f(x_n)-x_n \sim a x_n^{k}$. The "equivalent" for functions is the ODE $y'=a y^{k}$ which reads $(y^{1-k})' = a(1-k)$.

Then the "discrete equivalent" of that is that we can expect that : $$x_{n+1} ^{1-k} - x_n^{1-k} \to a(1-k).$$ And then we prove this by just the same way you did.

Please not that I do not pretend of a truly new proof, the most interesting part of it is the method. For example it apply to the problem of finding an equivalent (or asymptotic development) of $x_{n+1}=x_{n}^2 \log x_n$ which is more tricky ;)


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