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Given $$f(x)=1-x^2+x^3 \qquad x\in[0,1]$$ calculate $$ \lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx} $$ where $f^n(x)=\underbrace{f(x)·f(x)·\dots\text{·}f(x)}_{n\ \text{times}}$.

This is a question from CMC(Mathematics competition of Chinese)in $2017$. The solution provides an idea: given $s∈(0,\frac{1}{2}),$ prove:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$ The final result is $\ln2.$


My approach
For this:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$ I want to do piecewise calculation:$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$Here is the proof: when$\ \ n≥\frac{1}{s^2}$, $$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$Here is the proof: given $t,0<t<s<\frac{1}{2},$then$$f(t)>f(s)>f(1-s).$$Define $m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$ so$$m_t=f(t)>f(1-s)=M_s.$$$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$ In conclusion,we can get:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$

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    $\begingroup$ What have you tried? What are your thoughts on the problem? $\endgroup$
    – Gary
    Oct 19, 2021 at 8:41
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    $\begingroup$ @Piquancy Copy paste that comment into your question so people can directly see what you've been upto. Some people may not go through the comments. $\endgroup$
    – William
    Oct 19, 2021 at 8:52
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    $\begingroup$ Is $f^n(x)=f(f(\cdots f(x)\cdots))$ ($n$ times) ? $\endgroup$
    – Jean Marie
    Oct 19, 2021 at 9:32
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    $\begingroup$ What is $*$? Convolution? $\endgroup$
    – Marcos
    Oct 19, 2021 at 10:23
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    $\begingroup$ @Piquancy I think you just write $[f(x)]^n$ which is clear. $\endgroup$
    – River Li
    Oct 21, 2021 at 5:51

2 Answers 2

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Alternative proof:

Let \begin{align*} J_n &= \int_0^1 (1 - x^2 + x^3)^n \,\mathrm{d} x, \\ K_n &= \int_0^1 (1 - x^2 + x^3)^n \ln (x + 2)\,\mathrm{d} x. \end{align*}

First, we have $$K_n \ge \int_0^1 (1 - x^2 + x^3)^n \ln 2\,\mathrm{d} x = \ln 2 \cdot J_n. \tag{1}$$

Second, we have \begin{align*} J_n &\ge \int_0^{1/\sqrt{n}} (1 - x^2 + x^3)^n \,\mathrm{d} x \\ &\ge \int_0^{1/\sqrt{n}} (1 - 1/n)^n \,\mathrm{d} x \\ &= \frac{1}{\sqrt{n}}(1 - 1/n)^n. \tag{2} \end{align*}

Third, we have \begin{align*} K_n - \ln 2 \cdot J_n &= \int_0^1 (1 - x^2 + x^3)^n \ln (1 + x/2)\,\mathrm{d} x \\ &\le \int_0^1 (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x \\ &= \int_0^{1/2} (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x + \int_{1/2}^{5/6} (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x\\ &\qquad + \int_{5/6}^1 (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x \\ &\le \int_0^{1/2} (1 - x^2 + x^3)^n \, (2x - 3x^2)\,\mathrm{d} x + \int_{1/2}^{5/6} (191/216)^n \, \frac{x}{2}\,\mathrm{d} x\\ &\qquad + \int_{5/6}^1 (1 - x^2 + x^3)^n \, (-2x + 3x^2)\,\mathrm{d} x \\ &= \frac{-(1 - x^2 + x^3)^{n + 1}}{n + 1}\Big\vert_0^{1/2} + \frac19(191/216)^n + \frac{(1 - x^2 + x^3)^{n + 1}}{n + 1}\Big\vert_{5/6}^1\\ &= \frac{24n - 167}{216n + 216}(191/216)^n - \frac{7}{8n + 8}(7/8)^n + \frac{2}{n + 1} \tag{3} \end{align*} where we have used:
(i) $\ln(1 + u) \le u$ for all $u \ge 0$;
(ii) $x/2 \le 2x - 3x^2$ for all $x\in [0, 1/2]$;
(iii) $1 - x^2 + x^3 \le 191/216$ for all $x\in [1/2, 5/6]$;
(iv) $x/2 \le -2x + 3x^2$ for all $x\in [5/6, 1]$.

Fourth, from (1), we have $$\frac{K_n}{J_n} \ge \ln 2.$$ From (1), (2) and (3), we have $$\frac{K_n}{J_n} \le \ln 2 + \frac{\frac{24n - 167}{216n + 216}(191/216)^n - \frac{7}{8n + 8}(7/8)^n + \frac{2}{n + 1}}{\frac{1}{\sqrt{n}}(1 - 1/n)^n}.$$ Note that $$\lim_{n\to \infty} \frac{\frac{24n - 167}{216n + 216}(191/216)^n - \frac{7}{8n + 8}(7/8)^n + \frac{2}{n + 1}}{\frac{1}{\sqrt{n}}(1 - 1/n)^n} = 0.$$ Thus, using the squeeze theorem, we have $$\lim_{n\to \infty} \frac{K_n}{J_n} = \ln 2.$$

We are done.

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Fisrtly,we prove that:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$ Perform piecewise calculation:$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$When$\ \ n≥\frac{1}{s^2}$, $$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$Given $t,0<t<s<\frac{1}{2},$then$$f(t)>f(s)>f(1-s).$$Define $m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$ so$$m_t=f(t)>f(1-s)=M_s.$$$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$ In conclusion,we can get:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$ Secondly,we calculate the result:$\ln2$.
For $\varepsilon\in(0,\ln\frac{5}{4})$,take $s=2(e^\varepsilon-1)$,than $s\in(0,\frac{1}{2}),\ln\frac{2+s}{2}=\varepsilon.$
From the above conclusion,we can know:
$\exists N\geq1,s.t.$,when $n\geq N$,$$\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq \varepsilon.$$So$$\lvert\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}-\ln2\rvert=\frac{\int_{0}^{1}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{1}f^n(x)dx}$$$$\leq\frac{\int_{0}^{s}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{s}f^n(x)dx}+\frac{\int_{s}^{1}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{s}f^n(x)dx}$$$$\le\ln\frac{s+2}{2}+\frac{\ln\frac{3}{2}\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\varepsilon(1+\ln\frac{3}{2}).\\$$Let $\varepsilon\rightarrow0$,we can get$$\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}=\ln2.\\$$That's all.

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