東京理科大学 理工学部 2010年度 問3

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解答作成者: 山中 晴文

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大学名 東京理科大学
学科・方式 理工学部
年度 2010年度
問No 問3
学部 理工学部
カテゴリ 積分法 ・ 積分法の応用
状態 解答 解説 ウォッチリスト

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\documentclass[b5paper,fleqn,papersize]{jsarticle} \usepackage{emathP} \usepackage{custom_suseum} \hoffset=0pt \textwidth=420pt \voffset=-30pt \textheight=610pt \footskip=15pt \renewcommand{\theenumi}{\arabic{enumi}} \renewcommand{\labelenumi}{(\theenumi)} \renewcommand{\theenumii}{\roman{enumii}} \renewcommand{\labelenumii}{(\makebox[5.6pt]{\theenumii})} \setlength{\mathindent}{2zw} %%% 自作 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\kakkoi}{(\hspace{1.37pt}\text{i}\hspace{1.37pt})} \newcommand{\kakkoii}{(\text{ii})} \newcommand{\kakkoiii}{(\hspace{-1.41pt}\text{iii}\hspace{-1.41pt})} \newcommand{\kakkoiv}{(\hspace{-1.25pt}\text{iv}\hspace{-1.25pt})} \newcommand{\kakkov}{(\hspace{0.15pt}\text{v}\hspace{0.15pt})} \newcommand{\kakkovi}{(\hspace{-1.25pt}\text{vi}\hspace{-1.25pt})} \newcommand{\kakkoI}{[\hspace{2.6pt}I\hspace{2.6pt}]} \newcommand{\kakkoII}{[\hspace{1.3pt}I\hspace{-1pt}I\hspace{1.3pt}]} \newcommand{\kakkoIII}{[I\hspace{-1pt}I\hspace{-1pt}I]} \newcommand{\migirule}{\vspace{-20pt}\begin{flushright}\rule{1pt}{7pt}\end{flushright}} \newcommand{\CLrule}{\rule[3.5pt]{150pt}{0.3pt}\vspace{-24pt} \begin{flushright}\rule[3.5pt]{250pt}{0.3pt}\end{flushright}} \newcommand{\renrule}{\hspace{-3.5pt}\rule[3.5pt]{100pt}{0.3pt}\vspace{-24pt} \begin{flushright}\rule[3.5pt]{300pt}{0.3pt}\end{flushright}} \newcommand{\dumyeqhspace}{\hspace{31.827pt}} \newcommand{\QED}{\hspace{1zw}\rule[0pt]{4pt}{8pt}} \newcommand{\答}{ \Cdots\Cdots(答)} \def\maru#1{{\ooalign{\hfil\raise.102ex\hbox{\small #1}\/\hfil\crcr \raise.167ex\hbox{\mathhexbox 20D}}}} \newcommand{\inbe}{\def\arraystretch{0.75}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 自作終わり %%% % \def\Anscolor{red} \def\Anscolor{white} \begin{document} \begin{FRAME} 次の問いに答えなさい。 \vspace{1zw} \begin{enumerate}<apnenum={\leftmargin=2zw}> \item 0以上の整数$m,~ n=0,~ 1,~ 2,~ \cdots$に対し,$I$\retu(m,n)を \[ I\retu(m,n)=\dint{0}{1}x^m(1-x)^n\,dx \] で定める。 \vspace{1zw} \begin{enumerate} \item $m \geqq 1$のとき,部分積分法を用いて, \[ I\retu(m,n)=\bunsuu{m}{n+1}\dint{0}{1}x^{m-1}(1-x)^{n+1}\,dx \] が成り立つことを示しなさい。 \vspace{1zw} \item $I\retu(m,n)=\bunsuu{m\kaizyou\,n\kaizyou}{(m+n+1)\kaizyou}$を示しなさい。 \end{enumerate} \vspace{2zw} \item $f(x)$を区間\![0,\,1]\!で定義された連続関数とする。自然数$n=1,~ 2,~ 3,~ \cdots$に 対し, \\ 多項式$P_n(x)$を \[ P_n(x)=\tretuwa{k=0}{n}\kumiawase{n}{k}f\left(\bunsuu{k}{n}\right)x^k(1-x)^{n-k} \] で定める。ここで,$\kumiawase{n}{k}=\bunsuu{n\kaizyou}{k\kaizyou\,(n-k)\kaizyou}$である。 このとき, \[ \dlim{n \to \infty}\dint{0}{1}P_n(x)\,dx=\dint{0}{1}f(x)\,dx \] となることを示しなさい。 \end{enumerate} \end{FRAME} \newpage \pagenumbering{arabic} \begin{enumerate} \item \begin{enumerate} \item \hspace{2zw}{\color{\Anscolor}$\displaystyle I\retu(m,n)=\dint{0}{1}x^m(1-x)^n\,dx$} \vspace{-37.78pt} \begin{align*} I\retu(m,n) & =\dint{0}{1}x^m(1-x)^n\,dx \\ & =\dint{0}{1}x^m\left\{\bunsuu{-1}{n+1}(1-x)^{n+1}\right\}'dx \\ & =\teisekibun{x^m\cdot \bunsuu{-1}{n+1}(1-x)^{n+1}}{0}{1} -\dint{0}{1}(x^m)'\cdot \bunsuu{-1}{n+1}(1-x)^{n+1}\,dx \\ & =(0-0)-\dint{0}{1}mx^{m-1}\cdot \bunsuu{-1}{n+1}(1-x)^{n+1}\,dx \\ & =\bunsuu{m}{n+1}\dint{0}{1}x^{m-1}(1-x)^{n+1}\,dx \hspace{2zw}\textbf{(証明終り)} \end{align*} \vspace{2zw} \item \hspace{-5pt}\kakkoi\hspace{-3pt}の結果より, $I\retu(m,n)=\bunsuu{m}{n+1}I\retu(m-1,n+1)$が成り立つので, \begin{align*} I\retu(m,n) & =\bunsuu{m}{n+1}I\retu(m-1,n+1) \\ & =\bunsuu{m}{n+1}\cdot \bunsuu{m-1}{n+2}I\retu(m-2,n+2) \\ & =\bunsuu{m}{n+1}\cdot \bunsuu{m-1}{n+2}\cdot \cdots \cdot \bunsuu{1}{n+m}I\retu(0,n+m) \end{align*} ここで, \begin{align*} I\retu(0,n+m) & =\dint{0}{1}(1-x)^{n+m}\,dx \\ & =\teisekibun{\bunsuu{-1}{n+m+1}(1-x)^{n+m+1}}{0}{1} \\ & =0-\bunsuu{-1}{n+m+1} \\ & =\bunsuu{1}{n+m+1} \end{align*} となるので, \begin{align*} I\retu(m,n) & =\bunsuu{m}{n+1}\cdot \bunsuu{m-1}{n+2}\cdot \cdots \cdot \bunsuu{1}{n+m}\cdot \bunsuu{1}{n+m+1} \\ & =\bunsuu{m(m-1)(m-2)\cdots 1}{(n+1)(n+2)\cdots (n+m+1)} \\ & =\bunsuu{m\kaizyou}{\bunsuu{(n+m+1)\kaizyou}{n\kaizyou}} \\ & =\bunsuu{m\kaizyou\,n\kaizyou}{(n+m+1)\kaizyou} \hspace{2zw}\textbf{(証明終り)} \end{align*} \end{enumerate} \newpage \item \hspace{2zw}{\color{\Anscolor}$\displaystyle P_n(x)=\tretuwa{k=0}{n}\kumiawase{n}{k}f\left(\bunsuu{k}{n}\right)x^{k}(1-x)^{n-k}$} \vspace{-36.5pt} \begin{align*} P_n(x) & =\tretuwa{k=0}{n}\kumiawase{n}{k}f\left(\bunsuu{k}{n}\right)x^{k}(1-x)^{n-k} \\ & =\kumiawase{n}{0}f(0)(1-x)^n+\kumiawase{n}{1}f\left(\bunsuu{1}{n}\right)x(1-x)^{n-1} +\kumiawase{n}{2}f\left(\bunsuu{2}{n}\right)x^2(1-x)^{n-2}+ \\ & \hspace{27zw}\cdots +\kumiawase{n}{n}f\left(\bunsuu{n}{n}\right)x^n \end{align*} なので, \begin{align*} \dint{0}{1}P_n(x)\,dx & =\dint{0}{1}\kumiawase{n}{0}f(0)(1-x)^n\,dx +\dint{0}{1}\kumiawase{n}{1}f\left(\bunsuu{1}{n}\right)x(1-x)^{n-1}\,dx \\ & \hspace{4zw}+\dint{0}{1}\kumiawase{n}{2}f\left(\bunsuu{2}{n}\right)x^2(1-x)^{n-2} +\cdots +\dint{0}{1}\kumiawase{n}{n}f\left(\bunsuu{n}{n}\right)x^n \\ & =\kumiawase{n}{0}f(0)\dint{0}{1}(1-x)^n\,dx +\kumiawase{n}{1}f\left(\bunsuu{1}{n}\right)\dint{0}{1}x(1-x)^{n-1}\,dx \\ & \hspace{4zw}+\kumiawase{n}{2}f\left(\bunsuu{2}{n}\right)\dint{0}{1}x^2(1-x)^{n-2} +\cdots +\kumiawase{n}{n}f\left(\bunsuu{n}{n}\right)\dint{0}{1}x^n \\ & =\tretuwa{k=0}{n}\kumiawase{n}{k}f\left(\bunsuu{k}{n}\right) \dint{0}{1}x^{k}(1-x)^{n-k}\,dx~~ \Cdots\Cdots\maru{1} \end{align*} ここで,(1)\hspace{-3pt}の結果より,$\dint{0}{1}x^m(1-x)^n\,dx =\bunsuu{m\kaizyou\,n\kaizyou}{(n+m+1)\kaizyou}$となるので, \[ \dint{0}{1}x^{k}(1-x)^{n-k}\,dx=\bunsuu{k\kaizyou\,(n-k)\kaizyou}{(n+1)\kaizyou} \] よって,\maru{1}より, \begin{align*} \dint{0}{1}P_n(x) & =\tretuwa{k=0}{n}\kumiawase{n}{k}f\left(\bunsuu{k}{n}\right)\cdot \bunsuu{k\kaizyou\,(n-k)\kaizyou}{(n+1)\kaizyou} \\ & =\tretuwa{k=0}{n}\bunsuu{n\kaizyou}{k\kaizyou\,(n-k)\kaizyou} f\left(\bunsuu{k}{n}\right)\cdot \bunsuu{k\kaizyou\,(n-k)\kaizyou}{(n+1)\kaizyou} \\ & =\tretuwa{k=0}{n}\bunsuu{1}{n+1}f\left(\bunsuu{k}{n}\right) \\ & =\bunsuu{1}{n+1}f(0)+\tretuwa{k=1}{n}\bunsuu{1}{n+1}f\left(\bunsuu{k}{n}\right) \\ & =\bunsuu{1}{n+1}f(0) +\bunsuu{n+1}{n}\cdot \bunsuu{1}{n}\tretuwa{k=1}{n}f\left(\bunsuu{k}{n}\right) \\ & =\bunsuu{1}{n+1}f(0) +\left(1+\bunsuu{1}{n}\right)\cdot \bunsuu{1}{n}\tretuwa{k=1}{n}f\left(\bunsuu{k}{n}\right) \end{align*} \newpage となるので, \begin{align*} \dlim{n \to \infty}\dint{0}{1}P_n(x) & =\dlim{n \to \infty}\left\{\bunsuu{1}{n+1}f(0) +\left(1+\bunsuu{1}{n}\right)\cdot \bunsuu{1}{n}\tretuwa{k=1}{n}f\left(\bunsuu{k}{n}\right)\right\} \\ & =0\cdot f(0)+(1+0)\cdot \dint{0}{1}f(x)\,dx \\ & =\dint{0}{1}f(x)\,dx \hspace{2zw}\textbf{(証明終り)} \end{align*} \end{enumerate} \end{document}