Approximating integrals with step functions











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For $f colon [1,2] to mathbb{R}$ , $f(x) = 1/x$, Choose a sequence of step functions $phi_n$ approximating $f$ with partition $P_n := [r/n : n < r < 2n]$ to show that $ 1/(n+1) + cdots + 1/2n to int_1^2 x^{-1} dx$ as $ n to infty$



So far i have played about with the right hand side of the integral, which equals log2, to no avail. f is continuous on [1,2], hence regulated. So we know that for any sequence of step functions $(phi_n)_{n=1}^infty$ converging to f uniformly, the sequence $(int_a^b phi_n(x) dx)_{n=1}^infty$ also converges. So i know that if i am to choose a sequence of step functions $phi$ approximating f that it will converge, but im not sure how to choose $phi$










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    For $f colon [1,2] to mathbb{R}$ , $f(x) = 1/x$, Choose a sequence of step functions $phi_n$ approximating $f$ with partition $P_n := [r/n : n < r < 2n]$ to show that $ 1/(n+1) + cdots + 1/2n to int_1^2 x^{-1} dx$ as $ n to infty$



    So far i have played about with the right hand side of the integral, which equals log2, to no avail. f is continuous on [1,2], hence regulated. So we know that for any sequence of step functions $(phi_n)_{n=1}^infty$ converging to f uniformly, the sequence $(int_a^b phi_n(x) dx)_{n=1}^infty$ also converges. So i know that if i am to choose a sequence of step functions $phi$ approximating f that it will converge, but im not sure how to choose $phi$










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      For $f colon [1,2] to mathbb{R}$ , $f(x) = 1/x$, Choose a sequence of step functions $phi_n$ approximating $f$ with partition $P_n := [r/n : n < r < 2n]$ to show that $ 1/(n+1) + cdots + 1/2n to int_1^2 x^{-1} dx$ as $ n to infty$



      So far i have played about with the right hand side of the integral, which equals log2, to no avail. f is continuous on [1,2], hence regulated. So we know that for any sequence of step functions $(phi_n)_{n=1}^infty$ converging to f uniformly, the sequence $(int_a^b phi_n(x) dx)_{n=1}^infty$ also converges. So i know that if i am to choose a sequence of step functions $phi$ approximating f that it will converge, but im not sure how to choose $phi$










      share|cite|improve this question













      For $f colon [1,2] to mathbb{R}$ , $f(x) = 1/x$, Choose a sequence of step functions $phi_n$ approximating $f$ with partition $P_n := [r/n : n < r < 2n]$ to show that $ 1/(n+1) + cdots + 1/2n to int_1^2 x^{-1} dx$ as $ n to infty$



      So far i have played about with the right hand side of the integral, which equals log2, to no avail. f is continuous on [1,2], hence regulated. So we know that for any sequence of step functions $(phi_n)_{n=1}^infty$ converging to f uniformly, the sequence $(int_a^b phi_n(x) dx)_{n=1}^infty$ also converges. So i know that if i am to choose a sequence of step functions $phi$ approximating f that it will converge, but im not sure how to choose $phi$







      real-analysis analysis integration






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      asked Oct 18 '13 at 14:58









      user65972

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          Take $phi_{n}(x)=sum_{i=1}^{n}frac{n}{n+i}chi_{E_{i}}(x)$ where $E_{i}={x: 1+frac{i-1}{n}<x<1+frac{i}{n}}$.



          Actually, you can see that it is natural by drawing the graph of $f$ and $phi_{n}$.






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          • Sorry i dont understand what $chi$ exactly means in this context?
            – user65972
            Oct 20 '13 at 23:34










          • $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
            – Guillermo
            Oct 22 '13 at 8:51













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          up vote
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          Take $phi_{n}(x)=sum_{i=1}^{n}frac{n}{n+i}chi_{E_{i}}(x)$ where $E_{i}={x: 1+frac{i-1}{n}<x<1+frac{i}{n}}$.



          Actually, you can see that it is natural by drawing the graph of $f$ and $phi_{n}$.






          share|cite|improve this answer





















          • Sorry i dont understand what $chi$ exactly means in this context?
            – user65972
            Oct 20 '13 at 23:34










          • $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
            – Guillermo
            Oct 22 '13 at 8:51

















          up vote
          0
          down vote













          Take $phi_{n}(x)=sum_{i=1}^{n}frac{n}{n+i}chi_{E_{i}}(x)$ where $E_{i}={x: 1+frac{i-1}{n}<x<1+frac{i}{n}}$.



          Actually, you can see that it is natural by drawing the graph of $f$ and $phi_{n}$.






          share|cite|improve this answer





















          • Sorry i dont understand what $chi$ exactly means in this context?
            – user65972
            Oct 20 '13 at 23:34










          • $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
            – Guillermo
            Oct 22 '13 at 8:51















          up vote
          0
          down vote










          up vote
          0
          down vote









          Take $phi_{n}(x)=sum_{i=1}^{n}frac{n}{n+i}chi_{E_{i}}(x)$ where $E_{i}={x: 1+frac{i-1}{n}<x<1+frac{i}{n}}$.



          Actually, you can see that it is natural by drawing the graph of $f$ and $phi_{n}$.






          share|cite|improve this answer












          Take $phi_{n}(x)=sum_{i=1}^{n}frac{n}{n+i}chi_{E_{i}}(x)$ where $E_{i}={x: 1+frac{i-1}{n}<x<1+frac{i}{n}}$.



          Actually, you can see that it is natural by drawing the graph of $f$ and $phi_{n}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 18 '13 at 18:41









          Guillermo

          1,228922




          1,228922












          • Sorry i dont understand what $chi$ exactly means in this context?
            – user65972
            Oct 20 '13 at 23:34










          • $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
            – Guillermo
            Oct 22 '13 at 8:51




















          • Sorry i dont understand what $chi$ exactly means in this context?
            – user65972
            Oct 20 '13 at 23:34










          • $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
            – Guillermo
            Oct 22 '13 at 8:51


















          Sorry i dont understand what $chi$ exactly means in this context?
          – user65972
          Oct 20 '13 at 23:34




          Sorry i dont understand what $chi$ exactly means in this context?
          – user65972
          Oct 20 '13 at 23:34












          $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
          – Guillermo
          Oct 22 '13 at 8:51






          $chi_{E}(x) = 1$ for $x in E$ and $0$ for $x notin E$.
          – Guillermo
          Oct 22 '13 at 8:51




















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