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 $phi$
real-analysis analysis integration
asked Oct 18 '13 at 14:58
user65972
475413
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up vote
5
down vote
favorite
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
asked Oct 18 '13 at 14:58
user65972
475413
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
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
asked Oct 18 '13 at 14:58
user65972
475413
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
real-analysis analysis integration
asked Oct 18 '13 at 14:58
user65972
475413
asked Oct 18 '13 at 14:58
user65972
475413
asked Oct 18 '13 at 14:58
user65972
475413
asked Oct 18 '13 at 14:58
user65972
475413
asked Oct 18 '13 at 14:58
user65972
475413
475413
add a comment |
add a comment |
1 Answer
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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 $phi_{n}$.
answered Oct 18 '13 at 18:41
Guillermo
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $phi_{n}$.
answered Oct 18 '13 at 18:41
Guillermo
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
add a comment |
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 $phi_{n}$.
answered Oct 18 '13 at 18:41
Guillermo
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
add a comment |
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 $phi_{n}$.
answered Oct 18 '13 at 18:41
Guillermo
1,228922
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 $phi_{n}$.
answered Oct 18 '13 at 18:41
Guillermo
1,228922
answered Oct 18 '13 at 18:41
Guillermo
1,228922
answered Oct 18 '13 at 18:41
Guillermo
1,228922
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
add a comment |
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
add a comment |
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