Find a specific partition to make $U(f,P) < 2/100$
$begingroup$
" Suppose $f : [0,1] to mathbb{R}$ is given by
$f(1/n) = 1/n$ when $n in mathbb{N}$ and $f(x) = 0$ for all other $x in [0,1]$. Show that for some partition $P$ of $[0,1]$, we have $U(f,P) < frac{2}{100}$" . My attempt is to subdivide the interval $[0,1]$ into subintervals with equal length $d$. Then I will make this number $d$ agree with the inequality $U(f,P) < 2/100$. But it turns out not to work well. How can I approach to solve this?
(Let $M_k = sup { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$)
real-analysis riemann-integration
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
$endgroup$
add a comment |
$begingroup$
" Suppose $f : [0,1] to mathbb{R}$ is given by
$f(1/n) = 1/n$ when $n in mathbb{N}$ and $f(x) = 0$ for all other $x in [0,1]$. Show that for some partition $P$ of $[0,1]$, we have $U(f,P) < frac{2}{100}$" . My attempt is to subdivide the interval $[0,1]$ into subintervals with equal length $d$. Then I will make this number $d$ agree with the inequality $U(f,P) < 2/100$. But it turns out not to work well. How can I approach to solve this?
(Let $M_k = sup { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$)
real-analysis riemann-integration
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
$endgroup$
$begingroup$
You have not defined $U(f,P)$. What is it?
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:37
$begingroup$
It is the Upper Sum of the function $f$. Let $M_k = { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$
$endgroup$
– Dong Le
Dec 4 '18 at 3:42
$begingroup$
How could anybody answer the question without that information? Please edit it into your question.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:45
$begingroup$
I edited it to the question!!
$endgroup$
– Dong Le
Dec 4 '18 at 3:48
add a comment |
$begingroup$
" Suppose $f : [0,1] to mathbb{R}$ is given by
$f(1/n) = 1/n$ when $n in mathbb{N}$ and $f(x) = 0$ for all other $x in [0,1]$. Show that for some partition $P$ of $[0,1]$, we have $U(f,P) < frac{2}{100}$" . My attempt is to subdivide the interval $[0,1]$ into subintervals with equal length $d$. Then I will make this number $d$ agree with the inequality $U(f,P) < 2/100$. But it turns out not to work well. How can I approach to solve this?
(Let $M_k = sup { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$)
real-analysis riemann-integration
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
$endgroup$
" Suppose $f : [0,1] to mathbb{R}$ is given by
$f(1/n) = 1/n$ when $n in mathbb{N}$ and $f(x) = 0$ for all other $x in [0,1]$. Show that for some partition $P$ of $[0,1]$, we have $U(f,P) < frac{2}{100}$" . My attempt is to subdivide the interval $[0,1]$ into subintervals with equal length $d$. Then I will make this number $d$ agree with the inequality $U(f,P) < 2/100$. But it turns out not to work well. How can I approach to solve this?
(Let $M_k = sup { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$)
real-analysis riemann-integration
real-analysis riemann-integration
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
edited Dec 4 '18 at 4:02
Dong Le
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
asked Dec 4 '18 at 3:32
Dong LeDong Le
717
717
$begingroup$
You have not defined $U(f,P)$. What is it?
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:37
$begingroup$
It is the Upper Sum of the function $f$. Let $M_k = { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$
$endgroup$
– Dong Le
Dec 4 '18 at 3:42
$begingroup$
How could anybody answer the question without that information? Please edit it into your question.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:45
$begingroup$
I edited it to the question!!
$endgroup$
– Dong Le
Dec 4 '18 at 3:48
add a comment |
$begingroup$
You have not defined $U(f,P)$. What is it?
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:37
$begingroup$
It is the Upper Sum of the function $f$. Let $M_k = { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$
$endgroup$
– Dong Le
Dec 4 '18 at 3:42
$begingroup$
How could anybody answer the question without that information? Please edit it into your question.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:45
$begingroup$
I edited it to the question!!
$endgroup$
– Dong Le
Dec 4 '18 at 3:48
$begingroup$
You have not defined $U(f,P)$. What is it?
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:37
$begingroup$
You have not defined $U(f,P)$. What is it?
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:37
$begingroup$
It is the Upper Sum of the function $f$. Let $M_k = { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$
$endgroup$
– Dong Le
Dec 4 '18 at 3:42
$begingroup$
It is the Upper Sum of the function $f$. Let $M_k = { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$
$endgroup$
– Dong Le
Dec 4 '18 at 3:42
$begingroup$
How could anybody answer the question without that information? Please edit it into your question.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:45
$begingroup$
How could anybody answer the question without that information? Please edit it into your question.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:45
$begingroup$
I edited it to the question!!
$endgroup$
– Dong Le
Dec 4 '18 at 3:48
$begingroup$
I edited it to the question!!
$endgroup$
– Dong Le
Dec 4 '18 at 3:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If subdividing into equal intervals does not work, you need to subdivide into unequal intervals. This is a classic technique, which shows that the upper sum for any countable set like the rationals is zero. Order the set, then put an interval of length $k$ around the first, $frac k4$ around the second, $frac k9$ around the third, and $frac k{n^2}$ around the $n^{th}$. The sum of the intervals is $frac {kpi^2}6$, so by choosing $k$ small enough you can get the sum as small as you want.
For your problem, equal intervals will work fine as long as you choose $d$ large enough. The bottom subinterval will have infinitely many points in it, but will only contribute $d^2$ to the sum. There will be only finitely many intervals above that have a $frac 1n$ in them and each will contribute $frac 1{nd}$ to the sum. As long as $d$ is large enough you will be there. Given a $d$ you need to bound this sum and you will be there.
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
add a comment |
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$begingroup$
If subdividing into equal intervals does not work, you need to subdivide into unequal intervals. This is a classic technique, which shows that the upper sum for any countable set like the rationals is zero. Order the set, then put an interval of length $k$ around the first, $frac k4$ around the second, $frac k9$ around the third, and $frac k{n^2}$ around the $n^{th}$. The sum of the intervals is $frac {kpi^2}6$, so by choosing $k$ small enough you can get the sum as small as you want.
For your problem, equal intervals will work fine as long as you choose $d$ large enough. The bottom subinterval will have infinitely many points in it, but will only contribute $d^2$ to the sum. There will be only finitely many intervals above that have a $frac 1n$ in them and each will contribute $frac 1{nd}$ to the sum. As long as $d$ is large enough you will be there. Given a $d$ you need to bound this sum and you will be there.
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
add a comment |
$begingroup$
If subdividing into equal intervals does not work, you need to subdivide into unequal intervals. This is a classic technique, which shows that the upper sum for any countable set like the rationals is zero. Order the set, then put an interval of length $k$ around the first, $frac k4$ around the second, $frac k9$ around the third, and $frac k{n^2}$ around the $n^{th}$. The sum of the intervals is $frac {kpi^2}6$, so by choosing $k$ small enough you can get the sum as small as you want.
For your problem, equal intervals will work fine as long as you choose $d$ large enough. The bottom subinterval will have infinitely many points in it, but will only contribute $d^2$ to the sum. There will be only finitely many intervals above that have a $frac 1n$ in them and each will contribute $frac 1{nd}$ to the sum. As long as $d$ is large enough you will be there. Given a $d$ you need to bound this sum and you will be there.
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
add a comment |
$begingroup$
If subdividing into equal intervals does not work, you need to subdivide into unequal intervals. This is a classic technique, which shows that the upper sum for any countable set like the rationals is zero. Order the set, then put an interval of length $k$ around the first, $frac k4$ around the second, $frac k9$ around the third, and $frac k{n^2}$ around the $n^{th}$. The sum of the intervals is $frac {kpi^2}6$, so by choosing $k$ small enough you can get the sum as small as you want.
For your problem, equal intervals will work fine as long as you choose $d$ large enough. The bottom subinterval will have infinitely many points in it, but will only contribute $d^2$ to the sum. There will be only finitely many intervals above that have a $frac 1n$ in them and each will contribute $frac 1{nd}$ to the sum. As long as $d$ is large enough you will be there. Given a $d$ you need to bound this sum and you will be there.
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
$endgroup$
If subdividing into equal intervals does not work, you need to subdivide into unequal intervals. This is a classic technique, which shows that the upper sum for any countable set like the rationals is zero. Order the set, then put an interval of length $k$ around the first, $frac k4$ around the second, $frac k9$ around the third, and $frac k{n^2}$ around the $n^{th}$. The sum of the intervals is $frac {kpi^2}6$, so by choosing $k$ small enough you can get the sum as small as you want.
For your problem, equal intervals will work fine as long as you choose $d$ large enough. The bottom subinterval will have infinitely many points in it, but will only contribute $d^2$ to the sum. There will be only finitely many intervals above that have a $frac 1n$ in them and each will contribute $frac 1{nd}$ to the sum. As long as $d$ is large enough you will be there. Given a $d$ you need to bound this sum and you will be there.
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
answered Dec 4 '18 at 3:59
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
add a comment |
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
How do you deal with the $M_k = sup ( f(x) ) $ where $x in [x_k, x_{k-1} ] $?
$endgroup$
– Dong Le
Dec 4 '18 at 4:03
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
$begingroup$
In the equal interval case, by making $d$ large enough that the length of the interval is very small. Most of the elements in the sum will be $0$ because there is not a $frac 1n$ in the interval. You get a harmonic sum that is about $frac 1d ln(n)$
$endgroup$
– Ross Millikan
Dec 4 '18 at 4:07
add a comment |
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$begingroup$
You have not defined $U(f,P)$. What is it?
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:37
$begingroup$
It is the Upper Sum of the function $f$. Let $M_k = { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$
$endgroup$
– Dong Le
Dec 4 '18 at 3:42
$begingroup$
How could anybody answer the question without that information? Please edit it into your question.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:45
$begingroup$
I edited it to the question!!
$endgroup$
– Dong Le
Dec 4 '18 at 3:48