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How can I see that $sum_2^n frac1j = int_1^n frac1j dj$?
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Can your show how this comes about or give me pointers of investigation so can I work it out? Thank you!
EDIT1: Various pointed out the equation is false, but that it can be taken as approximation. Since no answered were added, I'd like to ask how I could start from the left hand side and construct an approximation to the sum by using the right hand side. What steps should I take to see how to do that?
calculus algebra-precalculus summation riemann-sum
asked Jul 16 at 20:30
Joep Awinita
537
 |Â
show 3 more comments
up vote
1
down vote
favorite
Can your show how this comes about or give me pointers of investigation so can I work it out? Thank you!
EDIT1: Various pointed out the equation is false, but that it can be taken as approximation. Since no answered were added, I'd like to ask how I could start from the left hand side and construct an approximation to the sum by using the right hand side. What steps should I take to see how to do that?
calculus algebra-precalculus summation riemann-sum
asked Jul 16 at 20:30
Joep Awinita
537
1
You can't because as written it's false. Unless you put in a ceiling function on $1/lceil x rceil$.
– Alex R.
Jul 16 at 20:33
That's not true. It is approximately true, You can see why by seeing that the sum is a Riemann sum for the integral on the right.
– Ethan Bolker
Jul 16 at 20:34
It already fails for $n=2$.
– Saucy O'Path
Jul 16 at 20:36
I'm interested in an approximation indeed. I'm editing the question.
– Joep Awinita
Jul 16 at 20:39
I don't understand the question. Each of those already is an approximation to the other. There;s no need to "construct an approximation". You can find a better approximation to the integral by refining the Riemann sum. Perhaps the answer from @user108128 is what you want.
– Ethan Bolker
Jul 16 at 20:51
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Can your show how this comes about or give me pointers of investigation so can I work it out? Thank you!
EDIT1: Various pointed out the equation is false, but that it can be taken as approximation. Since no answered were added, I'd like to ask how I could start from the left hand side and construct an approximation to the sum by using the right hand side. What steps should I take to see how to do that?
calculus algebra-precalculus summation riemann-sum
asked Jul 16 at 20:30
Joep Awinita
537
Can your show how this comes about or give me pointers of investigation so can I work it out? Thank you!
EDIT1: Various pointed out the equation is false, but that it can be taken as approximation. Since no answered were added, I'd like to ask how I could start from the left hand side and construct an approximation to the sum by using the right hand side. What steps should I take to see how to do that?
calculus algebra-precalculus summation riemann-sum
asked Jul 16 at 20:30
Joep Awinita
537
edited Jul 16 at 20:47
asked Jul 16 at 20:30
Joep Awinita
537
asked Jul 16 at 20:30
Joep Awinita
537
asked Jul 16 at 20:30
Joep Awinita
537
537
1
You can't because as written it's false. Unless you put in a ceiling function on $1/lceil x rceil$.
– Alex R.
Jul 16 at 20:33
That's not true. It is approximately true, You can see why by seeing that the sum is a Riemann sum for the integral on the right.
– Ethan Bolker
Jul 16 at 20:34
It already fails for $n=2$.
– Saucy O'Path
Jul 16 at 20:36
I'm interested in an approximation indeed. I'm editing the question.
– Joep Awinita
Jul 16 at 20:39
I don't understand the question. Each of those already is an approximation to the other. There;s no need to "construct an approximation". You can find a better approximation to the integral by refining the Riemann sum. Perhaps the answer from @user108128 is what you want.
– Ethan Bolker
Jul 16 at 20:51
 |Â
show 3 more comments
1
You can't because as written it's false. Unless you put in a ceiling function on $1/lceil x rceil$.
– Alex R.
Jul 16 at 20:33
That's not true. It is approximately true, You can see why by seeing that the sum is a Riemann sum for the integral on the right.
– Ethan Bolker
Jul 16 at 20:34
It already fails for $n=2$.
– Saucy O'Path
Jul 16 at 20:36
I'm interested in an approximation indeed. I'm editing the question.
– Joep Awinita
Jul 16 at 20:39
I don't understand the question. Each of those already is an approximation to the other. There;s no need to "construct an approximation". You can find a better approximation to the integral by refining the Riemann sum. Perhaps the answer from @user108128 is what you want.
– Ethan Bolker
Jul 16 at 20:51
1
1
You can't because as written it's false. Unless you put in a ceiling function on $1/lceil x rceil$.
– Alex R.
Jul 16 at 20:33
You can't because as written it's false. Unless you put in a ceiling function on $1/lceil x rceil$.
– Alex R.
Jul 16 at 20:33
That's not true. It is approximately true, You can see why by seeing that the sum is a Riemann sum for the integral on the right.
– Ethan Bolker
Jul 16 at 20:34
That's not true. It is approximately true, You can see why by seeing that the sum is a Riemann sum for the integral on the right.
– Ethan Bolker
Jul 16 at 20:34
It already fails for $n=2$.
– Saucy O'Path
Jul 16 at 20:36
It already fails for $n=2$.
– Saucy O'Path
Jul 16 at 20:36
I'm interested in an approximation indeed. I'm editing the question.
– Joep Awinita
Jul 16 at 20:39
I'm interested in an approximation indeed. I'm editing the question.
– Joep Awinita
Jul 16 at 20:39
I don't understand the question. Each of those already is an approximation to the other. There;s no need to "construct an approximation". You can find a better approximation to the integral by refining the Riemann sum. Perhaps the answer from @user108128 is what you want.
– Ethan Bolker
Jul 16 at 20:51
I don't understand the question. Each of those already is an approximation to the other. There;s no need to "construct an approximation". You can find a better approximation to the integral by refining the Riemann sum. Perhaps the answer from @user108128 is what you want.
– Ethan Bolker
Jul 16 at 20:51
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
More tenuous, you can say
$$
beginalign
sum_j=1^ndfrac1j
&=ln n+gamma+varepsilon_n\
sum_j=2^ndfrac1j
&=int_1^ndfrac1jdjcolorred+gamma+varepsilon_n-1
endalign
$$
where $varepsilon_nsimdfrac12n$. You may see the details here.
answered Jul 16 at 20:49
Nosrati
19.8k41644
add a comment |Â
up vote
1
down vote
$int_j=1^nfrac1xdx=sum_j=2^nint_j-1^jfrac1xdx$. However $int_j-1^jfrac1xdx approx fracj-frac12j(j-1)approx frac1j$, giving an approximation to the result you want.
The last step could be made a little clearer. $frac1j-1gt int_j-1^jfrac1xdxgt frac1j$. Therefore $int_j-1^jfrac1xdxapprox frac1j$.
answered Jul 16 at 21:40
herb steinberg
94129
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
More tenuous, you can say
$$
beginalign
sum_j=1^ndfrac1j
&=ln n+gamma+varepsilon_n\
sum_j=2^ndfrac1j
&=int_1^ndfrac1jdjcolorred+gamma+varepsilon_n-1
endalign
$$
where $varepsilon_nsimdfrac12n$. You may see the details here.
answered Jul 16 at 20:49
Nosrati
19.8k41644
add a comment |Â
up vote
4
down vote
accepted
More tenuous, you can say
$$
beginalign
sum_j=1^ndfrac1j
&=ln n+gamma+varepsilon_n\
sum_j=2^ndfrac1j
&=int_1^ndfrac1jdjcolorred+gamma+varepsilon_n-1
endalign
$$
where $varepsilon_nsimdfrac12n$. You may see the details here.
answered Jul 16 at 20:49
Nosrati
19.8k41644
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
More tenuous, you can say
$$
beginalign
sum_j=1^ndfrac1j
&=ln n+gamma+varepsilon_n\
sum_j=2^ndfrac1j
&=int_1^ndfrac1jdjcolorred+gamma+varepsilon_n-1
endalign
$$
where $varepsilon_nsimdfrac12n$. You may see the details here.
answered Jul 16 at 20:49
Nosrati
19.8k41644
More tenuous, you can say
$$
beginalign
sum_j=1^ndfrac1j
&=ln n+gamma+varepsilon_n\
sum_j=2^ndfrac1j
&=int_1^ndfrac1jdjcolorred+gamma+varepsilon_n-1
endalign
$$
where $varepsilon_nsimdfrac12n$. You may see the details here.
answered Jul 16 at 20:49
Nosrati
19.8k41644
edited Jul 16 at 21:08
answered Jul 16 at 20:49
Nosrati
19.8k41644
answered Jul 16 at 20:49
Nosrati
19.8k41644
answered Jul 16 at 20:49
Nosrati
19.8k41644
19.8k41644
add a comment |Â
add a comment |Â
up vote
1
down vote
$int_j=1^nfrac1xdx=sum_j=2^nint_j-1^jfrac1xdx$. However $int_j-1^jfrac1xdx approx fracj-frac12j(j-1)approx frac1j$, giving an approximation to the result you want.
The last step could be made a little clearer. $frac1j-1gt int_j-1^jfrac1xdxgt frac1j$. Therefore $int_j-1^jfrac1xdxapprox frac1j$.
answered Jul 16 at 21:40
herb steinberg
94129
add a comment |Â
up vote
1
down vote
$int_j=1^nfrac1xdx=sum_j=2^nint_j-1^jfrac1xdx$. However $int_j-1^jfrac1xdx approx fracj-frac12j(j-1)approx frac1j$, giving an approximation to the result you want.
The last step could be made a little clearer. $frac1j-1gt int_j-1^jfrac1xdxgt frac1j$. Therefore $int_j-1^jfrac1xdxapprox frac1j$.
answered Jul 16 at 21:40
herb steinberg
94129
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$int_j=1^nfrac1xdx=sum_j=2^nint_j-1^jfrac1xdx$. However $int_j-1^jfrac1xdx approx fracj-frac12j(j-1)approx frac1j$, giving an approximation to the result you want.
The last step could be made a little clearer. $frac1j-1gt int_j-1^jfrac1xdxgt frac1j$. Therefore $int_j-1^jfrac1xdxapprox frac1j$.
answered Jul 16 at 21:40
herb steinberg
94129
$int_j=1^nfrac1xdx=sum_j=2^nint_j-1^jfrac1xdx$. However $int_j-1^jfrac1xdx approx fracj-frac12j(j-1)approx frac1j$, giving an approximation to the result you want.
The last step could be made a little clearer. $frac1j-1gt int_j-1^jfrac1xdxgt frac1j$. Therefore $int_j-1^jfrac1xdxapprox frac1j$.
answered Jul 16 at 21:40
herb steinberg
94129
edited Jul 17 at 15:39
answered Jul 16 at 21:40
herb steinberg
94129
answered Jul 16 at 21:40
herb steinberg
94129
answered Jul 16 at 21:40
herb steinberg
94129
94129
add a comment |Â
add a comment |Â
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1
You can't because as written it's false. Unless you put in a ceiling function on $1/lceil x rceil$.
– Alex R.
Jul 16 at 20:33
That's not true. It is approximately true, You can see why by seeing that the sum is a Riemann sum for the integral on the right.
– Ethan Bolker
Jul 16 at 20:34
It already fails for $n=2$.
– Saucy O'Path
Jul 16 at 20:36
I'm interested in an approximation indeed. I'm editing the question.
– Joep Awinita
Jul 16 at 20:39
I don't understand the question. Each of those already is an approximation to the other. There;s no need to "construct an approximation". You can find a better approximation to the integral by refining the Riemann sum. Perhaps the answer from @user108128 is what you want.
– Ethan Bolker
Jul 16 at 20:51