Mixing
Changing order of partial sum and integral all under limit to infinity
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$$ lim_{n rightarrow infty} int_{a}^b sum_{k=1}^{n}f_k(x) mathrm dx= sum_{k=1}^{infty}int_{a}^b f_k(x) mathrm dx $$
Is this generaly true ? Integral is a sum , two sums can interchange, right?
I ve faced this in a proof of a theorem that says that integral of a uniformly convergent sum is equal to the sum of integral.
$int sum g_n = sum int g_n$ ( $ sum g_n $ converges uniformly )
The problem i am facing is that the stament in the title is used to prove the previous theorem.
I dont think this is a duplicate, this question is about a partial sum that changes order with an integral not an infinite
It is said the answer below is incorrect, can someone explain why
real-analysis integration summation uniform-convergence
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
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add a comment |
$begingroup$
$$ lim_{n rightarrow infty} int_{a}^b sum_{k=1}^{n}f_k(x) mathrm dx= sum_{k=1}^{infty}int_{a}^b f_k(x) mathrm dx $$
Is this generaly true ? Integral is a sum , two sums can interchange, right?
I ve faced this in a proof of a theorem that says that integral of a uniformly convergent sum is equal to the sum of integral.
$int sum g_n = sum int g_n$ ( $ sum g_n $ converges uniformly )
The problem i am facing is that the stament in the title is used to prove the previous theorem.
I dont think this is a duplicate, this question is about a partial sum that changes order with an integral not an infinite
It is said the answer below is incorrect, can someone explain why
real-analysis integration summation uniform-convergence
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
$endgroup$
$begingroup$
math.stackexchange.com/questions/83721/…
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– gunes
Jan 26 at 19:00
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It is not always true, but with uniform convergence you can.
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– Atmos
Jan 26 at 19:07
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@Atmos the first stament or the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:09
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The problem i am facing is that the stament in the title is used to prove the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:11
1
$begingroup$
The equality you wrote is not always true.
$endgroup$
– Atmos
Jan 26 at 19:19
add a comment |
$begingroup$
$$ lim_{n rightarrow infty} int_{a}^b sum_{k=1}^{n}f_k(x) mathrm dx= sum_{k=1}^{infty}int_{a}^b f_k(x) mathrm dx $$
Is this generaly true ? Integral is a sum , two sums can interchange, right?
I ve faced this in a proof of a theorem that says that integral of a uniformly convergent sum is equal to the sum of integral.
$int sum g_n = sum int g_n$ ( $ sum g_n $ converges uniformly )
The problem i am facing is that the stament in the title is used to prove the previous theorem.
I dont think this is a duplicate, this question is about a partial sum that changes order with an integral not an infinite
It is said the answer below is incorrect, can someone explain why
real-analysis integration summation uniform-convergence
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
$endgroup$
$$ lim_{n rightarrow infty} int_{a}^b sum_{k=1}^{n}f_k(x) mathrm dx= sum_{k=1}^{infty}int_{a}^b f_k(x) mathrm dx $$
Is this generaly true ? Integral is a sum , two sums can interchange, right?
I ve faced this in a proof of a theorem that says that integral of a uniformly convergent sum is equal to the sum of integral.
$int sum g_n = sum int g_n$ ( $ sum g_n $ converges uniformly )
The problem i am facing is that the stament in the title is used to prove the previous theorem.
I dont think this is a duplicate, this question is about a partial sum that changes order with an integral not an infinite
It is said the answer below is incorrect, can someone explain why
real-analysis integration summation uniform-convergence
real-analysis integration summation uniform-convergence
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
edited Jan 28 at 9:50
Milan Stojanovic
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
asked Jan 26 at 18:58
Milan StojanovicMilan Stojanovic
452513
452513
$begingroup$
math.stackexchange.com/questions/83721/…
$endgroup$
– gunes
Jan 26 at 19:00
$begingroup$
It is not always true, but with uniform convergence you can.
$endgroup$
– Atmos
Jan 26 at 19:07
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@Atmos the first stament or the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:09
$begingroup$
The problem i am facing is that the stament in the title is used to prove the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:11
1
$begingroup$
The equality you wrote is not always true.
$endgroup$
– Atmos
Jan 26 at 19:19
add a comment |
$begingroup$
math.stackexchange.com/questions/83721/…
$endgroup$
– gunes
Jan 26 at 19:00
$begingroup$
It is not always true, but with uniform convergence you can.
$endgroup$
– Atmos
Jan 26 at 19:07
$begingroup$
@Atmos the first stament or the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:09
$begingroup$
The problem i am facing is that the stament in the title is used to prove the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:11
1
$begingroup$
The equality you wrote is not always true.
$endgroup$
– Atmos
Jan 26 at 19:19
$begingroup$
math.stackexchange.com/questions/83721/…
$endgroup$
– gunes
Jan 26 at 19:00
$begingroup$
math.stackexchange.com/questions/83721/…
$endgroup$
– gunes
Jan 26 at 19:00
$begingroup$
It is not always true, but with uniform convergence you can.
$endgroup$
– Atmos
Jan 26 at 19:07
$begingroup$
It is not always true, but with uniform convergence you can.
$endgroup$
– Atmos
Jan 26 at 19:07
$begingroup$
@Atmos the first stament or the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:09
$begingroup$
@Atmos the first stament or the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:09
$begingroup$
The problem i am facing is that the stament in the title is used to prove the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:11
$begingroup$
The problem i am facing is that the stament in the title is used to prove the latter
$endgroup$
– Milan Stojanovic
Jan 26 at 19:11
1
1
$begingroup$
The equality you wrote is not always true.
$endgroup$
– Atmos
Jan 26 at 19:19
$begingroup$
The equality you wrote is not always true.
$endgroup$
– Atmos
Jan 26 at 19:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is true (only after my edits) directly by the linearity of the integral and the definition of infinite sum.
What you wrote is correct - An integral and a (finite) sum can be interchanged with no further conditions. This is exactly the linearity of the integral.
answered Jan 26 at 19:20
MOMOMOMO
717312
$endgroup$
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
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That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
1
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
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I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
1
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
add a comment |
$begingroup$
Consider the calculation attempt $$lim_{ntoinfty}int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx=sum_{k=1}^inftyint_a^bf_k(x)dx.$$The first $=$ sign works provided each $int_a^bf_k(x)dx$ is finite, since then$$int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$$and both sides have the same $ntoinfty$ behaviour (which might involve a limit not existing).
The second $=$ sign is true by the definition of the summation operator $sum_{k=1}^infty$ as a limit of partial sums. In other words:
- If each $int_a^bf_k(x)dx$ is finite then $int_a^bsum_{k=1}^nf_k(x)dx$ and $sum_{k=1}^nint_a^bf_k(x)dx$ are equal and have the same limit or each have no limit; and
$lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$ is either $sum_{k=1}^inftyint_a^bf_k(x)dx$ if that exists, or non-existent if it doesn't.
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
$endgroup$
$begingroup$
Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
$endgroup$
– Milan Stojanovic
Jan 27 at 9:56
1
$begingroup$
@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
$endgroup$
– J.G.
Jan 27 at 9:58
1
$begingroup$
@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
$endgroup$
– MOMO
Jan 27 at 17:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is true (only after my edits) directly by the linearity of the integral and the definition of infinite sum.
What you wrote is correct - An integral and a (finite) sum can be interchanged with no further conditions. This is exactly the linearity of the integral.
answered Jan 26 at 19:20
MOMOMOMO
717312
$endgroup$
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
$begingroup$
That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
1
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
$begingroup$
I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
1
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
add a comment |
$begingroup$
This is true (only after my edits) directly by the linearity of the integral and the definition of infinite sum.
What you wrote is correct - An integral and a (finite) sum can be interchanged with no further conditions. This is exactly the linearity of the integral.
answered Jan 26 at 19:20
MOMOMOMO
717312
$endgroup$
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
$begingroup$
That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
1
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
$begingroup$
I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
1
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
add a comment |
$begingroup$
This is true (only after my edits) directly by the linearity of the integral and the definition of infinite sum.
What you wrote is correct - An integral and a (finite) sum can be interchanged with no further conditions. This is exactly the linearity of the integral.
answered Jan 26 at 19:20
MOMOMOMO
717312
$endgroup$
This is true (only after my edits) directly by the linearity of the integral and the definition of infinite sum.
What you wrote is correct - An integral and a (finite) sum can be interchanged with no further conditions. This is exactly the linearity of the integral.
answered Jan 26 at 19:20
MOMOMOMO
717312
answered Jan 26 at 19:20
MOMOMOMO
717312
answered Jan 26 at 19:20
MOMOMOMO
717312
answered Jan 26 at 19:20
MOMOMOMO
717312
717312
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
$begingroup$
That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
1
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
$begingroup$
I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
1
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
add a comment |
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
$begingroup$
That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
1
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
$begingroup$
I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
1
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
$begingroup$
Yeah it should be f sub k
$endgroup$
– Milan Stojanovic
Jan 26 at 19:22
$begingroup$
That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
$begingroup$
That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:27
1
1
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
$begingroup$
Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both.
$endgroup$
– lcv
Jan 26 at 19:29
$begingroup$
I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
$begingroup$
I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral.
$endgroup$
– Ted Shifrin
Jan 26 at 19:34
1
1
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
$begingroup$
@TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense)
$endgroup$
– Milan Stojanovic
Jan 26 at 19:42
add a comment |
$begingroup$
Consider the calculation attempt $$lim_{ntoinfty}int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx=sum_{k=1}^inftyint_a^bf_k(x)dx.$$The first $=$ sign works provided each $int_a^bf_k(x)dx$ is finite, since then$$int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$$and both sides have the same $ntoinfty$ behaviour (which might involve a limit not existing).
The second $=$ sign is true by the definition of the summation operator $sum_{k=1}^infty$ as a limit of partial sums. In other words:
- If each $int_a^bf_k(x)dx$ is finite then $int_a^bsum_{k=1}^nf_k(x)dx$ and $sum_{k=1}^nint_a^bf_k(x)dx$ are equal and have the same limit or each have no limit; and
$lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$ is either $sum_{k=1}^inftyint_a^bf_k(x)dx$ if that exists, or non-existent if it doesn't.
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
$endgroup$
$begingroup$
Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
$endgroup$
– Milan Stojanovic
Jan 27 at 9:56
1
$begingroup$
@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
$endgroup$
– J.G.
Jan 27 at 9:58
1
$begingroup$
@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
$endgroup$
– MOMO
Jan 27 at 17:51
add a comment |
$begingroup$
Consider the calculation attempt $$lim_{ntoinfty}int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx=sum_{k=1}^inftyint_a^bf_k(x)dx.$$The first $=$ sign works provided each $int_a^bf_k(x)dx$ is finite, since then$$int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$$and both sides have the same $ntoinfty$ behaviour (which might involve a limit not existing).
The second $=$ sign is true by the definition of the summation operator $sum_{k=1}^infty$ as a limit of partial sums. In other words:
- If each $int_a^bf_k(x)dx$ is finite then $int_a^bsum_{k=1}^nf_k(x)dx$ and $sum_{k=1}^nint_a^bf_k(x)dx$ are equal and have the same limit or each have no limit; and
$lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$ is either $sum_{k=1}^inftyint_a^bf_k(x)dx$ if that exists, or non-existent if it doesn't.
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
$endgroup$
$begingroup$
Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
$endgroup$
– Milan Stojanovic
Jan 27 at 9:56
1
$begingroup$
@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
$endgroup$
– J.G.
Jan 27 at 9:58
1
$begingroup$
@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
$endgroup$
– MOMO
Jan 27 at 17:51
add a comment |
$begingroup$
Consider the calculation attempt $$lim_{ntoinfty}int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx=sum_{k=1}^inftyint_a^bf_k(x)dx.$$The first $=$ sign works provided each $int_a^bf_k(x)dx$ is finite, since then$$int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$$and both sides have the same $ntoinfty$ behaviour (which might involve a limit not existing).
The second $=$ sign is true by the definition of the summation operator $sum_{k=1}^infty$ as a limit of partial sums. In other words:
- If each $int_a^bf_k(x)dx$ is finite then $int_a^bsum_{k=1}^nf_k(x)dx$ and $sum_{k=1}^nint_a^bf_k(x)dx$ are equal and have the same limit or each have no limit; and
$lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$ is either $sum_{k=1}^inftyint_a^bf_k(x)dx$ if that exists, or non-existent if it doesn't.
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
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Consider the calculation attempt $$lim_{ntoinfty}int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx=sum_{k=1}^inftyint_a^bf_k(x)dx.$$The first $=$ sign works provided each $int_a^bf_k(x)dx$ is finite, since then$$int_a^bsum_{k=1}^nf_k(x)dx=lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$$and both sides have the same $ntoinfty$ behaviour (which might involve a limit not existing).
The second $=$ sign is true by the definition of the summation operator $sum_{k=1}^infty$ as a limit of partial sums. In other words:
- If each $int_a^bf_k(x)dx$ is finite then $int_a^bsum_{k=1}^nf_k(x)dx$ and $sum_{k=1}^nint_a^bf_k(x)dx$ are equal and have the same limit or each have no limit; and
$lim_{ntoinfty}sum_{k=1}^nint_a^bf_k(x)dx$ is either $sum_{k=1}^inftyint_a^bf_k(x)dx$ if that exists, or non-existent if it doesn't.
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
answered Jan 27 at 8:34
J.G.J.G.
31.4k23149
31.4k23149
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Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
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– Milan Stojanovic
Jan 27 at 9:56
1
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@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
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– J.G.
Jan 27 at 9:58
1
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@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
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– MOMO
Jan 27 at 17:51
add a comment |
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Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
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– Milan Stojanovic
Jan 27 at 9:56
1
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@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
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– J.G.
Jan 27 at 9:58
1
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@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
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– MOMO
Jan 27 at 17:51
$begingroup$
Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
$endgroup$
– Milan Stojanovic
Jan 27 at 9:56
$begingroup$
Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ?
$endgroup$
– Milan Stojanovic
Jan 27 at 9:56
1
1
$begingroup$
@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
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– J.G.
Jan 27 at 9:58
$begingroup$
@MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $lim_{ntoinfty}int_a^bsum_{k=1}^n f_k(x)dxneint_a^blim_{ntoinfty}sum_{k=1}^n f_k(x)dx=int_a^bsum_{k=1}^infty f_k(x)dx$.
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– J.G.
Jan 27 at 9:58
1
1
$begingroup$
@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
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– MOMO
Jan 27 at 17:51
$begingroup$
@MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite.
$endgroup$
– MOMO
Jan 27 at 17:51
add a comment |
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math.stackexchange.com/questions/83721/…
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– gunes
Jan 26 at 19:00
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It is not always true, but with uniform convergence you can.
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– Atmos
Jan 26 at 19:07
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@Atmos the first stament or the latter
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– Milan Stojanovic
Jan 26 at 19:09
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The problem i am facing is that the stament in the title is used to prove the latter
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– Milan Stojanovic
Jan 26 at 19:11
1
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The equality you wrote is not always true.
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– Atmos
Jan 26 at 19:19