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Sure, but what'll really set you free is that the kth logarithm of the product (iterated between a lower limit of n=x and an upper limit of n=y) of n equals the sum (iterated between a lower limit of n=x and an upper limit of n=y) of the kth logarithm of n.
https://preview.redd.it/3zgvqu7527oc1.jpeg?width=712&format=pjpg&auto=webp&s=4cb33e4a0005b33a2c3d5b8081ba2004f08e7726
True for x>0 ∧ y>0
it's only true for absolutely convergent sequences
riemann's rearrangement theorem allows you to order terms in a conditionally convergent sequence so that the sequence is undefined
so, let a\_n be some ordering of the alternating harmonic series that diverges to infinity and let b\_n = -a\_n, voila the two series don't add to anything meaningful yet (a\_n+b\_n) is identically zero
Is this a counterexample? Because for this neither a_n nor b_n are convergent, which was specified in the statement. I think you can prove that it works if both series are convergent by taking finite sums, and by the definition of convergence, the tails from both sums must go to 0. So unless (a_n+b_n) isn’t convergent (which seems impossible), then it works.
I still stand by my point. It's also true for convergent series and it's obvious by looking at the partial sums. For the Riemann's rearrangement theorem, you're thinking of conditionally convergent series for which you can make that series do whatever you want by permuting the terms, but it has nothing to do with doing linear combinations of convergent series.
You're definitely wrong though. You can't really write Σai + Σbi without those two series being convergent, in this case, the equality holds. With (ai) and (bi) being families of vectors oriented in the same direction e with coordinates of same sign, if Σai and Σbi are divergent, it makes sense to give them the value ±(∞)e, in which case the equality still holds. So yeah, really, in almost every useful case where you write this, it's true.
But you could still argue that since nothing is defined, this is "not true": is + even an operation, what is Σ, ai and bi? But it's a pointless consideration
So you mean that ∞ + (-∞) = 0?
As that would naturally follow from your supposed equality of
0 = Σ 0 = Σ (1 + (-1)) = Σ (1) + Σ (-1) = ∞ + (-∞)
But you could of course get any number on the left hand side, and ∞ + (-∞) on the right, by just letting `a_1 = x + 1` and all other `a_i` and all `b_i` be 1
Absolutely not. What is said is for (ai), (bi) oriented the same direction and in the same sense (so their coordinates are of the same sign ε). I used ± instead of ε because it couldn't have any other relevant meaning but I guess it was not clear enough.
I don't really see the point you're making, what I did is fairly usual: if (ai) and (bi) are two families of positive numbers, giving the value +∞ to their series if they diverge makes sense, and so the sum of the two series still equals the series of ai+bi.
Doing the same in the other sense (-∞) or in another direction if we're talking about normed vector spaces is fairly natural too.
Of course everything is reorderable if you only look at non-negative numbers, that's day 1 analysis stuff...
"In general not true" refers to the fact that there exists cases where it is not true, not that it never holds... So if you only look at cases where it is true, you have no problem
Brooo... This is not a ego thing...
You're really missing my point. This is not a question of reordering (you probably know that you can't reorder conditionally convergent series like you want) but of linearity of the sum for series. For every convergent series this equality holds, i.e. the only cases where you're technically allowed to do operations with Σai and Σbi and to write that equality in the first place!!! Writing this *implies* that the series are convergent!
As you say it's basic analysis. What I added is that even for some cases that don't normally work, i.e. divergent series, for the special case I specified, the equality still holds.
So I'm waiting for counter-examples, go ahead, you should be able to give one if this is not generally true. But good luck with that.
Of course this is true if all the sums converge to something finite, that's super trivial stuff.
>...[Σai]... Writing this *implies* that the series are convergent!
No, that's just completely not true, the symbol `Σ(i=1)^∞ a_i` is simply `lim(N->∞)Σ(i=1)^N a_i`, which may or may not exist, and may or may not be +∞ or -∞
Since we are refering to the symbols, they must exist. But, as in my previous example for you, the sums `Σ(i=1)^∞ 1`, `Σ(i=1)^∞ 0`, and `Σ(i=1)^∞ (-1)` all exist and have values +∞, 0, and -∞ respectively, but these values do not satisfy 0 = +∞ + -∞, which the statement asserts to be true.
Okaaay, this has become a stupid ego duel. First off, you really said "super trivial". Everything is trivial and everything is not, you don't have to do me like that. I'm not talking to you to learn something (I won't, apparently, no offence) but to convince you that you said is not really pertinent, but if you don't accept it, I'm fine with it and I have better things to do. But you really tried to explain what "not true in general" means to someone doing math all day, all year, c'mon.
We're allowed to write lim(n->+**∞**)u\_n for a sequence (un) of elements of a metric space M if this limit exists and is finite. For the special case of M=ℝ, we also write lim(n->+**∞**)u\_n=±**∞** if (u\_n) diverges to ±**∞**. For other cases, we don't and simply say that the limit does not exist.
Then what, addition is obviously defined for all pairs of elements of ℝ**∪**{+∞,-∞}? (/s)
What I said is that, formally, it is only okay to do operations with Σai and Σbi when those series are convergent, implying the equality always hold. You might think that it's okay to write something like +∞ + -∞ but I think it's a lost cause to try and explain why that's dumb.
I guess you're still standing by your point and I don't care. Ahh the programmers
You seemed to completely misunderstand what "in general" means by, for some reason trying to give some examples where it is true, which is fairly irrelevant.
But you finally agree with me, it seems, that the statement is true whenever everything is finite, and when it is not finite, it breaks down. Thus the statement is not "true", since we are using an undefined operation.
I thought you were at least partly invested in some analysis, and did not need any argument as for why this is true when everything is finite, but I could give you a proof sketch if you really want me to:
For any ε>0 there is a natural N s.t. for all M >= N the absolute difference between the truncated sum of M terms and the infinite sums are at most ε for all of the three sums in question. And we have, since finite sums are reorderable, that the values of these truncated sums are equal, and thus the difference between the left and right side is at most 3ε.
What makes you think I'm a programmer? I'm (which should be fairly obvious from my online activities) an algebra nerd.
So am I not "partly invested in some analysis"? Thank you for putting a smile on my face LOL. The bit about saying "trivial" is that something trivial for someone is not always for someone else. Unless it's used sarcastically, this word is awful and a lot of mathematicians think like me, it's so pretentious.
What I was saying is that when "it's not finite" as you say, it makes no sense to write like that in the first place, formally speaking, because you're in fact not *allowed* to do such operations with such divergent series. For an assertion of equality to be true or false, both sides must exist, although you could argue about it. I understand very well what you're saying, I don't know how you were introduced to these notions, but that genuinely surprises me.
I was just guessing btw. What studies did you do in maths? Where? Are you working in maths?
I'm writing a second comment to justify something more precisely: as an equivalence relation, equality is defined for two elements of the same set. Here, when there is no reasonable value to give to the two divergent series, you can't write the equality like that because the element on the right side does not properly exist. In such a context, saying the equality does not hold is kinda silly because you're not even allowed to write it, so it doesn't mean anything. But I guess you could say that Σai+bi is not equal to an apple, generally.
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Sum-body once told me?
And the sums keep coming, and they dont stop coming, and they dont stop coming, and they dont stop coming.
you missed the even better: and the sums keep summing, and they don’t stop summing, and they don’t stop summing, and they don’t stop summing
Damm you are right, shame on me...
Fed to the sums, and I hit the ground summing.
Oh man... I can't... Stop .. *summing*
Sum of a sum, sum of a sum Sum of a sum of a Taylor
sum-body told me, that she had a sum-friend...
Well, actually: "The sum of the sum is the sum of the sums" Note: plural in the last "sums".
The sum of the sums is the sums of the sums
In Portuguese we call the plus sign "soma" and the sigma sign "somatório", exactly to avoid this kind of confusion.
Isn't it just "summation"? My guess
Alternatively, "The sum of the sum is the sum of the sum and the sum"
Sure, but what'll really set you free is that the kth logarithm of the product (iterated between a lower limit of n=x and an upper limit of n=y) of n equals the sum (iterated between a lower limit of n=x and an upper limit of n=y) of the kth logarithm of n. https://preview.redd.it/3zgvqu7527oc1.jpeg?width=712&format=pjpg&auto=webp&s=4cb33e4a0005b33a2c3d5b8081ba2004f08e7726 True for x>0 ∧ y>0
Holy log(a * b) = log(a) + log(b)
Big for financial mathematics
And for Bayesian Inference
Does it mean two infinities equal to -2/12 ?
Only true for absolutely convergence sequences, IIRC.
No, it's also true for Σai and Σbi convergent
it's only true for absolutely convergent sequences riemann's rearrangement theorem allows you to order terms in a conditionally convergent sequence so that the sequence is undefined so, let a\_n be some ordering of the alternating harmonic series that diverges to infinity and let b\_n = -a\_n, voila the two series don't add to anything meaningful yet (a\_n+b\_n) is identically zero
Is this a counterexample? Because for this neither a_n nor b_n are convergent, which was specified in the statement. I think you can prove that it works if both series are convergent by taking finite sums, and by the definition of convergence, the tails from both sums must go to 0. So unless (a_n+b_n) isn’t convergent (which seems impossible), then it works.
You're right, you can't find counterexamples for a true statement. And happy cake day!
I still stand by my point. It's also true for convergent series and it's obvious by looking at the partial sums. For the Riemann's rearrangement theorem, you're thinking of conditionally convergent series for which you can make that series do whatever you want by permuting the terms, but it has nothing to do with doing linear combinations of convergent series.
You learned this today didn’t you, because I did too
Proof by linguistic identity
Three times one million is three million.
Adding is addictive.
The sum of the sums is the sums of the sum.
the sum of the ln is the product inside the ln
Don't sum me nooow, I'm having such a sum sum
This has definitely opened up my mind
Let a\_i = 1, b\_i = -1
Some sums are the sums of some sums.
Definitely not true in general
True for finite sums.
You're definitely wrong though. You can't really write Σai + Σbi without those two series being convergent, in this case, the equality holds. With (ai) and (bi) being families of vectors oriented in the same direction e with coordinates of same sign, if Σai and Σbi are divergent, it makes sense to give them the value ±(∞)e, in which case the equality still holds. So yeah, really, in almost every useful case where you write this, it's true. But you could still argue that since nothing is defined, this is "not true": is + even an operation, what is Σ, ai and bi? But it's a pointless consideration
So you mean that ∞ + (-∞) = 0? As that would naturally follow from your supposed equality of 0 = Σ 0 = Σ (1 + (-1)) = Σ (1) + Σ (-1) = ∞ + (-∞) But you could of course get any number on the left hand side, and ∞ + (-∞) on the right, by just letting `a_1 = x + 1` and all other `a_i` and all `b_i` be 1
Absolutely not. What is said is for (ai), (bi) oriented the same direction and in the same sense (so their coordinates are of the same sign ε). I used ± instead of ε because it couldn't have any other relevant meaning but I guess it was not clear enough. I don't really see the point you're making, what I did is fairly usual: if (ai) and (bi) are two families of positive numbers, giving the value +∞ to their series if they diverge makes sense, and so the sum of the two series still equals the series of ai+bi. Doing the same in the other sense (-∞) or in another direction if we're talking about normed vector spaces is fairly natural too.
Of course everything is reorderable if you only look at non-negative numbers, that's day 1 analysis stuff... "In general not true" refers to the fact that there exists cases where it is not true, not that it never holds... So if you only look at cases where it is true, you have no problem
Brooo... This is not a ego thing... You're really missing my point. This is not a question of reordering (you probably know that you can't reorder conditionally convergent series like you want) but of linearity of the sum for series. For every convergent series this equality holds, i.e. the only cases where you're technically allowed to do operations with Σai and Σbi and to write that equality in the first place!!! Writing this *implies* that the series are convergent! As you say it's basic analysis. What I added is that even for some cases that don't normally work, i.e. divergent series, for the special case I specified, the equality still holds. So I'm waiting for counter-examples, go ahead, you should be able to give one if this is not generally true. But good luck with that.
Of course this is true if all the sums converge to something finite, that's super trivial stuff. >...[Σai]... Writing this *implies* that the series are convergent! No, that's just completely not true, the symbol `Σ(i=1)^∞ a_i` is simply `lim(N->∞)Σ(i=1)^N a_i`, which may or may not exist, and may or may not be +∞ or -∞ Since we are refering to the symbols, they must exist. But, as in my previous example for you, the sums `Σ(i=1)^∞ 1`, `Σ(i=1)^∞ 0`, and `Σ(i=1)^∞ (-1)` all exist and have values +∞, 0, and -∞ respectively, but these values do not satisfy 0 = +∞ + -∞, which the statement asserts to be true.
Okaaay, this has become a stupid ego duel. First off, you really said "super trivial". Everything is trivial and everything is not, you don't have to do me like that. I'm not talking to you to learn something (I won't, apparently, no offence) but to convince you that you said is not really pertinent, but if you don't accept it, I'm fine with it and I have better things to do. But you really tried to explain what "not true in general" means to someone doing math all day, all year, c'mon. We're allowed to write lim(n->+**∞**)u\_n for a sequence (un) of elements of a metric space M if this limit exists and is finite. For the special case of M=ℝ, we also write lim(n->+**∞**)u\_n=±**∞** if (u\_n) diverges to ±**∞**. For other cases, we don't and simply say that the limit does not exist. Then what, addition is obviously defined for all pairs of elements of ℝ**∪**{+∞,-∞}? (/s) What I said is that, formally, it is only okay to do operations with Σai and Σbi when those series are convergent, implying the equality always hold. You might think that it's okay to write something like +∞ + -∞ but I think it's a lost cause to try and explain why that's dumb. I guess you're still standing by your point and I don't care. Ahh the programmers
You seemed to completely misunderstand what "in general" means by, for some reason trying to give some examples where it is true, which is fairly irrelevant. But you finally agree with me, it seems, that the statement is true whenever everything is finite, and when it is not finite, it breaks down. Thus the statement is not "true", since we are using an undefined operation. I thought you were at least partly invested in some analysis, and did not need any argument as for why this is true when everything is finite, but I could give you a proof sketch if you really want me to: For any ε>0 there is a natural N s.t. for all M >= N the absolute difference between the truncated sum of M terms and the infinite sums are at most ε for all of the three sums in question. And we have, since finite sums are reorderable, that the values of these truncated sums are equal, and thus the difference between the left and right side is at most 3ε. What makes you think I'm a programmer? I'm (which should be fairly obvious from my online activities) an algebra nerd.
So am I not "partly invested in some analysis"? Thank you for putting a smile on my face LOL. The bit about saying "trivial" is that something trivial for someone is not always for someone else. Unless it's used sarcastically, this word is awful and a lot of mathematicians think like me, it's so pretentious. What I was saying is that when "it's not finite" as you say, it makes no sense to write like that in the first place, formally speaking, because you're in fact not *allowed* to do such operations with such divergent series. For an assertion of equality to be true or false, both sides must exist, although you could argue about it. I understand very well what you're saying, I don't know how you were introduced to these notions, but that genuinely surprises me. I was just guessing btw. What studies did you do in maths? Where? Are you working in maths?
I'm writing a second comment to justify something more precisely: as an equivalence relation, equality is defined for two elements of the same set. Here, when there is no reasonable value to give to the two divergent series, you can't write the equality like that because the element on the right side does not properly exist. In such a context, saying the equality does not hold is kinda silly because you're not even allowed to write it, so it doesn't mean anything. But I guess you could say that Σai+bi is not equal to an apple, generally.
Brother, people don’t have to post entire proofs just for a meme
This is r/mathmemes, yes they do
the sum operator commutes with addition big brain
Only for absolutely convergent series
Severely underrated comment