Can you change a sum by rearranging its numbers? --- The Riemann Series Theorem
HTML-код
- Опубликовано: 2 июн 2024
- Normally when you add up numbers, the order you do so doesn't matter and you get the same sum regardless. And, of course, the same holds true even if you add up infinitely many numbers.....
Right?
=Chapters=
0:00 - Let's rearrange a sum!
1:48 - Investigation
6:32 - Riemann Series Theorem explained visually
13:58 - Resolving objections
18:52 - A step further and a challenge
20:07 - Significance of the Riemann Series Theorem
21:47 - Final thoughts
This video is a participant in the 3Blue1Brown First Summer of Math Exposition (SoME1). You can find out more about it here:
www.3blue1brown.com/blog/some1
#SoME1
===============================
Want to support future videos? Become a patron at / morphocular
Thank you for your support!
===============================
The animations in this video were mostly made with a homemade Python library called "Morpho".
If you want to play with it, you can find it here:
github.com/morpho-matters/mor...
![Eminem - Houdini [Official Music Video]](http://i.ytimg.com/vi/22tVWwmTie8/mqdefault.jpg)
I'm mesmerized by how intuitive you made the theorem seem. I always felt it was sort of like a "paradox", but your explanation made it look almost plain obvious. Great video!
It's a paradox in the sense that "conditional convergence" isn't really a valid mode of convergence.
It was obvious all the time.
The sign of a great teacher I would say !
@@0x6a09 No it wasn't. You were lucky enough to have a good teacher who made it obvious to you the first time the subject was introduced. *Not everyone is so lucky.*
@@General12th How are my math teachers related to this? They never talked about this. But they talked about limits, and i think this is enough to understand this theorem.
My strategies for the extra problems:
To make the series oscillate, instead of having one target sum, make it two, for example 1 and 0. First take enough positive terms to get the partial sum above 1, then take negative terms to get it below 0, then repeat. This way the series will have infinitely many partial sums both above and below the interval [0,1].
To make the series diverge to infinity, use the same strategy, but make the target sums increase by 1 each time they are reached. I.e. first take positive terms to get above 1, then take negative terms to get below 0, then get above 2, then below 1, then above 3, then below 2, etc. Since the negative terms converge to 0, the sequence of partial sums will have an increasing lower bound that will go to infinity.
Corollary: you can get S from infinitely many rearrangements.Just pick an arbitrarily long (finite) sub-sequence, and the remaining terms still form a conditionally convergent series, which you can rearrange to get a new target sum of S minus the partial sum from the sub-sequence.
I agree with your approach on the non-converging sequence. For the sequence converging to infinity I like your approach, but you can do a little better: whenever you cross your threshold switch to negative numbers until you go below it. Then increase your threshold (you added 1 but you could also multiply by 2, square it, or use any other sequence that will diverges to infinity) and repeat. You will use your negative terms much more slowly but that doesn't matter. As long as they all get used eventually...
Does that work for oscillation, though? The reason the rearrangement works normally is because the positive and negative values approach zero. The deviation of the partial sum from the target sum decreases and eventually converges to zero. To oscillate like that, the deviation is at LEAST one, and never gets smaller.
@@williamrutherford553 Since the sum of the positive terms diverges, even though the terms converge to zero, you can always take enough of them to pass the upper target. Same for negative terms and the lower target.
I think with diverging you can use the successive partial sums of any divergent series as the upper target sums, and any sequence of numbers that's bounded above and below as the difference between pairs of upper and immediately following lower target sums.
There may be more generalizations to be made here; it seems that for the upper and lower target sum series you have to pick two series such that the differences of the corresponding pairs of terms both don't converge to zero and don't diverge to infinity, and that one of the series themselves diverges to infinity (it is obvious that the other one does if these conditions are true)
It makes sense to a degree, the same way that ∞ - ∞ can equal whatever you want
That's an excellent comment!
Using the fact that
∞ + x = ∞
For any real number x, We can get
∞ = ∞ - x
∞ - ∞ = - x
And we can replace -x with y; We can say that y = -x and y belongs in the set of real numbers, now we have
∞ - ∞ = y.
And y can be any real number, because for any y you want to choose, there's an x that will get you that value.
∞−∞ is indeterminate
@@manioqqqq ∞−∞ is straight up undefined, because infinity is not a number
@@SlightSmile Incorrect
Sometimes I get frustrated by how you always state the obvious (you do it very slowly, of course). Then I realise that I would never have come up with what is "obvious" in the middle of the video had you not told me about what is "obvious" previously. And then I just realise that that's how math works! You just state the "obvious" and you come up with more "obvious" statements. This proves how good of an educator you are, great job.
There's a legend that Isaac Newton invented the cat door. A door within a door. So obvious in hindsight anybody could come up with it. But it took the world's greatest genius to actually come up with it. Now, this never happened, but this kind of thing happens all the time. Genius is putting together the novel out of the obvious.
@@angelmendez-rivera351 look at any body's top 5 or 10 list of greatest geniuses of all time l, and Newton will be on every one of them, often at the top. Who else compares? Arhimedes, Einstein, John von Neumann?
The list of geniuses throughout history with a great list of accomplishments is very extensive. However, if they were ordered by the magnitude of their accomplishments then Newton, Gauss, Euler, Einstein, and Von Neumann are at the top of the list (at least among those who lived recently enough for us to be sure of their accomplishments).
@@angelmendez-rivera351 All of them are tied at the very top of the list.
@@angelmendez-rivera351 that list is missing archimedes and Leonardo Da Vinci, Euler was superior to Gauss in terms of achievements, and John von Neumann may have been smarter, but didn't accomplish as much as Fermi or Planke. And Stephen Hawking certainly deserves a spot on the list.
I like how this is your only video and it's an absolute banger.
It being the only video is, I hope, only a temporary deficiency.
@@morphocular I was here before you blew up!!
This explanation video was on par with 3b1b, if not better. As a very loyal 3b1b viewer, I want to emphasize that it means a lot.
"loyal ... viewer"? What's that supposed to mean? Hahahahahahaha.
I remember learning about conditional convergent series when taking calculus class and it always felt "why do I care if a series is conditionally vs. absolutely convergent". Your video answered my long-standing question. Thank you!
fun fact: for the (-1)^n/(2n+1) sequence, the series exactly equals arctanh(p)/2 + pi/4 where p is the proportion of negative and positive terms, ranging from -1 (all negative terms) to 1 (all positive terms)
For example, the ++- pattern converges to arctanh(1/3)/2 + pi/4
I would love to see a proof for why this is the case, because it seems too simple not to have an elegant reason for being that way.
I don’t have a proof, but the Taylor series of arctanh(x) goes like x + x^3/3 + x^5/5 + … which must play a role in it somehow
@@zachteitler9622 wow thanks! you clearly did a lot of work to find all that out
You can also just say that the 1/1+1/3+1/5+1/7+... approaches 1/2+1/4+1/6+1/8+...+a finite constant=
C+ln(n/2)/2, after numbers lower than n.
If half of them are missing, n is half as small, so we are at C+ln(n/4)/3=C+ln(n/2)/2-ln(2)/2.
I instantly came up with this proof when I saw the video title, so this is very natural and simple.
@@caspermadlener4191 no your proof is bullshit
-1 isn't all negative terms, it's just positive terms are very rare and become more rare after more terms and they are 0% of negative terms.
And same for 1
A minor mistake in the argument at around 8:52 for about why the added terms must converge to 0. It is not because otherwise it must diverge to infinity, but because otherwise it must diverge to infinity OR by oscillating (e.g 1 -1 1 -1… series does not diverge to infinity.)
Yep, you are right. And good catch! I was actually aware of that before, but I decided to leave the argument as is (while gritting my teeth) since it was a sidenote focused on intuition that I wanted to keep brief, and I felt covering that edge case might distract from the main video's thrust.
A really impressively clear explanation. Thanks a lot for making this!
Thank you for watching! I really appreciate it.
I think this one might be my fav one so far! Amazing job! the animation you have made it really to follow along and helped me understand the concept a lot!
My idea for one oscillation algorithm:
Sort each subset (up and down arrows) by size, as done in the video, use the arrows in order of decreasing size.
Start at zero (use this as a new minimum), then use the first up arrow to create a new maximum.
From here on, after reaching a new maximum, add down arrows until you exceed the previous global minimum. And every time you reach a new global minimum, switch to adding up arrows until you exceed the previous global maximum.
This will have the series oscillate with increasing amplitude - because every time you go down, you go down to an all-time low, and every time you go up, you go up to an all-time high.
Because the series of up and down arrows each diverge when viewed independently (see video), it will always be possible to add enough arrows to exceed the previous maximum/minimum.
i belive the optimal version might be in np
Would that result in a series that switches between positive and negative infinity? For instance, imagine that each time you switch directions, your new y-coordinate becomes something like n=1-2+3-4+5-6+... The sum, n, keeps switching between increasingly large negative and positive numbers.
I think if you add a restriction like "|n| should always be approximately equal to a constant k", then you can also make it just oscillate between finite numbers. For instance, for k=1, you might oscillate between n=-1 and n=1 in the limit.
Similarly, to rearrange the series to make it blow up to infinity, I think you could just make n monotonically increasing
For the infinity rearrangement try this.
Suppose that the up and down arrows are sorted in terms of decreasing length.What you do is add on the first down arrow. Then add enough up arrows such that the total sum of all the arrows is bigger than 1. Add the second down arrow, then add enough up arrows to make the total sum bigger than 2. Add the third down arrow then add enough up arrows to make the total sum bigger than 3.
Add the 4th down arrow then add enough up arrows to make the total sum bigger than 4. And so on.
This was an incredibly lucid explanation. I think I’m actually going to try to teach this to the students in my math-for-art-students course. Before seeing this video, I wouldn’t have dreamed of trying to explain something like this to them, but if I replicate your explanation, I think some (hopefully most) of them might actually get it!
amazing. I felt I watched a 5 minute video since your presentation was smooth and intuitive, great job!!
How have i never came across you before? Excelent quality of content, far beyond your current 10k subscriber count.
Keep it up!
Wow i'm very impressed of the combination of such pretty visuals together with a very good explaination !
What an explanation!!! Absolutely the best Maths video I've seen this year!
Subscribed and watched every video. Great job with the visualizations and distilling the salient features and ideas of the proof into something not just manageable but intuitive, all without any handwaving. Keep going!
It is amazing how it has already been two years... I remember the time I watched your first video when it was 2 months old.... time flies by.
You must have put a whole lot of work into this and it was so good, thanks a bunch for this!
This is good content. Thank You. The idea of absolute convergence is much more clear to me now.
Great job making counterintuitive and difficult-to-grasp concepts seem natural!
The production quality for the first video is insane! You totally matched mathologer's video quality on this one. (He did a video on the same topic a few years ago)
Awesome presentation, hoping for more videos., Thank you !
❤This is my favorite theorem in the whole calculus, both for sounding so counterintuitive and with so intuitive a proof.
Wow, that's a great video! Step by step easy to follow, and yet not glossing over any important detail.
I don't know what about your channel is different, but you've helped me understand these paradoxical maths things better than anyone else! Both this and fractional derivatives, finally explained understandably
I think the strength of this channel is that the explanations don’t require the viewer to hold many unexplained pieces in place for unknown reasons, waiting for the thing that ties them together.
@@ericpmoss the video on fractional derivatives kind of does require the viewer to do so a little bit. The viewer is referred to another video for some things in the integration part, and the gamma function is not defined.
Stupendous work!
this video made me feel like I knew a lot about the subject by simply explaining the topic really well
A great video, 3lue1brown level quality, really nice!
A beautiful entry. Thank you
This explanation is so beautiful and awesome
Superb video! To summarize, the trick is we have changed the underlying distribution of the negative numbers vs. that of the positive numbers. In other words, how often the often occurs with respect to the other, leading to the different result.
A fantastic video, thank you!
That is such a vivid presentation
Amazing video, maybe the best maths video I saw on YT
this got me in stitchs awesome vid!
This was great. Thanks.
So so good! I'd love to see the other half of this, that ordering doesn't matter for absolutely convergent series
I just found this video a few seconds ago and i must respectfully give you a compliment by saying you have an outstanding sense of humor because i giggled and laughed and rewinded the intro several times. Are you also a comedian?
A very good explanation of what seemed impossible. Well done.
Very well articulated. Your channel is a great boon to your audience and burgeoning mathematicians the world over.
This is an absolutely impressive video
wow! this is mindblowing!
Very well done video!
19:40 My take on your problem:
1-So a series diverge by oscilation you will set 2 goal lines, one that need to be transpassed in each phase ( upward and downward ).
2-To diverge to infinity do the exactly same thing as diverging by oscilation, but move each target up or down by some amount on each cicle.
This was a really good video love your explanation
Very well explained 👏
Amazing quality content
Oscillating sequence: add positive terms until the partial sum is greater than 1, then add negative terms until the partial sum is less than 0
Positive infinity: add positive terms until the partial sum is greater than 1, then add the first negative term, then keep adding positive terms until the partial sum is greater than the next integer followed by the next single negative term (negative infinity can be reached by starting with the negative terms)
Awesome sir
nice intuitive explanation
Absolutely beautiful ❤️
Diverging to infinity:
Consider the length of the largest red arrow, and place a number of green arrows that add up to more than twice that length. Now place the longest red arrow. Continue in the same pattern.
Of course red and green can be switched here to make negative infinity
I think for positive infinity, all you need to do is add green arrows until it’s longer than the next red arrow, not twice as long. It doesn’t need to diverge *fast* it just needs to diverge.
@@titaniadioxide6133 nope you need the factor of 2, since the length of the red arrow decreases but by choosing this factor of 2, you're actually having sums of the lengths of the red arrows as bound, rather than individual lengths themselves
@@titaniadioxide6133 remember that you have to add the red arrows themselves also
@@helloitsme7553 but so long as you are always going farther up than the last arrow took and the next arrow will take you down, won't you always keep going up? Plotting all of your maximum values you might end up with a graph that looks a bit like ln(x) where it nearly flattens out, but so long as it never does you will still approach infinity.
@@GhostGlitch. Always increasing doesn't mean you'll diverge to infinity. For example the sequence 0/1, 1/2, 2/3, 3/4, 4/5, 5/6... is strictly increasing but converges to 1. He mentions this at 14:22 with a geometric series as an example.
Absolutely amazing & beautiful
Congratulations
thank you for your perfect explications.
For divergence to infinity, get the arrows to build a stair case by placing up arrows until they’re over 2 and then down arrows until under 1, then up arrows until over 3 to 2 to 4 etc. All the arrows must be used and it diverges to infinity.
For a divergence by oscillation, pick any two real numbers a,b with a>b (WLOG). Place up arrows until they’re sum is greater than ‘a’ and then place down arrows until the sum is less than ‘b’ and then repeat. The arrows are always enough to get from a to b and back again and oscillate between the two numbers.
What does (WLOG) mean in your comment🤔?
@@christopherrice891 without loss of generality, basically it doesn’t matter which of a and b is bigger because it works in both cases :)
I don't feel that it's particularly paradoxical to begin with. A (only slightly naive) definition of an infinite sum is the number that we approach as we take longer and longer partial sums. If you cram two times more positive terms into each sum as in the first example, it'll obviously make all partial sums bigger and since you changed the ordering for the whole infinite series, it hols for the infinite sum as well.
I agree, infinite sum is often defined as limit when number of terms tends to infinity. This means partial sums matter and moving terms makes two partial sums have different terms. So even if each infinite sum have same terms, partial sums do not have same collection of terms.
Very good!!!
Love this video
Imagine a dancer who takes one step forward followed by one step backward forever. Then rearrange their steps so instead they take two steps forward followed by one step backwards forever. Somehow in my example it is not surprising at all that the dancers end up in different places after you repeat their moves an arbitrary number of times, even though you can rearrange the infinite sequence of moves of one dancer into the other's.
A good point. But then you might find it surprising that this strategy doesn't work on all series. An absolutely convergent series like 1 - 1/4 + 1/9 - etc. consisting of alternating reciprocal squares will not change its sum no matter how you rearrange it. To pursue your analogy, whether rearranging the sequence of steps in a dance affects the long-term position of a dancer depends on the kind of dance they're doing!
that is the thought I had. after 10 000 steps, or as in the video, summing the first 10000 terms, you are no longer summing the same terms. For the infinite summation you could say the infinitely many terms are somewhere in both sequences, but for any intermediate value like adding the first 5 or 10 terms, you are not adding the same terms. Even if you took a google term, it is still very small amount of term compared to the infinitely many terms that exist in the sequence. Convergence is an hypothesis on the answer derived from observing how summing the starting terms behave, if you are not adding the same terms then you won't get the same result.
@@morphocular what would the sum be, if instead of doing + + -, you were to put every negative term first, followed by all the positive term ? I'm not sure you'd even place a single positive term as there are infinitely many negative term in that sequence to be placed first.
Wow this is really good
Elegant. The reason this seems counter intuitive is because we forget that two infinities are not always equal. If you pick one pos and one neg number each time, that is a different infinite series than taking two pos and one neg each time. They will not converge to the same number.
Where you showed π÷4=atan(1) was absolutely COMIC
That's actually quite important in quantum mechanics! When switching from Schrödinger's to Dirac's picture of QM, one substitutes the concept of "wavefunctions" with that of "state vectors" living in a Hilbert space that is (for most problems) infinite dimensional. In this case the usual representation of matrices and vectors is not very intuitive, since an infinite-dimensional matrix cannot even be written down. Nevertheless the maths behind it still works in the conventional ways, and the definition of scalar product between vectors of infinitely many components is coherent and most importantly finite. Turns out that the dot product of a state vector with itself is just the infinite sum of the square magnitudes of the Fourier coefficients needed to build the Fourier expansion of the wavefunction in the chosen basis. So yeah, a dot product between infinite dimensional vectors actually represents a converging series, and this fact has physical importance!
Change the number for when you switch, like every time you go up double the number, or every time you switch you switch between two values (or more). In fact you can probably do some fun things by switching the number randomly with different distributions. Like make it so it doesn't diverge to anything. What makes this weird is what happens if you pick a nonconstructable number? I suppose this would imply that there couldn't be a nice rule to pick the next set of arrows.
I added your video to my private list labeled '' treasure videos ''
And that's enough as a compliment
This has made me think of the axiom of choice and how careful one has to be to be confident of infinite proofs using just cardinality ignoring order, see Cantor and Gödel.
great video !
Alright. This guy deserves a sub
19:25 The trick here is to make the "target line" dynamic. To blow it up to plus infinity you move the line up a finite distance every time you cross it downwards, to blow it down you move it down a finite distance every time you cross it upwards. To make it oscillate you move the line up when crossing it downwards and move it back to its original location when crossing the moved line upwards. You can also make the line jump a random (but finite) distance against the crossing direction every time you cross it to make a sum that has no limit and does not oscillate.
The to infinity construction is fun. You have two types of steps: even steps and odd. At each even step you enumerate arrows to get to n - x where n is the step you're on and x is the next negative number. Then at each odd step you enumerate a negative number. The cleanest construction I can think of.
thanks for this, it was very interesting… subscribed after 5 minutes (watched the entire video)… good luck
I came up with one way to approach infinity. For each down arrow, pop enough up arrow until the magnitude is more than twice that of the down arrows. Since this meta-addition will result in more than the total magnitude of down arrows, the sum would approach positive infinity.
Same can be applied to approach negative infinity.
What strikes me is that while you can prove that there will always be some re-arrangement of a conditionally convergent sequence that will generate any real number you choose, it doesn't mean that finding that precise rearrangement will be that easy in reality, as virtually all real numbers will be generated by a rearrangement of the sequence with no regular pattern, in which you simply need to know an infinite number of terms, rather than be able to infinitely generate them from a pattern (i.e. if you randomly choose a target number to converge on then it is almost certain that the rearrangement solution will not be of the form 'a times up followed by b times down, repeated infinitely many times' or anything of that sort).
To make it diverge to infinity (or negative infinity), or to oscillate I feel like you could move the desired sum each time you cross it (but such that the addition/movement of the arrow is still sufficient to cross the line; for simplicity let's say you move it the maximum possible amount in your intended direction) in the direction of either positive or negative infinity, or in the case of oscillation, in the direction of your desired maximum or minimum (and have it be that whenever your sum reaches within some delta of the desired maximum or minimum, you instead choose to approach the other of the maximum or minimum instead). Since you cross it infinitely many times and because the series is only conditionally convergent, I believe that would mean that you move the line an infinite amount of times, and the sum of the movements of the desired sum would diverge because it's only conditionally convergent, meaning the movement of the desired sum would also diverge. The only aspect I'm not 100% sure on is you could possibly make a series where if you add only the absolute values of the terms which cross the threshold it would be absolutely convergent even if the total series is conditionally convergent, but I suspect that due to the nature of conditional convergence that that might not be the case, but I don't have the tools to prove that part.
The explanation that got me is that the serie is INF+ (-INF), when you arrange P then N - which can result in any number.
Fun fact: using the same construct you can actually find a sub-sequence of the partial sum sequence adjacent to any sequence u(n) you want. If you call S(n) the partial (rearranged) sequence it mean that you can find a rearrangement and a sequence k_n such that S(k_n) - u(n) tends to zero. The sum will therefore approache every term of the sequence, getting better and better every time. Even more stronger, you can control how it tends to zero: if you fix a threshold ε>0, then you can make so that |S(k_n) - u(n)|
Awesome!
If RUclips allowed double likes I would have given them to you....that passing comment about L_2 spaces made my day; it makes so much sense - invariance of inner product under space transformations
Your explanation made me go from "wtf that's impossible" to "oh this is so obvious"
I lost brain cells very early but this video is amazing. Never expected something like this to happen.
The sums always added up to the same value for the number of terms shown or upto 1/23 in all the three cases for me. When taken in the original order, when the adjacent terms are interchanged, and also when all the positive terms are taken together and all the negative terms grouped together.
You can get infinite or finite divergent sums by changing the target line to some other function which has a strictly smaller derivative than the lower bounding function of the terms (or the negative of the negative terms). If Aln(n+1) is strictly less than the sum of all positive terms, then Bln(n+1) is a valid target line iff B
This is undoubtedly a nice video and contains a lot of new ideas but I think the real challenge is how can these visual thoughts be turned into a formal proof . This is the real challenge I believe 👍❤
why would you need to? the theorem is already proven, so you don't really have to make up any new proofs based on youtube videos
Interesting video. Nice theorem. I'm curious about how it could (or could not) be generalized for complex series...
For all series that do not absolutely converge, any sum is possible after the rearrangement of terms
I'm not sure where you got this false idea from, but let me try to explain why the conditional convergence is, in a sense, necessary.
Conditional convergence can be seen as a combination of three conditions.
(1) The terms approach zero.
(2) The positive terms have divergent sum.
(3) The negative terms have divergent sum.
Not being absolutely convergent only requires one of (2) and (3) to hold.
So one might ask which of the conditions (1), (2) and (3) are necessary for the rearrangement property to hold.
Try to come up with infinite series for which only two of the conditions hold, but not the third, and try to see if the rearrangement property holds. You will find that any combination of two of the three conditions will not suffice for the rearrangement property to hold. Maybe you can even find a proof that any series with the rearrangement property needs to have all three properties.
@@SmileyMPV Sorry typo and the previous logic was incorrect. should be conditionally convergent implies rearrangement can make up any real number, not related to absolute convergence:)
@@angelmendez-rivera351 Right, I meant to say that not being absolutely convergent only requires one of (2) and (3) to hold. I edited my comment.
For a more general proof, replace the constant target number with an arbitrary infinite series: it literally doesn't matter what the series is as long as every term in the series is a real number. Starting with the first number in the target series, apply enough up or down arrows (one or three other; not both) to cross that value; then advance to the next value in the target series and repeat the process.
The difference between your partial sum and the value of that target series will converge to zero, which can also be phrased as your infinite sum converting to the infinite series. Meaning that you can make the infinite sum become infinitely close to the behavior of the target series. So if the target series converges, so will your infinite sum; if it diverges, so will your infinite sum, and in exactly the same way: positive infinity, negative infinity, or forever oscillating.
Or, if you have a more detailed means of measuring how the series behaves in the long run (e.g., hyperreal numbers or intervals), your partial sum can be made to match whatever the measurement of the target series is.
As long as you take all infinit 1 over uneven number, the sum is the same but when you take any even finite number of those added and subtracted up, you will have a ratio of 1to1 of subtracted and added numbers for the first one and around 2to1 for the second one.
The crux of this is that conditionally convergent series are just discrete ways to write out infinity - infinity, which you know to be indeterminate, and can take on any value when given a concrete representation. It's not entirely obvious this is the case initially, but once you realize that the conditional convergence implies P+N = finite, but P-N (or P + abs(N)) is infinite, it becomes pretty clear imo
For the end challenges, you could simply set a series of targets to reach, for +inf let's say we aim for 2, 1, 3, 2, 4, 3, 5, 4,... , for -inf simply the negation of all those terms, and for divergence we can do 1, -1, 1, -1,... . In all these cases, there's a total infinite upwards movement and downwards movement, so all the arrows will be used.
Its about reading expressions as combinations of associative and commutative sequences of symbols.
I tried to solve your challenge and here’s what I came up with:
To make a series converge to nothing, just add up-arrows until they exceed 1, then add down-arrows until they’re below 0, then add up-arrows again until they reach 1 and so on. (Obviously the numbers don’t have to be 1 and 0).
To blow up to positive infinity, add up-arrows until you hit an integer (k), then add down-arrows until you’re below k-1/2, then add up-arrows until you reach k+1, then down-arrows until you’re below k+1/2 and so on. There’s probably a much simpler solution, but this works fine. Negative infinite is just the opposite strategy.
I loved the way you explained little l2 space. Does that mean that big L2 is the space where the dot product is extended to infinite functions?
For convergent series (sums where the terms get smaller and smaller, approaching zero), the rearrangement is a problem when:
The nth term is moved m terms away from where it was, and m has no limit/bound to how large it can get. So, at 1:19, -1/15 should be in the 5th term, but gets rearranged to be in the 12th term. The sum shown is different because terms can be moved up to any distance from their starting positions, even an infinite distance.
For non-convergent series, such as 1-1+1-1+1-1..., almost any rearrangement will result in the sum value changing, because the sum doesn't have a consistent clear value.
Here's my ideas:
For +∞: Sum all the negative numbers, then add the positive ones. Vice versa for -∞. (At least I think that would work).
Divergent: First place the largest positive arrow (or number). Next place negative arrows (starting with the largest) such that they are longer than the positive one in total. Then place positive arrows such that they are longer then the previous negative ones together. Repeat for all the numbers.
This also provides a novel bijection from N^N to R (or R^k, as Lévy-Steinitz proves).
Does anyone know about random rearrangements of conditionally convergent series?
It's actually pretty simple. If the series converges absolutely, so do the positive and negative subseries. But in a series where all elements have the same sign, rearrangement can't change the sum. Conversely, in a conditionally convergent series, both the positive and the negative subsequence go to infinity. Hence, you can approach any value by just picking negative elements when the partial sum is above the target and positive values when below. You will always pass the target because you have an infinite budget and you will get closer and closer because the elements get arbitrarily small.
Brilliant
I was first having trouble accepting this, until I realised that we assume that we know the real number we're trying to make our series sum to, to infinitely many decimal digits, or equivalently, we always no precisely whether or not our partial sums are greater, lesser or equal to the number