1:01:27 I think there is confusion because the LHS can be written in 2 equivalent ways : (a0 + b0) + (a1 + b1) + (a2 + b2) + .... and also (a0 + a1 + a2 + ....) + (b0 + b1 + b2 + ....) This is because you can break up the summation across addition (which is the commutative law of addition in disguise). So in fact, he is commuting the terms but this is happening BEFORE he applies the summation machine. In other words, he is correct in what he is saying : he is not commuting the terms in the scope of the summation machine.
to not be confused you have to understand, that a_n and b_n by themselves are not summation terms of the LHS. So when he regroups a_n and b_n he is not commuting the terms of a series. (You can not rewrite the LHS as a_0 + b_0 + a_1 + b_1 + a_2 + ...) But you can split the summation by the assumption that S is a linear operator on series, where the sum of two series is defined as the series over their added terms. And this assumption is a definition at this point.
Does anyone has an answer of why doesn't Borel summation work for 1+2+4+8+16+... After doing all the stuff, I get in the integral e^t, so I think I am missing something (even if it is working for Euler summation and general summation machine, giving an answer of -1)
s0 = 1 - s1. any s0+s1=1 will generate an infinite sequence of (1-1+1-1+1-1+...). is there any reason why "s=1-s" is required? 1/3. 1-2/3. 1-1+1/3. 1-1+1-2/3. ... (1-1+1-1+...). sum from x=0: (-1)^x.
Absolutely stunning! However, isn't taking apart a few initial terms of a series and summing them separately from the rest of the series a mild form of application of associativity?
that is another possible "summation machine". And if it gives a result (as in this case it does), then it has to agree with all other summation machines (as long as they obey the two rules).
Begging your pardon, professor, BUT (1:13:20), you effectively SHIFT the series TWICE, and then you sum (vertically), in what seems to be a _term-by-term summation_ , in groups of non-corresponding terms of the series. That, looks to me, as SHIFTING the terms of the series, before adding them. And that, me thinks, is NOT permitted under your own rules.
It doesn’t matter that the series are related. This is axiom 2, but used in the inverse way he explained it, that is if you have 2 series s1 and s2 consisting of a(n) and b(n) then S(s1)+ S(s2)= S(a(n)+ b(n)). This can be obtained by setting alpha and beta equal to one. Without wanting to sound pedantic, I hope you understand that writing some words in capitals does not increase their truth and makes you look a bit crazy (even though I believe you had no wrong intentions and just wanted to communicate). This is a scientific lecture, and scientists in general prefer to raise valid arguments or explain how/why they’ve done something with others if necessary and can also reflect on their mistakes without someone virtually screaming at them for what they have (supposedly) done wrong.
@@remysmets Thank you for your reply. Sometimes, a cigar is just a cigar, as another scientist, Sigmund Freud, once noted :-). The capitalization, as you half-guessed, was simply meant to *emphasize* the main points of my argument, and not to represent yelling. Leaving aside the ad hominems, let us not get our nickers tied in a knot, at the drop of a hat. Axiom#1 appears to be a mathematical sleight of hand, to permit the shifting of the terms of the sequence without explicitly declaring so. And if all the zeros in the whole sequence were removed from it, the same way that the zeros resulted from the removal of first a0 and then a1, (applying axiom#1 twice), then the whole trick of the term-by-term vertical summation (dadada, dadada, dadada, ...) would have never worked at all. Maybe the answer lies into stating clearly the exact terms and conditions required, before applying axiom#1.
Elias Andrikopoulos alright! I think the problem with removing all the zero’s is that there are infinitely many of them. Therefore, I think you can’t apply axiom 1 to an infinite amount of numbers in the series and that’s why you can’t keep removing the zeros. Furthermore, axiom 1 applies to the first term only, and because 0 is not always the first term it is once again impossible to remove all of them
S(1 -1 +1 -1+...)=1/2 S(1 +0 -1 +1 +0 -1 +1 +0 -1+...)=2/3 S(1+0 +0 -1 +1 +0 +0 -1 +1 +0 +0 -1 +...)=3/4 S(1+0 +0 +0 -1 +1 +0 +0 +0 -1 +1 +0 +0 +0 -1 +...)=4/5 The value of this generic sum is a property of the repeating (cyclic) pattern of numbers and the assumption of linearity. Adding an infinite number of zeros would have no effect on a normal sum.
The bridge won't be infinitely long, just 2 bricks long: piling up the bricks it's like drawing the "graph" of 1/n^2 vertically, which never traverses its asymptote, which is 2 units of distance away from the outer edge of the first brick.
Mire ha mi yo tengo halgo ke exponer sobre Hoyente ha distansia local interviene el TAO mas todo los conponente X a ininsio anos generasion de paso X mujer no es igual ha honbre y con un pero si de mismo hablar con un ofisina de ufolojia se podia denunsiar primer caso primero guardia sin ofisina no vale
Prof. Bender obviously enjoys teaching this course. He makes it look like great fun.
+Ralph Dratman He's amazing. I wanna him to teach me!
Godlike. Now I understand meaning of this ridiculous summing.
Professor Bender's joy reminds me of the similar joy MIT professor Herb Gross showed teaching Complex Variables in the MIT OpenCourseWare lectures.
Agree. I love those Herb Gross lectures!
I am spellbound by this stuff! He makes it so easy! Thanks for posting.
Beautiful lecture, thanks a lot for uploading Bender's lecture.
This guy is hands-down the next best teacher to Feynman.
Orpheus kgdggg656 of course
Actually, I take it back. Prof. V. Balakrishnan from IIT Madras beats all of them.
@@bboyHarrypotter nope, I think quantum lectures are very fantastic by Balakrishnan sir but mathematical lectures are too fuzzy.
I think that he's much better than Feynman.
@@ndbchannellocustgroveva1952 Feynman could make anything complicated understandable to most. That's what made him the greatest teacher
Brilliant.
Going to go over more of your videos in my spare time after my exam tomorrow. You know how to teach.
1:01:27 I think there is confusion because the LHS can be written in 2 equivalent ways : (a0 + b0) + (a1 + b1) + (a2 + b2) + .... and also (a0 + a1 + a2 + ....) + (b0 + b1 + b2 + ....)
This is because you can break up the summation across addition (which is the commutative law of addition in disguise). So in fact, he is commuting the terms but this is happening BEFORE he applies the summation machine. In other words, he is correct in what he is saying : he is not commuting the terms in the scope of the summation machine.
I still don't understand, put:
a_n=1/(2n+1)
b_n=-1/(2n+2)
Alpha=Beta=1
Then obviously the lhs coverges, but the rhs does not...
A L0T 0F EINSTEINS IN CHAT
to not be confused you have to understand, that a_n and b_n by themselves are not summation terms of the LHS. So when he regroups a_n and b_n he is not commuting the terms of a series. (You can not rewrite the LHS as a_0 + b_0 + a_1 + b_1 + a_2 + ...)
But you can split the summation by the assumption that S is a linear operator on series, where the sum of two series is defined as the series over their added terms. And this assumption is a definition at this point.
what book do they use?
Love the hockey stick. Oh Canada!
It's ridiculous to believe that he's 68 in this video.
So what's he saying about=1,-1.....? Does it converge? Who knows. Very confusing. I emailed him and he referred me to his text?????
Does anyone has an answer of why doesn't Borel summation work for 1+2+4+8+16+... After doing all the stuff, I get in the integral e^t, so I think I am missing something (even if it is working for Euler summation and general summation machine, giving an answer of -1)
So fucking amazing!!!! OMFG!!!
s0 = 1 - s1. any s0+s1=1 will generate an infinite sequence of (1-1+1-1+1-1+...). is there any reason why "s=1-s" is required? 1/3. 1-2/3. 1-1+1/3. 1-1+1-2/3. ... (1-1+1-1+...). sum from x=0: (-1)^x.
I love it
Absolutely stunning! However, isn't taking apart a few initial terms of a series and summing them separately from the rest of the series a mild form of application of associativity?
I think that’s okay because you’re only re-associating finitely many terms, and addition is finitely associative
Thanks! You are right :)
49:58, doesn't it have something to do with analytic continuation?
1-1+1-1+1.... converges to 1/2 because 1/2 is the average value of the partial sums.
that is another possible "summation machine". And if it gives a result (as in this case it does), then it has to agree with all other summation machines (as long as they obey the two rules).
and it makes tschaka tschaka tschaka
Begging your pardon, professor, BUT (1:13:20), you effectively SHIFT the series TWICE, and then you sum (vertically), in what seems to be a _term-by-term summation_ , in groups of non-corresponding terms of the series.
That, looks to me, as SHIFTING the terms of the series, before adding them. And that, me thinks, is NOT permitted under your own rules.
It doesn’t matter that the series are related. This is axiom 2, but used in the inverse way he explained it, that is if you have 2 series s1 and s2 consisting of a(n) and b(n) then S(s1)+ S(s2)= S(a(n)+ b(n)). This can be obtained by setting alpha and beta equal to one. Without wanting to sound pedantic, I hope you understand that writing some words in capitals does not increase their truth and makes you look a bit crazy (even though I believe you had no wrong intentions and just wanted to communicate). This is a scientific lecture, and scientists in general prefer to raise valid arguments or explain how/why they’ve done something with others if necessary and can also reflect on their mistakes without someone virtually screaming at them for what they have (supposedly) done wrong.
@@remysmets
Thank you for your reply.
Sometimes, a cigar is just a cigar, as another scientist, Sigmund Freud, once noted :-).
The capitalization, as you half-guessed, was simply meant to *emphasize* the main points of my argument, and not to represent yelling. Leaving aside the ad hominems, let us not get our nickers tied in a knot, at the drop of a hat.
Axiom#1 appears to be a mathematical sleight of hand, to permit the shifting of the terms of the sequence without explicitly declaring so. And if all the zeros in the whole sequence were removed from it, the same way that the zeros resulted from the removal of first a0 and then a1, (applying axiom#1 twice), then the whole trick of the term-by-term vertical summation (dadada, dadada, dadada, ...) would have never worked at all.
Maybe the answer lies into stating clearly the exact terms and conditions required, before applying axiom#1.
Elias Andrikopoulos alright! I think the problem with removing all the zero’s is that there are infinitely many of them. Therefore, I think you can’t apply axiom 1 to an infinite amount of numbers in the series and that’s why you can’t keep removing the zeros. Furthermore, axiom 1 applies to the first term only, and because 0 is not always the first term it is once again impossible to remove all of them
Awesome
S(1 -1 +1 -1+...)=1/2
S(1 +0 -1 +1 +0 -1 +1 +0 -1+...)=2/3
S(1+0 +0 -1 +1 +0 +0 -1 +1 +0 +0 -1 +...)=3/4
S(1+0 +0 +0 -1 +1 +0 +0 +0 -1 +1 +0 +0 +0 -1 +...)=4/5
The value of this generic sum is a property of the repeating (cyclic) pattern of numbers and the assumption of linearity. Adding an infinite number of zeros would have no effect on a normal sum.
Is this well defined.
The bridge won't be infinitely long, just 2 bricks long: piling up the bricks it's like drawing the "graph" of 1/n^2 vertically, which never traverses its asymptote, which is 2 units of distance away from the outer edge of the first brick.
this is the sum of 1/n, not 1/n^2 so since 1/n diverges then the bridge is infinite.
Giorgio Scattolini Yeah it is 1/n. Also you mean 1/2^n ... 1/n^2 sums to pi^2/6.
Mire ha mi yo tengo halgo ke exponer sobre Hoyente ha distansia local interviene el TAO mas todo los conponente X a ininsio anos generasion de paso X mujer no es igual ha honbre y con un pero si de mismo hablar con un ofisina de ufolojia se podia denunsiar primer caso primero guardia sin ofisina no vale
Math plus physics is a disaster
1:10:58 Instead of using L'Hôpital, simply cancel the common factor of (1-x) to obtain (1+x)/(1+x+x²), which clearly approaches ⅔.