Was Ramanujan Wrong? | A Controversial Infinite Sum

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SyberMath

SyberMath

Күн бұрын

Пікірлер: 62
@knutthompson7879
@knutthompson7879 10 ай бұрын
When you are dealing with infinite sums, you need to define what you mean by “sum” and “equals” since infinite processes don’t really fit in our finite world. You can make a lot of interesting definitions that allow you to assign interesting values to infinite series. They are consistent under their definition even if they seem ridiculous using our everyday common sense and intuition. The problem with the proofs that start “Let’s call this sum S …” is it is only valid to do that if S exists. If the sum doesn’t exist, anything we do afterward will be nonsense.
@l.w.paradis2108
@l.w.paradis2108 10 ай бұрын
Infinite processes cannot be completed, but that does not have any bearing on whether such sums exist. The set of all natural numbers already exists; it does not need to be generated. It "fits" in our world just fine. We first encountered this in grammar school, when we divided 1/3: 0.333333 . . . = 3/10 + 3/100 + 3/1000 + ••• + 3/(10^n) + 3/(10^(n + 1)) + ••• I like your conclusion a lot. Don't bow to "common sense" with that other stuff. Infinity is cool.
@SyberMath
@SyberMath 10 ай бұрын
@@l.w.paradis2108 thank you!
@Chris-5318
@Chris-5318 4 ай бұрын
@knuttompson7879 there are no processes involved in this topic. 0.999... is not the result of a process. It just exists as a complete infinite series.
@yusufdenli9363
@yusufdenli9363 10 ай бұрын
Infinite is not a number. So, (Infinite) - (Infinite) is not equals zero.. It's undefined
@Music--ng8cd
@Music--ng8cd 10 ай бұрын
Depends on the infinity. .999999.... = 1
@Chris-5318
@Chris-5318 4 ай бұрын
@@Music--ng8cd That doesn't address @yusud...'s comment at all.
@bubblecast
@bubblecast 10 ай бұрын
This series proves the unity of the universe. Every number can be proved to be equal to every other number.
@JourneyThroughMath
@JourneyThroughMath 10 ай бұрын
He took the approach thatIve suggested. Bring in other series to help out than numberphile did and see if you get a different answer. Thank you so much for this video!
@SyberMath
@SyberMath 10 ай бұрын
Np. Thank you!
@NadiehFan
@NadiehFan 10 ай бұрын
For a brief moment I thought you were going to talk about the Riemann zeta function and analytic continuation. The series representation of ζ(s) = ∑ n⁻ˢ with n ≥ 1 does not converge for Re(s) ≤ 1 but for Re(s) < 0 we can use the identity ζ(s) = 2ˢ·πˢ⁻¹·sin(πs/2)·Γ(1 − s)·ζ(1 − s) Since Γ(n) = (n − 1)! and ζ(2) = π²/6 (the famous _Basel problem_ solved by Euler) this immediately gives ζ(−1) = −¹⁄₁₂. But in spite of all that he is hyped up to be Ramanujan would not have understood this because he had no understanding of complex analysis due to his lack of formal training in pure mathematics.
@SyberMath
@SyberMath 10 ай бұрын
Good point!
@BrijeshYadav-gq2cq
@BrijeshYadav-gq2cq 2 ай бұрын
Thank you for sharing your perspective! Your approach is quite interesting and appears to be a unique way of organizing the series. Let's break it down step by step to explore its implications: 1. Restating Your Argument You considered the series and grouped terms in a specific pattern: You identified that certain groups add up to multiples of 9: , , and . Based on this pattern, you proposed that the sum can be rewritten as: S = 1 + 9 \times (1 + 2 + 3 + 4 + 5 + 6 + \dots) 2. Algebraic Rearrangement From your equation, you get: S = 1 + 9S 8S = -1 S = -\frac{1}{8} 3. Analysis of the Result The value you derived, , differs from the conventional result of . However, it's important to note that manipulating divergent series like is highly sensitive to the methods used. In standard mathematical frameworks: The grouping and rearrangement of terms in an infinite series need to be handled carefully because they can lead to different results. Ramanujan’s original result of is consistent with analytic continuation and the Riemann Zeta function. Why Does Your Approach Differ? 1. Rearrangement Issue: In the series , rearranging or regrouping terms without a formal method (like analytic continuation) can lead to contradictions because the series is divergent. 2. Conventional Framework: The result comes from analytic continuation, a formal process that extends the domain of functions to assign finite values to divergent series. 3. Your Method: By grouping terms into multiples of 9, you’ve introduced a specific pattern, but such groupings are not generally accepted in formal summation methods. Conclusion Your proof highlights the difficulty and non-intuitive nature of dealing with divergent series. It challenges traditional results, showing that divergent series require rigorous frameworks to produce consistent answers. The accepted result of is grounded in methods like the Riemann Zeta function and regularization, which might not align with other groupings or interpretations. If you have more thoughts or another approach, feel free to share!
@SyberMath
@SyberMath 2 ай бұрын
Great analysis! Thank you
@lawrencejelsma8118
@lawrencejelsma8118 10 ай бұрын
The sums to infinity all violate Gauss's Law. Gauss's Law really show the indefinite form of infinity/infinity division and infinity difference with infinity errors. Gauss already said 1 + 2 + 3 + ... + n = n(n+1)/2. Out to infinity this is close to 1/2(n)^2 or 1/2(infinity)^2 or infinity because infinity is too huge. The error results because you are assigning a sum to infinity to also say 1/2(infinity + 1)(infinity) = infinity or 1/2(s + 1)s = s or after dividing out s, 1/2(infinity + 1) = 1. In this case infinity = 1. In the 1/2s^2 = s approximation where infinity + 1 was infinity = 2. Without dividing out and using the quadratic equation we see 0 or 2 being infinity. We are sort of creating the building blocks of infinity mathematics making as much sense that since 1 times 0 = 2 times 0 then 1 = 0 or 2 proofs. That is why they all are failing just by how Gauss's Law proves 1 = 0 or 2. It also says approximate proof infinity and infinity + 1 not being approximately equal because you then could subtract infinity from infinity. In that case you can get any finite number as you tried to do to extract 0 times a finite number equivalent expressions.
@jasonthomas2908
@jasonthomas2908 10 ай бұрын
Your statement of the theorem at 8:50 is enough to settle this. If the sequence doesn't go to the limit of 0 then the infinite series doesn't converge absolutely, top marks. People should review their Real Analysis :D It's fun to think about how the converse isn't true: if a sequence converges to a limit of 0 then that doesn't necessarily mean that the infinite series converges absolutely. The Harmonic Series is a fun example to think about. Clearly the series S = 1/x -> 0 as x -> \inf, but the infinite series diverges to \inf. Thinking about this now, Real Analysis was a very rewarding course.
@SyberMath
@SyberMath 10 ай бұрын
Absolutely!
@GetMeThere1
@GetMeThere1 10 ай бұрын
I wrote a comment to mathologer's first video and said that if a sum of positive integers can equal a negative number then ALL mathematical proofs which depend on reduction to an absurdity are questionable. He responded and said "Well, mathematicians have all said it was right." (paraphrased). I like to think that perhaps my comment caused him to reconsider (and so the second video).
@calculus988
@calculus988 10 ай бұрын
This is one of my favorite videos. Nice!
@SyberMath
@SyberMath 10 ай бұрын
Glad you enjoyed it!
@Qermaq
@Qermaq 10 ай бұрын
Funny, because there are two actual values you can plug into the triangular number formula that spit out - 1/12, they're -1/2 +/- sqrt(3)/6. So maybe they are infinity too. ;)
@sundareshvenugopal6575
@sundareshvenugopal6575 10 ай бұрын
Those two series certainly cannot be subtracted when they are finite cause they are not identical. Neither can they be subtracted when they are infinite. This does not mean mathematics is flawed. By definition a series is a sequence of partial sums. So two series are always identical if and only if, iff, the sequence of partial sums are identical or equal. This would be a better definition to use when dealing with equality of series in general for both finite and infinite series. Otherwise, if two different infinite series converge to the same limit, then we would have to consider them to be equal just cause they infact reach or arrive at the same result or conclusion but only by adopting different means, ways, methods, approaches, routes , paths, call it what you will, some perhaps more direct, others perhaps rather more circuitous. Are means just as important as the ends ? It clearly seems to be. Are some means always better than others ? Possibly or could be only in some cases but not in others, depending on what else needs to be achived during the intervening stages, or exactly how a computation in fact plans to use each and every individual term of the sequence of partial sums, the shorter or shall I say the faster sequence may not necessarily generate a lot more intermediate data, for a computation to use. These are all options to always bear in mind, to keep open and to carefully consider in computing.
@Music--ng8cd
@Music--ng8cd 10 ай бұрын
If you can find a practical use for an infinite sum, then the equation holds true. Otherwise, it's just a possibility. This summation has uses in physics, so it is valid. The -1/12 is actually the y-intercept of the parabola which forms the smoothed asymptote of the summation of the series: en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Physics
@philkeyouz2157
@philkeyouz2157 5 ай бұрын
You cannot take infinity as a real number, you must use imaginary number (complex number). And this sum has a real value because is the square of i/2×sqrt(3) , as the sum of the reciprocal of the square which is also a complex square : (2×ln(i)/sqrt(2)×sqrt(3)×i)^2 which as a real value of pi^2/6.
@NickCager
@NickCager 6 ай бұрын
He only missed it by infinity.
@Mehrdad_Basiry-fj4rl
@Mehrdad_Basiry-fj4rl 10 ай бұрын
So amazing... learn alot...❤❤❤.
@SyberMath
@SyberMath 10 ай бұрын
Thank you so much 😀
@jalma9643
@jalma9643 10 ай бұрын
So i would like to ask a website where i could get to see problem and their solution. I am preparing for some olympiad and i need some exercise to sharpen my mind
@SyberMath
@SyberMath 10 ай бұрын
artofproblemsolving.com
@wtspman
@wtspman 10 ай бұрын
I was prepared to be annoyed when I clciked on this video. After watching it, I am pleased that you revealed the nonsense of -1/12. Thank you!
@SyberMath
@SyberMath 10 ай бұрын
Glad to hear that!
@shamilbabayev8405
@shamilbabayev8405 10 ай бұрын
so idea is that you can not denote the sum with finite number of s ......
@RSLT
@RSLT 7 ай бұрын
In many instances, people see the video below regarding this Numberphile video and tell me that 1+2+3...=-1/12 is debunked by mathematicians. I left the comment below; please read my comment on the video below, before making any inquiries and assertions. kzbin.info/www/bejne/j6asep2Cp5upi6M In Mathologer video, I observe three distinct parts: Firstly, Mathologer demonstrates a lack of understanding regarding the meaning of 1+2+3+...=-1/12 and attempts to use the limit definition and a high schooler's method for divergent series. Consequently, everything presented in Numberphile is deemed incorrect. Secondly, Mathologer returns to continue the video and realizes that 1-1+1-1...=1/2 makes sense, but the assertion that 1+2+3+...=-1/12 is definitely incorrect. Lastly, it is delightful to witness Mathologer discovering the actual method behind the true meaning of 1+2+3+...=-1/12, which is Ramanujan's method. However, he neglects to revisit his initial argument from the first day. It is unfair to a genius like Ramanujan to not be appreciated, and the mockery of his work is truly disheartening. I must acknowledge that Mathologer demonstrates exceptional genius throughout this video. He acquires knowledge that typically takes people much longer to grasp. Here you can see several proofs for 1+2+3… = -1/12 . kzbin.info/door/Bm0bVo59PQgGUCcuWXyHfA Ignoring the Identity Theorem is not a good method. It is interesting that even after so many years, Ramanujan is still greatly misunderstood. Numberphile presents Ramanujan's work, and it is unfortunate that some individuals, have formed a mathematical 'religion' and have disregarded figures like Ramanujan, assuming their own mathematical beliefs as infallible truths. Treating a mathematical concept as a rigid doctrine akin to the Bible, which must be strictly followed, is not helpful. Mathematics is not just a collection of theorems; it is a set of rules that we define and occasionally modify. For a long time, we have known that the limit definition is poorly defined. Let's consider the series 1-1+1-1... as a switch that we can turn on and off repeatedly. If we do it long enough, we will perceive a light with half intensity. Similarly, in our homes, we have alternating current (AC) power, but we don't observe divergent light. This debunking is comparable to someone claiming that complex analysis is wrong because the square root of -1 does not exist. While that statement may have some meaning, it is completely incorrect. We utilize complex numbers because they are extremely useful. Therefore, constructing arguments based on the poorly defined limit definition is not correct. The limit definition itself has its limitations, just like real numbers, and we need to employ logical reasoning to extend them beyond their obvious range. Overall, it is highly recommended for everyone to watch this video completely. If viewers possess the same level of intelligence as Mathologer, they will undoubtedly comprehend the true meaning behind the equation 1+2+3...=-1/12.
@SyberMath
@SyberMath 7 ай бұрын
The sum diverges and therefore cannot equal a finite value. Case closed! 😄
@antonelloluzzi5661
@antonelloluzzi5661 10 ай бұрын
So, why is Ramanujan' s result however considered valid on history of mathematics?
@gregorysmathchannel5357
@gregorysmathchannel5357 10 ай бұрын
Are you saying that you think Ramanujan's result is thought to be valid?
@yoav613
@yoav613 10 ай бұрын
The sum diverges but still it has some meaning,for example there ate definite integrals that converges and when you solve them you get crazy infinite sum like this.
@vascomanteigas9433
@vascomanteigas9433 10 ай бұрын
Analytical continuation
@GsBabu-sk6iv
@GsBabu-sk6iv 10 ай бұрын
What S=0 proves that infinite n numbers on 'Number Line' ends up where it began. Geometrically, the Number Line is a Circle. Because only a circle can represent Infinite points as all numbers in existence and the summation of Infinite Number becomes circumference. This is just my personal Mathematical Conjecture, which I treat as Postulate. P. S. Ramanujan' summation is proven as fact when applied in Quantum physics.
@heraldoftruth9522
@heraldoftruth9522 5 ай бұрын
Do yall even know indeterminate forms?
@user-og9nl5mt1b
@user-og9nl5mt1b 10 ай бұрын
It's the same as 2+2 = 5 videos .
@SyberMath
@SyberMath 10 ай бұрын
😮
@charleskrueger5523
@charleskrueger5523 10 ай бұрын
Cmon, don’t waste our time with this!
@scottleung9587
@scottleung9587 10 ай бұрын
Ha, we proved him wrong once and for all!
@SyberMath
@SyberMath 10 ай бұрын
😁
@Yogut3k
@Yogut3k 10 ай бұрын
🧐🤔🙃😉🤓😂👍👏
@habeebalbarghothy6320
@habeebalbarghothy6320 10 ай бұрын
Zero & infinity are not numbers
@bsmith6276
@bsmith6276 10 ай бұрын
Zero is a number
@OK-ei7io
@OK-ei7io 10 ай бұрын
Zero is most certainly a number
@RSLT
@RSLT 7 ай бұрын
Very good explanation. I wish you were more clear that Ramanujan was 100% correct and, in fact, ahead of his time, still admired by many.
@SyberMath
@SyberMath 7 ай бұрын
Thank you! He was definitely ahead of his time but he either got this wrong or was just messing around 😁
@mircoceccarelli6689
@mircoceccarelli6689 10 ай бұрын
S( n ) = 1 + 2 + ... + ( n - 1 ) + n S( n ) = n ( n + 1 ) ÷ 2 n > / = 0 => n ( n + 1 ) ÷ 2 > / = 0 => S( n ) > / = 0 => Ramanujan sbaglia ! 😊👍👋
@SyberMath
@SyberMath 10 ай бұрын
La serie infinita i cui termini sono i numeri naturali 1 + 2 + 3 + 4 + ⋯ è una serie divergente. Caso chiuso. 😁
@GsBabu-sk6iv
@GsBabu-sk6iv 10 ай бұрын
@@SyberMath Tu italini..
@erikroberts8307
@erikroberts8307 10 ай бұрын
This is the same stupid nonsense that Einstein used when coming up with his theories of relativity. Not only is his math wrong, but also his conclusions concerning the nature of physics are wrong as well; such as there are no time dilations, no length dilations, no mass increases, no multiple universes, and no light dependencies on a moving object. And yet, if you try and point out these mistakes to the average physist, they will fight you all the way as though you are trying to kill their religion or something. Weird. I am just glad that we have an excellent channel on KZbin that can point out this type of mistake correctly and have some fun doing it. Yes, the mistake comes into play when dealing with divergent series. I learned about the 2nd method on redpenblackpen.
@SyberMath
@SyberMath 10 ай бұрын
Thank you!
@robertveith6383
@robertveith6383 10 ай бұрын
@ erikroberts8307 -- No, it is not. You are just running your ignorant, dismissive mouth about Einstein, his theories, and real-world applications. He has no bearing in this subforum. Do much Internet browsing on many sites about how what you referred to has been tested out. You need to delete your first paragraph.
@erikroberts8307
@erikroberts8307 10 ай бұрын
@robertveith6383 Regarding Einstein's theories, in truth, nothing has ever been tested out or verified by any legitimate scientist or scientific organization. For example, what is spacetime? Sure, he talks about it all the time. He even has the audacity to come up with equations based on it, but what the hell is it? Where is it? How does the Universe co-exist with it? The thing here is this: All of Einstein's theories begin with a false premise. Anything that comes afterward, whether true or false, is just bullshit like the mathematical example in this episode. Sure, the math may look to be correct, but when the original premise is completely wrong, the conclusion is wrong, too. You can't find the sum of a series once it has gone into divergent mode. It's that simple.
@starlord2633
@starlord2633 2 ай бұрын
1 + 2 + 3 + ... + ∞ = -1/12✅✅✅
@SyberMath
@SyberMath 2 ай бұрын
False! Ramanujan was wrong 😁
@starlord2633
@starlord2633 2 ай бұрын
@SyberMath Ramanujan was always ✅✅✅ you are wrong 😊
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