This uses a very advanced concept that is not at all well explained. Even knowing about umbral calculus, I struggled. The problem is, that most viewers will not catch the subtlety, that these don't behave like exponents at all - otherwise, you could just simply solve for B. Instead, powers are not actually powers, we are abusing notation and using powers to compute indices, so "powers" are actually different numbers (that is, (B^1)^2 is not equal to B^2). It's a very nice piece of math, but I'm afraid most viewers will leave more confused than before, which is not usually happening in Numberphile videos.
I felt like the notation here was sometimes very confusing. It could get through the basics, but I think a seperate video just on the bernoulli numbers would have helped
That is called the umbral method and is a very deep and fascinating thing. You can represent relations between elements in a sequence as a polynomial equation.
Yes, it's ok if you don't explain how/why you can transform powers to superscripts to subscripts, but I feel like the fact that that was what was going on could've been explained a lot more clearly. Call attention to the fact that what you're doing is weird (applying polynomial expansion to subscripts), don't just say "we do this". If you're going to introduce "scare quotes" that are the notation that allows this, mention that they're the notation that allows this, don't just put things in quotes and then don't say what the quotes are for.
I couldn't help but notice how the final equation presented in this video closely aligns with my own research, which revolves around similar concepts of numbers, their powers, and the operations that connect them. I've been working on a paper that delves into these ideas, and I'm excited to share that my formula has even helped me minimize a function on the OEIS sequence A259569.Interestingly, I've found additional connections to the concepts in "The Book of Numbers" by Sir John Conway, specifically on page 54. I believe this discovery has the potential to be a significant game-changer in number theory, as it can generate numerous number sequences and quadratic equations that are pivotal across various mathematical disciplines. Below are links to a few OEIS entries that illustrate these patterns, which span dimensions, topology, and other fields of mathematics:A001117, A000920, A000918, A001117, A000051, A000919 These patterns found in the OEIS represent just a fraction of my discovery. The equation I formulated encompasses far more, and I believe that publishing this research could lead to even more significant findings. More importantly, I have developed a theory suggesting that Pascal's Triangle possesses multiple layers or dimensions, and our current operations are limited to its first layer. My formula provides an opportunity to explore these additional layers, potentially revolutionizing number theory by unveiling deeper structural insights and new mathematical relationships.I've been trying to reach out to Ellen Eischen for feedback, but unfortunately, I haven't received a response yet. I truly believe that this work has the potential to make a huge impact on the mathematical community and beyond. Any guidance or assistance in getting this research published would be greatly appreciated.
It was a struggle to read, wish I hadn't. Seems just kind of weird. Maybe the intention is to show that mathematicians can be goofy and mess around which is a pleasant notion, I guess, but I don't really care about the lore of some professor, I just want to understand how tf to do my assignment.
The Bernoulli numbers were also the subject of the first computer program ever written, by Ada, Countess of Lovelace, for Babbage's unbuilt Analytical Engine.
Great video! The use of the notation "B" as a formal symbol such that its "powers" are the Bernoulli numbers is called "umbral calculus". To me it feels like it shouldn't work but it somehow does. It's such a neat "turn the idea on its head" encapsulation of the more standard way to solve this problem, namely: noticing that (n+1)^k = sum_i [(i+1)^k - i^k] (from i = 0 to n), expanding out (i+1)^k - i^k as powers of i, and thus relating the formula for the sum of (k-1)th powers to the formulas for all the lower sums of powers.
Thanks for sharing. Some of it went way over my head I don't understand the meaning of some of the symbols but when Ellen cancelled things out it made sense.
There's a mistake in the list of the Bernoulli's numbers (7:59), you skipped N=60 and as a result the apparent rule that odd Ns yield B_N == 0 no longer seems to apply. But it's just a typo ;)
This is awesome, I have the same stuff in my old notebooks! I manually calculated the formulas up to around the power of 14 before writing a program to generate what I now know are the Bernoulli numbers 😊 I have a notebook somewhere where I filled it up to around the power of 30. I don't do much math anymore, but these occupied years of my life and bring a lot of nostalgia.
I did the same, in my freshman year as a physics student 🙂I don't recall exactly how many powers I worked out manually, but I realized at some point there was a recurrence relation, and then pretty much left it at that. Amazing that quite a lot of people would independently tackle this problem as an exercise, almost like a pastime. Then again, it's not that surprising when you think about it, since this kind of problem crops up in many practical calculations in math and physics.
I feel like this was left unsaid, so I'll try my best to guess: When we say B_k := B^k, and "For k > 1, (B - 1)^k = B^k", we don't actually want to calculate B and raise it to the power k (by using quadratic formula or whatever), but instead we want to express B^k in terms of B^(k - 1), B^(k - 2), ..., B^0, and then replace each occurrence of B^j with B_j. Those B_j can then be calculated in the same way, and we use their values to find B_k = B^k = ... This twisted my brain. Is this right?
There definitely seems to be some important steps missing from this video half of this is incomprehensible, but the interesting looking subscript on the sigma at 11:07 was enough to keep me entertained
I agree and it seems poorly explained as to the importance of this. It may be that the channel has reached a point where more confusing and esoteric subjects have to be explored in order to avoid repetition
Great content! I won a competition for the best mathematical text for gymasium-student in sweden many year ago, writing about precicely this sum. I love that you bring this up. Great problem with many lovely surprises! I can send you my work if you like to see my thoughts as an 18 year old. Funny enough half a life time ago, I am 36 today. Thank you for this!
The only time that I have encounter the Bernoulli numbers is one time. I teach my students how to get the tangent function (Taylor's version) long dividing the Taylor series of sin(x) and cos(x). I got the first 4 terms and normally move on... but one time a student asked me if tan(x) has a compact form, I told them that I didn't know, I proceeded to google it and there they were!!!! Bernoulli numbers appear in the Taylor expansion of tan(x) :D
I think a clear explanation why the quotation marks were used would have helped. Cause the algebra worked without them. Generally really liked the video though. Stumbled upon Faulhaber's formula and Bernoulli numbers a couple years back when I looked more mathematically into the 1+2+3+... = -1/12 story. There is a nice blog post of Terry Tao on it.
The quotation marks are used because "B" does not behave like a number. In this case, at least, the quotation marks are instructing you to expand the expression and then replace the powers of B with the corresponding Bernoulli numbers. The important thing to note is that the Bernoulli numbers don't have the relation between them that they would have if they were actually powers of a single number B.
Funnily enough I was just looking into this like 1 hour ago myself. I got to using a trick where I summed x^-(x-1)^k from x=1 to x=n. Where I expanded the (x-1)^k term on the right side of the =-sign. This yields a binomial factor in front of all the sums of x^i from i=0 to i=k-1, meaning you can use previous summation formulas to find the next one, recursively.
I also worked on this a couple of months ago with a similar thought process. I assumed the formula is a polynomial with degree k+1. Although I remember finding equations that can be solved for coefficents of that polynomial. I am not sure whether they were recursive or not but they included summations and I couldn't find a way to get rid of them.
@@divadus2487 That is what this boils down to. You expand (x-1)^k on the right side and take the x^k - (x-1)^k. Sum both sides from x=1 up to x=n and you will be left with n^k on the LHS. The right-handside will require formula for smaller ks, including the k-1 formula. Sorry for explaining it badly.
@@DolphyWind guessing a solution like you did works too. You can even work out some of the first few sums by hand so you can determine the coefficients. You have the advantage that you don't need previous solutions for the next Sk.
I like that numberphile is a place where someone can mention their "childhood math's journal" and it doesn't need to be qualified with a joke or self deprecation.
Back in High School, I was more interested in the coefficients of the formula f(n) for each k. The first coefficient is one over k, the next is one half. I believe the next was one over (n times (n minus 1)). Then lots more... seems like I had about 20. I put this into a Westinghouse competition. In the process, since I was working in floating-point numbers, I made a process to change a decimal into a fraction of small integers.
This is perfect example of why I consider Numberphile to be the best mathematics channel on TH-cam. The problem of summing a power series of integers as been around for thousands of years and has been worked on by mathematicians from all over the world. The only criticism that I would make is that Jakob Bernoulli was responsible for the formula presented. Bernoulli numbers are attributed to him, even though his Japanese contemporary, Seki Takakazu, beat him to publication.
After a long time this is a numberphile video where i did not understand anything anymore after some point in the middle of the video. But the Formula surely has cool name.
also to understand Bernoulli numbers start by taking (B-1)^k = B^k and expanding out the left hand side of the equation and then subtract B^k from both sides, then for each power of B replace it with the Bernoulli number (B^1 is replaced with B(1), B^2 is replaced with B(2) , B^3 is replaced by B(3) etc) (btw I using B(n) as the nth Bernoulli number) to find them use induction, start with k = 2 and when you subtract B^2 from both sides and done the replacement step, the only variable is B(1) so solve for that to find B(1). next do k = 3, once you subtract B^3 from both sides and replaced the powers with the B(n) function, you have 2 variables: B(2) and B(1), you already know B(1) having already computed it so substitute it in and solve for B(2),. next do k = 4, once you have subtracted B^4 from both sides and done the conversion, you have 3 variables: B(3), B(2) and B(1) you already know B(1) and B(2) so substitute those in and solve for B(3) as so on
Using this B object, you can show that the Taylor series of the function f(x) = xe^x/(e^x-1) is the Bernoulli numbers over k!, and this utilizes exp(x*B). Instead of using quotation marks, I'll consider this an operator T from the span of B^k to the real numbers. We know that T[(n+B)^k - B^k] = k * [1^(k-1) + ... + n^(k-1)], k=0,1,2,3,... specifically, T[(1+B)^k - B^k] = k Consider the following: exp(x*(1+B)) = exp(nx + xB) = e^x*exp(xB) T[exp(x(1+B)) - exp(xB)] = T[(e^x - 1)exp(xB)] = (e^x - 1)T[exp(xB)] on the other hand, T[exp(x(1+B)) - exp(xB)] = sum{k=0 to infinity}x^k/k! * T[(1+B)^k - B^k] = sum{k=0 to infinity}x^k * k/k! = sum{k=1 to infinity}x^k /(k-1)! = xe^x putting this together, (e^x - 1)T[exp(xB)] = xe^x T[exp(xB)] = xe^x / (e^x - 1) and since T[exp(xB)] = sum{k=0 to infinity}x^k/k! * T[B^k] = sum{k=0 to infinity}B_k * x^k/k! this gives us a nice Taylor series.
Finally, some umbral calculus! There are lots of other weird formulas one can "prove" by replacing powers with subscripts. The Bernoulli numbers are usually defined by x/(e^x-1)=sum B_n/n!*x^n, and then replacing the subscripts with powers, the formula becomes x/(e^x-1)=e^(Bx). Then you can clear denominators to get x=e^((B+1)x)-e^(Bx). For powers of x beyond x^1, the right side must have coefficients that are all 0's. Expanding the right-hand side as a Taylor series again gives x=sum ((B+1)^n-B^n)/n! x^n, so (B+1)^n=B^n for n>1, as in the video.
@@leif1075 It's umbral calculus, but yes, it comes from the same root as "umbrella" because of the "shady" methods used to prove things. The basic idea is the trick of switching subscripts and exponents. It definitely doesn't always work, but it does give valid results in a surprising number of cases.
Why would anyone create a list of numbers with tbat definition is the key quesruon it seems everyone else is ignoring. Its arbitrary to some extent..why not e^× plus 1or minus 2or anything else..see whar i mean..to tnat extent itnos arbitrary..and hiw was it derivedtobe included in tbis for.ula is not re.otrly addressed here..
@@leif1075 I'm not exactly sure of the original use for them, but they come up often when looking at trig functions, for example. When tan x, cot x, sec x, and csc x are written using complex exponentials, each of them has something like e^(ix)-1 in the denominator. That gives the connection with Bernoulli numbers. The expansion of cot x also leads to Euler's formula for values of the zeta function at even positive integers.
I would love to see a follow-up video about the Bernoulli numbers as they relate to the zeta function... if it can be rendered accessible to us math hobbyists. I love number theory and the big unsolved problems like the Riemann hypothesis, but I haven't taken a formal mathematics course in over 30 years.
When i was in college i remember solving the intial sum using openblas. I created a matrix with certain values and then mldivided, and it spat back coefficients that would equal the sum to N for some power K.
I have no idea how formulas of this complexity are discovered, but is both depresssing and interesting that someone can discover formulas like this or the cubic and quartic equations which are so complicated, they are impractical. We definitely need to engineer better brains.
I know less about Bernoulli numbers now than before watching this and I haven't a clue what the Fabulous Formula is. But, I do love your videos, they can’t all be gems I suppose.
using the information in the video I was able to use the Formula to complete the sums of power sequences up the the 6th power: (1/2)N(N+1) (1/6)N(N+1)(2N+1) (1/4)N^2(N+1)^2 (1/30)N(N+1)(2N+1)(3N^2+3N-1) (1/12)N^2(N+1)(2N^3+4N^2+N-1) (1/42)N(N+1)(2N+1)(3N^4+6N^3-3N+1) (as far as I know they can't be factorised any further but I could be wrong)
There's an even neater version of FFF that you can write as $$\sum_{k=1}^n k^p = \int_b^{b+n} x^p dx$$ which of course makes sense, since the sum should be approximated by an integral if you fudge the endpoints of integration a bit.
The notation used is incredibly confusing. From B_k = B^k it would follow that B_{k+1} = B_k * B which is not what is intended. Also, what's with the quotes?
As I understand, this trick of "lowering the exponents into subscripts" is really a shortcut to building and handling a generating function (formal series) for the Bernoulli Numbers, right?
Before going to university I studied the exact same problem, I never got to the Bernoulli numbers but I got and algorithm to compute a closed formula for each k, even now I really intrigued by the how different the problem is when k is negative
Gotta be honest, this wasn't numberphile's finest work. I think there was a glimmer of something there, but a zillion characters later, "it all cancels out". I don't know. I think it could've been very interesting, maybe it just needs another run at it, or 2 parts, or well, I don't know.
So S(0)=N, S(1)=N*(N+1)/2, S(2)= N*(N+1)*(2*N+1)/6, S(3) = (N*(N+1)/2)^2 Do any other S(k) factor? Do infinitely many factor? Do all factor? I mean do their numerators factor over integers.
Here is my sum of power formula without using Bernoulli number in LaTeX format \sum_{k=1}^{n} k^{m}=\sum_{b=1}^{m+1} \binom{n}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i
If anyone wants more info about the functions S_k = 1^k+...+n^k as functions of n, I recommend Donald Knuth. Johann Faulhaber and sums of powers. Mathematics of Computation 61, no.203 (Jul 1993), p.277--294. And no invalid messing around with B's.
I feel like i'm missing something here. Are they asserting that the sum of (1^2+2^2+...100^2) is 1,015,050? That can't be right. Surely squaring the numbers from 1 to 100 can't be more than 100 squared times 100. (100^2+100^2+...100^2) = 100^3 = 1,000,000 so (1^2+2^2+...3^2) has to be less than 1,000,000. (plugging it into excel confirms that... the sum is 338,350 not 1,015,050. What am I missing?
Look a step back to see that 1^2+2^2+...+100^2 = 1/3 of the computation including the Bernoulli number, which in this case is 1,015,050. Then the sum of squares up to 100 is indeed 1,015,050/3 = 338,350.
![ลองสุ่มไอโฟน 990 โกงมั้ย? [ โกงมั้ยครับ ep.101 ] | DOM](http://i.ytimg.com/vi/bfsWe0j492k/mqdefault.jpg)
This uses a very advanced concept that is not at all well explained. Even knowing about umbral calculus, I struggled. The problem is, that most viewers will not catch the subtlety, that these don't behave like exponents at all - otherwise, you could just simply solve for B. Instead, powers are not actually powers, we are abusing notation and using powers to compute indices, so "powers" are actually different numbers (that is, (B^1)^2 is not equal to B^2).
It's a very nice piece of math, but I'm afraid most viewers will leave more confused than before, which is not usually happening in Numberphile videos.
This was supposed to be a 30 mins video. Bernouli numbers aren't a joke.
I felt like the notation here was sometimes very confusing. It could get through the basics, but I think a seperate video just on the bernoulli numbers would have helped
Agree. The Bernoulli Coefficients still haven't yet be explained how to derive ?
That is called the umbral method and is a very deep and fascinating thing. You can represent relations between elements in a sequence as a polynomial equation.
Yes, it's ok if you don't explain how/why you can transform powers to superscripts to subscripts, but I feel like the fact that that was what was going on could've been explained a lot more clearly. Call attention to the fact that what you're doing is weird (applying polynomial expansion to subscripts), don't just say "we do this". If you're going to introduce "scare quotes" that are the notation that allows this, mention that they're the notation that allows this, don't just put things in quotes and then don't say what the quotes are for.
@@SeanTBarrett Also don't yell as a compensation for not going into the details.
You did better than me. I was lost about 1-2 minutes into this video.
What an awesome story about John Conway. Clearly a brilliant mathematician but wonderful to know he was an equally brilliant teacher.
The story is about Ellen Eischen.
Yes, he was. A man who could say "its obvious" and then get you to the point where he is right, it is obvious (now).
John Conway's Game of Life was the only thing that kept me sane during many many many hours of boredom in school.
can we stop obsessing over singular persons, please?
A whole video on Bernoulli numbers seems needed.
I couldn't help but notice how the final equation presented in this video closely aligns with my own research, which revolves around similar concepts of numbers, their powers, and the operations that connect them. I've been working on a paper that delves into these ideas, and I'm excited to share that my formula has even helped me minimize a function on the OEIS sequence A259569.Interestingly, I've found additional connections to the concepts in "The Book of Numbers" by Sir John Conway, specifically on page 54. I believe this discovery has the potential to be a significant game-changer in number theory, as it can generate numerous number sequences and quadratic equations that are pivotal across various mathematical disciplines. Below are links to a few OEIS entries that illustrate these patterns, which span dimensions, topology, and other fields of mathematics:A001117, A000920, A000918, A001117, A000051, A000919 These patterns found in the OEIS represent just a fraction of my discovery. The equation I formulated encompasses far more, and I believe that publishing this research could lead to even more significant findings.
More importantly, I have developed a theory suggesting that Pascal's Triangle possesses multiple layers or dimensions, and our current operations are limited to its first layer. My formula provides an opportunity to explore these additional layers, potentially revolutionizing number theory by unveiling deeper structural insights and new mathematical relationships.I've been trying to reach out to Ellen Eischen for feedback, but unfortunately, I haven't received a response yet. I truly believe that this work has the potential to make a huge impact on the mathematical community and beyond. Any guidance or assistance in getting this research published would be greatly appreciated.
For anyone who didn't pause to read the journal entries at 1:20, take the time to do so now. Definitely worth it.
It was a struggle to read, wish I hadn't. Seems just kind of weird. Maybe the intention is to show that mathematicians can be goofy and mess around which is a pleasant notion, I guess, but I don't really care about the lore of some professor, I just want to understand how tf to do my assignment.
The Bernoulli numbers were also the subject of the first computer program ever written, by Ada, Countess of Lovelace, for Babbage's unbuilt Analytical Engine.
Daughter of George Lord Byron.
> when I was a kid I kept a math journal...
> when I got to Princeton...
Sounds about right
I would guess "when I was a kid I kept a math journal" must be the origin story of Fermat.
I've never seen someone use quotation marks in a formula before.
_"I've never seen someone use quotation marks in a formula before,"_ he added, with equal parts disdain and enthusiasm. 👀 - j q t -
Great video! The use of the notation "B" as a formal symbol such that its "powers" are the Bernoulli numbers is called "umbral calculus". To me it feels like it shouldn't work but it somehow does. It's such a neat "turn the idea on its head" encapsulation of the more standard way to solve this problem, namely: noticing that (n+1)^k = sum_i [(i+1)^k - i^k] (from i = 0 to n), expanding out (i+1)^k - i^k as powers of i, and thus relating the formula for the sum of (k-1)th powers to the formulas for all the lower sums of powers.
Ut this doesnt negma to exain what bwenoulli numbwrs are so no oneno matter how smart could fully understand this formula just from this vjdep.
@@leif1075 you ok, bud?
Thank you! I was wondering what the equation (B-1)^k = B^k was.
Thanks for sharing. Some of it went way over my head I don't understand the meaning of some of the symbols but when Ellen cancelled things out it made sense.
That amazing astounding alliteration though.
There's a mistake in the list of the Bernoulli's numbers (7:59), you skipped N=60 and as a result the apparent rule that odd Ns yield B_N == 0 no longer seems to apply. But it's just a typo ;)
This is awesome, I have the same stuff in my old notebooks! I manually calculated the formulas up to around the power of 14 before writing a program to generate what I now know are the Bernoulli numbers 😊 I have a notebook somewhere where I filled it up to around the power of 30.
I don't do much math anymore, but these occupied years of my life and bring a lot of nostalgia.
I did the same, in my freshman year as a physics student 🙂I don't recall exactly how many powers I worked out manually, but I realized at some point there was a recurrence relation, and then pretty much left it at that.
Amazing that quite a lot of people would independently tackle this problem as an exercise, almost like a pastime. Then again, it's not that surprising when you think about it, since this kind of problem crops up in many practical calculations in math and physics.
I feel like this was left unsaid, so I'll try my best to guess:
When we say B_k := B^k, and "For k > 1, (B - 1)^k = B^k", we don't actually want to calculate B and raise it to the power k (by using quadratic formula or whatever), but instead we want to express B^k in terms of B^(k - 1), B^(k - 2), ..., B^0, and then replace each occurrence of B^j with B_j. Those B_j can then be calculated in the same way, and we use their values to find B_k = B^k = ...
This twisted my brain. Is this right?
Yes, that's correct.
Are we all just going to ignore the 1:25 John Conway leap-frog journal entry? Because that's hilarious.
Pausing the video there is both fun and worthwhile.
I can imagine any other professional mathematician saying "This is trivial and left as an exercise for the reader"
Fermat was an amateur mathematician and he left a famous theorem as an exercise for the reader in his copy of Diophantus. You may have heard of it...
There definitely seems to be some important steps missing from this video half of this is incomprehensible, but the interesting looking subscript on the sigma at 11:07 was enough to keep me entertained
I agree and it seems poorly explained as to the importance of this. It may be that the channel has reached a point where more confusing and esoteric subjects have to be explored in order to avoid repetition
A little hard to wrap my head around it, but a great video nonetheless. Will definitely be looking more into this
Yes, I'm not a fan of 'Algebra' explanations. Can get those outside numberphile
Great content! I won a competition for the best mathematical text for gymasium-student in sweden many year ago, writing about precicely this sum. I love that you bring this up. Great problem with many lovely surprises! I can send you my work if you like to see my thoughts as an 18 year old. Funny enough half a life time ago, I am 36 today.
Thank you for this!
Would love to watch your discovery as a video❤
@@PMA_ReginaldBoscoG i might do that!
Always a joy to be early to a Numberphile video!
The only time that I have encounter the Bernoulli numbers is one time. I teach my students how to get the tangent function (Taylor's version) long dividing the Taylor series of sin(x) and cos(x). I got the first 4 terms and normally move on... but one time a student asked me if tan(x) has a compact form, I told them that I didn't know, I proceeded to google it and there they were!!!! Bernoulli numbers appear in the Taylor expansion of tan(x) :D
Taylor's version omg I'll be thinking about that next time I Taylor expand!
@@awaredeshmukh3202 yeahhh!!!!!
I think a clear explanation why the quotation marks were used would have helped. Cause the algebra worked without them. Generally really liked the video though. Stumbled upon Faulhaber's formula and Bernoulli numbers a couple years back when I looked more mathematically into the 1+2+3+... = -1/12 story. There is a nice blog post of Terry Tao on it.
The quotation marks are used because "B" does not behave like a number. In this case, at least, the quotation marks are instructing you to expand the expression and then replace the powers of B with the corresponding Bernoulli numbers.
The important thing to note is that the Bernoulli numbers don't have the relation between them that they would have if they were actually powers of a single number B.
Funnily enough I was just looking into this like 1 hour ago myself. I got to using a trick where I summed x^-(x-1)^k from x=1 to x=n. Where I expanded the (x-1)^k term on the right side of the =-sign. This yields a binomial factor in front of all the sums of x^i from i=0 to i=k-1, meaning you can use previous summation formulas to find the next one, recursively.
I also worked on this a couple of months ago with a similar thought process. I assumed the formula is a polynomial with degree k+1. Although I remember finding equations that can be solved for coefficents of that polynomial. I am not sure whether they were recursive or not but they included summations and I couldn't find a way to get rid of them.
You can also use telescopic sums.
And the formulas for smaller ks.
@@divadus2487 That is what this boils down to. You expand (x-1)^k on the right side and take the x^k - (x-1)^k. Sum both sides from x=1 up to x=n and you will be left with n^k on the LHS. The right-handside will require formula for smaller ks, including the k-1 formula.
Sorry for explaining it badly.
@@DolphyWind guessing a solution like you did works too. You can even work out some of the first few sums by hand so you can determine the coefficients. You have the advantage that you don't need previous solutions for the next Sk.
I actually proved by myself the solution for k=4 in middle school. Couldn’t remember how I did that but I got a cookie from my math teacher
What a character Mr. Conway was!
I like that numberphile is a place where someone can mention their "childhood math's journal" and it doesn't need to be qualified with a joke or self deprecation.
Back in High School, I was more interested in the coefficients of the formula f(n) for each k. The first coefficient is one over k, the next is one half. I believe the next was one over (n times (n minus 1)). Then lots more... seems like I had about 20. I put this into a Westinghouse competition. In the process, since I was working in floating-point numbers, I made a process to change a decimal into a fraction of small integers.
This is perfect example of why I consider Numberphile to be the best mathematics channel on TH-cam.
The problem of summing a power series of integers as been around for thousands of years and has been worked on by mathematicians from all over the world. The only criticism that I would make is that Jakob Bernoulli was responsible for the formula presented. Bernoulli numbers are attributed to him, even though his Japanese contemporary, Seki Takakazu, beat him to publication.
This one went over my head pretty quickly, I have to refresh my knowledge on Bernoulli numbers first.
Okay your not going to understand but when Ellen wrote "k=2," it was so beautiful I cried. ❤
I think you need to elaborate here. Allow us to cry along with you.
I understood everything right up until 0:01
I can’t believe I was finally here for the start of a Numberphile!
You were still a couple minutes late. Sorry.
@@MathsMadeSimple101you're one to talk 🥩🍇🤠
The Contrabulous Fabtraption of Professor Horatio Hufnagel
After a long time this is a numberphile video where i did not understand anything anymore after some point in the middle of the video. But the Formula surely has cool name.
well i guess you broke the record of beating the 301 views 💀
It's great to find the origins and proofs of math puzzles you've busied yourself with
also to understand Bernoulli numbers start by taking (B-1)^k = B^k and expanding out the left hand side of the equation and then subtract B^k from both sides, then for each power of B replace it with the Bernoulli number (B^1 is replaced with B(1), B^2 is replaced with B(2) , B^3 is replaced by B(3) etc)
(btw I using B(n) as the nth Bernoulli number)
to find them use induction, start with k = 2 and when you subtract B^2 from both sides and done the replacement step, the only variable is B(1) so solve for that to find B(1).
next do k = 3, once you subtract B^3 from both sides and replaced the powers with the B(n) function, you have 2 variables: B(2) and B(1), you already know B(1) having already computed it so substitute it in and solve for B(2),.
next do k = 4, once you have subtracted B^4 from both sides and done the conversion, you have 3 variables: B(3), B(2) and B(1) you already know B(1) and B(2) so substitute those in and solve for B(3)
as so on
Using this B object, you can show that the Taylor series of the function f(x) = xe^x/(e^x-1) is the Bernoulli numbers over k!, and this utilizes exp(x*B).
Instead of using quotation marks, I'll consider this an operator T from the span of B^k to the real numbers.
We know that
T[(n+B)^k - B^k] = k * [1^(k-1) + ... + n^(k-1)], k=0,1,2,3,...
specifically,
T[(1+B)^k - B^k] = k
Consider the following:
exp(x*(1+B)) = exp(nx + xB) = e^x*exp(xB)
T[exp(x(1+B)) - exp(xB)] = T[(e^x - 1)exp(xB)] = (e^x - 1)T[exp(xB)]
on the other hand,
T[exp(x(1+B)) - exp(xB)] = sum{k=0 to infinity}x^k/k! * T[(1+B)^k - B^k]
= sum{k=0 to infinity}x^k * k/k! = sum{k=1 to infinity}x^k /(k-1)! = xe^x
putting this together,
(e^x - 1)T[exp(xB)] = xe^x
T[exp(xB)] = xe^x / (e^x - 1)
and since T[exp(xB)] = sum{k=0 to infinity}x^k/k! * T[B^k] = sum{k=0 to infinity}B_k * x^k/k!
this gives us a nice Taylor series.
Finally, some umbral calculus! There are lots of other weird formulas one can "prove" by replacing powers with subscripts. The Bernoulli numbers are usually defined by x/(e^x-1)=sum B_n/n!*x^n, and then replacing the subscripts with powers, the formula becomes
x/(e^x-1)=e^(Bx).
Then you can clear denominators to get x=e^((B+1)x)-e^(Bx). For powers of x beyond x^1, the right side must have coefficients that are all 0's. Expanding the right-hand side as a Taylor series again gives
x=sum ((B+1)^n-B^n)/n! x^n,
so (B+1)^n=B^n for n>1, as in the video.
Umbrella calculus? Sounds shady..why and what is that
@@leif1075 It's umbral calculus, but yes, it comes from the same root as "umbrella" because of the "shady" methods used to prove things. The basic idea is the trick of switching subscripts and exponents. It definitely doesn't always work, but it does give valid results in a surprising number of cases.
Why would anyone create a list of numbers with tbat definition is the key quesruon it seems everyone else is ignoring. Its arbitrary to some extent..why not e^× plus 1or minus 2or anything else..see whar i mean..to tnat extent itnos arbitrary..and hiw was it derivedtobe included in tbis for.ula is not re.otrly addressed here..
@@leif1075 I'm not exactly sure of the original use for them, but they come up often when looking at trig functions, for example. When tan x, cot x, sec x, and csc x are written using complex exponentials, each of them has something like e^(ix)-1 in the denominator. That gives the connection with Bernoulli numbers. The expansion of cot x also leads to Euler's formula for values of the zeta function at even positive integers.
We are interested. Do it again WITHOUT skipping details. ;)
Pace was just a bit quick on this one, I would have to rewatch a couple of times.
I would love to see a follow-up video about the Bernoulli numbers as they relate to the zeta function... if it can be rendered accessible to us math hobbyists. I love number theory and the big unsolved problems like the Riemann hypothesis, but I haven't taken a formal mathematics course in over 30 years.
you can actually define these sums in this video using nested integrals which is a really fun way of solving it
bobber's fabulous formula and the bruli numbers
Mathologer's video on Bernoulli numbers is by far the best so far on youtube.
When i was in college i remember solving the intial sum using openblas. I created a matrix with certain values and then mldivided, and it spat back coefficients that would equal the sum to N for some power K.
RIP Conway. What a Man. What a Mind.
sum of k^n can be derived from the formula (k + 1) ^n - k^n it seems bernoulli number is a list of precomputated values involved in this derivation.
I have no idea how formulas of this complexity are discovered, but is both depresssing and interesting that someone can discover formulas like this or the cubic and quartic equations which are so complicated, they are impractical. We definitely need to engineer better brains.
Mathologers video was amazing
I know less about Bernoulli numbers now than before watching this and I haven't a clue what the Fabulous Formula is. But, I do love your videos, they can’t all be gems I suppose.
It seems like the Bernoulli numbers were conceived specifically to have the properties that it has which is really cool
i wish i watched this video when i was more awake, because it seems really fascinating with the math
Fantastic..... During my school days I found recurrence formula using integration
using the information in the video I was able to use the Formula to complete the sums of power sequences up the the 6th power:
(1/2)N(N+1)
(1/6)N(N+1)(2N+1)
(1/4)N^2(N+1)^2
(1/30)N(N+1)(2N+1)(3N^2+3N-1)
(1/12)N^2(N+1)(2N^3+4N^2+N-1)
(1/42)N(N+1)(2N+1)(3N^4+6N^3-3N+1)
(as far as I know they can't be factorised any further but I could be wrong)
"We both agree that re-inventing the wheel is both fun and worthwhile." 1:29
There's an even neater version of FFF that you can write as
$$\sum_{k=1}^n k^p = \int_b^{b+n} x^p dx$$
which of course makes sense, since the sum should be approximated by an integral if you fudge the endpoints of integration a bit.
This one has gone over my head! I'm not a mathematician though so I think I can be forgiven! :D
Faulhaber's Fabulous Formula is fantastic!
The notation used is incredibly confusing.
From B_k = B^k it would follow that B_{k+1} = B_k * B which is not what is intended.
Also, what's with the quotes?
This is a watch twice video. Doing that to superscripts and subscripts definitely deserves quotation marks.
Is that the one piece
It seems like there should be some cool animation of Pascal's triangle involving algebra autopilot to generate the Bernoulli Numbers.
@10:47 the first N should be lowercase to match the others.
It's the conference room that has light switches built into the back of bookcases again.
As I understand, this trick of "lowering the exponents into subscripts" is really a shortcut to building and handling a generating function (formal series) for the Bernoulli Numbers, right?
A reference to Pascal would be nice as the method for deriving the was partly due to him.
Before going to university I studied the exact same problem, I never got to the Bernoulli numbers but I got and algorithm to compute a closed formula for each k, even now I really intrigued by the how different the problem is when k is negative
Love seeing another numberphile video. Ellen is awesome! But not at UO - no bueno. FTD - GO DAWGS
there is also a formula using stirling’s numbers
Unusually picky comments here. Great video!
Gotta be honest, this wasn't numberphile's finest work. I think there was a glimmer of something there, but a zillion characters later, "it all cancels out". I don't know. I think it could've been very interesting, maybe it just needs another run at it, or 2 parts, or well, I don't know.
"Faulhaber's Fabulous Formula" sounds positively Vancian
Out of all this, I notices that the email server has misspelled Ellen's first name! "ELLLEN EISCHEN" 1:27
I didn't understand the notation here, seems very confusing how a power can be an index and vice-versa.
7:53 B_36 🤨🧐🤌❓
BTW, Johann Faulhaber was a 17th century German mathematician.
I think you might need to talk more about these numbers if you could.
For a second, I thought we were getting -1/12 again and I was going to have a meltdown.
Near the end I started hearing - _n_ to the _B_ - _n_ to the _B_ - and my eyes went funny
Obvious next question, what do Bernoulli numbers encode?
5:08 The formula is covered IN B's!
12:55 IS REAL!
"So little Ellen what do you want to be when you've grown up?
*Keeps a secret math journal* "Well..." 😂
Faulhaber's fabulous formula, peresented by Ellen E. Eischen (where the E. stands for Ellen E. Eischen).
Ellen Ellen Ellen Ellen Ellen Ellen Ellen Ellen ..... Eischen Eischen Eischen Eischen Eischen Eischen
So S(0)=N, S(1)=N*(N+1)/2, S(2)= N*(N+1)*(2*N+1)/6, S(3) = (N*(N+1)/2)^2
Do any other S(k) factor? Do infinitely many factor? Do all factor? I mean do their numerators factor over integers.
Definitely one of my favourite problems! I remember finding out about these myself when I was twelve, I actually still have my notebook! :)
Here is my sum of power formula without using Bernoulli number in LaTeX format
\sum_{k=1}^{n} k^{m}=\sum_{b=1}^{m+1} \binom{n}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i
The only part of this video that I can't understand is how somebody who is obviously knowledgeable about Numberphile asks if there is more paper.
If anyone wants more info about the functions S_k = 1^k+...+n^k as functions of n, I recommend Donald Knuth. Johann Faulhaber and sums of powers. Mathematics of Computation 61, no.203 (Jul 1993), p.277--294. And no invalid messing around with B's.
Gotta love a purple sharpie :3
Bernoulli numbers, ah yes, the every other 0 numbers, with total abject chaos for the other half.
I don't understand how 3B^2 becomes 3B-sub-2.
Please do a video on why Log base a of b times Log base b of a equals 1?
I can understand that they produce reciprocals but why?
Ugh :D We still don't know what B is, but by some magic, this crazy abuse of notation somehow seems to work out to result in something meaningful.
Just a reminder you can easily make things less complicated without using letters in math 😅
I feel like i'm missing something here. Are they asserting that the sum of (1^2+2^2+...100^2) is 1,015,050? That can't be right. Surely squaring the numbers from 1 to 100 can't be more than 100 squared times 100. (100^2+100^2+...100^2) = 100^3 = 1,000,000 so (1^2+2^2+...3^2) has to be less than 1,000,000. (plugging it into excel confirms that... the sum is 338,350 not 1,015,050. What am I missing?
I wondering that also. funny enough 1015050/3 =338350 and, the sum of even's squared to 100 - 1015050=338350
Look a step back to see that 1^2+2^2+...+100^2 = 1/3 of the computation including the Bernoulli number, which in this case is 1,015,050. Then the sum of squares up to 100 is indeed 1,015,050/3 = 338,350.
@@DOTvCROSS Oh - I see - 1,050,050 is the result of "(100+B)^3-B^3". In the formula you multiply that by 1/k (1/3 in this case)
8:26 they start with a multiplier if 1/3 but lost track of that. Take that into account and it matches.
Oh and now I see all the other replies :)
9:28 You forget to divide the answer by 3
The same John Conway that created "The Game of Life." ?
301 views gang where you at?