Faulhaber's Fabulous Formula (and Bernoulli Numbers) - Numberphile
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7 ай бұрынFeaturing Ellen Eischen from the University of Oregon.
More links & stuff in full description below ↓↓↓
Ellen Eischen: www.elleneischen.com
John Conway videos: bit.ly/ConwayNumberphile
CORRECTION: Sum of the first 100 squares is 338350, not 1015050 as depicted. Sorry!
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Video by Brady Haran and Pete McPartlan
Thanks to Debbie Chakour and Chip Rollinson from The Numberphile Society for some useful error spotting.
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@huleboermannhule44 7 ай бұрын
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
@kauffmann101 7 ай бұрын
Agree. The Bernoulli Coefficients still haven't yet be explained how to derive ?
@viliml2763 7 ай бұрын
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.
@SeanTBarrett 7 ай бұрын
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.
@zstanojevic9574 7 ай бұрын
@@SeanTBarrett Also don't yell as a compensation for not going into the details.
@dodaexploda 7 ай бұрын
You did better than me. I was lost about 1-2 minutes into this video.
@sloppycee 7 ай бұрын
A whole video on Bernoulli numbers seems needed.
@dieago12345 7 ай бұрын
What an awesome story about John Conway. Clearly a brilliant mathematician but wonderful to know he was an equally brilliant teacher.
@divadus2487 7 ай бұрын
The story is about Ellen Eischen.
@andrewharrison8436 7 ай бұрын
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).
@soilnrock1979 7 ай бұрын
John Conway's Game of Life was the only thing that kept me sane during many many many hours of boredom in school.
@yonaoisme 7 ай бұрын
can we stop obsessing over singular persons, please?
@Simon-fg8iz 7 ай бұрын
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.
@primenumberbuster404 7 ай бұрын
This was supposed to be a 30 mins video. Bernouli numbers aren't a joke.
@zh84 7 ай бұрын
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.
@joegillian314 Күн бұрын
Daughter of George Lord Byron.
@cczpzrjq5348 7 ай бұрын
For anyone who didn't pause to read the journal entries at 1:20, take the time to do so now. Definitely worth it.
@KEKW-lc4xi 4 ай бұрын
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.
@petervillano3484 7 ай бұрын
> when I was a kid I kept a math journal... > when I got to Princeton... Sounds about right
@vigilantcosmicpenguin8721 7 ай бұрын
I would guess "when I was a kid I kept a math journal" must be the origin story of Fermat.
@johnchessant3012 7 ай бұрын
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.
@leif1075 7 ай бұрын
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.
@gregd2243 7 ай бұрын
@@leif1075 you ok, bud?
@SunnyTheGentleFox 7 ай бұрын
Thank you! I was wondering what the equation (B-1)^k = B^k was.
@FirstLast-gw5mg 7 ай бұрын
I've never seen someone use quotation marks in a formula before.
@quill444 7 ай бұрын
_"I've never seen someone use quotation marks in a formula before,"_ he added, with equal parts disdain and enthusiasm. 👀 - j q t -
@ivarkrabol 7 ай бұрын
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?
@TheReligiousAtheists 7 ай бұрын
Yes, that's correct.
@MathsMadeSimple101 7 ай бұрын
That amazing astounding alliteration though.
@DusanPavlicek78 7 ай бұрын
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 ;)
@nathankane464 7 ай бұрын
Are we all just going to ignore the 1:25 John Conway leap-frog journal entry? Because that's hilarious.
@vigilantcosmicpenguin8721 7 ай бұрын
Pausing the video there is both fun and worthwhile.
@nutsnproud6932 7 ай бұрын
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.
@kumomelody8305 7 ай бұрын
A little hard to wrap my head around it, but a great video nonetheless. Will definitely be looking more into this
@user-wh2vu9my4d 7 ай бұрын
Yes, I'm not a fan of 'Algebra' explanations. Can get those outside numberphile
@pearceburns2787 7 ай бұрын
I can imagine any other professional mathematician saying "This is trivial and left as an exercise for the reader"
@zh84 7 ай бұрын
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...
@meerjt11 7 ай бұрын
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
@jcf3996 7 ай бұрын
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
@ZedaZ80 7 ай бұрын
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.
@Nikolas_Davis 7 ай бұрын
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.
@sebastiandierks7919 7 ай бұрын
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.
@tom.prince 7 ай бұрын
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.
@robin9740 7 ай бұрын
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.
@DolphyWind 7 ай бұрын
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 7 ай бұрын
You can also use telescopic sums. And the formulas for smaller ks.
@robin9740 7 ай бұрын
@@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.
@robin9740 7 ай бұрын
@@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.
@svenviktor1234 7 ай бұрын
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!
@PMA_ReginaldBoscoG 7 ай бұрын
Would love to watch your discovery as a video❤
@svenviktor1234 7 ай бұрын
@@PMA_ReginaldBoscoG i might do that!
@Charliehuangmagic 7 ай бұрын
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
@joyboricua3721 7 ай бұрын
What a character Mr. Conway was!
@JeffKaylin-ft5cx 7 ай бұрын
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.
@mimasweets 7 ай бұрын
Okay your not going to understand but when Ellen wrote "k=2," it was so beautiful I cried. ❤
@harrysvensson2610 7 ай бұрын
I think you need to elaborate here. Allow us to cry along with you.
@PotatoMcWhiskey 7 ай бұрын
Always a joy to be early to a Numberphile video!
@luismijangos7844 7 ай бұрын
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
@awaredeshmukh3202 7 ай бұрын
Taylor's version omg I'll be thinking about that next time I Taylor expand!
@luismijangos7844 7 ай бұрын
@@awaredeshmukh3202 yeahhh!!!!!
@theCDGeffect 7 ай бұрын
you can actually define these sums in this video using nested integrals which is a really fun way of solving it
@Kramer-tt32 7 ай бұрын
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.
@hrithikgeorge4751 7 ай бұрын
It's great to find the origins and proofs of math puzzles you've busied yourself with
@claytonarg5947 7 ай бұрын
I can’t believe I was finally here for the start of a Numberphile!
@MathsMadeSimple101 7 ай бұрын
You were still a couple minutes late. Sorry.
@aceman0000099 7 ай бұрын
@@MathsMadeSimple101you're one to talk 🥩🍇🤠
@breathless792 7 ай бұрын
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
@akswrkzvyuu7jhd 7 ай бұрын
This is perfect example of why I consider Numberphile to be the best mathematics channel on KZfaq. 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.
@NousSpeak 6 ай бұрын
This one went over my head pretty quickly, I have to refresh my knowledge on Bernoulli numbers first.
@jacksonstarky8288 7 ай бұрын
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.
@MathFromAlphaToOmega 7 ай бұрын
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 7 ай бұрын
Umbrella calculus? Sounds shady..why and what is that
@MathFromAlphaToOmega 7 ай бұрын
@@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.
@leif1075 7 ай бұрын
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..
@MathFromAlphaToOmega 7 ай бұрын
@@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.
@bomberdan 7 ай бұрын
The Contrabulous Fabtraption of Professor Horatio Hufnagel
@MooImABunny 7 ай бұрын
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.
@Nikolas_Davis 7 ай бұрын
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?
@dr.bogenbroom894 7 ай бұрын
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
@Tarsonis42 7 ай бұрын
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.
@gosuf7d762 7 ай бұрын
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.
@justaguywhoisanidiot159 7 ай бұрын
I wrote about this in my summar vacation school maths project !
@breathless792 7 ай бұрын
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)
@BB13580 7 ай бұрын
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?
@apburner1 7 ай бұрын
I understood everything right up until 0:01
@Etudio 7 ай бұрын
RIP Conway. What a Man. What a Mind.
@matze9713 7 ай бұрын
bobber's fabulous formula and the bruli numbers
@deltalima6703 7 ай бұрын
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.
@01binaryboy 7 ай бұрын
Fantastic..... During my school days I found recurrence formula using integration
@xyzbesixdouze 4 ай бұрын
Is there a known proof that in the Pascal triangle numbers are prime in the second culumn, if and only if all numbers in that line are divisable by that number?
@nikykovalski5869 7 ай бұрын
Is that the one piece
@user-xi6oy9xi4r 7 ай бұрын
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.
@DonRedmond-jk6hj 7 ай бұрын
A reference to Pascal would be nice as the method for deriving the was partly due to him.
@drmathochist06 3 ай бұрын
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.
@nightowl9512 7 ай бұрын
I didn't understand the notation here, seems very confusing how a power can be an index and vice-versa.
@lucasdepetris5896 7 ай бұрын
there is also a formula using stirling’s numbers
@PretzelBS 7 ай бұрын
It seems like the Bernoulli numbers were conceived specifically to have the properties that it has which is really cool
7 ай бұрын
@10:47 the first N should be lowercase to match the others.
@inscitia 7 ай бұрын
"We both agree that re-inventing the wheel is both fun and worthwhile." 1:29
@brettgoodroad7747 7 ай бұрын
amazing
@martinkunev9911 7 ай бұрын
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?
@bananatassium7009 7 ай бұрын
i wish i watched this video when i was more awake, because it seems really fascinating with the math
@kianushmaleki 7 ай бұрын
That is an interesting story
@DeclanMBrennan 7 ай бұрын
It seems like there should be some cool animation of Pascal's triangle involving algebra autopilot to generate the Bernoulli Numbers.
@Android480 7 ай бұрын
Obvious next question, what do Bernoulli numbers encode?
@gachadogeyt7724 7 ай бұрын
what's the actual view count on the 301 vid?
@karlboud88 7 ай бұрын
I have a weird question: What is the relation between the numbers 360 and 66? for some reason a right triangle with sides 360 and 66 have a hypothenuse of 366 and I feel like it's very odd. Is it just coincidence?
@mokovec 7 ай бұрын
Yes, but since they are all multiples of 6, they appear more related, since the numbers are bigger. If you compare 60, 11 and 61, do you get the same vibe? Two of them are prime already, so one can't reduce further. If you want to be fancy, you can call them Diophantine.
@PhilBagels 7 ай бұрын
@@mokovec Yes, it's a primitive Pythagorean triple. Not really related to the video (but something I understand better). 11 squared is 121, which is 60+61. Just like 3 squared = 4+5 5 squared = 12+13 7 squared = 24+25 etc. And all of those are Pythagorean triples.
@karlboud88 7 ай бұрын
@PhilBagels thank you! I knew there was a simple explanation! So it was a coincidence that I happened to check out those numbers XD
@masonwheeler6298 7 ай бұрын
Love seeing another numberphile video. Ellen is awesome! But not at UO - no bueno. FTD - GO DAWGS
@Alex-ik8pr 7 ай бұрын
This one has gone over my head! I'm not a mathematician though so I think I can be forgiven! :D
@user-sz1gc1hj3m 3 күн бұрын
awesome
@Nethershaw 7 ай бұрын
Did anyone else notice that the definition of the formula is recursive and there was not one mention of recursion through the whole video? Can we go deeper?
@MooImABunny 7 ай бұрын
really makes me wonder what other sequences can be encoded in an algebric object like B
@IsoYear 7 ай бұрын
very cool story
@nycgus 7 ай бұрын
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?
@DOTvCROSS 7 ай бұрын
I wondering that also. funny enough 1015050/3 =338350 and, the sum of even's squared to 100 - 1015050=338350
@JackieTeutch 7 ай бұрын
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.
@nycgus 7 ай бұрын
@@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)
@hrysp 7 ай бұрын
8:26 they start with a multiplier if 1/3 but lost track of that. Take that into account and it matches.
@hrysp 7 ай бұрын
Oh and now I see all the other replies :)
@bergpolarbear 6 ай бұрын
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.
@MarloTheBlueberry 7 ай бұрын
Faulhaber's Fabulous Formula is fantastic!
@kpopalitfonzelitaclide2147 7 ай бұрын
Mathologers video was amazing
@highseassailor 7 ай бұрын
What a beast!
@rajivjha8066 7 ай бұрын
Small notation issue at 10:32. The first bracket should also have n instead of N
@the_eternal_student 7 ай бұрын
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.
@leif1075 7 ай бұрын
HOW exactly did Falhauberir Bernoulli or whoever derive or deduce this formula and WHY??
@rosiefay7283 7 ай бұрын
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.
@Impatient_Ape 7 ай бұрын
I have never heard "easy/easily" pronounced so strongly as "izzy/izzily" before. I'm aware of lots of American shibboleths, but this is new to me. Where's in the USA is this common?
@TheEternalVortex42 7 ай бұрын
To me it sounds a bit in-between /izi/ and /ɪzi/. I'm going to guess it's a midwest thing.
@anakimluke 7 ай бұрын
Gotta love a purple sharpie :3
@theultimatereductionist7592 7 ай бұрын
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.
@jetzeschaafsma1211 7 ай бұрын
So what happened to the formula then? Because the answer of 1015050 is correct?
@hedger0w 7 ай бұрын
5:08 The formula is covered IN B's!
@frankwilliamabagnale3303 7 ай бұрын
Faulhaber's fabulous formula, peresented by Ellen E. Eischen (where the E. stands for Ellen E. Eischen).
@molybd3num823 7 ай бұрын
Ellen Ellen Ellen Ellen Ellen Ellen Ellen Ellen ..... Eischen Eischen Eischen Eischen Eischen Eischen
@dgrandlapinblanc 3 ай бұрын
Ok. Thanks.
@invisibledave 7 ай бұрын
It's the conference room that has light switches built into the back of bookcases again.
@johnboyer144 7 ай бұрын
I don't understand how 3B^2 becomes 3B-sub-2.
@Sdp40fguy 7 ай бұрын
well i guess you broke the record of beating the 301 views 💀
@aachucko 7 ай бұрын
Out of all this, I notices that the email server has misspelled Ellen's first name! "ELLLEN EISCHEN" 1:27
@daniellassander 7 ай бұрын
Im pausing at 4:31, what if K is 100? 1^99 + 2^99 + 3^99...+ N^99 and that is supposedly 1/100, something doesnt work out here at all.
@user-gd9vc3wq2h 7 ай бұрын
@daniellassander When pausing, you can't hear the voice which at that very moment is saying "I haven't told you what the B is..."
@caspermadlener4191 7 ай бұрын
Definitely one of my favourite problems! I remember finding out about these myself when I was twelve, I actually still have my notebook! :)
@patrick.c.nwaoduah8639 4 ай бұрын
What happened to the 1/3 in the formula. The answer gotten as 1,015,050 must be wrong then. Please reconfirm.
@andrewharrison8436 7 ай бұрын
This is a watch twice video. Doing that to superscripts and subscripts definitely deserves quotation marks.
@elaaboudikhalid4963 7 ай бұрын
Checking wikipedia bernoulli numbers B_1 = -0.5 , but in the video it is 0.5 , anyone can explain ? Anyway Awesome video
@wesleydeng71 7 ай бұрын
BTW, Johann Faulhaber was a 17th century German mathematician.
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