ZERO DOUBLE FACTORIAL

[@AndDiracisHisProphet]

I'm proposing the third solution. sqrt[2/pi] = 1

[@blackpenredpen]

Much better!!!

[@Fematika]

Breaking news: Although engineers have widely believed pi = 3, a new ground-breaking discovery shows that pi = 2! This is great news, because now it'll be easier to multiply by pi.

[@mmukulkhedekar4752]

Fematika how??.......ok understood

[@blackpenredpen]

Fematika Mathematicians hate him! Click the link below and start to learn how to multiply by pi the new easy way. Www.piequalstotwo.com

[@__8023]

blackpenredpen is the link a virus?!?!?!

[@heliocentric1756]

Pi=2. Problem solved

[@aniruddhvasishta8334]

Heliocentric that’s what I thought

[@super-awesome-funplanet3704]

π=3. 14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679...

[@ilkerberkeelcioglu1130]

@@super-awesome-funplanet3704 you must be fun at parties

[@dudono1744]

There is actually a proof of pi=2

[@maxhaibara8828]

@@dudono1744 this is the proof of pi=2

[@milos_radovanovic]

You have defined Γ generalization with odd numbers and then used it on even numbers without showing that formula works for both. And in fact, by that definition: for n=1.5 you would have 2!! = 2*sqrt(2/pi)*Γ(2) = 2*sqrt(2/pi) ≠ 2 for n=2.5 you would have 4!! = 4*sqrt(2/pi)*Γ(3) = 8*sqrt(2/pi) ≠ 8 etc.

[@blackpenredpen]

Miloš Radovanović yes. "Extend the double factorial". I didnt state 0!! has to be that. I asked which one should it be.

[@Idran]

blackpenredpen but if f(n) != g(n) for n in the domain of f, then g isn't an extension of f. This isn't an extension of the double factorial over integers in the first place because it doesn't hold for even integers, it's only a extension of double factorial over odd integers.

[@blackpenredpen]

TehAvenger29 true, and this is KZbin. :)

[@angelmendez-rivera351]

Idran No, you fail to understand the true issue. The problem is that there are two different definitions of the double factorial at use here, and technically, as such, two different functions. Every function with the set of integral numbers as its domain can be defined by a recursion formula and a base case chosen axiomatically. The recursion formula holds for both cases, but the base case has been defined differently in secret. There is a resolution to this. The formula for any general odd n > 0 differs from that any general even n > 0, but both formulas can be combined into a single formula which is valid for any n > 0, and this formula yields 0^0 for the case n = 0, demonstrating the fact that the base case is ambiguous and arbitrary. The formula can be extended to the negative integers as well, and there two possible extensions. One extension consists of generalizing the recursion formula itself for negative n, which would allow for any integer value to have a well-defined value for the double factorial. The more traditional choice of extension consists instead of preserving the recursion, at the expense of having undefined values at the negative even integers. Both extensions are related to each other via a simple multiplication. The general explicit formula I’m talking about can be written in terms of other elementary functions, but it is quite troublesome to write it on the phone so I will refrain from it. However, I will give hints to solve the problem. For n = 2k - 1, k > 1 or k = 1, n!! = (2k)!/(2^k*k!), while for n = 2k, n!! = 2^k*k!, where k is a natural number in every case. This can be merged into a single equation which works for all natural n, expressive in terms of only n. Then, a recursion formula can yield an extension to negative integers, and then one may continue it to complex numbers if the operations are analytic and well-defined for complex numbers.

[@angelmendez-rivera351]

kajacx Oh, but you can, and that is where you are wrong. I’ve read enough math papers to know this with 100% certainty. Good luck convincing me of a falsehood

[@ffggddss]

0!! =1 The first (and simplest) way you did this, was the right one. The second way is invalid, because the formula you got for Γ(n+½), is valid only for integers n; so you can't use it for n=½. [In particular, try using it to find 2!! or 4!! or ... by letting n = 3/2 or 5/2 or ...] Before watching this, I did a little scribbling myself, using the formulas you derived in your previous video for n!! when n is even and when n is odd. I got the same pair of results. That's because the two definitions of n!! in terms of ordinary factorials - one for odd n, one for even n - are mutually incompatible; they disagree when you try to force an n of the wrong parity on either of them. So you must fall back on the rule that a vacuous product = 1, and so, 0!! = (-1)!! = 1; and n!! for any smaller even integer is undefined; for smaller odd integers, the recursion will give reciprocals of the positive odd double factorials, but with alternating signs. Fred

[@takyc7883]

Thanks Fred! Tak

[@xinpingdonohoe3978]

But it's just a bit of fun. Of course 0!!=1 but he's simply messing with the numbers because he can. XD

[@kishankumar37373]

But @ffggddss told the correct way

[@kishankumar37373]

Everyone does not know this

[@hardtimes2597]

Thank you a lot!!!

[@AdasiekkkTrzeci]

8:45 Two times three is six minus one that's five ｑｕｉｃｋ　ｍａｔｈｓ

[@TheRealFlenuan]

I thought of that too lmao

[@MateFacilYT]

:O Excellent video! I like it!

[@adrianhdz138]

Español

[@ralfbodemann1542]

First things first: Congratulations to the LA 2018 Marathon Finisher! That's just amazing. Re 0!!: As long as we can't come up with an unambiguous definition, just leave it undefined!

[@blackpenredpen]

Ralf Bodemann Thank you!!

[@angelmendez-rivera351]

We can find an unambiguous definition for 0!!. It is the case that 0!! = 1. The reason the calculations yielded a different value is because the formula is not valid for even integers.

[@AlbertTheGamer-gk7sn]

That's what people are doing to 0! and 0^0 as well.

[@m1n3c4rt]

undefined factorial

[@HelmutNevermore]

9:56 Your mic went German

[@blackpenredpen]

Pavel Zubkov LOLLLL

[@somerandomdragon558]

Why german?

[@James-zs3vm]

@@somerandomdragon558 his words start to sound german for some reason

[@MrTobify]

As a german I can approve that we have to learn to speak with triple echo at the lowest possible pitch by the age of 4

[@shayanmoosavi9139]

@@MrTobify also don't forget rolling the "r" and saying the "ch" (like nacht) very harshly :) P.S : I'm not a german myself but I like to learn german.

[@henrikljungstrand2036]

The way i see it, we have TWO double factorial functions, one defined for even natural numbers, and one defined for odd natural numbers. When taking the analytic continuation (through Gamma e.g.), they don't completely match, but are off by a factor. Which means we'll either have to redefine 0!! = sqrt(2/π), or otherwise redefine 1!! = sqrt(π/2), in any case we'll get a constant factor involving π, multiplied into one of the original formulas for double factorial.

[@jannis8218]

The second approach is actually wrong, if you plug in 3/2 you will get something different than 2!! which contradicts with the definition.

[@pengiiin]

Jannis Just let pi be equal to 2

[@elliottmanley5182]

Aren't you a bit early for April Fool's Day? Or are you trying to do a Numberphile -1/12 controversy?

[@blackpenredpen]

Elliott Manley maybe both? :)

[@dlevi67]

It's a Parker factorial (parkorial?)

[@blackpenredpen]

dlevi67 it's double factoreo

[@elliottmanley5182]

I spent ages searching for the definition of 'chen lu' before I got the joke.

[@dlevi67]

Chen lu is why you can never stop when factoreos appear: you have to get to the end of the packet.

[@KurdTillDeath]

you lost me at "okay as we all know"

[@twwc960]

As many have pointed out, your second approach is simply wrong as it yields wrong values for double factorials of other even numbers, like 2, 4, 6, etc. Zero double factorial is generally defined to be 1, for example, in the Online Encyclopedia of Integer Sequences (sequence A000165).

[@balijosu]

Whoever came up with that notation needs their math license revoked.

[@Freakinawesome333]

0!!=sqrt(2/pi) contradicts the original rule that n!!=n(n-2)!! as it would imply that 2!!=2sqrt(2/pi). When finding an extension of the OG factoreo function, we decided that the extension should follow the rule n!=n(n-1)!. So we should do the same here with the double factorial. But this means 0!! must be 1.

[@akshat9282]

After you're done with the double factorial series, maybe generalise for n-factorials. That seems legitimate, izzint?

[@Chiinox]

Akshat K Agarwal what if n is complex

[@akshat9282]

Nox let us define it for real numbers first, maybe? 😂 Well defining it for complex would generalise it even more and it'd be better but you know *another video*

[@pepebriguglio6125]

Akshat K Agarwal 😂😂😂😂😂

[@NotYourAverageNothing]

Let’s define more numbers first, such as |x| = -1 and ln -1.

[@akshat9282]

Not Your Average Nothing no

[@mmukulkhedekar4752]

can you do pi!! or pi!

[@danielmencl2764]

Yes, by using the Gamma function. Γ(π+1) = π! ≈ 7.188

[@somename2803]

You must be pretty excited about pi

[@maxweinstein1537]

In your last double factorial video, you found (2k)!!=(2^k)k! But, plugging in n+0.5 to the gamma definition in this video, you get (2n)!!=[(2^(n+0.5))/sqrt(pi)]gam(n+1) = sqrt(2/pi)(2^n)n! This is in direct contradiction with your previous definition

[@angelmendez-rivera351]

Max Weinstein This is true. However, the advantage to the latter definition is that it can easily be defined for any complex number as long as the number is not an even negative number, whereas for the original definition, no obvious extension is actually possible.

[@sheppsu7353]

If you ask me, then I think that the double factorial of 0 is 1. The generalization of Gamma that you constructed was written so that it made sense with the factorials of fraction with denominators of 2, however, when you actually test the generalization with a case where n=3/2, you get that 4=sqrt(8/pi). This completely defies the logic that 4!! = 4 * 2, so then technically, the generalization is proven wrong to work for all positive integers and we can also assume that it's answer on 0!! is wrong. Therefore it is more likely that 0!! is 1.

[@adamel-sawaf4045]

I have never seen anyone alternate pen colors and write in the same hand so smoothly

[@henryclc920]

利用gamma(1.5),gamma(2.5)…的歸納觀查，只得到奇數雙階乘與根號pi相乘除以2＾n的結果，用此奇數雙階乘的定義套入分數以求出0!!，當然得到錯誤的結果。

[@thegrb93]

Ahhh, so that's how you evaluate the gammas. Thanks, I need to try that!

[@yoavshati]

So pi=+-2?

[@ffggddss]

No, it can't be negative; it's under a radical which is set equal to a real number. Fred

[@bamberghh1691]

Complex pi value when?

[@Kalmakka]

While the odd double factorial expression (n!! = n×(n-1)×…×3×1) extends to x!! = √(2/π)√(2^x)Γ(x/2 + 1), the even double factorial expression (n!! = n×(n-1)×…×4×2) extends to x!! = √(2^x)Γ(x/2 + 1) (just note that (2n)!! = n!2^n and substitute x = 2n) The √(2/π) factor difference shows up everywhere, not just at 0!!.

[@tusharkaushalrajput]

√2/√π is amazing

[@johannesvalks]

But there is the error ... the equation is only valid for positive integers, not for fraction like 1/2 or 3/2 Put in n=3/2 and you would get 2!! = sqrt( 8 / pi)

[@3manthing]

looking this video at night, from part at 9:58 made me think my computer was possessed :D :D other than that nice video :]

[@rikschaaf]

So in short, if you base your answer on the definition for double factorial for even numbers, then you get 1, but if you base it on the definition for odd numbers, you get sqrt(2/pi). Since 0 is an even number, I feel like 1 should be the correct answer.

[@sansamman4619]

I think the 9:08 case should be reported to the bprp department as quick as possible.

[@jonasdaverio9369]

San Samman Illuminati confirmed

[@jkn6644]

No, as I expected an en.wikipedia.org/wiki/Ood wanted his translation sphere back!

[@tobiasreckinger2212]

if 0!!=1 & 0!!= sqrt(2/pi) then: 1=sqrt(2/pi) => 1=2/pi => pi=2 As an engineer I can`t see any problem here

[@aweebthatlovesmath4220]

🤣🤣

[@AlbertTheGamer-gk7sn]

That's why people choose to keep 0! undefined, like some people do with 0^0.

[@teekayanirudh]

It was only a matter of time before this video appeared in my notifications. Masterpiece

[@twinpeaks8680]

9:57 demon summoned

[@factsheet4930]

But by the second definition, if you plug in n=1.5 then you get that 2!!=1.59577... which is also not the same as the original definition... n has to be an integer!

[@hetsmiecht1029]

n!! for even numbers can be written as (n/2)! * 2^(n/2). (No gamma or pi function required) Filling in n=0 gives 0!! = 0!*2^0 which is 1

[@spaceman392001]

0!! = 1. You used the formula for odd !!, but plugging in n = 1/2 made the operand 0, which is an even number. So that formula no longer holds

[@_DD_15]

Lol at minute 2 i was like: he is gonna use the gamma function, watch out, watch out.. Boom gamma function coming out lol

[@verainsardana]

Never heard about double factorials what are they used for?

[@gian2kk]

Series

[@OonHan]

9:57 EAR RAPEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

[@aashsyed1277]

now what about 0!!!?(0 triple factorial)

[@Ostup_Burtik]

+

[@_DD_15]

Ohhh what a wonderful conclusion. You made my day with this video. Imo 0!!=1. Here is why. We define the factorial of a generic positive integer n as n! = n(n-1)(n-2)...(1). At the same time there is a function called gamma function which happens to have the same properties of the factorial for the positive integers. But, and here is my doubt which hasn't allowed me to sleep for years :D what guarantees that if you use the gamma function for the negative integers or positive fractions it's going to give you a factorial of the given number in the way we intend it. I mean the problem of not being able to evaluate the factorial of negative numbers is conceptual: let n be - 6, we should multiply -6 per some numbers in order to obtain the factorial of -6. What guarantees me that the gamma function has the same interpretation for the negative numbers as it is for the positive ones? Also about the fractions applies the same rule. Just because the integral is evaluable and for the positive integers coincides with the factorial of the same i don't think you can affirm the same thing about the rest. Or maybe I'm missing something?

[@Risu0chan]

And now I introduce to you.... Purplepen.

[@yoavboaz1078]

but when you introduced the gamma function you used the odd number double factorial and 0 is even!

[@quenchize]

Your original definition using Γ assumes N is an integer. If N is a real your definition of !! does not apply. or all consistent axiomatic formulations of number theory include undecidable propositions

[@Risu0chan]

It makes sense that 0!! = 1, though. Making a product of no factor is, by default, the "neutral" factor, aka 1. And the n!! = (n/2)! 2^(n/2) formula is consistent with it (the number 0 being even). As Miloš Radovanović already pointed out in the comments, trying to use a formula for integers with half integers can only lead to inconsistent, albeit amusing, results.

[@TheLeviathan1293]

0!! is still an empty product, so it can be defined as 1 without any issue. The problem appears when we think that any analytic function agreeing with a combinatorially defined function in all but one of the points in its domain, must also agree in that point. At the end, the definition we use will be the most convenient one. If we do combinatorics or algebra, we probably want to define it as $1$. If we do analysis and the double factorial comes from an analytic expression, we probably need the new value.

[@teekak7949]

This is so cool man, I am honestly blown away

[@karinano1stan]

It is. For every positive integer it states: n!!=2^k*k! where 2k=n, and k is bigger or equal to zero.

[@comradepeter87]

I clicked thinking you were saying why ( 0 ! ) ! wasn't 1. I was confused as shit lmao XD

[@sivad1025]

F(x)=x*F(x-2) is recursive. Recursive equations can be scaled and shifted. Because this recursive equation has only one determining function, it can only be scaled to fit one solution. Therefore, you can create a line given that it passes through one point. The problem is that you defined double factorial as 1!!=1 and 2!!=2. You gave two independent solutions. Because of this, your odd function and even function aren't the same line. They are perfect scales of eachother, but not the same. So 0 will have a different solution depending on the graph you decide to use

[@lalitverma5818]

Wow amazing I can't imagine!

[@mmukulkhedekar4752]

omg you are really going crazy now!!!

[@donaldhayman8294]

At 9:15 you said that there 2^n n's in the denominator, shouldn't that be just 2n

[@obi6822]

0!! is an empty product (i.e. with no terms) and thus is equal to one, the multiplicative neutral element. Just as any empty sum is equal to zero, the additive neutral element.

[@YTBRSosyalEmre]

9:01 he is speaking the language of gods

[@veganwolf3268]

The second one is the most convincing because you proved it.

[@ethanpfeiffer7403]

What about negative double factorial?

[@Fematika]

Just use n!! = (n+2)!!/(n+2)

[@OonHan]

definitely sqrt(2/pi) because the method to it is more intutitive

[@Fematika]

Actually, doing n=1.5 to get 2!! yields 2*sqrt(2/pi), and no (2n)!! will actually yield the correct result, as the generalization only worked for odd numbers.

[@blackpenredpen]

Oon han, i messaged u on hangout

[@younesabid5481]

Please do (i)! And (i)!! BPRP ❤❤

[@jacksainthill8974]

So - just to be clear - you have not _fully_ extended the double factorial using the gamma function.

[@pj4717]

Do you mind explaining? Does this resolve the apparent paradox?

[@angelmendez-rivera351]

8 13 Yes, this resolves the paradox. The extension only extends the double factorial to odd integers of any sign, it does not extend them to even parity or any other real number. The formula does not apply to even numbers, so when you substitute z = 0, an even number, you obtain a contradiction which we avoid by using the formula correctly only for odd numbers. The resolution is to multiply the even double factorial by some patching function s(n) such that it will return 1 for even n and it will return the correct function such that when multiplied with the even double factorial, it will yield the odd double factorial for odd n in s(n). Only then can we appropriately extend to all complex numbers.

[@MrRishik123]

Taking the limit of your subscriber count. Getting dangerously close to 100k 😀

[@alessandroarmenti5562]

The first definition is on Natural numbers, the second is on Real numbers. If you took the floating of√2/√π you get 1 on Natural, so they should be the same answer just on different fields. I think the best way of actually thinking it is that 0.99999999=1 as mutch as 1.00000...1=1 and this is true in the Reals but not in the Naturals. A cool proof of this statement could be finding all the complex solutions and see the floating value on Naturals. (Sorry for the bad English)

[@jackmaibach8316]

pi = 2

[@maximilianarold]

0!! should be 1 as should all further iterations of factorial (tripple quadruple and so on). It's nice to have some constant values

[@mujilbtoq1827]

Thank so much keep it on you will come with your own theorem.

[@srpenguinbr]

10:00 nice sound effects

[@georget8008]

Is there a continuous function which equals the n!!. ? Something like the Γ(x) for the n! ?

[@HaxClassic35]

9:57 is the devil

[@pandabearguy1]

What is 0 infinity factorial?

[@jayjay4752]

Hi, could you do a video on index notation, like Einstein's sum. Maybe prove some curl properties?

[@Broody90]

9:58 that is robo blackpenredpen and robo blackpenredpen will give you a bad mark in your exam if you dont like him

[@erikolsen1333]

i love all of your videos keep up the good work, just graduated with a degree in pure math At UNCA (University of North Carolina at Asheville)

[@eddiecurrent7721]

Maybe the consistency of the definitions of ! & !! are to blame?

[@xXpr0d0fRSXx]

Would the same value be given if you had used the pi function instead?

[@alucard8955]

Regular factorial can be use it on, rational numbers, and if we follows the result equation 3!! =/= 3 but = 3*2², and that's a big problem.

[@GurkiratSingh-ds8dq]

Nice vid bprp! Though I must say, I wasn't too fascinated when I saw the ✓2/✓pi thing. Personally, I prefer the value 1 because the formula (gamma one) that you derived works for double factorials of odd only (2n-1), and not for any even (2n). It would be cooler if it gave the correct value for evens as well Btw are double factorials your own invention or something?

[@ffggddss]

I can testify that double factorials are older than Mr. pen. Fred

[@elliottmanley5182]

You could use the same faulty logic to show that 0!!=1/3 Start with Gamma(1-1/2) = rt(pi) Gamma(2-1/2) = (1/2)rt(pi) Gamma(3-1/2) = (3/2)(1/2)rt(pi) .... leading to Gamma(n-1/2)=(n+1)!!rt(pi)/2^(n-1) Setting n=-1 and rearranging 0!!=Gamma(-3/2)/4rt(pi) where Gamma(-3/2)=4rt(pi)/3 Hence 0!!=4rt(pi)/3x4rt(pi), which cancels to 0!!=1/3

[@aljuvialle]

it's not clear how you able to use n=1/2. Could you explain this? and how does this fit into (2)!! recursive formula you started with.

[@jamesknapp64]

The issue I have is the formula you derive with double factorial and gamma function is only valid for n being a natural number. The formula doesn't work at 3/2 for example. Thus you can't use it to derive a formula for 0!!

[@lucashoffses9019]

If you set n equal to 3/2, so that way you end up computing 2!!, you'll get 2sqrt(2/pi), and not 2, which is the actual value of 2!!. I think this function you created (which is very elegant and beautiful by the way) only works with odd numbers, and you need a different function for even numbers.

[@fakeaccount5827]

If we accept the method showing that 0!! = sqrt(2/pi), then we can plug in n=3/2 to get 2!! = sqrt(8/pi) So I prefer 0!!=1

[@aniketsaha7455]

can you bring videos on number theory and functional equations(problems with some theory) for olympiads...

[@Archik4]

even and odd double factorial is different function for this reason zero does not converge

[@angelmendez-rivera351]

Archik4 This is not a convergence issue. The function converges for 0. 0 is an even number.

[@bryangohmppac6417]

special integral please

[@Koisheep]

Maybe the mic glitches because you are summoning an ancient demon of maths Also, so tired in Chinese sounds like "joé" to me, which means fuck (as in "fuck, how tired I am!") In Spanish Lol

[@asserzayed1902]

@8:30 QUICK MAFS, sorry I had to

[@PrincessEev]

My personal take on the debate: since n!! for even n has the multiplication end at 2, it's not appropriate to directly use it to derive values for n!! if n < 2. The √(2/π) method makes more sense in that respect since it seems to follow from a (however incomplete) generalization of n!! (in the same respect as how the gamma function generalizes n!). So I'm more on the √(2/π) side of this. Though I'd have to look up / derive a full and proper generalization for n!! to be sure (rather than necessarily one that relies on a convenient pattern). Maybe you would know, BPRP, if one exists? At least on the face of it, it doesn't seem too difficult to derive. (Then again I thought Dr. Peyam's video on the integral of the fractional part of tan(x) wouldn't be all that difficult and look where that got us, so what do I know? xD)

[@angelmendez-rivera351]

Daniel Chaviers All you proved is that your argument is invalid by admitting that the odd double factorial is not a complete generalization of the double factorial, since it obviously only works for odd numbers. 0 is not an odd number.

[@goodplacetostart9099]

Bro can you say which degree have you graduated in ? Just curious to know And well what work do you do ? Remember just curious to know!!

[@indranathmukherjee6164]

Too good.

[@noway2831]

perhaps an exclamation mark with a 2 in subscript would be a better notation. n!! would imply n factorial factorial. To clarify, fact(fact(n)) or (n!)!

[@sugarfrosted2005]

Combinatoric interpretation gets precedent and we should realize our equation doesn't generalize to nonintegers.

[@mujilbtoq1827]

So by substitute n=1/2 is a mistake because is half is not an interger

[@CCequalPi]

How can you just use yhe gamma function for double factorials. I thought it was used for single factorial

[@angelmendez-rivera351]

Jake Montgomery Watch his other double factorial video.

[@younesabid5481]

Please do (i)! And (i)!! BPRP

[@owenmath1146]

Isn't the formula derived for (2n-1)!! just an extension formula which only works well when n is a positive integer? Since if you put n=k+0.5 into it, it ruins the original definition of double factorial.

[@davidspencer3726]

sqrt(2/pi)= about 0.79 so you could say they are both equal to 0.9 +- 13%, so you get the same answer both ways.

[@AlbertTheGamer-gk7sn]

The same scenario with 0^0, that's why 0! and 0^0 are the same.

[@kkanden]

so you were a robot this whole time

[@addityasinghal897]

That initial doraemon music

[@andrewpod5693]

Is the first definition correct? Cause you define n!! goes only to two before(for even numbers) . And then try to calculate n!! as if it goes to 0. But wolfram define 0!! as 0!

[@wkingston1248]

Im not exactly sure what double factorial is used for but i would assume its something to with permutations or a similar action 0!!=1 is likely defined as such to get an intuitive ansrer based on the application of the operation. Could also be like not calling 1 prime just because stuff breaks if you so excluding it from the definition is better.

[@alefdias4468]

WiSpKing hi, I have seen a place where doble factorial would show up, there is a method for solving ordinary differential equations I think is called Frobenius, it consists of supposing your answer is a infinite sum of the type Σan*x^n, the double factorial shows up a lot there with some types of differential equations.

[@balazsb2040]

Hi Sir! I just saw an interesting math problem on a competition, and I didn't manage to solve it. I think it's an awsome problem, so I would like to share it with you. Prove that the sum 1!+2!+3!+4!+5!+...+2016! is divisible by 9. This is by the way a problem from a high school, I believe 9th grade math competition. I'm sure there's a trick to solve it, but I didn't find that.

[@ThAlEdison]

1!+2!+3! = 9. 4!+5! = 144 which is divisible by 9. Every factorial from 6! up is divisible by 9. Therefore the entire sum has a common factor of 9.

[@balazsb2040]

Well, I didn't expect it to be that easy. Thanks! :)