A Factorial Sum Produces the Factorial Number System (visual proof)

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This is a short, animated visual proof demonstrating a finite sum involving products of factorials. The proof exploits the classic inductive proof of the formula in question. As a bonus, we discuss the factorial number system and show how the formula can be used to count up in this system.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoff.... Thanks!
This animation is based on a proofs by Tom Edgar from the December 2016 issue of Mathematics Magazine page 338 (doi.org/10.416....
If you want to know more about the factorial number system, check out en.wikipedia.o...
#mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #factorial #squares #factoradic #discretemath #inductionproof #induction #factorialnumbersystem #finitesums #sum #series
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Опубликовано:

14 окт 2024

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Комментарии : 68

@Dalroc 4 месяца назад

Since 5! = 120 we know that 77 will be a four digit number in base factorial. How many 4! can we fit in 77? 4! = 24, so we can fit three of those for a total of 24*3 = 72. We're still missing 5 to get 77. 3! = 6 and thus we have none of those. 2! = 2, so we can fit two of those. 2*2 = 4. We're still missing 1. 1! = 1 and that fits exactly once. 77 is thus written as 3021 in base factorial.

@MathVisualProofs 4 месяца назад

greedy strategy ftw :)

@theseusswore 3 месяца назад

my though t process was 60% of the way there and reading this comment made me feel like I got that last Tetris block to complete the board. nice!

@clemdelaclem 4 месяца назад

I think that's the very first time I've seen only an ! as a subscript and have it not be nonsense

@narfharder 4 месяца назад

Wait, did you see it somewhere that it _was_ nonsense?

@jakobthomsen1595 4 месяца назад

Great to see the factorial number system here! Combinatorial number system next?

@MathVisualProofs 4 месяца назад

Would be fun but I don’t know a good visual to go with it…. You know one?

@jakobthomsen1595 4 месяца назад

@@MathVisualProofs Hm, not yet. A really nice visualization of the binomial coefficient is as cut through a hypercube, as shown e.g. in the video PBS Infinite Series "in Dissecting Hypercubes with Pascal's Triangle". If I find more I'll let you know.

@RandyKing314 4 месяца назад

your geometrical animation is nice and shows a neat result, but you took it to “a whole ‘nother level” by introducing a related concept with the symbolic animation. excellent work as usual! perhaps future early childhood education will use animation to motivate and demonstrate concepts … this would no doubt change the way those children will perceive knowledge … not only as static pictures and numbers and words, but as elements of the fluid of reason. could be an interesting research project on learning…

@MathVisualProofs 4 месяца назад

👍

@leif1075 4 месяца назад

@@MathVisualProofsThanks for sharing but wouldnt yiu agree I don't see 2jy anyone would ever.tgink of writing the minus 1 likenthat..why wojld they? Thanks

@sdf420 4 месяца назад

@hex.enigma 3 месяца назад

"We can keep counting up by adding one each time" was the only part I understood.

@Proud_Kuffar 4 месяца назад

You are doing a great job. I don't know why this types of videos are not famous.

@MathVisualProofs 4 месяца назад

Thanks!

@narfharder 4 месяца назад

I'm not sure how you could consistently extend this to fractions, since the n!-th place "digit" d represents the product of (d >= 0, d < n) and (n!), where n is negative. Does the undefined nature of these two factors somehow "cancel out"? Also note: even in that case, the 0!-th place must be undefined, since only one of the above factors is undefined.

@wendolinmendoza517 3 месяца назад

Very interesting!! I did not expect the sum to lead to a number system.

@fdileo 4 месяца назад

I didn't know this theorem. It's amazing

@MathVisualProofs 4 месяца назад

👍😀

@oida10000 4 месяца назад

The first 5 digit factorial number is 120 as 5!=120 and the digit representation of 77 in base factorial is 3021 (3*4!+0*3!+2*2!+1*1!) as 2*4!=48

@alucs6362 4 месяца назад

Since the nth decimal position corresponds to a value of 1/(n-1)! (such that, for example, 0.1*10=1) then e = 1x2!+0x1!+1/2!+1/3!+...=10.11111...

@MathVisualProofs 4 месяца назад

What is 0.1 here though. Is it 1/1! Or 1/2’? It’s a weird system but I think you want 10.011111….

@alucs6362 4 месяца назад

@@MathVisualProofs 0.1 is generically defined as 1/10 (e.g., in base ten, 0.1=1/(10^1), instead 1/(10^0)). Even if you don't define it like that, base changes shouldn't affect the truth of algebraic relationships; since 0.1*10=1 independently of basis, that means 0.1 would be 1/2!. This also preserves the idea that a purely decimal number is smaller than or equal to 1 and maintains unique finite numerical representations!

@MathVisualProofs 4 месяца назад

agree. I’ve just seen it a number of places as I mentioned. :)

@Dalroc 4 месяца назад

@@MathVisualProofs It all comes down to whether we care about redundancy or not. If we used your suggested definition in this comment e could also be written as: 1.11111....., since 1! + 1/1! = 2!. Positional values in regular bases are b^n, where n=0 is the last position before the decimal point and n increases to the left. In base factorial the positional values are (n+1)!. We need to add 1 in the factorial to avoid redundancy since 0! = 1!. If we did not add 1 we'd get 10 = 1. If we were to simply extend this system to the right of the decimal point we'd get that the first decimal would be 0!, again resulting in redundancy. All other decimal values would be ±∞ and if the system wasn't broken before it surely is now. Thus there has to be an alternative definition for decimals in this system. Defining them as (abs(n)+1)!^-1 seems most logical and gives e = 10.1111... as the one and only representation. If we do not care about redundacy we'd have to ask ourselves: why skip 0! and 1/0! ? Depending on how you resolve this issue you can have multiple answers, but only one resolves the redundancy. If we include 1/0! we could also write e as 0.111111...

@MathVisualProofs 4 месяца назад

@@Dalroc for sure. I was thinking about the 1.11111... version but hadn't thought about 0.1111... :)

@nickm3694 3 месяца назад

Fun fact: this numbering system only becomes more efficient than the decimal system after 25 digits (assuming you don't factor in that you need as many symbols for the value of a digit as the number of digits in the largest number you'll use) Also xkcd 2835

@LeftGuard 4 месяца назад

Great video and two nice bits of homework. 👍🏻

@MathVisualProofs 4 месяца назад

👍😀

@alvargd6771 4 месяца назад

fun algebraic proof: n(n!)=(n+1-1)n!=(n+1)n!-n!=(n+1)!-n!, and hence you get a telescoping sum for 77 just count as always, 24 is in there thrice so u have 4, then u have a 3 left u can do with 2! and 1! so u get 3021 and e is just 10+1+1/10+1/100+1/1000+1/10000... (which in number systems with a constant base is just 11.111111111...)

@alanthayer8797 4 месяца назад

Thanks for da VISUALS Visuals visuals! Life w/o them Ain’t da Same!

@MathVisualProofs 4 месяца назад

😀

@ajayvarshney2117 Месяц назад

my good guess for writing the e = (1.111111….){in factorial base) Why? Here it is, i recall the base-2 logic for writing 0.5 (= 0.1{in base-2}) which is also 1/2, so, 1 {in factorial base) = 1*1! 0.1 {in factorial base) = 1*(1/1!) 0.01 {in factorial base) = 1*(1/2!) and so on so e = 1 + 1/1! + 1/2! + … = 1*1 + 1*1/1! + 1*1/2! + … = (1.11…) {in factorial base}

@3_14pie 4 месяца назад

I will have to worldbuild a whole culture to use this number system

@Simpson17866 4 месяца назад

I'm not sure how practical it would be ;) but there have been plenty of cultures that used straightforward systems alongside idiosyncratic systems :D Like how Romans were expected to be experts in political history in order to know what year something happened in. (WALL OF TEXT INCOMING) If the first digit of is either 0 or 1 And the 2nd digit is either 0, 1, or 2 And the 3rd digit is either 0, 1, 2, or 3 ... and the 9th digit is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 and the 10th digit is either 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or A etc... then you run into the worst of both problems with a small base system (needing many digits per number) AND a large base system (needing many possible symbols for certain digits), and you also can't learn arithmetic intuitively because different place values play by different rules 1 + 1 = 10, but 10 + 10 = 20 11 + 11 = 100, but 110 + 110 = 220 21+ 21 = 120, but 210 + 210 = 1020

@3_14pie 4 месяца назад

@@Simpson17866 @Simpson17866 nah, not everything needs to be practical, it's nice to do things just for fun sometimes, but thanks for the advice

@vanshsingh7270 4 месяца назад

Damn! Math is beautiful

@MathVisualProofs 4 месяца назад

No disagreement here :)

@Smartas599 3 месяца назад

Amazing.Thanks

@MathVisualProofs 3 месяца назад

Thanks for checking it out!

@yusufdenli9363 4 месяца назад

Factorial number system is very cool 💥

@MathVisualProofs 4 месяца назад

for sure!

@konstantindrumev8036 3 месяца назад

Hi, can you please recommend any books you found useful in your journey of math? Ty❤

@MathVisualProofs 3 месяца назад

I will see about making a video describing some of the books I have found helpful, especially in this particular endeavor. The short answer, though, is to check out books by Roger B. Nelsen :)

@David-bh7hs 4 месяца назад

3:20 - does this mean each n must be equal to or less than its index i?

@MathVisualProofs 4 месяца назад

Yes. Each digit is less than or equal to its position.

@KrasBadan 4 месяца назад

5:19 I think that the entended answer was either 10.011111... or 1.11111... This is because e is both 1•2!+0•1!+0•0!+1/1!+1/2!+... and 0•2!+1•1!+1•0!+1/1!+1/2!+... But this answer is wrong in my opinion. You see, in our number system the base function is 10ⁿ, and it makes sense that for fractions we just put negative n. But for factorial number system the base function is n!, and there is no reason to assign (-n)! to 1/(n!). In fact, the factorial function is undefined for negatives (and even gamma function is undefined for negative integers). So the actual answer is there does not exist a representation of number e in this system. Also, there is actually no ambiguity with the fact that 0!=1!. One may think that in this system 1=0.1 or 10=1.1, but actually we can't put any digit in place with the position less than this digit, so just like we can't have a number 2.0 because 2 is greater than 1, we also can't have a number 0.1 because 1 is greater than 0.

@SumanYadav-wr3cn 3 месяца назад

Please make videos on sieve theory

@Izzythemaker127 3 месяца назад

Oh I see, so instead of digit x being kˣ with k being the place value constant, its x! instead. Does a system like that use an infinite amount of digits? If not what is the minimum?

@MathVisualProofs 3 месяца назад

yes. this is a problem with this system. you need more and more digits for each position.... so the number of digits you need is unbounded.

@stevehines7520 4 месяца назад

"by two" limitless from be-ginning!

@mr.bastolas.7478 4 месяца назад

Can you do a visual representation of sir ramanujan's sum of integers from 1 to infinity. ?

@eonasjohn 4 месяца назад

Goes beyond my comprehension.

@YT-AleX-1337 3 месяца назад

77 = (3, 1, 0, 1)

@MathVisualProofs 3 месяца назад

Might be just off. This looks like 79…

@RenanMoreira1728 3 месяца назад

(m+1)!-1=n.(n-1)!

@SeanSkyhawk 4 месяца назад

wait.... you can do that?

@MathVisualProofs 4 месяца назад

I hope so, 'cause I just did :) What part are you referring to?

@SeanSkyhawk 4 месяца назад

@@MathVisualProofs the part where one basically represents all natural numbers solely on the basis of linear combinations of factorials

@MathVisualProofs 4 месяца назад

@@SeanSkyhawkyep. That’s one possible positional system. There’s a link in description to the wikipedia entry on it. Nice connections to enumerating permutations.