double factorial vs. regular factorial
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- Опубліковано 14 бер 2018
- We will find the connection between the double factorial and the regular factorial. We will see how to simplify double factorial n!! when n is even and when n is odd.
Here are 7 factorials that you probably did not know: 👉 • 7 factorials you proba...
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Logic: A double factorial gives you a smaller number than a single factorial.
Ethanol 314 maybe it should be called a fractional factorial, notated with n factorial to the power of one half
n! = n!! × (n-1)!!
Logic: (1/2)^2 < (1/2)^1
yeah this should be n(!^(1/2)) imo.
this would be consistent with defining n(!^0) to just equal n
and higher powers should get bigger faster while fractional powers should approach n as the fraction approaches 0.
How about calling it a fractorial?
The given problem transforms to a limit of (2k)!/(2^k*k!)^2 where k approaches infinity. Using Ramanujan's or maybe Stirling's approximation on factorials, the limit reduces to 1/(pi*k)^0.5 where k approaches infinity so we get the answer as zero.
Thank You blackpenredpen
bprp Please pin this comment.
Isn't it just (2k+1)!!/(2k)!! ?
@@clockfixer5049 Almost... actually it's the limit of (2k-1)!!/(2k)!! but how do you calculate that ? You have to develop it as Abhimanyu said.
Great job! I wonder if there is a way to prove it using logic or induction.
Double factorial looks really excited, like you should shout it. SIX!!
Lucroq lol yea. There also the triple one haha
I love to see that my "FactOREO" is still being used :D
Yay!!! That was a great one! I loved it!
Hey. I really love your videos. I’ve always loved experimenting with maths and numbers, not just learning simple things in school. Your content is really good so keep up the good work!
Here's something interesting. Substitute k=0 into the odd case to find that (-1)!!=1.
Ben G lol nice!
Jinger McBlabbersnitch you're right. In that case, you could define n!! for any negative integer and 0
@@blackpenredpen But, as n has to be positive, 2k and 2k-1 have to be positive too
@@beng2995 nice one
It has been said that k should be positive integer (natural number)
Never thought about this. Fascinating!
robin tremblay thank You!
See what you did in the title there.
Duncan Schaafsma 2!!!
That's a triple factorial.
I was wrong the title is actually”(!!)!!”
6!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Spotted this straight away when you wrote 6!! = 6×4×2 and 5!! = 5×3×1
Recursive/inductive definition: n!! = n! / (n-1)!!
Luca yup!!
How I think about whether you should use 2k-1 or 2k+1 is whether you want to start counting at 1 or 0, it's how you pair up the odds and evens with their "internal" integer.
never heard of such a function!! thanks for the knowledge.
Another observation that can be made: n! = n!!*(n-1)!! for all n in N+
Also, if n is even, you can also say n!! = n!/(n-1)!! for all n in N+, but that probably needlessly complicates matters a bit. It's still an observation of an equivalent expression at the least.
really elegant notation. Thank you very much.
the second person to say that with the first one being whoever decided to notate as that
First, we can express the denominator as 2^n *(n)! where n -> oo. Then, multiply both numerator and denominator by 2*4*6*.... Thus, we have: n! / [(2^n * n!) ^ 2]. After further simplication, we obtain: 1/ [2^(2n) * n!], which approaches 0 as n -> oo. Therefore, the answer is 0.
I’d be interested in seeing an analytic expansion of the double factorial using something similar to the pi or gamma function definition of factorial which maintains the same recursive relationships on the even and odd integers for the double factorial definition in the video. In other words give a single smooth continuously differentiable function that maintains the same recursive relationships that the positive integers hold.
wolfram alpha gives one
now n!!!
AndDiracisHisProphet N!!!!!!!!
:)
That would have been my nect suggestion
n!^m where !^m is m factorials
It's called the "mth" step falling factorial and it has a general expression in gamma even for real-step :)
cool
Shadows of fractional operators á la Dr. Peyam!
Expression so good 🙏🙏🙏
Encore! Encore :-D
Your videos made me fall in love again with maths
thank you . this video is very helpful
The given product can be written as (using a bit of algebra)
[(n!)/(((2^(n/2))((n/2)!))^2)]
Since this was inf/inf, I used L'H with the gamma expansion of the factorial and after applying limits, you'll get zero. So yea the product is 0
Edit: I'm not sure if bprp intended on using the Ramanujan or Stirling expansion because honestly, I have no idea what it is(or maybe I'm just missing out on something here). I saw the pinned comment and well I was a bit late but at least my solution isn't as complicated xD
The limit is equal to 0 indeed, but for simpler reason: in the fraction the numerator, n!/((n/2)!(n/2)!) is in fact the biggest binomial coefficient C(n;n/2) and the divisor is 2^n. But we know that the sum of all binomial coefficients at the n-th line of the Pascal's triangle is exactly 2^n. Thus, our fraction is less than 1 (obviously), but why the limit=0? There is connection with harmonic series 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/n + ... which diverge with n -> inf.
VibratorDefibrilator oh yes I get your point. I couldn't really think about that.
++Love your username, man
Thank you for the swift reply, Akshat. Nevertheless, I think that this problem needs a bit more detailed explanation. Once we've spotted that binomial coefficient C(n;n/2), we might conjecture that its part of the sum 2^n will decrease constantly until it reach 0 with n->inf. If we compare two consecutive values: C(n;n/2) and C(n+2;(n+2)/2) indeed we'll have C(n+2;(n+2)/2) / C(n;n/2) = (2n+2)*(2n+1) / (n+1)^2 = 4*(1 - 1/(n+1)) when at the same time the corresponding sum will be quadrupled - 2^(n+2) / 2^n = 2^2 = 4. So, the temp of decreasing of that binomial coefficient will be (1 - 1/(n+1)). But if we multiply all these, we'll have our infinite product 1/2*3/4*5/6*7/8*... It turns out that our argument will be circular and finally we have to try the approach with harmonic series. At least now we know how to do it:
We are taking reciprocal fraction of every term of the infinite product: 1/2 -> 2/1; 3/4 ->4/3; 5/6 ->6/5;...
1/2*3/4*5/6*7/8*... = ((2/1)*(4/3)*(6/5)*(8/7)*...)^(-1) = ((1+1/1)*(1+1/3)*(1+1/5)*(1+1/7)*...)^(-1)
Now, this infinite product contains (after uncovering the brackets) a sum 1/1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ... and we can easily prove that it diverges. It is part of the harmonic series 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...
If we pair every two consecutive terms of that sum we'll get:
1/3 + 1/5 = 8/15 > 1/2
1/7 + 1/9 = 16/63 > 1/4
1/11 + 1/13 = 24/143 > 1/6
1/(2*k-1) + 1/(2*k+1) = 4*k/(4*k*k - 1) > 4*k/(4*k*k) =1/k, k is even
1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ... > 1/2 + 1/4 + 1/6 + 1/8 + ... = (1/2)*(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...)
Thus, we have P = 1/2*3/4*5/6*7/8*... = 1/(1+(1/2)*(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...))=0
Frankly, I don't see what the double factorial has to do with all this proof, but, on the other hand, it is good exercise.
P.S.: My username came up after a hilarious conversation with a fellow female doctor and somehow sticked with me.
VibratorDefibrilator tell me about that conversation sometime haha
Oh, the exact situation is already in the mist of the past, but one have to know and remember that the doctors are one of the most cynical and dirty-minded people in the world, no exceptions, I think. Slowly, during the years, I came to the conclusion that they have to be... fitted... that way in order to preserve their sanity and to do their noble job. After all, they are close to all human beings from their birth to their deathbed, in illness and health, so nothing sacred and unknown remains there... It is enormous benefit for anybody who is fortuned enough to befriend even one of those noblest of cynics out there.
As you may already noticed, I'm inclined to be painfully meticulous when I have to explain something. Don't bother even to answer. Wish you good luck.
The infinite product can be simplified to lim n->inf of (2n)!/(2^2n * (n!)^2) which is simply the probability of X=n for X~Bin(2n,1/2) which approaches X~Norm(n,n/2) by the Central Limit Theorem implying that P(X=n) approaches 0.
Thanks sir world's best teacher
I think that the answer is 1. The reason is that as n approaches infinity, n/(n+1) approaches 1 and thus by Cauchy's Theorem of the limit of product of terms in a series the answer is 1. I have however worked out the solution by expressing the product as
(2k)!/(((2k)!!)squared) which can be simplified to (2k)!/(((2 power k).k!)squared) and then further expanding and simplifying as k approaches infinity.
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why the LoL comment on math video?
We don't have to calculate all these or convert it to a limit, just by intution ( numerator )< (denominator) => 0< ratio < 1 but ratio -> 0 as elements goes to oo in both num. and den..
I could be very wrong here, since I haven't been in math classes in ages, and follow this channel due to my interest in math, but I did it as such. Rewrote it as a limit as n approaches infinity of ((2n-1)!/2n!), which then reduces to limit as n approaches infinity of 1/2n, which seems to approach zero.
problem at the end can be rewritten as prod(n=1 to k of (2n-1)/2n), which, as n tends to infinity, the product tends towards 0.
Very good job....
Can you explain me why you write the def. of an odd number as: (2k-1) instead of (2k+1)? I mean, what if (k=0)? then it is going to be -1 and you cannot make negative number factorials right? Sorry if I am asking something stupid, but I need to know it.
For the final question, an infinite product of n!!/((n+1)!!) approaches zero because the denominator is generally larger, and multiplication makes the resulting fraction so much smaller.
14:03
Well, I’m not gonna use factorials, but I am going to use the capital Pi notation, like Signa, but for multiplication, I’m gonna name it CapPi, and so your expression is: lim{x-> +Inf} CapPi{n=1; n=x}((2n+1)/(2n+2)).
This can be simplified into; lim{x-> +Inf} CapPi{n=1; n=x}(-1/(2n+2)), which means that you multiply more and more with smaller and smaller fractions, which means it will reach zero.
Thank you Sir. It was good.
Now if you can please explain !n and n !!!
solution: (1 * 3 * 5 ...) / (2 * 4 * 6 ...) can be created with big PI notation: \prod_{n=1}^{x}\frac{2n-1}{2n}; the solution goes to 0. (1 * 3 * 5 ...) / (2 * 4 * 6 ...) = 0.
Importantly? n!!! = 3^k * k! also given that n is divisible by 3. It works for any power? of factorial actually: n !^p = p^k * k! given n=pk where k is a whole number. It even works for 1!
can you do a video on (1/2)!
You can put it in a calculator and get an answer, but how?
KeroseneAndFire The Gamma Function.
He's already done it
Oh, ha, never saw that video. Thanks!
Wait, I thought I pasted the link for you for it after I hearted the comment. But the link never showed up...
Did u find it? If not, I will go find it again.
blackpenredpen Haha, yeah I found it. I didn’t see the video when it came out I guess. Thanks!
Sir, I have a doubt about definite integrals. My question is if I consider a piece wise defined function f(x) which give 0 if x is Rational, f(x) = x if x is irrational. So for some boundaries, is the definite integral of f(x) dx = definite integral of x ? Or not?
I don't know - is it always possible to find a rational number between any pair of irrational numbers? I suspect that it is. Even if your range is infinitessimally small, you can stack your two bounds on top of each other and compare them a decimal at a time. Because an irrational number's decimal expansion is infinite, close numbers must start with digits that match exactly, until a point of first difference. That first difference gives us an opportunity to construct a rational number - use the same string of starting digits that matches your bounds, then at the difference, take the digit from the upper end of the bound, then terminate your constructed number. You've constructed a rational number which must be larger than the lower bound, but must also be smaller than the lower bound. In fact, you can now create an infinite number of rational numbers between these two irrationals by using your rational number and adding digits. As long as the digits added remain less than or equal to the corresponding digits in the irrational upper bound. Without continuity - especially the discontinuity that results from an infinite count of missing places - the proof of equivalence of those two integrals would probably prove that the indefinite integrals are equal as well. Which... I'm just not sure how to even think about.
Francis Sirizzotti Even then. Countability maybe an important factor here. Since it is an integral, we can actually separate the rational points in a particular order, which when multiplied with dx gives 0. The same cannot be said about an uncountable number of irrationals, because they cannot be separated per se.
Yeah, I definitely agree this is a possibility. And when you couch it in terms of "does there exist some boundary where integral of f(x) = integral of x", it definitely seems plausible. But my reasoning was more about proving that there are a countably infinite number of rational numbers between _any_ pair of irrational numbers, and since there are also uncountably many irrational numbers in that same interval, and reasoning from differences in infinity would apply to the the space of all real numbers - so your question might be the same as "is Integral f(x) = 1 + C".
I think Riemann would say the integral is undefined, and Lebesgue would say that removing the rationals makes no difference to the integral, since they have measure zero. See Dr. Peyam's videos on Lebesgue integration.
Greg Ewing is right. From a "Riemann" point of view, you can't integrate the function since it isn't continuous at ANY point x. However, since the number of discontinuities is equal to the cardinality of the set of rationals, it is countable and therefore has measure zero; so, from a "Lebesgue" point of view you could technically remove all of the discontinuities and end up with a function identical to f(x) = x, which you could then integrate over a definite interval.
@blackpenredpen What about double factorials for something that is a non-integer, 0, and negative numbers? Also, why can’t double factorials be inputted on a calculator?
I have another question for you
I tried what ratio the 2 formulas for even and odd double factorials have compared to each other, so I put the the even function over the odd function and I somehow got seemingly a straight line parallel to the x axis at around y=1.253. Yet I dont get where this number suddenly comes from, do you know the answer? Is the equation somehow simplifiable?
Here the equation in written form if my descriotion wasnt understandable
(2^(x/2)*(x/2)!)/(((x+1)!)/(2^((x+1)/2)*((x+1)/2)!))
Well, this is a brilliant move!
Eres el mejor
The given problem can be very coolly written as the limit (as k tends to infinity) of 2^(-2k) * (2k choose k). It's very clear this is 0, as the LHS term dominates.
Every factor in the top can be canceled with its double in the bottom leaving you with 1/(2*4*2*8*2*12*2*16...) which you can multiply pairwise to rewrite as 1/(8*16*24*32...). Using the same trick in the video, this becomes lim x->inf of 1/(8^x * x!), which is pretty trivially, zero.
... I think? The pairwise multiplication/canceling on an infinite series seems pretty fishy to me, but I'm not really sure why I should disallow it.
Hi i am from Cambodia 👋👋
I like u blackpenredpen😍😚
i’m from the country next to you, thailand
At 9:31 my mind was like😍
Nice!!
We have: n!!=n!/(n-1)!!. setting a_n=n!! we get the recursive relation: a_n=n!/a_(n-1) with a_1=1, putting this into walframAlpha we get:
a_n = exp((-1)^(n + 1) ( sum_(k=-1)^(n - 1) (-1)^(-k) log((k + 1)!)))=n!!
I would like to see indefinite integration, integration carried out an indefinite number of times, just to see the symbolism not with ellipses (the language kind not the conic section) therefore, or indefinite composition, again indefinite number of layers of composition, for the compact symbolism. I know the derivative of the indefinite composition would use the TT, product, operator, product of the derivative of the outermost layer, successive factors being derivatives of one more layer in until the innermost layer.
Is it....zero? I rewrote it as e^ln(1*3* .... / 2*4* ....) = e^( ln1 - ln2 + ln3 - ln4 + ...). That series diverges and approaches negative infinity, so the whole thing goes to zero
It tends to 0 [1/(2^n)].
Thanks for made a great channel like that, I really like it
BTD thanks!!
Wow, this is great! Thanks to you :^)
Not necessarily. If you ignore the one and compare term by term:
3>2
5>4
7>6
And so on. So, seeing it this way, the numerator would be larger and it would not make sense that it should tend to zero.
Chris Sekely I Will try to explain it...
You can write the denominator like
2*(2*2)*(2*3)*(2*4)*(2*5)....
Eventaully all the impar (2K-1) numbers will be simplificated with all the numbers in the numerator.
Then the denominator will be conformed just for par number (2K) which can be written like 2*2*2*2*2*2.... And that is why the denominator will be (2^n).
[1/(2^n)].
I hope that I can explain it well...
Tell me if you have any question
You can't rearrange terms in an infinite product arbitrarily. The product does go to 0 but it actually goes as 1/sqrt(k).
Using the relations presented in the video
(2k-1)!!/(2k)!! = (2k)!/(2^(2k)*(k!)^2)
Then using asymptotic Stirling's approximation on this expression
sqrt(4*pi*k)*(2k/e)^(2k)
-----------------------------------------------------------
2^(2k) * ( sqrt(2*pi*k) * (k/e)^k )^2
Which simplifies to
1/sqrt(pi*k)
so it goes to 0 as 1/sqrt(k)
Using the bounded form of Stirling's formula (en.wikipedia.org/wiki/Stirling%27s_approximation) this holds
2*sqrt(pi)/e^2 * 1/sqrt(k)
what is the relation recurrence for 0!! and (-1)!! teacher?
The best description of multiple factorials that I've seen is "the louder you yell at the number the more scared it gets and the smaller it shrinks in on itself"
Everyone loves the good ol' definition of factorial
We write the product as limit as k approaches infinity of (2k-1)!!/(2k)!!, and plugging in the formulas, we get (((2k)!)/2^k(k)!)/2^k(k)!=((2k)!/(2^{2k}(k!)^2), and applying Stirling's Approximation yields 0. :)
it arises from integral of x^(2n) exp(-x^2/2) over the real line
I went down for the odd version, so a little bit uglier but same result. Great explanation.
I tried doing the problem at the end but I could only go so far as to say that it was less than or equal to ((k+sum(1->k,1/n))/k)^k (using the fact that the geometric mean is always less than or equal to the arithmetic mean) (then I would need to take the limit as k goes to infinity which I am just not able to do)
However since the terms approach eachother (2n-1 approaches 2n as n goes to infinity) I do know that that 'less than or equals to' sign would eventually become an equals to, so that would be a solution if knew how to do it.
Some basic observations I had
The product always reduces until infinity since (2n-1)/2n = 1-1/n = 1 for n->inf
The product will always be positive (possibly including zero) and less than 1/2
For odd you can just cancel the 6s to get 5!/2^2*2!. In simple you can say n=2k+1 and then n!!=n!/(2^k*k!) or n!!=n!/k!! to be even simpler.
I saw this solution instantly! Your videos must be rubbing off.
n!! = n! / (n-1)! * (n-2)! / (n-3)! * (n-4)! ... 1!
Dividing or multiplying by 1! gives the same answer so it doesn't matter if n was odd or even.
Not the shortest but I like how it looks. :P
Noticing each pair of factors is less than 1, the sequence is strictly decreasing, and since it is bounded by 0, we know there is convergence. Now, we can write both the numerator and the denominator as double factorials, (2k-1)!! / (2k)!!, applying the formulas in this video we get a formula with usual factorials, and applying the Stirling equivalence for large numbers, it naturally yields to 1/sqrt(pi*n). The sequence converges to zero.
Hey, check this out
After calculating the even case (2k!!), you could just have noticed that
(2k-1)!!*(2k)!!=(2k)!
because if you alternated odd numbers with 2k-1and then even numbers with 2k you covered all of them, and then you can divide by 2k!!
(2k-1)!!=(2k!)/(2k!!) and everything on the right hand side is known.
Miguel Alcubierre that’s literally what he did
n!!=n(n-y)(n-2y)(n-3y)...
where y = 2.
6!!=6(6-2)(6-4)
xy=z'(the one that ends the series)
it ends at n-z'>0
if odd, then n-z'=1
if even, then n-z'=2
two examples for reference.
6 and 5
6!!=6(6-2)(6-4)=6x4x2
6-4>0
since 6 is even then the series ends with a two
5!!=5(5-2)(5-4)
odd series so it ends with 1.
well done sir from Pakistan
Idk man but I think the answer for the question at the end is 4/2^n . n!
Where n is equal to the denominator
gReat work
I believe the problem would equal 0. I did some algebra and got the fraction (2k)!/2^(2k)(k!)^2. I wrote out (2k)! and factored out the 2k. It turned out to be 2k(1*(1-1)*(1-2)...). 1-1 = 0, so the whole expression must equal 0, Since the numerator is 0, the whole number is 0. (Hope my logic was right here)
I know this doesn't have anything to do with double factorials, but I have a question about that limit at the end. If we were to write the fraction as the product of corresponding terms (i.e. 1/2 * 3/4 * 5/6 * 7/8...), then we can conclude that the limit must be less than 1/2. But if we dismiss the 1 in the numerator because it doesn't change the overall product, and we now chunk the fractions as 3/2 * 5/4 * 7/6..., then we can conclude that the limit must be greater than 3/2. But this is a contradiction. What's the mistake in my reasoning?
Sponge . That's what I concluded
You can't reorder infinite series. The correct way to remove the 1 would be to extract the 1/2 otherwise you are effectively multiplying the entire series by 'n' (which is being pushed to infinity)
Ah that makes sense. Thanks for the explanation!
Magic Gonads yes
Yeah magic gondas is correct, u just can't reorder any terms in multiplication in case of infinite series.Ex:- (1*2*3*.......)/10² = oo but if we chunk into fraction like (1/10²) *(2*3*5*...) < 1 and another form if we take 10^10000 instead of 10² we have (1/10^10000) * (2*3*5....)= 0*oo indeterminate form, hope u get it.. 😊
We have(2n-1)!!/(2n)!! = (2n)!/(4^n * n!n!) = (2n choose n)/4^n. Now, taking the limit, we have 1/sqrt(πn) -> 0.
Question: When you worked out the double factorial for odd numbers, why even bother multiplying by 6 on top and bottom? Leaving it out would make the generalized form in terms of n somewhat nicer, as the (n+1)! on the top would simplify to just n!, and the two (n+1)/2's on the bottom would flip to (n-1)/2's.
how about !!! for . 6!!!= . 6*3=18
Carlton Johnson yup
no dont just scrap this notation and use !! for actual iterated factorial
For dealing with n!! when n is odd, I prefer to just observe that n! = n!! * (n-1)!! and substitute the formula of n!! with n even.
same I did !!!
Arthur Schuster also called this an "alternate factorial" which I think is a much better name, because "double factorial" implies you are doing something twice. If I had to come up with a name, I'd call it a "semi-factorial". But hey, that's just me.
Plz
How to do this 2^(n+1) factorial ?
We know the limit of the product, (1/2)(3/2)(3/4)(5/4)... approaches 2/π (Wallis Product)
The question can be rephrased as the following:
What is the limit of the product (2/π)/{(3/2)(5/4)(7/6)...}. Since, each term of the denominator is clearly greater than 1, the denominator approaches infinity and hence the product asked in the question approaches 0.
But I am getting another answer too!
The question asks to find the limit of (1/2)(3/4)(5/6)... ,which on rearranging can be written as (1/1)(3/2)(5/4)(7/6)...
And if you look at the rephrased version, the denominator is exactly this product. This unsure rearrangement gives the limit to be √(2/π).
So, is the answer √(2/π) or 0????
1.) Lim [100(tan^-1(x)/x)]
(x-->0)
where [.] represents the greatest integer function
2.) lim [100(tan(x)/x)]
(x-->0)
where [.] represents the greatest integer function
Its intresting
Yay!
This will be limit ,when x go to inf, of ((x+1)!/((4^x)*(x!)^2)) and this is the same as ((x+1)/((x)!*4^(x))) that must go to zero.
My answer - 0
Thanks for your channel!
Wrong
Now I wonder if there is an analytic continuation for this one😅
Don’t know why this clip popped up among your great videos, but I know the music played at the end ;-)
Thank's very much
From Palestine 🌍♥️🇵🇸
If double factorial is factorial every other number, then what is half a factorial? Going down by 1/2 each time? Then, what is a factorial'ed factorial? For example: when n is an integer, what about n(!-n times) more specific example.. so for n = 4, what is n!!!!, is it just n? So, then 4? So then what about n = 0? Just 0?
(n+1)!/{2^[(n+1)/2].[(n+1)/2]!}/{2^n.[n!]}
i start from here and do the simplifications and call n+1=2k then go on simplificate more after all is done i turn to then world n=2k-1 and it call n infinity cuz it is ''isnt it'' i take the limit of it and its 1/inf so its 0
Just for fun, exist something like gamma function for double factOreo?
Oscar Aguilar
You will see in my coming video! :)
is the aswer to the question at the end "(2k)!"?
Sir, Please answer my doubt
ABC = A!! + B!! + C!! A≠B≠C≠0
Harith
So 6!!! Would equal 6x3x1=18? Also how would a half factorial work?
Isnt that reciprocal squareroot of euler identity (1st order)? Then its sqrt of 2/phi ~~
Just a guest
I think lim n tends to infinity (2n-1)/2n=1 may be right answer....
So I write it down as lim(k->inf)((2k)!/(2^k * k!)^2) then i factor out the (2k)!/(k!)^2 as the binomial coefficient of 2k choose k. So the limit looks like this 1/2^2k * 2k choose k. We know that 2^2k is a sum of all values in 2kth row in Pascal's triangle and 2k choose k is a middle value in that row. One term of a sum divided by the sum in a limit to infinity is approaching to zero.
Sorry for any linguistical mistakes. ( I hope theese are the only ones)
the question at the end... for every term in the top, 2x that is in the denominator, along with a lot of other stuff, so it would go to zero? hmm... let me try to math-ize this...
numerator is: (2k-1)!! as k approaches infinity
denominator: 2*(2k-1)!! * 4*(2k-1)!! * 8*(2k-1)!! * ... = 2^(k(k+1)/2) * ((2k-1)!!)^k as k approaches infinity
seems to me like the bottom is exponentially infinitely bigger than the top, so it approaches 0.
On the other hand, I'm not certain I can ignore the fact that if you ignore the 1 at the front of the numerator, then every term in the numerator is bigger than the denominator... but i dont think that kind of logic works with infinite things...
I am just a middleschooler, please show mercy
Please I need to the subject (Gaussian jorden and elimination)
if the total number of terms = n then the solution is n!/[2^(n/2)*(n/2)!]^2 if n is even and if n is odd then the solution is [n!*2^((n+1)/2)*((n+1)/2)^2]/(n+1)!
I have made an observation.
Can anyone figure out the differences between 12:40 and 12:48?
Papai Pal lol
sly dog
No. That edited part is shown in the beginning of the video.
Is there something similar to the factorial function except it’s something like 6+5+4+3+2+1
It’s_Robbie Time well sort of. the series of adding consecutive natural numbers happens to be the triangular numbers. the formula for the nth triangular number is n(n+1)/2
@@adeelsumar3721 ahh okay
n!! is not (n!)!, if x! = f(x), then x!! must be equal to f^2(x) (square root of function(x)), while x!! =/= g(f(x)), right?
The product is divergent. You could write it as an infinite product of fractions that are all bigger or equal to one (1 * 3/2 * 5/4 * ...). At the same time you could write it as an infinite product of fractions, that are all smaller than one (1/2 * 3/4 * 5/6 * ...).
That's why you're not allowed to rearrange the terms.
Also, for odd n, n!! = n!/([n-1]!!)
¿What is the answer? I'm very interested please:)
The more factorials, the louder you scream the number
oh so just like 1! = 1 and n!=n*(n-1)! we can define !! in a similar sense. So 1!! = 1 and n!!=n!/(n-1)!!. Then I did !!! aswell for fun. It would be 1!!! = 1 and n!!!=n!/(n-1)!!!/(n-2)!!!. To extend it to even more factorials you just change all the triples to whatever you want and make sure you divide on the bottom by all factorials of that type from n-1 to n-(order factorial-1).