Finding the closed form for a double factorial sum
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- Опубліковано 4 сер 2022
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So the even term bit is pretty trivial and the odd term bit requires an integral, a change of variable, a trigonometric substitution, another couple of integrals and changes of variables, and ends up with an expression in the error function. Nice.
Reminds me of the difference between odd and even Riemann zeta values
Is like Inception movie...
Sounds like life. Every other problem seems trivial compared to the next one.
I think that the less cumbersome method would be differentiation of these series and getting two ODEs: f'-x*f=0 and f'-x*f=1. The solution for the first on can be easily guessed exp(x^2/2), the second one may solved by representation of solution in a form g(x)*exp(x^2/2), then we get equation for g by substituting f=g*exp(x^2/2) into ODE: g'=exp(-x^2/2). From that we get erf related integral
wow, nice
Yep, that's exactly what I did. Obviously, the even part is pretty simple (as it doesn't really require using a differential equation), but getting the odd part is definitely more difficult; it helps to think about the generating function f(x)=x+x^3/(3*1)+x^5/(5*3*1)+... ; I was reminded of the differentiation trick when I thought about how the generating function for the Catalan numbers was derived.
Cool idea
You don't need to split out the odd and even series. If
G(z) = sum(n=0 to inf) z^n/n!!
G = 1/1 + z/1 + z^2/2 + z^3/3 + z^4/8 + z^5/15 + z^6/48 + z^7/105...
G' = 1/1 + z/1 + z^2/1 + z^3/2 + z^4/3 + z^5/8 + z^6/15 ...
z(G-1) = z^2/1 + z^3/2 + z^4/3 + z^5/8 + z^6/15 ...
= G' - z - 1
-z cancels
zG = G' - 1
G' - zG = 1
This is a first order linear ODE. The integrating factor is exp(-z^2/2)
G' exp(-z^2/2) - zG exp(-z^2/2) = exp(-z^2/2)
(G exp(-z^2/2))' = exp(-z^2/2)
G = [c + integ 0 to z exp(-t^2/2) dt] exp(z^2/2)
Boundary condition z=0, G=1 => c=1
Sum S = G(1) = [1 + integ(0 to 1) exp(-t^2/2) dt]exp(1/2)
Substitute u = t/sqrt(2)
dt = du sqrt(2)
t=1 => u=1/sqrt(2)
S = [1 + sqrt(2) integ(0 to 1/sqrt(2)) exp(-u^2)) du] exp(1/2)
What do you z exponential.terks mean..whybdonyou have exponents?
@@leif1075 It's a standard method for solving this type of differential equation. The integrating factor is to get a perfect differential on the left hand side
@@pwmiles56 what do you mean by perfect differential.sorry? And would you agree unless you have already seen something likex^n/n! There's no reasonbtonthinkbkfbtbat since in this expression we always have one in the numerator nkt anything like x much less x raised to anything?
@@leif1075 I made the left hand side into
G' exp(-z^2/2) - zG exp(-z^2/2)
This is the derivative of
G exp(-z^2/2
Now I can integrate both sides
G(z) is called a generating function. It's a standard approach in this kind of problem.
Don't worry if this is new to you. It's first or second year bachelors level maths
How did you determine the boundary conditions z = 0, G = 1?
Awesome video!
17:01
Great video!
Not sure if it will be much of helpful, but we can express it with Incomplete Gamma Function, with pleasantly looking Gamma(1/2,1/2) within it :)
sqrt(2) times our final integral can be written as sqrt(pi/2) erf(1/sqrt(2))
Thus, the solution is equal to:
sqrt(e)[1+sqrt(pi/2) erf(1/sqrt(2))]
Now, using the connection with Incomplete Gamma Function
erf(x) = 1 - 1/sqrt(pi) Gamma(1/2,x^2)
We can rewrite the solution as:
sqrt(e)[1+sqrt(pi/2)-sqrt(1/2)Gamma(1/2,1/2)]
which is approximately: 3.059407405342576144539475499233278612...
thanks so much
great video!
Your videos are GREAT ... thx alot
I noticed that the Taylor series for sqrt(x) uses the odd double factorial, no wonder I couldn’t easily find a closed form for it. I ended up simply using the product definition. It was a little clunky but it works. Is there a link between variations of factorials and nth roots? Is the rising factorial relatated to tower functions, is the triple factorial related to cube roots and a cubed root version of the error function?
Beautiful!
ahhh i wish you showed the erf(x) correction directly!!
If f(x) = sum(x^n/n!!) then if take a derivative of the sum it looks like 1 + sum(x^(n-1)/(n-2)!!) = 1 + xf(x) (we also know that f(0) = 1 from the sum formula). So rearranging (f’ - xf) = 1 implies (exp(-x^2/2)f(x))’ = exp(-x^2/2)
So exp(-x^2/2)f(x) = Int(exp(-x^2/2)) +c
f(x) = exp(x^2/2)*(c+int(exp(-x^2/2)) (the c allows you really to pick the integral either from -inf or from zero) integrating from zero to x we look at
F(x) = exp(x^2/2)*(c+int(0,x,-u^2/2,du))
Put x =0
1 = 1 * c
C = 1
The root 2 you had was from picking the z^2 form rather than the normal dist form i guess - nice problem anyway u = x then u/root 2 = x/root 2 suggesting a substitution z = u/root 2 - i think your formula emerges
That's how I did it too
we need more factorials in the denominator that would be fun
Though quite difficult!!
@@GrahamPointer1972 uh i'm sure michael can do it. there might even be a stack overflow math stackexchange question on this out there somewhere
@@panagiotisapostolidis6424 I hope so!!
@@angelmendez-rivera351 No why do you ask angel?
@@angelmendez-rivera351 As always, unnecessarily hostile. I don't have a proof that a "neat" solution exists (given that "neat" seems to include the error function, which is hardly neat), but another commenter suggested an approach. With the n-tuple factorial (triple, quadruple etc) one can break the sum up into n pieces, such as even and odd, or 3n,3n+1,3n+2. Then these series can be differentiated term by term (ignoring your incoming complaints about term by term differentiation) to give n differential equations, which can be solved instead. The differential equations for this video would be f' - xf =0 for the even sum, and g'-xg = 1 for the odd sum. Then f is easy to get, and g can be found too. I don't think it's hard to suggest that this process works for larger n-tuple factorials.
Tens o meu respeito doutor... Obrigado pelo seu excelente trabalho!
Fun! Neat to see how related it is to these more famous integrals
How about 1/(n!)! ?
Cant imagine that having a nice form
it's about 1801/720
According to OEIS A336686, it has a lot of early 8s: 2.50138888888888888888889...
@@Alex_Deam that’s probably because 1/(n!)! decreases VERY quickly, so the sum is vey close to (1/(0!)! +1/(1!)! +1/(2!)! +1/(3!)!), the 1/(3!)! Is 1/720 which ends in 8s and the other terms is just 2.5.
why bother with splitting evens and odds, just use that n/n!! = 1/(n-2)!!, from this you derive the ODE f'(x)=1+xf(x) so dy/dx - xy = 1 which by multiplying through by exp(-x^2/2) you can solve and get y=exp(-x^2/2)(1+1/2sqrt(2pi)erf(x/sqrt(2)))
1/n!! = (n+2)/(n+2)!! implies the generating function satisfies G(x) = (G'(x) - 1)/x, with G(0) = 1.
soooooooo cool to think that Fijians can now learn for free all sort of things including cool maths like yours.... no more need to pay big money to universities to get at least some background in different fields. this is like a National Geographic for me.
Regards!
MN
and we can even freeze the video to have a beer and watch some 👙 then come back.
You would get more views if you Hired a chick like Scarlett Johansson in bikini to explain the whole thing... but the viewers would struggle to learn EXCEPT those who are really serious. Sometimes think about New strategies.... hide serious stuff behind a apparent yes no joke.
p.s. your English is very clear and good.
I calculated sum of odd and even terms separately
Hi Dr.!
Do we have some approximate value?
Well, but then, what's the sum of the 1/(n!)! . . . ? Which is what I thought the "double factorial" was? This series should be NICE and convergent, and rather less than e. GO MICHAEL GO!!
I thought that's what the video was about and I was very excited until I saw the !! definition :
Interestingly a close approximation to the final answer is sqrt(2) + sqrt(e). It is about .001 greater than the actual sum.
I wonder if sum(1/n!) is proportional to sum(1/n!!) ??
It's all about your definition of proportionality. Two non zero numbers are always gonna be proportional
@@panadrame3928 Yeah true that's wasn't really a good question lol. I meant are they proportional by an interesting constant. As in k*sum(1/n!) = sum(1/n!!). Is k an interesting value depending on e or something? Could be an interesting insight to the relation of these two sums.
Yeah, I saw that approximation too. I wonder if there's a more rigorous reason that the error function of 1/sqrt(2) is so close to 1/sqrt(e)
okay so can someone explain to me why the double factorial is SMALLER THAN THE REGULAR FACTORIAL
Isn't this way easier using the relations of the double factorial with the regular factorial?
Thanks for the video. I have another question. For a number n, what is the least amount of selections between (1, n) whose sum will cover the maximum number of values between (1,n)?
Sounds about powers of 2, but should be proved (may not work for small n numbers)
@@panadrame3928 I guess this is a problem in linear combination.
You can represent any number as sum of powers of another (basically writing in that base); I have not solved for the general case but for 1000 writing in base 2 was most optimal. So we'll have the powers of 2 plus the difference between the highest power of 2 smaller than n and n. This I think covers ALL values
That was a very interesting problem but the integral at the end is very sad 😪😪
crazy
Thinking 🤔
i have a doubt at 15:04
how could you subtract both integrals the z is different for both the integral one z is x/√2-y and other is x/√2+y they are different isn't it so how could you subtract that ???
z is nothing. It's a "dummy variable". The two things that determine the value of an integral are the function being integrated and the bounds.
Think about the Riemann sum definition of integral[a to b] f(z) dz. z doesn't appear anywhere in this definition.
@@martinepstein9826 thank you
what about the sum of 1/(n!)! ?
I prefer the differential equation method already mentioned. At least I can handle that.
Is the (integral) result easier to evaluate accurately than the original series? I think not, in which case this is just a mathematical exercise with no particular value.
for Comparison Test just compare it to Sum(1/n!), why pick Sum(1/n^2) ?
The sum given in the problem is greater than what you mentioned since n!! < n! for n > 2 so can't have comparison with it. Whereas, n!! > n^2 for n > 5.
Looks very complicated. Next sum of triple factorial 1/n!!!.
i think there is a fault, the gamma function should have t to the power n+1 , m not so sure thoo. i didnt finish the video tho maybe its correct somehow haha!!
I got the even part really easily, but man, that odd part is absurd
It's not that difficult but it takes a lot of steps, requires a bit more litteracy and is rather inelegant a calculation. I think I felt more satisfaction about discovering the notion of double factorial than I did following the derivation!
Before watching the vkdeondodnt everyone else think n!! Meant (n!)! Like 3!! Would be (3!)!=6! ..that is a valid interpretation too..but I suppose there is no closed form for that expression?
@@professionalprocrastinator8103 It’s pretty difficult. No need to stroke your ego
Cool problem
Does anyone else want to know what he saw at around minute 12?
Now do the triple factorial
Can you explain me, why 6!!=6•4•2?! Who is the reason?! ...
Definition of the double factorial. Instead of going down one each step, you go down two
Oh, I thought this was the sum 1+1+1/2+1/24+1/(24!)+1/(120!)+..., so not the double factorial but the factorial of n!, so the product n!(n!-1)(n!-2)... 2 1
It should have been (n!)! then
@@IoT_ I understand what you mean, however writing it as n!! is also possible, though leads to misunderstandings
@@JonathanMandrake No you are wrong, there is no misunderstanding but only your limited knowledge
N!! Seems different than assumed by Michael.
Hold on, at 15:00 aren't the Z's from both intervals actually different variables? One z is x/√2-y, the other is x/√2+y. You can't just subtract the two integrals then, right?
Edit: nvm, you never went back to x or y in the end. Thus it doesn't matter.
Definite integral, doesn’t matter. Once you do a substitution in a definite integral, the original variable is gone.
Shows some thinking 🤔🤔🤔
why can u set z both equal to "x/sqrt(2)-y" and "x/sqrt(2)+y" at the same time? arent they different meaning?
2 differents integrals means 2 differents z variables : they are "silent", since their name isn't relevant
@@panadrame3928 but after a while he seems to make a one integral instead of two before
@@panadrame3928 he subtracts the two integrals (~15:00)
Write a sentence. Spell out "you."
The Rock at 12.17
one would think that n!! should be equal to (n!)! why is it so unintuitive tho
you: sum(x^2n/(2n)!!)
the guy she told you not to worry about: sum(x^2n+1/(2n+1)!!)
Nice t shirt
Is that a Lynel from Breath of the Wild on your shirt?
I think Michael called 'e' Euler's constant in this video. Euler's constant (also called the Euler-Mascheroni constant) is denoted by lower case gamma (γ) and is not the same as 'e' which is called Euler's number.
"e" is sometimes called Euler's constant. Euler just has a lot named for him.
Fussy.
So needlessly pedantic…lol
It is so pendatic that falls into inaccuracy zone.
J3rk
Closed form? We got series transformed into an integral.. do not see any simplification :)
it was transformed into something that can be plugged into most statistical calculators.
You don't use the power series in x at any point. You don't differentiate it nor find closed expressions. Thus I don't see why you introduced the x. You could have set x=1 for the whole time and get the same answer. Great video though
asnwer=(1+n)/1 🤣😂
The value of this sum differs by e at least 0.1124...
I only saw the thumbnail and thought it was about (n!)! If so, what is the sum?
Is the integral not just the complementary error function Erfc(x)?
Sum over odd part was cooler.
🤔
Awesome sum. Fun fact: the double factorial of a negative odd integer is well defined.
how?
I thought n!! meant (n!)! ?
Why integral can be considered as more "closed form" than infinite sum? Usual definition of "closed form" denies infinite sums, integrals and special functions (like error function). That's why there is no closed form for a double factorial sum.
it's what we can call mathematical surgery. Hhhhhhhhhh. Thank you a lot.
That is most definitely not a good place to stop! How can Mike be so incurious about what the actual answer is in real money. I think there must be something up with his wiring that he can take so much evident pleasure in the algebra but fail to go that small extra step to compute from the final formula. I worked it out to be 3.059407 of which the even terms contribute 1.648721 and the odd terms 1.4107. I guess the evens have the advantage of ending at 2 while the odd ones finish with an n=1 so always have that small advantage as both head up to infinity. The error function can be readily evaluated from Normal probability tables because erf(1/sqrt(2)) is obtainable from the Normal integral at x=1 (it equals 2*N(1)-1)..
really, the numerical approximation to this series is not an interesting result by itself. you didn't even need michael's answer to the problem to find it, you could've just taken partial sums of the original question to see a decimal representation.
numerical approximations are a helpful tool: indeed, euler numerically verified his solution to the basel problem. the reason these problems are interesting hardly arises from "where they are approximately on the real line". the reason is that we regularly see very simple summands for these series evaluating to well-known mathematical constants (that often appear unrelated to the original series.)
The double factorial notation is so bad. That should real mean n factorial factorial.
At no point did you tell us what 0!! is, and it seemed to be ignored when the even terms were computed.
as always with factorials, 0 gets to be an empty product. 0!! = 1.
3.5346
Had no idea that that's how n!! was defined. I would have guessed it would mean (n!)! instead. How would n!!! be defined then?
n!!! = n(n-3)(n-6)...3, 2 or 1 depending on the value of n mod 3.
a silly and unhelpful way to remember this: if you yell at a number, it gets bigger (n -> n!). but if you yell any louder, it gets scared and goes smaller (n! -> n!_(a) for a > 1).
"That's about as simple as you can get for this sum". Sir, you lost me about 10 minutes ago... I was hoping for some nice, clean result... Disappointed, would watch again ;)
I had sent you a mail to you . Just to confirm can you share your email id
Beautiful!