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.

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

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?

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.

That's how I did it too

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.

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 thought that's what the video was about and I was very excited until I saw the !! definition :

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)

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

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

"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

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

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

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.

Definite integral, doesn’t matter. Once you do a substitution in a definite integral, the original variable is gone.

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."

Definition of the double factorial. Instead of going down one each step, you go down two

how?

it was transformed into something that can be plugged into most statistical calculators.

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.)

as always with factorials, 0 gets to be an empty product. 0!! = 1.

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).