So good!: ONE Integral, TWO Solutions! [ Desmos and WA Insights included ]
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Beta Function: • Eulerian Integral of t...
Beta trig version:
Leibniz Rule: • The Leibniz rule for i...
Let us take a look at a nice integral, which is going to yield two different solutions depending on the method of evaluation being used! With the solution being not elementary, i.e. it's with respect to hypergeometric functions, we are going to see, how their asymptotic behaviour is! Using Desmos and Wolfram Alpha, we can conclude, that both solutions are the same, and are evaluating to the same numerical value. Enjoy! =)
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![Eulerian Integral of the First Kind - Deriving the BETA FUNCTION! [ The non-trigonometric Version ]](/img/1.gif)
1. You can indeed convert from one series to the other. The first and second terms of the second series you derived are equal to 0, thus you can start at κ = 2. Shift the summation index such that κ = 0, and you recover the first series you derived. They are the same series. As for the difference of the hypergeometric functions shown in Wolfram Alpha, you can convert the one to the other via some identities those functions satisfy.
2. You can simplify the first series you derived significantly further from what you did, getting rid of Gamma functions in favor of the double factorial, using some Gamma function properties and identities. For starters, Γ(κ/2) = π^(1/2)·(κ - 2)!!/2^((κ - 1)/2). This is the well-known half-integer identity for the Gamma function. Similarly, if you use the Legendre duplication theorem for the Gamma function, which I hope you will prove eventually in a future video, then you can derive that Γ((κ + 1)/2) = π^(1/2)·2^(1 - κ)·Γ(κ + 1)/Γ(κ/2). Applying these substitutions to the summand of the series, you can algebraically simplify the summand in a trivial manner, and you would obtain (κ + 1)Γ(κ + 1)/[κ·κ!·(κ - 2)!!^2·(κ + 2)]. Now, you would have no fractional arguments for the Gamma function in this series. Moreover, Γ(κ + 1) = κ!, which simplifies the summand to (κ + 1)/[κ·(κ + 2)·(κ - 2)!!·(κ - 2)!!]. By the definition of the double factorial, (κ + 2)!! = (κ + 2)·κ!!. Therefore, the summand is κ(κ + 1)(κ + 2)/[(κ + 2)!!]^2, accomplished by multiplying numerator and denominator by κ(κ + 2). Since κ(κ + 1)(κ + 2) = 0 whenever κ = 0, the summation index can be shifted so that κ = 1. You can now choose to split the series into a summation over even κ and over odd κ, since the double factorial for both simplifies differently. Once this is done, you can eliminate the double factorials in favor of factorials and exponentials only. For odd κ = 2n - 1, n = 1 is the initial bound of summation, and the same is true for even κ = 2n. Therefore, the summand is (2n - 1)(2n)(2n + 1)/[(2n + 1)!!]^2 + (2n)(2n + 1)(2n + 2)/[(2n + 2)!!]^2. It is well-known that (2n + 1)!! = (2n + 1)!/[2^n·n!] and (2n + 2)!! = 2^(n + 1)·(n + 1)!. Applying these substitutions give the sum starting at n = 1 of 2^(n + 1)·n·n!·(2n - 1)(2n + 1)/[(2n + 1)!]^2 + 2^2·n(n + 1)(2n + 1)/[2^2(n + 1)·(n + 1)!^2]. (2n + 1)(2n)(2n - 1)/(2n + 1)! = 1/(2n - 2)! & n(n + 1)/(n + 1)! = 1/(n - 1)!. Therefore, the summand is 2^n·n!/[(2n + 1)!·(2n - 2)! + (2n + 1)/[4^n·(n - 1)!·(n + 1)!]. Since we are dealing with factorials, you could even decide to shift the summation index to start it at n = 0 to get a summand of 2^(n + 1)·(n + 1)!/[(2n + 3)!·(2n)!] + (2n + 3)/[4^(n + 1)·n!·(n + 2)!]. I think this simplification is beautiful. You could even start converting into binomial coefficients and get even juicier expressions.
Thanks for the book
You think you can just steal good morning mathematicians from me and I wouldn’t notice?
Wow, two solutions in one video! Youre spoiling us papa
Autocaptions on Papa's intro: "whats going on smut people my fermentations way come back to Nvidia"
I did not know where to post this and it is more to the channel in general, but I was struggling, started my maths degree 3 years ago and then had 1,5 years health issues and in the end I thought that I have lost my love for mathematics and I had no idea what to do. Then I started to watch you solve some cool integrals and just have fun with maths and my passion for maths returned. Thank you so much.
"This is just how maths works"
Problem solved :)
Mathologer in the thumbnail....... Instant click
Now with glasses (100% more proofs in one video!)
Do you have to wear them glasses or are you trying to s seduce the fan base?
3:08 smart boy ? what about smart girl?
Grill
Please talk about the hypergeometric function and its use to evaluate integrals with polynomials to arbitrary rational powers! I’ve scoured the web for information but cannot find anything.
No, make minecraft videos instead
I love so much that your channel grows up ❤ idk if u remember me, but that doesnt matter because u really changed my life. Almost every my small success of this year started with watching Papa's videos 😉 and u look cute in those glasses as well
Is this an old video? You didn't write the infinity boys with the 'greater or equal' sign
papa flammy's and Andrew have chemistry to each other, this is a clear proof of "1=6"
Trivial enough problem that your mic in the beginning and the glasses you wore outshine it :)
Euler integrals are not necessary because reduction formula can be derived by parts using pythagorean identity
Int(cos^{n}(x)dx)=Int(cos(x)cos^{n-1}(x)dx)
Int(cos^{n}(x)dx)=sin(x)cos^{n-1}(x)-Int(sin(x)(n-1)cos^{n-2}(x)(-sin(x))dx)
Int(cos^{n}(x)dx)=sin(x)cos^{n-1}(x)+(n-1)Int(cos^{n-2}(x)sin^{2}(x)dx)
Int(cos^{n}(x)dx)=sin(x)cos^{n-1}(x)+(n-1)Int(cos^{n-2}(x)(1-cos^{2}(x))dx)
Int(cos^{n}(x)dx)=sin(x)cos^{n-1}(x)+(n-1)Int(cos^{n-2}(x)dx)-(n-1)Int(cos^{n}(x)dx)
nInt(cos^{n}(x)dx)=sin(x)cos^{n-1}(x)+(n-1)Int(cos^{n-2}(x)dx)
Int(cos^{n}(x)dx)=\frac{1}{n}sin(x)cos^{n-1}(x)+\frac{n-1}{n}Int(cos^{n-2}(x)dx)
Hi Papa, mir ist aufgefallen, dass die Graphen bis auf eine Verschiebung in x-Richtung gleich aussehen. Und Tatsache das sind gleiche Reihen nur um 2 verschoben. Lol... deswegen konvergieren die Dinger auch gegen das gleiche
Lässt sich schnell nachrechnen
3:05 surely you can't bring the cos^2 into the summation. Wouldn't it change the value
Benjamin Brady Multiplication distributes over addition. c(a + b) = ca + cb. Assuming this holds for any size sum you should indeed be able to bring the cos^2 inside.
To make it rigorous you might have to replace the infinite sum with a limit over a finite sum but it’ll work out to the same thing if it converges.
17:20 you say they look different, so let's compare directly. Notice that they share sqrt(pi)g((1+k)/2)/k! terms. factoring them out of both gives:
k(k+1)/g(k/2 +1) on the top board, and (1+k)/(k g(k/2)(2+k)) on the bottom. The right one can become 1/(2k g(k/2 + 1) (2+k)).
Now (in our comparison of the terms) we can 'factor out' the common 1/g(k/2 + 1) term to get:
k(k+1) on top, and 1/2k(2+k) on bottom. they're almost reciprocals of each other, one is basel-ish, the other is a squared infinity boi
The series is much nicer if you separate the odd and even terms!
For the second solution you doubly differentiate t^k from k=0 to infinity, but surely the second derivative for k=0 and 1 is 0? That would change the bounds on your sum to from k=2 to infinity.
Zac Harrold Yeah, without trying to work it out myself it does look like the two series only differ by the two leading zeros which don’t change the values at all.
Flammable Maths You can indeed convert one series to the other, by using using a shift k |-> k + 2. They are the same series, except the second one you derived includes two addends that are 0, but this leaves the values unchanged.
Also, I believe you can simplify either series even further using some Gamma function identities.
I hope LetsSolveMathProblems notices your Inteh-geral shenanigans
Imitate Dr Peyam in some way lmao
"Alright! Thanks for watching!!!"
What is the music that sounds when you put your memes at the start?
In the one with (k-2)! if you set k-2=t (or start the sum from k=2) you get the other series I believe
Very nice video papa! But why ur mocking with Andrew? He’s mah sweet boi
This is just how papa works
*_Beta-male gets trig-gered_*
The two solutions are different.The gamma terms in the two solutions are the same.The k terms are different.Is it because of the cos^2 term in the integrand?
Why we follow BODMAS rule during calculations spacialy DMAS
because people are DuMb AS*
Papa Flamy, my school's mascot is the Engineer, what do I do?
Please do the integrals from the book "Almost impossible series and integrals". That book contains no "breh" material everything is so lit!
See his community post.
oʞoʎos I'm referring to that post!
@@8ball437 Ok! I think that's his plan.
*Ok, so these two functions behave identically asymtotically & they both converge to the same value of 01.86 (to three sig. figs.), but, what on Earth is the significance of this discovery?*
Integrate the Gamma Function next
Ye
14:40 shouldnt the be 1/2 beta function
Nah nvm.
I thought (1/2)! was sqrt(pi)/2
Nidhi Aggarwal Yes, but (1/2)! = Γ(3/2), not Γ(1/2).
integeral
3 dislikes are from MYD, Bprp and Peyam
What a stupid hater comment, bprp and peyam are by far more structured and better at what they do, also their channels specially bprp are much better than this one.....
@@dannygonzalez687 I would totally agree with you that bprp and Peyam are better maths channels, I really like both of them. I'm simply memeing while on this one, which I enjoy doing separately. Bprp, Peyam and Flammy all have amazing maths channels as far as I'm concerned but only flammy can have a laugh at the same time
I'm thirsty, gi'mme more
8/10 not enough doujins
So area 51?
Algún chavalito más por aquí?
@@PapaFlammy69 OMG do you speak spanish?
@@alexc.r2793 obviously google translate.....I BET PAPA IS NOT THAT TALENTED... :)...jk flammy, i know you are a gifted mathematician, also a gifted comedian
@@PapaFlammy69 Sprachen Sie auch Spanisch Mr. Papa?
Si, de argentina
@@iskanderherboso9664 Conocías los cursos del MIT de OPENCOURSEWARE?
I see Papa Flammy has a taste for bikini-clad Figürchen. :)
hes so cute
Papa, the smart daddy look is sexy! 🔥
This is what I got
www.wolframalpha.com/input/?i=sum_(k+%3D+0)%5Einfinity+(((k+%2B+1)*(2k+%2B+1)*(pi)%2F((((k+%2B+1)!)%5E2)+*+(4%5E(k+%2B+1))))+%2B+((4%5E(k+%2B+1))*(((k+%2B+1)!)%5E2)%2F(((2k+%2B+1)!)*((2k+%2B+3)!))))
The issue of getting infinity is your fault!!! When you simplified k*(k-1) on top with (k!) at the bottom to (k-2)!; you should have increased the lower index to 2 because the first two terms are 0. That was embarrassing from you :))