An awesome calculus result I cooked up
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- Опубліковано 16 жов 2024
- Strange things happen when math nerds get bored.....you start off with an integral....things can escalate quite quickly in a multitude of ways from there....this video is an example of that.
My complex analysis lectures:
• Complex Analysis Lectures
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CORRECTION: At the 2:00 minute mark, I wrote cos(2x)/sin(2x) as tan(2x) instead of cot(2x) and forgot to do a phase shift to show the results are equal. Terribly sorry for the inconvenience, however, it doesn't affect the result.
Can you write a small summary of how you can show that they are equivalent on the interval please? I'm having trouble.
@@flippinamazing1423 let 2t = 𝜋/2 - 2x; then the integration bounds change to from x = (0, 𝜋/4) to t = (𝜋/4, 0). Then dx = -dt and cot(2x) = cot(𝜋/2-2t) = tan(2t). Finally, use the - sign to flip the integration bounds.
Babe wake up, Maths 505 has posted another amazing integral calculus result.
The first word is kinda unrealistic considering you seem to be a mathematician Kappa
The maths cook is back, and this time its so pure you get two times the solution for one times the integral
3 and a half ladies is crazy.
either way, we "okay, cool"
i wonder how hard all these integrals would get if we didn't have the amazing beta function, it is so useful sometimes
bro cooked the most delicious meal ive ever seen in my life
thank you for your fantastic program !! could you explain imaginary integrals like Mellin Barnes etc ...Mario
Fantastic!
Keep the boredom going so the math keeps on coming.😅
I enjoy so much your videos, man. Thank you
This is interesting. Thank you indeed.
Hi,
"ok, cool" : 3:53 , 5:19 , 6:09 , 6:33 , 11:09 ,
"terribly sorry about that" : 4:18 , 9:09 , 12:20 , 13:06 .
can we tell him that cos2x / sin2x = cot2x not tan2x yet it doesn't change the answer
How does it not changes the answer, please reply
Nvm I got it 😁
Good that I throw an eye over the comments before adding my own. I was going to do the same remark!
@@Player_is_I Well if you use cot2x instead of tan2x . After the substitution theta = 2x. You can write sqrt(cotθ) = sqrt(cosθ/sinθ)=cosθ^1/2 . sinθ^-1/2. We use the bêta function again which has the property that B(u,v)=B(v,u). And notice that the exponents just switched places compared to the video. Thus it doesn't change the answer.
@@wassimaabiyda Yup, saw the beauty in it
5:23 I also talk to calculus results 🤣
Sorrys....me not know English too good.
I feel that this summation may have a richer general formula for ratios of gamma function or at least a general formula resembling the summaton of ratios, and, the function or the formula equals 2√π at a certain value of x which is hiding in the inner values 1/4 and 7/4
What countries watch your videos the most? What are the percentages
Mostly the US, India and European countries.
Please elaborate on your activities when you're bored 🤔
My parole officer is also a subscriber to my channel and has also advised me not to expand on that💀
I discovered this: 2^(x-y) = product from 0 to infinity [{(2k+x)(2k+1+x)(k+y)} /{(2k+y)(2k+1+y)(k+x)}]
This question looking similar to ramanujan's paradox by taking same question in both x and thetha form and getting 1 value as pi/2 and using it to form a summation of gamma function k value as 2 × sqrt(pi) 😎😎😌
2:09 it's cot(2x)
what app do you use to do math?
cool identity.
also, it's time to abolish the Gamma function. it's supposed to extend the factorial, why the hell would it be shifted like this??
Instead use Π(z) = prod_1^∞ (1+1/n)^z / (1+z/n)
Π(z) = int_0^∞ x^z e^-x dx
I believe in Pi function supremacy
In the last week or so so I read the proofs (not the most rigorous ones)
of C-R equations,cushy theorem,cushy integral formula for nth degree pole,residue theorem,liouville theorem,the fact that holomorphic implies analytic, the Reflection Principle
and some basic facts besides them
now on the lest I have the identiy theorem,laurent series
now I am going to start watching your videos on contour integrals that I have always been avoiding because I don't understand or know the proof of what you where doing
and my summerbreak started about a week ago so I am going to read Thomas calculus book
it cover calculus up to calc 3 and vector calc with keeping up with liner algabra on UA-cam
before I start my second year of college
so hopefully in a year I would be ready for some heavy stuff with real analysis,abstract algabra and topology and some other courses and then hopefully then I would be ready for a real dive in complex analysis and not the basics I know right now and to have a strong basis in mathematics not some random stuff I know from watching UA-cam videos
Great work bro 🔥
if I'm not wrong the Thomas book u talking is brown colour right ?
i only have vector calculas in that book to complete it .
i don't remember exactly the book edition but I think it's 11 or 10 idk
because I went to library and took a heavy and large book😅
i also remember it has a lot of simple integral results in the back as a shortcut.
correct me if I'm talking about the wrong book. 😅
@@aravindakannank.s. it's more like yellow
I think it's the one
I have the 12th edition
About a 1000 page
@@illumexhisoka6181 ok bro i will check that out if it is available in library
thanks for clarifying
Is there any one who can help solve this problem, even my teacher is having problem with it , “given that h(x)=integral of ((f’(x)x-f(x))/x^2) , where 2h(2)=f(2)+4 and f(-1)=5 , what is the value of h’(-1) ?
There are some steps that i have done, h(x)=f(x)/x+C and i found the value of C which is 2, thus h(x)=f(x)/x+2 and h(-1)=-3 , now how can i find the value of h’(-1) ?
Appreciate your help
2:13 it's actually cot(2x) instead of tan(2x)
Fixed it in the pinned comment. Technically not an error since the integrals are equal anyway.
Gamma(-1/2)=-2√π
So this sum is -gamma(-1/2)
i have a nice integral
int(from 0 to pi/2) ln(cos(x))/1 + x² dx
It should be from 0 to infinity. Then it's possible by Fourier series and evaluation of the integral of the form cos(ax)/(1+x^2) dx from 0 to infinity.
@@SussySusan-lf6fk My friend just mixed numbers and functions randomly and gave me this integral
@@wassimaabiyda you did good that you gave this unconventional integral. There is no problem except for that it should be from 0 to infinity instead of from 0 to pi/2. OK I'll write the solution to you here tomorrow. Thanks for the integral Brother.
(Edit - I wrote the solution brother)
int 0 to ♾️ , ln|cosx| /(1+x^2) dx
the Fourier series of ln(cosx) for x ranging from 0 to pi/2 is -ln2 -$(k=1 to ♾️) (-1)^k cos(2kx) /k
But cosx is sometimes negative on values ranging from 0 to infinity, so to use the Fourier series,we need to make sure cosx is positive. That's only possible when we use absolute value of cosx,thus|cosx| . That's why the suitable integral is
int 0 to ♾️ , ln|cosx| /(1+x^2) dx
Now we use Fourier series of ln|cosx| = -ln2 -$(k=1 to ♾️) (-1)^k cos(2kx) /k
We have
-pi ln2 /2 - $(k=1 to ♾️) (-1)^k /k int 0 to ♾️, cos(2kx) /(1+x^2) dx
We know integral of cos(ax)/(1+x^2) dx from 0 to ♾️ = pi/2 e^-a , we can easily prove it through Laplace transformation.
So we have,
-pi ln2 / 2 - pi/2 $(k=1 to ♾️) (-1)^k (e^-2)^k /k
We know ln(1+x) = (-1)^(k-1) x^k /k , we use it to get
-pi ln2 / 2 + pi/2 ln(1+e^-2)
I don't understand why you didn't do it straight away. It is possible straightaway from the sum itself . You'll get an integral , 2/sqrt(pi) int 0 to 1 , x^-3/4 sqrt(1 - x) /(1+x) dx. Now do a thing let sqrt(x)=t ,
We have, int 0 to 1 , 4/sqrt(pi) sqrt(1-t^2) / { sqrt(t) (1+t^2) } dt
int 0 to 1 , 4/sqrt(pi) sqrt(1/t - t) / (1+t^2) dt
t=tanu
int 0 to pi/4 , 4 /sqrt(pi) sqrt(cotu - tanu) du
Now easy to do. Cute problem.
As soon as I see square root of pi, I immediately think of a Gaussian distribution. Probably.
Can you solve infinite series ((-1)^n)/n^2 and ((-1)^n)/n^3. I'm looking for their resolution everywhere but i can't find🤔🧐😒😮💨
These are pretty well known results in terms a dirichlet eta function, one evaluated at 2 and another one evaluated at 3.
lool, why did you put cos(2x)/sin(2x) = tan(2x), need to be corrected lol.
Doesn't affect the result tho
can we tell him that cos2x / sin2x = cot2x
Already fixed it in a comment. Technically it's not an error since the integrals are equal anyway.
clicked on video thinking that 7/4 was gamma/4
woah.....
hi from one of the 7/2 ladies :D
Greetings
Now I'm just waiting for the hobbits to reveal themselves in the comments too.
@@maths_505They will reveal themselves if you integrate a function that happens to look exactly like the outline of a hobbit hole’s roof.
Fortunately for us, we can approximate it well enough with a Fourier transform, giving us a series, which we can then integrate.
This will lead us to the hobbits.
Doing this is left as an exercise to the channel’s resident video creator. 😂
crazy result but it should be cot2x sadly
Nah it's cool....doesn't affect the result.
The new math update dropped
I edge to your video btw, just thought I'd let a homie know 😁
🤨
3:40 i am not very good at math 😂😂 are u serious, actually yeah people's usually good at calculus, complex analysis are are actually poor in fraction and applying simple mathematical operation
Sono arrivato fino a ..S=(1/π√2)Σ(1/(k+3/4))Γ(1/4+k)Γ(1/4-k)...poi nin riesco a proseguire...
Ooooooook cooooool
it's me, i'm the half lady
Greetings
"31 secs ago"
So bored he changed the font on the thumbnail
You have a lawyer?
It was more out of necessity than want💀💀
Bro thinks hes walter white cooking up 95% pure math
I am the danger
I am the one who knocks