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.

@flippinamazing1423

19 күн бұрын

Can you write a small summary of how you can show that they are equivalent on the interval please? I'm having trouble.

@dbmalesani

17 күн бұрын

@@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.

@LeoX2

21 күн бұрын

The first word is kinda unrealistic considering you seem to be a mathematician Kappa

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

The maths cook is back, and this time its so pure you get two times the solution for one times the integral

bro cooked the most delicious meal ive ever seen in my life

5:23 I also talk to calculus results 🤣

@maths_505

21 күн бұрын

Sorrys....me not know English too good.

can we tell him that cos2x / sin2x = cot2x not tan2x yet it doesn't change the answer

@Player_is_I

20 күн бұрын

How does it not changes the answer, please reply

@Player_is_I

20 күн бұрын

Nvm I got it 😁

@trelosyiaellinika

20 күн бұрын

Good that I throw an eye over the comments before adding my own. I was going to do the same remark!

@user-yg5zb9gk6f

20 күн бұрын

@@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.

@Player_is_I

20 күн бұрын

@@user-yg5zb9gk6f Yup, saw the beauty in it

Fantastic! Keep the boredom going so the math keeps on coming.😅

I enjoy so much your videos, man. Thank you

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 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 .

As soon as I see square root of pi, I immediately think of a Gaussian distribution. Probably.

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) 😎😎😌

Gamma(-1/2)=-2√π So this sum is -gamma(-1/2)

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

clicked on video thinking that 7/4 was gamma/4

Please elaborate on your activities when you're bored 🤔

@maths_505

21 күн бұрын

My parole officer is also a subscriber to my channel and has also advised me not to expand on that💀

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

@maths_505

20 күн бұрын

Mostly the US, India and European countries.

2:09 it's cot(2x)

can we tell him that cos2x / sin2x = cot2x

@maths_505

21 күн бұрын

Already fixed it in a comment. Technically it's not an error since the integrals are equal anyway.

what app do you use to do math?

i have a nice integral int(from 0 to pi/2) ln(cos(x))/1 + x² dx

@SussySusan-lf6fk

18 күн бұрын

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.

@user-yg5zb9gk6f

18 күн бұрын

@@SussySusan-lf6fk My friend just mixed numbers and functions randomly and gave me this integral

@SussySusan-lf6fk

18 күн бұрын

@@user-yg5zb9gk6f 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)

@SussySusan-lf6fk

18 күн бұрын

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)

2:13 it's actually cot(2x) instead of tan(2x)

@maths_505

21 күн бұрын

Fixed it in the pinned comment. Technically not an error since the integrals are equal anyway.

Ooooooook cooooool

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 KZread 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 KZread videos

@maths_505

21 күн бұрын

Great work bro 🔥

@aravindakannank.s.

20 күн бұрын

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. 😅

@illumexhisoka6181

20 күн бұрын

@@aravindakannank.s. it's more like yellow I think it's the one I have the 12th edition About a 1000 page

@aravindakannank.s.

20 күн бұрын

@@illumexhisoka6181 ok bro i will check that out if it is available in library thanks for clarifying

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🤔🧐😒😮‍💨

The new math update dropped

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

crazy result but it should be cot2x sadly

@maths_505

21 күн бұрын

Nah it's cool....doesn't affect the result.

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...

woah..... hi from one of the 7/2 ladies :D

@maths_505

21 күн бұрын

Greetings Now I'm just waiting for the hobbits to reveal themselves in the comments too.

@IshayuG

19 күн бұрын

@@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. 😂

I edge to your video btw, just thought I'd let a homie know 😁

@birdbeakbeardneck3617

20 күн бұрын

🤨

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.

it's me, i'm the half lady

@maths_505

20 күн бұрын

Greetings

"31 secs ago"

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

Bro thinks hes walter white cooking up 95% pure math

@maths_505

21 күн бұрын

I am the danger

@maths_505

21 күн бұрын

I am the one who knocks

So bored he changed the font on the thumbnail

You have a lawyer?

@maths_505

20 күн бұрын

It was more out of necessity than want💀💀