Again, many thanks for the shout out. I think I speak for everyone here when I say you have a fantastic channel. The graph is what started me thinking of trying to find the nth term formula. First I tried multiples of pi and then it hit me that cosine of multiples of 2pi are much nicer to deal with. No - 1. And the +/- area pairs seemed oddly symmetrical. I did notice what looked like a Laplace Transform but only the next day - good spotting Mokou. The infinite geometric series with 0

This is not one of the Fresnel integrals. It is the Dirichlet integral.

@ianfowler9340

Жыл бұрын

That is true. Honest mistake.

Feynman's technique, or Laplace transform of sinc function. A textbook example, now done with slightly different way, cool.

He just confirmed all the stereotypes about math guys 😂 But still he's cool

This has to be my favorite way to integrate sinx/x with the bounds

So beautiful. Thanks for your proof. It really shows the elegance and beauty of analysis.

This is just so cool.math works out in so many ways

Thank you for all your wonderful videos this year Dr Peyam!

@drpeyam

Жыл бұрын

Thank you!!

Very cool solution. Thanks a lot👍

Très élégant, merci

Thank u doctor , we learn from you . Exactly it's a new way and should teach in Integration section

Merry Christmas Dr peyam. You are always great.

@drpeyam

Жыл бұрын

Thank you so much, you too!

Merry Christmas and happy new year, Dr. Peyam, you inspire us!

@drpeyam

Жыл бұрын

Thank you so much, you too :)

Laplace Transforms are wonderful.

It's a hot integral!

Very Cool

Good!

Thank you , lower limit of Im (t) . Excellent presentation 👌

0:32 impressive french !

@drpeyam

Жыл бұрын

Merci!!!

gracias por este ingenioso regalo de Navidad feliz año nuevo para ti y tus espectadores otro gran amigo 🎁🌲

Echt coole Methode merry christmas and a happy new year

Excellent votre français. Bonnes fêtes et bonne année 2023.

@drpeyam

Жыл бұрын

Merci beaucoup!! A vous aussi 😁

Nice video and nice way to compute this integral ! You need the continuity of the function I at t=0 at the end to justify that the integral of I' between 0 and infinity is "I(infinity) - I(0)".

@SimsHacks

Жыл бұрын

the bigger sin is not verifying the conditions to use the Feynman's trick though 😆

Happy holidays doctor!

@drpeyam

Жыл бұрын

Thank you so much!! You too :)

Is Fresnel Integral Int(sin(x^2),x) This is sine integral and only thing which we can do to calculate indefinite integral is integrate power series term by term

Sir, thanks for the amazing video. Could you please also teach me the French you mentioned about simplicity and complexity?

Hi Dr. Peyam! What a Christmas miracle!

@drpeyam

Жыл бұрын

Hahaha awwww!!!

Wow and Wow...Wow n times

Sir can u please tell how to solve the integration [(t-1)^3/2 ]/t dt ? Also please mention the method used for the same

This isn't the fresnal integral. This is dirichlet's integral.

The thumbnail's integral lacks the integration limits, so it looks like a mathematical fairytale (a formula for the indefinite integral)!!! XD

et en bon français Doc ! Happy Holidays from France.

@drpeyam

Жыл бұрын

Merci beaucoup :)

Integral from me to me +2pi 👏

i was watching til the feymann part then i thought "wait isn't this just taking the laplace transform of the integrand then plug in s=0?" and then you divided it into smaller integrals

@ianfowler9340

Жыл бұрын

It took me til the next day to see that. Good spotting.

2:43 integral of two Peyam . . .

Hi doctor, your videos are awesome. Can you please let me know the source book from which you’ve studied the explanation of the video “Life changing quadratic formula” video?

@drpeyam

Жыл бұрын

No book, I learned it from po shen loh himself

@prasanthkumar1770

Жыл бұрын

@@drpeyam great 👍🏼 thank you for the prompt response.

Feynman crying in a corner

Where did (e) come firstly 😢?

Pourquoi faire compliqué si on peut faire encore plus compliqué? Great !

Is there a simpler way to find the answer?pourquoi faire simple quand on peut faire compliqué

Are you Persian? It would be cool if you make a video teaching in “Farsi”

@drpeyam

Жыл бұрын

I already did!

Dirichlet integral?

Ton français a l'air vraiment propre.

@drpeyam

Жыл бұрын

Merci :)

comment ca s'ecrit la phrase que tu as dit ?

@drpeyam

Жыл бұрын

Pourquoi faire simple si on peut faire compliqué?

@michaelempeigne3519

Жыл бұрын

@@drpeyam merci et ​Joyeux noel

I'm sorry doctor, you made a mistake, two PiM is an undefined value, there could only be one Dr.PiM ;),

@drpeyam

Жыл бұрын

Awwwww

Sir I got a question Can you please help on it The question says " find the locus of the centre of curvature of a given function , eg. f(x)= sinx

@drpeyam

Жыл бұрын

No idea

@kumarashutosh9554

Жыл бұрын

@@drpeyamwith all respect sir! inspite being a PhD student , you shouldn't give up like that

@drpeyam

Жыл бұрын

I’m not a PhD student

@ianfowler9340

Жыл бұрын

That poses an interesting question. So given a general point (a,sin(a)) on the graph y = sin(x) we need to: (1) Find the curvature, k, at the point (a,sin(a)). This is usually done using k = y ' '/[ (y ')^2 + 1]^(3/2). We can talk about where this formula comes from using the definition of curvature and some properties of y ', y ' ' and arc length, but that is a discussion better left for another day. Let's just accept it for now. And we don't need to bother with the absolute value . Either the + or - will do so let's just take the +. It will not change the general description of the locus. (2) The curvature of a circle is constant and = 1/radius. So we construct a circle having radius = 1/k as calculated in (1). This gives a unique circle but with an unspecified center. (3) Now we take that circle and move it so that it it is tangent to y = sin(x) at the point (a,sin(a)). There are actually 2 possible circles with radius 1/k tangent to sin(x) - one on each side of (a,f(a)). That's where the absolute value of k comes in. But as I said, it won't change the general description of the locus of the center. Let's just pick the side of the curve that is concave down. (4) By moving the circle so that it is tangent to (a,sin(a)) we have now fixed the center - called the "center of curvature". Say (h,k). Our job now is to find the values of h and k as a function of a. We will need the fact that f ' (a) is the slope perpendicular to the radius (negative reciprocal stuff) and that (a,f(a)) lies both on the radius and on f(x). Then solve a quadratic system in h and k .Then we let "a" roam freely over the real numbers, replace h and k with x and y and we get parametric equations of the locus of the center. (5) This sounds reasonable, in principle, but the parametric equations you get for h and k are wildly complicated. I tried it and obtained nothing easily recognizable - for me anyway. (6) I'm not trying to trash the idea - I think it's a good one. Maybe I'll try a much simpler function. I'll let you know.

@kumarashutosh9554

Жыл бұрын

@@ianfowler9340 I have still doubt that if there are two possible circles , then how did you insure that it will not effect the locus of required centre ( the concave down)

I am a grade 12 student, but I did not understand it, why 😭😭😭😭🥺🥺🥺