Feynman’s Technique is Unstoppable!!!

Feynman’s Technique is Unstoppable!!! | Fresnel Integrals

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Күн бұрын

Today I solve the Fresnel integrals using Feynman’s integration technique. This video teaches you how to apply Feynman’s method to the fresnel integrals, leading to an amazing result.
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#maths #advancedmath #calculus #feynmantechnique #JEE #feynman #fresnel

Пікірлер: 33

@Jagoalexander Ай бұрын

I hope you guys enjoyed that intro 😌

@CeoOfSkate Ай бұрын

Loved it

@on_God_ Ай бұрын

It was like I'm watching a series and it jus said " Previously on ..."😂

@Jagoalexander Ай бұрын

@on_God_ I totally see that hahahah

@Jagoalexander Ай бұрын

It has also just occurred to me that I didn’t explicitly state the integrals we will be solving, we are trying to solve Sin(x^2) and Cos(x^2)

@AndyGoth111 Ай бұрын

Seems you do state them by name, with the formulas onscreen, at 1:07

@abhinavjuly Ай бұрын

Nice video i knew it this prob is going to solve by fey.. method but i was stuck into some steps.you explained well now i understood where i was making mistake.😊

@Jagoalexander Ай бұрын

Glad it helped mate 🙂

@heyathereboi 28 күн бұрын

Hey there, loved the part where you substitute the (2t*exponential) by the derivative, really reminded me of solving for the expectation value of x^2 in the introductory QM, but in this case specifically is taking the derivative with respect to t justified, as u has a t dependence itself (u=x/t) so the derivative also acts on it, doesn't it? Otherwise - great video!

@YaleQuatts Ай бұрын

I’m missing that last bit…. How did the imaginary bit (attributed to sin) become real? What happened to the “ i “ in the last step?

@Jagoalexander Ай бұрын

@@YaleQuatts so when dealing with complex numbers, they are split into 2 parts, a real part (a) and an imaginary part (b). Forming complex numbers of the form: a +ib . When you ask “what is the imaginary part of the complex number 3 + 2i?” For example, you strictly look at the numbers that have got the imaginary constant i next to them. So the imaginary part of 3 + 2i would be 2. When we look at the video and reach the end part, we stated that the solution to the sine integral was the imaginary part of whatever result we got for the e^-ix integral. The result we got was root(π/8) - i x root(π/8) . So as we discussed, looking at the imaginary part of this complex number would be strictly looking at the numbers that have an imaginary constant. Therefore, we take -root(π/8) as our answer. We then had to make it positive due to an earlier step. Hope this helped !

@NickNicholas398 24 күн бұрын

When you plugged in I(0) wouldnt it also make U zero since it is a function of I?

@christoffel840 Ай бұрын

I don’t see how the integral goes to zero when you take the infinite limit. If the i wasn’t in the exponential it definitely would, but with the i you would just get oscillations. The integral of the magnitude as far as I can tell doesn’t go to zero either.

@attica7980 Ай бұрын

Intuitively, you can approach the probem of calculating the limit as t tends to infinity as follows. Break up the integral as a sum, each term corresponding to half a period, that is a change in u that corresponds to a change of the exponent of i pi. Then you get a finite alternating series of decresing terms, and the sum of the series will be between the first term and the sum of the first two terms (sort of, if the terms were real -- they are not, but they are still of the opposite signs). as t goes to infinity, both these quantities tend to 0, so the integral will tend to 0. I did not do rigorous calculations, but I am sure that these ideas can be turned into a rigorous argument. Note added on Mon Aug 12 19:40:19 EDT 2024 Using the Riemann-Lebesgue lemma suggested in the replies is much simpler than the argument above.

@maximiliansvensson6205 Ай бұрын

Check out the "Riemann-Lebesgue Lemma"

@attica7980 Ай бұрын

@@maximiliansvensson6205 You are right, and I have to correct my earlier reply to you (now deleted). After the change of variable v=u^2+1, the Riemann-Lebesgue lemma is directly applicable to the resulting integral.

@lucapolidori8817 Ай бұрын

When a SIN is extended to infinity its area is equal to the COS, just looking at the waveform. Having the first peak of the curve at 0 or at 1, at the infinity doesn't change anything. Am I wrong?

@BoringExtrovert 22 күн бұрын

Hey man, you should give a good argument on why you only took the principle square root. Other than that good video

@philipframpton9428 Ай бұрын

Can you explain 5:47, why did the derivative of the integral simplify so easily.

@Jagoalexander Ай бұрын

You can prove it using the Leibniz Rule and the fundamental theorem of calculus. It just is a lot to explain so I skipped past it. Basically the derivative with respect to t of the integral with the upper bound at t results in you substituting t into the function and that's your answer.

@jimcameron6803 Ай бұрын

And all the people rejoiced.

@user-kp2rd5qv8g Ай бұрын

Couldn't we just say I = Im \int^{\infty}_{0} dx e^-x^2/i = Im 1/2 √(pi i), (using the result for integrating a Gaussian) = √(pi)/2 Im e^(i pi/4) = √(pi)/2 sin(pi/4) = √(pi)/2 (1/√2) = √(pi/8)?

@faustinus23 27 күн бұрын

Yes, because there are no poles and integral vanishes over the arc at Infinity. So we can simply move the contour and piggy back on the real gaussian integral.