Feynman’s Integration Technique is Overpowered…
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- เผยแพร่เมื่อ 20 มี.ค. 2024
- In this video I use maths / the internets most favourite integration technique known as Feynman’s technique or differentiation under the integral sign to evaluate a difficult integral of sins / x from zero to infinity.
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We could use the Laplace transform after introducing Feynman's technic I'(a)= - L(sin(bt))=-b/(a^2 +b^2) , with b=1, we can easily reach the solution faster.
Excellent idea
Sin x = 1/1! x - 1/3! x³ + 1/5! x⁵ - ....
So sin x /x = 1/1! - 1/3! x² + 1/5! x⁴ - ....
And the I tergal is just :
∫ Sin x/x . dx = 1/1! x - 1/3! x³/3 + 1/5! x⁵/5 - .... + (-1)^n 1/n!. x^n /n +...
You can also integrate many functions including the naughty function f(x)= e^x² which is otherwise does not have an explicit formula of integral. Off course with a condition of being an analytic function.
Integral of sin(x) / x dx from 0 to infinity is a classic.
Here's an algebraic approach. It does extend into the patterns of series, binomial theorem - identities, as well as the complex plane:
This may not be a complete proof or solution, but it illustrates the point.
I find this to be another decent approach towards evaluating or trying to solve it.
Setup and a few basic common principles: Not all of them may be directly used but are good and useful to keep in mind.
Slope-Intercept form of a line y = mx + b.
Slope formula: m = (y2-y1)/(x2-x1) = deltaY/deltaX = sin(t)/cos(t) = tan(t) where t is the angle theta between the line y = mx+b and the +x-axis.
Initial Conditions: m = 1, b = 0.
Constraints: b will always be 0.
Simplification: y = 1*x + 0 y = x.
Substitutions: y = mx == y = sin(t)/cos(t) * x == x * tan(t).
We can write this as sin(t) / t. The thing to recognize here is that the integration here is in relation to the angle, as opposed to the x - dx form.
We know that 90 degrees or PI/2 radians is a Right Angle. We know that, multiplying by the imaginary unit i vector has the same exact effect of rotating by 90 degrees and by multiplying any value by i^(4*N) where N is an +Integer is the same as multiplying by 1 since it rotates it by 360 degrees or 2PI radians.
Taking the graph of this function and looking for the area under the curve can be broken down into intervals based on the properties and relationships between PI/2 and i within the context of the summation of their series that converges to PI/2 or 90 Degrees.
The Series: n=0 |--> +infinity of: (2n)!! / (2n+1)!! * (1/2)^n = PI/2
The double factorial (!!) is define by 0!! = 1!! and n!! = n(n-2)!!
Then: f(t) = Series: n=0 |--> +infinity of: (-1)^n / (2n+1) * t^n
Note that f(1) = PI/4. We can take the Euler Transform of the series:
1/(1-t) * f( t / (1-t) = OuterSum: n=0 |--> +infinity { InnerSum: k=0 |-->n ( n : choose k) ( -1)^k / (2k + 1) } * t^n
Then:
Sum: k = 0 |--> n (n: choose k) (-1)^k/(2k+1) = (2n)!! / (2n + 1)!!
Proving the above just refer to proving a binomial sum identity.
We can see that:
The Integral from 0 to infinity of Sin(x)/x dx is equal to:
The Series: n=0 |--> +infinity {(2n)!!/(2n+1)!!}*(1/2)^n = PI/2.
Forgive me if there's any typos in the math... "Y.T." isn't very friendly with their parsing of comments.
Here's a link for the above Series: math.stackexchange.com/a/14116/405427
Absolutely brilliant
I wanted to stress test Feynman's method, so I used sin(ax) instead of introducing exp(-ax). What happens is, you get the integral of cos(ax) over [0,∞) which is undefined. But if you regularize it by introducing the regulator exp(-tx), then solve that equation, you find the regularized integral is atan(a/t) which goes to 𝜋/2 for all a>0.
That regularization is basically the exact thing you did right from the beginning, just with an extra step. So, it seems introducing exp(-ax) is the "canonically correct" way to use Feynman's method here.
Still feels kinda interesting to me two different choices for where to put the Feynman parameter end up giving similar results, if you grant the use of regularizing divergent terms as a tool.
is the function you used exp(-ax) just e^(-ax)? i've seen it used before but couldn't find what it was but when i did this same method i used e^(-ax) and got the same thing
Yes exp(x) is e^x
Interesting. My first time here. I look forward to more.
Wonderfully explained. Loved It!
Loved the explanation
Thanks for your interesting video.
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Your subscribers might want to see this 1:29 minutes video showing under the right conditions, the quantization of a field is easily produced.
The ground state energy is induced via Euler’s contain column analysis. Containing the column must come in to play before over buckling, or the effect will not work. The sheet of elastic material “system”response in a quantized manor when force is applied in the perpendicular direction.
Bonding at the points of highest probabilities and maximum duration( ie peeks and troughs) of the fields “sheet” produced a stable structure when the undulations are bonded to a flat sheet that is placed above and below the core material.
Some say this model is no different than plucking guitar strings. You can not make structures with vibrating guitar strings or harmonic oscillators.
th-cam.com/video/wrBsqiE0vG4/w-d-xo.htmlsi=waT8lY2iX-wJdjO3
At this time in my research, I have been trying to describe the “U” shape formed that is produced before phase change.
In the model, “U” shape waves are produced as the loading increases and just before the wave-like function shifts to the next higher energy level.
Over-lapping all frequencies together using Fournier Transforms, can produce a “U” shape or square wave form.
Wondering if Feynman Path Integrals for all possible wave functions could be applicable here too?
If this model has merit, seeing the sawtooth load verse deflection graph produced could give some real insight in what happened during the quantum jumps between energy levels.
The mechanical description and white paper that goes with the video can be found on my LinkedIn and TH-cam pages.
You can reproduce my results using a sheet of Mylar* ( the clear plastic found in some school essay folders.
Seeing it first hand is worth the effort!
Wow. That’s a great video.
Thank you bro loved it
I think this is a two-part trick method. First trick part comes from the need to “temperate” the integrand, which is swinging wildly. (Btw, we need to state, right off the bat, that a is positive.) This way we get a function, the temperated integrand, whose integral on [0,infinity) is finite. The second trick part is to use differentiation (under integral) wrt the parameter of the “temperator”.
I was upset about the dx until you fixed it! I found this derivation in high school, and loved it, and only now as an undergrad can I appreciate this technique's similarity to the Laplace or Fourier transform.
You know a guy is genius when he invents a new way to integrate!
What’s funny is Feynman had an average IQ the guy just developed extraordinary thinking techniques. So it’s possible for you and I as well!
he popularized it, clearly not invented. would not be surprised if even Euler used it
Love the bow
I was always taught , in the absence of x or ln, that e would be chosen as u in integration by parts. But it gives the same result.
Wonderful.
Very cool.
Solved easily with Laplace transform
Love the « in the a world »
a=0 has a problem. At 6:50, you show 1/a as a factor in the formula. So, if you repeat all the steps with "a" having a fixed value of zero, the whole thing breaks down at that formula..
A can be any number you choose, however when we go ahead and say a=0 is our integral, it’s more the limit as a-> 0. You’re right it could not be a=0 it should be a limit, sorry if I didn’t make that clear.
what app is this?? the one u are writing on??
Goodnotes on IOS on my ipad
Thaks a lot, however , I lost you at 17:39, integral of 0 is not 0, is it? Its some contstant.
Actually not. There he tried to find the values when 'a' approaches infinity. Then the Left side of the equation would result into zero while the right side would be -π/2 + c. That just gives us the equation c -π/2 = 0 or c = π/2. Zero is just equal to zero and he has actually integrated zero as zero.
Thank you, All I know is that int of 0 is constant as diff of constant is zero.
@@giorgibliadze1151 Yes, you’re right that the integral of 0 is a constant. When dealing with definite integrals, like the one in the video, we evaluate the integral over a specific range. This process gives a specific value, so we don’t need to add a constant. It’s similar to differentiation: when you differentiate a function at a specific point, you get a single value, not a function
I love obi wan teaching me calculus!
Haha what?
"Feynman Technique" is TH-cam's favorite moniker for "The method of integration by-parts". Just open up any elementary calculus book written before Feynman was born.
incorrect, this is a different technique involving the introduction of a completely new variable
@@Cow.cool. again this "trick" was known way before Feyman. Only physics enthusiasts with insufficient math background would refer to it as "Feynman technique", Feynman himself never claimed over such a thing. Some Feyman's pupils may have referred to it as Feyman's teqnique because they did not see this method prior to taking Feynman's lecture. He's a great scientist and had great contributions to the field of QFT, but this is not his "technique".
You took so much time in just applying by parts your way of doing is such a waste of time
Who asked
Vulgar abuse contributes nothing to the discussion. If you can't make a useful contribution please refrain from antisocial comments.
well how do you do it then genius
You might consider it to be a waste of time, but the presenter of this video didn't...
Kudos: respect and appreciation to the Content Creator!
@@robert-skibeloit’s his video…….if you don’t like it, scroll away you fossil 🦖