There are a lot of contents on UA-cam that explains these maths in layman's terms. For math lecture type of explanation, your contents are the best I have seen! Awesome work!!!
I guess most people think that the phrase “the Gaussian integral trick” is a well-defined and fully unambiguous notion, but it’s a special case of various more general approaches. For example, one could consider integration problems in which a multiple-integral (I’ll illustrate, like you did, with only a double integral, but there are certainly higher-dimension iterated integral examples) integrand has the following property: f(x)f(y)=g(x+y^3). In this case, one can again use differentiation to see that if such a function f is differentiable (it’s far from clear that all integrable such functions are differentiable, but let us roll with it…), and such a function g also is differentiable, then they satisfy the equations f’(x)f(y)=g’(x+y^3) and f(x)f’(y)=3y^2g’(x+y^3), and then it follows that 3y^2f’(x)f(y)=f(x)f’(y), and then the ratio of f’ to f must be a constant so that f is an exponential function but is not the Gaussian kernel. I’m skipping several steps; hopefully I’ve not miscomputed it, but my main point is that this “trick” is similar enough to “the Gaussian integral trick” to see that they’re both members of a family of approaches to computing integrals by squaring the original integral and rewriting it as an iterated integral where the new integrand has a “nice enough” form to exploit some algebraic or functional properties that may lead to a nice result for a definite integral. It should be noted that with “the Gaussian integral trick”, and many others like this, the integral in question is a definite integral, and often these “tricks” don’t do much at all for symbolic evaluation of a related antiderivative; in fact, in the case of the Gaussian integral, one has very little information about the symbolic evaluation of the corresponding antiderivative from this approach, and much better insight is gained from supplementing one’s “toolbox” beyond elementary functions by including a new function defined using the antiderivative of the Gaussian kernel. Such a function is the error function, and is well known to analysts, physicists, and engineers. In fact, in some undergraduate heat and mass transfer textbooks, tables of values of the error function were often provided (at least they were in the 1980s) just like in a trigonometry textbook of that era, tables of sines and cosines and tangents, etc, were provided, so students could use such tables to do problems that use approximate values of the error function at various arguments in its domain.
Interesting, I never wondered if the Gaussian integral trick (squaring the integral then convert to polar coordinates) is unique and indeed, it only shows up in this integral as far as I know. By the way, there is another approach to the Gaussian integral which also takes the square but does not use polar coordinates. Basically, consider the square of the integral from 0 to t. Take the derivative in respect to t, substitute x by x/t, etc. In the end, the main trick is actually squaring the original integral. So I wonder if there is another type of integral that can be easily solved using the squaring trick.
Alternatively, you could divide both sides by x y f(x) f(y) to get f'(x)/(x f(x)) = f'(y)/(y f(y)). As the LHS is solely a function of x and likewise the RHS of y, they can only be always equal if they're in fact constant. It's then easy to show that f(x) = C exp(a x²) for some constants C and a. Still looking forward to you lecturing from 201 E Bridge or 22 Gates...
Wouldn’t this also work if f(x)f(y)=g(tan^-1(y/x))=g(arctan(y/x))=g(theta) that way g is a function of theta, since in polar coordinates theta= tan^-1(y/x)?
I mean it’s just a change of variables. That’s nothing new to anyone who’s taken calculus III. Also, it’s not the only “Gaussian integral trick,” since Laplace used a double integral as well and didn’t change to polar coordinates and still got root(pi) as the answer.
Man, hope you come back. But take your time. The video is great! Love your videos. But I need to add something. This does not show that it only works for f(x)=K exp(c x^2), it just provide a solution. You can see this if you try the trick on 2^(-x^2) or any w^(-x^2). Again, hope you the best
Where did that r come from? Adding that r changes the value of the equation. I have seen its addition justified in the x and in the y directions, but I get the impression they are trying to justify something they have been taught, not something they believe.
If you're referring to the factor of r that appears in the integral after substitution, that is the Jacobian. I have a video explaining it here: ua-cam.com/video/-nH7lme2by0/v-deo.html
@@MuPrimeMath I have a degree in biochemistry. My senior research specifically focused on inputs to chemical experiments and whether the results (and conclusions) were justified. Switching a coordinate system does not justify changing the input data - ever. The most difficult class I have ever taken is Quantum Theory. I kind of remember the Jacobian from then. I do not recall it ever being utilized to justify the introduction of new data. I don't mean to be harsh, I really appreciate your explanations. Maybe you might want to take another look to see why you think that is what it is saying?
Please start uploading again. I am missing your videos man. Your content is awesome.
Wish granted :)
@@isavenewspapers8890He really came back after more than a year, lol
Broo we need u back♥️
There are a lot of contents on UA-cam that explains these maths in layman's terms.
For math lecture type of explanation, your contents are the best I have seen!
Awesome work!!!
I guess most people think that the phrase “the Gaussian integral trick” is a well-defined and fully unambiguous notion, but it’s a special case of various more general approaches. For example, one could consider integration problems in which a multiple-integral (I’ll illustrate, like you did, with only a double integral, but there are certainly higher-dimension iterated integral examples) integrand has the following property: f(x)f(y)=g(x+y^3). In this case, one can again use differentiation to see that if such a function f is differentiable (it’s far from clear that all integrable such functions are differentiable, but let us roll with it…), and such a function g also is differentiable, then they satisfy the equations f’(x)f(y)=g’(x+y^3) and f(x)f’(y)=3y^2g’(x+y^3), and then it follows that 3y^2f’(x)f(y)=f(x)f’(y), and then the ratio of f’ to f must be a constant so that f is an exponential function but is not the Gaussian kernel. I’m skipping several steps; hopefully I’ve not miscomputed it, but my main point is that this “trick” is similar enough to “the Gaussian integral trick” to see that they’re both members of a family of approaches to computing integrals by squaring the original integral and rewriting it as an iterated integral where the new integrand has a “nice enough” form to exploit some algebraic or functional properties that may lead to a nice result for a definite integral.
It should be noted that with “the Gaussian integral trick”, and many others like this, the integral in question is a definite integral, and often these “tricks” don’t do much at all for symbolic evaluation of a related antiderivative; in fact, in the case of the Gaussian integral, one has very little information about the symbolic evaluation of the corresponding antiderivative from this approach, and much better insight is gained from supplementing one’s “toolbox” beyond elementary functions by including a new function defined using the antiderivative of the Gaussian kernel. Such a function is the error function, and is well known to analysts, physicists, and engineers. In fact, in some undergraduate heat and mass transfer textbooks, tables of values of the error function were often provided (at least they were in the 1980s) just like in a trigonometry textbook of that era, tables of sines and cosines and tangents, etc, were provided, so students could use such tables to do problems that use approximate values of the error function at various arguments in its domain.
Interesting, I never wondered if the Gaussian integral trick (squaring the integral then convert to polar coordinates) is unique and indeed, it only shows up in this integral as far as I know.
By the way, there is another approach to the Gaussian integral which also takes the square but does not use polar coordinates. Basically, consider the square of the integral from 0 to t. Take the derivative in respect to t, substitute x by x/t, etc.
In the end, the main trick is actually squaring the original integral.
So I wonder if there is another type of integral that can be easily solved using the squaring trick.
I just wrote a comment about how this would work.
Gaussian integral trick? More like "Great; your vids are sick!" 🔥
What is the music that is playing at the end of the video? It's interesting.
Check the description!
great analysis
Alternatively, you could divide both sides by x y f(x) f(y) to get f'(x)/(x f(x)) = f'(y)/(y f(y)). As the LHS is solely a function of x and likewise the RHS of y, they can only be always equal if they're in fact constant. It's then easy to show that f(x) = C exp(a x²) for some constants C and a.
Still looking forward to you lecturing from 201 E Bridge or 22 Gates...
Dear mu prime, Wishing you a milestone, happiest birthday !
that's actually really cool
Wouldn’t this also work if f(x)f(y)=g(tan^-1(y/x))=g(arctan(y/x))=g(theta) that way g is a function of theta, since in polar coordinates theta= tan^-1(y/x)?
I just wrote a comment on a related point.
nice follow up to 3b1b's last video on motivating the gaussian integral trick
Nice vid.
r^2*(sec^2(θ)-tan^2(θ)^2=r^2
I might add that there is a quadratic formula in that if you look hard enough.
Thanks a lot!!❤🌼
My man has grown a lot
I mean it’s just a change of variables. That’s nothing new to anyone who’s taken calculus III. Also, it’s not the only “Gaussian integral trick,” since Laplace used a double integral as well and didn’t change to polar coordinates and still got root(pi) as the answer.
Man, hope you come back. But take your time. The video is great! Love your videos. But I need to add something. This does not show that it only works for f(x)=K exp(c x^2), it just provide a solution. You can see this if you try the trick on 2^(-x^2) or any w^(-x^2). Again, hope you the best
Note that
2^(-x^2) = (e^(ln 2))^(-x^2) = e^(-(ln 2)x^2).
Thus 2^(-x^2) is simply e^(c x^2) with c = -ln 2. The same applies to other bases.
@@MuPrimeMath You're right I did not thought this way
Instead of setting y=y0 use partial derivatives. Secondly the integral only makes sense for c1
Where did that r come from? Adding that r changes the value of the equation. I have seen its addition justified in the x and in the y directions, but I get the impression they are trying to justify something they have been taught, not something they believe.
If you're referring to the factor of r that appears in the integral after substitution, that is the Jacobian. I have a video explaining it here: ua-cam.com/video/-nH7lme2by0/v-deo.html
@@MuPrimeMath I have a degree in biochemistry. My senior research specifically focused on inputs to chemical experiments and whether the results (and conclusions) were justified. Switching a coordinate system does not justify changing the input data - ever. The most difficult class I have ever taken is Quantum Theory. I kind of remember the Jacobian from then. I do not recall it ever being utilized to justify the introduction of new data.
I don't mean to be harsh, I really appreciate your explanations. Maybe you might want to take another look to see why you think that is what it is saying?
pls collab with qncubed3