How to integrate with respect to any function!
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- Опубликовано: 17 авг 2023
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for the case of differentiable function g, this is integral of f(x) times differential of g(x) which is g'(x)dx, which you use all the time when doing substitutions.
It's actually how we were taught substitutions in high school, which I didn't really like. For instance, if we had the integral of e^(x²)*2x, our teacher would write 2x as d(x²) and write the integral as ∫e^(x²) d(x²) which equals e^(x²) + C.
As a physicist, I would say dg/dx is a sum of delta functions, and be done with it.
You ever watch Andrew Dotson? He's had some gems, like how every matrix (even non square ones) are invertible. "If it's in physics, it's in-vertible."
(That was from a joke video with Flammable Maths.)
As an astronomer and data scientist I would also say that.
We always have underdetermined problems once we've taken our data. That's no problem to handle for linear problems with the general inverse of non-square matrices with some regularisation.
I love physics for that fact that it never ceases to amaze me… it never ceases to amaze me how much physicists disrespect and disregard the lengths mathematicians have gone to prove/discover something and simply choose to summarize it in as few, undescriptive, incomplete words that convey as little information and make things as confusing as possible…
No hate to the original commenter, but all hate to physicists who do yhos
we kinda use that all the time with differentiable functions. it's just usually called substitution in that case. like you integrate du with u=u(x)
I like the blue and pink border surrounding the white text of the definition
“Stieltjes” definitely seems like a Dutch name, in which case it would be pronounced like “steeltyes” (/stiːltjəs/ for those who can read IPA). It could be one of those cases where a Dutch last name was ported somewhere else like the USA though, like “vandenberg” and such.
It’s very neat that the floor function takes the integral to a summation, feels very intuitive!
No, not *any* g:[a,b]->R. terms and conditions apply.
You have one of the best channels for any high undergrad/graduate/ post graduate mathematician. Its great to refresh old high level concepts and learn new high level concepts in different areas of mathematics.
Thank you Professor Penn. For me the timing of this video couldn't be better. I'm building intuition on how Fredholm integral equations are solved when the kernel is singular, and afterwards how those are computationally calculated via Gaussian Quadrature. It can be said that the topics are not exactly the same, but for my currents needs/wants, they are related closely enough, which is great
Thank you so much for this video
Cool video and good timing. Just recently watched a lecture on the mathematical formulation of Quantum Mechanics, where that pops up too.
If you wanted to cover the Stiltjies integral completely, you would have to do at least 10 videos! 😂
23:45 good place to stop ❤️
@20:15 I believe the interval should be (a+(k-1), a+k-(1/N)
Which is still within an integer but a different interval than what Michael said
I've been putting the g under the d since forever, since it's much easier to mentally visualize the u dv formula
Amazing stuff.
Do you really need the integral to be from integer to integer? At first glance I think you could have any real numbers and you'd just get an extra f(a) at the end.
Many thanks for this video! An interesting and meaningful special case for g is a Brownian motion, which is continuous everywhere bu not differentiable anywhere. This leads to the non-intuitive Ito integral which has scary applications in financial engineering.
Thanks for introducing this important concept. Maybe it would have been even better if a couple minutes had been dedicated to actual applications. Also, even though the exemple detailed at length uses the beautiful floor function, perhaps the max(x, 0) would have been more useful (and easier to structure) considering the applications for neural networks.
g(x) = max{x,0} is not interesting. This is simply the usual ∫ f(x)dx, where you start from 0 instead of a, when a is negative.
Another method: notice that the "derivative" of floor(x) is just the sum of dirac distributions at every integer between a and b. So when you have discontinuities, you could account for them that way.
In other words, it’s just a sum of compactly supported differential forms (dg = g’dx), integrated over half-open intervals. Stokes’ theorem generalizes this idea to arbitrary manifolds
U can also use the Lebesgue integral with respect to the counting measure to get a summation.
My favourite non-continuous function is Thomae's function, but since that is zero except on a set of measure zero I presume taking an interval with respect to it would also give zero. Intuition, however, is dangerous in mathematics.
I think signed measure can be quite usefull with regards to the defentition of such an integral.
Hi, I've just learned about Pade Approximation. Do you have any videos on that?
Thanks for this video. It would be interesting to see solutions to an integral involving a function of “dx”, i.e. g(dx). An example I have seen is in the application of equity options securities pricing (the Black-Scholes formula). This was a case of “sqrt(dx)”.
"Stieltjes" is a Dutch name, pronounced like the English phrase "STEAL chess".
Fun fact: This integral is equivalent to the integral with respect to the signed measure G where G((a,b]) = g(b) - g(a).
*cough differential 1-form *cough.
Please next video: how to differentiate with respect to any function. I really need it. It might be one of the most useful videos of my life.
I don't mean to be rude, but wouldn't that just be the chain rule?
@@ardan981 I think it is a question about the functional derivative
df / dg = (df / dg) * (dx / dx) = df * dx / dg / dx = (df / dx) / (dg / dx) = f' / g' note this is not the same as d(f/g)/dx which would need the quotient rule
@@Yoshinoyo1 But this video doesn't talk about calculus of variations at all, which the functional derivative is about. I don't think they meant functional derivative.
Well, df/dg = (df/dx)/(dg/dx) = f'/g'. Yes, it takes a shortcut through infinitesimals, but that always gives the right answer anyway.
Interesting. The general formula is what we use for change of variables (like polar coordinates for instance). The case of the floor function is also interesting. Once you get the intuition of what's going on on the graph the proof is not that cool to write as it starts to be a little bit messy and you've actually cheated by redefining N = Nm as the value for summation. You would have to show that the behavior when b-a doesn't divide N is actually the same. It would be very technical and rather boring but needed for a full proof.
But actually, it would be much easier to show this definition of integral is compatible with Lebesgue integration. Derivative of floor function is then a distribution of Diracs at natural numbers values, hence the result.
Why is it not valid to use Nm instead of N?
The limit is N to “infinity”, and both N and Nm go to arbitrarily large integers. At a certain point the fact that N doesn’t divide the interval perfectly and Nm does becomes insignificant because the divisions (perfect or not) are so tiny any error is minuscule.
Unless I’ve missed something?
@@kikivoorburg In case of doubt, the general rule is go back to the definition. In this case, we need to remember what taking a limit means. And the definition of a limit says for all epsilon, there is a No such that for all N > No, etc. We only have proven the result for specific values of N, not all Ns.
@@fuxpremier I suppose that makes sense, there’s probably some sort of step missing from the video for a truly rigorous argument. Calling it cheating seems harsh, but I can see your argument now
its the ubiquitous scenario in probability and statistics.
What if g is decreasing or not monotone?
So I know that Cauchy has a repeated integral named after him, where you can express n nested integrals as a single integral. Has the idea been generalized to repeated integration WRT a continuous function g?
Small detail, but in 2:20 if we go from a=x0 to b=X_N, didn't we divide the interval into N+1 instead of N?
You could have saved a ton of work if you used the dirac comb instead of looking at the details of the comb. The answer becomes immediate after establishing the comb and finishing the integral as a sum of delta integrals
6:35 favourite function? Oh sh. here we go again
wtf is going on on 06:09 ??? there is a x_n* inside g' and not x_n, we don't know anything about the monotony of g so it's not that obvious to say that the sum reduces to the integral of fg'
I think this can be proved using the uniform continuity of g (Heine's theorem), but this would require g to be C1
@@giovanni1946 actually, to say that the integral exists in riemann sence, and if it has to be for all f , g' has to be continuous
For N going to infinity, the size of the intervals (x_n-1;x_n) is going to zero. And x_n* is an element of such an interval. So x_n* has to converge to x_n, which becomes equal to x_n-1 in the limit. Monotony is rather irrelevant here, as far as I can see. (You only need continuity of g', as others already have pointed out.)
@@bjornfeuerbacher5514 i said monotony as a sufficient condition to see it as an obvious conclusion, if it s not the case, and if we only have the continuity of g' the result is not obvious at all to state it without even mentioning it
@@xaxuser5033 Sorry, I really don't understand your problem. I explained how the result follows, and that line of reasoning is totally obvious.
You only consider a subsequence (mN instead of N) of the sequence of Riemann sums, so you cannot be sure that the integral actually exists
Yeah, and I think it turns out that the integral doesn't exist if f isn't continuous at the integers, but this derivation is more of an "it can't exist and be anything other than this" calculation, rather than exploring when it exists.
for the continuous case you can simply multiply the integral by 1=dx/dx so you get \int f(x) dg * dx / dx = \int f(x) dg / dx * dx = \int f(x) g'(x) dx
Differential geometers are crying at your comment
I have always wondered what is the graphical meaning of the integration of a function with respect to another function analogous to what we usually say about a ordinary integral with respect to x. I would be grateful if someone tell me.
make a plot with f(x) on the y-axis and g(x) on the x-axis. This integral then has the usual interpretation of area under the curve
@@praharmitra What do you mean exactly by "g(x) on the x-axis"? Because simply relabeling the x-axis to say "g(1), g(2)" etc. doesn't give any more geometrical intuition as the definition does.
f(x)dx means to multiply f(x+delta-x)delta-x and taking the limit for delta-x to 0, roughly speaking. So in the end adding up rectangles that re infinitly small in width and have the height of f(x). But for this "funtional" integration the width of the rectangle "changes" like the function while still being infinitesimal small. f(x)d(sin(x)) would mean the infinitesimal width of the rectangle "wobbles". This is how I imagine it graphially. But if that has any deeper meaning or where it is useful: no idea.
@@praharmitra you mean that instead of selecting x from [a,b] as x-axis I should replace it with g(x) from [g(a), g(b)]. yes that's a way to look at it. But then g(x) should by any function so why we always restrict it to be a monotonically increasing function while defining Stieltjes Integral.
@@johannmeier6707 actually this created another question for me... how would you imagine the geometrical significance of taking the derivative of a function f with respect to another function g?
By your similar approach, could it be the slope of the function f with respect to g?( I don't know how to even think about that)
No mystery here derivative of a floor function is just a sum of chronicer deltas at all integers
g'(x) = dg /dx, so replace dg = g'(x) dx, it is that simple...
I use this trick of integrating f(x) wrt g(x) all the time Micheal but my peers deem it too outlandish until I justify it with a Substitution. They don't seem to share the intuition that when we Substitute we are basically integrating with respect to some function instead. (I do it with nice functions which are cont and diff, but your video with the floor function now that's what makes your vids enticing and exciting 💫)
My ways don't always sync with the standard rigourous ways but I make sure I know what I am doing when I'm doing something unconventional
int f(x) dg = int f(x) g'(x) dx. simple as that. we learn differentials for a simple reason