Dr. Peyam, the product integral exists and was discovered in the late 1880s. It has many uses, from population modeling to solving PDE, to image analysis and even Quantum physics analysis. Thank you for introducing this to us!
@@Vishakh_Patel "Product integral" is an umbrella term that describes a few different approaches to what is essentially the same thing. For specifics, you can look into the Volterra integral, the geometric integral, or the biometric integral. These are all product integrals.
The symbol for a classical summation integral is long letter S (for sum), so the symbol for a product integral could be elongated letter P (for Peyam), then it would be possible to write the bounds
Oh my gosh. Yesterday I had the idea for this kind of integral and had exactly the same thoughts as you about notation. Surprised to know that the concept has been studied
Yeah no like that's the only reason I can't watch this channel all the time is because I have to think really hard as to what he's writing XD... It's very hard to see sometimes.
The inverse of this integral is called "geometric derivative". This definition is natural in the sense of "geometric". Then the product integral becomes also quite natural. Very interessting in this context is the "fundamental theorem" and the relation to the "law of big numbers". :)
This is very cool. I alway like to think of "increasing the order of the operations" and having all the analogies to the "normal" (i. e. first) operations. Also you should write Pintegral f(x)^dx, not *dx
23:00 the best analogy is the fact that sum integrals are distributive over summation operations, and therefore the nature of a product integral makes it distributive over multiplication operations. Sum integrals can be multiplied safely by constants, and this leads me to believe that product integrals can be exponentiated (including the features you posed with reciprocation and such) by constants
Normal Rieman integral is similar to weighted arithmetical mean, but weights do not sum up to 1, but to lenght of an interval. We do not have to use arithmetical mean we can use different meand. You used geometric mean, but we can just as well do harmonic mean = 1/(Σ_i Δx/x*_i) or do quadratic mean = sqrt(Σ_i Δx(x*_i)^2 ).
I see from comments that I've not been alone in coming up with this (idk, a few decades ago, as I recall); so I'm sure your opening statement that it "probably exists already," must be right. What I did some years ago, when I had this idea, was to just take the "∫" sign and close it at the top like a "P." Because, as at least one other commenter (Qubix) points out, "∫" is a stretched-out "S;" so what we should have for this is a stretched-out "P." Here I'll juxtapose two symbols. Let's say, like, jᵖ. [Just imagine those as overwritten.] I think my purpose at the time, was to formulate a sort of "continuous factorial," by multiplying incrementally changing terms and raising the product to the (small) increment power, then letting that increment go to 0. So what I wrote for that was: jᵖ f(x)^dx Then I realized that it can be put in terms of existing terminology this way: jᵖ f(x)^dx = e^[ ∫ ln(f(x)) dx ] Now having watched, yes, that was very good! As far as another use, I could suggest that "continuous factorial" idea: x! ≈ jᵖ[½, x+½] t^dt = e^[ ∫[½, x+½] ln(t) dt ] where I've put the individual integers, n, in the factorial product, at the centers of subintervals, (n-½, n+½); and which, when you work it through, is not a half-bad approximation for the factorial function. Fred
@@sebaitor Yes I did. You speak from ignorance. Both of me, and of what it means to "come up with something." I never claimed to have invented it; my comment even admits that. Fred
What about an exponent integral? (im not sure if it can be boiled down to the product integral again but yeah... maybe a sine integral? Idk stuffs pretty interesting)
I wonder if there are any ways to combine aspects of geometric/logarithmic derivatives and fractional/complex derivatives get more interesting results, like getting interesting, particular cases of continuous Leibniz's Rule.
I think this can be thought of as a special case of Bochner-integration, if we take V = (R+, *) as an R - vectorspace (where scalar multiplication is exponentiation). I quick-checked this in my head and if I'm not mistaken it fullfills all properties required. This would give a natural explanation as to why any measurable f must be strictly positive. Also, since V is one-dimensional it's isomorphic to our regular R, for example by taking {e} as our basis for V. I wonder whether one could try to use this fact to prove your transformation formula en this route. Schöne Grüße aus Deutschland EDIT (follow-up now that I have seen the full video): The properties shown in step 4 should definitely be able to be shown this way, too. You said this kind of integral isn't linear, but it is, just for different operations. Also the product integral having all these various properties pretty much confirms my previous suspicion.
Flewn but if you take exponentiation as scalar multiplication then your thing is no longer a vector space because 0*v=0 is not true (a real number to the power of 0 is 1). Viele Grüße aus Köln :)
Hey Dr. πm, this video has spiked my curiosity... Is there such a thing as a product derivative? That is, is there any way to "differentiate" this product integral to get back the original function? If so, what is that way?
some days ago I was thinking about derivatives with division instead of subtraction, like df/dx defined as lim h->0 (f(x+h)/f(x))^(1/h). I found out that this is very similar to this type of integral: u get e^(d(ln(f))/dx), with is e^(f'/f).
so the regular integral between a and b of f(x) gives the area below the curve of f(x). what gives the product integral in respect of f(x) ? anything tangible?
This kind of reminds me of the 3B1B's oplus operation from his triangle of power video where we try something a little different and find it is expressible in other terms.
Assume there is, let it be f(x). Then for f(x): e^int(ln(f(x)))=f(x), so int(ln(f(x)))=ln(f(x)), and so ln(f(x))=e^x, hence f(x)=e^(e^x) is your function (pretty fun stuff I gotta admit)
@@drpeyam I go to a small school, CSU East Bay, so I try to find resources from bigger schools to prep myself for grad school, so I looked at finals for all of the top UCs (berk, ucla, ucsd, etc). I remember seeing your finals back when I was looking for calc stuff and your extra credit problems were really interesting!!! On a side note, I'm actually trying to read on how product integrals apply to solving systems of linear ODEs right now. It's a book called "Product Integration with Applications to Differential Equations". I'm dying.
The product integral can also be defined for non-positive functions under some assumptions. Let f(t) = r(t) e^{i v(t)}. Then P f(t)^dt = ( P r(t)^dt ) e^{i S v(t) dt}, where P denotes the product integral and S ordinary integral. Thus, if f is complexvalued and its argument (the angle v) is assumed to be welldefined (e.g. smooth), then the product integral of f is defined.
This reminds me, 3years ago, I figured out, instead of transforming a function to a infinite powerseries (Taylor): f(x) = a0 + a1 x + a2 x² + a3 x³ + ..., you can also transform functions to a infinite product: f(x) = a1^x * a2^x² * a3^x³ * .... And there's no easy relationship between 'sum' and 'product'.
How you transform a function to a infinite product? Without talking about convergence a1*x * a2*x^2 * a^3*x^3 * ... = a1*a2*a3 * x^(1+2+3+4+5+...) = some number * x^(some power) I guess it is not meaningful to talk about "Taylor series product". Maybe with another definition there is a meaningful way.
@@easymathematik The trick is, to extend your equation in a way, that doesn't change the result of the equation, but let you transform the entire thing. It took me a while to figure it out. I even made a paper with an example. It worked very well. I numerically proved it, on a computer, for many x. But I'm just a hobbymathematician. There are much smarter guys out there ^^
could be any base, but consider the mapping: e^int(ln(f)) = 10^(log(e)*int(ln(f))) = 10^(int(ln(f))/ln(10)) = 10^(int(ln(f)/ln(10))) = 10^int(log(f)), so changing base doesn't change anything about it besides mathematical ease, since e has the special property (e^x)' = e^x and ln(x)' = 1/x without any constant multiples
You could go backward here and define another operation which is ln(integral(exp(f(x)dx))), then you wouldn’t be constrained to positive functions. I don’t know what this corresponds to though, haha. Maybe int dx^f?
regular integral: arithmetic mean of the height, multiplied by the width product integral: geometric mean of the height, raised to the power of the width
Thank you very much indeed Dr. Peyam. I was looking for a good presentation on the subject and you did it just perfect. Perhaps you will never see this, but if you do, I want to ask you a question. The Riemann sum has a geometric interpretation of calculate the area between fixed values of some function with respect to some axis or another function, on single variable calculus of curse. So, what is the geometrical interpretation of the product integral?
@@drpeyam So nice to get a response for you Dr. Peyam! I calculated the product integral of x^(dx) - 1 and the answer is x ln (x) - x +c, and we know that the derivative of that is the ln (x). So the expression x^(dx) -1 should be somehow equal or equivalent to this function. What I see for now, is that the product integral can be very useful in the study of particle decay, for having this strong relation to the ln. I believe, that it is not a matter of have a geometrical meaning, but to have a mathematical meaning in something that is already justified by some Newtonian Calculus. I am not a Mathematician, I am just a Physicist who loves to go always beyond! That is what I think that we can say for this technique. Again, thank you for answering me! Have a lovely night (if you are in the Europe region)
@@drpeyam But a negative number to a small index is a positive number slightly off one, plus a small imaginary number, right? But I think I see what you mean, ln can be multi-valued. Maybe you can just take the one with the smallest arg? Interesting to consider.
I think that the product integral is more interesting when it is used for a noncommuting algebra like a Lie group/algebra. The solution to u'(t) = A(t) u(t) can be written as u(t) = P e^{A(s) ds} u(0), where the product integral P is taken over s running from 0 to t, and the factors are sorted so that higher values of s comes to the left of lower values of s. Here A(t) belongs to a Lie algebra and e^{A(s)} to the corresponding Lie group.
As far as I know, yes, it already is a thing. I've read in Wikipedia about "geometric calculus" and couldn't really figure out how to do geometric integrals, but I did figure out that the geometric derivative of x is the xth root of e. It's a pretty interesting concept, but I honestly kinda feel a little disappointed it boils down to exponentiating e to the power of a normal integral or a normal derivative, albeit with some modification [in the case of the integral, it's the integral of the natural log of the function, and in the case of the derivative, it's the derivative divided by the function]. I think it would have been more exciting if there wasn't such an easy connection and it was instead something completely new with weirder rules, but that's just me. Edit: Apparently it's useful in probability.
If you allow f(x) to be negative by analytically continuing ln(product integral of f(x)), you could define a complex exponential derivative that has a period of 2pi. But you still can’t find the product integral/derivative of 0 because of the irremovable discontinuity.
I also thought of a similar thing, I thought of a product derivative, where instead of calculating the slope, you would calculate the "continuous factor". So e^x gives you e for example. Maybe your product integral is my product anti derivative. Instead of df/dx you'd have the dx root of qf (the quotient of f)
another idea I had was a harmonic operation, where arithmetic operation is addition (arithmetic series), geometric operation is multiplication (geometric series). So this arithmetic operation is what happens to every subsequent member of a harmonic series
@@drpeyam The video or the solution.. I mean.. I watched I few weeks ago a guy who made a video giving some explanations about "Tipical Day of a PHD student" .. and he shows some weird integral.. and also crazy.. one of those It was the integral of Sin(sin(x))dx .. he didn't solve it by the way
From linear algebra point of view: Let Integral be a linear operator on a vector space (R+, +, *) where f(x) + g(x) is defined as e^(f(x) + g(x)) and f(x) * g(x) as e^(f(x) * g(x)) ... not that useful, but still ....
Hey Dr.Peyam. NOTICE ME PLEASE! I challenge you to derive the integral formula for a reverse quotient rule in such a way that a form similar to the derivative of a quotient of two functions is equivalent to the quotient of that two functions when integrated. Thanks.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Dr. Peyam, the product integral exists and was discovered in the late 1880s. It has many uses, from population modeling to solving PDE, to image analysis and even Quantum physics analysis. Thank you for introducing this to us!
Wen Fisher where can I study it ??
is there a name for it that is more widely used?
@@Vishakh_Patel "Product integral" is an umbrella term that describes a few different approaches to what is essentially the same thing. For specifics, you can look into the Volterra integral, the geometric integral, or the biometric integral. These are all product integrals.
@@Vishakh_Patel For more general info, research multiplicative calculus.
Yes! The man volterra came up with it, and so they’re called volterra integrals
The symbol for a classical summation integral is long letter S (for sum), so the symbol for a product integral could be elongated letter P (for Peyam), then it would be possible to write the bounds
Oh my gosh. Yesterday I had the idea for this kind of integral and had exactly the same thoughts as you about notation. Surprised to know that the concept has been studied
I had the same idea for notation
i still prefer dx in the exponent
You could start writing closer to the left edge of the whiteboard to give yourself more readable space.
Portsmouth FC, yes
Yeah no like that's the only reason I can't watch this channel all the time is because I have to think really hard as to what he's writing XD... It's very hard to see sometimes.
I can't read it by the time he's half way across the board. A better camera angle and lighting would help too.
17:36
@@andrewsantopietro3526 Obscure math 😂
The inverse of this integral is called "geometric derivative". This definition is natural in the sense of "geometric".
Then the product integral becomes also quite natural.
Very interessting in this context is the "fundamental theorem" and the relation to the "law of big numbers". :)
it’s there anywhere I can read about that
All websites talk about Clifford algebras
This is very cool. I alway like to think of "increasing the order of the operations" and having all the analogies to the "normal" (i. e. first) operations.
Also you should write Pintegral f(x)^dx, not *dx
23:00 the best analogy is the fact that sum integrals are distributive over summation operations, and therefore the nature of a product integral makes it distributive over multiplication operations. Sum integrals can be multiplied safely by constants, and this leads me to believe that product integrals can be exponentiated (including the features you posed with reciprocation and such) by constants
Normal Rieman integral is similar to weighted arithmetical mean, but weights do not sum up to 1, but to lenght of an interval. We do not have to use arithmetical mean we can use different meand. You used geometric mean, but we can just as well do harmonic mean = 1/(Σ_i Δx/x*_i) or do quadratic mean = sqrt(Σ_i Δx(x*_i)^2 ).
That's so cool!
I see from comments that I've not been alone in coming up with this (idk, a few decades ago, as I recall); so I'm sure your opening statement that it "probably exists already," must be right.
What I did some years ago, when I had this idea, was to just take the "∫" sign and close it at the top like a "P."
Because, as at least one other commenter (Qubix) points out, "∫" is a stretched-out "S;" so what we should have for this is a stretched-out "P."
Here I'll juxtapose two symbols. Let's say, like, jᵖ. [Just imagine those as overwritten.]
I think my purpose at the time, was to formulate a sort of "continuous factorial," by multiplying incrementally changing terms and raising the product to the (small) increment power, then letting that increment go to 0. So what I wrote for that was:
jᵖ f(x)^dx
Then I realized that it can be put in terms of existing terminology this way:
jᵖ f(x)^dx = e^[ ∫ ln(f(x)) dx ]
Now having watched, yes, that was very good!
As far as another use, I could suggest that "continuous factorial" idea:
x! ≈ jᵖ[½, x+½] t^dt = e^[ ∫[½, x+½] ln(t) dt ]
where I've put the individual integers, n, in the factorial product, at the centers of subintervals, (n-½, n+½); and which, when you work it through, is not a half-bad approximation for the factorial function.
Fred
Very cool!
You didn't come up with anything
@@sebaitor Yes I did. You speak from ignorance. Both of me, and of what it means to "come up with something."
I never claimed to have invented it; my comment even admits that.
Fred
What about an exponent integral? (im not sure if it can be boiled down to the product integral again but yeah... maybe a sine integral? Idk stuffs pretty interesting)
@@angelmendez-rivera351 Quaternionic analysis is possible though, but quaternions don't commute.
I wonder if there are any ways to combine aspects of geometric/logarithmic derivatives and fractional/complex derivatives get more interesting results, like getting interesting, particular cases of continuous Leibniz's Rule.
This was great! It's awesome how well it related too your other videos
This reminds me of geometric integrals. Very interesting topic.
Haha, it is a geometric integral
darn good observation - I totally missed this until you pointed this out. Thanks both of U!
I think this can be thought of as a special case of Bochner-integration, if we take V = (R+, *) as an R - vectorspace (where scalar multiplication is exponentiation). I quick-checked this in my head and if I'm not mistaken it fullfills all properties required.
This would give a natural explanation as to why any measurable f must be strictly positive.
Also, since V is one-dimensional it's isomorphic to our regular R, for example by taking {e} as our basis for V. I wonder whether one could try to use this fact to prove your transformation formula en this route.
Schöne Grüße aus Deutschland
EDIT (follow-up now that I have seen the full video): The properties shown in step 4 should definitely be able to be shown this way, too. You said this kind of integral isn't linear, but it is, just for different operations. Also the product integral having all these various properties pretty much confirms my previous suspicion.
Flewn but if you take exponentiation as scalar multiplication then your thing is no longer a vector space because 0*v=0 is not true (a real number to the power of 0 is 1).
Viele Grüße aus Köln :)
the problem is the sum, we are considering the usual sum of R, if we define (a+b) as the product (a.b), then we have a vector space.
Jonas Alves I think you both misunderstood, that with R+ I meant the set of positive Reals, addition was already meant to be multiplication
Okay, I thought R was referring to real numbers, I'm accustomed to Z for integers.
Jonas Alves I meant positive reals😂
Brain-AFK me
reminds me of the integration factor for first order odes. didn't have the ln but seems like it could be useful.
Please do 100 product integrals 😂
In one take.
Hahahaha
-Fitzy-
Please do 100 product integrals!
Ask blackredpen I bet he will do it lol
@@blackpenredpen well well well, how the turntables
if c < 0 all that happens is a trivial + i pi term in the integral, and you get an extra +(b-a)i pi term
Positively delightful! What a neat little exploration.
Hey Dr. πm, this video has spiked my curiosity... Is there such a thing as a product derivative? That is, is there any way to "differentiate" this product integral to get back the original function? If so, what is that way?
Indeed, just reverse the results, I think, so ln of derivative of e^f or something like that
I thought of that a while back. Nice that you have made it way more complete than I could ever have.
some days ago I was thinking about derivatives with division instead of subtraction, like df/dx defined as lim h->0 (f(x+h)/f(x))^(1/h).
I found out that this is very similar to this type of integral: u get e^(d(ln(f))/dx), with is e^(f'/f).
so cool! this relation is so simple i love it. Seems to me like a vital/basic observation for the riemann hypothesis
so the regular integral between a and b of f(x) gives the area below the curve of f(x). what gives the product integral in respect of f(x) ? anything tangible?
This kind of reminds me of the 3B1B's oplus operation from his triangle of power video where we try something a little different and find it is expressible in other terms.
Does this formulation of a product integral also hold for complex product integrals? I see a potential application for such a thing.
SUCH CONJUGATION, SUCH FUN! GOOD STUFF MY DUDE
is there a functions which if integrated using the product integral will be itself? Like e^x? for regulär integrals
Assume there is, let it be f(x). Then for f(x): e^int(ln(f(x)))=f(x), so int(ln(f(x)))=ln(f(x)), and so ln(f(x))=e^x, hence f(x)=e^(e^x) is your function (pretty fun stuff I gotta admit)
Nice topic! Was wondering if there can exist something like the "product" derivative, an inverse function of the product integral.
Geometric derivative
@@sermuns it’s there anywhere I can read about that
All websites talk about Clifford algebras
I recall you putting a problem related to this on a midterm/final you gave for calc II back when you were a GSI at Berkeley :D
I did!!! How did you know? 😄
@@drpeyam I go to a small school, CSU East Bay, so I try to find resources from bigger schools to prep myself for grad school, so I looked at finals for all of the top UCs (berk, ucla, ucsd, etc). I remember seeing your finals back when I was looking for calc stuff and your extra credit problems were really interesting!!!
On a side note, I'm actually trying to read on how product integrals apply to solving systems of linear ODEs right now. It's a book called "Product Integration with Applications to Differential Equations". I'm dying.
The product integral can also be defined for non-positive functions under some assumptions. Let f(t) = r(t) e^{i v(t)}. Then P f(t)^dt = ( P r(t)^dt ) e^{i S v(t) dt}, where P denotes the product integral and S ordinary integral. Thus, if f is complexvalued and its argument (the angle v) is assumed to be welldefined (e.g. smooth), then the product integral of f is defined.
Is it possible to do a convolution of a product integral? In other words, what happens if you Fourier transform the product integral of a(X)b(x-xo)
That’s a really interesting question! I’m not sure
@@drpeyam thanks for your honesty. Do you know where I may look to hopefully find an answer to this question?
This reminds me, 3years ago, I figured out,
instead of transforming a function to a infinite powerseries (Taylor): f(x) = a0 + a1 x + a2 x² + a3 x³ + ...,
you can also transform functions to a infinite product: f(x) = a1^x * a2^x² * a3^x³ * ....
And there's no easy relationship between 'sum' and 'product'.
Against NAZO! Are you Euler reincarnated? Euler was the first one to find that out for Basel problem.
@@danielgates7559
I prefere Archimedes.
Euler was swiss, and swiss are usually hypocrite.
How you transform a function to a infinite product? Without talking about convergence
a1*x * a2*x^2 * a^3*x^3 * ... = a1*a2*a3 * x^(1+2+3+4+5+...) = some number * x^(some power)
I guess it is not meaningful to talk about "Taylor series product". Maybe with another definition there is a meaningful way.
@@easymathematik
The trick is, to extend your equation in a way, that doesn't change the result of the equation, but let you transform the entire thing.
It took me a while to figure it out. I even made a paper with an example. It worked very well. I numerically proved it, on a computer, for many x.
But I'm just a hobbymathematician. There are much smarter guys out there ^^
@@Handelsbilanzdefizit No problem.
Can u show me your paper? :)
I can't imagine how this product works.
I noticed if you evaluate this product integral from 0 to x of x you get (x/e)^x. I wonder if this has any connection to stirling's approximation
Interesting! Not sure
but why base e? e is like a secret lover of calculus, simple and always fun, but you could do it with any log base.
could be any base, but consider the mapping: e^int(ln(f)) = 10^(log(e)*int(ln(f))) = 10^(int(ln(f))/ln(10)) = 10^(int(ln(f)/ln(10))) = 10^int(log(f)), so changing base doesn't change anything about it besides mathematical ease, since e has the special property (e^x)' = e^x and ln(x)' = 1/x without any constant multiples
What books do you recommend reading to become like you, man?
You could go backward here and define another operation which is ln(integral(exp(f(x)dx))), then you wouldn’t be constrained to positive functions. I don’t know what this corresponds to though, haha. Maybe int dx^f?
regular integral: arithmetic mean of the height, multiplied by the width
product integral: geometric mean of the height, raised to the power of the width
Thank you very much indeed Dr. Peyam. I was looking for a good presentation on the subject and you did it just perfect. Perhaps you will never see this, but if you do, I want to ask you a question. The Riemann sum has a geometric interpretation of calculate the area between fixed values of some function with respect to some axis or another function, on single variable calculus of curse. So, what is the geometrical interpretation of the product integral?
I’m not really sure, but thank you :)
@@drpeyam So nice to get a response for you Dr. Peyam! I calculated the product integral of x^(dx) - 1 and the answer is x ln (x) - x +c, and we know that the derivative of that is the ln (x). So the expression x^(dx) -1 should be somehow equal or equivalent to this function. What I see for now, is that the product integral can be very useful in the study of particle decay, for having this strong relation to the ln. I believe, that it is not a matter of have a geometrical meaning, but to have a mathematical meaning in something that is already justified by some Newtonian Calculus.
I am not a Mathematician, I am just a Physicist who loves to go always beyond! That is what I think that we can say for this technique.
Again, thank you for answering me!
Have a lovely night (if you are in the Europe region)
I don't know if this idea makes any sense but I'd love to see you explore the integral of dx^-1 even if its just explaining why it makes no sense.
I wonder if this has any uses in the gaussian integral since the product integral of e^x from 0 to x should be similar to the integral, no?
If you consider complex numbers, I think this might kind of work for negative functions? Not sure.
I mean, maybe, but ln basically behaves weirdly for negative numbers, and things might not converge as nicely
@@drpeyam But a negative number to a small index is a positive number slightly off one, plus a small imaginary number, right?
But I think I see what you mean, ln can be multi-valued. Maybe you can just take the one with the smallest arg? Interesting to consider.
I think that the product integral is more interesting when it is used for a noncommuting algebra like a Lie group/algebra. The solution to u'(t) = A(t) u(t) can be written as u(t) = P e^{A(s) ds} u(0), where the product integral P is taken over s running from 0 to t, and the factors are sorted so that higher values of s comes to the left of lower values of s. Here A(t) belongs to a Lie algebra and e^{A(s)} to the corresponding Lie group.
Product integrals include many types, including this type shown in this video called the *geometric integral* .
As far as I know, yes, it already is a thing. I've read in Wikipedia about "geometric calculus" and couldn't really figure out how to do geometric integrals, but I did figure out that the geometric derivative of x is the xth root of e.
It's a pretty interesting concept, but I honestly kinda feel a little disappointed it boils down to exponentiating e to the power of a normal integral or a normal derivative, albeit with some modification [in the case of the integral, it's the integral of the natural log of the function, and in the case of the derivative, it's the derivative divided by the function].
I think it would have been more exciting if there wasn't such an easy connection and it was instead something completely new with weirder rules, but that's just me.
Edit: Apparently it's useful in probability.
Sungguh bagus penjelasannya. Terimakasih ya
Don't we need f to be just non-zero? Because ln(-x)=ln(x)+i*pi. (Using the principal branch and not having a problem with complex numbers).
@Zyzzyzus Ok, thanks. Than it works just for f>0 only.
Awesome, could you attach some graphical/physical meaning to this like the normal integral?
What about complex contour product integrals?
Integral sign with a letter P under it means Cauchy principal value.
If you allow f(x) to be negative by analytically continuing ln(product integral of f(x)), you could define a complex exponential derivative that has a period of 2pi. But you still can’t find the product integral/derivative of 0 because of the irremovable discontinuity.
I also thought of a similar thing, I thought of a product derivative, where instead of calculating the slope, you would calculate the "continuous factor". So e^x gives you e for example. Maybe your product integral is my product anti derivative.
Instead of df/dx you'd have the dx root of qf (the quotient of f)
And then you could have fractional and complex product derivatives lol
Thank you Blackpenredpen for sharing this!!
Awesome! :D
is there product derivative?
Yeah, there’s a video on that
That thumbnail I think this is a Flammable Math video LOL
Not enough memes in it tho.
Can you maybe do the integral of sin(dx)?
This might actually be useful in Quantum Field Theory if you did it for Path Integrals... 🤔
Great video, I really enjoyed it!
Sympathic. Thanks for all.
glad I wasnt the only one who had this idea, but I was kind of disappointed that its just e to the integral of that f
another idea I had was a harmonic operation, where arithmetic operation is addition (arithmetic series), geometric operation is multiplication (geometric series). So this arithmetic operation is what happens to every subsequent member of a harmonic series
I came up with this by myself too when thinking about factorials until I realized it already exists
This is stunning 😮
ほげぇー
ついさっき微積階数を複素数まで拡張したと思ったら、今度はべき乗？だと？
これはまた全然定義が異なりそう･･･だな？ ふんぐー
Can you make a video about how to solve the integral of sin(sin(x)) .. plis!?
I don’t think it’s possible....
@@drpeyam The video or the solution.. I mean.. I watched I few weeks ago a guy who made a video giving some explanations about "Tipical Day of a PHD student" .. and he shows some weird integral.. and also crazy.. one of those It was the integral of Sin(sin(x))dx .. he didn't solve it by the way
Both! The integral has no closed form
But it’s a great idea, I’ll consider it!
*Alright Thanks for watching*
Edit : Thanks πm sir soo much for heart.
Thanks for the heart, Respected sir.
3:45 hum. Probably just the wind.
Holy 60fps Batman when did that happen?
Now let's do the alpha product integral! 😂🎉
MAGNIFIQUE !!!!
don't forget to write "• C" after every integral without bounds
Hahaha
Rieman integration is better than Lebesgue integration in every way.
Change my mind
From linear algebra point of view: Let Integral be a linear operator on a vector space (R+, +, *) where f(x) + g(x) is defined as e^(f(x) + g(x)) and f(x) * g(x) as e^(f(x) * g(x)) ... not that useful, but still ....
Hey Dr.Peyam. NOTICE ME PLEASE!
I challenge you to derive the integral formula for a reverse quotient rule in such a way that a form similar to the derivative of a quotient of two functions is equivalent to the quotient of that two functions when integrated. Thanks.
Sounds like homework lol
Nice & strange video D peyam السلام عليكم
Nice :)
I also saw your video on product derivatives. See my comment there.
yay peyaaaam
so int_0^1 x^dx = 1/e
how neat
you have to do the limit as the lower bound approaches 0
What is an integral in the first place and why should people care
Check out my video Integration Sucks
@@drpeyam oh no no no more integrals. Don't you have a vid of girls in bikinis singing karaoke. No more mathematics please
are u single? u're pretty cute
I am single! And really? ☺️
Nice... Sorry. Sympathic is not good.
Haha, I got what you meant, sympathique :)