@@maths_505 i've checked it out on google right after you said it has applications on statistical mechanics, i still didn't find any but i've noticed that they use a product symbol instead of a regular integral symbol. cool stuff i'll tell you that
@@wassimaabiydabro u made me shit myself i just don't completely understood the physical meaning of dx in the power of a continuous function in constant bounds but u already begun to ask like what if the bounds where dx why whyyyy whyyyyyyyy? if it exists already then.... nevermind im not able to.... im having mental breakdown 🤪
How about the derivative? I actually derived that myself a few days ago. I have started with a loose formulation of the fundamental theorem of calculus - "the sum of all small changes on a shape is the total change on the shape boundary". I've started applying it do different ideas of "change". Standard derivative defines the change as the arithmetic difference normalized with respect to the change in x. I thought to myself - why don't we use normalized ratio as the change? So I have came up with this: [f(x + dx)/f(x)]^(1/dx) (the brackets here only to show clearly the order of operations) Using this as my starting point I had written down what the fundamental theorem should look like - the product of all small ratios on a segment is ratio of the borders. Through this I have derived the integral you are showing here. So the question is - what about derivative? P.S. the fundamental theorem holds for many more notions of change. You can describe it for discreet functions (from Z or N to R usually). That's called "forward difference operator". You can also use graphs. For the graph (V, E), where V is the set of vertices and E is the set of directed edges connecting them we can describe some function f: N -> V that would describe something traveling along the graph. Than the derivative would the function df: N -> E that defines the edges on the path. Than the "integral" from a to b where a and b are natural would be the path from f(a) to f(b) that the function had took. Idk if this useful or not, but it works.
Multiplicative calculus does have some interesting applications. A Google search and a few pdfs will yield lots of wonderful concepts including the product derivative you're discussing.
I took graduate level stat. mech. and absolutely loved the math that was involved; there are some really intricate manipulations which result in simple expressions for such complex systems. I ain't never seen no exponentiated dx though, but my first thought was to take log of it.
It can probably be interpreted in terms of a geometric mean! A normal integral is the arithmetic mean of the function on an interval (it's average value) times the size of the interval. The product integral is the geometric mean of a function on an interval to the power of the size of the interval. The first one is the run-of-the-mill area under the function, the second is like a "logarithmic area" of sorts? Here's my best attempt at a geometric interpretation: This "logarithmic area" over an interval is the product of the logarithmic areas of its subintervals and for finite products, the value is the n-dimensional volume of a hyper-cuboid whose sides are the values you are multiplying. (If you divide the size of the sides by the number of things you are multiplying the hyper volume is literally just the geometric mean). Okay, so, for the product integral the value will be the limit of the n-volume of this hypercuboid as its sides become infinitesimal sides but the number of dimensions become infinite. Not an amazing picture but it's something!
The right-hand side looks like an integrating factor for a linear differential equation. If you have dy/dx+p(x)y=q(x), the integrating factor to multiply both sides by is e^int(p(x))dx, which results in the equation d/dx(ye^P(x))=q(x)e^P(x). Integrating both sides gives ye^P(x)=int(q(x)e^P(x))dx y=e^-P(x)•int(q(x)e^P(x))dx y=(1/int(p(x))^dx)int(q(x)int(p(x))^dx)dx that's not a geometric interpretation, but that's what I think of.
However, the biggest disadvantage of the integral is that only positive functions f(x) > 0 can be considered, as the ln is only defined for these. This is a very, very, very strong restriction.
@maths_505 The logarithm is also only defined holomorphically on $C \(-\infty, 0]$ in the complex and therefore the transition does not help here and it remains with my statement that we can only evaluate functions that are not negative the integral.
and if this is a path integral, then we don't need it to be holomorphic on the whole domain, just along each neighbourhood in the path (up to a constant difference, which would a constant factor?)
The exponential form of the product integral kinda reminds me of the formula for the factor of integration for first order, linear and nonhomogeneous differential equations, wonder if there's something there.
I remember I once wondered what the equivalent of Π is in continuous functions just as Σ has ∫ I thank this video for finally answering that question many years later. However, what would be the inverse function of this "product integral"?
Another really cool weird integral is the normal integral (not product) of f(x)^dx-1 this quantity approaches 0 so it would make sense when you apply an integral it reaches a number, you can actually find out that it’s equivalent to the integral of ln(f(x)) if you multiply and divide by dx and notice the inside is a limit, to be extra sure you can put it into the summation definition for the integral and evaluate it to get the same result.
So can we generalize this by saying: Given an operator X and a product integral Pi, X Pi = Pi X I.e. the operations commute? Would this be the connection to Statistical mechanics?
How exactly did you get a product from taking the integral of your function raised by dx? Shouldn’t it just be the limit of the sum from i=1 to n of f(x_i)^(delta-x_i) as n goes to infinity?
This doesn’t appear to be the same as the kinds of product integrals mentioned in Wikipedia, and I seem to recall that your previous video agreed with the Wikipedia version. Maybe I’m misremembering? To be sure, Wikipedia indicates that there are multiple versions of the notion of a product integral, but I don’t see the definition you’re using there. Here is a link to the Wikipedia article for your convenience. en.wikipedia.org/wiki/Product_integral I wish you had included a reference so I can trace its provenance Anna so we can all learn more about this. The Wikipedia article cites Volterra for their version, and that’s quite a traditional origin. Who introduced the world to this product integral you’ve discussed?
@@mtz4821 Please correct me if I’m wrong, but not once in that wikipedia article do they show the use of dx in the exponent while simultaneously using the integral symbol. Whenever dx is in the exponent, they use the product symbol instead. Whenever an integral symbol is used, dx is a factor rather than an exponent. I feel there is good reason for this.
I think this is a mistake in the video. The wikipedia page for product integrals makes no mention of turning integrals into products, and I suspect that it is not possible to do as it has been shown in the video.
@@omarsherif5659 Until the person who posted this video gives a valid reference, this video is clickbait. This video is interesting in that it shows the relationship between discrete arithmetic means and continuous integrals, and it shows how one can use this same idea to generalize discrete geometric means into a continuous analogue, but the premise of taking the integral of some function raised to the power of dx feels like clickbait to me on an otherwise already interesting video.
@Null_Simplex pretty sure all the algebra is correct but just the turning the integral into a product part is what I'm not getting. Could it have anything to do with raising it to the power of dx
@@omarsherif5659 You and I agree. Everything after he turns the integral into a product is fine. The issue is I’m pretty confident that you cannot turn an integral into a product via raising the function to the power of dx. No where in the wikipedia article for product integrals is this notation used. The integral of f(x)^dx should be the limit as n goes to infinity of the sum from i=1 to n of f(x_i)^(delta-x_i)
It makes no sense to use the integral symbol for this. The integral symbol is a vertically elongated S (and this is also how the lowercase s was written a few centuries ago) which recalls the word "sum", which is the operation that appears, discrete, in its definition. ALL types of integrals can be traced back, in some way, to a direct sum of values of the integrating function. Here, however, we have an object that arises from a product, so it should be indicated with a symbol that recalls the letter P (capital or lowercase).
This video is clickbait until you give a valid reference for your use of dx in the exponents. No where in the wikipedia article for Product Integrals is the integral symbol used while simultaneously the dx is an exponent. When the integral symbol is used, dx is a factor. When dx is an exponent, a product symbol is used instead of the integral symbol. The video is interesting in that it shows how arithmetic mean correlates with the integral, and how we can use that correlation to generalize the discrete geometric mean into a continuous analogue. But an integral with dx as an exponent is clickbait until proven otherwise.
can we say (f(x))^(d/dx)=lim_h->0((f(x+h)/f(x))^(1/h) then ln(f(x)^(d/dx))=ln(lim_h->0((f(x+h)/f(x))^(1/h))=lim_h->0((ln(f(x+h))-ln(f(x)))/h)=f'(x)/f(x) f(x)^(d/dx)=e^(f'(x)/f(x)) ... (Ac^x)^(d/dx)=e^(ln(c))=c No idea if this is legit, and the notation is questionable.
It’s not. It’s an interesting additional topic, so if you had seen it in calculus class, it would have been because your teacher was a more creative teacher who goes outside the core topics to give their students more. If they’d done that, some students would have complained, and then the administration would have come down on that teacher for “too much math in the math classes”.
Please avoid using language like wtf for describing work which is pathfinding. Even I am irritated that I am not even following. So respect. What is the emoji for respect?
You can't just casually mention that this madness has application in statistical mechanics and not be prepared to show some of them.
Something like a geometric mean. The regular integral is related to the arithmetic mean so a product integral is related to geometric mean.
@@maths_505 Fascinating. I don't think I've ever encountered the geometric mean in my university physics. It'd be so cool to see it in action.
@@mcalkis5771 there's actually a wide variety of product integrals. One Google search yeilds some really interesting papers 🔥
@@maths_505 i've checked it out on google right after you said it has applications on statistical mechanics, i still didn't find any but i've noticed that they use a product symbol instead of a regular integral symbol. cool stuff i'll tell you that
Yeah I am also very curious, as I've had a thermodynamics and statistical mechanics course, but didn't encounter these kind of integrals 😅
the integral sign ∫ is an elongated "S", for sum
using it for product integrals is a clash of flavors, like using salt in place of pepper
Great point!
eh, just slap a pi after it and it's fine. That is, unless you want a definite product integral.
@@maxvangulik1988
Why don’t we use + for both addition and multiplication in elementary school?
@@writerightmathnation9481 because that would be completely indistinct
I thought he was using a stylized “P” for the first minute until I realized it was just his really bad handwriting
I lost it at sin^dx(x) lmfao
Cool video btw
Therapist: Cursive π/2 isn’t real, he can’t hurt you, it’s all in your head
Cursive π/2:
Mathematicians: God damn, physicists! Stop treating dy/dx as a fraction, that's not how it works!
Also mathematicians:
I’m not entirely convinced he is allowed to turn an integral into a product the way he does.
well, if you define anything pretty well, you can do math with it
Engineers be like idgaf if it's right or wrong as long as it works
This is the absolutely best integral you solved. Thank you for this innovative integral and proofs.
Its inverse function is the product derivative f*(x) = lim (f(x+h)/f(x))^(1/h), which is df^(1/dx) (x), and is equal to exp(f'(x)/f(x)) 😊
NOT THE DX IN THE EXPONENT 😭😭😭😭
@@kingzenoiii this is for real- I mean for complex
@@maths_505 lmaoo
imagine dx in the limits of integration
@@wassimaabiyda we need that next
@@wassimaabiydabro u made me shit myself
i just don't completely understood the physical meaning of dx in the power of a continuous function in constant bounds
but u already begun to ask like what if the bounds where dx
why
whyyyy
whyyyyyyyy?
if it exists already then.... nevermind
im not able to....
im having mental breakdown 🤪
How about the derivative? I actually derived that myself a few days ago. I have started with a loose formulation of the fundamental theorem of calculus - "the sum of all small changes on a shape is the total change on the shape boundary". I've started applying it do different ideas of "change". Standard derivative defines the change as the arithmetic difference normalized with respect to the change in x. I thought to myself - why don't we use normalized ratio as the change? So I have came up with this:
[f(x + dx)/f(x)]^(1/dx)
(the brackets here only to show clearly the order of operations)
Using this as my starting point I had written down what the fundamental theorem should look like - the product of all small ratios on a segment is ratio of the borders. Through this I have derived the integral you are showing here. So the question is - what about derivative?
P.S. the fundamental theorem holds for many more notions of change. You can describe it for discreet functions (from Z or N to R usually). That's called "forward difference operator". You can also use graphs. For the graph (V, E), where V is the set of vertices and E is the set of directed edges connecting them we can describe some function f: N -> V that would describe something traveling along the graph. Than the derivative would the function df: N -> E that defines the edges on the path. Than the "integral" from a to b where a and b are natural would be the path from f(a) to f(b) that the function had took.
Idk if this useful or not, but it works.
Multiplicative calculus does have some interesting applications. A Google search and a few pdfs will yield lots of wonderful concepts including the product derivative you're discussing.
Big fan ❤❤❤
I took graduate level stat. mech. and absolutely loved the math that was involved; there are some really intricate manipulations which result in simple expressions for such complex systems. I ain't never seen no exponentiated dx though, but my first thought was to take log of it.
@@peterdyszel2382 check out the pinned comment, there's a whole discussion going on about that 😂
Cool, now do the integral of f(x) tetrated by dx
Is there any geometric interpretation of a product integral??? what can we evaluate with it? great video btw
It can probably be interpreted in terms of a geometric mean! A normal integral is the arithmetic mean of the function on an interval (it's average value) times the size of the interval. The product integral is the geometric mean of a function on an interval to the power of the size of the interval.
The first one is the run-of-the-mill area under the function, the second is like a "logarithmic area" of sorts? Here's my best attempt at a geometric interpretation:
This "logarithmic area" over an interval is the product of the logarithmic areas of its subintervals and for finite products, the value is the n-dimensional volume of a hyper-cuboid whose sides are the values you are multiplying. (If you divide the size of the sides by the number of things you are multiplying the hyper volume is literally just the geometric mean).
Okay, so, for the product integral the value will be the limit of the n-volume of this hypercuboid as its sides become infinitesimal sides but the number of dimensions become infinite.
Not an amazing picture but it's something!
The right-hand side looks like an integrating factor for a linear differential equation. If you have dy/dx+p(x)y=q(x), the integrating factor to multiply both sides by is e^int(p(x))dx, which results in the equation d/dx(ye^P(x))=q(x)e^P(x). Integrating both sides gives ye^P(x)=int(q(x)e^P(x))dx
y=e^-P(x)•int(q(x)e^P(x))dx
y=(1/int(p(x))^dx)int(q(x)int(p(x))^dx)dx
that's not a geometric interpretation, but that's what I think of.
@@alucs6362 thank u so much!!
@@maxvangulik1988 thanks!!
The integral is interesting. Thanks.
Absolutely awesome! I'm so happy now. I will expect next video!
Since the integral sign represents a long S, the product integral/ geometric integral is representent by a long PP
Wonderful. 👏🏻
Awesome 👍
After a long time I will see something else NEW in the integration related vedio ....!!!
Thanks for sharing with us ....!!!
nice one! rediculously awesome indeed Kamal!
I enjoyed this video indeed
that's sooo cool! especially the distributive property (if you will)
However, the biggest disadvantage of the integral is that only positive functions f(x) > 0 can be considered, as the ln is only defined for these. This is a very, very, very strong restriction.
The only restriction needed is a non zero function as we can use the principle branch of the logarithm from complex analysis.
@maths_505 The logarithm is also only defined holomorphically on $C \(-\infty, 0]$ in the complex and therefore the transition does not help here and it remains with my statement that we can only evaluate functions that are not negative the integral.
@@noctis7359 you can choose a different place to put the branch cut
and if this is a path integral, then we don't need it to be holomorphic on the whole domain, just along each neighbourhood in the path (up to a constant difference, which would a constant factor?)
Hi,
"ok, cool" : 1.42 , 6:26 , 6:45 , 8:19 ,
"terribly sorry about that" : 8:49 ,.
The exponential form of the product integral kinda reminds me of the formula for the factor of integration for first order, linear and nonhomogeneous differential equations, wonder if there's something there.
Yeah I do know the statistical mechanics meme! Now it's everyone else's turn to know it.
I remember I once wondered what the equivalent of Π is in continuous functions just as Σ has ∫
I thank this video for finally answering that question many years later.
However, what would be the inverse function of this "product integral"?
What we need here is a product derivative video
Dr. Peyam discussed the sqrt(dx) a long time ago. Is there anything you can add to this?
@@insouciantFox I shall see to it
Every calculus student's dream,
Integral of f.g = intgeral of f times integral of g!
Fun stuff 🎉
infinite products are awesome
Another really cool weird integral is the normal integral (not product) of f(x)^dx-1 this quantity approaches 0 so it would make sense when you apply an integral it reaches a number, you can actually find out that it’s equivalent to the integral of ln(f(x)) if you multiply and divide by dx and notice the inside is a limit, to be extra sure you can put it into the summation definition for the integral and evaluate it to get the same result.
And I thought I discovered this operator😅😅😅.
Ty for letting me know it's already done😊😊😊.
Ok, but he gave no references or citations, so maybe he got it from you? Please send me your preprint.
@@writerightmathnation9481well I didn't published it anywhere so there's no way for him to get it from me. Thanks for your concern😊😊😊
@@writerightmathnation9481i was just playing with a lim which turned out to be the exp(f'(x)), I just reversed it to get integral_π, nothing special.
@@writerightmathnation9481sorry for the delay in replying to your comment.🙏🏻
@@spinothenoooob6050
Which comment?
Wonder if there would be a good definition for ∫f(dx), something like ∫sin(dx)
So can we generalize this by saying: Given an operator X and a product integral Pi,
X Pi = Pi X
I.e. the operations commute? Would this be the connection to Statistical mechanics?
F = int(f^dx)
F' = f^dx
Yeah bad notation, you could integrate f(x)^dx -1, and that's just the integral of ln(f(x))dx, but for a product integral draw a p instead of the s...
on who's regard did you take the initiative to replace the summation with a product?
Can we replace the numbers a and b with the functions f and g in normal integal if so would the formula change ?
Now do product derivatives
Why didn't you use the Pi symbol (related to the product) instead of the elongated S symbol (related to the sum?
How exactly did you get a product from taking the integral of your function raised by dx? Shouldn’t it just be the limit of the sum from i=1 to n of f(x_i)^(delta-x_i) as n goes to infinity?
This doesn’t appear to be the same as the kinds of product integrals mentioned in Wikipedia, and I seem to recall that your previous video agreed with the Wikipedia version. Maybe I’m misremembering?
To be sure, Wikipedia indicates that there are multiple versions of the notion of a product integral, but I don’t see the definition you’re using there.
Here is a link to the Wikipedia article for your convenience.
en.wikipedia.org/wiki/Product_integral
I wish you had included a reference so I can trace its provenance Anna so we can all learn more about this. The Wikipedia article cites Volterra for their version, and that’s quite a traditional origin.
Who introduced the world to this product integral you’ve discussed?
Under commutative case, type 2: geometric integral. It is there.
@@mtz4821 Please correct me if I’m wrong, but not once in that wikipedia article do they show the use of dx in the exponent while simultaneously using the integral symbol. Whenever dx is in the exponent, they use the product symbol instead. Whenever an integral symbol is used, dx is a factor rather than an exponent. I feel there is good reason for this.
@@mtz4821
Not with the integral sign notation.
If you had to do this integral in the reverse direction, would you have xd in the exponent or the base?
Is it possible to request an integral?
why does the sum turn into the product for the riemann sum
I think this is a mistake in the video. The wikipedia page for product integrals makes no mention of turning integrals into products, and I suspect that it is not possible to do as it has been shown in the video.
@@Null_Simplex so this whole proof is wrong?
@@omarsherif5659 Until the person who posted this video gives a valid reference, this video is clickbait. This video is interesting in that it shows the relationship between discrete arithmetic means and continuous integrals, and it shows how one can use this same idea to generalize discrete geometric means into a continuous analogue, but the premise of taking the integral of some function raised to the power of dx feels like clickbait to me on an otherwise already interesting video.
@Null_Simplex pretty sure all the algebra is correct but just the turning the integral into a product part is what I'm not getting. Could it have anything to do with raising it to the power of dx
@@omarsherif5659 You and I agree. Everything after he turns the integral into a product is fine. The issue is I’m pretty confident that you cannot turn an integral into a product via raising the function to the power of dx. No where in the wikipedia article for product integrals is this notation used. The integral of f(x)^dx should be the limit as n goes to infinity of the sum from i=1 to n of f(x_i)^(delta-x_i)
What software do you use to create those drawings?
I wonder if Mathematic can compute this forms
У меня сразу появилась подспудная мысль, что без логарифма здесь не обойтись.
Can you state the source of the work? Thanks.
Could you please tell me what application you use for this? I've been trying to find a nice whiteboard like app for a while.
So can you take the dx root of dy?
Hello, how can i start watching your videos. I'm a class 12th student, us there anything i can understand. If you know plz tell me😢
@@jyotsanabenpanchal7271 you can keep trying to solve the problems and research stuff as you go on.
Okay sir 💪👍🏻
Wow interesante
0:15 LIES! Well okay, maybe it is a thing but it makes my head hurt.
It makes no sense to use the integral symbol for this. The integral symbol is a vertically elongated S (and this is also how the lowercase s was written a few centuries ago) which recalls the word "sum", which is the operation that appears, discrete, in its definition. ALL types of integrals can be traced back, in some way, to a direct sum of values of the integrating function. Here, however, we have an object that arises from a product, so it should be indicated with a symbol that recalls the letter P (capital or lowercase).
to be even more ridiculous in the last step you should've turned e^-sin(x) into 1/e^sin(x), so you would have A*(sin(x)/e)^sin(x)
Bad notation. The integral sign is a typographic variant of the letter S, as the sum sign Sigma also. Product should use different symbols.
but why is it an integral sign if it is not defined as a sum?
This video is clickbait until you give a valid reference for your use of dx in the exponents. No where in the wikipedia article for Product Integrals is the integral symbol used while simultaneously the dx is an exponent. When the integral symbol is used, dx is a factor. When dx is an exponent, a product symbol is used instead of the integral symbol.
The video is interesting in that it shows how arithmetic mean correlates with the integral, and how we can use that correlation to generalize the discrete geometric mean into a continuous analogue. But an integral with dx as an exponent is clickbait until proven otherwise.
Where did you meet such in stat. mech.?
See the pinned comment...there's a whole conversation going on about that 😂
Why product and not sum? Isn't integral is always a sum?
What if divide by dx?
int((f(x))/dx)
I tried that but it doesn't converge
Absolutely crazy 🤣
How does product term becomes summation term ??
Log properties
@@maths_505true….didnt know it applies to series terms too 😊
❤
Product derivatives?
Say no more fam
At 2:55, it is said: “we know that this product now turns into a sum”.. Could someone please explain why ?
It's one of logarithmic properties
@@ethanperret9644 log(xy)=log(x)+log(y)
@@maths_505 oh of course… I didn’t look at it this way.. thank you ☝🏻
can we say (f(x))^(d/dx)=lim_h->0((f(x+h)/f(x))^(1/h)
then ln(f(x)^(d/dx))=ln(lim_h->0((f(x+h)/f(x))^(1/h))=lim_h->0((ln(f(x+h))-ln(f(x)))/h)=f'(x)/f(x)
f(x)^(d/dx)=e^(f'(x)/f(x))
...
(Ac^x)^(d/dx)=e^(ln(c))=c
No idea if this is legit, and the notation is questionable.
purple is hard to read
I never learner this in calc? When is this even taught?
It’s not. It’s an interesting additional topic, so if you had seen it in calculus class, it would have been because your teacher was a more creative teacher who goes outside the core topics to give their students more.
If they’d done that, some students would have complained, and then the administration would have come down on that teacher for “too much math in the math classes”.
wWhat does that even mean?
Third
just take a log
Poor Ehrenfest. He had some demons to fight. He lost.
firsties
My legendary friend 🔥
"and it's properties"
AAAH, it should be "and its properties".
@@RandomBurfness sorry about that - I mean - terribly sorry about that.
@@maths_505 You are forgiven. xD
Avoid using words like wtf.
Please avoid using language like wtf for describing work which is pathfinding. Even I am irritated that I am not even following. So respect. What is the emoji for respect?
8th 😎 wait I mean 7th
what is the equivalent of this property in the physical world
BRO BE WATCHING THAT ONE MATH GUY BRI! 🗣️🗣️🐐💀☠️🔊🗿💥💥🗣️🗣️☠️🐐💀🐐🐐💀💀🔊🗿🗿💥💀📢💀☠️💀🗣️💥🔊🔈💀