The magical geometric derivative.
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I love how the geometric derivative of e^x=e
That makes sense, it's basically the limit of the common ratio of successive terms as the difference goes to 0
@@replicaacliper Exactly and the common ratio is constant by definition for any exponential with arbitrary base.
f(x)=a^x
f(x+h)/f(x)=a^(x+h)/a^(x)=a^h
taking the h'th root we get a:
(a^h)^(1/h) = a
In the general case, for any polynomial p(x), the geometric derivative of e^p(x) is e^p'(x).
All the geometric derivative essentially does is just to apply the regular derivative in logarithmic space.
@@Zxymr logarithmic space?
more generally, the geometric derivative of b^x is b.
Enjoyed your presentation. Please do one about the bigeometric derivative. It is also magical because in addition to being multiplicative, its derivative is scale-free. Each of these is a "non-Newtonian calculus", created by my husband Michael Grossman and Robert Katz decades ago! Now they are hot topics. Thanks for giving lessons.
Let me see if I understand, from what I just read. While the geometric derivative is given by
F*(x) = e^{F'(x)/F(x)}
the bigeometric derivative is given by
F°(x) = e^{xF'(x)/F(x)}
So, if we change the scale of values or arguments, meaning for r, s non zero real numbers we take the function
(r•F•s)(x) = rF(sx)
this means
F° = (r•F•s)°
Is that it???
What is the intuition behind the geometric derivative? Is it the rate at which the function increases by a factor? If the geometric derivative is 2, is it effectively doubling for each unit of change in x?
EDIT: Using your formula for the function f(x) = a^x, we get f*(x) = exp{(a^x)ln(a) / a^x} = a.
So the geometric derivative of a^x is a, and the function a^x is increasing by a factor of _a_ for each unit of x. I guess that means the intuition is that the geometric derivative is the rate at which the factor of the function increases for unit x.
Keep in mind exp(y'/y) is always positive. So "the rate at which the factor of the function increases" would mean it could decrease for another function.
Another point is to look at the function cosx, which geometric derivative would be exp(tanx). Though its true that when cosx decreases exp(tanx)>1, the opposite is also true.
@@DaniMadridDaniMineCraft It makes sense that the geo. derivative is always positive. I don't understand how a function can increase by a "negative factor". If the function is decreasing, then it is "increasing by a factor less than one". E.g. if the geo. der. is 1/2, then the function is halving for every unit of x.
The geo derivative of cosx is actually exp(-tanx). And if you plot it, it is 1 in (Pi/2, Pi), and IDK why.)
EDIT: NVM it is consistent all the way through. If a negative function is getting increasingly negative, it means its factor is increasing, and geo derivative should be >1.
"Instantaneous exponential growth rate" instead of "Instantaneous slope"
It's the instantaneous exponential growth rate, or the growth rate of the exponential function tangent to the curve at each point. We can construct the tangent exponential via the equation y = f(a) [ f*(a) ]^(x-a) at each point a.
Take a function that is bijective, say b(x) over some contiguous domain (say b(x) = ln x for Real x in the range (0, Infinity)). Define a binary operator & by (let bi be the inverse of b), Define: x & y = bi (b(x) + b(y)).
Apply to definition of derivivative = geometric derivative.
Geometric Derivative: **completed**
Exponential Derivative: *[must unlock Tetration to play this level]*
Lol jokes aside, I wonder how this could be extended to "exponential derivatives" if we already had tetration defined across real inputs...
Just don't. You don't want it
you need to be able to invert and differentiate your map in order to work with it this way, tetration is not easy to invert nor easy to differentiate and if you go further it will get less scrutable
The problem with the closed form is that you might divide by zero.
For instance, the function f(x) == 1 is not included in the family exp(exp(x+c)) because we had a ln(y) in the denominator, which can't be zero. However, f*(x) = lim(h->0) (1/1)^(1/h) = 1 = f(x).
So the multiplicative identity solves y*=y just like how the additive identity solves y’=y. That’s satisfying
This is an interesting video. I've spent a good deal of time studying a similar type of multaplicative derivative, expect with x+h replaced by a*x and 1/h replaced by 1/ln(a) in the limit as a -> 1. I think this definition gives a much nicer relation between the multaplicative and tradition derivative and allows for an infinite family of different calculuses to be defined by steadily raising or lower operators. It's a really interesting field of study that I wish would get more attention.
Just to put the pieces together you're saying your definition is like f*(x) = lim (a->1) (f(xa)/f(x))^(1/ln(a)) ?
@@Mr4thdimention That is exactly right. This definition has some nice properties that the version used in this video misses, like a rule reminiscent of the product rule for the traditional derivative in which if f(x)=e^(ln(g(x))*ln(h(x))) for some functions g(x) and h(x) then f*(x)=e^(ln(g*(x))*ln(h(x))+ln(g(x))*ln(h*(x))).
If you want to know something really cool about the geometric derivative but with the extension of the factorials(gamma functiony stuff)...
Well you know how we can express 1+2+3+...+x=f(x) and then we can express f(x)=x(x+1)/2? Then we can think of the normal derivative as a sort of "additive derivative" or "arithmetic derivative" and the arithmetic derivative f'(x)=x+1/2.
Now what if we think about this but in a multiplicative context???
Let's say g(x)=1*2*3*...*x. Then this is just x factorial or x!=g(x). And of course using the gamma function we can extend the definition to non-integer values of x, allowing us to have a geometric derivative g*(x). And the cool thing is that the difference between g*(x) and x+1/2 actually approaches 0 as x goes to infinity, so g*(x) is approximately x+1/2 or f'(x).
And so the arithmetic derivative of the function associated w 1+2+3+...+x is approximately the same as the geometric derivative of the function associated w 1*2*3*...*x, and the approximation gets better as x approaches infinity.
but we already have an infinite family of differential operators using composition of differentiable invertible maps
suppose g is a differentiable invertible map, d is a differential operator, then we can generate a differentiable operator D[f] = g^-1 of d[f] of g or D[f] = g of d[f] of g^(-1) (the choice is arbitrary)
so if we start with the usual derivative as d, then we can generate a unique D for every differentiable invertible map g, so the number of differential operators we obtain this way is the same as the number of differentiable invertible maps, which is uncountably infinite
and it goes without saying that uncountably infinite is bigger than countable (as the hyperoperation degree is a natural number)
further this easily allows us to define the corresponding antiderivative and integrals which I cannot see easily from your definition
(and notions about differentiability like smoothness, twice differentiable, holomorphic etc have easy definitions via the transformation)
btw if g is ln or g is exp depending on the choice of D formula then that obtains the multiplicative derivative here, so it is a natural choice
it has the natural interpretation as the instantaneous growth rate
another way to generate uncountably infinite differential operators is using fractional exponents on the operators which still has a neat definition for their corresponding antiderivatives and integrals
and a much lamer way to generate uncountably infinite differential operators it is to just consider a polynomial with real coefficients over the differential operator (although taylor series over operators are still very interesting)
And this would be the inverse operator of a instantanios profuct(the same way integral is a instantenios sum) and we could make a reverse operator of the product which would simply be this but instead of aproaching 0,it would be aproaching 1(aka it would be f(x+1)÷f(x))
For the last part, you probably could have just write y'/y as d/dx(ln(y)) and use the fact that e^x is the only function that is its own derivatives to get the same answer
@Khoo Zu Actually c*e^x is more general, but e^x is the only solution which satisfies the initial conditions
@@ethannguyen2754 no... U still can put lny=Ae^(x+c). Then apply the initial condition, u will get c=ln(1/A) and final solution y=e^e^x
@@khoozu7802 If you object to c*e^x being the most general solution to f’(x) = f(x), A*e*(x+c) = A*e^c * e^x, which is just an arbitrary constant times e^x, meaning all solutions of the form A*e^(x+c) are equivalent to solutions of the form C*e^x
@@ethannguyen2754 you're right. They are the same thing.
You forgot about the awesome function f(x) = 0
That was different. It's interesting to see some math that is not from the standard math education. Thanks.
You can include an infinite product as an analogy to the integral, and a "primitive product" as well.
If F(x)=e^(primitive of(Ln(f(x))), then the "geometric derivative" of F is f.
By the way if f has some roots, this infinite producte won't notice them
12:20 You could also note here that y'/y=ln(y)'. This makes it so you can very easily solve for ln(y). In general you can say that f*=exp(ln(f)') which may be easier for some problems than f*=exp(f'/f) and also IMHO makes it more clear why you call this a geometric derivative.
That's exactly what I thought, in fact I could even guess in advance that this would be the formula he would end up with. It fits right in with many similar cases of multiplicative equivalents of other concepts:
- A product is equal to e^(the sum of the logarithms)
- The geometric mean is equal to e^(the arithmetic mean of the logarithms)
- The geometric derivative is equal to e^(the derivative of the logarithms)
You are, or were, absolutely right. Man, I can't understand why he did the way he did.
Hell never heard of the geometric derivative in 5 decades!!!
Of course I know I the obvious geometric interpretation of the derivative(tangent).
I find it also very interesting, that there is a similar multiplicative analogue to the integral (Riemann products) that is even the inverse of the geometric derivative. I'm a little interested in this stuff because I made the same definitions one or two years ago but didn't yet know how to really prove any interesting properties, or even what to look for
The geometric integral of f is e^(integral lnf(x))
@@johnk7025 yep, that was pretty much the last thing I found out before I stopped
I didn't know the name tho
@@li__on6403 it is pretty interesting. I think they are to products what integrals are to sums.
@@johnk7025 yes! That's the way I discovered them First. I Just took the Definition of a Riemann sum and replaced the sum with a product and instead of multiplying by Delta x I raised the sum to its power
One would expect the geometric derivative to have x*h instead of x+h in the numerator, as well as h->1 in the limit.
If you did that then it wouldn’t be so easily related to the derivative
So it's the exponential of the logarithmic derivative; the logarithmic derivative does have an application in areas where relative change is relevant. Doesn't really surprise me too much because the geometric mean is the arithmetic mean of the logs exponentiated back.
Any applications of this? It kind of reminds me of the hazard function in survival analysis or the likelihood function in statistics, where we usually work with log-likelihoods. Thus the geometric derivative would represent the derivative of the likelihood rather than the log-likelihood.
Nice! Are there any applications for this? And, of course, does geometric integration also exist? What would the application or meaning be of that?
I suppose if there were a geometric integral, then it would just be an inverse operator on the mult. derivative, analogous to normal derivatives and integrals.
As an example, let f*(x) = F(x), then define the geometric integral, \[gint], as follows: \[gint] F(x) dx = cf(x), where c is an arbitrary constant. This is just an intuitive guess, I don't really know what the general notations and definitions are.
Since the geometric derivative is just the regular derivative in log space, the geometric integral is just the regular integral in log space
i.e. if f*(x) = y, then f(x) = e^(∫ ln(y) dx)
@@Zxymr isn't that the regular product integral?
Is there any example where geometric integrals are used for the probability of independent events (P(A^B)=P(A)P(B), (for a continuous version) ?
For the f(x) = x thing, notice that ((x+h)/x)^(1/h) = (1+h/x)^(1/h) = (1+(1/t)/x)^t as t goes to infinity which is e^t = e^(1/h).
Similar to the product rule, we get a nice quotient rule and chain rule:
(f/g)*(x) = f*(x)/g*(x)
(f o g)*(x) = f*(g(x))^g'(x)
Another nice thing is (a^f(x))* = a^f'(x)
Sum and difference formulas:
(f+g)* = f*^(f/(f+g)) times g*^(g/(f+g))
(f-g)* = f*^(f/(f+g)) / g*^(g/(f+g))
Wow! I presented this idea in my first year in college, i thought i made a discovery...
Since that it has been always the same, every time i invent something new someone from the past steals my ideas... mostly Euler.
Haha yeah I am researching the equation a^x + b^x = c since it shares similarities with the sinh and cosh function, but because its so simple I'm concerned someone else already wrote an entire paper on it lol
@@fantiscious what are you trying to do?
Do you mean integer values for x or a b c?
Do you want to relate the solution to other problem?
@@EneldoSancocho Generally real numbers for x,a,b, and c with a,b,c > 0. I'm looking at the inverse(s) of a^x + b^x, their properties, and how they can easily be computed without the use of typical methods such as newton's method and whatnot. It is inspired from Riemann's zeta function since it's a series of infinite exponentials, while this is a sum of only two, which could potentially give more insight on it. Instantaneously it would also give information to a^x + b^x = c^x since that is the same as (a/c)^x + (b/c)^x = 1.
Or Gauss. Those two ruined recreational math for the rest of us forever. /jk
will you do the harmonic derivative or logarithmic derivative next? (taking inspiration from the various types of means (geometric means, harmonic means, and logarithmic means))
I've experimented a bit with harmonic derivatives (or at least with the way I'd define them) and at some points it was resembling integrals a little bit. Like the harmonic derivative of a^x is a^x/ln(a) and for x^n it's (x^n+1)/n (which is not the integral of x^n but looks close enough).
@@seeeeeelf Oh, that's cool, I wonder if it could be useful for anything
So what's the next type of derivative? The exponential derivative? Would that involve f(x+h)^(1/(f(x)) and then inverse tetration of h? Might cause problems with exponentiation not being associative?
Also: e^x is the function whose normal derivative is itself. e^e^x is the function whose geometric derivative is itself. Perhaps e^e^e^x is the function whose exponential derivative is itself.
And thus we also should have a harmonic derivative. Will brood on that one (OK, OK and a generalised derivative, based on the generalised mean).
The sum rule for geometric derivative is: (f + g)* = exp[(f log(f*) + g log(g*))/(f + g)]
So for f(x) = x, the derivative is defined for all x, but the geom. derivative is undefined at x=0. As x=> 0+, f* => infinity, while x=> 0-, f* =0.
Actually by Michael's definition, f*(x) is not defined for x where f(x)=0 (division by zero). Zeroes of f are non-derivable per se.
would be fun if you used L'H rule with the geometric derivative instead of the regular derivative. I wonder if it would actually work.
Wow... I will look for its applications
I love watching your videos, sometimes I don't even follow along I just enjoy listening to you explain stuff in an entertaining and accessible way, thanks for that
up next: P A R T I A L G E O M E T R I C D I F F E R E N T I A L E Q U A T I O N.
I feel this Geometric Derivative is somewhat related to the Product Integral... I don't know just a hunch, should probably look into that.
They are related; they're inverses, which you could call 'the fundamental theorem of multiplicative calculus'
This lit a lot of fireworks in my head.
Can we also define an inverse action for geometric derivative (likewise differentiating and integrating)? It could be very useful for calculating some difficult integrates, I think.
Let y = f* on [a,b] where f(x) ≠ 0. Then y = e^(f'(x)/f(x)) and lny = ln'(f(x)). Then lnf(x) = int{ln(f*) dx} and f(x) = e^(int{ln(y) dx} + c) So for example for y = 1/x we have e^(int{dx/x}+c) = e^(lnx + c) = cx
Multiplicative Functions and your Number Theory videos brought me here 😊 Great explanation!
Before I watch:
It's basically the derivative of ln(f(x)) which is f'(x)/f(x)
but then you exponentiate it so exp(f'(x)/f(x))
I remember thinking I discovered something new
Another nice geom equation is y*=a which has y=ca^x as a solution
For geometric differential equations involving physical quantities, you need to worry about reconciling units (if it even makes physical sense.)
I know that this is of no value to anyone
But I derived this totally independently a few years ago when trying to work out how to describe a curve in terms of its complex curvature, and called it the multiplicative derivative
Like not even bragging or anything I just think it's a cool feeling to have a problem you're trying to solve, "invent" some math definition, and then find out some dude in California in 1972 invented it already
Isn't it just exp( d/dx ln(f(x)) ) by definition?
He shows as much, at 10:45.
Not by definition, but your result follows directly from his definition
can't wait until geometric differential equation finds application in physics, as often does for obscure math
3:54 take the log? Who does that??? Just use the definition
e := lim (1+h)^(1/h)
Ok, I am at the closed general form f*(x) = exp(f'(x)/f(x), and thinking polynomials
So when f(x) = x^2, then f*(x) = exp(2x/x^2) = exp(2/x), and f(x)=x^3 gives us exp(3/x), etc. So for the polynomial f(x) = x^n, then f*(x) = exp(n/x). Then again, that is consistent with the multiplication rule with says the derivative of f(x)g(x) = f*(x)timesg*(x) (so f*(x^2) = f*(x) times f*(x) = exp(1/x) * exp(1/x) = exp(2/x)
Unfortunately, the wrinkle comes in terms of the addition rule, which you didn't address, so I don't know how to do polynomial equations....
I would need to sit and think about how to do f*(x^2 + x)
@ 3:56, he just could have written ( ln(x+h) - ln(x) )/h inside the limit, which is just the (normal) derivative of ln( x ), which is 1/x. Same result directly, no fancy hospital rules needed.
Edit: ...just like he does in the general case @ 9:30... 😉
Edit2: I really should refrain from commenting before watching the whole thing...
The geometric derivative has to have some sort of nexus to the arithmetic mean - geometric mean inequality if there is is a such thing as mathematical justice!
Will explore that.
So what was so magical about this derivative versus the traditional difference quotient derivative that has been taught for centuries?
I've seen a proof using Taylor series that the shift operator E = e^D, where Ef(x) = f(x + 1) and D is the standard differentiation operator. Could there be any relationship to this geometric differentiation?
Very interesting, as usual. But in solving the geom DE, isn't it simpler to observe that (ln y)' = y' / y and that if y(0) = e, then ln y (0) = 1; whence the initial value prob becomes: find y given (ln y)' = In y and ln y (0) = 1
There is, of course, only one function satisfying these conditions x -> e^x; hence ln y = e^x and the desired conclusion follows. [Penn may have mentioned this; I was watching on the tube with the sound off]
i like the 2000's style mathematician pages
Hey when is your challenge video (..the one that you mentioned in your community post....) coming??
Really excited for it....
I love this video, thank you... But is there a generalization for the geometric derivatives? Ive never seen them b4 and am trying to sus out when its useful etc
Can you go one stop higher and do an exponential derivative?? I’m playing around with it and I can’t figure out how to get an indeterminate form that the derivatives have
Hm, I rather find the following "geometric differential equation" to be more interesting:
y* = 1/y or y·y* = 1 with y(0) = exp(-C), where C is a positive real number.
This is then the solution of that "geometric differential equation":
*y(x) = exp(-C·exp(-x))*
which has the following nice properties:
*(I) lim(x→-∞)[y(x)] = lim(t→∞)[exp(-C·t)] = 0.*
*(II) y(x) = exp(-C·exp(-x)) > 0 for any x∈ℝ.*
*(III) lim(x→∞)[y(x)] = exp(-C·lim(x→-∞)[exp(-x)]) = exp(-C·0) = 1.*
*(IV) y'(x) = C·exp(-x-C·exp(-x)) > 0 for any x∈ℝ,*
such that y(x) is a monotonically increasing bounded positive function with values being only between 0 and 1,
*which makes y(x) to be a continous empirical distribution function all in all,*
which satisfies the following differential equations:
y* = 1/y or y·y* = 1 with y(0) = exp(-C), where C is a positive real number
⇒ y' = -y·ln(y) with y(0) = exp(-C), where C is a positive real number,
which further means, that the probability density function y'(x) relates to some sort of an "entropy distribution of the original empirical distribution" here given the ordinary derivative function of an empirical distribution function to be a probability density function.
Just an amazing concept but trying to wrap my mind around the physical meaning of it?
Is there any sense of continuity with continued geometric differentiation? Or simple patterns?
For "normal" (arithmetic?) differentiation, taking the derivative of a polynomial always results in a lower degree polynomial. Is there a similar pattern for geometric differentiation?
How would you construct a basis for the set of once/twice/... n times differentiable functions (arithmetic or geometric)?
you can do it by analogy:
we have an invertible, differentiable map g on some domain
g = id gives us the usual derivative, g = exp or g = ln gives us the multiplicative derivative (depending on which direction of the map we consider the inverse)
so we have a map from our usual world of functions to a new world, then we apply our constraints relative to the usual derivative to obtain all our specified functions, then we take this map back to our original world to find the functions that obey this constraint for our fancy g-derivative.
so for example g-smooth functions in the usual space are those which are smooth in the g-space of functions
specifically multiplicative-smooth functions f are those where exp of f (or ln of f) is smooth
since g is invertible, if g is smooth (such as exp) then g of f is smooth if and only if f is smooth, so g-smooth and smooth are the same thing for smooth g (so multiplicatively smooth is the same thing as usual smooth)
similarly for smooth g other constraints like once/twice/etc differentiable functions are the same class of functions (so again for multiplicative derivative)
PS: "arithmetic derivative" is already a thing which is defined on rationals Q whereas the usual derivative is defined on (differentiable) R(x) (univariate functions from reals to reals (R -> R))
14:36 - probably a phrase to be avoided
Can you share a book in which the subject of the geometric derivative appears?
Why don’t you use the notable limit lim x->0 (1+alpha x)^(1/x)= e^alpha?
That was my first thought, too, but I think the way he did it generalizes more nicely to the next thing he presented.
Im late to the comments here but could you get a geometric derivative analog to a Taylor series using an infinite product? The first factor would be a function, then the geometric derivative raised to x, then the second order geometric derivative raised to either 2x or x^2, etc.
So what is the La Place Transform of the multiplicative derivative?
11:00 this formula reminds me of integrating f’/f and retrieving a log. Can anyone explain this connection
Really nice! Does this have some meaning / use / interepretation? Or is it a calculation the you just do and thats it?
Not sure about "use". But here's my stab at an interpretation. The "normal" derivative gives the instantaneous slope of a function, that is, the instantaneous change in y as measured by how much you ADD to y divided by how much you ADD to x. So maybe we can think of this as giving the "instantaneous exponential growth rate" of a function, that is, the instantaneous change in y as measured by how much you MULTIPLY into y divided by how much you ADD to x.
thank you for explaining this subject, should we assume given any 2 or 3 operators, a calculus can be created from them, and we can expect society to create a very large set of calculus to study?
💥TREMENDOUS💥
I wonder if you can generalize this further to tetration
The first hurdle would be finding a continuous and differentiable extension of tetration to the real numbers. There is no unique solution to this to my knowledge.
However, my understanding is that there is a unique analytic extension of tetration to the complex plane that can be used.
The geometric derivative is already a thing in geometric calculus, much different from this
Very cool. But why is it still f(x + h) and not f(hx)? I wonder if that would also be an interesting concept to define…
see Sean Fancher's answer
As a physicist, I am wondering whether there is any part of physics where this geometrical derivative might make our equations simpler, and therefore more intuitive.
why is it not f(x*(1 + h)) that would feel more in line with the multiplicative. wouldn't make much of a difference but would look cooler
Interesting story, very recently, I discovered this same "Geometric Derivative" in my thoughts as sort of a shower thought, and then I came across this video. Very nice how the world just aligns like that
Would be interesting to know where the geometric derivative came from. Why it's called the geometric derivative.
Like everything else, I could probably Google it.
As a math student, I always asked myself, is there a "level" of operation in math ? Like, if "+" is the first "level" , "*" the second, "^" the third... is there a way to study them, or find others "levels" like decimal level or something ? If someone know a thing that could relate to this, I would be very grateful to be informed.
Arrow notation
If + is the "first" level, them I'd suggest that succession (increasing an integer by one) is the "zeroth" level. Addition can be regarded as repeated succession.
Why is it not (f(hx)/f(x))^(1/h)?
This is what I thought it should be as well
Then exponent function would not have geometric derivative. Linear function except constant, too. Not nice!
As h goes to zero in that expression, the numerator goes to f(0) - that doesn't seem quite right, right? Seems like we want the two "f" expressions to approach being equal like they do in a derivative. Perhaps an interesting tweak would be (f(x(1+h))/f(x))^(1/h) ?
@@Mr4thdimention You could change the additive identity in h→0 to the multiplicative identity, so h→1 which would make f(hx) approach f(x), however this unfortunately doesn't play nice with the h'th root because we would get lim_{h→1} (f(hx)/f(x))^(1/h) which (if f is continuous) would approach (f(x)/f(x))^1 = 1.
Maybe you could use a different operation instead of the h'th root, perhaps a logarithm to the base h (idk I haven't thought about this)
@@schweinmachtbree1013 That sounds like it could make sense, approaching the multiplicative identity. In another comment someone suggested (1/ln(h)) for the power in this setup, since, as h goes to 1 ln(h) will go to zero. I wonder if these are equivalent or different. If they're different we sort of have a thing where we can make the "input side" additive or multiplicative AND the "output side" additive or multiplicative, which generates even more questions...!
My intuition is failing me here. Is there a graphical interpretation of the geometric derivative that mirrors the "slope of the tangent line" interpretation of the regular derivative?
Its the exponential of the derivative of ln(f(x)),with whatever it implies geometrically
I'll take a stab at it. When we learn normal derivative we learn that it gives the "instantaneous rate of change" and we form an analogy to the concept of slope of straight lines. Another concept of "rate of change" can be found in any exponential function, where we talk about the "exponential growth rate". For instance 2^x has an exponential growth rate of 2. So this is extracting an "instantaneous exponential growth rate". Is that a useful intuition?
One geometric formulation is that f*(a) = b means that the unique exponential function that is tangent to y=f(x) at x=a is of the form y=k b^x. So as Allen say, it is an instantaneous relative growth rate at x=a.
What's the chain rule for the magical geometric derivative?
i'll try a way
(e^g(x))* = g(x), this kind of feels like lacking the chainrule itself, but i think it works if g(x) != 0 in a certain domain.
so f(g(x))* = e^(f(g)' / f(g))* = f(g)'/f(g) =f'(g)*g'(x)/f(g).
I tried several ways and ended up with this result.
it's a consequnce of the ln properties.
woah i love this derivative.
i calculated (x+a)! * for postiive x, and it approximates very well to x+a+0.6 ^^, woah linearity
Where can I read more about this?
There are some more examples (the multiplicative derivatives of constant functions, linear functions, power functions, exponentials, logarithms, the gamma function, and the function x^x) at the wikipedia article "List of derivatives and integrals in alternative calculi". The multiplicative derivative doesn't have its own article but the multiplicative integral does ("Product integral") and there's a bit about the multiplicative derivative a.k.a. geometric derivative there
What is the geometric interpretation of the geometric derivative?
amazing
How would you program this for time series? :O
This is cool and all, but it is a shame you didn't talk about the inverse, the geometric/multiplicative integral. It is arguably way cooler. It provides an explanation of the origin of the name geometric. It is closely related to the geometric mean just as the standard integral is closely related to the standard mean. It leads to a continuous form for the definition of the geometric mean.
The other cool results are is it provides a duality between exterior point and interior point methods in optimization. Certain interior point methods can be constructed as the standard integral of a function. There is no natural extension to the exterior point version due to addition causing weird behavior near the intersection of two constraints. Using a geometric formulation instead avoids this issue. This is due to the duality between the two methods. The use of the the exponential and the natural log to convert between the two hint at this duality and relation
This is really interesting but is there any current application in physics or engineering or is it simply an interesting mathematical structure?
it's the instantaneous growth rate (so the exponential function tangent to the function at a point or the best local exponential approximation for the function at that point) so it's useful for anything that models exponential growth (but you will just use the usual derivative in a log-plot instead of explicitly using this derivative)
What does it measure?
What is this normally called? Geometric derivative, multiplicative derivative? You said it is somewhere on Wikipedia? I can't find ist. :-/ But isn't simply the exponentiation of the logarithmic derivative?
There's a lovely video about the product integral by Dr Pyam which is basically the inverse geometric derivative th-cam.com/video/shdK9DAiDBE/w-d-xo.html
I guess there's also a harmonic derivative?
Maybe the harmonic derivative of f should be: ((f⁻¹)')⁻¹
We can write the geometric derivative of a sum in terms of regular derivatives but there doesn’t seem to be a nice “Sum Rule”.
also, like what are the applications of this?
Nice that the geometric derivative of sin(x) is e^tan(x)
cot(x)
you made it overcomplicated. If you take log you will (get ln(x+h) -ln(x))/h what is derivative of ln(f(x))
so back to original, f*(x) = e^(ln(f(x))'
I have a question - Why is substituting y' for dy/dx abusing notation - is it simply because it mixes different ways of writing the derivative? Is there any situation this can lead to wrong answers?
It's not that substituting dy/dx is abusing notation, it's that treating dy/dx as a fraction where you can manipulate the dy and dx separately is considered an abuse of notation. It works in situations like this, but typically not when there's higher derivatives or multiple variables. Aiui, differentials as their own thing can be rigorously treated at a higher level as differential forms or even using non-standard analysis, but that's the limit of my knowledge.
@@Alex_Deam To expand:
In geometry, differentials (of independent variables) are defined as the dual basis of the tangent space (and so a basis for the cotangent space). Essentially, they are functions that eat vectors and spit out their component in the corresponding basis direction. For real numbers, there is only one direction, so dx is actually just the identity function -- dx(r) = r for all r in R. For an independent variable, we treat it as a scalar function (0-form) and use the exterior derivative to turn it into a differential (1-form). So, if y is a function of x alone, we have dy(r;p) = f'(p) dx(r), where p is the point you take the differential at.
Since dy and dx are functions whose values are real numbers, the algebra of real numbers applies to them, and so we can write dy(r;p)/dx(r) = f'(p). Here, the left-hand side is the ratio of two functions, but since that ratio is constant (taking p as a parameter), we can think of it as just a number. The "abuse of notation" is where we drop the arguments and just refer to the functions directly, and we typically take the parameter as a variable x, so we write dy/dx = f'(x). While this kind of relationship is implied by the full notation, the shorthand does not capture the full picture. Still, we are always able to manipulate this as a fraction, and there's little risk of the shorthand notation causing issues.
Higher derivative are handled much differently in geometry (because higher dimensions make derivatives a different type of object from the function itself), so there's not really an analogy to be made there. Instead, the notation used for higher derivatives is better thought of as a condensed operator notation: (d/dx) is the differentiation operator, and so (d/dx)^n f = d^nf/dx^n is the notation used for nth derivatives. This applies to first derivatives as well, but this approach makes dy/dx decidedly NOT a fraction, and so any uses in that regard are abuses. However, for calculus of real numbers, those manipulations are not necessary, only convenient. So, a rigorous real analysis course will never treat dy/dx as a fraction (and may avoid the notation altogether), and still prove all the results of the subject.
What is the use of it?
So the multiplicative derivative is identically zero. And that’s where we’re going to stop
I tried out some elementary geometric derivatives and liked that x^x becomes ex 😊
Now sentence like "[x^x]* wife" could be totally misunderstood :D
Is there a even bigger derivative?
Next video "Harmonic derivative" 😜😜😜
Mathemagical 🙂
Nice, except you didn't mention what the geometric derivative is useful for
sites.google.com/site/nonnewtoniancalculus/
But... why?
Pointless definition. You can just take the usual derivative of ln(f(x)) and get the same result.
well it just ordinary deivative... d(ln(g(x)))/dx=
(ln(g(x+dx))-ln(g(x)))/dx=ln(g(x+dx)/g(x))/dx=ln((g(x+dx)/g(x))^(1/dx)) therefor e^([ln(g(x))]')=(g(x+dx)/g(x))^(1/dx)
example: let g(x)=x [ln(g(x))]'=[ln(x)]'=1/x so (g(x+dx)/g(x))^(1/dx)=e^([ln(g(x))]')=e^(1/x) =[(x+dx)/x]^(1/dx)
edit: i just read the a index pic
i found this by myself some time ago... i don't think this is useful
asnwer=1 os isit
Hey first!!! Incredible.
Noice
@@fenilpatel417 Noise
This video was crap. He introduces a mathematical concept, and immediately dives into some details, without giving any hint of what the concept can be used for or what it means intuitively. This video seems to be mainly for advertising.
lol
Schwarz Derivative Next?????
I think he already did it, I googled schwarz derivative and his channel popped up first, go figure...