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Leibniz Notation

by Moose352
Tags: leibniz, notation
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dextercioby
#19
Jan4-05, 09:29 AM
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Quote Quote by danne89
Hey! How's about the second derivate in this notation. Is this right:
[tex]\frac{d(dy/dx)}{dx} = \frac{ \frac{d(dy)}{d(dx)}}{dx} =\frac{d^2y}{d^2y} \frac{1}{dx}= \frac{d^2y}{d^2dx^2}[/tex]
No.It should be:
[tex] \frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{dx^{2}} [/tex]

Your notation makes no sense.
The derivative of "n"-th order wrt to "x" of the function "y" in the notation of Joseph Louis Lagrange is
[tex] y^{(n)}(x) [/tex]
In the notation of Gottfried Wilhelm Leibniz the same "animal" is
[tex] \frac{d^{n}y(x)}{dx^{n}} [/tex]
and it can be seen as appying the operator [itex] \frac{d}{dx} [/itex] "n" times on the function "y(x)".

Daniel.
danne89
#20
Jan4-05, 09:38 AM
P: 181

Hmm. I've a hard time understanding this. Can you point me in right direction?
dextercioby
#21
Jan4-05, 09:53 AM
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Okay,let's assume for simplicity that our functions depend only on variable and they are infinite times derivable on an arbitrary interval.
Then,we adopt Lagrange's notation and we say that the first order derivative of the function called y(x) is [tex] y'(x) [/tex].We can look on it as applying a linear operator called derivative which we'll denote by "'"(prime).So applying this operator on the function y(x),we get the derivative of the function "y(x)":[itex] y'(x) [/tex].The same operator (called 'derivative') is denoted by G.W.Leibniz as [itex] \frac{d}{dx} [/itex],since the result is the same (namely the first derivative of 'y' wrt to 'x'),no matter the notation,we can say
[tex] y'(x)=\frac{d}{dx}y(x)=\frac{dy(x)}{dx} [/tex]

For the second derivative,u apply this operator on the first derivative and u get,in the Lagrange notation:
[tex] [y'(x)]'=y''(x)=y^{(2)}(x) [/tex]
,and in the Leibniz notation
[tex] \frac{d}{dx}[\frac{dy(x)}{dx}]=\frac{d^{2}y(x)}{dx^{2}} [/tex]

You equate the results,since they represent the same thing;the second-order derivative of the function "y(x)" wrt to "x".
[tex] y^{(2)}(x)=\frac{d^{2}y(x)}{dx^{2}} [/tex]

And the same u do for every order of the derivative.For"n"
[tex] y^{(n)}(x)=\frac{d^{n}y(x)}{dx^{n}} [/tex]

Daniel.
Calculuser
#22
Feb6-12, 05:42 AM
P: 26
"dx" is inifinitesimal but I'm confused that it's dx>0 or dx<0 ??
chiro
#23
Feb6-12, 05:48 AM
P: 4,573
Quote Quote by Calculuser View Post
"dx" is inifinitesimal but I'm confused that it's dx>0 or dx<0 ??
dx is usually considered a positive quantity because we are considered the tangent as the function changes as a result of positive change (ie delta > 0 which means dx > 0).

You could make dx negative, but by doing that you have to adjust the other definitions even if you want to consider how the tangent changes when you are going 'backwards' (dx < 0).

In numerical calculus we do have situations where we consider these kinds of things, but for normal calculus its almost always assumed that dx is an infinitesimal quantity with dx > 0
Calculuser
#24
Feb6-12, 06:36 AM
P: 26
Okay, thanks.

But in this notation [itex]\int^{b}_{a}{f(x).dx}[/itex] I know it means area under the curve on interval [a,b] what if I take dx<0 how can I explain it? I mean I can take limit (lim [itex]\frac{Δy}{Δx}[/itex] as x approximates either side of x-axis (Δx→0[itex]^{+}[/itex] and Δx→0[itex]^{-}[/itex]) I realize that if dx>0 or dx<0 at derivative of a function but I'm not sure about integral??
lavinia
#25
Feb6-12, 07:37 AM
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P: 1,716
A differentiable function can be approximated by the first term in its Taylor series

y = y0 + dxy'

This approximation gets increasingly accurate as dx approaches zero.

So the ratio (y-y0)/dx is arbitrarily close to y'

I think of dy as the change in y for a small change in x,dx. In physics books you will find this way of looking at it as well.
Calculuser
#26
Feb6-12, 10:53 AM
P: 26
In Physics, we know that what "Δx" means. (Δx=x[itex]_{2}[/itex]-x[itex]_{1}[/itex]) It can also be Δx>0 or Δx<0.In Calculus, Leibniz Notation shows dx=Δx and dy≠Δy I got that part why it's so.But if "Δx" can be both Δx>0 and Δx<0, for dx and dy it must be the same (dx,dy>0 or dx,dy<0) right??

Finally, in my opinion.We use from right and from left derivative to recognize that the function is differentiable.If we have a differentiable function, don't need to do this step.Therefore, Leibniz thought that "dx" infinitesimal number (dx=[itex]\frac{1}{∞}[/itex]>0) and used it a differentiable funtion, so that he didn't need to approximate the funtion to observe whether it's differentiable or not and he admitted his notation to find derivative of a funtion which doesn't need to be observed whether it's differentiable or not.

Is that all right?
lavinia
#27
Feb6-12, 12:14 PM
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P: 1,716
Quote Quote by Calculuser View Post
In Physics, we know that what "Δx" means. (Δx=x[itex]_{2}[/itex]-x[itex]_{1}[/itex]) It can also be Δx>0 or Δx<0.In Calculus, Leibniz Notation shows dx=Δx and dy≠Δy I got that part why it's so.But if "Δx" can be both Δx>0 and Δx<0, for dx and dy it must be the same (dx,dy>0 or dx,dy<0) right??

Finally, in my opinion.We use from right and from left derivative to recognize that the function is differentiable.If we have a differentiable function, don't need to do this step.Therefore, Leibniz thought that "dx" infinitesimal number (dx=[itex]\frac{1}{∞}[/itex]>0) and used it a differentiable funtion, so that he didn't need to approximate the funtion to observe whether it's differentiable or not and he admitted his notation to find derivative of a funtion which doesn't need to be observed whether it's differentiable or not.

Is that all right?
dx and dy mean deltas for sufficiently small dx. delta by itself does not imply the increasingly accurate estimation of dy from dx. The small d implies the existence of an unobserved - in some sense transcendental quantity - which is the scale factor that translates dx into dy. I imagine that this is Leibniz's idea of derivative. Perhaps Mathwonk an comment on that.
Calculuser
#28
Feb6-12, 02:33 PM
P: 26
dx=Δx??
chiro
#29
Feb6-12, 08:02 PM
P: 4,573
Quote Quote by Calculuser View Post
dx=Δx??
In terms of things like Riemannian integration and standard differentials we consider that dx is the limit of Δx going to zero, but it never actually becomes zero because if it did we would get absolute nonsense in terms of answers and analysis.


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