My Math Forum Rigorous definition of "Differential"

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 January 6th, 2017, 08:45 AM #11 Senior Member   Joined: Dec 2015 From: Earth Posts: 147 Thanks: 20 $\displaystyle dy=\sqrt{f'(C_1)dy + \int \limits_{C_1}^{z}\int \limits_{C}^{y}d^{2}y d{z}}$
 January 6th, 2017, 11:26 AM #12 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 Given f'(x) By definition, u and v are differentials if they are related by: u=f'(x)v for given x. For example, dy=f'(x)dx. If f'(x) =10, and dx=.001, dy=.01. dx=1000, dy =10000 There are properties of the differential stemming from considerations of calculus, but the above is the basic definition. EDIT If z=f(x,y), u,v,w are differentials if they are related by: w = $\displaystyle \frac{\partial z}{\partial x}u + \frac{\partial z}{\partial y}v$ Last edited by zylo; January 6th, 2017 at 11:41 AM.
January 6th, 2017, 12:42 PM   #13
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Quote:
 Originally Posted by HamedBegloo Do you think it's possible?
To improve math education in this and in many other ways? Most definitely. When the math educators of the world ask my opinion, I'll surely set them straight. Till then, the calculus students are stuck with the system we've got.

January 6th, 2017, 03:44 PM   #14
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 Originally Posted by zylo Given f'(x) If z=f(x,y), u,v,w are differentials if they are related by: w = $\displaystyle \frac{\partial z}{\partial x}u + \frac{\partial z}{\partial y}v$
If, for example, z=f(x,y) and we use the standard designation dx, dy, dz for u,v,w

$\displaystyle dz = f_{x}dx + f_{y}dy$

$\displaystyle dx = (1/f_{x})dz - (f_{y}/f_{x})dy = \frac{\partial x}{\partial z}dz + \frac{\partial x}{\partial y}dy$

$\displaystyle \frac{\partial x}{\partial z} = 1/f_{x}$
$\displaystyle \frac{\partial x}{\partial y} = - (f_{y}/f_{x})$

Unfortunately, the trend seems to be to replace useful basic concepts with unintelligble jargon which becomes the basis of unintelligible discussion for the "in" crowd.

 January 7th, 2017, 09:58 AM #15 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 w = w(x,y,z) = 0 is equation of a surface. Small excursions dx,dy,dz from a point on the surface are related approximately by $\displaystyle dw = w_{x}dx + w_{y}dy + w_{z}dz = 0$, which is equation of the tangent plane at the point. As points on the plane, the differentials dx,dy,dz can have any value. If you solve w = 0 for y = y(x,z), $\displaystyle dy = -(w_{x}/w_{y})dx-(w_{z}/w_{y})dz = \frac{\partial y}{\partial x}dx + \frac{\partial y}{\partial z}dx$ $\displaystyle \frac{\partial y}{\partial x} = -w_{x}/w_{y}$ $\displaystyle \frac{\partial y}{\partial z} = -w_{z}/w_{y}$
 January 8th, 2017, 09:55 PM #16 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 SUMMARY: $\displaystyle \Delta y = y'\Delta x + \epsilon$ dy = y'dx, any dx. Definition of derivative y' at x: y' = Lim $\displaystyle \frac{\Delta y}{\Delta x}$ Having defined y', increments $\displaystyle \Delta y= f(x+\delta x)-f(x)$, and $\displaystyle \Delta x$ are related by: $\displaystyle \Delta y = y'\Delta x + \epsilon$, Where Lim $\displaystyle \epsilon / \Delta x$ = 0. Definition of differentials dy and dx at x: dx = $\displaystyle \Delta x$ dy = y'dx Last edited by zylo; January 8th, 2017 at 10:21 PM.
 January 9th, 2017, 11:00 AM #17 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 $\displaystyle dy = \frac{dy}{dx}$dx is ambiguous. It is either an identity, b = $\displaystyle \frac{b}{a}$a, or a formula relating variables (differentials), db = b'da, where da is the independent variable (differential) and db is the dependent variable (differential), by definition, da arbitrary (.0001 or 10000 for ex) Partial Derivatives and Differentials. All d's in the following are differentials. Let u=u(x,y). Then $\displaystyle du = u_{x}dx + u_{y}dy$, The dependent variable (differential) du is a function of the variables (differentials) dx and dy, regardless of their source of variation: either independently assigned or functions of other variables. In the latter case, let x=x(v,w) and y=y(v,w). Then u = u(v,w) and $\displaystyle dx = x_{v}dv + x_{w}dw$ $\displaystyle dy = y_{v}dv + y_{w}dw$, and $\displaystyle du(v,w) = u_{x}(x_{v}dv + x_{w}dw) + u_{y} (y_{v}dv + y_{w}dw)$ where now v and w are indepentent and $\displaystyle du(v,w) = u_{v}dv + u_{w}dw$, so that: $\displaystyle u_{v}dv + u_{w}dw = u_{x}(x_{v}dv + x_{w}dw) + u_{y} (y_{v}dv + y_{w}dw)$ v,w,dv,dw independent, u dependent. Now let dw=0 and divide by dv $\displaystyle \neq$ 0, and let dv=0 and divide by dw $\displaystyle \neq$ 0, respectively. $\displaystyle u_{v} = u_{x}x_{v} + u_{y}y_{v}$ $\displaystyle u_{w} = u_{x}x_{w} + u_{y}y_{w}$ Shortcut (chain rule) Under above condtions: $\displaystyle du = u_{x}dx + u_{y}dy$ divide by dv and dw respectively: $\displaystyle u_{v} = u_{x}x_{v} + u_{y}y_{v}$ $\displaystyle u_{w} = u_{x}x_{w} + u_{y}y_{w}$

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