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StevensElectronicAccount 
Posted: March 16, 2011 02:48 am

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
The derivative of a curve y=f(x) is equal to (y1  y0)/(x1  x0) where x1 and x0 are brought close together.
For example to get the derivative of x^2 one starts with (x1^2  x0^2)/(x1x0) which equals x0 + x1. Then one brings x0 and x1 close and one gets 2x. If one doesn't apply the limit then one get's the divided difference of a function. I'm wondering how the rules of divided difference apply to calculus? I've already worked out the sum rule and the chain rule; they are obvious. I've also worked out the product rule. duv/dx = u0*dv/dx + v1*du/dx I've figured out the pattern for the power rule but don't know how to notate it. How do other rules like the rule for differentiating by logarithms work out? What are trigonomic divided differences? What's the divided difference of e^x? P.S. What's the opposite of the divided difference called? 
Faraday's Cage 
Posted: March 16, 2011 06:53 pm

Forum Addict ++ Group: Cleanup Taskforce Posts: 1,848 Member No.: 7,707 Joined: January 06, 2007 
I think the opposite of a divided difference is a definite integral...

Village Idiot 
Posted: March 17, 2011 01:03 am

Forum Addict ++ Group: Trusted Members Posts: 1,685 Member No.: 11,398 Joined: October 08, 2007 
Actually, it would be a summation as usually denoted by an upper case Sigma ∑.

StevensElectronicAccount 
Posted: March 18, 2011 04:10 am

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
Oh right.
As a side note I was just learning about the limit of a summation method for integration in class today. So then calculus rules for integration would apply to summations then right? I don't understand how to interpret how some of those rules apply. For example could someone explain how this rule applies to summations? uv = E v du + E u dv 
StevensElectronicAccount 
Posted: March 21, 2011 01:07 am

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
I've been working on divided differences a little more, and I'm wondering if I'm using them correctly?
Does this look right? s is position, t is time, v is velocity, a is acceleration. (s 1  s 0)/(t 1  t 0) = v 0 1 2*(v 1 2  v 0 1)/(t 2  t 0)= a 0 1 2 v 0 1 = (v 1 2)  (a 0 1 2) * (t 2  t 0) / 2 s 1  s 0 = (t 1  t 0) * ((v 1 2)  (a 0 1 2)*(t 2  t 0)/2) ds = dt * (v  a(dt)/2) s=vt  a(t^2)/2 
tashirosgt 
Posted: October 28, 2011 06:19 am

Sr. Member Group: Trusted Members Posts: 252 Member No.: 21,543 Joined: December 17, 2008 
StevensElectronicAccount,
I think you are rediscovering the mathematics that is called "The Calculus Of Finite Differences" ( rather than "divided differences"). In the Calculus Of Finite Differences, there are analogies for the differentiation formula and summation is view as "antidifferencing". A good book on the Calculus of Finite Differences was written by George Boole himself. You can probably find much information elsewhere on the web, but we can discuss the details here if you can't. 
millwood 
Posted: October 28, 2011 12:22 pm


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
more precisely, dy/dx=df(x)/dx=(f(x)f(x0))/(xx0) where x>x0. for some functions, you may have two derivatives at x where x>x0 from the left (x<x0) or x>x0 from the right (x>x0).
simple: d(uv)/dx=d((u0+du)*(v0+dv))/dx =d(u0*v0+u0*dv+v0*du+du*dv)/dx =d(u0*v0)/dx + u0*dv/dx+v0*du/dx+du*dv/dx =0+u0*v'+v0*u'+0 no need to remember anything. 

tashirosgt 
Posted: October 28, 2011 04:25 pm

Sr. Member Group: Trusted Members Posts: 252 Member No.: 21,543 Joined: December 17, 2008 
Some examples from the Calculus Of Finite Differences
For a given constant h, define the operator delta() to act on functions by: delta f(x) = f(x+h)  f(x) Define the operator incr() to be incr f(x) = f(x+h) An analogy to the product rule for derivatives is: delta( f(x)g(x)) = f(x+h)g(x+h)  f(x)f(x) = f(x+h)g(x+h)  f(x)g(x+h) + f(x)g(x+h)  f(x)g(x) = ( f(x+h)  f(x)) g(x+h) + f(x)(g(x+h)  g(x)) = delta (f(x)) incr(g(x)) + f(x) delta( g(x)) Similarly, delta( f(x)g(x)) is also = delta( g(x)) incr( f(x)) + g(x) delta( f(x)) In the calculus of finite differences, you don't let h go to zero. It stays constant.  An analogy to higher derivatives is to apply the delta operator several times to a function. Since delta(f(x)) = incr(f(x))  f(x) = (incr  1) f(x) , where we interpret "incr 1" as an operator. So delta ( delta(f(x)) ) = (incr  1) (incr 1 ) f(x) = ( (incr)(incr)  2 incr + 1) f(x) = incr ( incr(f(x))  2 incr( f(x)) + f(x) = incr ( f(x+h) )  2 f(x+h) + f(x) = f(x + 2h)  2 f(x+h) + f(x) The same formula could be derived from the definition of delta without using incr. delta( delta f(x)) = delta( f(x+h)  f(x) ) = f(x+2h)  f(x+h)  ( f(x+h)  f(x) ) etc.  The fundamental theorem of calculus has the analog: If delta( F(x)) = f(x) then sum from i = M to i = N of f(i) = F(N+h)  F(M) 
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