Disproving that One is Equal to Two
Date: 11/13/95 at 10:20:15 From: Anonymous Subject: Calculus (Derivatives) I've been struggling with finding the error in my logic for quite some time now, but can't seem to quite get where I messed up. If anyone could point out the error I would appreciate it. 1. Assumption: if a(x) = b(x)+c(x)+d(x)+ ... z(x) a'(x) = b'(x) + c'(x) + d'(x)+ ...z'(x) (I'm not 100% sure of this but I think its valid.) 1 = 1 = 1^2 (1 squared) 2 + 2 = 2 x 2 = 2^2 (2 squared) 3 + 3 + 3 = 3 x 3 = 3^2 (3 squared) 4 + 4 + 4 + 4 = 4 x 4 = 4^2 (4 squared) X + X + X + ... + X (X times) = X x X = X^2 (X squared) Now take the derivative of both sides of the last equations 1 + 1 + 1 + ... + 1 (X times) = 2X X = 2X 1 = 2 X isn't 0, I'm not dividing by 0, not taking any plus roots on one side and minus roots on the other? What is the flaw? Tony Chamberlain firstname.lastname@example.org
Date: 11/13/95 at 17:41:10 From: Doctor Ethan Subject: Re: Calculus (Derivatives) Hello, Well your example is very neat and the flaw is pretty subtle, but I think I can convince you. 1. Your assumption is correct but we need to clarify it a little more before we can really use it well. When you say, " if a(x) = b(x)+c(x)+d(x)+ ... z(x)" you are assuming that this is true for some range of x values so a(1) = b(1) + ... + z(1) and a(1.347) = b(1.347) + ... + z(1.347) and all the numbers in between for example. So the problem with your equation is that it is not a well defined function. It is only defined at the one integer value. What you are saying is for any x add x to itself x times. Either you have to use x^2 to mean that or a chart, as you have done. But no point on that chart is differentiable because that expression is only good at that point. I hope that this has made some sense. If it hasn't, then write back and I will try to help more. -Doctor Ethan, The Geometry Forum
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.