My Math Forum x^2 (1, 4, 9, 16, ...), and 2x (2, 4, 6, 8, ...), and 2x+1 (3, 5, 7, 9, ...)

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 December 4th, 2015, 10:42 AM #1 Newbie   Joined: Dec 2015 From: Palestine, Gaza Posts: 1 Thanks: 0 x^2 (1, 4, 9, 16, ...), and 2x (2, 4, 6, 8, ...), and 2x+1 (3, 5, 7, 9, ...) Is there a relation between the fact that the derivative of x^2 is 2x and that the difference between 1,4,9,16, ... is 3, 5, 7, 9, ...? And why is the difference always 2?
 December 4th, 2015, 11:01 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra The difference quotient $Q(f;h)$ is $$Q(f;h) = {f(x+h)-f(x) \over h} = {(x+h)^2 - x^2 \over h} = 2x + h \qquad \text{where f(x) = x^2}$$ Now, when $h=1$ we get $$Q(x^2;1) = (x+1)^2 - x^2 = 2x + 1$$ Last edited by v8archie; December 4th, 2015 at 11:05 AM.
 December 4th, 2015, 11:23 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Yes. Generally there is a strong analogy between calculus and its discrete version using difference quotients in place of derivatives.

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