 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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 Calculus Calculus Math Forum

 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. Tags relation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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