Posted 19 Nov 2013 · Report post Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n. 0 Share this post Link to post Share on other sites
0 Posted 20 Nov 2013 · Report post Let f(n+1) = sec(arctan(f(n))) and define that f(0) = 0 ...that will work, however, that "definition" technically uses 2 zeros in defining the base case. I will try my best to get the meaning across: We know that sec(arctan(x)) = sqrt(x^{2}+1)...and that means sec(arctan(sqrt(x))) = sqrt(x+1) With that, we can see this pattern form: sec(arctan(0)) = sqrt(1) sec(arctan(sqrt(1))) = sec(arctan(sec(arctan(0)))) = sqrt(2) sec(arctan(sqrt(2))) = sec(arctan(sec(arctan(sec(arctan(0)))))) = sqrt(3) sec(arctan(sqrt(3))) = sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))) = sqrt(4) ...etc... From that, we can see that with enough iterations of "sec(arctan())" we will be able to produce all positive numbers (and all of their square roots, too!) using just secants, inverse tangents, and a single instance of 0: 1 = sec(arctan(0)) 2 = sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))) 3 = sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))))))))))))) etc... Maybe I'm just not thinking clearly right now...but I'm still trying to think of a good mathematical definition of that "function" that only uses one zero, however... 0 Share this post Link to post Share on other sites
0 Posted 3 Dec 2013 · Report post Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n. Bumping this thread. Does that include the argument the function takes? If not, f(x) = 0 + tan(tan-1(x)) where x is assigned the desired integer n. Or even simpler f(x) = 0 + x. No trig needed. I know that I'm missing something. Could you give an example as to form, one that is not the answer?. 0 Share this post Link to post Share on other sites
0 Posted 6 Dec 2013 (edited) · Report post this one breaks the op because it uses "floor command" and also sneaks in an implied (-1) but since you asked for a bad example, here you go: (floor(arccos(-cos(0))))! using the main branch of the arccos [0,pi]. Edited 6 Dec 2013 by BMAD 0 Share this post Link to post Share on other sites
0 Posted 6 Dec 2013 · Report post Let f(n+1) = sec(arctan(f(n))) and define that f(0) = 0 ...that will work, however, that "definition" technically uses 2 zeros in defining the base case. I will try my best to get the meaning across: We know that sec(arctan(x)) = sqrt(x^{2}+1)...and that means sec(arctan(sqrt(x))) = sqrt(x+1) With that, we can see this pattern form: sec(arctan(0)) = sqrt(1) sec(arctan(sqrt(1))) = sec(arctan(sec(arctan(0)))) = sqrt(2) sec(arctan(sqrt(2))) = sec(arctan(sec(arctan(sec(arctan(0)))))) = sqrt(3) sec(arctan(sqrt(3))) = sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))) = sqrt(4) ...etc... From that, we can see that with enough iterations of "sec(arctan())" we will be able to produce all positive numbers (and all of their square roots, too!) using just secants, inverse tangents, and a single instance of 0: 1 = sec(arctan(0)) 2 = sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))) 3 = sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))))))))))))) etc... Maybe I'm just not thinking clearly right now...but I'm still trying to think of a good mathematical definition of that "function" that only uses one zero, however... on the right track! 0 Share this post Link to post Share on other sites
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Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n.
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