BMAD 64 Posted November 19, 2013 Report Share Posted November 19, 2013 Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n. Quote Link to post Share on other sites

0 Pickett 13 Posted November 20, 2013 Report Share Posted November 20, 2013 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... Quote Link to post Share on other sites

0 bonanova 84 Posted December 3, 2013 Report Share Posted December 3, 2013 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?. Quote Link to post Share on other sites

0 BMAD 64 Posted December 6, 2013 Author Report Share Posted December 6, 2013 (edited) 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 December 6, 2013 by BMAD Quote Link to post Share on other sites

0 BMAD 64 Posted December 6, 2013 Author Report Share Posted December 6, 2013 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! Quote Link to post Share on other sites

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## BMAD 64

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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