Trigonometry Formula to calculate positive integer n

[BMAD]

Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n.

[Pickett]

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(x²+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...

[bonanova]

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?.

[BMAD]

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

[BMAD]

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(x²+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!