Math Puzzle [num 17]

...Wouldn't you just press cosine? Unless there's something I'm not getting here..
Edit: I see your edit. Will ponder now.

Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.

I can't vouch for the truth/falsehood of the above. But if you can create a system for 1/a, then simply typing 'b' before the operation should yield the desired fraction.

barccot(tan(a)) = b/a

Unless it's cheating to use the implied multiplication

As I stated above, sqrt(a/1) is possible.
But sqrt(a/2) is also possible.
If a is divisible by 2, then it just reduces to sqrt((a/2)/1), this gives us sqrt(2/2), sqrt(4/2), forever
If a is not divisible by 2, well
apply operation 1 to sqrt(2/1) to get sqrt(1/2)
apply operation 5 to that repeatedly to get sqrt(3/2), sqrt(5/2), forever

Now let us induct on n, using the same process.
Say we can get sqrt(a/1), sqrt(a/2), sqrt(a/3), ... ,sqrt(a/n) for any n
then, take sqrt((n+1)/1), sqrt((n+1)/2), ..., sqrt((n+1)/n) and apply operation 1 to them all
this gives sqrt(1/(n+1)), sqrt(2/(n+1)), ..., sqrt(n/(n+1)).
If we apply operation 5 to all of these, we can get sqrt(a/(n+1)) for any n.

But this gives us the square root of any rational number, so we can obtain any m/n by obtaining sqrt(m^2/n^2).

You get sqrt(2/2), sqrt(4/2), sqrt(6/2), etc. because they are equal to sqrt(1), sqrt(2), etc.

You get sqrt(2/3) by flipping sqrt(3/2)

You can't. You don't have an "=" button.

Pressing the operator also shows the answer. Old school way.

oh yeah that's right...we've filled in a/1 and a/2, and the different values for (a mod 3)/3 can be obtained by flipping previous values. so 1/3 can be gotten from flipping 3/1, and 2/3 can be gotten from flipping 3/2, and therefore a/3 is filled. and this pattern stacks on top of itself recursively.

just wanted a little clarification, sorry. I missed that little jump in the logic.

0: already exists

begin with cos 0 = 1.

now, notice that

f(x) = cot(arctan(x)) = 1/x
g_1(x) = cos(arctan(x)) = 1/sqrt(x^2+1)
g_2(x) = sin(arctan(x)) = x/sqrt(x^2+1)
h_1(x) = tan(arccos(x)) = sqrt(1-x^2)/x
h_2(x) = tan(arcsin(x)) = x/sqrt(1-x^2)
h_3(x) sin(arccos(x)) = sqrt(1-x^2)

so we can get any composition of these functions, applied to 1 or 0. call a number that can be reached by a sequence of keypresses "good." If n is good, so is 1/n, since 1/n = f(n). We will show that all numbers x such that x^2 is rational are good; this clearly implies the problem statement.

we only need to show that all rationals q less than 1 (and 1 is good) are good, since every rational greater than 1 is the reciprocal of a rational less than 1.

starting at 1, we find g_2(1) = 1/sqrt(2), g_2(g_2(1)) = 1/sqrt(3), and by induction, if g_2^(k)(1) = 1/sqrt(k+1) [where ^(k) means the function is composed k times], then g_2^(k+1)(1) = g_2(1/sqrt(k+1)) = (1/sqrt(k+1)) / (sqrt( (k+2)/(k+1) ) ) = 1/sqrt(k+2).

so all 1/sqrt(n), n a positive integer, can be reached.

furthermore, we find that all g_1(1/sqrt(n)) = sqrt(n/(n+1)) can be achieved.

Lemma. If sqrt(a/b) is good, so is sqrt(a/(ka+b)) for all nonnegative integers k.

Proof. Notice that g_2(sqrt(a/b)) = sqrt(a/(a+b)); by composing this repeatedly, by induction, we can achieve all sqrt(a/(ka+b)) for nonnegative integer k.

next, we use strong induction on the numerator of sqrt(a/b) to show that all square roots of rationals are good.

We can already construct all sqrt(1/b), so the base case is true.

Assume that for k = 1, 2, ..., n-1, sqrt(k/j) is good for all integers j.

Now, we need only show that all sqrt(n/j) are good, which would complete the induction. If gcd(j,n) = g > 1, then we can cancel g from the top and bottom and get a numerator between 1 and n inclusive, so the statement is true. Hence, we only need to consider cases where gcd(j,n)=1.

Otherwise, let j' be the smallest positive integer so that (j-j' ) is divisible by n; clearly j'<= n, and j' is not n because we don't need to consider when gcd(j,n)>1. Then sqrt(j'/k) is good by the induction hypothesis, so f(sqrt(j'/k)) = sqrt(k/j' ) is also good. Then by the lemma, we can construct all sqrt(k/(j'+k)), so we get all denominators that are j' mod k.

Since we have considered all residues mod k, the denominator can therefore be any positive integer, and the induction is complete.

method 1:
tan(cot-1(n)) = 1/n
so for any n=1/(a/b) you can also get to n=a/b

method 2:
cos(tan-1(n))) = 1/sq(n^2+1)
combined with method 1
tan(cot-1(cos(tan-1(sq(a)) = sq(a+1)
starting with 0, this pattern allows for all positive integers - but only for integers for now

now for some modulus:
some x where
x mod b is equivalent to a mod b
(x/b can be expressed as x/b+C where C is an integer)
x^2 mod b^2 is equivalent to x^2 mod b^2
also where
x mod b is equivalent to b - (x mod b)
x^2 mod b^2 is equivalent to x^2 mod b^2
this is less important though

in these cases n=x/b can be express as sq(x^2/b^2 + D) where D is some integer
combined with method 2, the conclusion is that given a/b you can get to any n=a/b+C where C is a positive integer

anyways the final piece of the puzzle is continued fractions where we want to express all rational numbers as
1/1/(1/ (a +C)+D) +E)....

I haven't gotten there yet.