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Math Puzzle [num 17]

Blogs > evanthebouncy!
Post a Reply
1 2 Next All
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
Last Edited: 2010-08-18 19:58:37
August 18 2010 19:51 GMT
#1
It's been awhile huh!! Here goes!

Again, hide answer in spoilers. And good luck!

You have a broken calculator with only these buttons working:
sin
cos
tan
cot
sin^-1
cos^-1
tan^-1
cot^-1

where when I say sin^-1 I mean the inverse of sine, not one over sine.
i.e. sine(sine-1(x)) = x for x between -1 and 1.

Your calculator now show 0. How do you, after a finite number of pressing buttons, make ANY positive rational number to appear on your screen?

Ahaha maybe there's some confusion, by ANY positive rational number, I mean this:
If I want to make 15/23 appear, I can do it.
If I want to make any p/q appear, where p, q in natural number, and p, q has no common divisors, I can make it happen after finite number of moves.

sorry! xD

*****
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
KumquatExpress
Profile Joined October 2009
United States344 Posts
Last Edited: 2010-08-18 20:03:35
August 18 2010 19:54 GMT
#2
+ Show Spoiler +
...Wouldn't you just press cosine? Unless there's something I'm not getting here..
Edit: I see your edit. Will ponder now.
Speedythinggoesin, speedythingcomesout.
blabber
Profile Blog Joined June 2007
United States4448 Posts
August 18 2010 19:56 GMT
#3
+ Show Spoiler +
maybe I don't understand the question correctly, but wouldn't just pressing cos give you 1?
blabberrrrr
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2010-08-18 20:29:19
August 18 2010 20:01 GMT
#4
a start
+ Show Spoiler +
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.


Edited, made progress
Translator:3
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
August 18 2010 20:10 GMT
#5
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
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.)

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.


Edited, made progress


we're getting somewhere
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
BajaBlood
Profile Joined August 2009
United States205 Posts
August 18 2010 20:33 GMT
#6
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
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.


Edited, made progress


+ Show Spoiler +

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


Empyrean
Profile Blog Joined September 2004
16981 Posts
August 18 2010 20:35 GMT
#7
The calculator doesn't have number buttons.
Moderator
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2010-08-18 20:51:55
August 18 2010 20:35 GMT
#8
On August 19 2010 05:33 BajaBlood wrote:
Show nested quote +
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
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.


Edited, made progress


+ Show Spoiler +

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




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

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).
Translator:3
Roggles
Profile Joined December 2009
United States38 Posts
August 18 2010 21:42 GMT
#9
On August 19 2010 05:35 infinitestory wrote:
Show nested quote +
On August 19 2010 05:33 BajaBlood wrote:
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
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.


Edited, made progress


+ Show Spoiler +

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




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

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).


how can you assume that you can get sqrt(a/n) for any n?

applying 1+5 only gets you sqrt(1+x^2). if you apply it again, you will get sqrt(2+x^2) etc. your claim is that by taking sqrt(2) and applying 1/x to make it sqrt(1/2), you can obtain sqrt(3/2), sqrt(5/2) etc as well, which is perfectly fine. However, if you tried it with sqrt(3), it would become sqrt(1/3), and then you would have a sequence of sqrt(1/3), sqrt (4/3), sqrt(7/3). combined with the natural numbers, you would get sqrts of 1/3, 3/3, 4/3, 6/3, etc. but we're missing the 2/3, 5/3, etc.

the 1+5 algorithm, when applied recursively, will take a fraction sqrt(a/b) and turn it into sqrt(1+a/b). therefore you will get the sqrt of a/b, (b+a)/b, (2b+a)/b, etc. which does not properly fill in the fractions between a/b and (b+a)/b.

tell me if I'm missing something stupid though...
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
August 18 2010 21:53 GMT
#10
On August 19 2010 06:42 Roggles wrote:
Show nested quote +
On August 19 2010 05:35 infinitestory wrote:
On August 19 2010 05:33 BajaBlood wrote:
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
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.


Edited, made progress


+ Show Spoiler +

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




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

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).

+ Show Spoiler +


how can you assume that you can get sqrt(a/n) for any n?

applying 1+5 only gets you sqrt(1+x^2). if you apply it again, you will get sqrt(2+x^2) etc. your claim is that by taking sqrt(2) and applying 1/x to make it sqrt(1/2), you can obtain sqrt(3/2), sqrt(5/2) etc as well, which is perfectly fine. However, if you tried it with sqrt(3), it would become sqrt(1/3), and then you would have a sequence of sqrt(1/3), sqrt (4/3), sqrt(7/3). combined with the natural numbers, you would get sqrts of 1/3, 3/3, 4/3, 6/3, etc. but we're missing the 2/3, 5/3, etc.

the 1+5 algorithm, when applied recursively, will take a fraction sqrt(a/b) and turn it into sqrt(1+a/b). therefore you will get the sqrt of a/b, (b+a)/b, (2b+a)/b, etc. which does not properly fill in the fractions between a/b and (b+a)/b.

tell me if I'm missing something stupid though...


+ Show Spoiler +

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)
Translator:3
jalstar
Profile Blog Joined September 2009
United States8198 Posts
August 18 2010 21:58 GMT
#11
+ Show Spoiler +
You can't. You don't have an "=" button.
Hypnosis
Profile Blog Joined October 2007
United States2061 Posts
August 18 2010 22:01 GMT
#12
where do you get these problems? i feel like i missed something in calc or something because im done with calc 3 already and idk those identities for shit! Maybe i should review my trig proofs haha
Science without religion is lame, Religion without science is blind
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2010-08-18 22:03:17
August 18 2010 22:01 GMT
#13
On August 19 2010 06:58 jalstar wrote:
+ Show Spoiler +
You can't. You don't have an "=" button.

+ Show Spoiler +
Pressing the operator also shows the answer. Old school way.


EDIT: @hypnosis: this is a classic contest math algebra problem (I'm kinda cheating since I solved it a couple years ago, then forgot the solution)
Translator:3
Roggles
Profile Joined December 2009
United States38 Posts
August 18 2010 22:04 GMT
#14
On August 19 2010 06:53 infinitestory wrote:
Show nested quote +
On August 19 2010 06:42 Roggles wrote:
On August 19 2010 05:35 infinitestory wrote:
On August 19 2010 05:33 BajaBlood wrote:
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
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.


Edited, made progress


+ Show Spoiler +

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




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

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).

+ Show Spoiler +


how can you assume that you can get sqrt(a/n) for any n?

applying 1+5 only gets you sqrt(1+x^2). if you apply it again, you will get sqrt(2+x^2) etc. your claim is that by taking sqrt(2) and applying 1/x to make it sqrt(1/2), you can obtain sqrt(3/2), sqrt(5/2) etc as well, which is perfectly fine. However, if you tried it with sqrt(3), it would become sqrt(1/3), and then you would have a sequence of sqrt(1/3), sqrt (4/3), sqrt(7/3). combined with the natural numbers, you would get sqrts of 1/3, 3/3, 4/3, 6/3, etc. but we're missing the 2/3, 5/3, etc.

the 1+5 algorithm, when applied recursively, will take a fraction sqrt(a/b) and turn it into sqrt(1+a/b). therefore you will get the sqrt of a/b, (b+a)/b, (2b+a)/b, etc. which does not properly fill in the fractions between a/b and (b+a)/b.

tell me if I'm missing something stupid though...


+ Show Spoiler +

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)


+ Show Spoiler +

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.
icystorage
Profile Blog Joined November 2008
Jollibee19346 Posts
August 18 2010 22:11 GMT
#15
On August 19 2010 05:35 Empyrean wrote:
The calculator doesn't have number buttons.

i think that's the challenge of the problem
LiquidDota StaffAre you ready for a Miracle-? We are! The International 2017 Champions!
FiBsTeR
Profile Blog Joined February 2008
United States415 Posts
August 18 2010 22:13 GMT
#16
I remember solving this when preparing for USAMO... I think this was on a past one.
]343[
Profile Blog Joined May 2008
United States10328 Posts
Last Edited: 2010-08-18 22:19:03
August 18 2010 22:18 GMT
#17
this is an old usamo problem

and VERY ANNOYING

+ Show Spoiler +

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.

Writer
Surrealz
Profile Blog Joined May 2010
United States449 Posts
August 18 2010 22:37 GMT
#18
On August 19 2010 07:18 ]343[ wrote:
this is an old usamo problem

and VERY ANNOYING

+ Show Spoiler +

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.



thank you sir
1a2a3a
palanq
Profile Blog Joined December 2004
United States761 Posts
August 18 2010 22:45 GMT
#19
since you have all the inverse functions, the problem is equivalent to taking any arbitrary rational number ( = a/b for some integers a and b) and turning it into zero. might be a useful approach going at it from the other direction
time flies like an arrow; fruit flies like a banana
TanGeng
Profile Blog Joined January 2009
Sanya12364 Posts
August 18 2010 22:59 GMT
#20
Incomplete
+ Show Spoiler +

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.





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