Best Answer Prime, 29 January 2013 - 06:46 AM

But they don't need to be integers, those 11 periodic values.

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Started by bushindo, Jan 27 2013 04:53 PM

Best Answer Prime, 29 January 2013 - 06:46 AM

But they don't need to be integers, those 11 periodic values.

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10 replies to this topic

Posted 27 January 2013 - 04:53 PM

This is based on a previous puzzle.

Find a closed form expression for function f(N) such that

f( N ) = 0 for 1 <= N <= 11

f( N ) = 1 for 12 <= N <= 22

f( N ) = 2 for 23 <= N <= 33

f( N ) = 3 for 34 <= N <= 44

f( N ) = 4 for 45 <= N <= 55

...

No modulus or rounding operators are allowed.

Posted 27 January 2013 - 08:38 PM

Spoiler for How about

- Bertrand Russell

Posted 28 January 2013 - 02:36 AM

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Posted 28 January 2013 - 10:14 PM

I am assuming N must be an integer.

As for how to slice two pi into 11 pieces, which would yield the required progression, I am stumped.

I just want to throw a different thought into this subject:

Spoiler for multiples

Past prime, actually.

Posted 29 January 2013 - 02:26 AM

Spoiler for My 2 cents.

Posted 29 January 2013 - 06:46 AM Best Answer

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Past prime, actually.

Posted 29 January 2013 - 07:24 AM

Wouldn't a polynomial be continuous?

- Bertrand Russell

Posted 29 January 2013 - 07:37 AM

Oops, messed up a little...

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Spoiler for CORRECTION

**Edited by Prime, 29 January 2013 - 07:37 AM.**

Past prime, actually.

Posted 29 January 2013 - 07:44 AM

Wouldn't a polynomial be continuous?

No. We have exactly 11 pairs of points (x_{k},y_{k}). Where x_{k} is one of the tan(...) values and y_{k} is one of the integer values 0 through 10. It should be big and messy, but a regular polynomial of 10th degree.

And another correction to start output with zeros instead of ones, just subtract another 1 from that whole thing.

**Edited by Prime, 29 January 2013 - 07:50 AM.**

Past prime, actually.

Posted 29 January 2013 - 06:08 PM

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Excellent answer

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