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Guest Message by DevFuse

# Closed form expression on steroids

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

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

Spoiler for one messy solution

Go to the full post

10 replies to this topic

### #1 bushindo

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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.

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### #2 bonanova

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Posted 27 January 2013 - 08:38 PM

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Vidi vici veni.

### #3 bushindo

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Posted 28 January 2013 - 02:36 AM

Spoiler for

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### #4 Prime

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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

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Past prime, actually.

### #5 Rob_Gandy

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Posted 29 January 2013 - 02:26 AM

Spoiler for My 2 cents.

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### #6 Prime

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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

• 0

Past prime, actually.

### #7 bonanova

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Posted 29 January 2013 - 07:24 AM

Wouldn't a polynomial be continuous?
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Vidi vici veni.

### #8 Prime

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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.

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Past prime, actually.

### #9 Prime

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Posted 29 January 2013 - 07:44 AM

Wouldn't a polynomial be continuous?

No. We have exactly 11 pairs of points (xk,yk). Where xk is one of the tan(...) values and yk 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.

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Past prime, actually.

### #10 bushindo

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Posted 29 January 2013 - 06:08 PM

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

Spoiler for one messy solution