today's puzzle

PT

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Oct 23, 2012, 5:02:46 PM10/23/12

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g(n) is a function of any integer n, positive or negative, which
produces an integer value, with conditions:

a) g(g(n)) = n

b) g(g(n + 2) + 2) = n

c) g(0) = 1

1. Determine g(n)
2. Prove your solution is unique.

---
Paul T.

Ilan Mayer

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Oct 23, 2012, 10:44:03 PM10/23/12

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SPOILER

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The function is g(n) = 1 - n

It satisfy all three requirements by inspection.

Proof of uniqueness by induction:

Since g(0) = 1, g(g(0)) = g(1) = 0 (from a and c).

Assuming that g(n) = 1 - n for n = 1, 0, ..., m where m is a negative integer:

g(g(m + 1) + 2) = m - 1 (from b)
Since g(m + 1) = 1 - ( m + 1 ) = -m, g(2 - m) = m - 1
g(m - 1) = g(g(2 - m)) = 2 - m = 1 - (m - 1) (from a)

This proves that g(n) = 1 - n for any negative n.

For positive n >= 2 g(1 - n) = 1 - ( 1 - n ) = n since 1 - n is negative.
Now g(n) = g(g(1 - n) = 1 - n (from a).

Please reply to ilan dot mayer at hotmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
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William Elliot

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Oct 23, 2012, 11:38:45 PM10/23/12

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On Tue, 23 Oct 2012, PT wrote:

> g(n) is a function of any integer n, positive or negative, which
> produces an integer value, with conditions:
>
> a) g(g(n)) = n
>
> b) g(g(n + 2) + 2) = n
>
> c) g(0) = 1

Let f(n) = 1 - n.

ff(n) = f(1 - n) = 1 - (1 - n) = n
f(f(n + 2) + 2) = f(1 - (n + 2) + 2) = f(1 - n) = n
f(0) = 1

> 1. Determine g(n)

g = f.

William Elliot

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Oct 24, 2012, 5:28:39 AM10/24/12

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f(0) = 1 = g(0)
f(1) = 0 = gg(0) = g(1)

If f(n) = g(n), then

n = g(g(n + 2) + 2)
1 - n = f(n) = g(n) = g(n + 2) + 2
g(n + 2) = 1 - n - 2 = 1 - (n + 2) = f(n + 2)

By induction, for all n >= 0, f(n) = g(n).

g(-0) = 1 = f(-0)

g(g(-1 + 2) + 2) = -1
g(-1) = g(-1 + 2) + 2 = g(1) + 2 = 2 = f(-1)

If n >= 0 and f(-n) = g(-n), then
g(g(-(n+2) + 2) + 2) = -(n + 2)
g(-(n + 2)) = g(-n) + 2 = 1 + n + 2 = 1 - (-(n + 2)) = f(-(n + 2))

Again by induction, for all n >= 0, g(-n) = g(-n). Done.

Herman Rubin

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Oct 24, 2012, 12:24:14 PM10/24/12

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From (a), g(g(g(n))) = g(n), so g is 1-1.
Then from (b), g(n+2) = g(n) - 2.

From (a) and (c), g(1) = 0. so we have g(0) and g(1), and increasing
n by 2 decreases g(n) by 2, so g is uniquely defined.

--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

William Elliot

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Oct 24, 2012, 10:39:32 PM10/24/12

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On Wed, 24 Oct 2012, Herman Rubin wrote:
> On 2012-10-24, William Elliot <ma...@panix.com> wrote:
> > On Tue, 23 Oct 2012, PT wrote:
>
> >> g(n) is a function of any integer n, positive or negative, which
> >> produces an integer value, with conditions:
>
> >> a) g(g(n)) = n
> >> b) g(g(n + 2) + 2) = n
> >> c) g(0) = 1
>
> > Let f(n) = 1 - n.
>
> > ff(n) = f(1 - n) = 1 - (1 - n) = n
> > f(f(n + 2) + 2) = f(1 - (n + 2) + 2) = f(1 - n) = n
> > f(0) = 1
>
> >> 1. Determine g(n)
> > g = f.
>
> >> 2. Prove your solution is unique.
>
> From (a), g(g(g(n))) = g(n), so g is 1-1.

Only if g is surjective.

> Then from (b), g(n+2) = g(n) - 2.
>
> From (a) and (c), g(1) = 0. so we have g(0) and g(1), and increasing
> n by 2 decreases g(n) by 2, so g is uniquely defined.
>

Ok, provided g is surjective, you've determined g(n) for all n >= 0.
What about g(n) for n < 0?

Randy Yates

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Oct 30, 2012, 11:06:02 PM10/30/12

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PT <ptane...@consultant.com> writes:
> g(n) is a function of any integer n, positive or negative, which
> produces an integer value, with conditions:

> [...]

This is in response to the recent domain/range discussion (including my
errant conclusion recently in this thread). Dilip is right: read the
problem carefully. g: Z --> Z (in the notation of Marsden and Tromba).
--
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com