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Take $0\leq r < m$, and let all values be nonnegative and integer. Consider the function on a sequence ${x(n)}$, $\Phi_mx(mn+r)=mx(n)+\frac{r}{m}(x(n+1)-x(n))$, where we consider $x(0)=0$.

As an example, $\Phi_3$ maps $1,1,2,3$ to $1,2,3,3,3,3,4,5,6,7,8,9$. Would anyone be able to walk me through why this is? I end up with different values.

114
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  • What is $x$, $m$, $n$ and $r$? – Luiz Cordeiro May 23 '13 at 13:53
  • @Luiz Edited to correct ambiguity. – 114 May 23 '13 at 14:21
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    Are you sure there's not a typo? If the formula were $(\Phi_mx)(mn+r)=m\left(x(n)+\frac{r}{m}(x(n+1)-x(n))\right),$ then you would get the result stated. – Will Orrick May 27 '13 at 00:53
  • Can you say something about the context in which this function arises? I might be missing something, but without further clues it's hard to say... – Will Orrick May 27 '13 at 02:24
  • Unfortunately there is not much additional context for it that leads up to the function it is merely stated. – 114 May 27 '13 at 03:06
  • Is $\Phi_m$ supposed to map integer sequences to integer sequences? What is it that you are meant to do with this function? Does this come from a textbook, an article, a homework problem? – Will Orrick May 27 '13 at 08:58
  • Would you be willing to provide a reference? – Will Orrick May 27 '13 at 15:25
  • Unfortunately I only have a hard copy of the article, it is rather old. I would need to type it out. – 114 May 27 '13 at 15:36
  • I gather it's an unpublished article? – Will Orrick May 27 '13 at 15:45
  • Yes, part of broader work. – 114 May 27 '13 at 16:01
  • Well, I'm convinced there's a typo somewhere. It's clear that, as written, the function does not map integer sequences to integer sequences, except in special cases. Presumably the author introduces this function in order to apply it to something. Knowing how it gets applied might provide some clue as to the correct interpretation. Short of that, I think it's going to be pretty hard to get an answer to your question... – Will Orrick May 27 '13 at 16:08
  • Thanks, I believe you were correct initially. – 114 May 27 '13 at 16:16
  • OK. I've expanded my comment into an answer. – Will Orrick May 27 '13 at 17:24

1 Answers1

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It appears that the definition of $\Phi_m$ was written incorrectly in the original article, and that the correct definition is $$\begin{aligned}(\Phi_mx)(mq+r)&=m\left(x(q)+\frac{r}{m}\left(x(q+1)-x(q)\right)\right)\\ &=mx(q)+r(x(q+1)-x(q)).\end{aligned}$$ Here $x$ is a sequence with initial term $x(0)=0,$ $m$ is a fixed positive integer, and $r$ is an integer satisfying $0\le r<m.$

If $y=\Phi_mx,$ then the $(mq)^\text{th}$ term of $y$ is $m$ times the $q^\text{th}$ term of $x.$ The $(mq)^\text{th}$ through $(mq+m)^\text{th}$ terms of $y$ form an arithmetic progression. So $y$ can be thought of as a piecewise linear function on the nonnegative integers.

To perform some explicit computations, let $x$ be the sequence $0,1,1,2,3.$ Then $(\Phi_3x)(0)=0,$ $(\Phi_3x)(3)=3x(1)=3,$ $(\Phi_3x)(6)=3x(2)=3,$ $(\Phi_3x)(9)=3x(3)=6,$ and $(\Phi_x)(12)=3x(4)=9.$ The full sequence $\Phi_3x$ contains two terms between each two successive terms in the sequence $0, 3, 3, 6, 9,$ and these two terms interpolate linearly between successive terms. Hence, one gets $$\mathbf{0}, 1, 2, \mathbf{3}, 3, 3, \mathbf{3}, 4, 5, \mathbf{6}, 7, 8, \mathbf{9}.$$ Alternatively, we may just apply the formula. For example, $(\Phi_3x)(7)=(\Phi_3x)(3\cdot2+1)=3x(2)+1\cdot(x(3)-x(2))=3\cdot1+1\cdot(2-1)=4,$ which agrees with the result above.

To take another example, $(\Phi_2x)$ is the sequence $$\mathbf{0}, 1, \mathbf{2}, 2, \mathbf{2}, 3, \mathbf{4}, 5, \mathbf{6}.$$ The even indexed terms (in boldface) are simply $2$ times the sequence $x,$ while the remaining terms interpolate linearly between the terms in boldface. An explicit computation of $(\Phi_2x)(3)$ is $$(\Phi_2x)(3)=(\Phi_2x)(2\cdot1+1)=2\cdot x(1)+1\cdot(x(2)-x(1)=2\cdot1+1\cdot(1-1)=2.$$

The functions $\Phi_m$ have the sequence $z(0)=0,$ $z(1)=1,$ $\ldots,$ $z(n)=n,$ $\ldots,$ that is, the sequence of nonnegative integers, as a fixed point. In other words, $(\Phi_mz)(n)=z(n)=n$ for all integers $m>0$ and $n\ge 0$. To see this, simply compute $$(\Phi_mz)(mq+r)=mz(q)+r(z(q+1)-z(q))=mq+r((q+1)-q)=mq+r.$$

Will Orrick
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