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Is it possible to create a function that maps x to y as follows:

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...

y 0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 ...

So far, this is what I have been able to come up with:

|5 sin ((1/10)*pi*x)|

Which will give the following y's for x = 0 to 10:


This sequence will repeat every 10 steps, with the peak (5) at step 5. This is getting close to what I want so all that is left is to somehow adjust the function so that the required integers appear (1,2,3,4), but I don't know how.

Since the function is cycling through a series of numbers over and over, I figured that it might be possible to write the function in terms of a periodic function and not piece by piece.

Please comment.


How would I go about writing a function that produce the following plot:(x is an integer >=0 , the pattern repeats after x = 8)

alt text

Can this plot be expressed using a single formula or must it be defined piecewise? Thank you.

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A piecewise linear function will do of course. Are there any extra requirements, like smoothness for example? – Hans Lundmark Oct 31 '10 at 16:06
Youre example of $|5\sin(\pi x/10)|$ doesn't actually work. – Robin Chapman Oct 31 '10 at 16:15
You probably should have asked "part 2" as a separate question... – J. M. Nov 20 '10 at 16:09
up vote 12 down vote accepted

$\frac{5}{\pi}\arccos\left(\cos\left(\frac{\pi x}{5}\right)\right)$ seems to do what you want, unless you need something smoother.

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...and this is in fact the "natural" piecewise linear function passing through the given points, but written in a clever way. – Hans Lundmark Oct 31 '10 at 16:50
I should say @Hans, I thought of it immediately after I saw your comment. :) – J. M. Oct 31 '10 at 16:58
@J.M: How did you find that function? Very impressive. – Robert Smith Oct 31 '10 at 17:17
+1 Very Good and very clever! – Américo Tavares Oct 31 '10 at 18:48
@JM: Yes,that is exactly what I wanted. Thank you for your help. – Sara Oct 31 '10 at 20:04

You ask for a function. A function is a mapping from one set to another set, but you neglect to specify which sets. Your list of $x$-values consists of small nonnegative integers together with an ellipsis. From this, one might surmise that your intended domain is $\mathbb{N}_0$, the set of nonnegative integers. Likewise you do not specify a codomain, but let's say for argument's sake that it is $\mathbb{N}_0$, the same as the domain.

Of course there are uncountably many functions from $\mathbb{N}_0$ to $\mathbb{N}_0$ containing your specified values. But you mention it has to be a "sine" function. The term sine refers to a particular function on the real numbers, taking all values between $-1$ and $1$ and having period $2\pi$. Your function cannot have all of these properties. Perhaps by a "sine" function you mean a periodic function? If so here is one defined on $\mathbb{N}_0$ which extends your table of values: $$f(n)=\begin{cases} 0&\text{if }n\equiv 0 \pmod {10},\\\\ 1&\text{if }n\equiv\pm1 \pmod {10},\\\\ 2&\text{if }n\equiv\pm2 \pmod {10},\\\\ 3&\text{if }n\equiv\pm3 \pmod {10},\\\\ 4&\text{if }n\equiv\pm4 \pmod {10},\\\\ 5&\text{if }n\equiv 5 \pmod {10}. \end{cases}$$ Of course, unlike the actual sine function, this never takes negative values.

Perhaps as in Hans's comment, you want a function on the reals restricting to your table of values. Here is a piecewise linear example as Hans suggests: $$g(x)= \begin{cases} 10\{x/10\}&\textrm{if }\{x/10\} < 1/2,\\ 10-10\{x/10\}&\textrm{if }\{x/10\}\ge 1/2 \end{cases}$$ where $\{t\}$ denotes the fractional part of a real number $t$.

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PART 1 You were close: Look at your values. They rise with slope $1$ till $5$, and then continue with slope $-1$. This may be described by the function $h(x)'=\text{sgn}(\sin(\pi x/5))\,$.

$\hskip1.7in$enter image description here

(from W|A).

Integrate this, $h(n)=\int_0^n \text{sgn}(\sin(x))\,dx\,$, and you're done.

Instead of that, you might adapt the Fourier Series for the Triangle Function to your problem: $$ \begin{align} x_\mathrm{triangle}(t) & {} = \frac {8}{\pi^2} \sum_{k=0}^\infty (-1)^k \, \frac{ \sin \left( (2k+1)\omega t \right)}{(2k+1)^2} \\ & {} = \frac{8}{\pi^2} \left( \sin (\omega t)-{1 \over 9} \sin (3\omega t)+{1 \over 25} \sin (5\omega t) - \cdots \right) \end{align} $$

Part 2 Starting from $(n,h(n))=(9,5)$, you have $6$ values decreases, followed by a jump of $+3$ and then $6$ values increasing. So this is essentially the same problem with an additional offset function $\frac32 \left(\text{sgn}(\sin(\pi x/5))+1\right)$...

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