# Is there any function which produces at least 26 recognizably distinct graphs?

Is there any function which produces at least 26 recognizably distinct graphs?

For example, $f(x) = x^n, n\geq0$ produces distinct graphs for for all positive integers $n < 6$. I'd like it to be obviously distinct without having to look at a scale.

$n$ should start at 0 or 1 (it should not be negative) and increment normally.

If this doesn't make sense, please ask for clarification.

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What about $f(x)=x^n$ for $1 \leq n \leq 26$? –  N. S. Nov 6 '11 at 2:44
It's hard to tell, without looking at a scale, $x^5$ from $x^7$ say. –  Aaron Yodaiken Nov 6 '11 at 2:47
What do you mean by "scale" ? ;) If you think you need to increase the scale, look at $f(x)=1000x^n$ ;) –  N. S. Nov 6 '11 at 2:52
Your example is a bit questionable in my opinion because one could argue that $x^3$ and $x^5$ are not markedly different either. But if consider $f_n(x) = x(x-1)\ldots(x-n)$, you can tell their graph apart instantly by counting their roots ($f_n$ has $n+1$ roots). –  Joel Cohen Nov 6 '11 at 2:57
Like, $x^2$ and $x^3$ are obviously different graphs, without looking at any numbers, right? Any way that you graph them? Same, to a lesser extent, with $x^2$ and $x^4$—one of them is perfectly parabolic, the other is more squarish. –  Aaron Yodaiken Nov 6 '11 at 2:59

If you're looking for an elementary family parametrized by nonnegative integers, you might try $f(x) = \sin(n x)/\sin(x)$ (defined to be equal to $n$ at the removable singularities $x = 0,\ \pi, \ldots$). Note that there are $n-1$ zeros between two high peaks (if $n \ge 3$ is odd) or between a high peak and a deep valley (if $n\ge 2$ is even).