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I'm trying to jog my memory of calculus from 10 years ago and it's proving a bit difficult. I'm hoping someone can give me a push in the right direction here. I've found a number of good resources online, but they're all either too far abstracted from my problem or too advanced. The concrete problem that I'm trying to solve is the following:

I'm writing a pagination function that uses $13$ discrete steps, i.e. there will be a maximum of $13$ links shown, regardless of number of pages. Now say I have $n$ total pages, and I'm on page $p$ where $1 \leq p \leq n$ (and for the sake of this discussion, $n > 13$). I need an equation that will yield $13$ links from $1$ to $n$, with the $6$ links immediately surrounding $p$ being pretty much linear - i.e., $45, 46, 47, 48, 49, 50, 51$ where $p = 48$ - and with the first and last step being the bounds - i.e., $1$ and $13$, respectively.

It looks to me like this would be something along the lines of $(x-s)^3/u + p$, where $s$ is the current discrete step (between $1$ and $13$) and $u$ is some variable that I don't know how to derive. I managed to come up with a quadratic equation that I could force to intersect with $(s,n)$, but I then I got lost trying to integrate it to achieve my cube equation..... I don't know, maybe I'm going way off in the wrong direction with all this.

Any help?


Really, this specific problem isn't the focus here. I've come up against this type of problem a number of times now and what's really bugging me is that I know there's an elegant mathematical way to solve it, but I can't figure out how. I'm basically trying to "mash" a graph into the given boundaries. The simplest way to think of it is to start with $f(x) = x^2$. This intersects $(-3,9)$ and $(3,9)$. Now say I want to move the vertex up to $(0,5.5)$, but I still want the graph to intersect those same two points. Then I want to move the vertex to $(1.2,5.5)$, but still intersect those points. See what I mean? It seems like the kind of thing where someone will come along and say "Oh, that's easy; you just do this, then plug in your numbers, then take the derivative of that" or something, but so far I haven't found that to be the case. Maybe I'm assuming this is much more trivial than it actually is. Am I wrong in thinking this shouldn't be terribly complicated?

Thanks, Kael

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If you're looking for a simple function that goes through particular points in the plane, you should start by checking out Lagrange interpolating polynomials: – Greg Martin Apr 19 '13 at 7:10
While this will require a little more study on my end, I'm pretty sure Greg Martin's this is the path I was looking for. I'll be in the jungle for a while catching up on some of the background here, but I think this will lead me out. Please post it as an answer so that I can select it and mark the question as answered! – kael Apr 21 '13 at 17:49
up vote 0 down vote accepted

If you're looking for a simple function that goes through particular points in the plane, you should start by checking out Lagrange interpolating polynomials.

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To me this isn't a calculus problem at all. You want 13 links, and you've already told us what 9 of them are: you want the links to be 1, ?, ?, $p-3$, $p-2$, $p-1$, $p$, $p+1$, $p+2$, $p+3$, ?, ?, $n$. Why not just have the missing links interpolate linearly between the known values? So the sequence starts 1, $(p-1)/3$, $(2p-5)/3$, $p-3$ (rounding off).

You haven't told us much about what you want your link choices to achieve; this solution seems to match what you have told us.

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Very true, this solution would work fine, and this is what I've implemented for lack of knowledge on how to create the function. See my update above. – kael Apr 19 '13 at 4:26

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