Thread: interpolation of countably infinite sequence

Hey, I was wondering if you can prove the following in the following way.

Given any infinite sequence, you can find a continuous function that interpolates each point in the sequence.

Proof (sketch):

Break the infite sequence into blocks of n, where n is finite. By a block of n I mean the first n terms, then the next n terms and so on. The union of these 'blocks' is the entire sequence.

Now, for each n block, we can applay the lagrangian interpolation formula to get a contious polynomial that intersects the n points. Then, you can do the same for the next block. And so on. Say we have 2 blocks, then we have 2 continuous functions. Now, we can turn them into one continuous function by keeping the parts of each function that intersect the points in its corospoding block and then erase the sections of the functions that are on the interval (n, n+1). Then define the new function in that area as the max of those 2 on the interval (n,n+1) if they don't intersect any more times in that interval. If they do, then you just apply the max on the intervals demarked by the intersections. In any area if the 2 functions overlap (not infinite intersection but both fill space above x axis), then just take the max.

Continue by induction.

Does this work. Please bare with me as I don't know a great deal about interpolation. If there is anonther theorem that states that you can interpolate a finite sequence with a continuous function, I would really appreciate learning about it.

If we are to do this, what if we just linearly interpolate between all the points. Say, the points are p1= (x1,y1), p2 =(x2,y2),..., the line segment from point [LaTeX ERROR: Convert failed] to [LaTeX ERROR: Convert failed] is just [LaTeX ERROR: Convert failed] for [LaTeX ERROR: Convert failed] , from the slope-intercept form of the line passing through [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] . You can see that this will be a continuous function if [LaTeX ERROR: Convert failed] .

Yeah, that works nicely. Do you know if what I describe above works though? I thought about connecting the points using lines, but I figured that the max would be nicer.

The line of reasoning that you chose has the blocks disjoint if I am not mistaken. This won't work. Consider this example:

You have four points p1 = (0,0), p2 = (1,0), p3 = (2,1) and p4 = (3,1). Now, take the first two points and find the interpolating function: it's just y = 0. Now take the second 2 points. The interpolating function in this case is just y =1. So you've got your two functions. Now you take the open interval ]1,2[, you erase whatever is in there from those two y's and then take the maximum of the two y's. Then on this interval your function is just y = 1, since the function for the 2nd 2 points is y = 1. Overall your interpolating function for the four points is just

y = 0 if x is in [0,1[; and 1 if x is in ]1,2]

And this function is not continuous. What if you choose another method:

Choice 1: Instead of having disjoint blocks, just pick the blocks so that they overlap at 1 point. Now if you have your 2 functions for the two blocks, they intersect at the common point and then you are done.

Choice 2: If you insist on having disjoint blocks, you can just put a line in between (n, y_n) and (n+1, y_{n+1}).