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Let's have the discrete periodic function $f(t)$ which has only 5 non-zero values $f(t_0) = -10, f(t_1) = -5, f(t_2) = -10, f(t_3) = -15$ and $f(t_4) = -10$, all other points within the period $[t_0,t_4]$ being zero. The obvious interpolating function which recovers these values is $f(t) = 5sin(t) - 10$. Is this a unique interpolating function or you can propose other interpolating functions, such as polynomial functions with constraints? Any concrete examples, please?

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1 Answer 1

Your $f$ only interpolates these values if $t_i=\frac{i\pi}{2}$. Also, $f$ is clearly not zero at all the other points in $[t_0,t_4]$.

Moreover note that you can, for any finite number of value pairs $(t_i,v_i)_i$, find arbitrarily many smooth functions $F$ s.t. $F(t_i)=v_i$ for all $i$. Look for example at Lagrange polynomials/Newton polynomials etc. You can always change the function values in between the prescribed points so that your function is still continuous or even smooth.

When you drop the continuity/smoothness requirement (which you do if you say that it should have only 5 non-zero values) then finding an interpolation function is completely trivial (just take the function that is 0 everywhere except on the prescribed points where it takes the prescribed values). If you necessarily want to write it down as a formula you can always use a linear combination of Kronecker deltas:

$$f(t)=\sum_i y_i \delta^t_{t_i}.$$

where $(t_i,y_i)_i$ is your data and $\delta^t_s$ is 1 if $t=s$ and 0 if not.

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So, then, what is $f'(t)$ of that function? Will it differ from $f'(t)$ when it is defined as $f(t) = 5sin(t_i) - 5$ where $i = 0, 1, 2, 3, 4$? –  ganzewoort Oct 20 '12 at 14:03
It doesn't make sense to write down the derivative of a function that is only nonzero at finitely many points. It's not differentiable. However, its differentiable almost everywhere and the derivative is 0 almost everywhere. –  Your Ad Here Oct 20 '12 at 14:04
This is the main point of my asking. If the interpolating function indeed is $f(t) = 5sin(t) - 10$ then it does have first derivative which is $5cos(t)$ and can be evaluated at the said 5 points. Also, from your answer I take it that there is no polynomial which would recover exactly the 5 points. All one can achieve through a polynomial is an approximation, right? –  ganzewoort Oct 20 '12 at 14:09
Then you should have mentioned that in your question. Your function $f$ does NOT have only 5 non-zero values by the way. I think you got some things confused. –  Your Ad Here Oct 20 '12 at 14:12
Of course there is a polynomial which recovers exactly these points. How did I say otherwise? Please look up Lagrange polynomials. Interpolating functions are not unique in the sense you seem to have in mind. –  Your Ad Here Oct 20 '12 at 14:13

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