I have written some code in Matlab to polynomially interpolate functions with Chebyshev nodes using the Chebyshev base ( I calculate the coefficients with respect to the Chebyshev base and multiply them). For higher degrees of the polynomial, it will approach the function.

I've calculated the errors for certain functions and plotted these in function of the degree of $p$. The functions I'm interpolating are $f_1 (x) = \frac {1}{1+25x^2}$, $f_2 (x) = |x|$, $f_3 (x) = |\sin (5x)|^3 $ and $f_4 (x) = \cos (20x) $. I've separated their errors into two graphs depending on the order of magnitude. Figure 1 Figure 2 Seeing these however has raised some questions.

First of all I'm confused about how these polynomials converge towards the actual function. According to my textbook the should either converge with $\mathcal{O}(C^{-n})$ or $\mathcal{O}(n^{-v})$ with $n$ the degree of the polynomial. But I'm lost as to how one would see this from the graph (especially since it's y-axis is plotted logarithmically) and how one would go about calculating $C$ or $v$.

Secondly I'm also confused about the oscillations of the error. Why does the error change with an even or odd degree polynomial?

And lastly I was also wondering what causes the change in convergence between functions. The graphs show that the functions that have an absolute value converge a lot slower than the other two. Is this because they aren't as smooth?


The oscillation of the error is typical for your functions. In general, any bounded function of $n$ is $\mathcal{O}(1)$ so multiplying by a bounded oscillating function is consistent with the error estimates. In your first graph the error estimates look like $\mathcal{O}(C^{-n})$ except for small $n$. In your second graph the error estimates look like $\mathcal{O}(n^{-v})$ where $v$ is $1$ or $2$. To calculate the constants you should skip the oscillations by using only even $n$ or odd $n$ and skipping values of $n$ less than $20$. Then use best linear fit for logarithm of the error.

  • $\begingroup$ Thanks for the reply, I get most of what you're saying but I'm still a little confused about de cos(20x) graph if it's exponential wouldn't plotting it with a logarithmic scale make it linear? The converging part doesn't quite look linear to me. $\endgroup$ – Jarne Renders May 7 '17 at 21:47
  • $\begingroup$ Well, the $cos(20x)$ is a difficult case. You have to go beyond $n=20$ and beyond $n=48$ the error curve flattens due to limited precision of the computation. After ignoring the oscillations, the log plot may have a small quadratic term. $\endgroup$ – Somos May 8 '17 at 2:42
  • $\begingroup$ Also, just to be clear, the big-Oh notation allows for the error estimate to be better than what is stated. That is, for example, an error estimate of $C^{-n^2}$ is $\mathcal{O}(C^{-n})$. There is a related but different notation for an estimate which is not just an asymptotic upper bound but also lower found. See the Wikipedia article en.wikipedia.org/wiki/Big_O_notation for this. $\endgroup$ – Somos May 9 '17 at 11:30

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