I am looking at approximating my function $f(x)$ using a Chebyshev and Legendre series and I ran into this question.

Is interpolation using $n+1$ Chebyshev nodes the same as representing the function using the first $n+1$ Chebyshev coefficients in its Chebyshev series, i.e., is $$f(x) = \sum_{k=0}^n f(x_k) l_k(x)$$where $x_k$ are the Chebyshev nodes and $l_k(x)$ is the appropriate Lagrange polynomial associated with the $k^{th}$ node, the same as $$f(x) = \sum_{k=0}^n a_k T_k(x)$$where $T_k(x)$ is the $k^{th}$ Chebyshev polynomial and $a_k = \dfrac{\langle f, T_k\rangle}{\langle T_k, T_k\rangle}$?

Here the inner product is the inner product using the appropriate weight function for the polynomials, i.e., $$\langle f, T_k \rangle = \int_{-1}^1 \dfrac{f(y)T_k(y)}{\sqrt{1-y^2}}dy$$

Is the same true for Legendre interpolation and Legendre expansion as well, where the inner product is $$\langle f, P_k \rangle = \int_{-1}^1 f(y)P_k(y)dy$$? If so, could someone direct me to the proof of this?


What you are doing here is projecting a function onto a basis of a finite dimensional space of polynomials. All you use in the formulas you show is that the basis $T_k$ is orthogonal, i.e., that $\left<T_i,T_j\right>=0$ if $i\neq j$ with regard to some scalar product $\left<\cdot,\cdot\right>$. All of this naturally generalizes to other orthogonal bases of polynomial spaces.

  • $\begingroup$ -1. This doesn't answer my question at all. My question asks if Lagrangian interpolation using $n$ Chebshev nodes is same as Chebsyhev series expansion with the first $n$ terms. $\endgroup$ – Adhvaitha Oct 13 '15 at 9:14
  • $\begingroup$ It depends on the scalar product you use. If your scalar product involves the evaluation of the function at the same point where you want to do the Lagrange interpolation, then whether you use one approach or the other, you end up with a polynomial of degree $n$ that interpolates the function at the same points. Because two polynomials of degree $n$ that coincide at $n+1$ points are equal, the two approaches lead to the same function. They simply use a different basis. $\endgroup$ – Wolfgang Bangerth Oct 13 '15 at 11:41
  • $\begingroup$ The scalar product is with the weight function associated with the polynomials, i.e., for the case of Chebsyhev, $$\langle T_m,T_n \rangle = \int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}}dx$$ and for Legendre it is $$\langle P_m,P_n \rangle = \int_{-1}^1 P_m(x)P_n(x)dx$$ $\endgroup$ – Adhvaitha Oct 13 '15 at 11:45
  • $\begingroup$ In that case you compute different projections onto the same polynomial space, and you will get different results. $\endgroup$ – Wolfgang Bangerth Oct 13 '15 at 13:45

The two are different. Taking $n=0$, we see that the interpolation is $f(0)$, whereas in the second one, the approximation is $\displaystyle\int_{-1}^1 \dfrac{f(y)}{\sqrt{1-y^2}}dy$.


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