关于傅里叶插值唯一性证明验证与非唯一性示例的技术问询

Hey there! Let's walk through this Fourier interpolation problem step by step, since you're working through a past exam question.

唯一性证明（当有N个互异插值点时） #

When we have exactly $N$ distinct interpolation points $(x_j, y_j)_{j=0}^{N-1}$, we can build the Fourier interpolant using a Lagrange-style basis tailored to the exponential functions ${e^{inx}}_{n=0}{N-1}$.

First, define the Lagrange basis functions:
$$l_m(e^{ix})=\prod_{j=0, j\neq m}^{N-1} \frac{e^{ix} - e^{ix_j}}{e{ix_m} - e^{ix_j} }$$
Each $l_m(e^{ix})$ is designed to be 1 at $x=x_m$ and 0 at all other $x_j$. Using these, the interpolating polynomial is:
$$\Pi_N(x)=\sum_{k=0}^{N-1} y_k ,l_k(e^{ix})$$
By construction, this satisfies $\Pi_N(x_l)=y_l$ for every $0\le l\le N-1$.

To prove uniqueness, suppose there's another valid interpolant $\Pi'_N(x)$ that also passes through all $N$ points. Let's look at the difference:
$$\Delta(x) = \Pi_N(x) - \Pi'_N(x)$$
This $\Delta(x)$ is a polynomial of degree at most $N-1$ in $z=e^{ix}$ — since both interpolants are linear combinations of the first $N$ exponential basis functions, which correspond to $z^0, z^1, ..., z^{N-1}$.

Now, for each $j$, $\Delta(x_j) = \Pi_N(x_j) - \Pi'_N(x_j) = y_j - y_j = 0$. That means $\Delta(x)$ has $N$ distinct roots (since the $x_j$ are distinct, so $z_j=e^{ix_j}$ are distinct points on the unit circle).

Here's the key point: a non-zero polynomial of degree $d$ can have at most $d$ distinct roots. Since our $\Delta(x)$ is degree $N-1$ but has $N$ roots, it must be the zero polynomial. Therefore, $\Pi_N(x) = \Pi'_N(x)$, so the interpolant is unique.

非唯一性示例 #

Uniqueness only holds when we have exactly $N$ distinct interpolation points matching the size of our basis. If we break this condition, interpolation can be non-unique. Let's use a simple concrete example:

Let’s take $N=2$, so our basis is ${1, e^{ix}}$. Suppose we only have 1 interpolation point: say $(x_0, y_0)$.

We can make multiple valid interpolants that pass through this point:

• $\Pi_2(x) = y_0$ (a constant function, which is a linear combination of the basis)
• $\Pi'_2(x) = y_0 + C(e^{ix} - e^{ix_0})$, where $C$ is any non-zero complex constant.

Both of these are valid Fourier interpolants (linear combinations of our basis functions) and satisfy $\Pi_2(x_0) = y_0$, but they're clearly not the same. This shows that when we have fewer interpolation points than the basis size, the interpolant isn't unique.

Another case of non-uniqueness is when we have repeated interpolation points. For example, if $N=2$ and our points are $(x_0, y_0)$ and $(x_0, y_1)$:

• If $y_0 \neq y_1$, there's no solution at all.
• If $y_0 = y_1$, there are infinitely many solutions (since the system reduces to $a_0 + a_1 e^{ix_0} = y_0$, which has infinitely many $(a_0,a_1)$ pairs).

备注：内容来源于stack exchange，提问作者Diffusion