Approximation by exponential polynomials

Question

Let $u(t) = \Sigma_{k=1}^n c_k e^{\lambda_k t} (c_k \in \mathbb C, \lambda_k \in \mathbb C) $ be an exponential polynomial of order $n$.

Define $E_n$ to be the collection of all exponential polynomial of order $n$, i.e.,

$$ E_n:= \{ u : u(t) = \sum_{k=1}^n c_k e^{\lambda_k t}, c_k \in \mathbb C, \lambda_k \in \mathbb C \}. $$ Notice, that two elememts of $E_n$ may have different (disjoint) set of exponents. Only requirement for $u$ to be in $E_n$ is that it has order at most $n$.

Let $\mathbf{P}_n$ be the collection of all polynomials of degree at most $n$.

Consider a function $f = \sum_{j=1}^{M} p_{m_j}(t) e^{\lambda_j t}, p_{m_j} \in \mathbf{P}_{m_j}, \sum_{j=1}^{M} (m_j+1) \leq n$.

My question is, can one find a sequence $u_m \in E_n$ such that, $$\sup_{x\in[0,1]}| f(x)- u_m(x) |\rightarrow 0. $$

If possible, then how should one go about constructing such a sequence ?

Answer 1

This follows from the fact that the set of $n\times n$ matrices with simple spectrum is dense in the space of all $n\times n$ matrices ${\bf M}_n(\mathbb C)$ (or that the set of polynomials of degree $n$ with simple roots is dense in the set of all complex polynomials of degree $n$).

The function $f$ solves the Cauchy problem for a linear ODE with constant complex coefficients $$y^{(N)}+a_1y^{(N-1)}+\dots+a_Ny=0,$$ $$y^{(k)}(0)=f^{(k)}(0),\quad k=0,1\dots,N-1,$$ where $N=\sum_{j=1}^{M} (m_j+1) \leq n$. The exponents $\lambda_j$, $j=1,\dots,M$ are the roots of the corresponding characteristic equation $$P(\lambda)=\lambda^{N}+a_1\lambda^{N-1}+\dots+a_N\lambda=0$$ with the multiplicities, respectively, $m_j+1$, $j=1,\dots,M$. The coefficients $a_1,\dots a_N$ can be found from the relation $$P(\lambda)=\prod\limits_{j=1}^{M}(\lambda-\lambda_j)^{m_j+1}.$$

Now, a small generic perturbation of the ODE's coefficients will produce an ODE $$y^{(N)}+a_1^{\varepsilon}y^{(N-1)}+\dots+a_N^{\varepsilon}y=0,$$ such that the roots of the corresponding characteristic equation are all simple. This implies that a solution to the latter ODE belongs to $E_n$. Finally, we may use a standard result that solutions to linear ODEs depend continuously on the coefficients (in the topology of uniform convergence on finite time intervals).

Answer 2

By induction on $n$. If $n=1$, no problem. If $n \geq 2$, if the polynomials $p_{m_j}$ are all constant, still no problem. So assume $p_{m_1}$ nonconstant. Approximating $f$ is the same as approximating $t \mapsto e^{-\lambda_1 t} f(t)$, so we may assume $\lambda_1=0$. Then we approximate $f'$, using induction (since $\lambda_1=0$, $\sum_{j=1}^M (m_j+1)$ is less for $f'$). Now if $f'-u_0$ is small, with $u_0(t) = \sum_{k=1}^{n-1} c_k e^{\mu_k t}$, then $f-u$ is also small, where $u(t)=f(0)-\sum_{k=1}^{n-1} c_k/\mu_k + \sum_{k=1}^{n-1} \frac{c_k}{\mu_k} e^{\mu_k t}$ (using integration). The only problem here is that maybe $\mu_k=0$ for some $k$. But we can take $\mu_k$ very small but nonzero instead, $u_0$ is still close to $f'$.

This method gives an algorithm: everything is computable, and the only approximations are in fact the choices of "replacements" for zero $\mu_k$ (if you take it to be $1/n$, you get a sequence).

Answer 3

I think the following is another proof. It suffices to approximate a polynomial of degree n by an exponential polynomial of degree n + 1. Now, define $$ g_{\lambda}(x) = \frac{e^{\lambda x}}{\lambda} - 1. $$ It is easy to check that $\sup_{x\in [0,1]} |g_{\lambda} - x| \leq \frac{1}{2} \lambda$. Also $g_{\lambda}$ has order $2$. Furthermore, we can compute $$ |g_{\lambda}(x)^n - x^n| \leq |g_{\lambda}(x) -x| \cdot n \leq \frac{n}{2} \lambda. $$ And note that $g_{\lambda}(x)^n$ has order $n+1$. So a polynomial of degree $n$ given by $$ P_n(x) = \sum_{k=0}^{n} a_k x^k $$ can be approximated by $$ \sum_{k=0}^{n} a_k (g_{\lambda}(x))^k $$ up to order $n^2 \lambda$. Letting $\lambda \to 0$ implies the claim.

Answer 4

The following exploits the negative integers $k$ as $\lambda_k$ (I will go for the simpler, not necessarily for the best bounds, and make it explicit). Take a uniform approximation by polynomials $p_m$ for the function $f(t):=\log\left(\frac{1}{1-t}\right)$ on the interval $J:=[0,1-1/e]$. Assume further $0\leq p_m(t)\leq 1$ for all $t\in J$ (you can always get this). Since $$\epsilon_m:=\|f-p_m\|_{\infty,\,J}=o(1),$$ as $m\to \infty,$ we get, putting $t:=1-e^{-x}$, that $\|x-p_m(1-e^{-x})\|_{\infty,[0,1]}=\epsilon_m$ and, more in general, a bound on the uniform distance on [0,1] between the polynomial $q(x)$ and the exponential polynomial $$q_m(x):=q\left(\,p_m(1-e^{-x})\right).$$ Indeed, by the mean value theorem applied to $q$ and the two points $x$ and $p_m(1-e^{-x}),$ both belonging to the interval $[0,1],$ $$ \| q - q_m \|\leq \|q'\|_{\infty,\, [0,1]}\, \epsilon_m .$$

For instance, the Taylor expansion of order $m$ of $f$ gives $$p_m(t):=\sum_{k=1}^{m}\frac{t^{\,k}}{k}$$ with $0\leq p_m(t)\leq 1$ for all $t\in J$ as said, and with $\epsilon_m\leq\frac{1}{m+1}.$