Tag Info

Hot answers tagged

7

The space $L^2(a,b)$ is a Hilbert space of infinite dimension. Therefore there is an element $f\neq 0$ of this space which is orthogonal to $e^{i\lambda_j x}$. Take this $f$. You can also find such $f$ is any finite dimensional subspace whose dimension is $>n$. Just choose a basis and solve a system of linear equations.


4

Consider the polynomial $ P(\xi)=\prod_{1\le j\le n}(\xi-\lambda_j). $ The inverse Fourier transform of $(\xi-\lambda_j)$ is $$ \int(\xi-\lambda_j) e^{2iπ x\xi} d\xi=(D_x-\lambda_j)(\delta_0)=\frac{\delta'_0}{2iπ}-\lambda_j\delta_0=T_j, \quad\text{support } T_j=\{0\}, $$ Let $F$ be the inverse Fourier transform of $P$: we have $$ F=T_1\ast\dots\ast ...


2

The space of $W$ compactly supported continuous functions is infinite dimensional, and the map $f\mapsto (\widehat{f}(\lambda_1),\cdots,{\widehat f}(\lambda _n))$ which is a map from $W$ into ${\mathbb C}^n$ has a nonzero element in the kernel.


1

There is no $O(N \log N)$ algorithm, because your computer will take $N^2$ computations to even read the entries of the matrix $z$. However, you definitely get $O(N^2 \log N)$ since this is the usual DFT done $N$ times and suitably restricted.


1

The explicit answer is the formula 2.5.24.1 on page 433 from Brychkov, Marichev, Prudnikov Integral and Series, vol. 1. Note that for the odd function the FT reduces to sinT, and take $\alpha=1, \delta=1$. And yes, the answer is via Bessel and Macdonald functions.


1

If the numbers $b_k=\sum_{n=1}^\infty a_nn^k$ are all fnite, you can find a power series solution: plug $$f(x)=\sum_{k=0}^\infty c_k x^k,$$ and you obtain a recurrency which determines $c_k$: $$c_{k+2}=\frac{1}{(k+1)(k+2)}\left((\lambda-b_k)c_k-c_{k-1}\right).$$ So you can set $c_0$ and $c_1$ arbitrarily, and then determine all $c_k$. To obtain a convergent ...



Only top voted, non community-wiki answers of a minimum length are eligible