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Define the Dirichlet Kernel by

$$D_{N}(\mathbf{x})=\sum_{k=-N}^{N}...\sum_{k=-N}^{N}\phi_{k}(\mathbf{x})= \prod_{i=1}^{d}\frac{\sin((N+\frac{1}{2})x_{i})}{\sin(\frac{x_{i}}{2})}$$

I am asked to derive this product form.

I begin by taking the sum of a finite geometric series

$$S_{N}=\frac{a-r^{N}}{1-r}$$

whence, in terms of the function above,

$$S_{N}(\mathbf{x})=\frac{a-\phi_{k}^{N}(\mathbf{x})}{1-\phi_{k}(\mathbf{x})}$$

Note that $\phi_{\mathbf{k}}(\mathbf{x}) = \exp(i \mathbf{kx})$; in particular, $a=\phi_{\mathbf{k}}(0) = \exp(0) = 1$. Hence

$$S_{N}(\mathbf{x})=\frac{1-(\exp(iN\mathbf{kx})}{1-(\exp(i\mathbf{kx})}$$

I am not sure how to proceed here. I have tried different things but I sort of just cycle around. I feel like I am supposed to use a trigonometric identity after expanding $\exp(\cdot)$ into Euler's formula but I have tried different identities and I feel like I am going around in circles.

Edit: Sorry for the weird ****. My computer doesn't seem to format latex characters on StackExchange properly and where it stays bolded when I ask the question, after posting it, it reverts to ****. Can someone please edit the question to make it display $x$ and $k$ bolded properly.

Abolfazl Chaman motlagh
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therickster
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    How are the $\phi_k$ defined ? – Jean Marie Mar 07 '23 at 19:21
  • these are the fourier coefficients. sorry i forgot to mention. but it is explicitly given as exp(ikx) – therickster Mar 07 '23 at 20:12
  • Now, 1) how $x$ is related to the $x_k$ ? 2) how is $d$ defined in the RHS ? Besides, can you express the LHS without the strange successive $\Sigma$s ? – Jean Marie Mar 07 '23 at 20:29
  • Why haven't you answered my questions ? – Jean Marie Mar 08 '23 at 18:58
  • sorry about the late response. The other person had given a reply that helped me get the answer. But $x$ is a $d$-dimensional vector with each $x_k$ being one of the components. $\phi_k = exp(ikx)$ is the fourier term for each set of $x \in R^d$ and $k \in N^d$. – therickster Mar 10 '23 at 06:49
  • Thanks for you answer : your notation $k=... -N \cdots N$ meaning that $k$ is an integer and $\phi_k(x)=\exp(ikx)$ or more precisely $\exp(i(k \ . \ x))$ where $k$ becomes a vector has been confusing to me... – Jean Marie Mar 10 '23 at 09:12
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    Yes sorry about that, that was the notation I was given, that k is one integer and k (boldface) refers to the vector of all k's for d dimensions. I was trying to show that k . x was a dot product of two vectors but my computer did not format it on StackExchange properly for some reason. – therickster Mar 12 '23 at 02:37

1 Answers1

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You can express each dimension independently knowing the property of $e^{x+y}=e^x e^y$.
But first we should solve it for one dimension. the summation is from $-N$ to $N$ so it has $2N$ terms not $N$ terms! and first term is $\phi_{-N}(x)$ not $\phi_N(0)$! since it's a geometric series $\sum_{k=-N}^{N} ar^k$ with $a=\phi_{-N}(x)$ and $r=e^{ix}$ we have : $$ \sum_{k=-N}^{N}e^{ikx} = e^{-iNx}(\frac{1-e^{i(2N+1)x}}{1-e^{ix}} )= \frac{e^{-iNx}-e^{i(N+1)x}}{1-e^{ix}} \times \frac{e^{-ix/2}}{e^{-ix/2}} = \frac{e^{-i(N+1/2)x}-e^{i(N+1/2)x}}{e^{-ix/2}-e^{ix/2}} \\ = \frac{\sin\big((N+1/2)x\big)}{\sin(x/2)} = S_N(x). $$ Now we can solve it for multidimensional kernel: $$ S=\sum_{k_d=-N}^{N}\dots \sum_{k_1=-N}^{N} e^{i \mathbf{k.x}} = \sum_{k_d=-N}^{N}\dots \sum_{k_1=-N}^{N} e^{i \sum_{j=1}^{d}k_jx_j} = \sum_{k_d=-N}^{N}\dots \sum_{k_1=-N}^{N} \prod_{j=1}^d e^{ik_jx_j} \\ = \sum_{k_d=-N}^{N}\dots \sum_{k_2=-N}^{N} \prod_{j=2}^d e^{ik_jx_j} \underbrace{(\sum_{k_1=-N}^{N} e^{ik_1x_1})}_{S_N(x_1)} = S_N(x_1) \bigg( \sum_{k_d=-N}^{N}\dots \sum_{k_2=-N}^{N} \prod_{j=2}^d e^{ik_jx_j} \bigg) \\ = S_N(x_1) \sum_{k_d=-N}^{N}\dots \sum_{k_3=-N}^{N}\prod_{j=3}^d e^{ik_jx_j} \underbrace{(\sum_{k_2=-N}^{N} e^{ik_2x_2})}_{S_N(x_2)} =S_N(x_1)S_N(x_2) \bigg( \sum_{k_d=-N}^{N}\dots \sum_{k_3=-N}^{N} \prod_{j=3}^d e^{ik_jx_j} \bigg) $$ and you can keep going to $j=d$ and you get : $$ S = \prod_{j=1}^{d} S_N(x_j) = \prod_{j=1}^{d} \frac{\sin\big((N+1/2)x_j\big)}{\sin(x_j/2)} $$

Abolfazl Chaman motlagh
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