Embedding of $\sqrt{|i-j|}$ distance into $(\mathbb{R}^n,\lVert \cdot\rVert_2)$

Question

Consider the metric space $(X=\{1,\ldots,n\},d)$ such that:

$$d(i,j)=\sqrt{|i-j|}$$

Can $(X,d)$ be isometrically embedded in $(\mathbb{R}^n,\lVert \cdot\rVert_2)$? If that is the case, can we find some natural isometry $\phi:X\to \mathbb{R}^n$?

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To add some context, I consider the random walk:

$$S_n=\sum_{i=1}^n X_i$$

where the $X_i$'s are independent standard Gaussians. $S=(S_1,\ldots,S_n)$ is normally distributed (because so are its projections along each direction), so there should exist $a_1,\ldots,a_n\in\mathbb{R}^n$ such that:

$$S\equiv (\langle a_1,g\rangle,\ldots,\langle a_n,g\rangle)$$

where $g$ is a $n$-dimensional standard Gaussian. But it turns out: $$|i-j|=\mathbb{E}(S_i-S_j)^2=\lVert a_i-a_j\rVert^2$$

which implies the existence of the embedding. I was wondering whether there exists an explicit proof of that (surely, there must be!).

Answer 1

Here's such an embedding: define $$ \phi(j) = (\overbrace{1,\dots,1}^j,0,\cdots,0). $$