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How to show if $cos^n(x^2-y^2)$ is a valid mercer kernel function if $n$ is positive?

For $cos(x^2-y^2)$ I would assume that: $cos(x^2-y^2) = sin(x^2)sin(y^2)+cos(x^2)cos(y^2)$ Is a valid mercer kernel with feature map $\phi(x) = (cos(x^2), sin(x^2))^T$

user1727556
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    What's your definition of a valid kernel? Does this satisfy that definition? – Sycorax Apr 07 '20 at 22:22
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    $\cos(.)$ can be negative, is this an issue? – Xi'an Apr 08 '20 at 07:51
  • kernel function is explained here: https://stats.stackexchange.com/questions/152897/how-to-intuitively-explain-what-a-kernel-is – user1727556 Apr 08 '20 at 09:32
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    I'm not asking what a kernel is *abstractly.* I'm asking what **your definition** is. The word "kernel" is used differently in different places, and knowing what definition **you** use is important. – Sycorax Apr 08 '20 at 12:48
  • It is same as explained in the link. – user1727556 Apr 08 '20 at 12:59
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    The question you've linked asks for intuitive explanations about what a kernel is; none of the answers provide a definition. For example, a Mercer kernel is (1) symmetric and (2) positive semidefinite. Neither of these criteria is expressly manifest on the linked page. Are you asking about Mercer kernels? Or something else? – Sycorax Apr 08 '20 at 14:13
  • Yes you are correct, Mercer kernels. – user1727556 Apr 08 '20 at 18:17
  • I've updated the question description and title to be more specific. – user1727556 Apr 08 '20 at 20:00
  • Mercer kernels are positive semi-definite. This means that the sum $\sum_{i=1}^n \sum_{j=1}^n k(x_i, x_j) c_i c_j \ge 0$ for all finite sequences $x_i, x_j$ and all choices of real numbers $c_i, c_j$. We can demonstrate (2) is not satisfied if we can find any example satisfying these hypotheses which violates the inequality. Suppose we have $n=2$. Can you choose $x_i, x_j, c_i, c_j$ such that this inequality is violated? – Sycorax Apr 10 '20 at 19:32

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