If my chalkboard scribblings are correct, if $f(x) = \cos(\alpha x)$ and $g(x) = \sin(\alpha x)$ for $\alpha \neq 0$, then $$ \mathscr{T}_k f(x) = \alpha^{-2}(f(x) + \alpha x\sin(\alpha T) - 1), \quad \mathscr{T}_k g(x) = \alpha^{-2}(g(x) - \alpha x \cos(\alpha T))$$ (up to some inconsequential signs), so that your orthonormal basis of eigenfunctions is $$ \psi_k(x) = \sqrt{\frac{2}{T}} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) $$ with corresponding eigenvalues $$ \lambda_k = \frac{T^2}{\pi^2(k+\tfrac{1}{2})^2}. $$ Hence, $$ k(x,y) = \sum_{k=0}^\infty \frac{2T}{\pi^2(k+\tfrac{1}{2})^2} \sin \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}y\right).$$ I hope this works!
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If my chalkboard scribblings are correct, if $f(x) = \cos(\alpha x)$ and $g(x) = \sin(\alpha x)$ for $\alpha \neq 0$, then $$ \mathscr{T}_k f(x) = \alpha^{-2}(f(x) + \sin(\alpha alpha x\sin(\alpha T) - 1), \quad \mathscr{T}_k g(x) = \alpha^{-2}(g(x) - \alpha x \cos(\alpha T))$$ (up to some inconsequential signs), so that your orthonormal basis of eigenfunctions is $$ \psi_k(x) = \sqrt{\frac{2}{T}} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) $$ with corresponding eigenvalues $$ \lambda_k = \frac{T^2}{\pi^2(k+\tfrac{1}{2})^2}. $$ Hence, $$ k(x,y) = \sum_{k=0}^\infty \frac{2T}{\pi^2(k+\tfrac{1}{2})^2} \sin \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}y\right).$$ I hope this works! |
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If my chalkboard scribblings are correct, if $f(x) = \cos(\alpha x)$ and $g(x) = \sin(\alpha x)$ for $\alpha \neq 0$, then $$ \mathscr{T}_k f(x) = \alpha^{-2}(f(x) + \sin(\alpha T) - 1), \quad \mathscr{T}_k g(x) = \alpha^{-2}(g(x) - \cos(\alpha T))$$ (up to some inconsequential signs), so that your orthonormal basis of eigenfunctions is $$ \psi_k(x) = \sqrt{\frac{2}{T}} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) $$ with corresponding eigenvalues $$ \lambda_k = \frac{T^2}{\pi^2(k+\tfrac{1}{2})^2}. $$ Hence, $$ k(x,y) = \sum_{k=0}^\infty \frac{2T}{\pi^2(k+\tfrac{1}{2})^2} \sin \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}y\right).$$ I hope this works! |
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