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 Jun 20 awarded Notable Question Oct 28 awarded Popular Question Oct 27 awarded Curious Apr 17 accepted convolution computation involving $e^{-x^2}$ Apr 16 accepted differential equation - solving a second-order ODE with variable coefficients Apr 16 asked convolution computation involving $e^{-x^2}$ Apr 16 asked Analysis - Fourier Transforms - show that convolution of characteristic functions is continuous Mar 30 awarded Tumbleweed Mar 23 accepted If $\mu$ is a complex measure, every set $E$ has $A \subset E$ so that $|\mu(A)| \ge \frac{1}{\pi}|\mu|(E).$ Mar 23 comment If $\mu$ is a complex measure, every set $E$ has $A \subset E$ so that $|\mu(A)| \ge \frac{1}{\pi}|\mu|(E).$ thank you very much Mar 23 asked differential equation - solving a second-order ODE with variable coefficients Mar 23 asked If $\mu$ is a complex measure, every set $E$ has $A \subset E$ so that $|\mu(A)| \ge \frac{1}{\pi}|\mu|(E).$ Mar 3 asked solve a system of linear differential equations with variable coefficients Nov 20 awarded Scholar Nov 20 accepted inner product and adjoint operator Nov 20 accepted self-adjoint operator and unitary orthogonal matrix Nov 20 accepted Find an adjoint; inner product Nov 20 accepted dimension of orthonormal set Nov 19 asked self-adjoint operator and unitary orthogonal matrix Nov 12 comment Find an adjoint; inner product Thank you. I obtained this same result, as noted above, only I used constants m,n rather than a,b. But I am not sure if the adjoint can be expressed in terms of the linear function's coefficients. Note how alpha is originally defined; the definition does not depend on its particular coefficients, but rather p(0) and p(1).