Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 335 / 2 / ii

Use

(F g) (q) = \int_{R^{3}} g (r) e^{- i q \cdot r} d r, F^{- 1} h = \frac{1}{( 2 π ) ^{3}} \int h (q) e^{i q \cdot r} d q .

The adjoint operator is consequently

F^{*} = (2 π)^{3} F^{- 1} .

With a unitary Fourier normalization, this is simply

F^{*} = F^{- 1}

At fixed incident direction and wavenumber, define

(T V) (r) = - \frac{1}{4 π} \int_{D} e^{- i k_{0} (r - x_{0}) \cdot r} V (r) d r .

Taking the complex conjugate of the kernel gives

(T^{*} g) (r) = - \frac{1}{4 π} \int_{S^{2}} e^{i k_{0} (r - x_{0}) \cdot r} g (r) d S (r), r \in D .

The least-squares minimizer of

∥ T V - f_{\infty} ∥^{2}

T^{*} T V = T^{*} f_{\infty}

. Fourier inversion on the measured transfer-vector set is exactly the corresponding Moore--Penrose reconstruction

T^{†} f_{\infty}

; hence the formal solution in part (i) is the minimum-norm least-squares solution when the data are incomplete or inconsistent.

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