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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-8/2/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 8 / 2 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

First let

n \geq 0

. The ordinary Leibniz rule, applied to a test function

h

, gives

\partial^{n} (f h) = \sum_{r = 0}^{n} (r n) f^{(r)} h^{(n - r)} .

Equivalently, the composition of differential operators is

\partial^{n} m_{f} = r = 0 \sum n (r n) m_{f^{(r)}} \partial^{n - r} .

For completeness, this formula follows by induction: applying

\partial

differentiates either the coefficient or the remaining derivative of

h

, and the two contributions combine by

(r n) + (r - 1 n) = (r n + 1)

. The initial case is multiplication by

f

.

For all integers

n

(0 n) = 1

and

(r n) = n (n - 1) \dots (n - r + 1) / r!

. The formal pseudodifferential composition rule is

\partial^{n} m_{f} = r \geq 0 \sum (r n) m_{f^{(r)}} \partial^{n - r}, n \in Z .

For

n = - s < 0

,

(r - s) = (- 1)^{r} (r s + r - 1)

, so this is usually an infinite series. In particular

\partial^{- 1} m_{f} = m_{f} \partial^{- 1} - m_{f^{'}} \partial^{- 2} + m_{f^{''}} \partial^{- 3} - \dots .

Applying

\partial

on the left gives

m_{f}

: all later coefficient derivatives cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in

n

, gives every negative power. More explicitly, applying

\partial

to the proposed

n

-series yields coefficient

(r n) + (r - 1 n) = (r n + 1)

at

f^{(r)} \partial^{n + 1 - r}

, so it is consistent with the already determined

(n + 1)

-series. Negative powers here are formal; differentiation on all of

C^{\infty} (R)

has a nontrivial kernel and has no genuine two-sided inverse there.

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