Simpson's Dissection

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Theorem

Let $\omega = e^{2 i \pi / q}$ be a primitive $q$th root of unity.

Let $p \not \equiv 0 \pmod q$.

Let:

$\ds \map f x = \sum_{n \mathop = 0}^\infty a_n x^n$

Then:

$\ds \sum_{n \mathop = 0}^\infty a_{n q + p} x^{n q + p} = \dfrac 1 q \sum_{j \mathop = 0}^{q - 1} \omega^{- j p} \map f {\omega^j x}$


Proof

Expanding the sum on the right hand side, we obtain:

\(\ds \dfrac 1 q \sum_{j \mathop = 0}^{q - 1} \omega^{- j p} \map f {\omega^j x}\) \(=\) \(\ds \frac 1 q \times \omega^0 \times \paren {a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 1 q \times \omega^{-p} \times \paren {a_0 + a_1 x \omega + a_2 x^2 \omega^2 + a_3 x^3 \omega^3 + \cdots}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 1 q \times \omega^{-2 p} \times \paren {a_0 + a_1 x \paren {\omega^2} + a_2 x^2 \paren {\omega^2}^2 + a_3 x^3 \paren {\omega^2}^3 + \cdots}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cdots\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 1 q \times \omega^{-\paren {q - 1} p} \times \paren {a_0 + a_1 x \paren {\omega^{q - 1} } + a_2 x^2 \paren {\omega^{q - 1} }^2 + a_3 x^3 \paren {\omega^{q - 1} }^3 + \cdots}\)

Summing down the $q$ values of the term with exponent n in the array shown above, we obtain:

$\ds \dfrac 1 q a_n x^n \times \paren {1 + \omega^{1 \paren {n - p} } + \omega^{2 \paren {n - p} } + \omega^{3 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} } }$


From Sum of Powers of Primitive Complex Roots of Unity, we have:

\(\ds \sum_{j \mathop = 0}^{q - 1} \omega^{j \paren {n - p} }\) \(=\) \(\ds 1 + \omega^\paren {n - p} + \omega^{2 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} }\)
\(\ds \) \(=\) \(\ds \begin {cases} q & : q \divides \paren {n - p} \\ 0 & : q \nmid \paren {n - p} \end {cases}\)


When $n \equiv p \pmod q$:

$\ds \dfrac 1 q a_n x^n \times \paren {1 + \omega^{1 \paren {n - p} } + \omega^{2 \paren {n - p} } + \omega^{3 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} } } = a_n x^n$

When $n \not \equiv p \pmod q$:

$\ds \dfrac 1 q a_n x^n \times \paren {1 + \omega^{1 \paren {n - p} } + \omega^{2 \paren {n - p} } + \omega^{3 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} } } = 0$


Hence:

\(\ds \dfrac 1 q \sum_{j \mathop = 0}^{q - 1} \omega^{- j p} \map f {\omega^j x}\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty a_{n q + p} x^{n q + p}\)

$\blacksquare$


Source of Name

This entry was named for Thomas Simpson.


Sources

  • 1757: Thomas Simpson: The invention of a general method for determining the sum of every 2d, 3d, 4th, or 5th, &c. term of a series, taken in order; the sum of the whole series being known (Phil. Trans. Vol. 50: pp. 757 – 769)
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