The Maths Tailor

Whenever a sum or product of $cos k θ$ or $sin k θ$ (or a multiple angle relation) appears, write each term as the real or imaginary part of $e^{ik θ}$ , so the whole problem becomes a geometric series, a binomial expansion, or the roots of $z^{n} = 1$ , which are pure algebra.

Trigger: a finite sum of cosines or sines, a product of many sines or tangents, or a statement linking $cos n θ$ to $cos θ$ .

Instances:
(i) $\sum cos k θ = Re \sum e^{ik θ}$ , summed as a geometric series.
(ii) The roots of $z^{n} = 1$ are $e^{2 π ik / n}$ , so a polynomial whose roots are $tan (2 π k / n)$ or $cos (2 π k / n)$ falls out of $Im$ or $Re$ of $(cos θ + i sin θ)^{n}$ .
(iii) Binomial expansion of $(1 + e^{2 i β})^{n}$ gives a closed form for a binomially weighted cosine sum.
(iv) $(e^{x} \pm e^{- x})^{n}$ expands $cosh^{n} x$ or $sinh^{n} x$ into a sum of $cosh k x$ .

Let

n

be a positive integer. The polynomial

p

is defined by the identity

p (cos θ) \equiv cos ((2 n + 1) θ) + 1.

(i) Show that

cos ((2 n + 1) θ) = r = 0 \sum n (2 r 2 n + 1) cos^{2 n + 1 - 2 r} θ (cos^{2} θ - 1)^{r} .

(ii) By considering the expansion of

(1 + t)^{2 n + 1}

for suitable values of

t

, show that the coefficient of

x^{2 n + 1}

in the polynomial

p (x)

2^{2 n}

.

(iii) Show that the coefficient of

x^{2 n - 1}

in the polynomial

p (x)

- (2 n + 1) 2^{2 n - 2}

.

(iv) It is given that there exists a polynomial

q

such that

p (x) = (x + 1) [q (x)]^{2}

and the coefficient of

x^{n}

q (x)

is greater than

0

.

Write down the coefficient of

x^{n}

in the polynomial

q (x)

and, for

n \geq 2

, show that the coefficient of

x^{n - 2}

in the polynomial

q (x)

2^{n - 2} (1 - n) .

Part (i). De Moivre's theorem gives

(cos θ + i sin θ)^{2 n + 1} = cos (2 n + 1) θ + i sin (2 n + 1) θ

. Expanding the left side with the binomial theorem and taking real parts, each term with

k

factors of

i sin θ

contributes to the real part only when

k

is even. Writing

k = 2 r

cos (2 n + 1) θ = r = 0 \sum n (2 r 2 n + 1) cos^{2 n + 1 - 2 r} θ (i sin θ)^{2 r} .

Since

(i sin θ)^{2 r} = (- 1)^{r} sin^{2 r} θ

and

sin^{2} θ = 1 - cos^{2} θ

, we have

(- 1)^{r} sin^{2 r} θ = (cos^{2} θ - 1)^{r}

, giving the required sum.

Part (ii). Every term in the sum for

cos (2 n + 1) θ

has the form

(2 r 2 n + 1) cos^{2 n + 1 - 2 r} θ (cos^{2} θ - 1)^{r}

. Expanding

(cos^{2} θ - 1)^{r}

and collecting the

cos^{2 n + 1} θ

contribution from each

r

, the coefficient of

x^{2 n + 1}

p (x)

equals

\sum_{r = 0}^{n} (2 r 2 n + 1)

.

To evaluate this sum, start from

(1 + t)^{2 n + 1} = \sum_{i = 0}^{2 n + 1} (i 2 n + 1) t^{i}

. Substituting

t = 1

and

t = - 1

2^{2 n + 1} = i = 0 \sum 2 n + 1 (i 2 n + 1), 0 = i = 0 \sum 2 n + 1 (i 2 n + 1) (- 1)^{i} .

Adding these two equations cancels all odd-indexed terms, leaving

2^{2 n + 1} = 2 \sum_{r = 0}^{n} (2 r 2 n + 1)

, so the coefficient of

x^{2 n + 1}

2^{2 n}

.

Part (iii). Tracking which term contributes a

cos^{2 n - 1} θ

factor: the

r

-th term of the sum, when

(cos^{2} θ - 1)^{r}

is expanded, provides a

cos^{2 n - 1} θ

contribution of

(- r) \cdot (2 r 2 n + 1)

. Summing over

r = 0

n

(the

r = 0

term contributes zero), the coefficient of

x^{2 n - 1}

p (x)

equals

\sum_{r = 0}^{n} (- r) (2 r 2 n + 1)

.

Differentiate

(1 + t)^{2 n + 1} = \sum_{i = 0}^{2 n + 1} (i 2 n + 1) t^{i}

with respect to

t

(2 n + 1) (1 + t)^{2 n} = i = 0 \sum 2 n + 1 i (i 2 n + 1) t^{i - 1} .

Substituting

t = 1

and

t = - 1

gives two equations. Taking their difference cancels even-indexed terms:

- (2 n + 1) \cdot 2^{2 n} = 2 r = 0 \sum n (- 2 r) (2 r 2 n + 1),

\sum_{r = 0}^{n} (- r) (2 r 2 n + 1) = - (2 n + 1) \cdot 2^{2 n - 2}

, which is the coefficient of

x^{2 n - 1}

.

Part (iv). Write

q (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + \dots

with

a_{n} > 0

. Expanding

(x + 1) [q (x)]^{2}

, the leading coefficient of

x^{2 n + 1}

a_{n}^{2} = 2^{2 n}

, giving

a_{n} = 2^{n}

(positive root).

Since

cos (2 n + 1) θ

contains only odd powers of

cos θ

, the polynomial

p (x)

has no even-degree terms, so the coefficient of

x^{2 n}

p (x)

is zero. From the expansion, this coefficient is

a_{n}^{2} + 2 a_{n} a_{n - 1} = 2^{2 n} + 2^{n + 1} a_{n - 1} = 0

, giving

a_{n - 1} = - 2^{n - 1}

.

Equating the coefficient of

x^{2 n - 1}

with the result from Part (iii):

a_{n - 1}^{2} + 2 a_{n} a_{n - 2} + 2 a_{n} a_{n - 1} = - (2 n + 1) \cdot 2^{2 n - 2} .

Substituting

a_{n} = 2^{n}

and

a_{n - 1} = - 2^{n - 1}

2^{2 n - 2} + 2^{n + 1} a_{n - 2} - 2^{2 n} = - (2 n + 1) \cdot 2^{2 n - 2} .

Dividing through by

2^{n}

2^{n - 2} + 2 a_{n - 2} - 2^{n} = - (2 n + 1) \cdot 2^{n - 2},

2 a_{n - 2} = 2^{n - 2} (4 - 1 - (2 n + 1)) = 2^{n - 2} (2 - 2 n),

a_{n - 2} = 2^{n - 2} (1 - n) .