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Thus the fixed-point iteration converges to the unique real root of a polynomial from every $x_{0} \in [0, 1/7]$ . This explicit error estimate proves the needed convergence directly.

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Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 5 / a / i / Solution 2026-10-25

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For the homogeneous linear recurrence relation, try

u_{n} = r^{n}

. Its characteristic equation of a linear recurrence is

r^{2} + 3 r - 4 = (r - 1) (r + 4) = 0.

(1)

The distinct roots of a polynomial

1, - 4

give the complementary terms

A + B (- 4)^{n}

. A constant cannot absorb the forcing because

1

is already a root of a polynomial of the characteristic equation; try a particular solution

c n

. Substitution gives

c = - 4 c + 1

, hence

c = 1/5

. The general solution is

u_{n} = A + B (- 4)^{n} + n /5

The initial conditions give

A + B = 0

and

A - 4 B = 4/5

. Hence

A = 4/25

B = - 4/25

, and

u_{n} = \frac{n}{5} + \frac{4}{25} (1 - (- 4)^{n}), n \geq 0.

(2)

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 2 / c / Solution 2026-10-20

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The quadratic formula gives

z = - b \pm b^{2} - 1

. For

- 1 \leq b \leq 1

, write this as

z_{\pm} = - b \pm i 1 - b^{2} .

(1)

Each root of a polynomial has modulus

1

, since

b^{2} + (1 - b^{2}) = 1

. As

b

increases from

- 1

1

, the upper root traverses the upper unit semicircle from

1

- 1

, and the lower root traverses the lower semicircle in the opposite angular direction. At

b = - 1

the roots coincide at

1

; at

b = 1

they coincide at

- 1

. For intermediate

b

they are complex conjugates.

Figure 1.
The conjugate roots travel along the two unit semicircles
. Arrows show increasing $b$ , with endpoints $b = - 1$ and $b = 1$ and the midpoint pair $b = 0$ . The figure is an original plot of the solution formula. **Details**.

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Past exam of the numerical analysis course of the University of Oxford / 2017 / 3 / b / iii / Solution 2026-10-31

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Fix

i

, write

α = x_{i - 1}

β = x_{i}

, and let

E = T - S

on this interval. The Hermite interpolation conditions give both

E (α) = E (β) = 0, E^{'} (α) = E^{'} (β) = 0.

(1)

Two applications of integration by parts therefore give

\int_{α}^{β} E T^{(4)} d x = [E T^{'''} - E^{'} T^{''}]_{α}^{β} + \int_{α}^{β} E^{''} T^{''} d x = \int_{α}^{β} E^{''} T^{''} d x .

(2)

On this single cubic piece, the same calculation also gives

\int_{α}^{β} E^{''} S^{''} d x = [E^{'} S^{''} - E S^{'''}]_{α}^{β} + \int_{α}^{β} E S^{(4)} d x = 0.

(3)

Here

S^{(4)} = 0

inside the interval, and all four endpoint values of

E, E^{'}

are zero. Since

T^{''} = E^{''} + S^{''}

, we obtain the exact identity

\int_{x_{i - 1}}^{x_{i}} (T - S) T^{(4)} d x = \int_{x_{i - 1}}^{x_{i}} (T^{''} - S^{''})^{2} d x \geq 0.

(4)

This is an integral inequality, with no assertion that the integrand on the left has one sign pointwise. The stronger nodal derivative conditions make the boundary terms vanish on each interval separately; part (i) only needed their cancellation after summing.

The stated Hermite interpolating polynomial exists uniquely in

Π_{2 n + 1}

. Indeed, a polynomial in the kernel of a linear map of the value-and-derivative data map has a double root of a polynomial at each of

n + 1

distinct interpolation nodes. A nonzero such polynomial would have polynomial degree at least

2 n + 2

, which is impossible. The map is therefore an injective function and a linear map between vector spaces of the same finite dimension of a vector space,

2 n + 2

, and hence is a linear isomorphism. No higher smoothness of the original function

f

is required to construct it.

An explicit illustration uses interpolation nodes

0, 1, 2

, values

0, 1, 0

, and zero derivatives at every interpolation node. The clamped cubic spline is

S (x) = 3 x^{2} - 2 x^{3}

[0, 1]

and its reflection on

[1, 2]

. Its Hermite interpolating polynomial is the quartic polynomial

T (x) = x^{2} (2 - x)^{2}

. On

[0, 1]

T - S = x^{2} (1 - x)^{2}

, and the corresponding reflected formula holds on

[1, 2]

. Direct integration gives

J (S) = 24

J (T) = 128/5

, and

J (T - S) = 8/5

, exactly as the energy identity predicts.

Figure 1.
A clamped cubic spline and a Hermite interpolating polynomial with the same nodal values and derivatives: the excess second derivative roughness penalty is $8/5$
. **Details**.

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Past exam of the numerical linear algebra and numerical solution of differential equations course of the University of Oxford / 2020 / 2 / c / Solution 2026-10-31

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For a constant-coefficient linear multistep method, zero-stability means uniform boundedness of the propagation of starting perturbations as

h \to 0

. Equivalently, every solution of the homogeneous linear recurrence

\sum_{j = 0}^{k} α_{j} v_{n + j} = 0

(1)

satisfies

∣ v_{n} ∣ \leq K max_{0 \leq j < k} ∣ v_{j} ∣

for all

n \geq 0

, with

K

independent of

n

and the starting values. The equivalent root condition for a multistep method is that every root of a polynomial

ζ

of the degree-

k

polynomial

ρ

satisfies

∣ ζ ∣ \leq 1

, and every root on

∣ ζ ∣ = 1

is simple. Indeed, a root of multiplicity

m

produces recurrence modes

n^{q} ζ^{n}

0 \leq q < m

: repeated roots strictly inside the disk are harmless, whereas roots outside it or repeated roots on its boundary produce unbounded modes.

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Past exam of the rings and modules course of the University of Oxford / 2021 / 3 / c / ii / Solution 2026-10-25

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Write

q = ∣ F ∣

. The unit group

U (F) = F ∖ {0}

is a finite abelian group of group order

q - 1

, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated module

U (F) ≅ Z_{d_{1}} \oplus \dots \oplus Z_{d_{r}}, 1 < d_{1} ∣ d_{2} ∣ \dots ∣ d_{r},

(1)

after omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put

D = d_{r}

. Every element

u \in U (F)

satisfies

u^{D} = 1

, because the order of a group element in each coordinate divides

D

. Hence all

q - 1

distinct nonzero elements of

F

are roots of a polynomial

X^{D} - 1

. The root bound for a polynomial over a field gives

q - 1 \leq D .

(2)

For completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial

f

vanishes at

a

, write

f = (X - a) g

; every other root of a polynomial is a root of

g

, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition is

q - 1 = d_{1} \dots d_{r} \geq d_{r} = D .

(3)

Equality is forced, so

d_{1} \dots d_{r - 1} = 1

. As every retained factor is greater than

1

, there can be only one. Consequently

U (F) ≅ Z_{q - 1} is cyclic .

(4)

The same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since

q - 1

need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 1 / d / ii / Solution 2026-10-31

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Try a linear function

f (x) = c x

. The functional equation becomes

c^{2} x = (c + 1) x

for all

x

, so

c

must be a root of a polynomial

c^{2} - c - 1

. Its two roots of a polynomial give

f (x) = \frac{1 + 5}{2} x or f (x) = \frac{1 - 5}{2} x .

(1)

For either choice,

c^{2} = c + 1

proves the required equality directly. The positive coefficient is the golden ratio.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 7 / c / iii / Solution 2026-10-31

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The characteristic polynomial of the homogeneous linear recurrence is

(r - 2)^{2}

, so its repeated-root solutions are

(A + B n) 2^{n}

. A constant particular solution is one. Thus

u_{n} = 1 + (A + B n) 2^{n}

. The initial conditions give

A = - 1

and

B = 1

, hence

u_{n} = 1 + (n - 1) 2^{n} .

(1)

The factor

n

is required by the repeated root of a polynomial; omitting it would lose one independent solution.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming / 1 / b / Solution 2026-10-25

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The coefficient assumptions imply a nonconstant polynomial with

p (0) < 0

and a positive leading coefficient. Thus

p (x) \to + \infty

x \to + \infty

: dividing by its leading power of

x

leaves the positive leading coefficient plus terms tending to zero. There is some

B > 0

with

p (B) > 0

. A polynomial is continuous, so the intermediate value theorem supplies at least one positive root of a polynomial in

(0, B)

This gives neither uniqueness nor monotonicity. Multiple crossings are possible. If the degree of a polynomial is an even integer, the same leading-term argument gives

p (x) \to + \infty

x \to - \infty

, so there is also a negative root of a polynomial. For an odd integer degree of a polynomial no negative root of a polynomial is forced; a linear polynomial with negative constant coefficient and positive slope is a counterexample.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming / 1 / c / Solution 2026-10-25

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Maintain opposite signs at the endpoints, rather than assuming that the polynomial is monotone:

def floorRoot(X: Int): Int = {
  require(X > 0 && eval(0.0) <= 0.0 && eval(X.toDouble) > 0.0)
  var lo = 0
  var hi = X
  while (hi - lo > 1) {
    val mid = lo + (hi - lo) / 2
    if (eval(mid.toDouble) <= 0.0) lo = mid
    else hi = mid
  }
  lo
}

The loop invariant is $0 \leq lo < hi \leq X$ and $p (lo) \leq 0 < p (hi)$ . Both branches preserve those signs, even when there are other roots of a polynomial or sign reversals elsewhere. When the interval width exceeds one, the midpoint is strictly between its endpoints, so the positive integer loop variant hi - lo strictly decreases. On exit hi == lo + 1, giving the required postcondition $p (lo) \leq 0 < p (lo + 1)$ . Continuity establishes a root of a polynomial between those adjacent integers, possibly at the left endpoint.

The width is at most halved with rounding upwards, so there are at most

⌈ lo g_{2} X ⌉

midpoint evaluations, plus two endpoint evaluations in the precondition check. The midpoint expression avoids adding two large endpoints. The explicit positive upper-bound precondition supplies the interval order assumed in the question. For the literal computed-eval contract, the same proof holds provided evaluated midpoints are not NaN; identifying its bracket with the real polynomial additionally requires trustworthy evaluation signs.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming / 1 / d / Solution 2026-10-25

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First find a sign bracket by exponential search, then bisect that bracket without evaluating its endpoints again:

def floorRoot(): Int = {
  require(eval(0.0) < 0.0)
  var lo = 0
  var hi = 1
  while (eval(hi.toDouble) <= 0.0) {
    lo = hi
    require(hi <= Int.MaxValue / 2, "doubling bound exceeds Int range")
    hi *= 2
  }
  while (hi - lo > 1) {
    val mid = lo + (hi - lo) / 2
    if (eval(mid.toDouble) <= 0.0) lo = mid
    else hi = mid
  }
  lo
}

For the first while loop the loop invariant is

p (lo) \leq 0

, with hi a positive power of two, and either

(lo, hi) = (0, 1)

hi = 2 lo

. Every previously tested power of two has nonpositive value. Let

t

be the least nonnegative integer for which

p (2^{t}) > 0

. It exists by the positive leading coefficient. If the current upper endpoint is

2^{k}

, a mathematical loop variant is

t - k

while the guard holds; the next doubling decreases it by one. The first while loop's postcondition is

p (lo) \leq 0 < p (hi)

. The second while loop has exactly the invariant, variant and adjacent-integer postcondition established in part c.

There are

t + 1

evaluations at powers of two. If

t \geq 1

, the final bracket has width

2^{t - 1}

and takes exactly

t - 1

midpoint evaluations. Including the initial sign check, the total is $2 t + 1$ evaluations for $t \geq 1$ , and two for $t = 0$ . Thus evaluation cost is

O (lo g U)

for the first positive dyadic bound

U = 2^{t}

; it is not a claim about the first positive root of a polynomial when the polynomial is nonmonotone.

The mathematical argument uses unbounded integer endpoints and reliable signs. The guarded Int implementation requires the successful dyadic bound to fit in Int; otherwise it explicitly fails rather than wrapping to a negative bound. The coefficient assumptions alone cannot guarantee that a requested integer result fits a fixed-width type. Floating-point overflow or unreliable signs likewise need a wider or certified evaluator for a fully general implementation.

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Polynomial Created 2026-09-24 Updated 2026-10-18

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A polynomial is a finite sum of monomials with coefficients in a ring. In one variable

X

over a commutative ring

R

, write

p (X) = a_{0} + a_{1} X + \dots + a_{n} X^{n}, a_{j} \in R .

(1)

The symbol

X

is an indeterminate: it records powers formally before we choose an input value. Finite expressions are easy to add, multiply, evaluate, and differentiate, making polynomials a common language for exact algebra and approximate numerical analysis.

Consider the real polynomial

p (X) = X^{3} - 3 X^{2} + 4 = (X + 1) (X - 2)^{2} .

(2)

The expanded form displays the coefficients and the degree of a polynomial; the factored form displays its roots of a polynomial. At

X = 3

, either description gives

p (3) = 4

. At

X = - 1

and

X = 2

, the factored form makes the zero values immediate. Keeping several descriptions of one object is useful because each makes a different question easy.

Figure 1.
A simple polynomial root crosses the axis, while a double root touches it; perturbing the double factor can produce real or complex roots
. **Details**.

The graph shows another connection. The factor

X + 1

changes sign across

- 1

, while

(X - 2)^{2}

stays nonnegative across

2

. Thus the graph crosses the axis at the simple root of a polynomial and touches it at the double root. Multiplicity of a root turns an algebraic repetition into visible local behavior.

For calculations, Horner's method evaluates a polynomial through nested multiplication. For modeling, a Taylor polynomial describes local behavior near a point, while polynomial interpolation constructs a polynomial matching prescribed data. Their usefulness comes with assumptions: a local approximation and an exact identity are different claims, and changing the coefficient ring can change which polynomial roots exist.

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Polynomial division Created 2026-09-24 Updated 2026-10-18

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Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given

f, g \in F [X]

over a field

F

, with

g \neq = 0

, there are unique polynomials

q, r

satisfying

f = g q + r, r = 0 or de g r < de g g .

(1)

Here

q

is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, take

f = 2 X^{3} + 3 X^{2} - 5 X + 6, g = X^{2} + X - 2.

(2)

Polynomial long division first cancels

2 X^{3}

by subtracting

2 X g

f - 2 X g = X^{2} - X + 6.

(3)

Subtracting one more copy of

g

cancels

X^{2}

(X^{2} - X + 6) - g = - 2 X + 8.

(4)

Thus

q = 2 X + 1, r = - 2 X + 8, f = (X^{2} + X - 2) (2 X + 1) + (- 2 X + 8) .

(5)

The final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term

- 2 + 8 = 6

Figure 1.
Polynomial long division cancels the cubic and quadratic terms in two steps, recording $q = 2 X + 1$ and $r = - 2 X + 8$
. **Details**.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of

g

can be discarded. At any root of a polynomial of

g

, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing

X

2 X + 1

with an integer polynomial quotient and a constant polynomial remainder would require

1 = 2 c

for an integer

c

, which is impossible. Over the rational numbers the division instead gives

X = (2 X + 1) \frac{1}{2} - \frac{1}{2} .

(6)

The obstruction is the nonunit leading coefficient

2

. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

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Polynomial remainder theorem 2026-10-18

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Dividing a polynomial