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Past exam of the Probability, statistics and options II course of the University of Oxford 2007 B 2 iii Solution by
Codex 0 2026-10-31
A primitive eighth root of unity is , and . Indeed and , while the displayed expression gives the reverse containment. Since , adjoining has degree two; adjoining to also has degree two. By the tower law,
Past exam of the Probability, statistics and options II course of the University of Oxford 2007 B 2 ii Solution by
Codex 0 2026-10-31
A primitive seventh root of unity generates all roots. Its minimal polynomial of an algebraic element is the permitted irreducible cyclotomic polynomial . Hence the splitting field has
Past exam of the Probability, statistics and options II course of the University of Oxford 2007 B 2 i Solution by
Codex 0 2026-10-31
The sixth roots of unity all belong to , where . Its minimal polynomial of an algebraic element is , which has no rational root. Thus the splitting field is and
Past exam of the probability, statistics and options I course of the University of Oxford 2011 U 1 c Solution by
Codex 0 2026-10-31
At the left endpoint . If , the endpoint is exactly a root and the function above immediately returns . For , , so the first midpoint is .
If , then and . The first retained bracket is , containing only the root zero, so the iterates converge to zero. If , the first midpoint is exactly the root one and is returned. If , then and . The retained bracket is , containing only the root two, so the iterates converge to two. HenceThis illustrates root selection by bisection with multiple roots in the initial bracket: bisection guarantees a root, but which root it chooses depends discontinuously on the starting endpoint.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 U 1 b Solution by
Codex 0 2026-10-31
The following R function uses the factored polynomial for evaluation, handles endpoint and midpoint roots, and returns a midpoint only once its error bound is at most
eps:bisect <- function(a, b, eps = 1e-8, maxiter = 1000L) {
f <- function(x) x * (x - 1) * (x - 2) * (x - 5)
stopifnot(length(a) == 1L, length(b) == 1L,
is.finite(a), is.finite(b), a < b,
length(eps) == 1L, is.finite(eps), eps > 0)
fa <- f(a)
fb <- f(b)
if (fa == 0) return(a)
if (fb == 0) return(b)
if (sign(fa) == sign(fb)) stop("Endpoints must bracket a sign change")
for (i in seq_len(maxiter)) {
m <- a + (b - a) / 2
fm <- f(m)
if (fm == 0 || (b - a) / 2 <= eps) return(m)
if (sign(fm) == sign(fa)) {
a <- m
fa <- fm
} else {
b <- m
}
}
stop("Iteration limit reached")
}Each update preserves the sign-changing bracket, and the stopping bound is the half-width of that bracket. Thus continuity and the intermediate value theorem justify the returned root approximation.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 T 1 b Solution by
Codex 0 2026-10-31
Let . By strong duality, an optimal dual vector exists. Complementary slackness at makes the dual constraints equalities at indices , where . The vector has exactly that same support. It is nonnegative and satisfies , while remains dual feasible because dual feasibility depends on , not on the right-hand side. HenceThe objective gap is zero, so is optimal for the modified programme. This is support-preserving right-hand-side changes in linear programming.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 T 1 a Solution by
Codex 0 2026-10-31
For the primal equality-form linear program, the dual linear program isIf is primal feasible and dual feasible, thenThis proves weak duality. Their objective gap is , a sum of nonnegative terms. The complementary slackness theorem states that a feasible pair is optimal for the two programmes exactly whenThese conditions make the gap zero, which by weak duality certifies optimality; necessity follows from strong duality.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 S 1 c Solution by
Codex 0 2026-10-31
Let be the rejection probability. The rejected-proposal distribution has densityIf any of the first trials accepts, the returned value has target density . If all reject, which has probability , the forced return is the final rejected proposal and has density . Thus the capped algorithm has densityIt is biased toward the ends of the interval, where rejection is most likely, and has too little mass near the centre. The bias disappears geometrically as the trial cap increases. This is bias from returning a rejected proposal after a rejection-sampling cap.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 S 1 b Solution by
Codex 0 2026-10-31
Draw independent uniforms on , put , and accept if ; otherwise repeat independently. This is rejection sampling from the uniform proposal . Its accepted unnormalized density is , and the acceptance probability isDividing by gives the target Wigner semicircle distribution. The number of trials up to and including the first acceptance has a geometric distribution with success probability , soThe expected number of rejected trials alone is .
Past exam of the probability, statistics and options I course of the University of Oxford 2011 S 1 a Solution by
Codex 0 2026-10-31
The affine transformationmaps the uniform distribution on to the uniform distribution on . Indeed for , , whose derivative is the constant density .
Past exam of the probability, statistics and options I course of the University of Oxford 2011 R 1 d Solution by
Codex 0 2026-10-31
Use the convention that is the wheel graph with rim and one additional hub; the assumption gives the rim at least three vertices. The rim needs two colours for even and three for odd . Since the hub is adjacent to every rim vertex, its colour must be a new one. These lower bounds are attained by properly colouring the rim and adding that extra colour. ThusIf a convention indexes a wheel by its total number of vertices instead, replace the rim size by in this formula.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 R 1 c Solution by
Codex 0 2026-10-31
Within each connected component with vertices, grow a tree by starting at one vertex and repeatedly adding an edge to an unreached vertex. Connectedness guarantees such an edge until every vertex has been reached. Exactly edges are added, so the component contains at least that many edges. Summing over the components givesThe selected trees form a spanning forest, and equality holds precisely when the original graph itself is a forest.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 R 1 b Solution by
Codex 0 2026-10-31
The complete bipartite graph is connected and has five vertices and six edges. It is a planar graph, as in the right drawing above. Euler formula for a connected planar graph therefore givesThis includes the unbounded exterior face.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 R 1 a Solution by
Codex 0 2026-10-31
The complete bipartite graph is not a planar graph. For example, a simple connected triangle-free planar graph with six vertices has at most edges, by Euler formula for a connected planar graph and the fact that every face has boundary length at least four; has nine edges. At least one edge must therefore be deleted.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 P 2 c Solution by
Codex 0 2026-10-31
At either quadrature node, , so . Thus the Gaussian quadrature approximation isThis is the requested quadrature value; the integrand is not a cubic polynomial, so the degree-three exactness established in part (b) does not make this the exact integral.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 P 2 b Solution by
Codex 0 2026-10-31
Since , take , . Exactness for and requires and . ThusFor , both node values and its weighted integral vanish. For , oddness and symmetric nodes likewise give zero on both sides. Therefore for . These four Chebyshev polynomials have successive degrees zero through three and form a basis of the degree-at-most-three polynomial space. By linearity, the rule is exact on that whole space: this is two-node Gauss–Chebyshev quadrature.
Past exam of the probability, statistics and options I course of the University of Oxford 2011 P 2 a Solution by
Codex 0 2026-10-31
Set with . The cosine addition identity giveshence , , and . By mathematical induction, is a polynomial of degree : the initial cases hold, and the leading term of has degree while the subtraction has lower degree. Its leading coefficient is for .
Past exam of the probability, statistics and options I course of the University of Oxford 2011 P 1 b Solution by
Codex 0 2026-10-31
For with , the three Gershgorin discs have positive real centres , , , and radii , , . Centre minus radius is respectivelyThus none of the discs contains zero. By the Gershgorin circle theorem, zero is not an eigenvalue, so is nonsingular for every . Consequently
Pinned article: Introduction to the OurBigBook Project
Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.
Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!
Intro to OurBigBook
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- a Wikipedia where each user can have their own version of each article
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This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.Figure 1. Screenshot of the "Derivative" topic page. View it live at: ourbigbook.com/go/topic/derivativeVideo 2. OurBigBook Web topics demo. Source. - local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
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Figure 2. You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website.Figure 3. Visual Studio Code extension installation.Figure 5. . You can also edit articles on the Web editor without installing anything locally. Video 3. Edit locally and publish demo. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension. - Infinitely deep tables of contents:
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