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The inverse function theorem states that a

C^{1}

map on an open subset of

R^{2}

has a

C^{1}

local inverse near a point where its derivative is invertible; the inverse derivative at the image point is the inverse matrix of the original derivative.

Put

s = 1 + x + y

. Quotient differentiation gives

D f (x, y) = \frac{1}{s ^{2}} (1 + y - y - x 1 + x), det D f (x, y) = \frac{1}{s ^{3}} .

(1)

The entries are smooth and the determinant is nonzero everywhere on the stated domain, so the theorem supplies the required local inverse at every point. The point

(1/3, 1/3)

is the image of

(1, 1)

, where

D f (1, 1) = \frac{1}{9} (2 - 1 - 1 2) .

(2)

Consequently

D f^{- 1} (1/3, 1/3) = (6336) .

(3)

Directly solving the two rational equations also gives

f^{- 1} (u, v) = (u / (1 - u - v), v / (1 - u - v))

u + v \neq = 1

, consistent with the derivative calculation.

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Past exam of the options I course of the University of Oxford / 2010 / E / E2 / b / iii by

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Past exam of the options I course of the University of Oxford / 2010 / E / E2 / b / ii by

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Past exam of the options I course of the University of Oxford / 2010 / E / E2 / b / i by

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Past exam of the options I course of the University of Oxford / 2010 / E / E2 / a / iii by

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Past exam of the options I course of the University of Oxford / 2010 / E / E2 / a / ii by

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Past exam of the options I course of the University of Oxford / 2010 / E / E2 / a / i by

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Past exam of the options I course of the University of Oxford / 2010 / E / E1 / b / Solution by

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For

(i) \Rightarrow (ii)

, suppose

B = U \cap F

with

U

open and

F

closed. Since

B \subseteq F

, also

\overline{B} \subseteq F

. Therefore

U \cap \overline{B} = B

, showing that

B

is relatively open in its closure.

For

(ii) \Rightarrow (iii)

, write

B = U \cap \overline{B}

with

U

open in

X

. Then

\overline{B} ∖ B = \overline{B} \cap (X ∖ U)

is an intersection of closed sets, hence closed in

X

For

(iii) \Rightarrow (i)

, set

F = \overline{B}

and

U = X ∖ (\overline{B} ∖ B)

. The assumed condition makes

U

open;

F

is closed and

U \cap F = B

. Thus the three conditions are equivalent. They characterize a locally closed subset.

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Past exam of the options I course of the University of Oxford / 2010 / E / E1 / a / iii by

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Past exam of the options I course of the University of Oxford / 2010 / E / E1 / a / ii by

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Past exam of the options I course of the University of Oxford / 2010 / D / D2 / 2 / Solution by

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For

x \geq 1

∣ sin (1/ x) ∣ \leq 1/ x

, so

∣ g (x) ∣ \leq x^{- 3/2}, \int_{1}^{\infty} x^{- 3/2} d x = 2.

(1)

Thus $g$ is integrable on this interval. But for

x > 1

x^{2} - 1 < x

, hence

h (x) > 1/ x

. The integral of

1/ x

diverges at infinity, so $h$ is not integrable. Defining

h (1) = 0

changes only a null set and has no effect on this conclusion.

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Both functions are integrable on $(0, 1)$ . The oscillatory function is measurable and bounded by

1

in absolute value, on an interval of finite measure. For the other function,

\int_{0}^{1} h (x) d x = \int_{0}^{1} \frac{d x}{1 - x ^{2}} = \frac{π}{2} < \infty.

(1)

In particular its singularity at the right endpoint is of integrable inverse-square-root type.

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Since

∣ f ∣ = f^{+} + f^{-}

, part (3) and closure under sums prove that

∣ f ∣

is measurable whenever

f

is.

The converse is false. Let

A \subseteq [0, 1]

be a non-Lebesgue-measurable set and define

f = 1

A

and

f = - 1

on its complement in

[0, 1]

. Then

∣ f ∣ = 1

is measurable, but

f^{- 1} ([0, \infty)) = A

is not. Taking absolute values can conceal a nonmeasurable sign choice.

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f

is measurable, then for a threshold

a \leq 0

{f^{+} \geq a} = E

, while for

a > 0

{f^{+} \geq a} = {f \geq a}

. The half-line criterion makes

f^{+}

measurable. By part (2),

- f

is measurable, and

f^{-} = (- f)^{+}

is therefore measurable as well.

Conversely, if

f^{+}

and

f^{-}

are measurable, part (2) and the permitted closure under sums show that

f = f^{+} - f^{-}

is measurable. Thus measurability is equivalent to measurability of both parts.

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For every real

a

{x \in E : - g (x) \geq a} = E ∖ {x \in E : g (x) > - a} .

(1)

Part (1), applied to measurable

g

, makes the set being complemented measurable. Its relative complement in measurable

E

is measurable too. The permitted half-line criterion therefore proves $- g$ is measurable.

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For every real

b

f^{- 1} ((b, \infty)) = n = 1 ⋃ \infty f^{- 1} ([b + 1/ n, \infty)) .

(1)

Each set on the right is measurable by hypothesis. Closure of a sigma-algebra under countable unions proves the assertion.

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Past exam of the options I course of the University of Oxford / 2010 / C / C1 / iv / Solution by

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Again set

ϵ = (a / p) \in {1, - 1}

. The Euler criterion applied to

ϵ

gives

χ (ϵ) = ϵ^{(p - 1) /2}

, as an equality of signs. If

p \equiv 1 (mod 4)

, the exponent is even and the result is

1

; if

p \equiv 3 (mod 4)

, it is odd and the result is

ϵ

. Hence

\frac{( \frac{a}{p} )}{p} = {1, (a / p), p \equiv 1 (mod 4), p \equiv 3 (mod 4) .

(1)

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Write

χ (a) = (a / p)

. If

p ∣ a

, the claimed identity has both sides zero. Otherwise let

ϵ = χ (a) \in {1, - 1}

. Applying the Euler criterion gives

χ (ϵ a) \equiv (ϵ a)^{(p - 1) /2} \equiv ϵ^{(p - 1) /2} χ (a) = ϵ^{(p + 1) /2} (mod p) .

(1)

Both sides are in

{1, - 1}

, which are distinct residue classes because

p

is odd. Therefore the modular congruence is an equality:

\frac{( \frac{a}{p} ) a}{p} = (\frac{a}{p})^{(p + 1) /2} .

(2)

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The Euler criterion states that for

p ∤ a

a^{(p - 1) /2} \equiv (\frac{a}{p}) (mod p) .

(1)

The same modular congruence holds for

p ∣ a

, with both sides zero.

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 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!

Video 1.

Intro to OurBigBook

. Source.

We have two killer features:

topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculus
Articles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
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/derivative
Video 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:
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
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.
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 4.
Visual Studio Code extension tree navigation
.
Figure 5.
Web editor
. 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.
Video 4.
OurBigBook Visual Studio Code extension editing and navigation demo
. Source.
Internal cross file references done right:
Infinitely deep tables of contents:
Figure 6.
Dynamic article tree with infinitely deep table of contents
.
Live URL: ourbigbook.com/cirosantilli/chordate
Descendant pages can also show up as toplevel e.g.: ourbigbook.com/cirosantilli/chordate-subclade

All our software is open source and hosted at: github.com/ourbigbook/ourbigbook

Further documentation can be found at: docs.ourbigbook.com

Feel free to reach our to us for any help or suggestions: docs.ourbigbook.com/#contact

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