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# cargo

## NAME

cargo - The Rust package manager

## SYNOPSIS

cargo [OPTIONS] COMMAND [ARGS]
cargo [OPTIONS] --version
cargo [OPTIONS] --list
cargo [OPTIONS] --help
cargo [OPTIONS] --explain CODE

## DESCRIPTION

This program is a package manager and build tool for the Rust language, available at https://rust-lang.org.

## COMMANDS

### Build Commands

cargo-bench(1)

Execute benchmarks of a package.

cargo-build(1)

Compile a package.

cargo-check(1)

Check a local package and all of its dependencies for errors.

cargo-clean(1)

Remove artifacts that Cargo has generated in the past.

cargo-doc(1)

Build a package’s documentation.

cargo-fetch(1)

Fetch dependencies of a package from the network.

cargo-fix(1)

Automatically fix lint warnings reported by rustc.

cargo-run(1)

Run a binary or example of the local package.

cargo-rustc(1)

Compile a package, and pass extra options to the compiler.

cargo-rustdoc(1)

Build a package’s documentation, using specified custom flags.

cargo-test(1)

Execute unit and integration tests of a package.

### Manifest Commands

cargo-generate-lockfile(1)

Generate Cargo.lock for a project.

cargo-locate-project(1)

Print a JSON representation of a Cargo.toml file’s location.

Output the resolved dependencies of a package, the concrete used versions including overrides, in machine-readable format.

cargo-pkgid(1)

Print a fully qualified package specification.

cargo-update(1)

Update dependencies as recorded in the local lock file.

cargo-verify-project(1)

Check correctness of crate manifest.

### Package Commands

cargo-init(1)

Create a new Cargo package in an existing directory.

cargo-install(1)

Build and install a Rust binary.

cargo-new(1)

Create a new Cargo package.

cargo-search(1)

Search packages in crates.io.

cargo-uninstall(1)

Remove a Rust binary.

### Publishing Commands

Save an API token from the registry locally.

cargo-owner(1)

Manage the owners of a crate on the registry.

cargo-package(1)

Assemble the local package into a distributable tarball.

cargo-publish(1)

Upload a package to the registry.

cargo-yank(1)

Remove a pushed crate from the index.

### General Commands

cargo-help(1)

cargo-version(1)

Show version information.

## OPTIONS

### Special Options

-V
--version

Print version info and exit. If used with --verbose, prints extra information.

--list

List all installed Cargo subcommands. If used with --verbose, prints extra information.

--explain CODE

Run rustc --explain CODE which will print out a detailed explanation of an error message (for example, E0004).

### Display Options

-v
--verbose

Use verbose output. May be specified twice for "very verbose" output which includes extra output such as dependency warnings and build script output. May also be specified with the term.verbose config value.

-q
--quiet

No output printed to stdout.

--color WHEN

Control when colored output is used. Valid values:

• auto (default): Automatically detect if color support is available on the terminal.

• always: Always display colors.

• never: Never display colors.

May also be specified with the term.color config value.

### Manifest Options

--frozen
--locked

Either of these flags requires that the Cargo.lock file is up-to-date. If the lock file is missing, or it needs to be updated, Cargo will exit with an error. The --frozen flag also prevents Cargo from attempting to access the network to determine if it is out-of-date.

These may be used in environments where you want to assert that the Cargo.lock file is up-to-date (such as a CI build) or want to avoid network access.

--offline

Prevents Cargo from accessing the network for any reason. Without this flag, Cargo will stop with an error if it needs to access the network and the network is not available. With this flag, Cargo will attempt to proceed without the network if possible.

Beware that this may result in different dependency resolution than online mode. Cargo will restrict itself to crates that are downloaded locally, even if there might be a newer version as indicated in the local copy of the index. See the cargo-fetch(1) command to download dependencies before going offline.

May also be specified with the net.offline config value.

### Common Options

-h
--help

Prints help information.

-Z FLAG…

Unstable (nightly-only) flags to Cargo. Run cargo -Z help for details.

## ENVIRONMENT

See the reference for details on environment variables that Cargo reads.

## Exit Status

0

Cargo succeeded.

101

Cargo failed to complete.

## FILES

~/.cargo/

Default location for Cargo’s "home" directory where it stores various files. The location can be changed with the CARGO_HOME environment variable.

$CARGO_HOME/bin/ Binaries installed by cargo-install(1) will be located here. If using rustup, executables distributed with Rust are also located here. $CARGO_HOME/config

.cargo/config

Cargo automatically searches for a file named .cargo/config in the current directory, and all parent directories. These configuration files will be merged with the global configuration file.

$CARGO_HOME/credentials Private authentication information for logging in to a registry. $CARGO_HOME/registry/

\$CARGO_HOME/git/

## EXAMPLES

1. Build a local package and all of its dependencies:

cargo build
2. Build a package with optimizations:

cargo build --release
3. Run tests for a cross-compiled target:

cargo test --target i686-unknown-linux-gnu
4. Create a new package that builds an executable:

cargo new foobar
5. Create a package in the current directory:

mkdir foo && cd foo
cargo init .
6. Learn about a command’s options and usage:

cargo help clean


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# Coloring sides of truncated triangular dihedral(bipiramid) into 3 colours

I need to find out the amount of ways to colour truncated triangular dihedron into 3 colours. So, the task will be easier if I had simple triangular dihedron.

First of all, do I understand right that truncated dihedron looks like : enter image description here The only things I know, that I must use Burnsside's lemma, calculate simmetries ( rotations ).

By the way, if you know some beginners algebra books with decent amount of examples in practice and more-or-less easy to understand for not english-speaker, thank you.

• As far as I remember, "dihedron" means "a body having two faces". The least possible number of faces of a body in $3D$ is however 4 (tetrahedron)... A dihedron would be just a two-sided plane figure. – user Jan 10 at 10:21
• My mistake, I meant bipyramid, sorry – Tovarisch Jan 10 at 10:26
• The next issue is meaning of "truncated". I would assume it means it is truncated by planes parallel to the base, so that the body has 8 faces (two equilateral triangles and 6 isosceles trapezoids). – user Jan 10 at 11:05
• @user I had this question too. In this case truncated means the dihedral consist of two truncated triangular piramids, alternatively saying, instead of lower and upper vericles there are triangles. So, the number of size is 6+2. – Tovarisch Jan 10 at 12:32

Note first of all that we obtain a cube when placing vertices at the centers of the hexagonal faces and connecting those where the faces share an edge. So we could solve this problem by asking about colorings of the vertices and the faces of the cube. We will work with the given solid however. (The cube was solved at this MSE link).

We evidently require the cycle index. There is the identity and three types of rotations. With $$a_q$$ for hexagonal faces and $$b_p$$ for the pyramids at the corners we get from the identity:

$$a_1^8 b_1^6.$$

The rotations about an axis passing through opposite pyramidal corners contribute

$$3 \times (2 a_4^2 b_1^2 b_4 + a_2^4 b_1^2 b_2^2).$$

The flips about an axis passing through the centers of two opposite edges ($$180$$ degree rotation) contribute

$$6 \times a_2^4 b_2^3.$$

Finally, the rotations by $$60$$ degrees and $$120$$ degrees about an axis passing through the centers of opposite hexagonal faces yields

$$4 \times 2 a_1^2 a_3^2 b_3^2.$$

We thus have for the cycle index

$$Z(T) = \frac{1}{24} (a_1^8 b_1^6 + 6 a_4^2 b_1^2 b_4 + 3 a_2^4 b_1^2 b_2^2 + 6 a_2^4 b_2^3 + 8 a_1^2 a_3^2 b_3^2).$$

We then get by Burnside for colorings with at most $$N$$ colors shared between hexagonal faces and pyramidal corners the closed form:

$$\frac{1}{24}(N^{14} + 6 N^5 + 3 N^8 + 6 N^7 + 8 N^6).$$

This gives the sequence

$$1, 776, 200961, 11198720, 254387500, 3265470936, 28260264606, \\ 183254654976, 953206454115, 4166682025000, \ldots$$

in particular for $$N=3$$ we obtain

$$\bbox[5px,border:2px solid #00A000]{ 200961.}$$

The formula when working with exactly $$N$$ colors as opposed to at most $$N$$ colors is (colors are shared between pyramids and hexagons):

$$\frac{N!}{24}\left({14\brace N} + 6 {5\brace N} + 3 {8\brace N} + 6 {7\brace N} + 8 {6\brace N}\right).$$

This gives the finite sequence

$$1, 774, 198636, 10399528, 200395755, 1903119150, 10359167700, \\ 35132702640, 77643165600, 113816102400, 109880971200, \\ 67199932800, 23610787200, 3632428800, 0, 0, \ldots$$

because there are only fourteen slots available. Note that the last one is $$14!/24 = 3632428800.$$

Remark. This cycle index also solves the problem where colors are not shared between hexagons and pyramids. We find:

$$\frac{1}{24} (N^8 M^6 + 6 N^2 M^3 + 3 N^4 M^4 + 6 N^4 M^3 + 8 N^4 M^2).$$

• I fucking love u man, best of luck to you, but as I see u did it for the figire of picture I provided in question . The figure I must solve is simple dihedral, but it consist of 2 truncated piramids, so we have 8 faces to colour ( imagine like we cut upper and lower part a bit). Am I right, that I need just do it for a18 without b? Sorry for bad english – Tovarisch Jan 10 at 15:48
• There are a number of examples at MSE Meta. I suggest you consult these and learn the technique in order to be able to solve this type of problem yourself. – Marko Riedel Jan 10 at 16:37
• Yep, I will have work for today's evening probably, thank you again mate – Tovarisch Jan 10 at 17:02
• for this dihedral picture I found: 1) 2 rotations by 120 degrees, which change only the vericles in the middle triangle + identity e( nothing changes) 2) 3 rotations by 180 degrees passing through medians of triangles ( so 2 triangles vericles exchange and also small triangles exchange) is it all? or I missed something? Because I found your solution where almost same figure and there is more cycles... – Tovarisch Jan 12 at 15:02
• As I am trying to understand I need to find group of rotations, for example, lets upper 1 triangle be 1, lower be 2, 3 faces of upper part 2,3,4 and lower part 5,6,7, so I can write rotations as permulations , for example for identity (1)(2)(3)(4)(5)(6)(7)(8), for rotations of triangles vericles (1)(234)(567)(8) and so on, and then the length of cycles will be degrees for N in burnside formula? Sorry for molestaion. – Tovarisch Jan 12 at 15:03

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# Generator of a Quotient Group

Along the same vein of another question I've posted is another question that merges group theory with number theory. I havn't taken a formal course on group theory, and is likely why I'm stumped at these questions posed from elementary number theory. The question goes as

Let $p$ be an odd prime and $g$ a generator of the group $(\mathbb{Z}/p \mathbb{Z})^*$. I need to show that either $g$ or $g+p$ is a generator of $(\mathbb{Z}/p^2 \mathbb{Z})^*$

I'm not sure where to begin. Just need a push in the right direction.

Thanks!

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Hint: reducing mod $p$ you would have to be a generator too. –  fretty Nov 5 '12 at 18:59
Show that if $g^{p-1} \equiv 1 \bmod p^2$ then $(g+p)^{p-1} \not\equiv 1 \bmod p^2$. –  Derek Holt Nov 5 '12 at 20:02
Note that an element of $(\mathbb Z/p^2\mathbb Z)^*$ has order $p(p-1)$ (that is, it is a generator) if it has order $>p$. Note further that is order $\pmod {p^2}$ cannot be smaller than its order $\pmod p$. –  martini Nov 5 '12 at 20:54

If $g$ is not primitive mod $p^2$, the fact that $\mathrm{ord}(g)$ divides $p(p-1)$ and is a multiple of $p-1$ implies that $\mathrm{ord}(g)=p-1$ exactly. The same holds for $g+p$.

So if neither $g$ nor $g+p$ is primitive mod $p^2$, we must have $g^{p-1} \equiv 1 \mod p^2$ and $(g+p)^{p-1} \equiv 1 \mod p^2$. (In particular, $g^{p-1} \equiv (g+p)^{p-1} \mod p^2$.) Yet $$(g+p)^{p-1} \equiv g^{p-1} + \binom{p-1}{1}g^{p-2}p \equiv g^{p-1}-pg^{p-2}\mod p^2,$$ and if this is congruent to $g^{p-1}$ mod $p^2$ it follows that $pg^{p-2} \equiv 0 \mod p^2$. This forces $g \equiv 0 \mod p$ as $\mathbb{Z}/p\mathbb{Z}$ is an integral domain. This is a contradiction, and our result follows.

-
sorry, I'm just re-reading this a couple days later and realized I don't quite understand the line "If g is not primitive mod p2, the fact that ord(g) divides p(p−1) *and is a multiple of p−1* implies that ord(g)=p−1 exactly. We know since <g> = (Z/pZ)* that |g| in (Z/pZ)* = (p-1). Is this what implies |g| in (Z/p^2 Z)* is a multiple of (p-1)? –  user45793 Nov 8 '12 at 19:29
or is it that since |(Z/p^2 Z)*| = p(p-1) that implies there exists a generator of order p(p-1) and hence every element of (Z/p^2 Z)* has order p(p-1) as well. –  user45793 Nov 8 '12 at 19:33
..but am I able to assume right away that (Z/p^2 Z)* is cyclic? –  user45793 Nov 8 '12 at 19:35
Let $k$ be the order of $g$ mod $p^2$. Then $g^k \equiv 1 \mod p^2$, so in particular $g^k \equiv 1 \mod p$. But this implies that $k$ is a multiple of the order of $g$ in $\mathbb{Z}/p\mathbb{Z}$, which we know is $p-1$ (as $g$ is a generator mod $p$). In other words, $p-1$ divides $k$. –  A Walker Nov 8 '12 at 19:51
More generally, let $\phi :G \to H$ be a group homomorphism. If $g^k =e_G$, then $\phi(g)^k = \phi(e_G)=e_H$, so $\mathrm{ord}(\phi(g))$ divides $\mathrm{ord}(g)$ (if both are finite). In our case, our map is projection $\pi: \mathbb{Z}/p^2 \mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$. –  A Walker Nov 8 '12 at 19:54


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# the fundamentals of abstract mathematics - U of L Personal Web Sites

2014 • 201 Pages • 1.05 MB • English
Posted April 14, 2020 • Submitted by adelia.doyle

PREVIEW PDF

## Summary of the fundamentals of abstract mathematics - U of L Personal Web Sites

Proofs and Concepts the fundamentals of abstract mathematics by Dave Witte Morris and Joy Morris University of Lethbridge incorporating material by P.D. Magnus University at Albany, State University of New York Preliminary Version 0.92 of December 2016 This book is oﬀered under a Creative Commons license. (Attribution-NonCommercial-ShareAlike 2.0)

Preface Unlike in earlier courses, success in advanced undergraduate mathematics classes (and beyond) does not depend nearly so much on being able to �nd the right answer to a question as it does on being able to provide a convincing explanation that the answer is correct. (Mathematicians call this explanation a proof.) This textbook is designed to help students acquire this essential skill, by developing a working knowledge of: 1) proof techniques (and their basis in Logic), and 2) fundamental concepts of abstract mathematics. We start with the language of Propositional Logic, where the rules for proofs are very straightforward. Adding sets and quanti�ers to this yields First-Order Logic, which is the language of modern mathematics. Being able to do proofs in this setting is the main skill necessary for success in advanced mathematics. It is also important to be familiar with (and be able to prove statements about) sets and functions, which are the building blocks of modern mathematics. In addition, a chapter on Cardinality provides an introduction to the surprising notion of \uncountable sets": in�nite sets with so many elements that it is impossible to make a list x1; x2; x3; : : : of all of them (even if the list is allowed to be in�nitely long).

to Harmony

Contents Part I. Introduction to Logic and Proofs Chapter 1. Propositional Logic 3 x1.1. Assertions, deductions, and validity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 x1.2. Logic puzzles - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 6 x1.3. Using letters to symbolize assertions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 x1.4. Connectives (:, &, _, ), ,) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 8 x1.5. Determining whether an assertion is true . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 x1.6. Tautologies and contradictions - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 19 x1.7. Logical equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 x1.8. Converse and contrapositive - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 24 x1.9. Some valid deductions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Summary - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 28 Chapter 2. Two-Column Proofs 29 x2.1. First example of a two-column proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 x2.2. Hypotheses and theorems in two-column proofs - - - - - - - - - - - - - - - - - - - - - - - - - - - 31 x2.3. Subproofs for )-introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 x2.4. Proof by contradiction - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 41 x2.5. Proof strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 x2.6. What is a proof? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 47 x2.7. Counterexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Summary - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 51 Part II. Sets and First-Order Logic Chapter 3. Sets 55 x3.1. Propositional Logic is not enough . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 x3.2. Sets, subsets, and predicates - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 56 x3.3. Operations on sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Summary - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 71 Chapter 4. First-Order Logic 73 x4.1. Quanti�ers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 x4.2. Translating to First-Order Logic - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 74 x4.3. Negations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 x4.4. The introduction and elimination rules for quanti�ers - - - - - - - - - - - - - - - - - - - - - - 84 x4.5. Some proofs about sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 x4.6. Counterexamples (reprise) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 91 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 i

ii Chapter 5. Sample Topics 95 x5.1. Number Theory: divisibility and congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 x5.2. Abstract Algebra: commutative groups - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 100 x5.3. Real Analysis: convergent sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .105 Summary - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 107 Part III. Other Fundamental Concepts Chapter 6. Functions 111 x6.1. Cartesian product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .111 x6.2. Informal introduction to functions - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 114 x6.3. Oﬃcial de�nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .117 x6.4. One-to-one functions - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 119 x6.5. Onto functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .123 x6.6. Bijections - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 126 x6.7. Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .129 x6.8. Composition of functions - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 132 x6.9. Image and pre-image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .135 Summary - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 138 Chapter 7. Equivalence Relations 139 x7.1. Binary relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .139 x7.2. De�nition and basic properties of equivalence relations - - - - - - - - - - - - - - - - - - - - 141 x7.3. Equivalence classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .144 x7.4. Modular arithmetic - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 145 x7.5. Functions need to be well-de�ned . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .147 x7.6. Partitions - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 148 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .150 Chapter 8. Proof by Induction 151 x8.1. The Principle of Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .151 x8.2. Other proofs by induction - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 157 x8.3. Other versions of induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .162 x8.4. The natural numbers are well-ordered - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 163 x8.5. Applications in Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .164 Summary - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 166 Chapter 9. Cardinality 167 x9.1. De�nition and basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .167 x9.2. The Pigeonhole Principle - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 171 x9.3. Cardinality of a union . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .174 x9.4. Hotel In�nity and the cardinality of in�nite sets - - - - - - - - - - - - - - - - - - - - - - - - - - 176 x9.5. Countable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .179 x9.6. Uncountable sets - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 183 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .186 Index of De�nitions 187 List of Notation 189

Part I Introduction to Logic and Proofs

Chapter 1 Propositional Logic The grand aim of all science is to cover the greatest number of empirical facts by logical deduction from the smallest number of hypotheses or axioms. Albert Einstein (1879{1955), Nobel prize-winning physicist in Life magazine For our purposes, Logic is the business of deciding whether or not a deduction is valid; that is, deciding whether or not a particular conclusion is a consequence of particular assumptions. (The assumptions can also be called \hypotheses" or \axioms.") 1.1. Assertions, deductions, and validity We will begin our discussion of Logic by introducing three basic ingredients: assertions, deduc- tions, and validity. Here is one possible deduction: Hypotheses: (1) It is raining heavily. (2) If you do not take an umbrella, you will get soaked. Conclusion: You should take an umbrella. (The validity of this particular deduction will be analyzed in Example 1.1.10 below.) In Logic, we are only interested in sentences that can be a hypothesis or conclusion of a deduction. These are called \assertions": DEFINITION 1.1.1. An assertion is a sentence that is either true or false. OTHER TERMINOLOGY. Some textbooks use the term proposition or statement or sentence, instead of assertion. EXAMPLE 1.1.2. � Questions The sentence \Are you sleepy yet?" is not an assertion. Although you might be sleepy or you might be alert, the question itself is neither true nor false. For this reason, questions do not count as assertions in Logic. � Imperatives Commands are often phrased as imperatives like \Wake up!," \Sit up straight", and so on. Although it might be good for you to sit up straight or it might not, the command itself is neither true nor false. � Exclamations \Ouch!" is sometimes called an exclamatory sentence, but it is neither true nor false. so it is another example of a sentence that is not an assertion. 3

4 1. Propositional Logic Remark 1.1.3. Roughly speaking, an assertion is a statement of fact, such as \The earth is bigger than the moon" or \Edmonton is the capital of Alberta." However, it is important to remember that an assertion may be false, in which case it is a mistake (or perhaps a deliberate lie), such as \There are less than 1,000 automobiles in all of Canada." In many cases, the truth or falsity of an assertion depends on the situation. For example, the assertion \It is raining" is true in certain places at certain times, but is false at others. In this chapter and the next, which are introductory, we will deal mostly with assertions about the real world, where facts are not always clear-cut. (For example, if Alice and Bob are almost the same height, it may be impossible to determine whether it is true that \Alice is taller than Bob.") We are taking a mathematical (or scienti�c) view toward Logic, not a philosophical one, so we will ignore the imperfections of these real-world assertions, which provide motivation and illustration, because our goal is to learn to use Logic to understand mathematical objects (not real-world objects), where there are no grey areas. Throughout this text, you will �nd exercises that review and explore the material that has just been covered. There is no substitute for actually working through some problems, because this course, like most advanced mathematics, is more about a way of thinking than it is about memorizing facts. EXERCISES 1.1.4. Which of the following are \assertions" in the logical sense? 1) England is smaller than China. 2) Greenland is south of Jerusalem. 3) Is New Jersey east of Wisconsin? 4) The atomic number of helium is 2. 5) The atomic number of helium is �. 6) Take your time. 7) This is the last question. 8) Rihanna was born in Barbados. DEFINITION 1.1.5. A deduction is a series of hypotheses that is followed by a conclusion. (The conclusion and each of the hypotheses must be an assertion.) If the hypotheses are true and the deduction is a good one, then you have a reason to accept the conclusion. EXAMPLE 1.1.6. Here are two deductions. Hypotheses: All men are mortal. 1) Socrates is a man. Conclusion: Socrates is mortal. Hypotheses: The Mona Lisa was painted by Leonardo da Vinci. 2) Neil Armstrong was the �rst man on the moon. Conclusion: Justin Trudeau went swimming yesterday. The �rst of these deductions is very famous (and was discussed by the ancient Greek philosopher Aristotle), but the second one is lame. It may seem odd to even call it a deduction, because the two hypotheses have nothing at all to do with the conclusion, but, given our de�nition, it does count as a deduction. However, it is is a very poor one, so it cannot be relied on as evidence that the conclusion is true. We are interested in the deductions that do provide solid evidence for their conclusions:


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# Cardiopulmonary Pharmacology: Drug Actions

cardiopulmonary pharm ppt, pgs 1-5
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Pharmacology
study of drugs; their origin, nature, properties and effects
Pharmaceutical Phase
method by which a drug is delivered (route of administration and drug form)
Pharmacokinetic Phase
the time required for drug absorption, drug action, drug distribution in the body and metabolization and excretion of drug.
Pharmacodynamic Phase
refers to the mechanism of action by which a drug causes its therapeutic effect
Tolerance
when increased amounts of a drug are needed to produce the desired effect, possibly due to increased enzyme levels related to enzymatic activity that develops over a period of time
Tachyphylaxis
rapid development of drug tolerance
Cumulative Effect
exaggerated response and a possibly toxic situation
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an exaggerated response, occurs when two or more drugs are administered with the same effect on the body
Synergy
when two or more drugs produce an effect or response that neither could produce alone
Potentiation
either when two drugs produce an effect that is greater than what they usually produce when given alone or when one drug enhances the effect of another drug
Antagonism
when two drugs have opposite effects


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# Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board chapter 2 - Applications of Derivatives [Latest edition]

## Chapter 2: Applications of Derivatives

Exercise 2.1Exercise 2.2Exercise 2.3Exercise 2.4Miscellaneous Exercise 2
Exercise 2.1 [Page 72]

### Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board Chapter 2 Applications of Derivatives Exercise 2.1 [Page 72]

Exercise 2.1 | Q 1.1 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : y = x2 + 2ex 2 at (0, 4)

Exercise 2.1 | Q 1.2 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : x3 + y3 – 9xy = 0 at (2, 4)

Exercise 2.1 | Q 1.3 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : x^2 - sqrt(3)xy + 2y^2 = 5 "at"  (sqrt(3),2)

Exercise 2.1 | Q 1.4 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : 2xy + π sin y = 2pi  "at" (1, pi/2)

Exercise 2.1 | Q 1.5 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : x sin 2y = y cos 2x at (pi/4, pi/2)

Exercise 2.1 | Q 1.6 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : x = sin θ and y = cos 2θ at θ = pi/(6)

Exercise 2.1 | Q 1.7 | Page 72

Find the equations of tangents and normals to the following curves at the indicated points on them : x = sqrt(t), y = t  - (1)/sqrt(t) at = 4.

Exercise 2.1 | Q 2 | Page 72

Find the point of the curve y = sqrt(x - 3) where the tangent is perpendicular to the line 6x + 3y – 5 = 0.

Exercise 2.1 | Q 3 | Page 72

Find the points on the curve y = x3 – 2x2 – x where the tangents are parllel to 3x – y + 1 = 0.

Exercise 2.1 | Q 4 | Page 72

Find the equation of the tangents to the curve x2 + y2 – 2x – 4y + 1 =0 which a parallel to the X-axis.

Exercise 2.1 | Q 5 | Page 72

Find the equations of the normals to the curve 3x2 – y2 = 8, which are parallel to the line x + 3y = 4.

Exercise 2.1 | Q 6 | Page 72

If the line y = 4x – 5 touches the curves y2 = ax3 + b at the point (2, 3), find a and b.

Exercise 2.1 | Q 7 | Page 72

A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.

Exercise 2.1 | Q 8 | Page 72

A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area increasing, when its radius is 5 cm?

Exercise 2.1 | Q 9 | Page 72

The surface area of a spherical balloon is increasing at the rate of (2"cm"^2)/sec. At what rate is the volume of the balloon is increasing, when the radius of the balloon is 6 cm?

Exercise 2.1 | Q 10 | Page 72

If each side of an equilateral triangle increases at the rate of (sqrt(2)"cm")/sec, find the rate of increase of its area when its side of length 3 cm.

Exercise 2.1 | Q 11 | Page 72

The volume of a sphere increases at the rate of 20 cm3/sec. Find the rate of change of its surface area, when its radius is 5 cm

Exercise 2.1 | Q 12 | Page 72

The edge of a cube is decreasing at the rate of( 0.6"cm")/sec. Find the rate at which its volume is decreasing, when the edge of the cube is 2 cm.

Exercise 2.1 | Q 13 | Page 72

A man of height 2 metres walks at a uniform speed of 6 km/hr away from a lamp post of 6 metres high. Find the rate at which the length of the shadow is increasing

Exercise 2.1 | Q 14 | Page 72

A man of height 1.5 metres walks towards a lamp post of height 4.5 metres, at the rate of (3/4)"metre"/sec.
Find the rate at which
(ii) the tip of shadow is moving.

Exercise 2.1 | Q 15 | Page 72

A ladder 10 metres long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 metres per second, find how fast the top of the ladder is sliding down the wall, when the bottom is 6 metres away from the wall.

Exercise 2.1 | Q 16 | Page 72

If water is poured into an inverted hollow cone whose semi-vertical angle is 30°, so that its depth (measured along the axis) increases at the rate of( 1"cm")/sec. Find the rate at which the volume of water increasing when the depth is 2 cm.

Exercise 2.2 [Page 75]

### Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board Chapter 2 Applications of Derivatives Exercise 2.2 [Page 75]

Exercise 2.2 | Q 1.1 | Page 75

Find the approximate values of : sqrt(8.95)

Exercise 2.2 | Q 1.2 | Page 75

Find the approximate values of : root(3)(28)

Exercise 2.2 | Q 1.3 | Page 75

Find the approximate values of : root(5)(31.98)

Exercise 2.2 | Q 1.4 | Page 75

Find the approximate values of : (3.97)4

Exercise 2.2 | Q 1.5 | Page 75

Find the approximate values of : (4.01)3

Exercise 2.2 | Q 2.1 | Page 75

Find the approximate values of : sin 61° , given that 1° = 0.0174°, sqrt(3) = 1.732

Exercise 2.2 | Q 2.2 | Page 75

Find the approximate values of : sin (29° 30'), given that 1°= 0.0175°, sqrt(3) = 1.732

Exercise 2.2 | Q 2.3 | Page 75

Find the approximate values of : cos(60° 30°), given that 1° = 0.0175°, sqrt(3) = 1.732

Exercise 2.2 | Q 2.4 | Page 75

Find the approximate values of : tan (45° 40°), given that 1° = 0.0175°.

Exercise 2.2 | Q 3.1 | Page 75

Find the approximate values of : tan–1(0.999)

Exercise 2.2 | Q 3.2 | Page 75

Find the approximate values of : cot–1 (0.999)

Exercise 2.2 | Q 3.3 | Page 75

Find the approximate values of : tan–1 (1.001)

Exercise 2.2 | Q 4.1 | Page 75

Find the approximate values of : e0.995, given that e = 2.7183.

Exercise 2.2 | Q 4.2 | Page 75

Find the approximate values of : e2.1, given that e2 = 7.389

Exercise 2.2 | Q 4.3 | Page 75

Find the approximate values of : 32.01, given that log 3 = 1.0986

Exercise 2.2 | Q 5.1 | Page 75

Find the approximate values of : loge(101), given that loge10 = 2.3026.

Exercise 2.2 | Q 5.2 | Page 75

Find the approximate values of : loge(9.01), given that log 3 = 1.0986.

Exercise 2.2 | Q 5.3 | Page 75

Find the approximate values of : log10(1016), given that log10e = 0.4343.

Exercise 2.2 | Q 6.1 | Page 75

Find the approximate values of : f(x) = x3 – 3x + 5 at x = 1.99.

Exercise 2.2 | Q 6.2 | Page 75

Find the approximate values of : f(x) = x3 + 5x2 – 7x + 10 at x = 1.12.

Exercise 2.3 [Page 80]

### Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board Chapter 2 Applications of Derivatives Exercise 2.3 [Page 80]

Exercise 2.3 | Q 1.1 | Page 80

Check the validity of the Rolle’s theorem for the following functions : f(x) = x2 – 4x + 3, x ∈ [1, 3]

Exercise 2.3 | Q 1.2 | Page 80

Check the validity of the Rolle’s theorem for the following functions : f(x) = e–x sin x, x ∈ [0, π].

Exercise 2.3 | Q 1.3 | Page 80

Check the validity of the Rolle’s theorem for the following functions : f(x) = 2x2 – 5x + 3, x ∈ [1, 3].

Exercise 2.3 | Q 1.4 | Page 80

Check the validity of the Rolle’s theorem for the following functions : f(x) = sin x – cos x + 3, x ∈ [0, 2π].

Exercise 2.3 | Q 1.5 | Page 80

Check the validity of the Rolle’s theorem for the following functions : f(x = x2, if 0 ≤ x ≤ 2
= 6 – x, if 2 < x ≤ 6.

Exercise 2.3 | Q 1.6 | Page 80

Check the validity of the Rolle’s theorem for the following functions : f(x) = x^(2/3), x ∈ [ - 1, 1].

Exercise 2.3 | Q 2 | Page 80

Given an interval [a, b] that satisfies hypothesis of Rolle's theorem for the function f(x) = x4 + x2 – 2. It is known that a = – 1. Find the value of b.

Exercise 2.3 | Q 3.1 | Page 80

Verify Rolle’s theorem for the following functions : f(x) = sin x + cos x + 7, x ∈ [0, 2π]

Exercise 2.3 | Q 3.2 | Page 80

Verify Rolle’s theorem for the following functions  : f(x) = sin(x/2), x ∈ [0, 2pi]

Exercise 2.3 | Q 3.3 | Page 80

Verify Rolle’s theorem for the following functions : f(x) = x2 – 5x + 9, x ∈ 1, 4].

Exercise 2.3 | Q 4 | Page 80

If Rolle's theorem holds for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3] with c = 2 + (1)/sqrt(3), find the values of p and q.

Exercise 2.3 | Q 5 | Page 80

If Rolle’s theorem holds for the function f(x) = (x –2) log x, x ∈ [1, 2], show that the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).

Exercise 2.3 | Q 6 | Page 80

The function f(x) = x(x + 3)e^(-(x)/2) satisfies all the conditions of Rolle's theorem on [– 3, 0]. Find the value of c such that f'(c) = 0.

Exercise 2.3 | Q 7.1 | Page 80

Verify Lagrange’s mean value theorem for the following functions : f(x) = log x on [1, e].

Exercise 2.3 | Q 7.2 | Page 80

Verify Lagrange’s mean value theorem for the following functions : f(x) = (x – 1)(x – 2)(x – 3) on [0, 4].

Exercise 2.3 | Q 7.3 | Page 80

Verify Lagrange’s mean value theorem for the following functions : x^2 - 3x - 1, x ∈ [(-11)/7 , 13/7].

Exercise 2.3 | Q 7.4 | Page 80

Verify Lagrange’s mean value theorem for the following functions : f(x) = 2x – x2, x ∈ [0, 1].

Exercise 2.3 | Q 7.5 | Page 80

Verify Lagrange’s mean value theorem for the following functions : f(x) = (x - 1)/(x - 3) on [4, 5].

Exercise 2.4 [Pages 89 - 90]

### Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board Chapter 2 Applications of Derivatives Exercise 2.4 [Pages 89 - 90]

Exercise 2.4 | Q 1.1 | Page 89

Test whether the following functions are increasing or decreasing : f(x) = x3 – 6x2 + 12x – 16, x ∈ R.

Exercise 2.4 | Q 1.2 | Page 89

Test whether the following functions are increasing or decreasing : f(x) = 2 – 3x + 3x2 – x3, x ∈ R.

Exercise 2.4 | Q 1.3 | Page 89

Test whether the following functions are increasing or decreasing : f(x) = (1)/x, x ∈ R , x ≠ 0.

Exercise 2.4 | Q 2.1 | Page 89

Find the values of x for which the following functions are strictly increasing : f(x) = 2x3 – 3x2 – 12x + 6

Exercise 2.4 | Q 2.2 | Page 89

Find the values of x for which the following functions are strictly increasing : f(x) = 3 + 3x – 3x2 + x3

Exercise 2.4 | Q 2.3 | Page 89

Find the values of x for which the following func- tions are strictly increasing : f(x) = x3 – 6x2 – 36x + 7

Exercise 2.4 | Q 3.1 | Page 89

Find the values of x for which the following functions are strictly decreasing : f(x) = 2x3 – 3x2 – 12x + 6

Exercise 2.4 | Q 3.2 | Page 89

Find the values of x for which the following functions are strictly decreasing : f(x) = x + (25)/x

Exercise 2.4 | Q 3.3 | Page 89

Find the values of x for which the following functions are strictly decreasing : f(x) = x3 – 9x2 + 24x + 12

Exercise 2.4 | Q 4 | Page 90

Find the values of x for which the function f(x) = x3 – 12x2 – 144x + 13 (a) increasing (b) decreasing

Exercise 2.4 | Q 5 | Page 90

Find the values of x for which f(x) = 2x3 – 15x2 – 144x – 7 is

(a) Strictly increasing
(b) strictly decreasing

Exercise 2.4 | Q 6 | Page 90

Find the values of x for which f(x) = x/(x^2 + 1) is (a) strictly increasing (b) decreasing.

Exercise 2.4 | Q 7 | Page 90

show that f(x) = 3x + (1)/(3x) is increasing in (1/3, 1) and decreasing in (1/9, 1/3).

Exercise 2.4 | Q 8 | Page 90

Show that f(x) = x – cos x is increasing for all x.

Exercise 2.4 | Q 9.1 | Page 90

Find the maximum and minimum of the following functions : y = 5x3 + 2x2 – 3x

Exercise 2.4 | Q 9.2 | Page 90

Find the maximum and minimum of the following functions : f(x) = 2x3 – 21x2 + 36x – 20

Exercise 2.4 | Q 9.3 | Page 90

Find the maximum and minimum of the following functions : f(x) = x3 – 9x2 + 24x

Exercise 2.4 | Q 9.4 | Page 90

Find the maximum and minimum of the following functions : f(x) = x^2 + (16)/x^2

Exercise 2.4 | Q 9.5 | Page 90

Find the maximum and minimum of the following functions : f(x) = x log x

Exercise 2.4 | Q 9.6 | Page 90

Find the maximum and minimum of the following functions : f(x) = logx/x

Exercise 2.4 | Q 10 | Page 90

Divide the number 30 into two parts such that their product is maximum.

Exercise 2.4 | Q 11 | Page 90

Divide the number 20 into two parts such that sum of their squares is minimum.

Exercise 2.4 | Q 12 | Page 90

A wire of length 36 metres is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum.

Exercise 2.4 | Q 13 | Page 90

A ball is thrown in the air. Its height at any time t is given by h = 3 + 14t – 5t2 . Find the maximum height it can reach.

Exercise 2.4 | Q 14 | Page 90

Find the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit, so that two vertices lie on the diameter.

Exercise 2.4 | Q 15 | Page 90

An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of pia^3cu cm of water. Find the dimensions so that the quantity of the metal sheet required is minimum.

Exercise 2.4 | Q 16 | Page 90

The perimeter of a triangle is 10 cm. If one of the side is 4 cm. What are the other two sides of the triangle for its maximum area?

Exercise 2.4 | Q 17 | Page 90

A box with a square base is to have an open top. The surface area of the box is 192 sq cm. What should be its dimensions in order that the volume is largest?

Exercise 2.4 | Q 18 | Page 90

The profit function P(x) of a firm, selling x items per day is given by P(x) = (150 – x)x – 1625 . Find the number of items the firm should manufacture to get maximum profit. Find the maximum profit.

Exercise 2.4 | Q 20 | Page 90

Show that among rectangles of given area, the square has least perimeter.

Exercise 2.4 | Q 21 | Page 90

Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.

Exercise 2.4 | Q 22 | Page 90

Find the volume of the largest cylinder that can be inscribed in a sphere of radius ‘r’ cm.

Exercise 2.4 | Q 23 | Page 90

Show that y = log (1 + x) – (2x)/(2 + x), x > - 1 is an increasing function on its domain.

Exercise 2.4 | Q 24 | Page 90

Prove that y = (4sinθ)/(2 + cosθ) - θ is an increasing function if θ ∈[0, pi/2]

Miscellaneous Exercise 2 [Page 92]

### Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 [Page 92]

Miscellaneous Exercise 2 | Q 1 | Page 92

Choose the correct option from the given alternatives :

If the function f(x) = ax3 + bx2 + 11x – 6 satisfies conditions of Rolle's theoreem in [1, 3] and  f'(2 + 1/sqrt(3)) = 0, then values of a and b are respectively

• 1, – 6

• – 2, 1

• – 1, – 6

• – 1, 6

Miscellaneous Exercise 2 | Q 2 | Page 92

Choose the correct option from the given alternatives :

If f(x) = (x^2 - 1)/(x^2 + 1), for every real x, then the minimum ue of f is

• 1

• 0

• – 1

• 2

Miscellaneous Exercise 2 | Q 3 | Page 92

Choose the correct option from the given alternatives :

A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of (1.5 "m")/sec. The length of the higher point of ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of

• 1

• 2

• 2.5

• – 1

Miscellaneous Exercise 2 | Q 4 | Page 92

Choose the correct option from the given alternatives :

Let f(x) and g(x) be differentiable for 0 x < 1 such that f(0) = 0, g(0), f(1) = 6. Let there exist a real number c in (0, 1) such that f'(c) = 2g'(c), then the value of g(1) must be

• 1

• 3

• 2.5

• – 1

Miscellaneous Exercise 2 | Q 5 | Page 92

Choose the correct option from the given alternatives :

Let f(x) = x3 – 6x2 + 9x + 18, then f(x) is strictly decreasing in

• ( - oo, 1)

• [3, oo)

• ( - oo,] ∪ [3, oo)

• (1, 3)

Miscellaneous Exercise 2 | Q 6 | Page 92

Choose the correct option from the given alternatives :

If x = – 1 and x = 2 are the extreme points of y = oo log x + betax^2 + x, then

• oo = - 6, beta = (1)/(2)

• oo = - 6, beta = -(1)/(2)

• oo = 2, beta = -(1)/(2)

• oo = 2, beta = (1)/(2)

Miscellaneous Exercise 2 | Q 7 | Page 92

Choose the correct option from the given alternatives :

The normal to the curve x2 + 2xy – 3y2 = 0 at (1, 1)

• meets the curve again in second quadrant

• does not meet the curve again

• meets the curve again in third quadrant

• meets the curve again in fourth quadrant

Miscellaneous Exercise 2 | Q 8 | Page 92

Choose the correct option from the given alternatives :

The equation of the tangent to the curve y = 1 - e^(x/2) at the point of intersection with Y-axis is

• x + 2y = 0

• 2x + y = 0

• x – y = 2

• x + y = 2

Miscellaneous Exercise 2 | Q 9 | Page 92

Choose the correct option from the given alternatives :

If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, then P is

• (4, 4)

• (– 1, 2)

• (3, 6)

• (9/4, 3/8)

Miscellaneous Exercise 2 | Q 10 | Page 92

Choose the correct option from the given alternatives :

The approximate value of tan (44° 30°), given that 1° = 0.0175, is

• 0.8952

• 0.9528

• 0.9285

• 0.9825

Miscellaneous Exercise 2 [Pages 93 - 94]

### Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 [Pages 93 - 94]

Miscellaneous Exercise 2 | Q 1 | Page 93

Solve the following : If the curves ax2 + by2 = 1 and a'x2 + b'y2 = 1, intersect orthogonally, then prove that (1)/a - (1)/b = (1)/a' - (1)/b'.

Miscellaneous Exercise 2 | Q 2 | Page 93

Solve the following : Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x2 drawn at the point (1, 2) and the coordinate axes.

Miscellaneous Exercise 2 | Q 3 | Page 93

Solve the following : Find the equation of the tangent and normal drawn to the curve y4 – 4x4 – 6xy = 0 at the point M (1, 2).

Miscellaneous Exercise 2 | Q 4 | Page 93

Solve the following : A water tank in the farm of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the water level when the depth is 6 feet.

Miscellaneous Exercise 2 | Q 5 | Page 93

Solve the following : Find all points on the ellipse 9x2 + 16y2 = 400, at which the y-coordinate is decreasing and the coordinate is increasing at the same rate.

Miscellaneous Exercise 2 | Q 6 | Page 93

Solve the following : Verify Rolle’s theorem for the function f(x)  (2)/(e^x + e^-x) on [– 1, 1].

Miscellaneous Exercise 2 | Q 7 | Page 93

Solve the following : The position of a particle is given by the function s (t) = 2t2 + 3t – 4. Find the time t = c in the interval 0 ≤ t ≤ 4 when the instantaneous velocity of the particle equal to its average velocity in this interval.

Miscellaneous Exercise 2 | Q 8 | Page 93

Find the approximate value of the function f(x) = sqrt(x^2 + 3x) at x = 1.02.

Miscellaneous Exercise 2 | Q 9 | Page 93

Solve the following : Find the approximate value of cos–1 (0.51), given π = 3.1416, (2)/sqrt(3) = 1.1547.

Miscellaneous Exercise 2 | Q 10 | Page 93

Solve the following : Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing.

Miscellaneous Exercise 2 | Q 11 | Page 93

Solve the following : Find the intervals on which the function f(x) = x/logx is increasing and decreasing.

Miscellaneous Exercise 2 | Q 12 | Page 93

Solve the following : An open box with a square base is to be made out of given quantity of sheet of area a2. Show that the maximum volume of the box is a^3/(6sqrt(3).

Miscellaneous Exercise 2 | Q 13 | Page 93

Solve the following : Show that of all rectangles inscribed in a given circle, the square has the maximum area.

Miscellaneous Exercise 2 | Q 14 | Page 93

Solve the following : Show that a closed right circular cylinder of given surface area has maximum volume if its height equals the diameter of its base.

Miscellaneous Exercise 2 | Q 15 | Page 93

Solve the following :  A window is in the form of a rectangle surmounted by a semicircle. If the perimeter be 30 m, find the dimensions so that the greatest possible amount of light may be admitted.

Miscellaneous Exercise 2 | Q 16 | Page 93

Solve the following : Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.

Miscellaneous Exercise 2 | Q 17 | Page 93

Solve the following:

A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least, if the radius of the circle is half of the side of the square.

Miscellaneous Exercise 2 | Q 18 | Page 94

Solve the following : A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares of equal area from all corners. If the total area of the removed squares is 100, the resulting box has maximum volume. Find the lengths of the rectangular sheet of paper.

Miscellaneous Exercise 2 | Q 19 | Page 94

Solve the following : Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is  (4r)/(3).

Miscellaneous Exercise 2 | Q 20 | Page 94

Solve the following : Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is (2"R")/sqrt(3). Also, find the maximum volume.

Miscellaneous Exercise 2 | Q 21 | Page 94

Solve the following :  Find the maximum and minimum values of the function f(x) = cos2x + sinx.

## Chapter 2: Applications of Derivatives

Exercise 2.1Exercise 2.2Exercise 2.3Exercise 2.4Miscellaneous Exercise 2

## Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board chapter 2 - Applications of Derivatives

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board chapter 2 (Applications of Derivatives) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Balbharati textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board chapter 2 Applications of Derivatives are Applications of Derivatives in Geometry, Derivatives as a Rate Measure, Approximations, Rolle's Theorem, Lagrange's Mean Value Theorem (Lmvt), Increasing and Decreasing Functions, Maxima and Minima.

Using Balbharati 12th Board Exam solutions Applications of Derivatives exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Balbharati Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board 12th Board Exam prefer Balbharati Textbook Solutions to score more in exam.

Get the free view of chapter 2 Applications of Derivatives 12th Board Exam extra questions for Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board and can use Shaalaa.com to keep it handy for your exam preparation


text:
## Luck Matters: Understanding Training Dynamics of Deep ReLU Networks

May 17, 2019 Blind Submission readers: everyone
• Keywords: deep ReLU network, theoretical analysis, lottery tickets
• TL;DR: A theoretical framework for deep ReLU network that can explains multiple puzzling phenomena like over-parameterization, implicit regularization, lottery tickets, etc.
• Abstract: We analyze the dynamics of training deep ReLU networks and their implications on generalization capability. Using a teacher-student setting, we discovered a novel relationship between the gradient received by hidden student nodes and the activations of teacher nodes for deep ReLU networks. With this relationship and the assumption of small overlapping teacher node activations, we prove that (1) student nodes whose weights are initialized to be close to teacher nodes converge to them at a faster rate, and (2) in over-parameterized regimes and 2-layer case, while a small set of lucky nodes do converge to the teacher nodes, the fan-out weights of other nodes converge to zero. This framework provides insight into multiple puzzling phenomena in deep learning like over-parameterization, implicit regularization, lottery tickets, etc. We verify our assumption by showing that the majority of BatchNorm biases of pre-trained VGG11/16 models are negative. Experiments on (1) random deep teacher networks with Gaussian inputs, (2) teacher network pre-trained on CIFAR-10 and (3) extensive ablation studies validate our multiple theoretical predictions.
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text:
# People.teaching.crafts

People.teaching.crafts
Short form
Pe.te.cr
Stat dependencies
Con Dex Int Str Wis
0 2 2 0 1
Skill tree
People
teaching
crafts
All skill trees
Skill in TM wiki

People.teaching.crafts is one of the skills used to determine a player's Effective Teaching Bonus (ETB) when teaching another player a skill from the crafts skilltree. The teacher's ETB is the average of their bonus in this skill and the skill they would like to teach.

## Primaries

No groups have this skill as a primary.

## TM opportunities

Bonus Action taken


text:
# geometric product between nonorthogonal basis and its reciprocal basis

This is related to the question of wedge product between nonorthogonal basis and its reciprocal basis in geometric algebra.

If {$e_i$} is a set of basis that are not necessarily orthogonal, and {$e^i$} is the corresponding reciprocal basis, we have $$e_i \cdot e^j = \delta_j^i$$

why do we also have the geometric product sum of
$$\sum_i e_i e^i = n$$

This would be obvious if $e_i$ are orthogonal, but is it also trivial for any basis? From the definition of reciprocal frame, we only know the inner product, why the outer product part is zero?

The above is used to show (4.106) in the book of Geometric Algebra for Physicists. ($e_i e^i \wedge A_r = e_ie^iA_r - e_ie^i \cdot A_r = (n - r) A_r$. Is this only true for orthogonal basis?)

• You ask later "Is this only true for orthogonal bases?" but earlier you phrase it as "why do we also have..." as if you knew it were true. Am I right in supposing that Doran and Lasenby always use orthonormal bases, and you are just curious about the general case? – rschwieb May 30 '14 at 15:41
• I've also forgotten which inner product is used there. Can you refresh my memory? – rschwieb May 30 '14 at 15:57


text:
7.11: Chapter Review

7.1 The Central Limit Theorem for Sample Means

In a population whose distribution may be known or unknown, if the size ($$n$$) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ($$n$$).

7.2 Using the Central Limit Theorem

The Central Limit Theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean $$\overline x$$ gets to $$\mu$$.

7.3 The Central Limit Theorem for Proportions

The Central Limit Theorem can also be used to illustrate that the sampling distribution of sample proportions is normally distributed with the expected value of $$p$$ and a standard deviation of $$\sigma_{\mathrm{p}^{\prime}}=\sqrt{\frac{p(1-p)}{n}}$$

7.11: Chapter Review is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.


text:
How can I apply Planck's law to calculate the photon energy density of a warm room?

I am trying to calculate the rate at which photons of any particular frequency will pass through a volume in a room illuminated by black body radiation only.

I've found a couple of starting points but now I'm stuck putting it together.

I could start with $B_\lambda(\lambda, T) =\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda k_\mathrm{B}T}} - 1}$ and pretend all the radiation is coming from one spot on the wall, then multiply it by the effective solid angle to the volume and the volume itself, but I'm not sure if this is reasonable.

Also since I'm counting photons it seems I want to take a definite integral over the relevant frequency range. In which case, setting $x=\lambda$, $k=2kc^2$, and $q=hc/(k_BT)$ I'll end up with this less-than-elegant answer:

I suspect there's an easier way. Is there a simple formula for the energy density of empty space when the volume under examination is "far" from the walls?

To restate the problem: take a small volume of space within a room whose walls are emitting black body radiation at a known temperature; the size and shape of the room is known (we can assume any shape which is convenient, if that matters) -- how many photons per second with wavelength $\lambda_{min}<\lambda<\lambda_{max}$ are passing through this volume?


text:
Q1: The odometer counts revolutions, and the bicycle moves a distance of 2$\pi \$R each turn. The "27 inch" refers to the wheel's diameter, so R27 = 13.5 inches and R24 = 12 inches. The ratio is R24/R27 = 0.889. So, if the odometer is calibrated for 27-inch wheels and you use it with 24-inch wheels, the bike has in fact traveled only 0.889 of what the odometer reads.

Q4: Of course, provided that the smaller force is applied sufficiently far from the rotational axis.

Q7: No: two equal but opposite forces can produce a torque if they are not co-linear.

No: An object can certainly be accelerated by several forces without being rotated if the net force acts along a line through the rotational axis.

Q9: Measure their rotational inertia: apply a torque and observe the resulting angular acceleration about a diameter. The hollow shell (of the same mass) will have a greater rotational inertia, and therefore a smaller angular acceleration for a given torque.

Q14: Greater. In fact, one can easily figure out the rotational inertia about this new axis using the parallel axis theorem (Equation 10-17 on p. 264). Since all quantities in this equation are positive, we can conclude that the rotational inertia about an axis through the CM (in a given direction) is always the least possible for any axis parallel to that direction.

P3: Approximate the diameter of the spot on the Moon as an arc-length of a circle of radius D. Then D = R$\theta \$ (with $\theta \$ in radians). The answer is D = 5320 m.

P6: The wheel radius is 0.34 m, and one rotation moves the bicycle forward a distance 2$\pi \$R = 2.136 m. So in traveling 7200 m, the wheel rotates (7200 m)/(2.136 m/rotation) = 3371 rotations.

P9: All three points have the same angular velocity, $\omega \$ = 2$\pi \$ radians/day (or 7.272x10-5 radians/sec). Therefore if the point moves in a circle of radius R, the total velocity will be v = R$\omega \$.

(a) at the Earth's equator, R = RE = 6.38x106 m, so v = 464 m/s.

(b) at 66.5 degrees latitude (N or S) the point moves in a circle whose radius is REcos(66.5) = 2.544x106 m. Thus v = 185 m/s.

(c) at 45 degrees latitude the point moves in a circle of radius REcos(45) = 4.490x106 so v = 327 m/s

At the north pole, Santa Claus moves in a circle of radius REcos(90) = 0, so his linear velocity is 0.

P13: Over a time interval of 720 seconds, we have $\omega \$o = 0, $\omega \$f = (1.0 rev/min)(2$\pi \$ rad/rev)(1 min/60 sec) = 0.105 rad/sec.

(a) So the angular acceleration $\alpha \$ is given by $\Delta \$$\omega \$/$\Delta \$t = 1.45x10-4 rad/s2.

(b) A point on the ship's skin moves in a circle of radius 4.25 m. The tangential component of the linear acceleration is

aT = R$\alpha \$ = (4.25)(1.45x10-4 = 6.16x10-4 m/s2

The radial component of the linear acceleration is our old friend v2/R, but v = R$\omega \$, so we can write aR = R$\omega \$2. We need a value for $\omega$ at a time 420 seconds after the start of the rotation:

$\omega \$ = $\omega \$o + $\alpha \$t = 0.0609 rad/sec

at which time

aR = R$\omega \$2 = 0.0156 m/s2

P19: We have $\omega \$o = 850 rev/min = 89.0 rad/sec and $\omega \$f = 0. During this time the wheel slows to a stop, it rotates through 2700$\pi \$ radians.

(a) We can obtain $\alpha \$ using Eqn 10-9(c): $\omega \$f2 = $\omega \$o2 + 2$\alpha \$($\Delta \$$\theta \$)2. The answer is $\alpha$ = - 0.467 rad/s2. (The minus denotes that the rotation is slowing down.)

(b) There are several equations that will give us the time required for the wheel to stop. The quickest is probably Eqn. 10-9(a), which gives (since $\omega \$f = 0):

$\Delta \$t = - $\omega \$o/$\alpha \$ = 191 seconds.

P25: There are three applied forces and therefore three torques to add. The axis is through the center of the wheel perpendicular to the page. To compute torque we can use any of Equations 10-10 on page 257. For each force, the "R" vector goes from the axis to the point where the force is applied, and for all three forces in this problem, R is perpendicular to F [and therefore in Eqn. 10-10c the angle sin($\theta \$) = sin(90$\circ \$) = 1].

We arbitrarily choose CCW to be the positive direction for rotation. Then

$\tau \$28 = (28 N)(0.24 m) = 6.72 N-m

$\tau \$35 = - (35 N)(0.12 m) = - 4.20 N-m

$\tau \$18 = - (18 N)(0.24 m) = - 4.32 N-m

So the net applied torque is 1.8 N-m CW. The frictional torque, then is 0.40 N-m, and will act CCW to oppose the CW rotation. The final net torque is, then, $\tau \$ = 1.4 N-m CW.

P30(a): The axis is at C perpendicular to the page. For each force, the vector R is from C to the point where the force is applied. We choose CCW as positive rotational direction. Then

$\tau \$56 = - (56 N)(1.0 m)sin(30$\circ \$) = - 28 N-m

$\tau \$65 = (65 N)(0 m) = 0

$\tau \$52 = (52 N)(1.0 m)sin(30$\circ \$) = +45 N-m

So $\tau \$Net = 17.0 N-m CCW.

(b) Do this for yourself! The answer is 10.0 N-m CW.

P40: The moment of inertia (or "rotational inertia") is defined as $\Sigma \$MiRi2. Here we have four point masses, each a known distance from the rotational axis.

(a) About the vertical axis shown:

Iy-axis = (m)(0.50)2 + (M)(0.50)2 + (m)(1.00)2 + (M)(1.00)2 = 1.25m + 1.25M = (1.25)(2.2) + (1.25)(3.1) = 6.625 kg-m2

(b) About the horizontal axis shown, the answer is Ix-axis = 0.66 kg-m2.

(c) Since $\tau \$Net = I$\alpha \$, the torque required for a specified angular acceleration is larger if I is larger. Here, Iy-axis is almost ten times Ix-axis. Thus, to have the same angular acceleration, we would need almost ten times as much torque about the y-axis as about the x-axis. Clearly it is "harder" to accelerate about the y-axis in case (a).

P46: Draw free-body diagrams for both blocks and for the pulley (as in Figure 10-21) There is no sliding friction, and therefore I will not write the equations for the forces perpendicular to the inclines, since we are not interested in the normal forces. The relevant forces are:

Gravity on MA: MAg downward

Tension (in the rope) on MA = FTA

Gravity on MB: Mbg downward

Tension (in the rope) on MB = FTB

Tension (in the rope) CCW on pulley = FTA, resulting torque $\tau \$CCW = FTAR

Tension (in the rope) CW on pulley = FTB, resulting torque $\tau \$CW = FTBR

The equations of motion that result are:

FTA - MAg sin(32$\circ \$) = MAa

- FTB + MBg sin(61$\circ \$) = MBa

FTBR - FTAR = I$\alpha \$ = I(a/R)

where I have been consistent with signs: "down-slope" is positive for MB, "up-slope" is positive for MA and CW is positive for rotation of the pulley. Thus, as the blocks slide "toward the right" the pulley rotates CW, and a is positive.

We are given R for the pulley, along with (on the Figure) the masses MA and MB, and the acceleration a. The unknowns in the three equations are FTA, FTB, and I, the rotational inertia of the pulley. The answers are:

FTA = 49.6 N

FTB = 75.7 N

I = 0.59 kg-m2

P51: This is almost the same problem as P46, but simpler since no angles are involved. We will get three equations: vertical motion of MA, vertical motion of MB, and rotational motion of pulley. You should draw free-body diagrams for all three objects. The equations of motion are:

FTA - MAg = MAa

MBg - FTB = MBa

FTBR - FTAR = I$\alpha \$ = I(a/R)

The unknowns are FTA, FTB, and a. The answers will contain the given quantities: the masses MA and MB. the rotational inertia I and the radius R of the pulley. The two forces are not requested and thus solving the equations involves eliminating them and solving for a. The answer(s) are on p. A-24 of Giancoli.

P58:(a) The rotational inertia of a point-mass M a distance Ro from the rotational axis AB is IAB = MRo2.

(b) The rotational inertia of a solid sphere of mass M and radius r1 rotating about a diameter is (from Figure 10-20 on p. 260):

ICM = (2/5)Mr12.

Then, by the parallel axis theorem, the rotational inertia about the axis AB is:

IAB = ICM + MRo2 = (2/5)Mr12 + MRo2

(c) The fractional error incurred in using the expression in part (a) would be:

fractional error = (Ib - Ia)/Ib = (2/5)MRo2/[(2/5)Mr12 + MRo2]

The mass divides out and the expression simplifies to

fractional error = 1/[1 + (5/2)(Ro/r1)2]

Using the values Ro = 1.0 m and r1 = 0.09 m, we get a fractional error of 0.0032 or a percent error of 0.32%.

P65: The amount of work on the MGR is, of course, the increase in its KE. Since it starts at rest, this is just the amount of the MGR's KE at the end. So

W = KE = (1/2)I$\omega \$2

For a "solid cylinder" rotating about its axis of symmetry (see Figure 10-20, p.260), I = (1/2)MR2. The rotational velocity given is 1.00 rev/8 seconds = $\pi \$/4 radians/sec. Using the values given for M and R, the answer is

W = 14,226 Joules

P67: This is the same physical system as P51, but different information is given and a different quantity required. It is also much easier because we are going to use conservation of energy.

Let the original energy be zero: nothing is moving, and we can set the zeros of gravitational PE at the starting heights of each mass (this means there is a different zero for each mass; this is OK).

After mass B has fallen a distance of 2.5 m, mass A has raised by this same height and they both have acquired KE. The pulley has also acquired KE. Conservation of energy requires that the total energy is still zero:

MAg(+2.5) + MBg(- 2.5) + (1/2)MAv2 + (1/2)MBv2 + (1/2)Ip$\omega \$2 = 0

Also, we know that

$\omega \$ = v/R

and, for a "uniform cylinder" rotating about its axis of symmetry

Ip = (1/2)MpR2

We can plug these into the energy equation and solve for v. The answer is

v = 1.40 m/s

P73: We are given values for ro, m, and the length and angle of the incline. We will use conservation of energy and the problem will be simple. The total vertical distance the sphere drops is h = (10.0 m)sin(30$\circ \$) = 5.0 m. Set the zero of gravitational PE at the bottom of the ramp. Then initial energy is mgh and the final energy is all KE (translational + rotational):

mgh = (1.2)mv2 + (1/2)I$\omega \$2

and we know (from Figure 10-20 on p. 260) that

I = (2/5)mr2

and we use the fact that

$\omega \$ = v/r

(a) Substitute these into the energy equation and solve for v. Note that both the mass m and the radius r of the sphere drop out. The answer is v = sqrt(10gh/7), and since h = 5.0 m

v = 8.37 m/s (independent of both m and r)

Then $\omega \$ = v/r = 34.2 rad/s (here r does matter)

(b) The ratio (KE)trans/(KE)rot = 5/2 (again, independent of m and r)

(c) As noted in the above answers, the answers for v and for the ratio of KEs do not depend on either m or r. The value for $\omega \$ does depend on r, but not on m.

P86: If the acceleration is 1.00 m/s2 and the starting speed is zero, then after 2.5 seconds, the speed of the axle must be

v = 0 + (1.00 m/s2)(2.5 s) = 2.5 m/s

This is the speed of the axle over the ground. But from the viewpoint of the axle, the ground is moving to the left at this speed, and a point at the top of the wheel is moving to the right at this same speed. To the speed of the top-most point is moving over the ground at a speed of 5.0 m/s.

P89: Let the distance between teeth on both front and rear sprockets be d. The circumferences are, then (using subscript F for "Front" and B for "Back"):

CF = 2$\pi \$RF = NFd

CB = 2$\pi \$RB = NBd

from which we obtain:

RF = NFd/(2$\pi \$)

RB = NBd/(2$\pi \$)

Now, for the two sprockets: the linear speed of the chain (call it v) around the rim of the sprockets must be the same. Thus vF = vB. But we know that, for circular motion, v = R$\omega \$. Thus we arrive at:

RF$\omega$F = RB$\omega$B

which, using the equations above for RF and RB, leads immediately to:

$\omega \$B/$\omega \$F = NF/NB

which answers part (a). The answers for (b) and (c) then are 4 and 1.5 respectively.

P93: A Challenge! As the wheel hits the step, the "corner" of the rise (where the wheel contacts the step), becomes an axis of rotation. As the wheel lifts off the bottom surface, only two forces can exert torque about this axis: F and the gravitational force, both of which act through the CM, but in different directions.

The torque exerted by F about the axis of rotation is F(R - h) in a CW direction.

The torque exerted by gravity is CCW about the axis of rotation, and equals Mg(d) where d is the distance from the lift-off point on the bottom surface to the axis of rotation. A little geometry (involving use of the Pythagorean Theorem) should convince you that d = sqrt[R2 - (R - h)2].

The minimum force needed is the one that makes the net torque zero. So, calling CW rotation positive

F(R - h) - (Mg)sqrt[R2 - (R - h)2] = 0

which can trivially be solved for F in terms of the other quantities (M, R, h, and g). The answer is in the textbook on page A-25.

P101: Comment: Another Challenge! Also, this problem illustrates why most of the problems we do are so "unrealistic:" realistic problems can be quite complicated!

(a) Take a horizontal line through the CM (perpendicular to page) as the axis of rotation. As the front wheel lifts off, only three forces act: gravity (Mg downward through CM), the normal force (FN upward at rear wheel point of contact with ground) and the frictional force (Ffr to the right at point of rear wheel contact with ground). The frictional and normal forces (but not gravity) provide torques about the rotational axis through the CM.

We apply Newton's laws of motion: we write an equation for the x-acceleration (a; the sought-for unknown), another equation for the y-acceleration (assumed to be zero), and one for rotation about the axis. Take CCW as positive rotational direction. Then the condition for lift-off of the front wheel (normal force on it is zero) is that the total torque produced by these two forces about the CM just equals zero. (Before lift-off the torque due to just these two forces was negative, but the normal force on the front wheel caused a balancing positive torque. A net torque - either CCW or CW - would cause a rotation. For the condition of "just barely" lift off of the front wheel, we need a net torque of zero, provided by just the two forces that remain.)

$\Sigma \$Fx = Ffr = Ma

$\Sigma \$Fy = FN - Mg = 0

$\Sigma \$$\tau \$CM = Ffrx - FNy = 0

Note that x and y are considered "given" here. The unknowns are Ffr, FN, and a. Solving is easy:

a = (x/y)g

(b) To get small value of a we want a small x and a large y. The rider should position himself as high and as far to the rear as possible for the smallest possible a that will provide lift-off of the front wheel.

(c)a = (x/y)g = (35/95)(9.8) = 3.6 m/s2


text:
# Math Help - Solve Each Equation

1. ## Solve Each Equation

Solve each equation. The letters a, b and c are constants.

(1) 1 - ax = b, where a DOES NOT = 0.

(2) (a/x) + (b/x) = c, where c DOES NOT = 0.

2. Originally Posted by symmetry
Solve each equation. The letters a, b and c are constants.

(1) 1 - ax = b, where a DOES NOT = 0.
Solve the equation for x, I presume?

It can bother students that there is more than one unknown in an equation. The thing to remember is that all these variables are simply numbers. If it bothers you, plug in a random number for the variable and solve the equation. The steps you have to use to solve it in the form with the variable in it will be almost identical.

$1 - ax = b$

$-1 + 1 - ax = -1 + b$

$-ax = -1 + b$

$\frac{-ax}{-a} = \frac{-1 + b}{-a}$

$x = \frac{-1 + b}{-a} = -\frac{-1 + b}{a} = \frac{b - 1}{a}$

where any one of the three expressions in the last line do the trick. (Most would probably pick the last form, but it's entirely up to you.)

-Dan

3. Originally Posted by symmetry
Solve each equation. The letters a, b and c are constants.

(2) (a/x) + (b/x) = c, where c DOES NOT = 0.
$\frac{a}{x} + \frac{b}{x} = c$

We want to get rid of the fractions, so I am going to multiply both sides by x:

$x \left ( \frac{a}{x} + \frac{b}{x} \right ) = cx$

$x \cdot \frac{a}{x} + x \cdot \frac{b}{x} = cx$

$a + b = cx$

$cx = a + b$

$\frac{cx}{c} = \frac{a + b}{c}$

$x = \frac{a + b}{c}$

I would just like to note that, looking at the original equation that x cannot be allowed to be 0. Thus we not only have the restriction that $c \neq 0$ as given in the problem statement, but also that $b \neq -a$.

-Dan

4. ## ok

What an interesting set of math notes.

Thanks


text:
# Monotonicity of skew information and its applications in quantum resource theory

We give an alternative proof of skew information via operator algebra approach and show its strong monotonicity under particular quantum TPCP maps. We then formulate a family of new resource measure if the resource can be characterized by a resource destroying map and the free operation should be also modified. Our measure is easy-calculating and applicable to the coherence resource theory as well as quantum asymmetry theory. The operational interpretation need to be further investigated.

## Authors

• 2 publications
• ### A new resource measure with respect to resource destroying maps

We formulated a family of new resource measure if the resource can be ch...
11/10/2018 ∙ by Weijing Li, et al. ∙ 0

• ### Symmetric distinguishability as a quantum resource

We develop a resource theory of symmetric distinguishability, the fundam...
02/24/2021 ∙ by Robert Salzmann, et al. ∙ 0

• ### Extendibility limits the performance of quantum processors

Resource theories in quantum information science are helpful for the stu...
03/28/2018 ∙ by Eneet Kaur, et al. ∙ 0

• ### Automatic Amortized Resource Analysis with the Quantum Physicist's Method

We present a novel method for working with the physicist's method of amo...
06/26/2021 ∙ by David M Kahn, et al. ∙ 0

• ### Efficiently computable bounds for magic state distillation

Magic state manipulation is a crucial component in the leading approache...
12/25/2018 ∙ by Xin Wang, et al. ∙ 0

• ### Quantum Rényi divergences and the strong converse exponent of state discrimination in operator algebras

The sandwiched Rényi α-divergences of two finite-dimensional quantum sta...
10/14/2021 ∙ by Fumio Hiai, et al. ∙ 0

• ### Finite Block Length Analysis on Quantum Coherence Distillation and Incoherent Randomness Extraction

We introduce a variant of randomness extraction framework in the context...
02/27/2020 ∙ by Masahito Hayashi, et al. ∙ 0

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## I Introduction

Wigner-Yanase skew information wigner1963information wigner1964positive was introduced originally to express the amount of information in density operator not commuting with the observables. For now there are a lot of generalizations and applications of skew information in quantum information sciences luo2003wigner luo2007skew luo2018coherence . In the other hand, Quantum resource theories (QRTs) chitambar2018quantum offer a highly versatile and powerful framework for studying different phenomena in quantum physics, typical examples are quantum entanglement horodecki2009quantum , quantum coherence winter2016operational , quantum reference frames gour2008resource and Quantum Thermodynamics brandao2015reversible . A general quantum resource theory consists of a class of “free” states along with a class of “free” or allowable operations coecke2016mathematical . The essential resource theoretic condition is that the set of free states is closed under the set of free operations. Hence, any state that is not free is a resource since it cannot be obtained using the allowable operations chitambar2016comparison . In the quantum coherence theory, for example, there are various free operations such as incoherent operation (IO), dephasing- covariant operation (DIO) and strictly-incoherent operation (SIO). In conjunction with each of the operational classes, one can define different measures of coherence. From a resource theoretic perspective, the crucial property of these measures is that they are monotonic under the specified class of operations. To give the measures physical meaning, one seeks to find some operational interpretation of the measure, thereby enabling the measure to quantify some particular physical property or process. Various resource measures and resource monotones had been formulated and some operational interpretations had bee provided chitambar2018quantum winter2016operational .

Among various type of resources, one can specify the resource using the resource destroying maps liu2017resource . Kollas Nikolaos showed a type of optimization-free measures based on resource destroying maps kollas2018optimization . Recently, Shunlong Luo luo2018coherence formulated a generalization of skew information luo2003wigner from the viewpoint of state-channel interaction. So it is naturally to consider whether we can use this state-channel interaction to quantifying the resource given the resource destroying channel.

In this paper we first give an alternative proof of the monotonicity of skew information via operator algebra approach, and then prove that the skew information is of strong monotonicity, which is an open problem in luo2007skew . Second we formulate a family of generalized skew information acted as proper resource measures in the framework of resource destroying map, the free operation should be also modified in the sense that it does not disturb the resource destroying map.

## Ii Monotonicity of skew information

Wigner and Yanase wigner1963information wigner1964positive introduced the notion of skew information of a density operator with respect to a self-adjoint observable ,

 I(ρ,H)=−12Tr[ρp,H][ρ1−p,H],

for and Dyson suggested extending this to . Recently Shunlong Luo luo2018coherence gave some new generalizations. Given a Hilbert space , let denote the set of all operator on . For any operator , which needs not to be Hermitian, the skew skew information of with respect to is defined as luo2018coherence :

 I(ρ,K)= ||[√ρ,K]||2=Tr[√ρ,K]†[√ρ,K] = Tr(ρK†K)+Tr(ρKK†)−2% Tr(√ρK†√ρK)

where for all and is the norm induced by the Hilbert-Schmidt inner product . And if is a Trace-Preserving and Completely Positive (TPCP) map with Kraus operators , i.e., , it is naturally to define the state-channel interaction luo2018coherence by

 I(ρ,λ)=∑iI(ρ,Ki).

The quantity enjoys some pleasant properties such as non-negativity, convexity, monotonicity, and is independent of the choice of Kraus operators of , see luo2018coherence for more details. Among various properties of , it is crucial that is monotone under some TPCP map , i.e.,

 I(ρ,λ)≥I(E(ρ),λ).

The original proof of monotonicity in luo2018coherence is based on Landau-von Neumann equation. In this paper we give an alternative proof of the monotonicity of the skew information via the operator algebra method. And furthermore, we show that is of strong monotonicity, i.e.,

 I(ρ,λ)≥ ∑npnI(ρn,λ),

where and . To this end, we first recall some lemmas.

Lemma 1 stormer2012positive : If is a unital positive map, then for every normal element in its domain, we have and .

Lemma 2 hiai2012quasi : Assume that is an operator monotone function with and is a unital Schwarz mapping, that is, for all , then

 SAf(α†(ρ1)||α†(ρ2))≥Sα(A)f(ρ1,ρ2)

for and for invertible density operators , where

 SAf(ρ1||ρ2) = ⟨Aρ1/22,f(Δ(ρ1/ρ2))(Aρ1/22) = Trρ1/22A†f(Δ(ρ1/ρ2))(Aρ1/22),

and is the linear mapping defined by

 Δ(ρ1/ρ2)(X)=ρ1Xρ−12.

Lemma 3: For a TPCP map acted as and an operator , if and only if and ; if and only if and , where is the adjoint of in the sense that .

Proof.—The necessity is very simple since if for all then and then , the conclusion follows.

For the sufficiency we only need to consider the equality

 ∑n[Mn,K]†[Mn,K] = K†K−E†(K†)K−K†E†(K)+E†(K†K).

The second statement is followed similarly.

We then give an proof of the monotonicity as well as strong monotonicity of .

Theorem 4: For a TPCP map , is well defined:

 I(ρ,λ)=∑iI(ρ,Ki).

If TPCP map satisfies and as well as , then is monotone and strong monotone under the action of :

 I(ρ,λ)≥ I(E(ρ),λ); I(ρ,λ)≥ ∑npnI(ρn,λ),

where and .

Proof.— Since is a TPCP map, then by Lemma 1 is a unital completely positive map and satisfies the Schwarz inequality: . For a fixed Kraus operator of , and , consider the function , it is easy to see that is operator concave as well as operator monotone bhatia2013matrix , by Lemma 2, we have

 Tr(Ki√ρK†i√ρ)≤Tr(Ki√E(ρ)K†i√E(ρ)).

Since and , by taking inner product with respect to , we have

 Tr(KiρK†i)= Tr(KiE(ρ)K†i), Tr(ρKiK†i)= Tr(E(ρ)KiK†i).

Since these two terms are both linear on , we can conclude that

 I(ρ,Ki) = Tr(ρK†iKi)+Tr(ρKiK†i)−2Tr(Ki√ρK†i√ρ) ≥ I(E(ρ),Ki),

by summing up all it must holds that

 I(ρ,λ)≥I(E(ρ),λ).

Thus we reproduce the monotonicity result of in luo2018coherence via the operator algebra viewpoint. Next we consider the strong monotonicity of .

For the free operation , consider the channel

 ~E(ρ⊗|0⟩⟨0|) = ∑nMn⊗Un(ρ⊗|0⟩⟨0|)M†n⊗U†n = ∑nMnρM†n⊗|n⟩⟨n|,

where is a unitary operator and is the number of Kraus operators . It is easy to verify that

 ~E†(Ki⊗I)=Ki⊗I

and

 ~E†(K†iKi⊗I)= K†iKi⊗I, ~E†(KiK†i⊗I)= KiK†i⊗I,

by the monotonicity of , therefore,

 I(ρ,λ) = I(ρ⊗|0⟩⟨0|,λ⊗I) ≥ I(∑nMnρM†n⊗|n⟩⟨n|,λ⊗I) = ∑npnI(ρn,λ).

This finished the proof.

In luo2007skew , the authors showed that for Hermitian operator , is monotone under a TPCP map :

 I(ρ,H)≥I(E(ρ),H)

as long as satisfies: and , or equivalently, for all , but they only proved that satisfies the strong monotonicity in two-dimensional case. In fact, by same line reasoning in Theorem 4, we can show that satisfies the strong monotonicity in general. Virtually,

 I(ρ,H) = I(ρ⊗|0⟩⟨0|,H⊗I) ≥ I(~E(ρ⊗|0⟩⟨0|,H⊗I)) = I(∑nMnρM†n⊗|n⟩⟨n|,H⊗I) = ∑npnI(ρn,H).

Hence we solve the open problem leaved in luo2007skew .

We can even do further. In jenvcova2010unified the authors formulated a family of functions

 gp(x)={1p(1−p)(x−xp)p≠1xlogxp=1,

where and . For strictly positive they define jenvcova2010unified

 Jp(K,A,B) ≡ ⟨(K√B),gp(Δ(A/B))(K√B)⟩ = ⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩1p(1−p)(TrK†AK−Tr(K†ApKB1−p))p∈(0,1)∪(1,2),TrKK†AlogA−TrK†AKlogBp=1,−12(TrK†AK−TrAKB−1K†A)p=2.

One can see that this definition generalizes the relative entropy as well as the skew information. In fact, when and , reduces to the usual relative entropy, i.e.,

 J1(I,A,B)=Tr(AlogA−AlogB).

And when , and ,

 J12(K,A,A)=−12p(1−p)Tr[K,Ap][K,A1−p],

which yields the original skew information up to a constant.

Since enjoys some pleasant properties like skew information, it is naturally to extend our previous discussion by utilizing this unified entropy.

Define

 Ip(ρ,K)=Jp(K,ρ,ρ)

for operator and density operator and , and when a TPCP map written as , we define

 Ip(ρ,λ)=∑iIp(ρ,Ki)

for . If , is very similar to in Theorem 4 but there are several subtleties due to the linear term. As a consequence, we only need to require that , or equivalently, for all . We must show that share the same properties as when . We first show that is independent of the Kraus operator representations and thus is indeed a quantity about and channel . In fact, if for all density operator

, then there exists a unitary matrix

such that . By some direct calculation one can see that , which is independent of the Kraus operators of . The non-negativity and convexity of can be found in jenvcova2010unified . For , the function is operator monotone as well as operator concave bhatia2013matrix . So the monotonicity and strong monotonicity of under TPCP map follows from the same line in Theorem 4. However, there does not exist any discussion about the monotonicity when . Since in this case, the function is a operator convex function, the method in Lemma 2 is not available anymore. We leave it as an open problem. For now we can conclude the following theorem.

Theorem 5: For a TPCP map with Kraus operators , the other TPCP map satisfies and for all and , the quantity

 Ip(ρ,λ)=∑iIp(ρ,Ki)

is nonnegative, convex on and monotonicity under the action of . That is, with equality if and only if , is convex on and more crucially,

 Ip(ρ,λ)≥Ip(E(ρ),λ).

Furthermore, it is of strong monotonicity:

 Ip(ρ,λ)≥∑npnIp(ρn,λ),

where and .

## Iii Applications in Resource framework

Given a Hilbert space , we say that a TPCP map is a resource destroying map liu2017resource if it satisfies the following two properties:

1. It maps any free state to itself; i.e., .

2. It maps any (possibly not free) density operator to a free state; i.e., .

From its definition, it is not clear at all that a resource destroying map exists for a given QRT. However, the full necessary and sufficient conditions for the existence of a resource destroying map were derived in gour2017quantum .

In this section we show that the skew information and its generalization can be acted as a proper resource measure with respected to resource destroying map . The typical example is in quantum asymmetry theory. The resource theory of asymmetry with respect to a given representation of a symmetry group has been used extensively to distinguish and quantify the symmetry-breaking properties of both the states and the operations marvian2012symmetry marvian2013theory . The asymmetry theory can be described by a resource destroying map liu2017resource , namely,

 λ(ρ)=∫dμUgρU†g,

where is the unitary representation of and is the Haar measure with respect to . For simplicity we confine our focus on finite group but note that the conclusions hold also for compact Lie group. The symmetric state of the free state are those invariant under , i.e.,

 F={ρ|λ(ρ)=ρ}.

An equivalently characterization is that marvian2012symmetry

 F={ρ|UgρU†g=ρfor allg∈G}.

The free operation we consider in this paper is the TPCP map such that for all . Since are unitary thus normal the second type of condition is satisfied automatically. By Lemma 3, it holds that for all and all . Therefore, the generalized skew information between state and channel can be served as a proper resource measure of asymmetry due to Theorem 5.

Another typical example is the resource theory of coherence winter2016operational streltsov2017colloquium marvian2016quantify yu2017quantum . In this case the resource destroying map is the dephasing map

 Δ(ρ)=∑i|i⟩⟨i|ρ|i⟩⟨i|

for a fixed basis . The Kraus operators are and for , reproduces the result in yu2017quantum up to a constant.

## Iv Discussions

In this paper we formulated a family of resource measure extended by skew information when the resource theory can be characterized by a resource destroying map and for a class of particular free operations. The new measure acted well in asymmetry theory and coherence theory. There are also some unsolved problems. The first one is how to generalize our measure to a more wide free operations. After all, the requirement for commute relation between Kraus operator seems too severe. The second problem is to consider the monotonicity of under the TPCP maps when . Recently, the author in vershynina2018quantum showed the monotonicity of quantum quasi-entropy under partial trace, but it is different from our open problem. It is expected that a similar operator algebra approach proof should be applied. And the last one is to figure out the relationship between our measure and the measure introduced in zhao2018coherence .

## V Acknowledgments

The author is very grateful to professor Shunlong Luo, professor Shaoming Fei and Wei Xie for insightful discussions.

## References

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