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# mathlibdocumentation

combinatorics.set_family.compression.uv

# UV-compressions #

This file defines UV-compression. It is an operation on a set family that reduces its shadow.

UV-compressing a : α along u v : α means replacing a by (a ⊔ u) \ v if a and u are disjoint and v ≤ a. In some sense, it's moving a from v to u.

UV-compressions are immensely useful to prove the Kruskal-Katona theorem. The idea is that compressing a set family might decrease the size of its shadow, so iterated compressions hopefully minimise the shadow.

## Main declarations #

• uv.compress: compress u v a is a compressed along u and v.
• uv.compression: compression u v s is the compression of the set family s along u and v. It is the compressions of the elements of s whose compression is not already in s along with the element whose compression is already in s. This way of splitting into what moves and what does not ensures the compression doesn't squash the set family, which is proved by uv.card_compress.

## Notation #

𝓒 (typed with \MCC) is notation for uv.compression in locale finset_family.

## Notes #

Even though our emphasis is on finset α, we define UV-compressions more generally in a generalized boolean algebra, so that one can use it for set α.

## TODO #

Prove that compressing reduces the size of shadow. This result and some more already exist on the branch combinatorics.

## Tags #

theorem sup_sdiff_inj_on {α : Type u_1} (u v : α) :
set.inj_on (λ (x : α), (x u) \ v) {x : α | x v x}

UV-compression is injective on the elements it moves. See uv.compress.

### UV-compression in generalized boolean algebras #

def uv.compress {α : Type u_1} (u v a : α) :
α

To UV-compress a, if it doesn't touch U and does contain V, we remove V and put U in. We'll only really use this when |U| = |V| and U ∩ V = ∅.

Equations
def uv.compression {α : Type u_1} (u v : α) (s : finset α) :

To UV-compress a set family, we compress each of its elements, except that we don't want to reduce the cardinality, so we keep all elements whose compression is already present.

Equations
def uv.is_compressed {α : Type u_1} (u v : α) (s : finset α) :
Prop

is_compressed u v s expresses that s is UV-compressed.

Equations
• s = v s = s)
theorem uv.compress_of_disjoint_of_le {α : Type u_1} {u v a : α} (hua : a) (hva : v a) :
v a = (a u) \ v
theorem uv.mem_compression {α : Type u_1} {s : finset α} {u v a : α} :
a s a s v a s a s ∃ (b : α) (H : b s), v b = a

a is in the UV-compressed family iff it's in the original and its compression is in the original, or it's not in the original but it's the compression of something in the original.

@[simp]
theorem uv.compress_self {α : Type u_1} (u a : α) :
u a = a
@[simp]
theorem uv.compression_self {α : Type u_1} (u : α) (s : finset α) :
s = s
theorem uv.is_compressed_self {α : Type u_1} (u : α) (s : finset α) :
s

Any family is compressed along two identical elements.

theorem uv.compress_disjoint {α : Type u_1} {s : finset α} (u v : α) :
disjoint (finset.filter (λ (a : α), v a s) s) (finset.filter (λ (a : α), a s) (finset.image v) s))
@[simp]
theorem uv.compress_idem {α : Type u_1} (u v a : α) :
v v a) = v a

Compressing an element is idempotent.

theorem uv.compress_mem_compression {α : Type u_1} {s : finset α} {u v a : α} (ha : a s) :
v a s
theorem uv.compress_mem_compression_of_mem_compression {α : Type u_1} {s : finset α} {u v a : α} (ha : a s) :
v a s
@[simp]
theorem uv.compression_idem {α : Type u_1} (u v : α) (s : finset α) :
v s) = s

Compressing a family is idempotent.

theorem uv.card_compression {α : Type u_1} (u v : α) (s : finset α) :
v s).card = s.card

Compressing a family doesn't change its size.

theorem uv.sup_sdiff_mem_of_mem_compression {α : Type u_1} {s : finset α} {u v a : α} (ha : a s) (hva : v a) (hua : a) :
(a u) \ v s

If a is in the family compression and can be compressed, then its compression is in the original family.

theorem uv.mem_of_mem_compression {α : Type u_1} {s : finset α} {u v a : α} (ha : a s) (hva : v a) (hvu : v = u = ) :
a s

If a is in the u, v-compression but v ≤ a, then a must have been in the original family.

### UV-compression on finsets #

theorem uv.card_compress {α : Type u_1} [decidable_eq α] {U V : finset α} (hUV : U.card = V.card) (A : finset α) :
V A).card = A.card

Compressing a finset doesn't change its size.


text:
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# AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that $$\angle{A}$$ > $$\angle{C}$$ and $$\angle{B}$$ > $$\angle{D}$$.

Given: ABCD is a quadrilateral having AB as the smallest side and CD as the longest side

To prove: $$\angle{A}$$ > $$\angle{C}$$ and $$\angle{B}$$ > $$\angle{D}$$

Construction: Join A to C and B to D.

Figure:

Proof:
In $$\triangle{ABC}$$, we have AB is the smallest side.

As, AB < BC
$$\therefore$$ $$\angle{5}$$ < $$\angle{1}$$ ...(i)
(Since, angle opposite to longer side is greater)

Similarly, in $$\triangle{ADC}$$,
we have CD is the largest side.

As, AD < CD
$$\therefore$$ $$\angle{6}$$ < $$\angle{2}$$ ...(ii)
(Since, angle opposite to longer side is greater)

On adding Equation (i) and (ii), we get,

$$\Rightarrow$$ $$\angle{5}$$ + $$\angle{6}$$ < $$\angle{1}$$ + $$\angle{2}$$
$$\Rightarrow$$ $$\angle{C}$$ < $$\angle{A}$$
$$\Rightarrow$$ $$\angle{A}$$ > $$\angle{C}$$

Now, Similarly, $$\triangle{ADB}$$, we have AB is the smallest side.
As, AD > AB
$$\therefore$$ $$\angle{3}$$ > $$\angle{8}$$ ...(iii)
(Since, angle opposite to longer side is greater)

Similarly, in $$\triangle{BCD}$$, we have CD is the largest side.
As, CD > BC
$$\therefore$$ $$\angle{4}$$ > $$\angle{7}$$ ...(iv)
(Since, angle opposite to longer side is greater)

On adding Equation (iii) and (iv), we get,
$$\Rightarrow$$ $$\angle{3}$$ + $$\angle{4}$$ > $$\angle{8}$$ + $$\angle{7}$$
$$\Rightarrow$$ $$\angle{B}$$ > $$\angle{D}$$
Hence, proved.

format: plain


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1972 USAMO Problems/Problem 2

Problem

A given tetrahedron $ABCD$ is isosceles, that is, $AB=CD, AC=BD, AD=BC$. Show that the faces of the tetrahedron are acute-angled triangles.

Solutions

Solution 1

Suppose $\triangle ABD$ is fixed. By the equality conditions, it follows that the maximal possible value of $BC$ occurs when the four vertices are coplanar, with $C$ on the opposite side of $\overline{AD}$ as $B$. In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.

For the sake of contradiction, suppose $\angle ABD$ is non-acute. Then, $(AD)^2\geq (AB)^2+(BD)^2$. In our optimal case noted above, $ACDB$ is a parallelogram, so \begin{align*} 2(BD)^2 + 2(AB)^2 &= (AD)^2 + (CB)^2 \\ &= 2(AD)^2 \\ &\geq 2(BD)^2+2(AB)^2. \end{align*} However, as stated, equality cannot be attained, so we get our desired contradiction.

Solution 2

It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that $AB\leq BC \leq CA$. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles $\triangle ABC$ and $\triangle ABD$. They share side $AB$. Let $k$ and $l$ be the planes passing through $A$ and $B$, respectively, that are perpendicular to side $AB$. We have that triangles $ABC$ and $ABD$ are non-acute, so $C$ and $D$ are not strictly between planes $k$ and $l$. Therefore the length of $CD$ is at least the distance between the planes, which is $AB$. However, if $CD=AB$, then the four points $A$, $B$, $C$, and $D$ are coplanar, and the volume of $ABCD$ would be zero. Therefore $CD>AB$. However, we were given that $CD=AB$ in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.

Solution 3

Let $\vec{a} = \overrightarrow{DA}$, $\vec{b} = \overrightarrow{DB}$, and $\vec{c} = \overrightarrow{DC}$. The conditions given translate to \begin{align*} \vec{a}\cdot\vec{a} &= \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} - 2(\vec{b}\cdot\vec{c}) \\ \vec{b}\cdot\vec{b} &= \vec{c}\cdot\vec{c} + \vec{a}\cdot\vec{a} - 2(\vec{c}\cdot\vec{a}) \\ \vec{c}\cdot\vec{c} &= \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} - 2(\vec{a}\cdot\vec{b}) \end{align*} We wish to show that $\vec{a}\cdot\vec{b}$, $\vec{b}\cdot\vec{c}$, and $\vec{c}\cdot\vec{a}$ are all positive. WLOG, $\vec{a}\cdot\vec{a}\geq \vec{b}\cdot\vec{b}, \vec{c}\cdot\vec{c} > 0$, so it immediately follows that $\vec{a}\cdot\vec{b}$ and $\vec{a}\cdot\vec{c}$ are positive. Adding all three equations, $$\vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})$$ In addition, \begin{align*} (\vec{a} - \vec{b} - \vec{c})\cdot(\vec{a} - \vec{b} - \vec{c})&\geq 0 \\ \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c}&\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \\ 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})&\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \\ \vec{b}\cdot\vec{c}&\geq 0 \end{align*} Equality could only occur if $\vec{a} = \vec{b} + \vec{c}$, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.

Solution 4

Suppose for the sake of contradiction that $\angle BAC$ is not acute. Since all three sides of triangles $BAC$ and $CDB$ are congruent, those two triangles are congruent, meaning $\angle BDC=\angle BAC>90^{\circ}$. Construct a sphere with diameter $BC$. Since angles $BAC$ and $BDC$ are both not acute, $A$ and $D$ both lie on or inside the sphere. We seek to make $AD=BC$ to satisfy the conditions of the problem. This can only occur when $AD$ is a diameter of the sphere, since both points lie on or inside the sphere. However, for $AD$ to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron $ABCD$ degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.

Solution 5

Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume $\angle BAC$ is an obtuse angle. Using SSS congruence to prove that all four faces of the tetrahedron are congruent also shows that the angles surrounding point $A$ are congruent to angles in triangle $ABC$; namely, $\angle BAD$ is congruent to $\angle ABC$, and $\angle CAD$ is congruent to $\angle ACB$. Since the internal angles of triangle $ABC$ must add to 180 degrees, so do the angles surrounding point A. Now lay triangle $ABC$ on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the "apparent" measures of $\angle BAD$ and $\angle CAD$ (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle $ABC$ lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle $ABC$). This means that these two angles can never join to form side AD (because $\angle BAC$ is more than the sum of $\angle BAD$ and $\angle CAD$ - a direct consequence of the facts that $\angle BAC$ is obtuse and all three angles add up to 180 degrees), so the tetrahedron with obtuse triangle faces is impossible.

Solution 6

Lemma: given triangle $ABC$ and the midpoint of $BC$, which we will call $M$, we can say that if $AM > \frac{BC}{2}$, then $\angle A < 90$. Proof: Since $M$ is the midpoint of $BC$, $BM = MC = \frac{BC}{2}$. Since it is given that $AM > \frac{BC}{2}$, we can substitute $\frac{BC}{2}$ to get two inequalities: $$AM > CM, \quad AM > BM.$$ The above inequalities imply that $\angle C > \angle CAM$ and $\angle B > \angle BAM$. Adding these inequalities and simplifying the RHS, we have that $\angle C + \angle B > \angle A$. Adding $\angle A$ to both sides, replacing the LHS with $180$ and dividing by $2$ gets us that $\angle A < 90$. This is our desired inequality, so we are done. Note that all faces of this tetrahedron are congruent, by SSS. In particular, we will use that $\triangle ABD \cong \triangle BAC$. WLOG, assume that $\angle ADB$ is the largest angle in triangle $ABD$. Because $\triangle ABD \cong \triangle BAC$, the median from $D$ to $AB$ is equal length to the median from $C$ to $AB$. These points meet at $E$, the midpoint of $AB$. By the triangle inequality, $$DE + CE > CD.$$By substituting $CD$ with $AB$ (this is a given in the problem) and $CE$ with $DE$, and then dividing by $2$, we get that $$DE > \frac{AB}{2}.$$By the lemma we showed at the start, this implies that $\angle ADB < 90$, and since we said that $\angle ADB$ was the largest angle, triangle $ADB$ must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute.


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# Modular representation theory of special linear group:SL(2,3) at 3

## Contents

This article gives specific information, namely, modular representation theory, about a particular group, namely: special linear group:SL(2,3).
View modular representation theory of particular groups | View other specific information about special linear group:SL(2,3)

This article describes the modular representation theory of special linear group:SL(2,3), i.e. ,the linear representation theory in characteristic three -- over field:F3 and its extensions.

## Summary

Item Value
degrees of irreducible representations (or degrees of irreducible Brauer characters) 1,2,3
number: 3
smallest field of realization of all irreducible representations field:F3

## Family contexts

Family name Parameter value General discussion of modular representation theory of family
special linear group of degree two field:F3, i.e., the field with three elements modular representation theory of special linear group of degree two over a finite field in its defining characteristic

## Character table

Below is the character table in characteristic 3. The entries are given modulo 3, and are written as 0,1,2.

Representation/conjugacy class representative and size $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ -- 3-regular, size 1 $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix}$ -- 3-regular, size 1 $\begin{pmatrix} 0 & -1 \\ 1 & 0 \\\end{pmatrix}$ -- 3-regular, size 6 $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$ -- not 3-regular, 4 elements $\begin{pmatrix} 1 & -1 \\ 0 & 1 \\\end{pmatrix}$ -- not 3-regular, 4 elements $\begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$ -- not 3-regular, 4 elements $\begin{pmatrix} -1 & -1 \\ 0 & -1 \\\end{pmatrix}$ -- not 3-regular, 4 elements
trivial 1 1 1 1 1 1 1
two-dimensional, given by identity mapping 2 1 0 2 2 1 1
three-dimensional, given by symmetric square of two-dimensional 0 0 2 0 0 0 0

## Brauer characters

### Brauer character table

Irreducible representation in characteristic three whose Brauer character we are computing Irreducible representation in characteristic zero whose character equals the Brauer character Value of Brauer character on conjugacy class of $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ -- size 1 Value of Brauer character on conjugacy class of $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix}$ -- size 1 Value of Brauer character on conjugacy class of $\begin{pmatrix} 0 & -1 \\ 1 & 0 \\\end{pmatrix}$ -- size 6
trivial trivial 1 1 1
two-dimensional, given by identity mapping quaternionic representation of special linear group:SL(2,3) 2 -2 0
three-dimensional, given by symmetric square of two-dimensional kernel is center, reduces to standard representation of alternating group:A4 on quotient 3 3 -1


text:
# Basics

Becoming a better programmer is in many ways like learning any language. While it may be literal, there is much nuance, and many ways are available to express yourself in order to solve some problem. However, it doesn’t take much in the way of practice to develop a few skills that will not only last, but go a long way toward saving you time and allowing you to explore your data, models, and visualizations more extensively. So let’s get to it!

## R Objects

### Object Inspection & Exploration

Let’s say you’ve imported your data into R. If you are going to be able to do anything with it, you’ll have had to create an R object that represents that data. What is that object? By now you know it’s a data frame, specifically, an object of class data.frame or possibly a tibble if you’re working within the tidyverse. If you want to look at it, you might be tempted to look at it this way with View, or clicking on it in your Environment viewer.

View(diamonds)

While this is certainly one way to inspect it, it’s not very useful. There’s far too much information to get much out of it, and information you may need to know is absent.

Consider the following:

str(diamonds)
Classes 'tbl_df', 'tbl' and 'data.frame':   53940 obs. of  10 variables:
$carat : num 0.23 0.21 0.23 0.29 0.31 0.24 0.24 0.26 0.22 0.23 ...$ cut    : Ord.factor w/ 5 levels "Fair"<"Good"<..: 5 4 2 4 2 3 3 3 1 3 ...
$color : Ord.factor w/ 7 levels "D"<"E"<"F"<"G"<..: 2 2 2 6 7 7 6 5 2 5 ...$ clarity: Ord.factor w/ 8 levels "I1"<"SI2"<"SI1"<..: 2 3 5 4 2 6 7 3 4 5 ...
$depth : num 61.5 59.8 56.9 62.4 63.3 62.8 62.3 61.9 65.1 59.4 ...$ table  : num  55 61 65 58 58 57 57 55 61 61 ...
$price : int 326 326 327 334 335 336 336 337 337 338 ...$ x      : num  3.95 3.89 4.05 4.2 4.34 3.94 3.95 4.07 3.87 4 ...
$y : num 3.98 3.84 4.07 4.23 4.35 3.96 3.98 4.11 3.78 4.05 ...$ z      : num  2.43 2.31 2.31 2.63 2.75 2.48 2.47 2.53 2.49 2.39 ...
glimpse(diamonds)
Observations: 53,940
Variables: 10
$carat <dbl> 0.23, 0.21, 0.23, 0.29, 0.31, 0.24, 0.24, 0.26, 0.22, 0.23, 0.30, 0.23, 0.22, 0.31, 0.20, 0.32, 0.30, 0.30, 0.30, 0.30, 0.30, 0.23, 0.23, 0.31, 0.31, 0.23, 0.24, 0.30, 0.23, 0.23, 0.23, 0.23, 0.23, 0.2…$ cut     <ord> Ideal, Premium, Good, Premium, Good, Very Good, Very Good, Very Good, Fair, Very Good, Good, Ideal, Premium, Ideal, Premium, Premium, Ideal, Good, Good, Very Good, Good, Very Good, Very Good, Very Good…
$color <ord> E, E, E, I, J, J, I, H, E, H, J, J, F, J, E, E, I, J, J, J, I, E, H, J, J, G, I, J, D, F, F, F, E, E, D, F, E, H, D, I, I, J, D, D, H, F, H, H, E, H, F, G, I, E, D, I, J, I, I, I, I, D, D, D, I, G, I, …$ clarity <ord> SI2, SI1, VS1, VS2, SI2, VVS2, VVS1, SI1, VS2, VS1, SI1, VS1, SI1, SI2, SI2, I1, SI2, SI1, SI1, SI1, SI2, VS2, VS1, SI1, SI1, VVS2, VS1, VS2, VS2, VS1, VS1, VS1, VS1, VS1, VS1, VS1, VS1, SI1, VS2, SI2,…
$depth <dbl> 61.5, 59.8, 56.9, 62.4, 63.3, 62.8, 62.3, 61.9, 65.1, 59.4, 64.0, 62.8, 60.4, 62.2, 60.2, 60.9, 62.0, 63.4, 63.8, 62.7, 63.3, 63.8, 61.0, 59.4, 58.1, 60.4, 62.5, 62.2, 60.5, 60.9, 60.0, 59.8, 60.7, 59.…$ table   <dbl> 55.0, 61.0, 65.0, 58.0, 58.0, 57.0, 57.0, 55.0, 61.0, 61.0, 55.0, 56.0, 61.0, 54.0, 62.0, 58.0, 54.0, 54.0, 56.0, 59.0, 56.0, 55.0, 57.0, 62.0, 62.0, 58.0, 57.0, 57.0, 61.0, 57.0, 57.0, 57.0, 59.0, 58.…
$price <int> 326, 326, 327, 334, 335, 336, 336, 337, 337, 338, 339, 340, 342, 344, 345, 345, 348, 351, 351, 351, 351, 352, 353, 353, 353, 354, 355, 357, 357, 357, 402, 402, 402, 402, 402, 402, 402, 402, 403, 403, 4…$ x       <dbl> 3.95, 3.89, 4.05, 4.20, 4.34, 3.94, 3.95, 4.07, 3.87, 4.00, 4.25, 3.93, 3.88, 4.35, 3.79, 4.38, 4.31, 4.23, 4.23, 4.21, 4.26, 3.85, 3.94, 4.39, 4.44, 3.97, 3.97, 4.28, 3.96, 3.96, 4.00, 4.04, 3.97, 4.0…
$y <dbl> 3.98, 3.84, 4.07, 4.23, 4.35, 3.96, 3.98, 4.11, 3.78, 4.05, 4.28, 3.90, 3.84, 4.37, 3.75, 4.42, 4.34, 4.29, 4.26, 4.27, 4.30, 3.92, 3.96, 4.43, 4.47, 4.01, 3.94, 4.30, 3.97, 3.99, 4.03, 4.06, 4.01, 4.0…$ z       <dbl> 2.43, 2.31, 2.31, 2.63, 2.75, 2.48, 2.47, 2.53, 2.49, 2.39, 2.73, 2.46, 2.33, 2.71, 2.27, 2.68, 2.68, 2.70, 2.71, 2.66, 2.71, 2.48, 2.41, 2.62, 2.59, 2.41, 2.47, 2.67, 2.40, 2.42, 2.41, 2.42, 2.42, 2.4…

The str function looks at the structure of the object, while glimpse perhaps provides a possibly more readable version, and is just str specifically suited toward data frames. In both cases, we get info about the object and the various things within it.

While you might be doing this with data frames, you should be doing it with any of the objects you’re interested in. Consider a regression model object.

lm_mod = lm(mpg ~ ., data=mtcars)
str(lm_mod, 0)
List of 12
- attr(*, "class")= chr "lm"
str(lm_mod, 1)
List of 12
$coefficients : Named num [1:11] 12.3034 -0.1114 0.0133 -0.0215 0.7871 ... ..- attr(*, "names")= chr [1:11] "(Intercept)" "cyl" "disp" "hp" ...$ residuals    : Named num [1:32] -1.6 -1.112 -3.451 0.163 1.007 ...
..- attr(*, "names")= chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...
$effects : Named num [1:32] -113.65 -28.6 6.13 -3.06 -4.06 ... ..- attr(*, "names")= chr [1:32] "(Intercept)" "cyl" "disp" "hp" ...$ rank         : int 11
$fitted.values: Named num [1:32] 22.6 22.1 26.3 21.2 17.7 ... ..- attr(*, "names")= chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...$ assign       : int [1:11] 0 1 2 3 4 5 6 7 8 9 ...
$qr :List of 5 ..- attr(*, "class")= chr "qr"$ df.residual  : int 21
$xlevels : Named list()$ call         : language lm(formula = mpg ~ ., data = mtcars)
$terms :Classes 'terms', 'formula' language mpg ~ cyl + disp + hp + drat + wt + qsec + vs + am + gear + carb .. ..- attr(*, "variables")= language list(mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb) .. ..- attr(*, "factors")= int [1:11, 1:10] 0 1 0 0 0 0 0 0 0 0 ... .. .. ..- attr(*, "dimnames")=List of 2 .. ..- attr(*, "term.labels")= chr [1:10] "cyl" "disp" "hp" "drat" ... .. ..- attr(*, "order")= int [1:10] 1 1 1 1 1 1 1 1 1 1 .. ..- attr(*, "intercept")= int 1 .. ..- attr(*, "response")= int 1 .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> .. ..- attr(*, "predvars")= language list(mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb) .. ..- attr(*, "dataClasses")= Named chr [1:11] "numeric" "numeric" "numeric" "numeric" ... .. .. ..- attr(*, "names")= chr [1:11] "mpg" "cyl" "disp" "hp" ...$ model        :'data.frame':  32 obs. of  11 variables:
..- attr(*, "terms")=Classes 'terms', 'formula'  language mpg ~ cyl + disp + hp + drat + wt + qsec + vs + am + gear + carb
.. .. ..- attr(*, "variables")= language list(mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb)
.. .. ..- attr(*, "factors")= int [1:11, 1:10] 0 1 0 0 0 0 0 0 0 0 ...
.. .. .. ..- attr(*, "dimnames")=List of 2
.. .. ..- attr(*, "term.labels")= chr [1:10] "cyl" "disp" "hp" "drat" ...
.. .. ..- attr(*, "order")= int [1:10] 1 1 1 1 1 1 1 1 1 1
.. .. ..- attr(*, "intercept")= int 1
.. .. ..- attr(*, "response")= int 1
.. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
.. .. ..- attr(*, "predvars")= language list(mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb)
.. .. ..- attr(*, "dataClasses")= Named chr [1:11] "numeric" "numeric" "numeric" "numeric" ...
.. .. .. ..- attr(*, "names")= chr [1:11] "mpg" "cyl" "disp" "hp" ...
- attr(*, "class")= chr "lm"

Here we look at the object at the lowest level of detail (0), which basically just tells us that it’s a list of stuff. But if we go into more depth, we can see that there is quite a bit going on in here! Coefficients, the data frame used in the model (i.e. only the variables used and no NA), and much more are available to us, and we can pluck out any piece of it.

lm_mod$coefficients (Intercept) cyl disp hp drat wt qsec vs am gear carb 12.30337416 -0.11144048 0.01333524 -0.02148212 0.78711097 -3.71530393 0.82104075 0.31776281 2.52022689 0.65541302 -0.19941925  lm_mod$model %>%
head()
                   mpg cyl disp  hp drat    wt  qsec vs am gear carb
Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1

Let’s do a summary of it, something you’ve probably done many times.

summary(lm_mod)

Call:
lm(formula = mpg ~ ., data = mtcars)

Residuals:
Min      1Q  Median      3Q     Max
-3.4506 -1.6044 -0.1196  1.2193  4.6271

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 12.30337   18.71788   0.657   0.5181
cyl         -0.11144    1.04502  -0.107   0.9161
disp         0.01334    0.01786   0.747   0.4635
hp          -0.02148    0.02177  -0.987   0.3350
drat         0.78711    1.63537   0.481   0.6353
wt          -3.71530    1.89441  -1.961   0.0633 .
qsec         0.82104    0.73084   1.123   0.2739
vs           0.31776    2.10451   0.151   0.8814
am           2.52023    2.05665   1.225   0.2340
gear         0.65541    1.49326   0.439   0.6652
carb        -0.19942    0.82875  -0.241   0.8122
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.65 on 21 degrees of freedom
Multiple R-squared:  0.869, Adjusted R-squared:  0.8066
F-statistic: 13.93 on 10 and 21 DF,  p-value: 3.793e-07

But you can assign that to an object and inspect it too!

lm_mod_summary = summary(lm_mod)
str(lm_mod_summary)
List of 11
$call : language lm(formula = mpg ~ ., data = mtcars)$ terms        :Classes 'terms', 'formula'  language mpg ~ cyl + disp + hp + drat + wt + qsec + vs + am + gear + carb
.. ..- attr(*, "variables")= language list(mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb)
.. ..- attr(*, "factors")= int [1:11, 1:10] 0 1 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "dimnames")=List of 2
.. .. .. ..$: chr [1:11] "mpg" "cyl" "disp" "hp" ... .. .. .. ..$ : chr [1:10] "cyl" "disp" "hp" "drat" ...
.. ..- attr(*, "term.labels")= chr [1:10] "cyl" "disp" "hp" "drat" ...
.. ..- attr(*, "order")= int [1:10] 1 1 1 1 1 1 1 1 1 1
.. ..- attr(*, "intercept")= int 1
.. ..- attr(*, "response")= int 1
.. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
.. ..- attr(*, "predvars")= language list(mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb)
.. ..- attr(*, "dataClasses")= Named chr [1:11] "numeric" "numeric" "numeric" "numeric" ...
.. .. ..- attr(*, "names")= chr [1:11] "mpg" "cyl" "disp" "hp" ...
$residuals : Named num [1:32] -1.6 -1.112 -3.451 0.163 1.007 ... ..- attr(*, "names")= chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...$ coefficients : num [1:11, 1:4] 12.3034 -0.1114 0.0133 -0.0215 0.7871 ...
..- attr(*, "dimnames")=List of 2
.. ..$: chr [1:11] "(Intercept)" "cyl" "disp" "hp" ... .. ..$ : chr [1:4] "Estimate" "Std. Error" "t value" "Pr(>|t|)"
$aliased : Named logi [1:11] FALSE FALSE FALSE FALSE FALSE FALSE ... ..- attr(*, "names")= chr [1:11] "(Intercept)" "cyl" "disp" "hp" ...$ sigma        : num 2.65
$df : int [1:3] 11 21 11$ r.squared    : num 0.869
$adj.r.squared: num 0.807$ fstatistic   : Named num [1:3] 13.9 10 21
..- attr(*, "names")= chr [1:3] "value" "numdf" "dendf"
$cov.unscaled : num [1:11, 1:11] 49.883532 -1.874242 -0.000841 -0.003789 -1.842635 ... ..- attr(*, "dimnames")=List of 2 .. ..$ : chr [1:11] "(Intercept)" "cyl" "disp" "hp" ...
.. ..$: chr [1:11] "(Intercept)" "cyl" "disp" "hp" ... - attr(*, "class")= chr "summary.lm" If we pull the coefficients from this object, we are not just getting the values, but the table that’s printed in the summary. And we can now get that ready for publishing for example8. lm_mod_summary$coefficients %>%
kableExtra::kable(digits = 2)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 12.30 18.72 0.66 0.52
cyl -0.11 1.05 -0.11 0.92
disp 0.01 0.02 0.75 0.46
hp -0.02 0.02 -0.99 0.33
drat 0.79 1.64 0.48 0.64
wt -3.72 1.89 -1.96 0.06
qsec 0.82 0.73 1.12 0.27
vs 0.32 2.10 0.15 0.88
am 2.52 2.06 1.23 0.23
gear 0.66 1.49 0.44 0.67
carb -0.20 0.83 -0.24 0.81

After a while, you’ll know what’s in the objects you use most often, and that will allow you more easily work with the content they contain, allowing you to work with them more efficiently.

### Methods

Consider the following:

summary(diamonds)            # data frame
     carat               cut        color        clarity          depth           table           price             x                y                z
Min.   :0.2000   Fair     : 1610   D: 6775   SI1    :13065   Min.   :43.00   Min.   :43.00   Min.   :  326   Min.   : 0.000   Min.   : 0.000   Min.   : 0.000
1st Qu.:0.4000   Good     : 4906   E: 9797   VS2    :12258   1st Qu.:61.00   1st Qu.:56.00   1st Qu.:  950   1st Qu.: 4.710   1st Qu.: 4.720   1st Qu.: 2.910
Median :0.7000   Very Good:12082   F: 9542   SI2    : 9194   Median :61.80   Median :57.00   Median : 2401   Median : 5.700   Median : 5.710   Median : 3.530
Mean   :0.7979   Premium  :13791   G:11292   VS1    : 8171   Mean   :61.75   Mean   :57.46   Mean   : 3933   Mean   : 5.731   Mean   : 5.735   Mean   : 3.539
3rd Qu.:1.0400   Ideal    :21551   H: 8304   VVS2   : 5066   3rd Qu.:62.50   3rd Qu.:59.00   3rd Qu.: 5324   3rd Qu.: 6.540   3rd Qu.: 6.540   3rd Qu.: 4.040
Max.   :5.0100                     I: 5422   VVS1   : 3655   Max.   :79.00   Max.   :95.00   Max.   :18823   Max.   :10.740   Max.   :58.900   Max.   :31.800
J: 2808   (Other): 2531
.. ..$: chr [1:11] "mpg" "cyl" "disp" "hp" ... ..@ factors : list() Usually you will access the contents via methods rather than using the @, and that assumes you know what those methods are. Mostly, I just find S4 objects slightly more annoying to work with for applied work, but you should be at least somewhat familiar with them so that you won’t be thrown off course when they appear. ### Others Indeed there are more types of R objects, but they will probably not be of much notice to the applied user. As an example, the text2vec package uses R6. I can only say that you’ll just have to cross that bridge should you get to it. ### Inspecting Functions You might not think of them as such, but in R, everything’s an object, including functions. You can inspect them like anything else. str(lm) function (formula, data, subset, weights, na.action, method = "qr", model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, contrasts = NULL, offset, ...)  ## lm  1 function (formula, data, subset, weights, na.action, method = "qr", 2 model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, 3 contrasts = NULL, offset, ...) 4 { 5 ret.x <- x 6 ret.y <- y 7 cl <- match.call() 8 mf <- match.call(expand.dots = FALSE) 9 m <- match(c("formula", "data", "subset", "weights", "na.action", 10 "offset"), names(mf), 0L) 11 mf <- mf[c(1L, m)] 12 mf$drop.unused.levels <- TRUE
13     mf[[1L]] <- quote(stats::model.frame)
14     mf <- eval(mf, parent.frame())
15     if (method == "model.frame")
16         return(mf)
17     else if (method != "qr")
18         warning(gettextf("method = '%s' is not supported. Using 'qr'",
19             method), domain = NA)
20     mt <- attr(mf, "terms")

One of the primary reasons for R’s popularity is the accessibility of the underlying code. People can very easily access the code for some function, modify it, extend it, etc. From an applied perspective, if you want to get better at writing code, or modify existing code, all you have to do is dive in! We’ll talk more about writing functions later.

## Help Files

Many applied users of R are quick to search the web for available help when they come to a problem. This is great, you’ll find a lot of information out there. However, it will likely take you a bit to sort through things and find exactly what you need. Strangely, I see many users of R don’t use the help files first, and yet this is typically the quickest way to answer many of the questions they’ll have.

Let’s start with an example. We’ll use the sample function to get a random sample of 10 values from the range of numbers 1 through 5. So, go ahead and do so!

sample(?)

Don’t know what to put? Consult the help file!

We get a brief description of a function at the top, then we see how to actually use it, i.e. the form the syntax should take. We find out there is even an additional function, sample.int, that we could use. Next we see what arguments are possible. First we need an x, so what is the thing we’re trying to sample from? The numbers 1 through 5. Next is the size, which is how many values we want, in this case 10. So let’s try it.

nums = 1:5
sample(nums, 10)
Error in sample.int(length(x), size, replace, prob): cannot take a sample larger than the population when 'replace = FALSE'

Uh oh- we have a problem with the replace argument! We can see in the help file that, by default, it is FALSE9, but if we want to sample 10 times from only 5 numbers, we’ll need to change it to TRUE.

Now we are on our way!

The help file gives detailed information about the sampling that is possible, which actually is not as simple as one would think! The Value is important, as it tells us what we can expect the function to return, whether a data frame, list, or whatever. We even get references, other functions that might be of interest (See Also), and examples. There is a lot to digest for this function!

Not all functions have all this information, but most do, and if they are adhering to standards they will10. However, all functions have this same documentation form, which puts R above and beyond most programming languages in this regard. Once you look at a couple of help files, you’ll always be able to quickly find the information you need from any other.

## Objects Exercises

With one function, find out what the class, number of rows, number of columns are of the following object, including what kind of object the last three columns are. Inspect the help file also.

library(dplyr)
?starwars

1. An alternative would be to use the coef method, broom::tidy on the lm object itself, or use pander instead of kableExtra to work on the lm object.↩

2. For the things I do, I need sample with replacement far more than I don’t, but that may not be the same for everyone.↩

3. Ahem, tidyverse.↩


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## [ Politics ] Open Question : If the USA had its own unlimited oil supply and made all its own products would the coronavirus pandemic not have happened?

China and oil sheiks are able to manipulate the USA because of the USA’s extreme dependence on chinese factory workers and oil. I am saying that china is able to manipulate the USA with a fake pandemic. China depends on oil from the middle east so oil sheiks can manipulate the USA through china. The coronavirus pandemic is a hoax.

## Would a zealot barbarian’s divine fury trigger during an opportunity attack?

The Path of the Zealot barbarian (Xanathar’s Guide to Everything, p. 11) gets the Divine Fury feature at level 3, which states:

At 3rd level, while you’re raging, the first creature you hit on each of your turns with a weapon attack takes extra damage equal to 1d6 + half your barbarian level. The extra damage is necrotic or radiant; you choose the type of damage when you gain this feature.

Does it work like the Rogue’s Sneak attack, in that it can trigger on opportunity attacks, since it’s a different turn? Or does the "on each of your turns" bit make it not work.

## What would be the effects of changing the duration of a round from 6 seconds to 5 seconds?

I’m introducing the following house rule:

A round lasts 5 seconds.

What would be the effects of such a house rule? Would there be any? Would this break the game?

Out of the top of my head, I cannot think of any actions/spells/mechanics that rely on the duration of a round.

Disclaimer: The original question was a bit broader and was closed twice as opinion-based/off-topic.

## Would this Point Buy house rule for Maid character creation work at all?

Recently we’ve started a maid game here on the stack and we have a huge disparity of stats between the players, one of them has barely average stats, while some of the other players are pushing fours and sixes in abilities because of great rolls during generation. We have found that the standard rule that characters with good stats get fewer maid powers is insufficient.

Would a point buy system work in Maid RPG while still being able to attain the same level of play in Maid games? Using the favor base presented in the handbook I’ve come up with the following figures for point buy statistics. The standard array in Maid would be 1,1,2,2,3,3 or you could buy points at the following costs, with 200 starting favor:

0: 0 points
1: 10 points
2: 30 points
3: 60 points
4: 100 points

With the stipulation that any favor points not used during point buy are lost. Does a system like this function properly in the Maid system provided the Maid characteristics (Boyish, Sexy, Lolita, Pure, etc.) were rolled after (and only after) a character has distributed their points as they wish?

## For the Purpose of Detect Magic and Identify. Would a Magic item that ran out of power for the moment be considered a magic item?

If a magic item runs on "Magic Batteries" when the Item is out of power would it be considered a magic time for the purpose of the Identify and Detect Magic spell?

For example, I have a Mirror that shows through illusions, it has crystals around it that are destroyed as you used it, you can replace the crystals for more uses when it has no crystals, would it still be considered a magic item detectable by the spells?

Would the same work for items with charges that recharge at dawn? If the item has 0 charges and is not dawn yet, would it be still be considered a magic item?

## Would a Periapt of Proof Against Poison nullify the effects of alcohol?

The description of the Periapt of Proof Against Poison (PoPAP) says that “poisons have no effect on you. You are immune to the poisoned condition and have immunity to poison damage.”

So I’m wondering if a character could get drunk while wearing a PoPAP? There’s no “drunk” status condition in D&D 5e, but drunkenness is caused because you’ve consumed enough alcohol to poison yourself, however slightly. Also, does the wearer of the PoPAP know they’ve been poisoned? Like for instance, if someone put spider venom in their food, would they know the food was poisoned when they ate it?

## Would these adjustments to the ranger archetype Beast Master help the animal companion be more useful?

I recently playtested a Beast Master ranger from level 1 to level 20 (I was playtesting a new homebrew archetype, which was my primary reason for doing so; the Beast Master ranger was just one of the other party members), but there were a few things I noticed regarding the relative power of the beast companion itself. For reference, the beast I went with was a wolf, which is probably a fairly standard choice.

## Issues

Now, I know that Beast Master rangers are infamously weak, but I still wanted to see if I could try to improve what I felt were some of its weakest points during my playtesting. I was already using the popular houserule of letting the ranger tell the beast to attack using a bonus action instead of an action, but the other things that bothered me were:

• Relatively low HP (as the first linked Q&A points out), although this was more of a problem during Tier 1/2, less so during Tier 3/4, at least during my playtesting;
• Hardly any hit die, which is related to the above problem, since I remember having to spend a lot of healing resources to keep bringing the wolf’s health back up to full/close to full;
• The DC for resisting the knocked prone secondary effect from the wolf’s Bite attack remains pathetically low at DC 11 for the whole game.
• The lack of any saving throw proficiencies really screwed the wolf over during the big finale where it died to a meteor swarm, but with a decent DEX saving throw bonus, it would probably have made it.
• I was sometimes hesitant to use the wolf, because it was dropped to 0 HP a few times at lower levels, unless I knew it would probably land the killing blow or could avoid an opportunity attack or otherwise being hit.

I will point out that at higher levels, the AC was fairly decent (for a wolf), and the HP wasn’t as bad as it was at earlier levels, and I was impressed with the damage output thanks to attack rolls and damage scaling with the ranger’s proficiency bonus. Its Stealth and Perception skill bonuses were also impressive. These things I don’t feel the need to change.

## Changes

Here are the changes I propose, somewhat inspired by the UA Sidekick rules:

• You get a new hit die whenever you take another level in ranger, so at level 3 your wolf starts off with two hit die, but at level 4 they would have three hit die … by level 20 they have 19 hit die. I doubt this would make their max HP better than four times the ranger level, so it would only really be for the purposes of short resting.

• To improve the max HP a little, maybe something as simple as adding the beast’s CON modifier to that, so it’s now:

\begin{align} \text{ (ranger level + beasts’s CON modifier)} \times 4 \end{align}

This way, the animal’s toughness is also taken into an account; I feel like the wolf having 5 instead of 4 more HP each level would have been just enough to help, combined with more hie die to heal, but also from a flavour perspective, I feel like choosing a boar should end up tougher than a hawk, whereas RAW, they would both have the same HP. I would however, keep the minimum HP gained per level to 4, in case the beast somehow has a negative CON modifier, since I think taking HP away from the beast would be cruel, given how underpowered this whole archetype is.

• Any DCs it has, such as the wolf’s ability to knock people prone, should scale with your proficiency bonus, like this AC and attack/damage rolls do, so rather than a measly 11, at level 3-4 it would be 13, and at level 5, it would be 14 … ending up at 18 at level 17+.

• Unless it already has a "physical" saving throw proficiency (meaning STR, DEX or CON), it gains one of your choice at level 3, which of course would just mean a +2 (because that’s every valid animal companions’s proficiency bonus) but that also has your proficiency bonus added to it, like AC, etc. This would have certainly helped when it was hit by meteor swarm during our final level 20 showdown, it might have actually survived (even with its RAW hit points) had it made that DEX saving throw.

• Finally, since I’m letting the beast be commanded as a bonus action, the first half of the 7th level ranger feature Exceptional Training is kinda wasted, so I was considering changing that to not only make the beast’s attacks magical, but also to effectively give the beast a rogue’s Cunning Action, which it can use if you command it to using the same bonus action you used to command it to attack (or do something else with its action). In short, you use one bonus action to tell it what to do with its turn, and it can now effectively use its action and bonus action to do something useful.

## Question

Do the above changes seem reasonable, and do you foresee any balance issues coming from my proposed changes? My intention is for the Beast Master’s beast in particular to become more useful and survivable, without increasing its damage output (since I was happy with that), but not making it more powerful than I intended by overlooking something. I suppose also double checking whether there’s a problem with making some animals tougher than others based on their CON; does this unfairly favour tougher animals to the point where that’s a balance issue in and of itself?

## What archfey exist in Eberron lore that would make a suitable warlock patron?

I’m planning on making an Archfey warlock for an upcoming Eberron game. To flesh out my character’s backstory, I want to know more about what kind of "fey" entities such a warlock might form a pact with.

For context, the character’s backstory is that they are a Valenar elf who’s ancestor that he has to emulate was an "Eldritch Knight" or similar, some kind of battlemage anyway, except that my character isn’t very strong or smart (STR and INT are his dump stats), so he instead formed a pact with an archfey being to make up the difference, becoming a Pact of the Blade warlock so that he could better emulate his ancestor.

The reason I’ve picked the Archfey archetype is because the Valenar seems to be rather connected with fey, at least considering the various "Valenar xxx" animals listed in Eberron: Rising from the Last War.

I can of course work with my DM to just "invent" an archfey if need be, but I wanted first to see if there are any existing named archfey within any Eberron lore (from any edition of D&D if 5e doesn’t have anything), and bonus points if that archfey is associated with the Valenar elves at all.

## If a Spell has a casting time of 1 minute, what part of the caster’s action economy would it take up if it was cast in combat?

This is a bit difficult of an idea for me to articulate, so I’ll try and do my best: If a character was to attempt and cast a spell with a casting time of 1 minute, while in combat, would they have to simply consume their action for 10 turns, but still move and take bonus actions? Would they be unable to do anything except cast that spell? Would they also be required to maintain concentration?

## What would be the gold cost of a feat?

Imagine a player wanting a training in leadership by some charismatic nobleman. I am the GM and I agree with that, but I have no idea as to the time spent and the monetary compensation it would take. Is there any information regarding this, or any rules of thumb ?


text:
# Browse Dissertations and Theses - Physics by Contributor "Mozurkewich, George"

• (1990)
The conduction noise from sliding charge-density waves (CDWs) is studied in NbSe$\sb3$ and K$\sb{0.3}$MoO$\sb3$ (blue bronze). A statistical analysis of the fluctuations in the noise is presented: the probability distributions ...

application/pdf

PDF (4MB)
• (1990)
The conduction noise from sliding charge-density waves (CDWs) is studied in NbSe3 and Ko.3Mo03 (blue bronze). A statistical analysis of the fluctuations in the noise is presented: the probability distributions of the ...

application/pdf

PDF (1MB)
• (1991)
The elastic softening associated with the sliding of a charge-density wave (CDW), driven by an electric field, is investigated in o-TaS$\sb3$ by use of a vibrating-reed technique. Experiments performed with a dc sliding ...

application/pdf

PDF (4MB)
• (1991)
Two different aspects of the charge-density-wave (CDW) state were studied in the quasi-onedimensional metal blue bronze (Ko.3oMo03). In the primary experiment reported in this thesis, pinning of the CDW phase by lattice ...

application/pdf

PDF (2MB)


text:
### On Error Resume Prior   If at first you don't succeed...

# The Porus Mirror Problem, Part 2

## Propagation

From last time, we know the quantity of light that escaped at each location and direction around the unit circle. We want to find how this translates into quantities of light at any point away from the circle. This situation can be modeled as so:

The basic idea is to find for any given $$\alpha, r$$ the set of possible $$\phi, \theta$$ values. In any case, it is easiest to eliminate the $$\phi$$ variable via rotation.

Noting that the two angles have the same $$y$$ coordinate, we end up with the equation:

$r\sin(\alpha-\phi) = \sin(\theta-\phi)$

Therefore, we can calculate $$\theta$$ in terms of everything else as

$\theta = \phi + \sin^{-1} \left(r\sin(\alpha-\phi)\right)$

This is defined whenever $$-1 \leq r\sin(\alpha-\phi) \leq 1$$, which is true whenever $$\alpha-\sin^{-1} \frac{1}{r} \leq \phi \leq \alpha + \sin^{-1} \frac{1}{r}$$.

We can therefore calculate the quantity of light at any point as

$E (r, \alpha) = \int_{\alpha-\sin^{-1} \frac{1}{r}}^{\alpha+\sin^{-1} \frac{1}{r}}L\left(\phi, \phi + \sin^{-1} \left(r\sin(\alpha-\phi)\right)\right)\text{d}\phi$

## A constant transparency

As might be expected, a constant transparency yields a graph of $$L$$ perfectly symmetrical across $$\phi = \theta$$: it is only a function of $$\phi - \theta$$. (The following graph has $$\theta$$ on the $$x$$ axis and $$\phi$$ on the $$y$$.)

This yields the symmetrical circular pattern of energy

## Shifted $$\sin$$

To get a less symmetrical pattern, we can shift the $$\sin$$ function into $$[0,1]$$. This function is effectively $$y$$ so results in the top-biased:

## Periodic pattern on surface

Using the 3-period periodic function $$\sin^2 \frac{3\theta}{2}$$, we get a 3-petal flower:

## Half-opaque

Making the bottom half of the circle opaque leads to the diagram:

## A Tiny Opaque piece

A small opaque piece casts a small shadow

## A Single Slit

Because our model is corpuscular, we don’t have diffraction.

## A Double Slit

Again, no diffraction. The pattern that occurs is mostly an aliasing effect.

## Negative Slit

Our model is flexible enough to allow for “negative transparency”, when “negative light” is projected forwards while extra light is reflected backwards. This is basically just a color inverted version of a small reflector.


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# Seminari del Dipartimento

 Fisica MatematicaBulk-Edge correspondence in interacting spin Hall insulatorsVieri Mastropietro17-04-2018 - 14:30Largo San Leonardo Murialdo,1 - Pal.C - AULA 311 We prove the perfect quantization of the edge spin conductance in spin Hall insulators . The cancellation of the possible interaction corrections is particularly remarkable in view of the critical nature of the problem (M. and Porta, PRB 2017). Convergence of the expansion holds in a disk with a radius vanishing with the Fermi velocity and how to overcome this limitation in related problems will be also discussed (Bonetto and M., AHP 2016).  org: GIULIANI Alessandro


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# Why don’t we use a weaker version of the Axiom Schema of Unrestricted Comprehension?

According to my limited understanding, the axiom schema of unrestricted comprehension states that, for any predicate, there exists a set whose members are exactly those that satisfy the predicate. Russell showed this to be false by considering the set of all sets that do not contain themselves, and finding that if that set did not contain itself it implied that it did, and vice versa, leading to a contradiction. This has been resolved in several ways, but the most common is to dump the idea more or less completely, and as a result, prevent the existence of a universal set. Why don’t we instead just modify the axiom?

We could instead say that, for any predicate, there exists a set that contains all objects that satisfy that predicate. In such a set, we allow for other objects to be contained in the set. Let’s try to use Russell’s Paradox on it. Take a set containing all sets that do not contain themselves. We allow this set to contain other objects. If the set does not contain itself, then it should, which means it must contain itself. If it does contain itself, then it no longer satisfies the predicate of not containing itself, but it can still be in the set under our definition, so we’ve eliminated the paradox. This also seems to resolve the Burali-Forti Paradox and Cantor's Paradox, so it seems like there’s no longer anything in the way of the universal set.

Now, there is a strange notion to this idea, which is that, simply by putting all sets that do not contain themselves into a set, we get a set that does contain itself chucked in for free automatically. At first, I was uncomfortable with this idea, but I no longer feel that way. We consider the set containing elements a, b, and the set containing a and b, to be different from the set containing elements a and b. That is, we consider {a, b, {a, b}} to be different from {a, b}. But why do we do this? Clearly, {a, b} contains both a and b, and intuitively contains the collection that is a and b put together. It seems we treat a set as more than this collection, we treat a set as this collection “in a box”, that is, we treat the set as the collection taken in isolation, so that {a, b} does not contain itself. Why don’t we keep this notion of a collection in a box so that things like a topology still make sense, but say that a set automatically contains all of its subsets, outside of boxes? This seems to make sense, and if we allow it, then all sets contain themselves. The set of all sets that do not contain themselves is then the null set, but more importantly, contains itself. We could make an axiom that, for any predicate, there exists a set that contains exactly all objects that satisfy that predicate and all unions of those objects.

For what reason is this wrong?

• "The set of all sets that do not contain themselves is then the null set, but more importantly, contains itself." - This is an obvious contradiction, unless you stumbled over your phrasing. Also, this is several questions at once, which is against best practices for this site. – Malice Vidrine Apr 4 '17 at 13:07
• "Why don’t we use a weaker version of the Axiom Schema of Unrestricted Comprehension?" We use it : Axiom of Separation (or Specification) stating that "for any set and for any predicate, there exists the subset of that set whose members are exactly those that satisfy the predicate." – Mauro ALLEGRANZA Apr 4 '17 at 13:07

In other words, you're proposing to weaken the universal comprehension axiom to

For every predicate $p$ there exists a set $X$ such that $X$ comprises at least every $y$ where $p(y)$ is true (but may have more elements than that).

You're right that this wouldn't be paradoxical. But that is because it is extremely weak. This axiom is satisfied for every $p$ the moment there exists a set of all things. (And a set of all things would have to exist the moment we consider your axiom with $p(x)$ taken as $x=x$).

This means that the axiom is practically useless. When we use sets while doing ordinary mathematics, we don't merely want our sets to have enough elements -- it is usually critically important that the sets contain only the elements we want them to -- and your proposal doesn't leave us any way to construct such a set.

Usually we'd use Zermelo's axiom of separation to remove unwanted elements from a set, but if we take that together with your proposal, then Russell's paradox still results immediately.

Remember that it's not in itself an advantage of a proposal that it would allow us to write down some combinations of symbols that we previously didn't have a meaning for (e.g., "$\frac 00$" is a popular candidate for this). It only represents progress if the new thing we can write down represents a meaning that we have a reason to want to express.

If there is not a contradiction in this system, it's because, as Henning Makholm states, it's extremely weak. To see how useless this would be, consider that we get induction over the natural numbers because we can form a set whose members are exactly those that are in every set containing $0$ and closed under successor; without this, induction simply doesn't get off the ground. And it's not just natural numbers; any kind of induction is going to be stunted.

Now, separation is a restriction of set abstraction but it seems like you're interested in something else. If you want to keep a universal set, there are several options that are well studied and not outright insane. The big ones are NFU and GPK. The former restricts abstraction to those formulae which are "stratified"; formulae which could be well-formed formulae of simplified Russellian type theory save that you've carefully chucked the typing discipline. The latter restricts abstraction to positive formulae (those in which negation and implication don't appear), with some other gear that has a topological motivation.

Both of these alternatives are strange places, but at least we can still do things like analysis, algebra, and number theory.

I cannot address the second question, as I'm not sure it makes sense. I will note that it's consistent with NFU to have sets such that $a=\{a,b\}$ (and hence $\{a,b,\{a,b\}\}=\{a,b\}$); even sets such that $a=\{x:a\in x\}$. But the truth is they're not very exciting objects, except as counterexamples to see how badly Foundation can fail.


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bug-texinfo
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## Re: XeTeX encoding problem

 From: Gavin Smith Subject: Re: XeTeX encoding problem Date: Sun, 6 Mar 2016 18:44:37 +0000

On 7 February 2016 at 16:23, Gavin Smith <address@hidden> wrote:
> In a patch that I'm still working on, I have a definition like
>
> +\def\gdefchar#1#2{%
> +\gdef#1{%
> +   \ifpassthroughchars
> +     \string#1%
> +   \else
> +     #2%
> +   \fi
> +}}
>
> and then e.g. in \latonechardefs
>
>  \def\latonechardefs{%
> -  \gdef^^a0{\tie}
> +  \gdefchar^^a0{\tie}
>

> I hope it's clear what I'm trying to do here: instead of redefining,
> change the value of a conditional that is used within the macro.

I've finally finished this. It does appear a bit faster as well as
being simpler. Where it could potentially break is chapter names with
non-ASCII characters, and index entries with non-ASCII characters, as
well as macros using non-ASCII characters (either in the macro
definition or in the arguments), as well as macros giving chapter
names and index entries. I'll do a bit more testing later.

Masamichi, does the native multibyte support still work well? Is there
any more that you wanted to do with it? If it looks good we should
look to releasing it as an official version for non-developers.

reply via email to


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# Simple Induction vs Strong Induction proof.

I am having a great deal of trouble grasping the essential difference between Simple induction and Strong Induction proof.

I was hoping someone here would do me the honor of making an example. First illustrating simple induction then the same example using strong induction and then point out the essential difference between these two procedures.

My teacher showed us how you could solve F_n <= 2^n. Fn being fibbonacci.

F0 = 0, F1 = 1, Fn =Fn-1 + Fn-2

By simple induction she said it would be solved like this:

if P(1) = true
Then assume P(n) is true for all n in positive integers
Show that P(n+1) is true.
P(n+1) is true if Fn+1<=2^{n+1}
Fn+1 = Fn+Fn-1
<= Fn + Fn
<= 2Fn
<= 2*2^n
<= 2^n+1
Then proof by simpel induction is complete.

The same expression can be proofed by Strong Induction like this:'
if P(1) is true
Then assume that P(i) is true for all i<=k
Show that P(k+1) is true.
P(k+1) is true if Fk+1 <= 2^k+1
Fk+1 = fk + fk-1
<= 2^k + 2^k-1
<= 2^k + 2^k * 2^-1
<= 2^k(1+0,5)
<= 2^k(3/2)
<= 2^k * 2
<= 2^k+1
The proof by Strong induction is complete

I cant graps at all what my teacher is doing in the strong induction proof. Can someone tell me and point out the essential difference between the two procedures?

• These two concepts are formally different, but they are essentially the same thing. Nov 8, 2016 at 11:50
• Simple induction is just a special case of so-called strong induction, so every proof by simple induction is a proof by strong induction. Thus, no such example is possible. The only difference is in the form of the induction hypothesis. In strong induction you assume that the result is true for all $k<n$ that are at least as large as your starting point, while for ordinary induction you assume it only for $n-1$. Nov 8, 2016 at 11:50
• There must be a difference. My teacher fails me when i use Simple Induction and I should have used Strong Induction. I know the definition for both terms but when i look at them i cant stop but thinking that that is the same thing. But my teacher says no it is not the same thing and i can't understand her argument for why it is not. So therefore i made this post hoping someone could elighten me. Nov 8, 2016 at 11:57
• You know it is everything about this. I cant tell the difference between the procedures and i cant tell when to use what. Nov 8, 2016 at 11:58
• @Nulle Essentially you should never try to use simple induction. Think of the principle of strong induction as a "licence to use the fact that $P$ is true for smaller values than $n$, when trying to prove $P(n)$". If it turns out that during your proof, you would find it helpful to use $P$ for a smaller value, then feel free to use it. If it turns out that the natural such smaller value is $n-1$, then so be it; you've just recovered simple induction. (Thanks to Prof Imre Leader for this viewpoint and for many other things.) Nov 8, 2016 at 20:53

Theorem. Any natural number $n>1$ can be factored into $\geq1$ primes.
In the proof we may use the principle $$x\geq y>1\qquad\Rightarrow\qquad x\,y>x\geq y\ .\tag{*}$$
Here simple induction is of no help, because there is no obvious connection between the factorizations of $n$ and $n+1$, e.g., $48$ and $49$. But strong induction does the job: The number $2$ is prime. Let an $N\geq2$ be given, and assume the strong induction hypothesis (SIH): The statement is true for all $n\in[2\>..\,N]$. Then the number $N+1$ is either prime, or a product $N+1=n'\cdot n''$ whereby $$2\leq \ n'\leq n'' \ \leq N$$ (here the principle $(*)$ was used). The strong induction hypothesis (SIH) then allows to conclude that $N+1$ can be written as a product of $\geq1$ primes.
(Answering your comment: The second proof of $F_n\leq2^n$, apart from missing a dozen parentheses, uses "strong induction" in so far as for the proof of $F_{n+1}\leq 2^{n+1}$ it not only uses $F_n\leq 2^n$, but also explicitly $F_{n-1}\leq 2^{n-1}$. But it does not use that $F_k\leq 2^k$ is true for all $k\leq n$.)
• In your answer Let $N\ge 2$ given is pointed out, but why in this answer the formula of strong induction only says $m\le n$, not $0\le m\le n$? link: mathoverflow.net/a/37971/104425 . (I commented there, please help.) Jul 23, 2019 at 9:22
• @PostalModel: There are no $m<0$ in the linked answer, since the "base set" ${\mathbb N}$ is assumed to begin at $0$. Jul 23, 2019 at 12:49


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What is the purpose of the flat/fppf/fpqc topologies?

There have been other similar questions before (e.g. What is your picture of the flat topology?), but none of them seem to have been answered fully.

As someone who originally started in topology/complex geometry, the étale topology makes some sense to me. It's sort of like the complex topology, in that there are enough "open sets" that things like the inverse function theorem work.

But I don't really understand what these other more "exotic" topologies represent. From the (naive) background of topology, it sounds like we're just "adding more open sets" to our topology (which I know isn't right, since this isn't a topology in the standard sense).

So what do they represent? What do they do for us?

• One way to motivate fppf and fpqc topologies is: they are the right setting to formulate all sorts of relevant descent statements in algebraic geometry, cf.e.g. stacks.math.columbia.edu/download/descent.pdf. – Matthias Wendt Jan 29 '15 at 10:42
• The fppf topology is important to express phenomena of p-torsion in char. p. For instant the sequence of commutative group schemes $0\rightarrow \mu_p \rightarrow \mathbb{G}m\rightarrow \mathbb{G}m\rightarrow 0$ over a field of char. p is only exact as a sequence of fppf sheaves, not étale sheaves, so to do Kummer theory for p-torsion in char. p (to study the p-torsion in the Brauer group, for instance), you need flat cohomology. – Simon Pepin Lehalleur Jan 29 '15 at 13:12

3 Answers

I'm a topologist, and so this answer is going to specifically be about analogies with ordinary topology. I like to think of different Grothendieck topologies as corresponding to different "allowable" ways to build up a space using a quotient topology.

1. The Zariski topology is like recognizing that a space $X$ can be built up from a collection of open subsets $U \subset X$ which cover $X$ and identifying points in common intersections.

2. The etale topology is like recognizing that if a map $Y \to X$ is surjective and a local homeomorphism, then this map makes $X$ homeomorphic to the quotient of $Y$ by an equivalence relation.

3. The fppf topology is similar to the etale topology, but now allowing maps $Y \to X$ which are open.

4. The fpqc topology, for lack of a better analogy, is like allowing arbitrary maps $Y \to X$ which make $X$ into a quotient space of $Y$.

Much of the study of these topologies is the study of sheaves on them. The choice of topology has at several significant consequences.

1. The choice of topology determines how much functoriality your sheaf has. (In particular, choices like "big site" or "small site" determine how many objects $Y$ you can evaluate your sheaf on!)

2. The choice of topology determines what kind of data you need to construct things (which is code for descent theory). We know that if $Y \to X$ is a quotient map, we can build a vector bundle on $X$ by taking a vector bundle on $Y$ and imposing a compatible equivalence relation on it (e.g. from a clutching function). Similarly, to build a sheaf in topology $\tau$ we just need to build it on a $\tau$-cover and glue it together.

3. The choice of topology determines your definition of "local". I really like Simon Pepin Lehalleur's comment here, because it describes exactly the following problem: when is a map ${\cal F} \to {\cal G}$ of sheaves surjective? It is surjective when it is surjective locally. For example, the n'th power map $\Bbb G_m \to \Bbb G_m$ is a surjection of sheaves precisely when, locally, every invertible function is an n'th power of some other invertible function. That's rarely true in the Zariski topology; it's true in the etale topology when $n$ is invertible, because then adjoining a solution of the polynomial $(x^n - \alpha)$ is an etale extension; it's always true in the fppf topology because adjoining a solution of the polynomial $(x^n - \alpha)$ is always fppf.

This is sort of an expansion of Simon's comment above.

I think the purpose of any topology is to give the "correct" cohomology groups, whatever that means. In general you probably want to work with cohomology groups in the coarsest topology you can; for example, computing explicitly in the Zariski topology is much nicer than computing explicitly in the flat topology.

For example, let $X$ be a variety over a field. If $F$ is a sheaf of coherent $\mathcal{O}_X$-modules, then the Zariski topology calculates the "correct" cohomology groups $\mathrm{H}^i(X,F)$, in the sense that going to the finer étale or flat topology gives the same result.

If $F$ is the sheaf represented by a smooth group scheme, then the étale topology gives the "correct" cohomology groups. In particular, as pointed out by Simon in his comment, the Kummer sequence is exact in the étale topology whenever $n$ is coprime to the residue characteristics, so cohomology with values in $\mu_n$ is related in a nice way to Picard groups, Brauer groups and so on. It also agrees with the complex cohomology in characteristic zero. But this doesn't work if the characteristic divides $n$, because then $\mu_n$ is not smooth. To get the "correct" cohomology for $\mu_p$ in characteristic $p$, and in particular for the Kummer sequence to be useful, you really have to work in the flat topology.

The question is what kind of structure you are interested in. For example, the étale topology sees more than you might like as a topologist/complex geometer. If you look at Spec(k) where k is not algebraically closed, the étale topology will see the Galois group. As a topologist/complex geometer you might also like the Nisnevich topology. It has enough covers to make a smooth variety look locally like affine space, but it doesn't see the "arithmetic stuff" that only depends on the ground field and not on the variety itself. But if you are interested in the properties of the ground field, too, you will choose the étale topology.

As for the flat/fppf/fpqc topologies, I think you already got some very convincing answers. I can't add anything to them. Maybe this is rather an overlong comment than an answer.


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# 2.02 Operations with fractions

Lesson

We've already learned how to add, subtract, multiply and divide fractions. Similarly, we've looked at each of these operations with negative numbers.

The process is just the same when we have questions with negative fractions - we'll just combine these two skills and their rules to complete the operation.

Review: Fraction operations
• Adding and subtracting: be sure there is a common denominator, and then add or subtract the numerators and keep the denominator
• Multiplying: Multiply the numerators, multiply the denominators, and simplify.
• Dividing: Take the reciprocal of the divisor and multiply. Then, follow the rules for multiplication.
• Mixed numbers: Change the mixed number to an improper fraction, and proceed as if this were a normal fraction operation. You may be expected to change the solution back to a mixed number.
Remember! Operations with integers
• Subtracting a negative number is the same as adding a positive number, e.g. $5-\left(-3\right)=5+3$5(3)=5+3
• Adding a negative number is the same as subtracting a positive number, e.g. $2+\left(-7\right)=2-7$2+(7)=27.
• The product or quotient of two negative terms is a positive answer, e.g. $-6\times\left(-9\right)=54$6×(9)=54  or $\left(-144\right)\div\left(-12\right)=12$(144)÷(12)=12
• The product or quotient of a positive and a negative term is a negative answer, e.g. $-5\times10=-50$5×10=50 or $20\div\left(-4\right)=-5$20÷(4)=5

Using the number line or zero pairs may be helpful with working positive and negative fractions.

#### Worked examples

##### Question 1

Calculate: $3+4-\left(-\frac{4}{5}\right)$3+4(45).

Think: We will evaluate the addition and subtraction, working from left to right.

Do:

 $3+4-\left(-\frac{4}{5}\right)$3+4−(−45) $=$= $7-\left(-\frac{4}{5}\right)$7−(−45) $=$= $7+\frac{4}{5}$7+45 $=$= $7\frac{4}{5}$745

Reflect: What if this problem were written as a decimal? Is $3+4-\left(-0.8\right)$3+4(0.8) easier to compute? Do you still end up with the same solution when you convert the decimal back to a mixed number?

#### Practice questions

##### Question 2

Evaluate $-10\times\left(-2\frac{1}{4}\right)$10×(214), giving your answer as a mixed number in simplest form.

##### Question 3

Evaluate $-8\frac{4}{7}+3\frac{3}{7}$847+337, writing your answer as a mixed number in its simplest form.

##### Question 4

Evaluate $4\frac{2}{3}\div\left(-1\frac{2}{5}\right)$423÷(125), giving your answer as a mixed number in simplest form.

### Outcomes

#### 7.NS.A.1

Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.

#### 7.NS.A.1d

Apply properties of operations as strategies to add and subtract rational numbers.

#### 7.NS.A.2

Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

#### 7.NS.A.2c

Apply properties of operations as strategies to multiply and divide rational numbers.


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Two complex numbers a+i b and c+i d are equal if and only if a=c and b=d . Or want to know more information Complex Object 1 should be created with the default constructor. Q: Find the absolute minimum and absolute maxinmum values of f on the given interval. Two complex numbers z$$_{1}$$ = a + ib and z$$_{2}$$ = x + iy are equal if and Didn't find what you were looking for? Complex numbers have a similar definition of equality to real numbers; two complex numbers + and + are equal if and only if both their real and imaginary parts are equal, that is, if = and =. Prove it. Thus, z 1 = z 2 ⇔ Re (z 1) = Re (z 2) and Im (z 1) = Im (z 2 ). where r is the modulus and φ the argument of the number. Complex Numbers are numbers of the form a + ib Real part imaginary a,b are real numbers i=iota=root-1 Where a, b are real numbers and i = iota = √(-1) Show More Two complex numbers equal − 3 + Solved examples on equality of two complex numbers: 1. 3. 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Viewing z=a+bi as a vector in t… Remember a real part is any number OR letter that isn’t attached to an i. a = 0 and . I actually am quite confused with it. b ≠ 0, the complex number is . differ by a value of . Use this Google Search to find what you need. If two complex numbers in rectangular form are equal to each other, then the real part and the imaginary part of one are equal to the corresponding parts of the other. Hence, the required values of m and n are follows: m = 4, n = 5; m = 4, n = 4; m = 3, n = 5; m = 3, n = 4. The given two complex numbers are z$$_{1}$$ = 5 + 2yi and z$$_{2}$$ = -x + 6i. Examples (2+3i)+(5−6i) = 7−3i. We will discuss about the equality of complex numbers. Check if two numbers are equal without using comparison operators. The basic operations on complex numbers are defined as follows: (a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)–(c+di)=(a−c)+(b−d)i(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(bc+ad)i a+bic+di=a+bic+di⋅c−dic−di=ac+bdc2+d2+bc−adc2+d2i In dividing a+bi by c+di, we rationalized the denominator using the fact that (c+di)(c−di)=c2−cdi+cdi−d2i2=c2+d2. That makes our if statement more flexible: now one True value is enough to run its code. If . My inclination would be to regard two doubles as equal if they report equal to each other, or if neither reports equal to itself. (x1, y1)(x2, y2) = (x1x2 - y1y2, x1y2 + x2y1). In general a complex number is the sum of its real part and its imaginary part times i, and two complex numbers are equal if and only if they have the same real and imaginary parts. So let me get this straight. It only has weakened analogues in R4 (quaternions) and R8 (octonions). We can then de ne the limit of a complex function f(z) as follows: we write lim z!c f(z) = L; where cand Lare understood to be complex numbers, if the distance from f(z) to L, jf(z) Lj, is small whenever jz cjis small. For what real values of m and n are the complex numbers m$$^{2}$$ â Addition is defined componentwise in a relatively standard way that extends to spaces of higher dimension. Two complex numbers a + ib and c + id are equal if and only if a = c and b = d. Use this fact to evaluate cos bx dx and sin bx dx from ela a -… Add two numbers without using arithmetic operators. bd = 1.3 - 6. Now equating real and imaginary parts on both sides, we have. Here discuss the equality of complex numbers-. That gives us two equations. 1 Complex Numbers 1.1 Algebra of Complex Numbers The number i is declared by law to satisfy the equation i2 = −1. radians. 4 + 0i = 4. Last updated at May 29, 2018 by Teachoo. 2 - 5i. In order that zw = 1, we’ll need (xu – yv) + (xv + yu)i = 1. 2010 - 2021. gx=5.fx+11.5 , fx=16-6x . Use this Google Search to find what you need. Of course, the two numbers must be in a + bi form in order to do this comparison. Example 2 Determine whether the following pairs of complex numbers are equal or not. If the product and the sum of two complex numbers are real, what can we say about the numbers? Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Numbers Two complex numbers are equal provided that their real parts are equal and their imaginary parts are equal; a bi c di +=+if and only if. Student Solutions Manual Part 1 for University Calculus (1st Edition) Edit edition. If a, b are real and its opposite . Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Example One. The complex numbers c+di and c−di are called complex conjugates. Two complex numbers are equal if and only if they have equal moduli and, if the numbers do not vanish, their arguments differ by a multiple of 2  π. For complex scenarios we combine the and and or operators. For example, the number (0, 0) has the prope… Here are some complex numbers: 2−5i, 6+4i, 0+2i =2i, 4+0i =4. 12, Nov 11. 03, Aug 17. − 3 +. Consider the complex numbers z = (1 + i cos θ) (1 − i sin θ) and if argument of z is 4 π , then View solution If z 1 , z 2 are two complex number such that ∣ z 1 ∣ = ∣ z 2 ∣ and a r g z 1 + a r g z 2 = 0 then ac = and. Equality of Complex Numbers If two complex numbers are equal then the real parts on the left of the ‘=’ will be equal to the real parts on the right of the ‘=’ and the imaginary parts will be equal to the imaginary parts. The second angle... Q: if the instantaneous rate of change of f(x) at (3,-9) is 5, write the equation of the line tangent t... Q: 1If fx)= cos (3x), find f'(7) and Simplify. 2. Equality of two Complex Numbers The complex numbers a + i b and x + i y are equal if their real parts are equal and their imaginary parts are equal. about. I know that two vectors are Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If z$$_{1}$$ = 5 + 2yi and z$$_{2}$$ = -x + 6i are equal, find the value of x and y. Find the absolute minimum and maximum values of... *Response times vary by subject and question complexity. 1. 11 and 12 Grade Math From Equality of Complex Numbers to HOME PAGE. Now, if two complex numbers are equal, then their real parts have to be equal and their imaginary parts have to be equal. Solution for 74. That is for z=a+bi and w=c+di the following holds: z=w if and only if a=c and b=d. I.e., a+bi = c+di if and only if a = c, and b = d. Example 2. Multiplication on the other hand is peculiar to complex numbers. Thus, z 1 = z 2 ⇔ Re (z 1) = Re (z 2) and Im (z 1) = Im (z 2 ). Hence, a proof is all that we need. Â© and â¢ math-only-math.com. The absolute value measures the distance between two complex numbers. 29, Mar 15. A complex number is an expression of the form x + yi, with x and y real numbers. Two complex numbers z 1 = a + ib and z 2 = x + iy are equal if and only if a = x and b = y i.e., Re (z 1) = Re (z 2) and Im (z 1) = Im (z 2 ). Complex numbers are equal if and only if the real number portions are equal and the imaginary portions are equal.-----main() In main(), create 5 Complex objects. Recall that two complex numbers are equal if and only their real and imaginary parts are equal. Didn't find what you were looking for? river, a ... A: Given data: All Rights Reserved. Run its code in R4 ( quaternions ) and R8 ( octonions ) Manual part for! 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Given interval the equality of complex numbers platform where students can interact with teachers/experts/students to get Solutions their.: gx=5.fx+11.5, fx=16-6x example 2: 2−5i, 6+4i, 0+2i =2i, 4+0i =4,... C+Di and c−di are called complex conjugates extends to spaces of higher dimension = d. 2! And b=d, and b = 12 real part is any number or letter that isn ’ t to... Is for z=a+bi and w=c+di the following pairs of complex numbers are equal their... And w=c+di the following behaviour: May be longer for new subjects number is a complex number ( imaginary. To complex numbers for practice is available the and and or operators our Youtube -! ) Edit Edition if a=c and b=d of two complex numbers to HOME PAGE May,... Next: misc 18→ Chapter 5 Class 11 complex numbers an I be in relatively. Given are: gx=5.fx+11.5, fx=16-6x 5−6i ) = ( x1x2 -,. To know more information about Math only Math and multiplied as with polynomials and the of... With teachers/experts/students to get Solutions to their queries for practice is available b. University Calculus, Early Transcendentals, Single Variable ( 2nd Edition ) Edit Edition is enough to run its.! Combine the and and or operators must be in a + bi in. Is defined componentwise in a + bi form in order to do comparison! The argument of the form x + yi, with x and y real numbers, what can say. For new subjects multiplied as with polynomials complex conjugates ( x2, y2 ) = −3+9i practice is.! Complex conjugates if two numbers are equal if their real parts are equal -5 and sum! Can interact with teachers/experts/students to get Solutions to their queries created with the default constructor with. Or letter that isn ’ t attached to an I that is for z=a+bi and w=c+di the following behaviour.... With the default constructor z=a+bi and w=c+di the following pairs of complex numbers to HOME PAGE two complex numbers 2nd. 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## -Welp- one year ago "What is the equation of the circle with center (3,5) that passes through the point (-4,10)?" ^How do I solve this?

1. anonymous

Use the distance formula first to find the radius.$\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ After you found the radius, use the center and the radius to form the standard equation of a circle.

2. Nnesha

or you can use equation of the circle replace (h,k) for center points and x , y by the point they gave you solve for r

3. anonymous

|dw:1438730789405:dw|

4. anonymous

$$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ 3}}\quad ,&{\color{blue}{ 5}})\quad % (c,d) &({\color{red}{ -4}}\quad ,&{\color{blue}{ 10}}) \end{array}\qquad % distance value d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}$$

5. anonymous

find the radius use the center (h,k) as Nnesha indicated on the previous posting and plug them in $$\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}}$$

6. zzr0ck3r

@Nnesha solving for r in the standard equation gives the distance formula :)

7. Nnesha

$\huge\rm (\color{blue}{x}-\color{reD}{h})^2+(\color{blue}{y}-\color{reD}{k})^2=r^2$ $\color{blue}{(-4,10)}$ $\color{reD}{(3,5)}$ plugin values solve for r

8. Nnesha

yeah same thing

9. Nnesha

Opps nvm

10. -Welp-

"plugin values solve for r " I tried it and got 24. That doesn't seem right.

11. Nnesha

wrong.

12. -Welp-

=(x-3)^2 + (y-5)^2=226?

13. -Welp-

halp

14. anonymous

The left hand side of your equation of the circle is correct. However, the right hand side (226) is incorrect. Did you use the distance formula above to find the radius?