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# Parity - "A Lethal Weapon"

This note is to show how $$‘PARITY’$$ can be used very effectively to solve olympaid problems .

$$'PARITY’$$ means that $$odd+odd=even,odd+even=odd,even+even=even,odd(odd)=odd,odd(even)=even,even(even)=even$$

To explain its significance a example of question of $$RMO 2016$$

QUESTION : $$f(x)=x^{3}-(k-3)x^{2}-11x+(4k-8)$$ Find all integers $$k$$ such that roots of $$f(x)$$ are integers.

Proof : let $$a,b,c$$ are integral roots of $$f(x)$$

This implies $$a+b+c=(k-3),ab+bc+ac=-11,abc=4(2-k)$$

Since $$a,b,c$$ are integers and $$abc=4(2-k)$$ therefore atleast one of them is even or $$k=2$$ but $$k=2$$ doesn’t give integral roots but $$ab+bc+ac=-11$$ therefore at a time 2 or 3 of $${a,b,c}$$ can’t be even .

Therefore only one of them is even .

Let $$a$$ is even .

Also $$a+b+c=even;k-3=even;k=odd$$

$$f(x)=(x-a)(x-b)(x-c)$$ , this implies for $$x$$ be an integer $$f(x)$$ is also an integer . Now put $$x=2$$(because at $$x=2$$ the $$k$$ term diappeas in $$f(x)$$) in $$f(x)$$ this gives $$f(2)=(2-a)(2-b)(2-c)=-10$$ . Now since $$a$$ is even therefore $$(2-a)={-2,2}$$ where $$2-a=-2$$ is the essential case as $$2-a=2$$ makes $$k=even$$.

So $$a=4$$ is a root of $$f(x)$$ . after putting values of $$x=4$$ in $$f(x)$$ we get $$k=5,b=1,c=-3$$ which matches our condition that $${k,b,c}=odd$$ .

1 year ago

Sort by:

Awesome use of parity!

- 1 year ago

Yes. Parity is of great help in many cases. It needs a little practice before we can make great use of it. I had used it in silution of quadratic equations. Thank you for the notes.

- 5 months, 1 week ago

Why didn't you consider the case 2-a=|10|?

- 1 year ago

ab+bc+ca=-11 not satisfied

- 1 year ago

Great solution!

- 1 year ago

- 1 year ago


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# Homework Help: CHEM QUESTION - answer fast thx alot

1. Jan 9, 2005

### xLaser

:::::: CHEM QUESTION - answer fast plz thx alot ::::::::

Hi, this maybe a very stupid question but I have a test coming up tomorrow and i'm not sure of this thing here:

Suppose you were given 1 M hydrochloric acid (I assume 1 M to mean a concentration of 1 mol/ L ) Calculate its [H3O]+ (square brackets mean concentratioN) and pH.

ok what did was using pH = - log [H3O]+

then u get -log (1) = 0

there pH =0 , which makes sense since HCl is a very strong acid.

and then apparently the [H3O]+ is 1 mol/L since it was given in the question.

BUT THEN IF IT LIKE THIS, THEN WHAT KIND OF STUPID QUESTION IS THIS???? LOL!

I doubt i did this right so can someone please confirm that i did it right or wrong? thanks.

2. Jan 9, 2005

### Pyrrhus

A strong acid diassociates completely in $H_{3}O^{+}$ or $H^{+}$ so yes the ph will be -log[concentration] and the amount of $H_{3}O^{+}$ or $H^{+}$ will be the HCL concentration.

format: plain


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# Quadratic Formula Calculator

The quadratic formula is used to solve the quadratic equation Ax2+Bx+C. A quadratic equation can be solved by other methods. These methods are

• Factorizing

• Completing the square

• Graphing

The quadratic formula determines whether the discriminant b2 - 4ac is less than, greater than, or equal to 0.

• If b2 - 4ac = 0, It shows that there is one real root.

• If b2 - 4ac > 0, It shows that there are two real roots.

• If b2 - 4ac < 0, It shows that there are two complex roots.

## Formula of Quadratic Formula Calculator

The formula for the quadratic equation is

$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Note:

To calculate fractions, you can use our Fraction Calculator.

### Example

Find the Solution for x2 + 4x + 0 = 0, where a = 1, b = 4 and c = 1, using the Quadratic Formula.

Solution

$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Putting the values, in the above equation we have

$x = \frac{{ - 4 \pm \sqrt {{4^2} - 4(1)(1)} }}{{2(1)}}$ $x = \frac{{ - 4 \pm \sqrt {16 - 4} }}{2}$ $x = \frac{{ - 4 \pm \sqrt {12} }}{2}$ $x = \frac{{ - 4 \pm \sqrt {4 \times 3} }}{2}$

Here, the discriminant b2-4ac > 0, so there are two real numbers. Simplifying the equation

$x = \frac{{ - 4}}{2} \pm \frac{{\sqrt {4 \times 3} }}{2}$ $x = \frac{{ - 4}}{2} \pm \frac{{2\sqrt 3 }}{2}$ $x = - 2 \pm \sqrt 3$ $x_1 = - 2 + \sqrt 3$ $x_2 = - 2 - \sqrt 3$

After more simplification it becomes

x1 = - 0.267 ,   x2 = - 3.732

### How to use the Quadratic Formula Calculator?

The steps to use the Quadratic formula calculator are as follows:

Step 1: Enter the value of an in the first required input.

Step 2: Enter the value of b in the second required input.

Step 3: Enter the value of c in the third required input.

Step 4: The calculator will automatically display an answer on the screen.

### Calculator use

Our Quadratic formula calculator is used to find the solution to quadratic equations. Quadratic equations can be solved with other methods but we preferred to solve them with quadratic equations.


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# Question #bf6ab

Apr 7, 2016

There is tradition to transform an improper fraction into a mixed form.
Example:
$\frac{28}{5} = 5 \frac{3}{5}$
But it's not compulsory. It all depends on the problem at hand.

#### Explanation:

Sometimes, it's more convenient to transform mixed fraction into improper form to perform intermediate calculations and then back to mixed fraction:
$\left(5 \frac{3}{5}\right) \cdot \left(2 \frac{2}{3}\right) = \frac{28}{5} \cdot \frac{8}{3} = \frac{28 \cdot 8}{5 \cdot 3} = \frac{224}{15} = 14 \frac{14}{15}$

It's not compulsory to convert a fraction into decimal format, but sometimes is desirable (not always possible to do it exactly, some approximation might be used):
$\frac{2}{3} \cong 0.67$

What is usually done and encouraged is reducing the fraction into smallest numbers in denominator and numerator possible:
$\frac{24}{16} = \frac{3}{2}$
This latter transformation is practically always performed, so you can consider it compulsory.


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An article is listed at 900 Rs. and two successive discounts of 8% and 8% are given on it. How much would the seller gain or lose, if he gives a single discount of 16%, instead of two discounts?
[A] Gain of 4.76 Rs.
[B] Gain of 5.76 Rs.
[C] Loss of 4.76 Rs
[D] Loss of 5.76 Rs


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# Transpose

The transpose of a matrix A is what you get when you rewrite the matrix with its rows as columns. Vectors can be transposed in the same way.

We can write the transpose of A using different symbols:

• AT
• A
• Atr
• At

## Examples

Here is the vector ${\displaystyle {\begin{bmatrix}1&2\end{bmatrix}}}$ being transposed:

• ${\displaystyle {\begin{bmatrix}1&2\end{bmatrix}}^{\mathrm {T} }\!\!\;\!=\,{\begin{bmatrix}1\\2\end{bmatrix}}.}$

Here are a few matrices being transposed:

• ${\displaystyle {\begin{bmatrix}1&2\\3&4\end{bmatrix}}^{\mathrm {T} }\!\!\;\!=\,{\begin{bmatrix}1&3\\2&4\end{bmatrix}}.}$
• ${\displaystyle {\begin{bmatrix}1&2\\3&4\\5&6\end{bmatrix}}^{\mathrm {T} }\!\!\;\!=\,{\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}}.\;}$
• ${\displaystyle {\begin{bmatrix}1&2&8\\3&4&3\\5&6&1\end{bmatrix}}^{\mathrm {T} }\!\!\;\!=\,{\begin{bmatrix}1&3&5\\2&4&6\\8&3&1\end{bmatrix}}.\;}$


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# Find a transformation from u,v space to x,y,z space that takes the triangle U = [(0,0),(1,0),(0,1)] to the triangle T=[(1,0),-2),(-1,2,0),(1,1,2)]

Question
Upper Level Math
Find a transformation from u,v space to x,y,z space that takes the triangle $$\displaystyle{U}={\left[\begin{array}{cc} {0}&{0}\\{1}&{0}\\{0}&{1}\end{array}\right]}$$ to the triangle
$$\displaystyle{T}={\left[{\left({1},{0}\right)},-{2}\right)},{\left(-{1},{2},{0}\right)},{\left({1},{1},{2}\right)}{]}$$

2021-01-03
Step 1
The triangle $$\displaystyle{U}={\left[\begin{array}{cc} {0}&{0}\\{1}&{0}\\{0}&{1}\end{array}\right]}$$ and the triangle
$$\displaystyle{T}={\left[\begin{array}{ccc} {1}&{0}&-{2}\\-{1}&{2}&{0}\\{1}&{1}&{2}\end{array}\right]}$$
Step 2
Consider $$\displaystyle{T}{\left({u},{v}\right)}={\left({a}_{{1}}{u}+{b}_{{1}}{v}+{c}_{{1}},{a}_{{2}}{u}+{b}_{{2}}{v}+{c}_{{2}},{a}_{{3}}{u}+{b}_{{3}}{v}={c}_{{3}}\right)}$$ .Obtain the transformation from u, v to x, y, z as
$$\displaystyle{T}{\left({0},{0}\right)}={\left({a}_{{1}}{\left({0}\right)}+{b}_{{1}}{\left({0}\right)}+{c}_{{1}},{a}_{{2}}{\left({0}\right)}+{b}_{{2}}{\left({0}\right)}+{c}_{{2}},{a}_{{3}}{\left({0}\right)}+{b}_{{3}}{\left({0}\right)}+{c}_{{3}}\right)}$$
$$\displaystyle{T}{\left({0},{0}\right)}={\left({c}_{{1}},{c}_{{2}},{c}_{{3}}\right)}$$
$$\displaystyle{T}{\left({0},{0}\right)}={\left({1},{0},-{2}\right)}$$
$$\displaystyle\Rightarrow{c}_{{1}}={1},{c}_{{2}}={0},{c}_{{3}}=-{2}$$
Step 3
Further evaluate as,
$$\displaystyle{T}{\left({1},{0}\right)}={\left({a}_{{1}}{\left({1}\right)}\right)}+{b}_{{1}}{\left({0}\right)}+{c}_{{1}},{a}_{{2}}{\left({1}\right)}+{b}_{{2}}{\left({0}\right)}+{c}_{{2}},{a}_{{3}}{\left({1}\right)}+{b}_{{2}}{\left({0}\right)}+{c}_{{3}}{)}$$
$$\displaystyle{T}{\left({0},{0}\right)}={\left({a}_{{1}}+{c}_{{1}},{a}_{{2}}+{c}_{{2}},{a}_{{3}}+{c}_{{3}}\right)}$$
$$\displaystyle{T}{\left({0},{0}\right)}={\left(-{1},{2},{0}\right)}$$
$$\displaystyle\Rightarrow{a}_{{1}}={c}_{{1}}=-{1},{a}_{{2}}+{c}_{{2}}={2},{a}_{{3}}+{c}_{{3}}={0}$$
$$\displaystyle\Rightarrow{a}_{{1}}=-{2},{a}_{{2}}={2},{a}_{{3}}={2}$$
Step 4
Also evaluate,
$$\displaystyle{T}{\left({0},{1}\right)}={\left({a}_{{1}}{\left({0}\right)}+{b}_{{1}}{\left({1}\right)}+{c}_{{1}},{a}_{{2}}{\left({0}\right)}+{b}_{{2}}{\left({1}\right)}={c}_{{2}},{a}_{{3}}{\left({0}\right)}+{b}_{{3}}{\left({1}\right)}={c}_{{3}}\right)}$$
$$\displaystyle{T}{\left({0},{0}\right)}={\left({b}_{{1}}+{c}_{{1}},{b}_{{2}}+{c}_{{2}},{b}_{{3}}+{c}_{{3}}\right)}$$
$$\displaystyle{T}{\left({0},{0}\right)}={\left({1},{1},{2}\right)}$$
$$\displaystyle\Rightarrow{b}_{{1}}+{c}_{{1}}={1},{b}_{{2}}+{c}_{{2}}={1},{b}_{{3}}+{c}_{{3}}={2}$$
$$\displaystyle\Rightarrow{b}_{{1}}={0},{b}_{{2}}={1},{b}_{{3}}={4}$$
Step 5
Therefore, $$\displaystyle{T}{\left({u},{v}\right)}={\left(-{2}{u}+{1},{2}{u}+{v},{2}{u}+{4}{v}-{2}\right)}$$

### Relevant Questions

Find a square number such that when twice its root is added to it or subtracted from it, one obtained other square numbers. In other words, solve a problem of the type.
$$\displaystyle{x}^{{2}}+{2}{x}={u}^{{2}}$$
$$\displaystyle{x}^{{2}}-{2}{x}={v}^{{2}}$$
Let Z be a cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eingenvalue lambda.Prove that span(Z) is a T-invariant subspace of V.
A bipolar alkaline water electrolyzer stack module comprises 160 electrolytic cells that have an effective cell area of $$\displaystyle{2}{m}^{{2}}$$. At nominal operation, the current density for a single cell of the electrolyzer stack is 0.40 $$\displaystyle\frac{{A}}{{c}}{m}^{{2}}$$. The nominal operating temperature of the water electrolyzer stack is $$\displaystyle{70}^{\circ}$$ C and pressure 1 bar. The voltage over a single electrolytic cell is 1.96 V at nominal load and 1.78 V at 50% of nominal load. The Faraday efficiency of the water electrolyzer stack is 95% at nominal current density, but at 50% of nominal load, the Faraday efficiency decreases to 80%.
Calculate the nominal stack voltage:
Calculate the nominal stack current:
Calculate the nominal power on the water electrolyzer stack:
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is $$\displaystyle\frac{{4}}{{3}}\pi{r}^{{3}}$$. The surface area is $$\displaystyle{4}\pi{r}^{{2}}$$.Set up the differential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted.
Approximate y(1,2), with h = 0.1 given that $$\displaystyle{x}{y}'={y}-{2}{x}^{{2}},{y}{\left({1}\right)}={2}$$, by RK-4 using 5-digits rounding.
Let $$\displaystyle\le{f}{t}{\left\lbrace{v}_{{{1}}},\ {v}_{{{2}}},\dot{{s}},\ {v}_{{{n}}}{r}{i}{g}{h}{t}\right\rbrace}$$ be a basis for a vector space V. Prove that if a linear transformation $$\displaystyle{T}\ :\ {V}\rightarrow\ {V}$$ satisfies $$\displaystyle{T}{\left({v}_{{{1}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n},$$ then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that $$\displaystyle{T}{\left({v}\right)}={0}$$ for every vector v in V.
(i) Let v be an arbitrary vector in V such that $$\displaystyle{v}={c}_{{{1}}}\ {v}_{{{1}}}\ +\ {c}_{{{2}}}\ {v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {v}_{{{n}}}.$$
(ii) Use the definition and properties of linear transformations to rewrite $$\displaystyle{T}\ {\left({v}\right)}$$ as a linear combination of $$\displaystyle{T}\ {\left({v}_{{{1}}}\right)}$$.
(iii) Use the fact that $$\displaystyle{T}\ {\left({v}_{{i}}\right)}={0}$$ to conclude that $$\displaystyle{T}\ {\left({v}\right)}={0}$$, making T the zero tranformation.
Guided Proof Let $${v_{1}, v_{2}, .... V_{n}}$$ be a basis for a vector space V.
Prove that if a linear transformation $$T : V \rightarrow V$$ satisfies
$$T (v_{i}) = 0\ for\ i = 1, 2,..., n,$$ then T is the zero transformation.
To prove that T is the zero transformation, you need to show that $$T(v) = 0$$ for every vector v in V.
(i) Let v be the arbitrary vector in V such that $$v = c_{1} v_{1} + c_{2} v_{2} +\cdots + c_{n} V_{n}$$
(ii) Use the definition and properties of linear transformations to rewrite T(v) as a linear combination of $$T(v_{j})$$ .
(iii) Use the fact that $$T (v_{j}) = 0$$
to conclude that $$T (v) = 0,$$ making T the zero transformation.
The function W defined by $$\displaystyle{W}{\left(\alpha\right)}={1.9}\alpha+{2.3}$$ models the weight of a kitten, in ounces, that is a weeks old for $$\displaystyle{0}\le\alpha\le{6}$$. According to this model, what is the weight, in ounces, of a kitten that is 3 weeks old?
Find $$\displaystyle{P}{r}{\left[{z}={3}\right]}$$
Consider $$\displaystyle{V}={s}{p}{a}{n}{\left\lbrace{\cos{{\left({x}\right)}}},{\sin{{\left({x}\right)}}}\right\rbrace}$$ a subspace of the vector space of continuous functions and a linear transformation $$\displaystyle{T}:{V}\rightarrow{V}$$ where $$\displaystyle{T}{\left({f}\right)}={f{{\left({0}\right)}}}\times{\cos{{\left({x}\right)}}}−{f{{\left(π{2}\right)}}}\times{\sin{{\left({x}\right)}}}.$$ Find the matrix of T with respect to the basis $$\displaystyle{\left\lbrace{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}},{\cos{{\left({x}\right)}}}−{\sin{{\left({x}\right)}}}\right\rbrace}$$ and determine if T is an isomorphism.


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# Tag Info

8

You have some misconception. Open and closed are not opposites of each other. More precisely, we can have sets that are (all as subsets of $\mathbb R$ in standard topology) neiter open nor closed: $(0,1]$ both open and closed: $\mathbb R$ open and not closed: $(0,1)$ closed and not open: $[0,1]$

4

Consider the connected open set $$U=\{\,(x,y)\in\mathbb R^2\mid x^2+y^2<4, (x-1)^2+y^2>1\,\}.$$ It's closure contains $(2,0)$ and there is no polygon leading away from it: Any line segment starting there leaves the outer or enters the inner circle. (For a simply connected example, consider the upper half of $U$ only)

3

I have never seen "closed subspace" used to mean "subspace closed wrt the vector operations". This makes sense because the property of being closed wrt the vector operations is part of the meaning of the word "subspace", so it would be redundant to specify it by an adjective, while the property of being topologically closed is not redundant, and is, ...

3

Consider the application $$f_k(M) = (\det M_{-i_1 \dots -i_k; -j_1 \dots -j_k})_{1\le i_1 <\dots < i_k\le n;1\le j_1 <\dots < j_k\le n}$$ where $M_{-i_1 \dots -i_k; -j_1 \dots -j_k}$ is the matrix $M$ without the lines $i_a$ and without the columns $j_b$. Hence $$\text{rank }M \le r\iff M\in \bigcap_{k = r+1}^n f_k^{-1}(\{0\})$$ Hence this ...

2

In the cofinite topology, '$A$ is closed' means '$A$ is finite or is $\mathbb{R}$' and '$A$ is open' means '$A$ has finite complement or is empty'. In particular... If $A \subseteq \mathbb{R}$ is infinite then the only closed set containing $A$ is $\mathbb{R}$, and hence $\operatorname{cl} A = \mathbb{R}$. In particular, if $A$ is infinite, every point of ...

2

It's neither open nor closed in standard topology of $\mathbb{R}$. You have argued it's not open. It's not closed since you can find a sequence in $A=(2,3]$ converging to $2$, which is not contained in $A$. "Since it is not open, it must be closed." is not correct in general. Closed is not opposite to open, "not open" is. But if you are considering ...

2

Let $S=\{2^{-n}:n\in\Bbb N\}$, and let $K=\{0\}\cup S$. Let $X=\{0,1\}\times K$. Points of $\{0,1\}\times S$ are isolated. For $i\in\{0,1\}$ let $p_i=\langle i,0\rangle$, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. A set $V\subseteq X$ is a nbhd of $p_i$ iff $p_i\in V$, $(\{i\}\times S)\setminus V$ is finite, and $\{n\in\Bbb N:\langle ... 2 Let$Y$be a totally disconnected space with topology$\tau_Y$, and for each$y\in Y$let$X_y$be a connected space with topology$\tau_y$. Without loss of generality assume that the spaces$X_y$are pairwise disjoint, and let$X=\bigcup_{y\in Y}X_y$. For each$y\in Y$fix a point$p_y\in X_y$, and let$\tau_y^*=\{U\in\tau_y:p_y\notin U\}$. If$y\in ...

2

We have $\lVert f\rVert_\infty\leqslant 1$ for each $f\in D$ hence $D$ is bounded. It is equicontinuous since $$|f(x)-f(y)|\leqslant |x-y|$$ for each $x,y\in X$ and each $f\in D$.

2

Hint: $\mathbb{Q}$ is dense in $\mathbb{R}$. It suffices to show $\forall x=(x_i)\in \mathbb{R}^n$, and $x\in I=(a_1,b_1)\times (a_2,b_2)\cdots \times (a_n,b_n)\cap \mathbb{Q}^n\neq \emptyset$. That is $\exists q=(q_i)\in \mathbb{Q}^n$, s.t. $q\in I$.

1

This works fine. It's probably easier in this case to use sequential compactness. Indeed, all you have to disprove is that a single point could have an infinite preimage: but then this preimage would have an accumulation point in the projective space, which would contradict the covering property.

1

a) If $\gamma\colon S^1\rightarrow \mathbb{R}P^2$ is a generator of the fundamental group, then $S^1\times S^1\rightarrow \mathbb{R}P^2,(x,y)\mapsto \gamma(x)$ will do the job. b)/c) It holds $$[X,S^1\times S^1]=[X,S^1]\times[X,S^1]=[X,K(\mathbb{Z},1)]\times[X,K(\mathbb{Z},1)]\cong H^1(X;\mathbb{Z})\times H^1(X;\mathbb{Z})$$ for any CW-complex X. Since ...

1

Usually a continuum means either (1) connected compact Hausdorff, or (2) connected compact metric. Authors should make it clear which definition they are using. In my papers, I use the first definition, then if I use "metric continuum" to describe a space satisfying (2). The theory of non-metric continua is very interesting, so that we should not limit ...

1

$\| f\|_1 \leq C \| f \|_2$ seems obvious because a pointwise value must be less than the sup. For the other direction, my hint is that $$\|f\|_{\infty} - |f(0)| = \sup_{x \in [0,1]} |f(x)| - |f(0)| \leq \sup_{x \in [0,1]} |f(x) - f(0)|$$ Now use the fundamental theorem of calculus and some obvious inequalities.

1

Hint: $\operatorname{rank}(A)\le r$ if and only if the determinant of any $k\times k$ submatrix of $A$ vanishes for $k>r$.

Only top voted, non community-wiki answers of a minimum length are eligible


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# The Stacks Project

## Tag 011K

Lemma 12.13.11. Notation and assumptions as in Lemma 12.13.10 above. Assume in addition that $\mathcal{A}$ is abelian. The morphism of complexes $\delta : C^\bullet \to A[1]^\bullet$ induces the maps $$H^i(\delta) : H^i(C^\bullet) \longrightarrow H^i(A[1]^\bullet) = H^{i + 1}(A^\bullet)$$ which occur in the long exact homology sequence associated to the short exact sequence of cochain complexes by Lemma 12.12.12.

Proof. Omitted. $\square$

The code snippet corresponding to this tag is a part of the file homology.tex and is located in lines 3137–3151 (see updates for more information).

\begin{lemma}
\label{lemma-ses-termwise-split-long-cochain}
Notation and assumptions as in
Lemma \ref{lemma-ses-termwise-split-cochain} above.
Assume in addition that $\mathcal{A}$ is abelian.
The morphism of complexes $\delta : C^\bullet \to A[1]^\bullet$
induces the maps
$$H^i(\delta) : H^i(C^\bullet) \longrightarrow H^i(A[1]^\bullet) = H^{i + 1}(A^\bullet)$$
which occur in the long exact homology sequence associated
to the short exact sequence of cochain complexes by
Lemma \ref{lemma-long-exact-sequence-cochain}.
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

Comment #2795 by Tim on September 3, 2017 a 5:13 pm UTC

Is there a reference, where the proof is carried out in detail?

Comment #2898 by Johan (site) on October 7, 2017 a 3:21 pm UTC

Why don't you try yourself?

## Add a comment on tag 011K

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).


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# eq lines and surd

• December 10th 2007, 10:05 AM
Chez_
eq lines and surd
have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

Iv now with the help of you great people being able to complete my practise paper and practised and found some new tips lol thnx
• December 10th 2007, 10:12 AM
topsquark
Quote:

Originally Posted by Chez_
have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

We need more information. Where does BC cross AB? There are any number of lines perpendicular to AB.

Quote:

Originally Posted by Chez_
Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

In LaTeX code, [ math ] 6 - 3 \sqrt{3} [ /math ] (without typing the spaces between the [ ].

$6 - 3\sqrt{3} = (6) + (-3)\sqrt{3}$

Is this what you are looking for?

-Dan
• December 10th 2007, 10:21 AM
janvdl
Quote:

Originally Posted by Chez_
have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

Iv now with the help of you great people being able to complete my practise paper and practised and found some new tips lol thnx

$BC: f(x) = - \frac{x}{4} + D$

Somewhere BC intercepts AB

So the equations of the two lines become equal at some point.

$- \frac{x}{4} + D = 4x + 5$

Simplify that and you'll find that the D-value in $f(x)$ is given by the following function:

$D = \frac{17}{4} x + 5$

Where the $x$ is the $x$-axis value where the two lines cross. From there you'll find the D-value of f(x).

When you have that, set $f(x) = 0$

Then solve for $x$ and you'll find the C-value that you're looking for.
• December 10th 2007, 10:27 AM
topsquark
Quote:

Originally Posted by janvdl
$BC: f(x) = \frac{x}{4} + D$

That would be the line $f(x) = -\frac{x}{4} + D$

-Dan
• December 10th 2007, 10:30 AM
janvdl
Quote:

Originally Posted by topsquark
That would be the line $f(x) = -\frac{x}{4} + D$

-Dan

Hehe, that had to happen sometime. Thanks Topsquark. :D


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amplify {dstat} R Documentation

Amplification of sensitivity analysis in observational studies.

Description

Uses the method in Rosenbaum and Silber (2009) to interpret a value of the sensitivity parameter gamma. Each value of gamma amplifies to a curve (lambda,delta) in a two-dimensional sensitivity analysis, the inference being the same for all points on the curve. That is, a one-dimensional sensitivity analysis in terms of gamma has a two-dimensional interpretation in terms of (lambda,delta).

Usage

amplify(gamma, lambda)

Arguments

 gamma gamma > 1 is the value of the sensitivity parameter, for instance the parameter in senmv. length(gamma)>1 will generate an error. lambda lambda is a vector of values > gamma. An error will result unless lambda[i] > gamma > 1 for every i.

Details

A single value of gamma, say gamma = 2.2 in the example, corresponds to a curve of values of (lambda, delta), including (3, 7), (4, 4.33), (5, 3.57), and (7, 3) in the example. An unobserved covariate that is associated with a lambda = 3 fold increase in the odds of treatment and a delta = 7 fold increase in the odds of a positive pair difference is equivalent to gamma = 2.2.

The curve is gamma = (lambda*delta + 1)/(lambda+delta). Amplify is given one gamma and a vector of lambdas and solves for the vector of deltas. The calculation is elementary.

This interpretation of gamma is developed in detail in Rosenbaum and Silber (2009), and it makes use of Wolfe's (1974) family of semiparametric deformations of an arbitrary symmetric distribuiton.

Strictly speaking, the amplification describes matched pairs, not matched sets. The senm function views a k-to-1 matched set with k controls matched to one treated individual as a collection of k correlated treated-minus-control matched pair differences; see Rosenbaum (2007). For matched sets, it is natural to think of the amplification as describing any one of the k matched pair differences in a k-to-1 matched set.

The curve has asymptotes that the function amplify does not compute: gamma corresponds with (lambda,delta) = (gamma, Inf) and (Inf, gamma).

A related though distict idea is developed in Gastwirth et al (1998). The two approaches agree when the outcome is binary, that is, for McNemar's test.

Value

Returns a vector of values of delta of length(lambda) with names lambda.

Note

The amplify function is also in the sensitivitymv package where a different example is used.

Author(s)

Paul R. Rosenbaum

References

Gastwirth, J. L., Krieger, A. M., Rosenbaum, P. R. (1998) Dual and simultaneous sensitivity analysis for matched pairs. Biometrika, 85, 907-920.

Rosenbaum, P. R. (2007). Sensitivity analysis for m-estimates, tests and confidence intervals in matched observational studies. Biometrics 63 456-64. (R package sensitivitymv) <doi:10.1111/j.1541-0420.2006.00717.x>

Rosenbaum, P. R. (2016) Using Scheffe projections for multiple outcomes in an observational study of smoking and periondontal disease. Annals of Applied Statistics, 10, 1447-1471. <doi:10.1214/16-AOAS942>.

Rosenbaum, P. R. and Silber, J. H. (2009) Amplification of sensitivity analysis in observational studies. Journal of the American Statistical Association, 104, 1398-1405. <doi:10.1198/jasa.2009.tm08470>

Rosenbaum, P. R. (2015). Two R packages for sensitivity analysis in observational studies. Observational Studies, v. 1. (Free on-line.)

Wolfe, D. A. (1974) A charaterization of population weighted symmetry and related results. Journal of the American Statistical Association, 69, 819-822.

Examples

amplify(4,c(5,6,7))

[Package dstat version 1.0.4 Index]


text:
# Controller

Output Controller of the company in the control of 50 randomly selected products found that 37 of them had no defect 8 has only one flaw, three had two defects, and two products had three defects. Determine the standard deviation and coefficient of variation.

Correct result:

s =  0.7746
c =  1.9365

#### Solution:

$c=s\mathrm{/}\mathrm{\mid }m\mathrm{\mid }=1.9365$

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text:
## Dragongame, sixth session

23 February, 2009 at 10:36 am (actual play, Dragongame)

Session $n-1$, everyone present.

# Play

Planning, idling, then on to the forest near the forthcoming elven tower. Vulcanus persuades (hypnotises) a small bird to act as his spy, forming a mind link to said bird so as to see through its eyes. After some complications it arrives to the desired location. There tower is somewhat taller, some wall of stones is being built in the usual way and there are three members lizardfolk also working, under watch but not maltreated. Some elf begins chirping at the birdie and Vulcanus leaves it.

Zaphádoranon takes the shape of an elf (without clothes) and some plans are as to how it would be best to approach the elves, who are now patrolling the forest. No decision is reached and the dragons decide to wait until night-time, as a squirrel told them that the griffons sleep in certain trees at night. At evening Hafnir notes that some magical pulse is being emitted from the upcoming elven tower, and that the auras of the dragons are responding to it. Not much later there are elves running at them from the forest and griffons circling above. Arrows are fired, Isla swoops back to threaten the elves, who jump to the ground at her icy breath, after which Isla quickly flies away, but not before an arrow hits her.

The others help Isla to the mountainfolk’s lair, wherein she is cared for, including a healing ritual or two. One of the rituals requires spring water, and the lizardman previously slighted who claimed to be called Blosk offers to do this, but the offer is quickly denied. Few bite wounds later the arrow is in Isla no more. The lizardfolk prepare for war, and drums are used to communicate this further.

The next day there are griffons flying above the valley where the dragons like to fly around; one of them is ridden by an elf. The lizardfolk are preparing for war. Elves are marching and souding horns. After the dragons show themselves (by peaking from a cave entrance) the gryphon-rider approaches. Zaphádoranon’s newly developed language skills show their worth, as soon enough it is clear that the elf wants to negotiate. A neutral ground is found and the elf even dismisses some of its gryphons so that the dragons come near. The elf rambles some nonsense about the world being once whole and now shredded and that the elves want to make certain a similar catastrophe won’t happen again by taking control of other peoples. All things considered, the elf seems a little off-kilter. There is posturing by both parties and eventually an agreement is reached: That the leaders of the lizardfolk and the dragons and the local elven leaders would meet and negotiate.

That done, word is taken to the three lizardfolk tribes and even to the dragon-that-is-a-hill, though anger is the only response therefrom, even after Vulcanus sang to it of dragons and their glory.


text:
Local and non local continuum limits of Ising type energies for spin systems

created by alicandr on 31 Jul 2014
modified on 23 Mar 2016

[BibTeX]

Published Paper

Inserted: 31 jul 2014
Last Updated: 23 mar 2016

Journal: SIAM J. Math. Anal.
Volume: 48
Number: 2
Pages: 895-931
Year: 2016

Abstract:

We study, through a $\Gamma$-convergence procedure, the discrete to continuum limit of Ising type energies of the form $F_\varepsilon (u)=-\sum_{i,j}c_{i,j}^\varepsilon u_i u_j,$ where $u$ is a spin variable defined on a portion of a cubic lattice $\varepsilon {\Bbb Z}^d\cap\Omega$, $\Omega$ being a regular bounded open set, and valued in $\{-1,1\}$. If the constants $c_{i,j}^\varepsilon$ are non negative and satisfy suitable coercivity and decay assumptions, we show that all possible $\Gamma$-limits of surface scalings of the functionals $F_\varepsilon$ are finite on $BV(\Omega;\{-1,1\}$ and of the form $\int_{S_u}\varphi(x,\nu_u)\, d{\cal H}^{d-1}.$ If such decay assumptions are violated, we show that we may approximate non local functionals of the form $\int_{S_u}\varphi(\nu_u)\, d{\cal H}^{d-1}+\int_\Omega\int_\Omega K(x,y)g(u(x),u(y))\, dxdy.$ We focus on the approximation of two relevant examples: fractional perimeters and Ohta-Kawasaki type energies. Eventually, we provide a general criterion for a ferromagnetic behavior of the energies $F_\varepsilon$ even when the constants $c_{i,j}^\varepsilon$ change sign. If such criterion is satisfied, the ground states of $F_\varepsilon$ are still the uniform states $1$ and $-1$ and the continuum limit of the scaled energies is an integral surface energy of the form above.


text:
# A friendly reminder that this site comes from the Science category

You can see that here. However, most of the questions here feel rather more on the technological side of artificial intelligence. Those questions are on-topic on Data Science. That site was created as a site for the technological aspect of machine learning and AI, and that is the site that is in the Technology category (see here), in spite of having "Science" in its name.

This was already emphasized by Robert Cartaino on Area 51:

Data Science is an applied site for all the programmers/statisticians/mathematicians who are trying to make this stuff work. [...]

Notice that this proposal is in the 'Science' category; not 'Technology'. [...]

It was convincing enough to give this site another try, but if this site were to simply start reiterating the implementation/tools questions that are already covered elsewhere, this site will not likely make it out of private beta.

I already tried to give a hint where we could find science questions here: Where can we find the science part of Artificial Intelligence? That is one thing we could do: ask more science questions. The other thing we can do, is closing questions. Please do close questions that are highly technological or asking for applications.

I'd like to link some questions that are, in my opinion (but I could be wrong), scientifical :

There are more questions around that are scientifical and high-quality (fortunately), I just picked a few from the first page of the highest voted list.

• Maybe, just maybe Data Science and Artificial Intelligence should fuse. That way we would not need to worry about where a more theoretical/practical question belongs. Sep 6 '16 at 19:25

I've seen this argument come up here and several times in other discussions about scope:

I don't see why both kinds of questions can't be on-topic

It's because the OPPOSITION against creating this site argued (correctly) that we already created sites to handle this subject explicitly. The argument FOR creating this site claimed that we have a missing socio-scientific angle that needed filling.

Private beta tests if that is a valid premise for creating a NEW site.

If the founding community does not live up to those expectations, it creates a strong argument for "I told you so" — that the initiative has failed.

Stick to the mission.

Don't give credence to arguments for closure.

• It's possible that questions can be ontopic on several SE sites in which case the asker can usually choose to ask wherever he likes. A bit of overlap in the scope of stackexchanges is okay, but I agree that Artificial Intelligence should clearly find it's niche and also define the limits. It just seems quite difficult to clearly define the limit between concepts and implementations (so pseudocode is okay, but any actual used language not?). Sep 6 '16 at 19:29

Data Science is an applied site

Yeah, for data science. Data science is not artificial intelligence. There is overlap around the statistical techniques for machine learning, but they just are not identical.

Please do close questions that are highly technological or asking for applications.

I'm sorry, but I think this is very misguided. Ignoring all aspects of implementation and technology on a se like this, is like a football team fielding an offense, but no defense (or vice versa). Or maybe I should say, it's like a Reese's Peanut Butter Cup without the chocolate.

The simple truth is, you can say "programming questions belong on xx.se (or so)" but there are programming questions which - in principle - would be best suited for this site. If somebody is asking about an AI specific technique or something highly specialized like rule induction using OPS5, this community is probably a better resource than datascience.se, stats.se, or possibly even so.

• I agree that I am exaggarating a bit in my post, it is just that it was very important for the site to not have too much overlap with other site to get out of private beta. Aug 25 '16 at 19:03
• @wythagoras you should add to your post why explicitly it is important to not overlap, and how this overlap will/might prevent getting out of beta Sep 10 '16 at 16:04

Also a friendly reminder that our tour page (also proposal) states:

A question and answer site for people interested in conceptual questions about life and challenges in a world where "cognitive" functions can be mimicked in purely digital environment. It's built and run by you.

So I don't see the reason why both kind of questions can be on-topic, conceptual and scientific or similar, otherwise we're limiting without any good reason.

Also please remember that it's run by us, so everybody can decide whether question should be on-topic by voting on it.

• Yes, conceptual questions are also on-topic. This are questions like ai.stackexchange.com/questions/17/… and ai.stackexchange.com/questions/10/what-is-fuzzy-logic, asking to explain the concept of Technological Singularity and Fuzzy logic respectively. But we must be careful that it is a concept related to AI and not to much form the machine learning side. Aug 12 '16 at 11:30
• Machine learning is a branch of AI, so it should be automatically included as well. Aug 12 '16 at 11:32
• Please see Will machine learning be considered as on-topic on Area 51. Aug 12 '16 at 11:33
• 'machine learning as far as implementation goes', so it's more about implementation being off-topic, not exactly 'machine learning'. Secondly the scope can be clarified/defined by community, as per comment on that post: 'Everything in the proposal is considered when evaluating whether the site would likely be viable.'. Aug 12 '16 at 11:35
• I don't see why both kind of questions can't be on-topic Because the OPPOSITION against creating this site argued (correctly) that we already created sites to handle this subject explicitly. The argument FOR creating this site claimed that we have a missing socio-scientific angle that needed filling. Private beta tests if that is a valid premise for creating a NEW site. If the founding community does not live up to those expectations, it creates a strong argument for "I told you so" — that the initiative has failed. Stick to the mission; don't give credence to arguments for closure. Aug 12 '16 at 18:50


text:
• ### Effects of Zn substitution on the electronic structure of BaFe$_2$As$_2$ revealed by angle-resolved photoemission spectroscopy(1304.5860)

May 18, 2013 cond-mat.supr-con
In Fe-based superconductors, electron doping is often realized by the substitution of transition-metal atoms for Fe. In order to investigate how the electronic structure of the parent compound is influenced by Zn substitution, which supplies nominally four extra electrons per substituted atom but is expected to induce the strongest impurity potential among the transition-metal atoms, we have performed an angle-resolved photoemission spectroscopy measurements on Ba(Fe_{1-x}Zn_x)_2As_2 (Zn-122). In Zn-122, the temperature dependence of the resistivity shows a kink around T~135 K, indicating antiferromagnetic order below the Neel temperature of T_N ~ 135 K. In fact, folded Fermi surfaces (FSs) similar to those of the parent compound have been observed below T_N. The hole and electron FS volumes are, therefore, different from those expected from the rigid-band model. The results can be understood if all the extra electrons occupy the Zn 3d state ~10 eV below the Fermi level and do not participate in the formation of the FSs.


text:
## Can e^x be thought of as a hyperbola?

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

arbiteroftruth
Posts: 476
Joined: Wed Sep 21, 2011 3:44 am UTC

### Can e^x be thought of as a hyperbola?

The more I look at it, the more an exponential curve seems to be the limit of a hyperbola in much the same way that a parabola is the limit of an ellipse. First off, the intuitive appearance of the curve resembles half of a hyperbola with a horizontal asymptote at y=0 and a vertical asymptote shoved off toward infinity, in much the same way that a parabola is an ellipse with one focus shoved to infinity.

A hyperbola and an ellipse can be described with one equation in which a real value for a given variable yields an ellipse and an imaginary value for the same variable yields a hyperbola. Similarly, eix forms a circle in the complex plane, while ex, as mentioned above, resembles a hyperbola.

Finally, let's look at the area under ex. If we think of the curve as a hyperbola, that would imply that, in a vague hand-wavy sense, there is a -e-x somewhere out beyond the vertical asymptote at infinity, forming the other branch of the hyperbola, and having exactly negative of the area under the more visible branch. Well, the integral of ex from 0 to +inf can be assigned a value in much the same way that a divergent sum can still be assigned a real value. In fact, if you convert the act of integration into a sum of a geometric series(where divergent sums can be easily and coherently defined), and find the limit of the summation as the interval approaches 0(equivalent to shrinking the width of rectangles in an approximation of integration), the integral of ex from 0 to +inf works out to -1. Since the integral from -inf to 0 is 1, this means the total area under ex is precisely 0, consistent with the notion of a hypothetical opposite branch of the hyperbola that cancels out the area.

[EDIT] Also of possible relevance: ln(x), the inverse of ex, is the integral of the right side of 1/x, which is a hyperbola. [/EDIT]

I realize the above analysis isn't exactly rigorous, but that's because my point is simply that this a conjecture/intuition I have about the subject, and I'm open to having all sorts of holes poked in it. What I'm really interested in is this: if ex can be thought of as the limit of a hyperbola as some property tends toward an extreme, how might we define what that property is, and thus define the family of shapes between hyperbolas and exponential curves? That is, if the exponential curve is the result of stretching(or whatever) a hyperbola until one of its asymptotes vanishes into infinity, what is the equation for a hyperbola that has been finitely "stretched" in the same manner?

And if this is something already well-known, somebody just tell me what it's called so I can know where to look to actually find stuff about it. Thanks.

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

### Re: Can e^x be thought of as a hyperbola?

I often find the most useful way to think about conic sections ie ellipses, parabolas and hyperbolas, as the intersection between a plane and a cone. That's what they are. If your cone is x^2+y^2=z^2, then a mostly horizontal plane, a plane with a normal vector (x, y, z) with x^2 + y^2 < z^2, will produce an ellipse, or a point if it passes through the origin. A mostly vertical plane, with a normal satisfying x^2 + y^2 > z^2, will produce a hyperbola, or a pair of lines if it passes through the origin. The final case, with a normal satisfying x^2 + y^2 = z^2 will produce either a parabola, or a line if it passes through the origin.

The relationship between an ellipse and a parabola is pretty easy to see from here: if you intersect a plane and a cone, and you get an ellipse, and then you tilt the plane in the right way, and the intersection between the plane and the ellipse becomes a parabola, and one of the foci hasn't changed through the entire tilt; if you tilt the plane the other way, and still fix a focus, you get a line. Similarly, if you start with a hyperbola, you can tilt the plane one way to get a parabola, or the other way to get a line.

On the other hand, none of what you've said is remotely as clear. If you want to say that e^x is a degenerate hyperbola, you might want to at least crack open an equation or two, and show that changing a particular parameter gets you e^x.

Also, this

arbiteroftruth wrote:a vertical asymptote shoved off toward infinity

makes baby jesus cry. Its pretty much meaningless, if not absolutely so. Please try to refrain from making these sorts of statements in the future.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

arbiteroftruth
Posts: 476
Joined: Wed Sep 21, 2011 3:44 am UTC

### Re: Can e^x be thought of as a hyperbola?

jestingrabbit wrote:On the other hand, none of what you've said is remotely as clear. If you want to say that e^x is a degenerate hyperbola, you might want to at least crack open an equation or two, and show that changing a particular parameter gets you e^x.

One may wish to consider the fact that I pretty explicitly stated that the entire point of this thread was to get help in working out precisely that issue.

jestingrabbit wrote:Also, this

arbiteroftruth wrote:a vertical asymptote shoved off toward infinity

makes baby jesus cry. Its pretty much meaningless, if not absolutely so. Please try to refrain from making these sorts of statements in the future.

See above.

But if you wish, I'll try to refrain from asking for help with something unless I already know how to fully explain it and thus don't need the help in the first place.

lightvector
Posts: 224
Joined: Tue Jun 17, 2008 11:04 pm UTC

### Re: Can e^x be thought of as a hyperbola?

No, e^x cannot be thought of as a hyperbola. Just about the only similarity is that the graph of e^x and a branch of a hyperbola both "look" like convex curves. Just like all of the dozens of other common functions that "look convex", their actual properties are wildly different, and it is not useful at all to think of one as the other.

One way in which a parabola can be considered a limit of ellipses is that for any parabola, there exists a sequence of ellipses that converges uniformly to the parabola within any bounded region. This is not true of hyperbolas and e^x. Every hyperbola whose asymptotes are horizontal and vertical is of the form y = c/(x-x_0) + y_0, for some constants c, x_0, and y_0. That is, nothing more than y = 1/x translated and scaled. It should then be obvious (and it should be easy to show rigorously too) that the graph e^x cannot be a limit of hyperbolas.

arbiteroftruth
Posts: 476
Joined: Wed Sep 21, 2011 3:44 am UTC

### Re: Can e^x be thought of as a hyperbola?

I should revise the relationship I'm trying to describe. It is indeed clear that e^x cannot be constructed strictly with true hyperbolas. It would be more accurate to say I think that both e^x and hyperbolas can be described as special cases of a more general family of curves.

lightvector, you bring up that all hyperbolas with horizontal/vertical asymptotes are linear transformations of 1/x, or x-1. One way to generalize this to a form that would include exponential(or in this particular case, logarithmic) curves would be to use A+B*x-C, rather than A+B*x-1).

The hyperbolas formed by transformations of 1/x would obviously be the special case where C=1. ln(x) is obtained by considering the family of cases in which A=-B=1/C, and taking the limit as C approaches 0. That is, (1-x-C)/C as C approaches 0.

I'm wondering what the more generalized version of that family would look like, in terms of focii/directrices etc.

EDIT: On a side note, I'd like to mention that the specific family of curves I mentioned, (1-x-c)/c, when c is positive, has a horizontal asymptote at 1/c, and this asymptote gets shoved toward infinity as c approaches 0.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Can e^x be thought of as a hyperbola?

Here is a relationship between hyperbolæ and exponentials.
wee free kings

Timefly
Posts: 56
Joined: Sun Nov 14, 2010 7:30 pm UTC

### Re: Can e^x be thought of as a hyperbola?

arbiteroftruth wrote:One way to generalize this to a form that would include exponential(or in this particular case, logarithmic) curves would be to use A+B*x-C, rather than A+B*x-1).

I don't know what you're trying to do here, but the function f(x) = k(1-x^(1/k)) is not equal to the log function for any k.

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

### Re: Can e^x be thought of as a hyperbola?

Timefly wrote:I don't know what you're trying to do here, but the function f(x) = k(1-x^(1/k)) is not equal to the log function for any k.

No, but

$\ln(x) = \lim_{h\to 0} \frac{x^h - 1}{h} \qquad \text{ and } \qquad e^x = \lim_{n\to \infty} (1+x/n)^n.$

arbiteroftruth wrote:On a side note, I'd like to mention that the specific family of curves I mentioned, (1-x-c)/c, when c is positive, has a horizontal asymptote at 1/c, and this asymptote gets shoved toward infinity as c approaches 0.

Sure, but its important to note that properties that a sequence has are not necessarily shared by the limit of the sequence. Talking about "asymptotes at infinity" isn't meaningful.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

chenille
Posts: 430
Joined: Tue Dec 20, 2011 7:25 pm UTC

### Re: Can e^x be thought of as a hyperbola?

jestingrabbit wrote:Also, this
arbiteroftruth wrote:a vertical asymptote shoved off toward infinity

makes baby jesus cry. Its pretty much meaningless, if not absolutely so. Please try to refrain from making these sorts of statements in the future.

Not at all. There are perfectly reasonable ways to extend things to infinity, you just have to be careful about what you mean. Here there is a natural idea of zeroes and poles at infinity which comes up all the time in complex analysis, where they are defined in terms of the Riemann sphere. For instance at x = 0 the function f(x)=x has a first-order zero and f(x)=1/x has a first-order pole, while at x = ∞ they are reversed. You can use rational functions to move these things around in a straightforward way.

The behavior of ex at infinity, or e1/x at the origin if you want to look at it there, is something very different from the pole of any x-c. Those give double-sided asymptotes when defined over the reals, whereas this goes to infinity from one side and zero from the other, and has periodic behavior over other complex numbers. So from this perspective, I don't think it's useful to think of it as anything like the pole of a hyperbola moved there. You can definitely look at it as a limit of poles of increasing degree, though, which is basically what's happening in the series ex = 1 + x + x2/2! + x3/3! + ...

Edit: It occurs to me you can do the same thing with the limit given above:
$e^x = 1/e^{-x} = \lim_{n\to \infty} (1-x/n)^{-n}$This sequence of functions starts with a hyperbola and has poles that move to infinity, so I think it answers the original question. Note, though, they also have zeroes of increasing degree at infinity, so the end result is really a very different thing from either. In the other limit, the zeroes get moved and the poles stay in place.


text:
These 10 questions should take 12 minutes?

Jamin2112
Let me know how long it takes you to solve these 10 questions that are supposedly representative of what 10 Quantitative Reasoning questions on the GRE would be like: http://www.majortests.com/gre/problem_solving_test03. The first time I took the GRE (2 years ago), I remember most of the QR questions being no-brainers. Maybe I need to get back in the groove, or maybe I've become stupider.

Homework Helper
9m20 here, though for some reason I messed up on one question. I hate those "choose all the numbers that cannot be..." questions, one has to be so careful with them. The hardest question was the addition one, it had a little trick.

Jamin2112
9m20 here, though for some reason I messed up on one question. I hate those "choose all the numbers that cannot be..." questions, one has to be so careful with them. The hardest question was the addition one, it had a little trick.

That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.

Gold Member
That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.

But D has to be 1 , since B+C +k =4 , where k=0 (if you don't carry) or k=1 ( if you carry), so B must be 8 or 9, so you must carry a 1 , which is D; you cannot have 50-something + a 2-digit number be 40-something; if 5+B+k =4, then B is either 8, or 9, so you carry a 1.

Homework Helper
Gold Member
I got all ten within 12 minutes, but I had to spend several minutes trying to interpret what the 5A + BC = D43 question really meant. The way it is worded, one solution should be A=0, C=3, B=9, D=1. Then all the digits being summed (i.e., 5, A, B, and C) are distinct. But 13 wasn't one of the answers. Then I realized that by "all the digits in the sum," they must have meant that all of 5, A, B, C, D, 4, 3 must be distinct, and I got 22 as an answer under that interpretation. Also, they didn't specify base 10, so the question as written is ill posed.

Staff Emeritus
That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.
Sure you can. The only way to get D=0 is if B is -1 (or -2 in the case A+C=13), and certainly neither -1 nor -2 counts as a digit.

Gold Member
I got all 10 in 13 minutes and some seconds. I made a dumb error on the digits one and added the wrong digits up at the end after figuring out it could be 57+86=143 (I added 7,8,6 and 3...instead of 7,8,6 and 1) so I got none of the answers...took me 2 minutes to figure out my error! lol

Jamin2112
Sure you can. The only way to get D=0 is if B is -1 (or -2 in the case A+C=13), and certainly neither -1 nor -2 counts as a digit.

I need to come up with a more organized structure of how I approach those problems to solve them efficiently. Right now I'm using a "greedy" approach: try to search for the first digit whose value I know; narrow it down from there.

Staff Emeritus
That problem isn't that hard. You know A+C is either 3 or 13. If A+C=3, that means B=9, D=1, and the sum is 13. If A+C=13, that means B=8, D=1, and the sum is 22. The first solution isn't on the list, but the second is.

The problem would have been trickier if the problem was 5A+BC=D40 instead of 5A+BC=D43 and if both 10 and 19 were listed as choices. With this change, A+C is 0 or 10. A+C=0 means B=9, D=1, and the sum is 10. A+C=10 means B=8, D=1 and the sum is 19. Both solutions are listed! Which one is wrong? A+C=0 isn't valid because that means A=C=0, and this violates the clause that A,B,C, and D are different digits.

That problem isn't that hard. You know A+C is either 3 or 13. If A+C=3, that means B=9, D=1, and the sum is 13. If A+C=13, that means B=8, D=1, and the sum is 22. The first solution isn't on the list, but the second is.

The problem would have been trickier if the problem was 5A+BC=D40 instead of 5A+BC=D43 and if both 10 and 19 were listed as choices. With this change, A+C is 0 or 10. A+C=0 means B=9, D=1, and the sum is 10. A+C=10 means B=8, D=1 and the sum is 19. Both solutions are listed! Which one is wrong? A+C=0 isn't valid because that means A=C=0, and this violates the clause that A,B,C, and D are different digits.

But if A+C = 3, as you point out, D = 1, B = 9, but then you're stuck with A+C = 3 where A and C must be distinct digits not equal to 1 or 9 (or 5 or 4 or 3). This is not possible - the only A and C such that A+C = 3 are A = 1 and C = 2 (or A = 2 and C = 1). D is already equal to 1 though, so this fails the criterion that all the digits in the sum must be different. Thus, A+C = 13 is the only valid answer.

Staff Emeritus
But if A+C = 3, as you point out, D = 1, B = 9, but then you're stuck with A+C = 3 where A and C must be distinct digits not equal to 1 or 9 (or 5 or 4 or 3). This is not possible - the only A and C such that A+C = 3 are A = 1 and C = 2 (or A = 2 and C = 1). D is already equal to 1 though, so this fails the criterion that all the digits in the sum must be different. Thus, A+C = 13 is the only valid answer.
Who says A and C must be between 1 and 9? Zero is a digit. There's nothing wrong with A or C being zero. If the question had both 22 and 13 as answers and had I checked 13, I would most certainly protest that selection as being marked wrong.

Medicol
it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet

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Medicol
Who says A and C must be between 1 and 9? Zero is a digit. There's nothing wrong with A or C being zero. If the question had both 22 and 13 as answers and had I checked 13, I would most certainly protest that selection as being marked wrong.
A+C can not be 3 (3 is present in the sum, D is 1 already), so A or C can not be either 0,1,2,3,4,5

johnqwertyful
it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet

##\sqrt{5}## percent of ##5\sqrt{5}## is ##\frac{\sqrt{5}}{100}5\sqrt{5}##

Staff Emeritus
it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet

The expression "a percent of b" means a·b/100. So in this case, with $a=\sqrt 5$ and $b=5\sqrt 5$, the answer is $\frac {(\sqrt 5)\cdot(5\sqrt 5)}{100} = \frac{5\cdot 5}{100}=\frac {25}{100} = \frac 1 4$.

A+C can not be 3 (3 is present in the sum, D is 1 already), so A or C can not be either 0,1,2,3,4,5
Sure they can. The question does not say that all of the digits in the problem are different from one another. It only says that "A,B,C and D represent different digits, and all the digits in the sum are different." The constraint against all of the digits in the sum being different just tells us that D is neither 3 nor 4, but we already know D has to be 1. The solution A=0, B=9, C=3, D=1 satisfies the sum problem and the constraints. A, B, C, and D are different digits. The only reason this is not the correct answer is because it is not one of the proffered choices.

This is somewhat common on multiple choice questions such as the GRE. There might well be multiple solutions to a problem, but only one will be listed.

Medicol
Thank you a lot.
Yes it is likely an issue with wording and comma; I think "all the digits in the sum are different" is also applied to 5A+BC=D43 not only to A+B+C+D

Gold Member
I only got 6 correct. Were we not supposed to do these in our head? I think I probably would have gotten most all correct if I'd just written them down.

I missed:
#2: Didn't read the question correctly. Picked the closest answer to "5 is what percent of 25" = 25. Ha!
#5: Can't divide in my head anymore. By this point I was wondering how anyone could do this quiz in 10 minutes. 668 / 4 = 100 with 268 left over. 268 / 4 = 60 with 28 left over. 28? 28 doesn't look divisible by 4. :tongue:
#8: The carry threw me off by one.
#10: Word problems... Blech! I totally guessed on that one.

Modest Learner
Got 7 correct in 9 minutes 48 seconds.

The 8th one was tricky.

Modest Learner
I got all ten within 12 minutes, but I had to spend several minutes trying to interpret what the 5A + BC = D43 question really meant. The way it is worded, one solution should be A=0, C=3, B=9, D=1. Then all the digits being summed (i.e., 5, A, B, and C) are distinct. But 13 wasn't one of the answers. Then I realized that by "all the digits in the sum," they must have meant that all of 5, A, B, C, D, 4, 3 must be distinct, and I got 22 as an answer under that interpretation. Also, they didn't specify base 10, so the question as written is ill posed.

That's exactly where I went wrong.

Mentor
10 correct, 4:45.

I missed the "A or C = 0" option, that saved some time.

#5 can be solved as "6^5 - 6^4 = 6^4 (6-1) = 6^4 * 5" then you don't have to divide or multiply anything (because the right answer is given as 6^4 as well).

The expression "a percent of b" means a·b/100. So in this case, with $a=\sqrt 5$ and $b=5\sqrt 5$, the answer is $\frac {(\sqrt 5)\cdot(5\sqrt 5)}{100} = \frac{5\cdot 5}{100}=\frac {25}{100} = \frac 1 4$.

Sure they can. The question does not say that all of the digits in the problem are different from one another. It only says that "A,B,C and D represent different digits, and all the digits in the sum are different." The constraint against all of the digits in the sum being different just tells us that D is neither 3 nor 4, but we already know D has to be 1. The solution A=0, B=9, C=3, D=1 satisfies the sum problem and the constraints. A, B, C, and D are different digits. The only reason this is not the correct answer is because it is not one of the proffered choices.

This is somewhat common on multiple choice questions such as the GRE. There might well be multiple solutions to a problem, but only one will be listed.

I interpret the statement "all of the digits in the sum are different" to mean that all of the digits in the problem are different, not simply that D is nether 4 nor 3. Stating that D is neither 4 nor 3 does not provide any information at all, since neither of those was a possibility from the start, so the only reasonable explanation for the introduction of that constraint is that it meant that all of the digits in the entirety of the problem (and because the problem is a summation, you could reasonably describe the problem as "the sum") are different. This also prevents A or C from being zero, since that would then require that the other one be 3, which violates the requirement that every digit is different.

Staff Emeritus
Arguing whether 13 is a valid solution is a bit silly because 13 was not on the list of answers.

Homework Helper
Gold Member
It only says that "A,B,C and D represent different digits, and all the digits in the sum are different."

It doesn't "only" say that, it says "In the following correctly worked addition sum, A,B,C and D represent different digits, and all the digits in the sum are different. What is the sum of A,B,C and D?"

There are two separate sentences. The first sentence refers to a "correctly worked addition sum" which is clearly the vertical sum including the digits 3, 4 and 5, and then states that all the digits in the sum (i.e. the sum which has already been referred to, not some other sum which has not yet been mentioned) are different. The second sentence then introduces a second sum - perhaps this is designed to be confusing, but by no correct interpretation can this new sum become the object of the preceding sentence.

Just under 7 minutes.

Jamin2112
Glad I'm not the only one who finds these easy to make mistakes on

Staff Emeritus
I didn't make a mistake. I got that one correct. (I got them all correct.) I saw that there were two solutions to that problem, solved for both, and saw that only one of my two solutions was on the list. Problem solved.

Staff Emeritus
Gold Member
It took me about 14 minutes 30 seconds. I'm not fast with these things, but at least I got them all right. I think I got overconfident after doing the first four in less than 4 minutes, and slowed down a bit. I don't think I would make that mistake twice.

I had completed 8 questions when the clock passed 12 minutes.

Are you allowed to use a pen and paper? I did for the equation in question 10. I can do these things without a pen, but then I need more time.

10 correct, 4:45.
Whoa. That's crazy fast.

#5: Can't divide in my head anymore. By this point I was wondering how anyone could do this quiz in 10 minutes. 668 / 4 = 100 with 268 left over. 268 / 4 = 60 with 28 left over. 28? 28 doesn't look divisible by 4. :tongue:
I lost some time on this one too, but after a while I realized that all you have to do to check if something is divisible by 4 is to divide by 2 and see if the result is even or odd.

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Gold Member
...
after a while I realized that all you have to do to check if something is divisible by 4 is to divide by 2 and see if the result is even or odd.

I figured that out after the quiz was over. :grumpy:

All those simple math tricks we knew when we were 10 years old...

All gone, like the rest of my marbles...

Medicol
...
Are you allowed to use a pen and paper?...
Try out several more free tests there, you'll not be able to solve some problems without them.

Some of the companies I took the IQ tests as what they called them used these similar problems which are a bit selectively harder. It took me a long time to solve 10. They evaluated me as an idiot of course.

Gold Member
10 correct, 4:45.

I missed the "A or C = 0" option, that saved some time.

#5 can be solved as "6^5 - 6^4 = 6^4 (6-1) = 6^4 * 5" then you don't have to divide or multiply anything (because the right answer is given as 6^4 as well).

dang, you are supa fast

Mentor
To test if a number is divisible by 4, you just need the last two digits - 100 is divisible by 4, so is every multiple of it. Therefore, 75654 is divisible by 4 if and only if 54 is.

20/4=5, so we can do the same trick again: At the next to last integer we just have to check if it is even or odd.
If it is even, we can remove it as multiple of 20, and the number is a multiple of 4 if the last digit is 0, 4 or 8.
If it is odd, we can replace it by 1, and the number is a multiple of 4 if the last digit is 2, or 6.

Examples:
647743534 -> 3 is odd, check if 14 is divisible by 4 -> no -> done.
762 -> 6 is even, check if 2 is divisible by 2 -> no -> done

There are tons of those tricks and they really make those tests and also real scientific problems easier.


text:
### Some Exceptional Phenomena in Multifractal Formalism: Part II

De-Jun Feng, Ka-Sing Lau, and Xiang-Yang Wang
Source: Asian J. Math. Volume 9, Number 4 (2005), 473-488.

#### Abstract

In Part I we showed that the $L^q$-spectrum of the 3-fold convolution of the Cantor measure has a non-differentiable point at a $q_0 < 0$ [LW], therefore the standard multifractal formalism does not hold. In this Part II, we prove a modified multifractal formalism for the measure.

First Page:

#### Related Works:

Full-text: Open access

Permanent link to this document: http://projecteuclid.org/euclid.ajm/1146673649


text:
9

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text:
# Planar mapping and normals

## Recommended Posts

Hi everybody :) Excuse my English... I'm implementing LightMaps in my 3D engine using the planar mapping algorithm. I will explain it for whoever is interested. Steps (for each polygon): 1- Obtain the polygon normal and choose the greater element to generate the texture coordinates using the remaining elements: For a polygon (suppose a triangle) [A B C], I get the normal [N] and compare:
if (Abs(N.x) > Abs(N.y)) (Abs(N.x) > Abs(N.z)) then
Make UV coordinates from the YZ elements of the polygon.
Set variable "Flag = 1"
if (Abs(N.y) > Abs(N.yx) (Abs(N.y) > Abs(N.z)) then
Make UV coordinates from the XZ elements of the polygon.
Set variable "Flag = 2"
else
Make UV coordinates from the XY elements of the polygon.
Set variable "Flag = 3"

2- The next step in the algorithm is to obtain the maximum and minimum of these coordinates, in order to clamp them to the range [0..1]: For each coordinate:
   Save the minimum U into U_Min
Save the minimum V into V_Min
PolygonVertex.U = (PolygonVertex.U - U_Min) / (U_Max - U_Min)
PolygonVertex.V = (PolygonVertex.V - V_Min) / (V_Max - V_Min)

- Then, the lightmap position vectors are transformed into world space by using the equation of the plane to obtain the remaining unknown element:
if Flag = 1 then
// Vertex 1
V1.x = - (N.y * U_Min + N.z * V_Min & D) / N.x; // Plane equation
V1.y = U_Min
V1.z = V_Min
// Vertex 2
V2.x = - (N.y * U_Max + N.z * V_Min & D) / N.x; // Plane equation
V2.y = U_Max
V2.z = V_Min
// Vertex 3
V3.x = - (N.y * U_Min + N.z * V_Max & D) / N.x; // Plane equation
V3.y = U_Min
V3.z = V_Max
if Flag = 2 then
// Vertex 1
V1.x = U_Min
V1.y = - (N.x * U_Min + N.z * V_Min & D) / N.y;
V1.z = V_Min
// Vertex 2
V2.x = U_Max
V2.y = - (N.x * U_Max + N.z * V_Min & D) / N.y;
V2.z = V_Min
// Vertex 3
...
if Flag = 3 then
...

Now, to achieve the standard per pixel lighting, I need to do the product N.L for every pixel (or lumel, in this case). I can easily get the L factor from the previous algorithm, just interpolating between through the vertices:
Edge1 = V2 - V1
Edge2 = V3 - V1
for each pixel U
for each pixel V
Pixel = V1 + Edge1 * U + Edge2 * V

Finally (!), here comes my problem: I don't know how to get the same interpolation factor for the N (normal) factor just like I did for the L factor. That's because I have only the normal of the plane, and what I need is the normal for each vertex V1, V2, and V3. The Planar Mapping algorithm projects the vectors into the UV plane and changes its orientation depending on the normal N, so I can't see how can I get my original normals. Is there another equation that allows me to get the original vectors from the equation of the plane, the normal of the plane and/or the transformed vectors (V1, V2, V3)? Better still, is there a matrix that represents the transformation done, so I can get the original vectors from the new ones (V1, V2 and V3)? I will appreciate any help :) Thanks guys!

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Quote:
 I don't know how to get the same interpolation factor for the N (normal) factor just like I did for the L factor. That's because I have only the normal of the plane, and what I need is the normal for each vertex V1, V2, and V3.

I don't quite understand what you're asking in this sentence. You want vertex normals?
Why can't you just sum up the plane normals around each vertex?

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Oops, I know sometimes (almost anytime :) ) I'm not clear...
I will be more graphical now:

After getting the texture coordinates with planar mapping, I have only one normal for the entire triangle, the normal of the plane the polygon lies on. When I compute the N.L product my result is:

Instead, I need the normals of the original vertices of the triangle, in order to make an interpolation between them for Phong Shading:

(This last capture uses dynamic lighting instead of lightmaps).

But after the world space transform I don't know how to get them.
If I make the interpolation from the normal 1 to 3 (in the order of the vertex buffer) my results are wrong:

I hope you can help me... Thanks!

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From what I remember:

For a triangle, the Phong (?) interpolated normal is:

N = N0*w0 + N1*w1 + N2*w2

where barycentric weights:
w0 = 1 - u - v
w1 = u
w2 = v

N0,N1,N2 are the vertex normals of your triangle.

[Edited by - JakeM on May 10, 2008 7:33:24 PM]

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Thanks for your answer, but I already knew that formula :(. In fact, that's the one I'm applying in the third screenshot.

I will try to show my problem in another way.

By creating the lightmap for each triangle, I get the minimum and maximum U and V, and the vertices of the plane for the lightmap are created from these values, which generates a projection that may be, for example:

Case 1:

Case 2:

Where N1, N2 and N3 are the vertex normals, and Np is the plane normal.

This makes me unable to interpolate the vertex normals, as the projection on the plane for the lightmap may vary depending on the position values for each vertex.

On the other hand, I know that it is possible to apply Phong Shading on a lightmap.
Does anyone could do this?
Is it necessary to implement another different technique than Planar Mapping?

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For a vertex normal, I average the normals of all the triangles that share the vertex. It comes out looking nice and smooth.

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Well, I kinda do that now, by using the plane normal (screenshot 1)... But it doesn't come out looking nice and smooth in my case :(

How do you achieve that with just an average? I assume that you're talking about cubic meshes or spheres with a high poly count...?

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Quote:
 Original post by jpventosoWell, I kinda do that now, by using the plane normal (screenshot 1)... But it doesn't come out looking nice and smooth in my case :(How do you achieve that with just an average? I assume that you're talking about cubic meshes or spheres with a high poly count...?

Well, on the first one it looks like the entire polygon is lighted evenly based on that polygon's normal.

What I was referring to was how to find the vertex normals, although you still have to interpolate between them. The entire process would work something like this:

//pseudocodefor every polygon p {  for every vertex i in polygon p {    for every polygon q which uses vertex i {      vertex_normal[i] += polygon_normal[q] / number_of_polygons_that_use_vertex_i    }  }for every point x in polygon p {  for every vertex i in polygon p {    point_normal[x] += weight[i]*vertex_normal[i]  }}[/code]Where weight[i] is inverse proportional to the distance between point[x] and vertex[i], and the sum of all weight[i] = 1.

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Wow, at a first glimpse I didn't understand a thing but that algorithm seems to be just the solution I'm looking for! :)

I'm going to print it and think about it carefully; Thanks for the help!

I'll post any news tomorrow. Thanks again!

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Quote:
 This makes me unable to interpolate the vertex normals, as the projection on the plane for the lightmap may vary depending on the position values for each vertex.

How do you plan to use this interpolated normal? Are you trying to project it to 2D? If so, why?

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Quote:
Original post by JakeM
Quote:
 This makes me unable to interpolate the vertex normals, as the projection on the plane for the lightmap may vary depending on the position values for each vertex.

How do you plan to use this interpolated normal? Are you trying to project it to 2D? If so, why?

He wants to know how bright a particular spot should be, using a particular lighting technique, and you need to know what the normal for that particular spot is. The more the normal points towards the light source, the brighter it should be. Even though the mesh isn't smooth, this method can give the illusion of smoothness.

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Quote:
 Original post by erissianHe wants to know how bright a particular spot should be, using a particular lighting technique, and you need to know what the normal for that particular spot is. The more the normal points towards the light source, the brighter it should be. Even though the mesh isn't smooth, this method can give the illusion of smoothness.

I know, you can do all of that without projecting the plane to 2D. I still don't see the problem he's talking about.

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Quote:
Original post by JakeM
Quote:
 Original post by erissianHe wants to know how bright a particular spot should be, using a particular lighting technique, and you need to know what the normal for that particular spot is. The more the normal points towards the light source, the brighter it should be. Even though the mesh isn't smooth, this method can give the illusion of smoothness.

I know, you can do all of that without projecting the plane to 2D. I still don't see the problem he's talking about.

He's doing this, ultimately, for a light map, so he needs to be able to take this information and use it in texture-space.

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Your algorithm greatly improved things (I'm now a step closer to a real phong lighting :) ), but as it can be noticed in this capture, a lighting difference between each triangle still remains:

I think this is due to the location of each triangle within the Lightmap (cases of the diagrams on my previous post), and I couldn't find a solution to this problem yet...

Erissian, do you use planar mapping to generate your Lightmaps (if you're using any)?

Thanks again!

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Quote:
 Original post by jpventosoYour algorithm greatly improved things (I'm now a step closer to a real phong lighting :) ), but as it can be noticed in this capture, a lighting difference between each triangle still remains:I think this is due to the location of each triangle within the Lightmap (cases of the diagrams on my previous post), and I couldn't find a solution to this problem yet...

This last bit is now a problem with the interpolation. The triangles appear discretely because they are still treated as a plane defined by the vertexes, although the light is generated smoothly across that plane. For the final step, it will be a little harder because you now have to interpolate not across the shape of the model, but across the shape that the model represents. This will require spherical linear interpolation (slerp) between the three vertexes.

Quote:
 Erissian, do you use planar mapping to generate your Lightmaps (if you're using any)?Thanks again!

Sometimes, yes. It depends on the project. For instance, if I were rendering planets I would use two lightmaps that represent the planet's surface and the light from the star(s), the latter of which rotates with time, and multiply their values. Shadows from most objects are a non-issue, and until you are very close to the surface, lighting just isn't that dynamic.

Also, building interiors with fixed lighting. There's no sense in calculating the lighting in a scene every time you render several dozen static objects.

Also, no problem!

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Quote:
 Original post by erissianThis last bit is now a problem with the interpolation. The triangles appear discretely because they are still treated as a plane defined by the vertexes, although the light is generated smoothly across that plane. For the final step, it will be a little harder because you now have to interpolate not across the shape of the model, but across the shape that the model represents. This will require spherical linear interpolation (slerp) between the three vertexes.

OK, but, when I use per pixel lighting with shaders, I thought that I was only doing a linear interpolation. I'm asking this because, when I use shaders, this problem does not occur.

Or perhaps it is happening due to the fact that lightmaps are calculated at a lower resolution and thus loses definition on the edges?

Quote:
 Original post by erissianAlso, building interiors with fixed lighting. There's no sense in calculating the lighting in a scene every time you render several dozen static objects.

I wanted to implement lightmaps for the same reason, to increase performance by reducing the lighting calculations when the environment is static, and also to generate soft shadows (I'm having problems with this last item but that will be a new topic :) ).

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Quote:
 Original post by erissianThis will require spherical linear interpolation (slerp) between the three vertexes.

What's the equation for slerping 3 vertex normals?

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Quote:
 Original post by JakeMWhat's the equation for slerping 3 vertex normals?

I'm interested in that as well...

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The trick isn't slerping between three points (you slerp between two and then slerp between that answer and the third point), it's doing it so that the points are evenly distributed. This is how you iteratively slerp through a triangle using two coordinates:

Let's call our verts A,B and C. We'll use two coordinates, t and u. n will determine the number of individual points you want to sample, per ½(n-2)(n-1)

//pseudocode, returns Rn = some integerf = 1/nfor t=0:n  if t==0    R = A  P = slerp(A,B,t*f)  Q = slerp(A,C,t*f)  g = 1/t  for u=0:t    R = slerp(P,Q,u*g)

It may help to look at a ternary plot to get an idea of how it walks through the triangle. Also, it would be foolish to implement this literally - you can optimize it quite a bit by working through the math and reducing.

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Quote:
 Original post by erissianLet's call our verts A,B and C.

Are ABC vertex normals or quaternions? If they're quats, how did they get from vector to quat?

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Let me add another question to it: In order to do the slerp, Do I need to know the angle between the edges right? How can I get it?

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Well, if nobody is going to step up, I'll try to answer it.
Yes, you can slerp vectors!

Apparently, you can convert 3D vectors into 4D quaternions, by simply setting w=0:
quat = VEC4( normal.x, normal.y, normal.z, 0 );

Then you can use this quat in any slerp function intended for quaternions.

To convert back, simply drop the w component:
vect = VEC3( quat.x, quat.y, quat.z );

Dropping the w component shouldn't be a problem, because if you're interpolating 3 normals,
then your result should still point relatively in the same direction. I hope.

So, to interpolate three vectors, v1,v2,v3:

result1 = slerp( v1, v2, t1 );
result2 = slerp( result2, v3, t2 );

I would suppose t1 and t2 relate to the barycentric weights chosen for each vertex normal.

But as erissian pointed out, this isn't the best way to do it. The final result from slerping 3 rotations
is dependent on the order! So, slerping r1,r2, and then r3, produces a different result than r3,r2 and r1.

The way to get slerping that is independent of rotation order, is to use an iterative slerp algorithm.

[Edited by - JakeM on May 19, 2008 2:19:19 PM]

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Oh! Ok, I imagined that the order would be a problem ...

That's why I think I'm going to ignore the slerping and use LightMaps only in cubic figures by now...

Besides it, another problem would be to determine whether each triangle will form a cubic or spherical joint, just to avoid using the slerp in cases that is not needed...

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# Motivic cohomology with finite coefficients for singular varieties

Let $X$ be a smooth variety over a field $K$ whose characteristic does not divide a positive integer $m$. Then the motivic cohomology of $X$ with coefficients in $\mathbb Z/m(j)$ can be computed in terms of the etale and Zariski topologies by the familiar rule $$H_M^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}),$$ where $\pi:Et\to Zar$ is the natural map between the (big or small) etale and Zariski sites, $\mu_m$ is the etale sheaf of $m$-roots of unity, and $\tau$ denotes the canonical truncation of complexes of Zariski sheaves. In fact, this result is a combination of two: the Beilinson-Lichtenbaum etale descent $$\mathbb Z(j) = \tau_{\le j}R\pi_*\pi^*\mathbb Z(j) = \tau_{\le j+1}R\pi_*\pi^*\mathbb Z(j)$$ and (a version of) the Suslin rigidity theorem $$\pi^*\mathbb Z/m(j) = \mu_m^{\otimes j}.$$ These results also remain true if one replaces the Zariski topology with the Nisnevich one.

For singular varieties, the above formula for motivic cohomology no longer holds. It suffices to consider the example when $X$ is the affine line with two points glued together. Then one has $H^1_M(X,\mathbb Z/m(0))=\mathbb Z/m$, though the formula would give $0$ as the answer for this motivic cohomology group. It appears that in this particular case ($X$ as above and $j=0$) the rigidity assertion still holds, so it must be the etale descent that breaks down.

Is there any way to make any or all of the above assertions true for singular varieties by changing the topologies one works with? E.g., replace the etale topology with the h topology and the Zariski/Nisnevich with cdh? If there is more than one way to do this, I would be greatly interested to hear about each and every of them.

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# distance between point and empty set

While playing with my little sister earlier we where inventing distances form earth to sun/stars/planets and who had the bigger distance wins. Now at some point she said "from earth to jesus" and since i'm not sure about his existence i was wondering , what if jesus does not exist? Then this distance is the distance form a point on earth to the empty set. Now i was wondering if such a distance is geometrically defined. Is there for example a value or definition for the distance on the real line between a number and the empty set? In euclidian space? In general? I have no knowledge of topology..maybe i could find an answer there?

I hope this is not a silly question , but i want to know who wins (:

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## 2 Answers

In genreal, only the distance between two points (in euclidean space or more generally in a metric space) is defined.

This notion can be extended to define the distance between two subsets $A.B$ of a metric space as $$d(A,B):=\inf\{\,d(a,b)\mid a\in A, b\in B\,\}.$$ But take care! It may happen that $d(A,B)=0$ even when $A\ne B$ (for example when $A\cap B$ is not empty), a property that would not be desired for a metric (read: it is not fully justified to call this $d$ a distance). Anyway, this extended notion of distance between two sets would tak ethe infimum of the empty set if $A$ or $B$ are empty. By careful definition, $\inf\emptyset$ should be the greatest real number that is a lower bound of $\emptyset$. However, every real number is a lower bound for $\emptyset$ and there is no greatest real number. Hence in a strict sense $\inf\emptyset$ is not defined (you win!). However, usually one extends the definition of infimum to allow two special values, namle $-\infty$ for a set that has no lower bound and $+\infty$ for the empty set. In this set, $d(A,\emptyset)=\inf\emptyset=+\infty$ (and you lose!)

So in a way there's atie between you two players (and isn't that WJWD?) On the other hand, as a Christian believer, your sister could not claim the win as God (and hence Jesus) is omnipresent, hence at distance $0$ from anywhere. :)

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Augh, you beat me to it :D –  Eric Stucky Aug 10 '13 at 22:22
@EricStucky Plus I have a twist to explain all possible outcomes of the game :) –  Hagen von Eitzen Aug 10 '13 at 22:22
nice answer!I have a little doubt , you say " every real number is a lower bound for $\emptyset$" and i don't know how you got there. Also couldn't we argue that no number is a lower bound for $\emptyset$? Then inf $\emptyset$=$\emptyset$ and...well here things get strange. –  sigmatau Aug 10 '13 at 22:39
@AmireBendjeddou: A lower bound $b$ for a set $S$ is a real number such that $b\leq s$ for all $s\in S$. Since there is nothing in $\varnothing$, the statement is true (vacuously) for any real number $b$. –  Eric Stucky Aug 11 '13 at 1:55

The typical definition of a distance function from a point $p$ to a set $S$ looks roughly like this:

$$d(p, S) = \inf\{d(p,s): s\in S\}$$

If $S$ is the empty set, then $d(p, S)=\inf\varnothing$. Under the usual rules of how to define a distance function, this is undefined.

However, there is a strong case that it "should be" $\infty$. It should be clear (after some thinking) that if $A\subseteq B$ then $d(p, B)\leq d(p, A)$ because $B$ simply has more points to choose from; that could decrease the minimum but it cannot possibly increase the minimum. Since $\varnothing$ is a subset of every set, it must be that $d(p,A)$ is greater than [or equal to] $d(p,B)$ for any set in the space. If the space is unbounded, then $d(p,B)$ can be arbitrarily large; so we need a value greater than any number— this is the definition of $\infty$.

[If the space is bounded, then we could do with smaller values, but we would never need a larger value.]

In short, you lose :)

Technical details: this all occurs in metric spaces which are a type of topological space where distance is defined in an intuitive way. In particular, all manifolds are metric spaces, so basically any physical model of the spacial dimensions of the universe. There are more general notions of distance that I am unfamiliar with, perhaps this intuition breaks down there.

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# How to color a sentence?

[Question edited on 5/28/16] I have read this link. But I still don't understand how to color the text a sentence. Can one actually do that?

• Could anyone give an example for coloring the text? a sentence?
• Where should I go when I have the similar formatting question?
• Use \color{blue}{text} to get $\color{blue}{text}$ – tatan May 29 '16 at 2:08

You cannot color text. That can be deduced from the post you linked to.

• $\rm \color{red}{BUT}\color{green}{\ YOU}\color{blue}{\ CAN!}$ – Bill Dubuque Jul 9 '11 at 1:25
• @Bill: I made the same point a few hours ago. But in order not to encourage it I removed my comment. (I agree that Mariano's answer invites such a comment but I'd be surprised if he hadn't known about that trick) – t.b. Jul 9 '11 at 2:09
• I even object to \rm, in fact :) A mismatch set of fonts is never nice on the eye... let alone if they are randomly colored! – Mariano Suárez-Álvarez Jul 9 '11 at 2:11
• @Bill: Mariano explicitly stated "text". You did not color "text". :) – Willie Wong Jul 9 '11 at 10:16
• @Willie But it is TeXt! For the rare occasions one needs such, this kludge usually suffices. – Bill Dubuque Jul 9 '11 at 13:24
• @Mar RGB is not random. Nor is anyone recommending fruit salad text or "mismatched" fonts. – Bill Dubuque Jul 9 '11 at 13:31
• @Bill: Did you do that in the  environment? – Jack Jul 21 '11 at 0:28
• @Jack Yes, I used the LaTeX \color command. – Bill Dubuque Jul 21 '11 at 1:27

$\color{blue}{\textrm{One can color text}}$ by writing

$\color{blue}{\textrm{One can color text}}$

• This is a poor way to do it. If you now already felt the need to bump this old thread for adding this answer you could at least include a solution that does not use a font different from the one used for actual text. – quid May 29 '16 at 1:26
• Actually, I don't understand the discussion under the answer of Mariano. And neither your comment. What do you mean by "a font different from the one used for actual text"? – Jack May 29 '16 at 1:29
• It is not the same font. What is unclear about this. Look closely and you will see it is not the same. And on the "text": you included a math environment, this is not text anymore. – quid May 29 '16 at 1:32
• Ah, so the point is that the answer depends on how I phrase my question. If one wants to color some words, then he/she can use the math environment. But no way to color text. I have never understood this until today. Thanks a lot for your comment! – Jack May 29 '16 at 1:41
• Yes. Coloring text is not supported. However, using MathJax it is possible to introduce color "through the backdoor." There are however several disadvantages to using it in this way, it is therefore discouraged to do what you recommend in your answer. A slightly better way is explained in meta.math.stackexchange.com/questions/20503 But one still should not used it to insert color all over. – quid May 29 '16 at 1:45