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• Apr 17th 2013, 05:44 PM
Fratricide
1) i: Find the value of c such that y = x + c is a tangent to the parabola y = x2 - x - 12.
ii: Find the values of m for which the straight line y = mx + 6 is tangent to the parabola y = -2x2 - 6x + 2.

2) Determine the equation of the following parabola:

Attachment 27999

• Apr 17th 2013, 05:47 PM
Prove It
What have you tried so far?
• Apr 17th 2013, 08:56 PM
Fratricide
1. x2 - x - 12 = x +c
x2 - 2x - 12 = c

What value would I substitute for x? I think it has something to do with the equations being tangental, but I can't remember exactly.

2. I know that I have to substitute coordinates into the equation ax2 + bx + c and solve simultaneously, but I don't know how to solve simultaneous equations with three variables.
• Apr 17th 2013, 10:22 PM
ibdutt
Now we have the curve y = x^2 - x -12. Slope of tangent to the curve is given by dy/dx. for this curve dy/dx = 2x -1 Let the point of contact be (h,k)
Thus slope of tangent = 2h-1. The slope of the given line y = x + c is 1. Hence 2h - 1 = 1 that gives h = 1. Form the equation of the curve , by pugging in value of h=1 for x we get ythe value of k since (h,k) lies on the curve.
That gives k = 12. Hence the point of contact is ( 1, - 12 ) and slope of tangent is = 1. Thus its equation is y + 12 = 1 ( x - 1 ) that is y = x - 13. Comparing it with the given equation we get c = -13.
• Apr 18th 2013, 12:52 AM
Fratricide
Quote:

Originally Posted by ibdutt
Slope of tangent to the curve is given by dy/dx. for this curve dy/dx = 2x -1

I don't think I quite understand? I got the rest of your post, but this doesn't fully make sense.
• Apr 18th 2013, 01:06 AM
MarkFL
Quote:

Originally Posted by Fratricide
I don't think I quite understand? I got the rest of your post, but this doesn't fully make sense.

He's using calculus, but we may use a pre-calculus technique instead, since you posted in the Algebra forum. You correctly found:

$x^2-2x-12=c$

If we arrange this in standard quadratic form, we get:

$x^2-2x-(12+c)=0$

Now, if the line and parabola are tangent, this means they share only one point in common, i.e., the above quadratic has only one root. What condition on a quadratic must hold in order for there to be only one root?
• Apr 18th 2013, 02:06 AM
Fratricide
The discriminant must be equal to zero?
• Apr 18th 2013, 02:08 AM
MarkFL
Yes, good! Now what is the discriminant, and what do you find when you equate it to zero?
• Apr 18th 2013, 02:13 AM
Fratricide
Is - (12 + c) the c value for the discriminant?
• Apr 18th 2013, 02:20 AM
MarkFL
Yes, it is the constant term. :D
• Apr 18th 2013, 02:45 AM
Fratricide
Alright, thanks. I'll have a go solving it tomorrow -- it's getting late here.

This may sound ridiculously stupid, but why does the equation become x^2 - 2x - (12 + c) = 0 as opposed to something like, say, x^2 - 2x + c -12 = 0? Specifically, why does the c get tied to the 12 when it's moved?
• Apr 18th 2013, 02:51 AM
MarkFL
Since c is a constant, it should be grouped with the constant 12, as the other two terms have a power of x as factors. Since it began with a positive sign on the right, when we move it to the left, we are essentially subtracting c from both sides.
• Apr 18th 2013, 07:27 PM
jpritch422
This is the solution to part ii of problem 1:

Since the tangent line and the given curve have a point in common we have:

$mx + 6 = -2x^2 - 6x + 2$

$2x^2 + (6 + m)x + 4 = 0$

slope of $m = \displaystyle\frac{dy}{dx}$ where $\displaystyle\frac{dy}{dx} = \displaystyle\frac{d}{dx} (-2x^2 - 6x + 2) = -4x - 6 = 0$

so $6 + m = 6 - 4x - 6 = -4x$

$2x^2 -4x^2 + 4 = 0$

$-2x^2 + 4 = 0$; $x^2 = 2$; $x = \pm \sqrt{2}$

if $x = \sqrt{2}$; $m = -4\sqrt{2} - 6 \approx -11.657$

if $x = -\sqrt{2}$; $m = 4\sqrt{2} - 6 \approx -.343$
• Apr 18th 2013, 07:46 PM
MarkFL


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# (6) Below is the hea experiment you performed in the ting curve for Experiment Temperature measurements-exactly as in the laboratory.

###### Question:

(6) Below is the hea experiment you performed in the ting curve for Experiment Temperature measurements-exactly as in the laboratory. You use this curve to calculate the time is curve correct?? Please write more explanations than" yes" or "no (S) ThermoCouptess Heating 2.00L+0 สุด

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# 9 – Exponential and Logarithmic Functions

• What happens when you or a loved one gets sick?
• What might have caused the illness, and what might you have to do to get well again?
• Imagine getting together with all of your friends.
• And all of their friends.
• And all of their friends’ friends.
• And all of their friends’ friends’ friends.
• Like putting rockets on rockets on rockets on rockets, exponential growth is all about going faster, faster, faster!
• From infectious outbreaks and nuclear reactions to medical treatments and computer algorithms, exponents and logarithms are essential for living in the 21st century.
• Logarithms are the inverses of exponential functions, and just as important. Logarithms help measure tremors deep within the earth.
• Your wireless devices uses logarithms to find the strongest wi-fi signal.


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Time complexity of several methods of reversing a domain name string

Note: I'm aware of this existing post which is similar. However, it does not address all my questions sufficiently.

I have the following three functions, all of which do the same thing: reverse a domain name string.

std::string reverse_domain(const std::string& s) {
std::stringstream ss(s);
std::string segment;
std::vector<std::string> breakList;

while (std::getline(ss, segment, '.')) {
breakList.push_back(segment);
}

std::string result;

auto size = breakList.size();

for (int i = size - 1; i >= 0; --i) {
result += breakList[i];

if (i != size - 1) {
result += ".";
}
}

return result;
}

std::string reverse_domain2(const std::string& s) {
std::string result;
auto index = 0;

for (const auto& c : s) {
if (c == '.') {
index = 0;
result.insert(result.begin() + index, '.');
continue;
}

result.insert(result.begin() + index, c);
index++;
}

return result;
}

std::string reverse_domain3(std::string& copy) {
std::reverse(copy.begin(), copy.end());
auto start = 0;

auto size = copy.size();

for (auto i = 0; i <= size; i++) {
if (copy[i] != '.' && i != size) continue;

std::reverse(copy.begin() + start, copy.begin() + i);
start = i + 1;
}

return copy;
}

My question is, what is the time complexity of all of these?

My educated guesses are:

• For reverse_domain: $$\O(n) + O(n^2) + ?\$$. The reasoning behind this is that I assume the getline loop is $$\O(n)\$$, and I assume appending to a string is $$\O(n)\$$, hence why I assume the for loop is $$\O(n^2)\$$. The last remaining piece is actually creating the string stream, hence the ? I have.

• For reverse_domain_2: $$\O(n^2)\$$. The reasoning behind this is that insert is linear time, and so the whole for loop would just be $$\O(n^2)\$$.

• For reverse_domain_3: $$\O(n) + O(n^2)\$$. The reasoning behind this is that reversal is linear time. And in the for loop, there's another reversal, hence $$\O(n^2)\$$. However, since its only reversing a piece of the string, I'm not too sure.

Are all these assumptions and time complexities correct?

I timed the performance of each of these methods on a multivariate plot, where the x axis is the length of the string, the y axis is the number of "." characters, and the z axis is the duration.

As you can see, reverse_domain is by far the slowest, and the other two are similar, with reverse_domain_3 being the fastest. However, I'm not exactly too sure why this is. I'm also not sure how to (or if I should) express the time complexity in terms of just the string length, or the length as well as the number of "." characters?

Thanks for any explanation(s).

• Big-O notation represents asymptotic complexity, so we don't normally write "O(n) + O(n²)" like that. As n gets larger, the linear term becomes insignificant compared to the quadratic term, so that becomes simply O(n²). Mar 9, 2022 at 8:37
• reverse_domain2(): What is the result of result.begin() + index immediately after index = 0;? Mar 10, 2022 at 7:52

reverse_domain()

This one has the largest extra space requirements, due to us creating a vector holding copies of all the components. It could be more efficient if the vector held string-view objects instead.

We could ensure that result doesn't need to reallocate if we reserve enough space - we know that it will be exactly the same size as the input string, because we're just rearranging the characters:

result.reserve(s.size())

We have a bug when we use int i to index the vector. Its size is a std::size_t, which has a different signedness and range to int. We can avoid the index variable altogether, though:

const char* sep = "";
for (const auto& component: std::views::reverse(breakList)) {
result += sep;
result += component;
sep = ".";
}

Runtime should scale as O(n) - appending to a string that has enough capacity should be linear in the quantity being appended, and independent on the string's length (unlike std::strcat(), which needs to examine the whole string to find the end).

reverse_domain2()

We don't need to force the caller to pass a std::string here - we could accept a std::string_view instead.

This function has the same problem of not reserving result capacity.

The type of index is deduced from 0 as int, but we need a std::size_t - it's better to be specific here. Alternatively, and more simply, we can use an iterator into the result string (once we've made sure it won't need to reallocate):

auto it = result.begin();

The implementation of this one doesn't require any extra storage beyond what's needed for the result, but the insert() operation is costly because it involves moving the existing string data to make room. As you state, that makes this version scale as O(n²).

reverse_domain3()

This is problematic, because of the misleadingly-named argument copy, which is a reference. We ought to accept std::string by value rather than by reference; if callers no longer need the original, they can use std::move() to make the copy cheaper.

Again, we use the wrong types for start and size, and for the loop iterator i. All these should be std::size_t.

Instead of looping over each character, we can offload that to our library, using std::find() or std::string::find() to locate each '.' in the string.

The asymptotic complexity here is linear, O(n), because each character is affected exactly twice by std::reverse(). And like reverse_domain2, we use no extra storage (and can re-use the caller's storage if the argument is move-constructed).


text:
zbMATH — the first resource for mathematics

Unbounded quasi-integrals. (English) Zbl 0957.28004
The aim of this paper is to extend the theory of quasi-measures in [J. F. Aarnes, Adv. Math. 86, No. 1, 41-67 (1991; Zbl 0744.46052)] to locally compact Hausdorff spaces $$X$$. For this purpose, the author introduces the notion of a quasi-measure in $$X$$ and the notion of a quasi-integral on $$C_0(X)$$ (resp. $$C_c(X)$$). A real-valued function $$\rho$$ on $$C_0(X)$$ (resp. $$C_c(X)$$) is called a quasi-integral on $$C_0(X)$$ (resp. $$C_c(X)$$) if the following conditions are satisfied: (1) $$b\geq 0\Rightarrow \rho(b)\geq 0$$ whenever $$b\in C_0(X)$$ (resp. $$C_c(X)$$). (2) $$\rho$$ is linear on $$A_0(a)$$ (the smallest uniformly closed subalgebra of $$C_0(X)$$ (resp. $$C_c(X)$$) containing $$a)$$ for each $$a\in C_0(X)$$ (resp. $$C_c(X)$$). A quasi-integral $$\rho$$ is said to be bounded if $$\sup\{\rho(a): 0\leq a\leq 1, a\in C_c(X)\}<+ \infty$$. Then it is shown that all quasi-integrals on $$C_0(X)$$ are bounded, a representation of quasi-integrals on $$C_c(X)$$ in terms of quasi-measures (a generalization of the Riesz representation theorem), and unique extensions of quasi-integrals on $$C_c(X)$$ to $$C_0(X)$$.

MSC:
 28A25 Integration with respect to measures and other set functions
Keywords:
quasi-measures; quasi-integrals
Full Text:
References:
 [1] Johan F. Aarnes, Quasi-states and quasi-measures, Adv. Math. 86 (1991), no. 1, 41 – 67. · Zbl 0744.46052 [2] J. F. Aarnes: “Image transformations and attractors,” Dept. of Math., University of Trondheim, Preprint no. 2 (1994). [3] J. F. Aarnes: “Quasi-measures in locally compact spaces,” Norwegian University of Science and Technology, Preprint no. 1 (1996). [4] J. F. Aarnes and A. B. Rustad: “Probability and Quasi-measures -a new interpretation,” To appear in Math. Scand. · Zbl 0967.28014 [5] Walter Rudin, Real and complex analysis, 3rd ed., McGraw-Hill Book Co., New York, 1987. · Zbl 0925.00005 [6] A. B. Rustad: “The multidimensional median as a quasi-measure,” Norwegian University of Science and Technology, Preprint no. 5 (1998). [7] G. Taraldsen: “Image-transformations and quasi-measures in locally compact Hausdorff spaces,” University of Trondheim, Preprint (1995).
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Why can't we say that all PDEs of a specified order require a fixed number of boundary conditions?

For an $n$th order ODE we always need $n$ boundary conditions (right?). But, as I've seen somewhere, for 2nd order PDEs there are many possible situations and a general answer to the question of how/what boundary conditions are needed doesn't exist. For example, Dirichlet or Neumann boundary conditions give unique solution for elliptic equations, while they don't for hyperbolic equations.

Isn't it really possible for a PDE with some specified boundary conditions to tell easily (with a rule of thump, like for ODEs) if it has a unique solution?

With PDEs is not as simple as with ODEs.

In ODEs, if the equation is of $n$th order, then you need $n$ initial conditions, to guarantee uniqueness.

In the case of hyperbolic PDEs of $n$th order, $$L_1\cdots L_n u=f,$$ where $L_j$ are first order linear operators, then you need $n$ initial condition, as in the case of the wave equation.

But in the case of elliptic equations, i.e., $L=\Delta^n$, then you need $n$ boundary conditions for $Lu=f$, with $L$ being an operator of order $2n$.

• It could be argued that for $\Delta^n$ you are supplying $n$ boundary conditions around the entire boundary, whereas for the hyperbolic PDE you are supplying $2n$ initial conditions on half of the boundary, and that the accounting is therefore somehow "the same." What is the existence&uniqueness situation if you supply $n$ boundary conditions, around the entire spacetime boundary, for a hyperbolic PDE? – user7530 Mar 20 '14 at 5:04

In fact for the "rule" that "An $n$th order ODE should need $n$ boundary conditions" is only true that for assuming the solutions which their continuous section(s) are maximized. Once allowing the solutions can be piecewise functions, the "rule" is no longer to be true.

For example for a simplest ODE $y'(x)=0$ , besides $y(x)=C$ are the solutions, e.g. $y(x)=Bu(x)+C$ , $y(x)=B[x]+C$ are also the solutions.

In fact allowing the solutions can be suitable types of piecewise functions, the ODEs can match more number(s) of the condition(s). So as for the PDEs.

In fact either ODEs and PDEs can have the concepts that "introducing suitable types of piecewise functions of the solutions so that the equations can match more number(s) of the condition(s)" , PDEs ones have much more discussions that ODEs ones only because of e.g. the real situations request.


text:
# Blocks in a room which is being accelerated

#### Like Tony Stark

Homework Statement
The two bodies shown in the picture have the same mass and are inside a room which is accelerating to the right with $2 \frac{m}{s^2}$. The coefficient of static friction is the same for both bodies. Determine the minimum value of the coefficient of static friction such that the blocks don't slide with respect to the room and the tension.
Homework Equations
Newton's equations
I wrote the equations for $m_2$ with respect to the room, which are:
$x) T-Fr_1-f_1 *=0$
$y) N_1-P_1=0$

For $m_1$ we hace:
$x) N_2=f_2 *$
$y) T+Fr_2 -P_2=0$

Where $f*_1$ and $f*_2$ are the pseudo-forces and $Fr_1$ and $Fr_2$ are the friction forces that contain $\mu$

Then I solved for $\mu$ and got $\mu =0.6$

Are my ideas correct? Because I wasn't sure about the sign of the friction.

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#### mfb

Mentor
x is in vertical direction? An unconventional choice.
You could have shown the intermediate steps. I get a different answer.

Working purely in the single degree of freedom: m*g is a force towards "m2 goes down", m*a is a force against it (where a=2 m/s^(2)), m*g*μ is against it (friction from m1), m*a*μ is against it (friction from m2).
m(g-a) = mμ(g+a), μ = 8/12 = 2/3.

"Blocks in a room which is being accelerated"

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# Find the values of the other trigonometric functions of theta if cot theta= -4/3and sin theta< 0.

Question
Trigonometric Functions
Find the values of the other trigonometric functions of theta if $$\displaystyle{\cot{\theta}}=-\frac{{4}}{{3}}{\quad\text{and}\quad}{\sin{\theta}}{<}{0}$$.

2021-02-10
$$\displaystyle{\cot{\theta}}=−{43}$$
therefore,
$$\displaystyle{\left|{\cot{\theta}}\right|}={\left|−\frac{{4}}{{3}}\right|}=\frac{{4}}{{3}}$$
as we know that $$\displaystyle{\left|{\cot{\theta}}\right|}=$$(base)/(perpendicular)
therefore,
$$\displaystyle{\left|{\cot{\theta}}\right|}=$$(base)/(perpendicular)=4/3ZSK
therefore,
base=4 and perpendicular=3
as we know that:
$$\displaystyle{\left(\text{ypotenuse})^2=(\text{perpendicular})^2+(\text{base})^}{y}{p}{o}{t}{e}\nu{s}{e}\right\rbrace}{)}^{{2}}={\left(\text{perpendicular}\right)}^{{2}}+{\left(\text{base}\right)}^{{2}}$$
therefore,
$$\displaystyle{\left(\text{ypotenuse})^2=(\text{perpendicular})^2+(\text{base})^}{y}{p}{o}{t}{e}\nu{s}{e}\right\rbrace}{)}^{{2}}={\left(\text{perpendicular}\right)}^{{2}}+{\left(\text{base}\right)}^{{2}}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}^{{2}}={\left({3}\right)}^{{2}}+{\left({4}\right)}^{{2}}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}^{{2}}={9}+{16}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}^{{2}}={25}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}=\sqrt{{25}}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}={5}$$
therefore hypotenuse=5
therefore,
$$\displaystyle{\left|{\sin{\theta}}\right|}$$=(perpendicular)/(hypotenuse)=$$\displaystyle\frac{{3}}{{5}}$$
$$\displaystyle{\left|{\cos{\theta}}\right|}$$=(base)/(hypotenuse)=$$\displaystyle\frac{{4}}{{5}}$$
$$\displaystyle{\left|{\tan{\theta}}\right|}$$=(perpendicular)/(base)=$$\displaystyle\frac{{3}}{{4}}$$
$$\displaystyle{\left|{\cos{{e}}}{c}{t}\hat{{e}}\right|}$$=(hypotenuse)/(perpendicular)=$$\displaystyle\frac{{5}}{{3}}$$
$$\displaystyle{\left|{\sec{\theta}}\right|}$$=(hypotenuse)/((base)=$$\displaystyle\frac{{5}}{{4}}$$
as $$\displaystyle{\cot{\theta}}=−{43}{\quad\text{and}\quad}{\sin{\theta}}{<}{0}$$
that implies both cotangent and sine function have negative values.
as we know that:
(1) in the first quadrant all trigonometric functions are positive.
(2) in the second quadrant sine and cosecant trigonometric functions are positive and rest of trigonometric functions are negative.
(3) in the third quadrant tangent and cotangent trigonometric functions are positive and rest of trigonometric functions are negative.
(4) in the fourth quadrant cosine and secant trigonometric functions are positive and rest of trigonometric functions are negative.
as both cotangent and sine function have negative values.
therefore θ is lying in the fourth quadrant.
therefore cosine and secant trigonometric functions will be positive and rest of trigonometric functions will be negative.
therefore,
$$\displaystyle{\left|{\sin{\theta}}\right|}$$=(perpendicular)/(hypotenuse)=$$\displaystyle-\frac{{3}}{{5}}$$
$$\displaystyle{\left|{\cos{\theta}}\right|}$$=(base)/(hypotenuse)=$$\displaystyle\frac{{4}}{{5}}$$
$$\displaystyle{\left|{\tan{\theta}}\right|}$$=(perpendicular)/(base)=$$\displaystyle-\frac{{3}}{{4}}$$
$$\displaystyle{\left|{\cos{{e}}}{c}{t}\hat{{e}}\right|}$$=(hypotenuse)/(perpendicular)=$$\displaystyle-\frac{{5}}{{3}}$$
$$\displaystyle{\left|{\sec{\theta}}\right|}$$=(hypotenuse)/((base)=$$\displaystyle\frac{{5}}{{4}}$$

### Relevant Questions

Given the values for sin t and cos t, use reciprocal and quotient identities to find the values of the other trigonometric functions of t.
$$\displaystyle{\sin{{t}}}=\frac{{3}}{{4}}{\quad\text{and}\quad}{\cos{{t}}}=\frac{\sqrt{{7}}}{{4}}$$

$$\sec \theta = -3, \tan \theta > 0$$. Find the exact value of the remaining trigonometric functions of
$$\theta$$.

Sketch a right triangle corresponding to the trigonometric function of the acute angle theta. Use the Pythagorean Theorem to determine the third side and then find the other five trigonometric functions of theta. $$\displaystyle{\cos{\theta}}=\frac{{21}}{{5}}$$
Given the following information about one trigonometric function, evaluate the other five functions.
$$\displaystyle{\cos{{u}}}=\frac{{5}}{{13}}$$ , where $$\displaystyle{0}\le{u}\le\frac{\pi}{{2}}.$$
If $$f(\theta) = \sin \theta = 0.2$$
Find $$f(\theta + \pi)$$
Write the trigonometric expression $$\displaystyle{\cos{{\left({{\sin}^{{-{1}}}{x}}-{{\cos}^{{-{1}}}{y}}\right)}}}$$ as an algebraic expression (that is, without any trigonometric functions). Assume that x and y are positive and in the domain of the given inverse trigonometric function.
Use the figures to find the exact value of the trigonometric function $$\displaystyle{\tan{{2}}}\theta$$.
$$\displaystyle{\sin{{\left(\frac{{{3}\pi}}{{4}}\right)}}}$$
Use a calculator to find the value of the trigonometric function $$\displaystyle{\sin{{\left(\frac{{{3}\pi}}{{10}}\right)}}}$$ to four decimal places.
Find derivative of trigonometric function $$\displaystyle{y}=\frac{{{3}{\left({1}-{\sin{{x}}}\right)}}}{{{2}{\cos{{x}}}}}$$


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# Rotating a sprite to look at mouse problem - SFML

## 4 posts in this topic

I just cannot fix this problem about rotating sprites around the mouse. The formula seems fine to me and it is similar to others I have googled. I just cannot figure out why it is behaveing in such a weird way.

void Animation::checkRotation(int x, int y, sf::Sprite& backSprite, int px, int py)
{
yOffset = y - py;
xOffset = x - px;
backSprite.SetCenter(px +200, py + 400);
a = (atan2(yOffset, xOffset)) * 180 / 3.141592;
backSprite.SetRotation( a );
}

where x and y are the mouse coordinates and px and py the sprite image coordinates. I thought that maybe when the sprite rotates, the x and the y also change dynamically, in that case, would there be a solution?

I just need a way to get my image looking at the mouse cursor. The sprite.getCenter() doesn't work so I cannot get the center of the image coordinates. Any ideas?

P.S: I am using SFML 1.6

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You should not be calling SetCenter.  SetCenter sets the center point of the object, and it's relative to the object; so, typically the "center" is the upper left of the sprite (0, 0), but you could set it to (ImageWidth/2,ImageHeight/2) and the "Center" would be the center of the image.  This would then be where the "position" of the sprite used.

So, don't use SetCenter here; it should be used (if at all) only once when the sprite is created, unless you're doing some really specilized stuff.  Just use setPosition() and getPosition() and assume the sprite uses upper left for it's position and you'll be good.

Otherwise, the code looks OK.  the x and y position of the sprite should not change at all.  BTW, the sfml-dev.org forums are very good, you can always ask the SFML specific questions there too.

Good luck.

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Instead of using atan2, use a regular arc tangent and give the slope.

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Instead of using atan2, use a regular arc tangent and give the slope.

Why?

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Register a new account


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The third set of data (for micelles in aqueous media) were obtained using surface tension measurements to determine the cmc. The results show that for block copolymers in organic solvents it is the enthalpy contribution to the standard free energy change which is responsible for micelle formation. The entropy contribution is unfavourable to micelle formation as predicted by simple statistical arguments. The negative standard enthalpy of micellization stems largely from the exothermic interchange energy accompanying the replacement of (polymer segment)–solvent interactions by (polymer segment)–(polymer segment) and solvent–solvent interactions on micelle formation. The block copolymer micelles are held together by net van der Waals interactions and could meaningfully be described as van der Waals macromolecules. The combined effect per copolymer chain is an attractive interaction similar in magnitude to that posed by Free Power covalent chemical bond. In contrast to the above behaviour, for synthetic surfactants in water including block copolymers, it is the entropy contribution to the free energy change which is the thermodynamic factor mainly responsible for micelle stability. Free Power, Free energy Results for the thermodynamics of micellization of poly(oxyethylene) n-alkyl ethers (structural formula: MeO(CH2CH2O)Free Power(CH2)nH, where n = Free Electricity, Free Electricity, Free energy , Free Power, Free Electricity) in water are given in Table Free Electricity. Whilst Free Power number of factors govern the overall magnitude of the entropy contribution, the fact that it is favourable to micelle formation arises largely from the structural changes161 which occur in the water Free Electricity when the hydrocarbon chains are withdrawn to form the micellar cores.
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Not Free Power lot to be gained there. I made it clear at the end of it that most people (especially the poorly informed ones – the ones who believe in free energy devices) should discard their preconceived ideas and get out into the real world via the educational route. “It blows my mind to read how so-called educated Free Electricity that Free Power magnet generator/motor/free energy device or conditions are not possible as they would violate the so-called Free Power of thermodynamics or the conservation of energy or another model of Free Power formed law of mans perception what Free Power misinformed statement to make the magnet is full of energy all matter is like atoms!!”
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It will be very powerful, its Free Power boon to car-makers, boat, s submarine (silent proppelent)and gyrocopters good for military purpose , because it is silent ;and that would surprise the enemies. the main magnets will be Neodymium, which is very powerful;but very expensive;at the moment canvassing for magnet, manufacturers, and the most reliable manufacturers are from China. Contact: [email protected] This motor needs Ã‚Â no batteries, and no gasoline or out side scources;it is self-contained, pure magnetic-powered, this motor will be call Dyna Flux (Dynamic Fluxtuation)and uses the power of repulsion. Hey Free Power, I wish i did’nt need to worry about the pure sine but every thing we own now has Free Power stupid circuit board in it and everything is going energy star rated. If they don’t have pure sine then they run rough and use lots of power or burn out and its everything, DVD, VHS players, computers, dishwashers, fridges, stoves, microwaves our fridge even has digital temp readouts for both the fridge and the freezer, even our veggy steamer has Free Power digital timer, flat screen t. v’s, you can’t get away from it anymore, the world has gone teck crazzy. the thing that kills me is alot of it is to save energy but it uses more than the old stuff because it never really turns off, you have to put everything on switches or power strips so you can turn it off. I don’t know if i can get away from using batteries for my project. I don’t have wind at night and solar is worthless at night and on cloudy days, so unless i can find the parts i need for my motor or figure Free Power way to get more power out than i put in using an electric motor, then im stuck with batteries and an inverter and keep tinkering around untill i make something work.
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Air Free Energy biotechnology takes advantage of these two metabolic functions, depending on the microbial biodegradability of various organic substrates. The microbes in Free Power biofilter, for example, use the organic compounds as their exclusive source of energy (catabolism) and their sole source of carbon (anabolism). These life processes degrade the pollutants (Figure Free Power. Free energy). Microbes, e. g. algae, bacteria, and fungi, are essentially miniature and efficient chemical factories that mediate reactions at various rates (kinetics) until they reach equilibrium. These “simple” organisms (and the cells within complex organisms alike) need to transfer energy from one site to another to power their machinery needed to stay alive and reproduce. Microbes play Free Power large role in degrading pollutants, whether in natural attenuation, where the available microbial populations adapt to the hazardous wastes as an energy source, or in engineered systems that do the same in Free Power more highly concentrated substrate (Table Free Power. Free Electricity). Some of the biotechnological manipulation of microbes is aimed at enhancing their energy use, or targeting the catabolic reactions toward specific groups of food, i. e. organic compounds. Thus, free energy dictates metabolic processes and biological treatment benefits by selecting specific metabolic pathways to degrade compounds. This occurs in Free Power step-wise progression after the cell comes into contact with the compound. The initial compound, i. e. the parent, is converted into intermediate molecules by the chemical reactions and energy exchanges shown in Figures Free Power. Free Power and Free Power. Free Power. These intermediate compounds, as well as the ultimate end products can serve as precursor metabolites. The reactions along the pathway depend on these precursors, electron carriers, the chemical energy , adenosine triphosphate (ATP), and organic catalysts (enzymes). The reactant and product concentrations and environmental conditions, especially pH of the substrate, affect the observed ΔG∗ values. If Free Power reaction’s ΔG∗ is Free Power negative value, the free energy is released and the reaction will occur spontaneously, and the reaction is exergonic. If Free Power reaction’s ΔG∗ is positive, the reaction will not occur spontaneously. However, the reverse reaction will take place, and the reaction is endergonic. Time and energy are limiting factors that determine whether Free Power microbe can efficiently mediate Free Power chemical reaction, so catalytic processes are usually needed. Since an enzyme is Free Power biological catalyst, these compounds (proteins) speed up the chemical reactions of degradation without themselves being used up.

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During the early 19th century, the concept of perceptible or free caloric began to be referred to as “free heat” or heat set free. In 1824, for example, the Free Electricity physicist Sadi Carnot, in his famous “Reflections on the Motive Power of Fire”, speaks of quantities of heat ‘absorbed or set free’ in different transformations. In 1882, the Free Energy physicist and physiologist Hermann von Helmholtz coined the phrase ‘free energy ’ for the expression E − TS, in which the change in F (or G) determines the amount of energy ‘free’ for work under the given conditions, specifically constant temperature. [Free Electricity]:Free Power.


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# Question regarding floor() function

Given x is a real number and c is a natural number. Prove that:

floor( floor(x) / c ) = floor( x / c )

When x > 0, I can prove it by let x = n + z where 0 < z < 1.
Then rhs = floor( n/c + z/c ) which equals to floor( floor(x) ) since z/c -> 0.
However, when x < 0, it's wrong when letting x = -( n + z ). Because it always give me a smaller negative number. Is my logic wrong here? Any hint?

Thanks,
Chan

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I suspect that you want $x=-n+z$ (no parentheses around $n+z$) as, for instance, $\text{floor}(-1.5)=-2$. –  Isaac Feb 4 '11 at 6:34
@Issac: Thanks, so if x = -1.67 then I would rewrite it as: -2 + 0.33? –  Chan Feb 4 '11 at 6:41

You can see it here:

-
@Chandru1: Thanks but you spoiled me :(! –  Chan Feb 4 '11 at 6:42
@chan: Sorry chan. –  anonymous Feb 4 '11 at 7:24

This is the special case $\rm\ m = 1\$ in a proof I presented here. See that thread for more on the universal viewpoint that explains the simplicity of this proof. For convenience, here is the proof.

LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \$ for $\rm\ \ n > 0$

Proof $\rm\quad\quad\quad\quad\quad\ \ \ k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor$

Compare the above trivial proof to more traditional proofs, e.g. the special case $\rm\ m = 1\$ here.

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Many thanks for a great solution. Very clever and succinct ;) ! –  Chan Feb 4 '11 at 15:34


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# Finding Volume using spherical coordinates

• March 26th 2010, 04:39 PM
Belowzero78
Finding Volume using spherical coordinates
Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere $\rho=4 cos\phi$ and the hemisphere $\rho=6, z\geq0$. Then evaluate the integral.

I found the integral with the correct limits for the solid in spherical coordinates as the following: $V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta$

I would like to check my volume, the answer which i got is $\frac{26\pi}{3}$.
• March 26th 2010, 07:04 PM
dwsmith
As long as all your bounds are correct for the figure, your answer is off. I obtained 400*pi/3 for that integral as is.
• March 26th 2010, 07:11 PM
Belowzero78
Alright, do you mind to show a few steps?
• March 26th 2010, 07:13 PM
dwsmith
If you show me your steps, it will be quicker for me to find your error.
• March 26th 2010, 07:35 PM
11rdc11
Quote:

Originally Posted by Belowzero78
Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere $\rho=4 cos\phi$ and the hemisphere $\rho=6, z\geq0$. Then evaluate the integral.

I found the integral with the correct limits for the solid in spherical coordinates as the following: $V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta$

I would like to check my volume, the answer which i got is $\frac{26\pi}{3}$.

$2\pi\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi$

= $2\pi\int_{0}^{\pi/2} \frac{\rho^3}{3}\sin\phi \bigg|^{6}_{4\cos\phi} d\phi$

= $2\pi\int_{0}^{\pi/2} 72\sin\phi -\frac{64\cos^3\phi\sin\phi}{3}\ d\phi$

= $2\pi( -72\cos\phi +\frac{64\cos^4\phi}{12}) \bigg|^\frac{\pi}{2}_{0}$

= $2\pi(72-\frac{16}{3}) = \frac{400\pi}{3}$


text:
# recognition of revenue

which of the follwing best describes the condition(s) that must be present for the recognition of revenue ?

1. the revenue must be earned, measurable and collected
2. the revenue must be earned, measurable and collectible
3. the revenue must be earned and collected
4. the revenue must be measurable and ...

There are actually 4 criteria, but if your book says 2, that's probably because you're in an intro class and they want to simplify things. Anyway, if your book says that the only two criteria are earned and collectible, you should go with choice number 3. the revenue must be earned and collected. However, the most accurate answer is number 2. the revenue must be earned, measurable and collectible. By measurable, it simply means the seller's price is fixed or determinable and not some vague estimate (for example, I might decide to rent you out a space in my grocery store so you can sell SuperGood candy bars and I will charge you 2% of sales. I have no way of determining how much that will be because you've never tried to sell this...

## Related Questions in Accounting Concepts and Principles

• ### Accounting problem: The adjusting Process

(Solved) February 12, 2011

Prior to the adjusting process, accrued revenue has Question 5 options: 1 ) been earned and cash received 2 )been earned and not recorded as revenue 3 ) not been earned but recorded as revenue 4 ) not been recorded as revenue but cash has been received

• ### 1. The revenue recognition principle states that revenue is recognized when: A. The revenue is...

(Solved) April 17, 2017

1 . The revenue recognition principle states that revenue is recognized when: A. The revenue is substantially earned. B. The revenue is realized. C. Performance is substantially complete. D. The related cash is received. 2 . Revenue from selling products is generally recognized: A. At the point

Dear Student, Please find attached solution, Thank...
• ### Final Team Project As we have discussed, revenue recognition does have great impact on quality of...

(Solved) December 16, 2014

statement. Revenue must be 1 ) realized or realizable, and 2 ) earned. Realized or realizable means that the seller’s net assets increase. That is, the seller receives an asset, such as cash or account receivable, or satisfies a liability, such as deferred revenues, as a result of a transaction. The...

The solution provides for the revenue recognition for the following: 1. A clothing retailer 2. A contractor that performs work under long-term government contracts; 3. A grocery store 4. A...
• ### ASAP! Journalize adjusting entry (Intro to accounting) I will Rate!

(Solved) June 09, 2009

• ### Week 4 Discussion

(Solved) January 24, 2012

Follow the guidelines! Include APA format references!


text:
# Revision history [back]

(1) For comparison using tpm, check the following data for selection.

 https://www.dell.com/support/article/kr/ko/krdhs1/sln312590/tpm-12-vs-20-features?lang=en

(2) Please use raid 1 because raid0 may have a problem with recovery. raid0 ca be quick access and extend storage capability, but I do not recommend it if you need stability.

(1) For comparison using tpm, check the following data for selection.

 https://www.dell.com/support/article/kr/ko/krdhs1/sln312590/tpm-12-vs-20-features?lang=en

(2) Please use raid 1 because raid0 may have a problem with recovery. raid0 ca can be quick access and extend storage capability, but I do not recommend it if you need stability.

(1) For comparison using tpm, check the following data for selection.

 https://www.dell.com/support/article/kr/ko/krdhs1/sln312590/tpm-12-vs-20-features?lang=en

In the case of tpm I think it is well explained on the web. For security reasons, you can increase the security of your connection through your firewall and security settings, and if necessary, encrypt the disk itself. As there are so many choices for security, these security issues are used in combination.

(2) Please use raid 1 because raid0 may have a problem with recovery. raid0 can be quick access and extend storage capability, but I do not recommend it if you need stability.

I asked the same question for one. raid1 can be restored in case of disk problems. However, it is not perfect if it is a soft raid setting, and if it is important equipment, it is set and used by hardware raid card. I usually use raid5 or higher to apply the system, and more stable if I use raid 10.

(1) For comparison using tpm, check the following data for selection.

 https://www.dell.com/support/article/kr/ko/krdhs1/sln312590/tpm-12-vs-20-features?lang=en

In the case of tpm I think it is well explained on the web. For security reasons, you can increase the security of your connection through your firewall and security settings, and if necessary, encrypt the disk itself. As there are so many choices for security, these security issues are used in combination.

(2) Please use raid 1 because raid0 may have a problem with recovery. raid0 can be quick access and extend storage capability, but I do not recommend it if you need stability.

(3) I asked the same question for one. raid1 can be restored in case of disk problems. However, it is not perfect if it is a soft raid setting, and if it is important equipment, it is set and used by hardware raid card. I usually use raid5 or higher to apply the system, and more stable if I use raid 10.

10

(4) If it is expensive, it can be economically managed by storing it on another storage device using an unused computer, and it is also desirable in terms of resource management. We do not have to stick to the latest equipment, and if you simply use a Fedora server to configure a file server, you can replace expensive equipment at a fraction of the cost.

(5) If you store valuable information over a long period of time, you can easily recover information in the event of an accident by installing a separate backup device in addition to the basic storage device. Likewise, important data is stored on a separate server, and such servers may even separate the network and power.


text:
# Necessary and sufficient condition

Necessary and sufficient condition

In logic, the words necessity and sufficiency refer to the implicational relationships between statements. The assertion that one statement is a necessary and sufficient condition of another means that the former statement is true if and only if the latter is true.

## Definitions

A necessary condition of a statement must be satisfied for the statement to be true. Formally, a statement P is a necessary condition of a statement Q if Q implies P (Q $\Rightarrow$ P).

A sufficient condition is one that, if satisfied, assures the statement's truth. Formally, a statement P is a sufficient condition of a statement Q if P implies Q (P $\Rightarrow$ Q).

## Necessary conditions

The sun being above the horizon is a necessary condition for direct sunlight; but it is not a sufficient condition as something else may be casting a shadow, e.g. in the case of an eclipse.

The assertion that P is necessary for Q is colloquially equivalent to "Q cannot be true unless P is true," or "if P is false then Q is false." By contraposition, this is the same thing as "whenever Q is true, so is P". The logical relation between them is expressed as "If Q then P" and denoted "Q $\Rightarrow$ P" (Q implies P), and may also be expressed as any of "P, if Q"; "P whenever Q"; and "P when Q." One often finds, in mathematical prose for instance, several necessary conditions which, taken together, constitute a sufficient condition, as shown in Example 5.

Example 1: In order for it to be true that "John is a bachelor" it is necessary that it be also true that he is
1. unmarried
2. male
since to state "John is a bachelor" implies John has each of those three additional predicates.
Example 2: For the whole numbers greater than two, being odd is necessary to being prime, since two is the only whole number that is both even and prime.
Example 3: Consider thunder, technically the acoustic quality demonstrated by the shock wave that inevitably results from any lightning bolt in the atmosphere. It may fairly be said that thunder is necessary for lightning, since lightning cannot occur without thunder, too, occurring. That is, if lightning does occur, then there is thunder.
Example 4: Being at least 30 years old is necessary of serving in the U.S. Senate. If you are under 30 years old then it is impossible for you to be a senator. That is, if you are a senator, it follows that you are at least 30 years old.
Example 5: In algebra, in order for some set S together with an operation $\star$ to form a group, it is necessary that $\star$ be associative. It is also necessary that S include a special element e such that for every x in S it is the case that e $\star$ x and x $\star$ e both equal x. It is also necessary that for every x in S there exist a corresponding element x” such that both x $\star$ x” and x$\star$ x equal the special element e. None of these three necessary conditions by itself is sufficient, but the conjunction of the three is.

## Sufficient conditions

That a train runs on schedule is often a sufficient condition for arriving on time (if the train arrives on time and one boards it, one will presumably arrive on time); but it is not always a necessary condition, since there are other ways to travel (if the train does not arrive on time, one might still arrive on time to one's destination through other means of transport).

To say that P is sufficient for Q is to say that, in and of itself, knowing P to be true is adequate grounds to conclude that Q is true. (It is to say, at the same time, that knowing P not to be true does not, in and of itself, provide adequate grounds to conclude that Q is not true either.) The logical relation is expressed as "If P then Q" or "P $\Rightarrow$ Q," and may also be expressed as "P implies Q." Several sufficient conditions may, taken together, constitute a single necessary condition, as illustrated in example 5.

Example 1: Stating that "John is a bachelor" implies that John is male. So knowing that it is true that John is a bachelor is sufficient to know that he is a male.
Example 2: A number's being divisible by 4 is sufficient (but not necessary) for its being even, but being divisible by 2 is both sufficient and necessary.
Example 3: An occurrence of thunder is a sufficient condition for the occurrence of lightning in the sense that hearing thunder, and unambiguously recognizing it as such, justifies concluding that there has been a lightning bolt.
Example 4: A U.S. president's signing a bill that Congress passed is sufficient to make the bill law. Note that the case whereby the president did not sign the bill, e.g. through exercising a presidential veto, does not mean that the bill has not become law (it could still have become law through a congressional override).
Example 5: That the center of a playing card should be marked with a single large spade (♠) is sufficient for the card to be an ace. Three other sufficient conditions are that the center of the card be marked with a diamond (♦), heart (♥), or club (♣), respectively. None of these conditions is necessary to the card's being an ace, but their disjunction is, since no card can be an ace without fulfilling at least (in fact, exactly) one of the conditions.

## Relationship between necessity and sufficiency

A condition can be either necessary or sufficient without being the other. For instance, being a mammal (P) is necessary but not sufficient to being human (Q), and that a number q is rational (P) is sufficient but not necessary to q‘s being a real number (Q) (since there are real numbers which are not rational).

A condition can be both necessary and sufficient. For example, at present, "today is the Fourth of July" is a necessary and sufficient condition for "today is Independence Day in the United States." Similarly, a necessary and sufficient condition for invertibility of a matrix M is that M has a nonzero determinant.

Mathematically speaking, necessity and sufficiency are dual to one another. For any statements P and Q, the assertion that “P is necessary for Q” is equivalent to the assertion that “Q is sufficient for P.” Another facet of this duality is that, as illustrated above, conjunctions of necessary conditions may achieve sufficiency, while disjunctions of sufficient conditions may achieve necessity. For a third facet, identify every mathematical predicate P with the set S(P) of objects for which P holds true; then asserting the necessity of P for Q is equivalent to claiming that S(P) is a superset of S(Q), while asserting the sufficiency of P for Q is equivalent to claiming that S(P) is a subset of S(Q).

## Simultaneous necessity and sufficiency

To say that P is necessary and sufficient for Q is to say two things, that P is necessary for Q and that P is sufficient for Q. Of course, it may instead be understood to say a different two things, namely that each of P and Q is necessary for the other. And it may be understood in a third equivalent way: as saying that each is sufficient for the other. One may summarize any—and thus all—of these cases by the statement "P if and only if Q," which is denoted by P $\Leftrightarrow$ Q.

For example, in graph theory a graph G is called bipartite if it is possible to assign to each of its vertices the color black or white in such a way that every edge of G has one endpoint of each color. And for any graph to be bipartite, it is a necessary and sufficient condition that it contain no odd-length cycles. Thus, discovering whether a graph has any odd cycles tells one whether it is bipartite and vice versa. A philosopher[1] might characterize this state of affairs thus: "Although the concepts of bipartiteness and absence of odd cycles differ in intension, they have identical extension.[2]

## References

1. ^ Stanford University primer, 2006
2. ^ "Meanings, in this sense, are often called intensions, and things designated, extensions. Contexts in which extension is all that matters are, naturally, called extensional, while contexts in which extension is not enough are intensional. Mathematics is typically extensional throughout." Stanford University primer, 2006

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• necessary and sufficient condition — condition that without which nothing will happen and upon whose fulfillment there is no need for any other conditions (Logic) …   English contemporary dictionary

• necessary and sufficient conditions — See condition, necessary/sufficient …   Philosophy dictionary

• Necessary — may refer to: Something that is a required condition for something else to be the case, see necessary and sufficient condition. A necessary truth, something that cannot fail to be true, see logical possibility. An important task or essential… …   Wikipedia

• Condition — or Conditions may refer to: Contents 1 Logic 2 Computer programming 3 Other 4 See also Logic Logical conditional …   Wikipedia

• condition — conditionable, adj. /keuhn dish euhn/, n. 1. a particular mode of being of a person or thing; existing state; situation with respect to circumstances. 2. state of health: He was reported to be in critical condition. 3. fit or requisite state: to… …   Universalium

• condition — noun 1 state of sth ADJECTIVE ▪ excellent, good, immaculate, mint, perfect, pristine ▪ reasonable ▪ bad …   Collocations dictionary

• Condition — • That which is necessary or at least conducive to the actual operation of a cause Catholic Encyclopedia. Kevin Knight. 2006. Condition     Condition      …   Catholic encyclopedia

• condition, necessary/sufficient — If p is a necessary condition of q, then q cannot be true unless p is true. If p is a sufficient condition of q, then given that p is true, q is so as well. Thus steering well is a necessary condition of driving well, but it is not sufficient,… …   Philosophy dictionary

• Necessary But Not Sufficient (novel) — Necessary But Not Sufficient is a 2000 novel authored by Dr. Eliyahu Goldratt and co authors Eli Schrangenheim and Carol A. Ptak and published by the North River Press, which has previously published most of Goldratt s works. Necessary but not… …   Wikipedia

• Condition-based maintenance — (CBM), shortly described, is maintenance when need arises. This maintenance is performed after one or more indicators show that equipment is going to fail or that equipment performance is deteriorating. Condition based maintenance was introduced… …   Wikipedia


text:
# siunitx: Warning “\textsuperscript invalid in math mode” and missing spaces [closed]

Is there a reasonable explanation for those warnings/errors?

How can I avoid them properly?

If I use pdfLaTeX, I get the same (incorrect) output, but no warning.

%!TEX TS-program = lualatex
\documentclass[fleqn]{scrartcl}

\usepackage{fontspec, polyglossia}
\setdefaultlanguage[spelling=new, babelshorthands=true]{german}

\usepackage{mathtools}
\usepackage{siunitx}

\begin{document}

%Warning: LaTeX Warning: Command \textsuperscript invalid in math mode on input line XYZ
%The non-marked lines are ok

$\SI{1}{\cm^3 \per \min}$
$\SI{1}{\cubic \cm \per \min}$ %warning
$\SI{1}{\square\cm\per\min}$ %warning
%$\SI{1}{cm^2 \per. min}$
$\SI{1}{\cm^2 \per \min}$
$\SI{1}{\N \cm\tothe{3} \kilo \mol\squared}$
$\SI{1}{\N \cm\tothe{3} \min}$ %warning and visibly missing space between N cm
$\SI{1}{\cm\tothe{3} \min}$ %warning
$\SI{1}{\cm\cubed \per \min}$ %warning
$\SI{1}{\N \cm \tothe{3} \kilo \gram \squared}$

$\SI{1}{\N \cm \tothe{3} \kilo \gram \squared}$
$\SI{1}{\N \cm \tothe{3} \kilo \mol \squared}$

\end{document}

## closed as off-topic by LaRiFaRi, Zarko, Svend Tveskæg, Kurt, barbara beetonMar 15 '16 at 14:57

• This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• The explanation is simple: fontspec loads xunicode that redefines \textsuperscript in a way that makes TeX issue that warning. – egreg Jun 1 '15 at 16:40
• I told Joseph Wright about this issue in chat:chat.stackexchange.com/transcript/message/21953530#21953530 – egreg Jun 1 '15 at 16:49
• While not a robust fix, you could use ~ when you are missing a space and can't seem to do much about it. $\SI{1}{\N~\cm\tothe{3} \min}$ %warning and visibly missing space between N cm – EngBIRD Jun 13 '15 at 20:15
• I did not think of that one yet. So far, i just used \SI{1}{N.cm^3.min} to avoid that warning and get the spaces. – Matthias Jun 15 '15 at 17:05
• I'm voting to close this question as off-topic because it is about an issue which has been solved in v2.6f (github.com/josephwright/siunitx/commit/…) – LaRiFaRi Mar 15 '16 at 14:28


text:
zbMATH — the first resource for mathematics

Fast numerical solution of nonlinear Volterra convolution equations. (English) Zbl 0581.65095
Authors’ summary: Numerical methods for general Volterra integral equations of the second kind need $$O(n^ 2)$$ kernel evaluations and $$O(n^ 2)$$ additions and multiplications. Here it is shown how the effort can be reduced for nonlinear convolution equations. Exploiting the convolution structure, most numerical methods need only O(n) kernel evaluations. With the use of fast Fourier transform techniques only O(n(log n)$${}^ 2)$$ additions and multiplications are necessary. The paper closes with numerical examples and comparisons.
Reviewer: J.Albrycht

MSC:
 65R20 Numerical methods for integral equations 45G10 Other nonlinear integral equations
Full Text:


text:
# Checking formality of a perfect complex Zariski-locally

Let $$X$$ be a locally Noetherian scheme and $$K^{\bullet}$$ a perfect complex of $$\mathcal{O}_X$$-modules. We say $$K^{\bullet}$$ is "formal" if it is quasi-isomorphic to the complex $$\bigoplus_{n}H_n(K^{\bullet})[n]$$.

Can formality of a perfect complex of $$\mathcal{O}_X$$-modules be checked Zariski-locally on $$X$$?

Remark. This seems dubious to me, unless there's some natural zigzag between $$K^{\bullet}$$ and $$\bigoplus_nH_n(K^{\bullet})[n]$$, in which case it'd be better to define formality in terms of that map in $$D(\mathcal{O}_X)$$ being an isomorphism. But maybe there's a procedure to build a formal representatives out of local data and I'm just unaware of it.

I'd be interested to know if it's possible. A quick literature search didn't turn up much.

Consider two vector bundles $$V,W$$ and $$\alpha \in {\rm Ext}^2(V,W)$$, $$\alpha \neq 0$$. Then $$C:={\rm cone}(V[1] \to^\alpha W[-1])$$ is a nonsplit complex with $$H^0(C) = V$$ and $$H^1(C) = W$$. But $$C$$ splits locally, because locally $$V$$ and $$W$$ are projective.
E.g. we can take $$X = \mathbb P^2, V =\mathcal O,$$ and $$W = \mathcal O(-3)$$.


text:
# How do you find vertical, horizontal and oblique asymptotes for  f(x)= (2x^3+11x^2+5x-1)/(x^2+6x+5)?

Dec 27, 2016

Vertical : $x = - 1 \mathmr{and} x = - 5$.
Slant: $y = 2 x - 1$.
See Socratic graph for the curve and the asymptotes.

#### Explanation:

y = f(x)=2x-1+(x-4)/((x+1)(x+5)

y = quotient = $2 x - 1$ gives the slant asymptote, with slope 2.

The factors ( x + 1 ) and ( x + 5 ) in the denominator of the remainder

equated to 0 give the vertical asymptotes.

The ad hoc graphs are not in uniform as-is scales.

They serve the purpose of revealing asymptoticity of the three

asymptotes, with respect to the three branches of the graphs, taken

in pairs..

graph{(y(x+1)(x+5)-2x^3-11x^2-5x+1)(x+5)(y-2x+1)(x+1)=0 [-10, 10, -30, `30]}

graph{(y(x+1)(x+5)-2x^3-11x^2-5x+1)(x+5)(y-2x+1)(x+1)=0 [-40, 40, -82.5, 57.5]}


text:
# Adjust vertical position of single character in WordPress' Latex

I'm trying to define a new operator (like \bigominus -- but it doesn't exist) to create an equation like

But the problem is that I can't shrink the lower indices after vertical aligning everything that follows. The code I have so far is

\mathop{\huge{\boldsymbol{\ominus}}}\limits_{\tiny{ijk}}\normalsize{\raisebox{0.7ex}{\text{x}[\,u\,][\,v\,][\,w\,]}}

If I remove \raisebox{0.7ex}{} it fixes the indices, but then the vertical alignment is gone.

Obviously it's noticeable with respect to the horizontal line of the symbol. Does anyone know how I may fix this? I would like to either move the \ominus symbol down so that the horizontal line is aligned with the cross of the \text{x}, or the \text{x} and everything that follows so that it aligns with the horizontal line of the \otimes.

I've spent a few hours trying to solve this problem but I cannot seem to find the right command in the rather limited available set from WordPress.

You can scale a bold \ominus; whether this works in WordPress I don't know, but it would be off-topic, because WordPress uses MathJax that's not TeX.

\documentclass{article}
\usepackage{amsmath,bm,graphicx}

\makeatletter
\newcommand*{\bigominus}{\DOTSB\bigominus@\slimits@}

\newcommand{\bigominus@}{\mathop{\mathpalette\bigominus@@\relax}}

\newcommand{\bigominus@@}[2]{%
\vcenter{\hbox{%
\sbox\z@{$\m@th#1\bigoplus$}%
\resizebox{\wd\z@}{!}{$\m@th#1\bm{\ominus}$}%
}}%
}
\makeatother

\begin{document}
$\bigominus_{ijk}x[u][v][w] \qquad \textstyle \bigominus_{ijk}x[u][v][w] \qquad \scriptstyle \bigominus_{ijk}x[u][v][w] \qquad \scriptscriptstyle \bigominus_{ijk}x[u][v][w]$
$\bigoplus_{ijk}x[u][v][w] \qquad \textstyle \bigoplus_{ijk}x[u][v][w] \qquad \scriptstyle \bigoplus_{ijk}x[u][v][w] \qquad \scriptscriptstyle \bigoplus_{ijk}x[u][v][w]$

\end{document}

\bigominus does exist, in the mathx font of the mathabx package. You can easily use it without having to change the default math font:

\documentclass{article}%
\usepackage[utf8]{inputenc}

\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{
<-6> mathx5 <6-7> mathx6 <7-8> matha7
<8-9> mathx8 <9-10> mathx9
<10-12> mathx10 <12-> mathx12
}{}
\DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
\DeclareMathSymbol{\bigominus}{\mathop}{mathx}{"C1}

\begin{document}

$A \bigominus B$%

\end{document}


text:
# Clipping lines and triangles in a software renderer

This topic is 2753 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I'm trying to get clipping and culling working for my software renderer, but am having a bit of trouble understanding how to do the actual clipping.

I've been following this paper on clipping but can't quite wrap my head around what space the clipping needs to be performed in. I was expecting to clip in normalised device coords, with their nice easy -1 to 1 range, but apparently that throws up some unpleasant edge cases, so it's better to clip in homogeneous space before the w divide to NDC space. Am I right in thinking this?

I'm starting with lines, so I figure I've got to do something like:

1. Transform vertices from world to homogeneous space by transforming by model, view and projection matrices.
2. Classify each vertex
3. Reject the whole line, accept the whole line or clip the line
4. Rasterise

1 and 4 I've got, but how do I start on 2 and 3? What planes am I actually classifying/clipping against?

Pointer's appreciated, I'm a unsure where I'm supposed to be going next...

##### Share on other sites
So I also found this reference, which seems to be the same technique. My understanding is that the equation for the right plane in clip space is (w + x = 0). So using the rearranged equation I can find 'a' which gives me the intersection distance of the plane against my line segment to be clipped.

However when I do this, my 'a' ends up being a negative number. Surely this isn't correct as it means the intersection point is not actually on the line.

Test code currently looks like this:
[font=Monaco][size=2] if ((startClassification & RIGHT_CODE) != 0[/font] || (endClassification & RIGHT_CODE) != 0) { // For right plane: final float top = clipStart.w + clipStart.x; final float bottom = (clipStart.w + clipStart.x) - (clipEnd.w + clipEnd.x); final float a = top / bottom; final float invA = 1f - a; // Clip start or end? final float dx = clipEnd.x - clipStart.x; final float dy = clipEnd.y - clipStart.y; final float dz = clipEnd.z - clipStart.z; final float dw = clipEnd.w - clipStart.w; final float len = (float)Math.sqrt(dx*dx + dy*dy + dz*dz + dw*dw); clipEnd.x = clipStart.x * invA + clipEnd.x * a; clipEnd.y = clipStart.y * invA + clipEnd.y * a; clipEnd.z = clipStart.z * invA + clipEnd.z * a; clipEnd.w = clipStart.w * invA + clipEnd.w * a;

Is my understanding of this 'a' value correct? And is a negative number valid or does it mean I've done something else wrong?

1. 1
2. 2
3. 3
frob
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4. 4
Rutin
18
5. 5

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text:
# n-fold symbolic integral in Mathematica

I am trying to compute symbolically a n-fold integral (n is a parameter of a function) over, say, the cube [0,a]^n. My code looks like this

intCube[n_, a_] := Module[{xvars, range, expression},
xvars = Table[Symbol["x" <> ToString[i]], {i, 1, n}];
range = Table[{xvars[[i]], 0, a}, {i, 1, n}];
range = Row[Riffle[range, ","]];
Print[range];
expression = Product[xvars[[i]], {i, 1, n}];
Print[expression];
Integrate[expression, range]
];

The range and the expression to integrate look both ok, however, I get an awkward result. Any ideas why ?

intCube[2, a]

{x1,0,a},{x2,0,a}
x1 x2
x1 x2 Row[{{x1, 0, a}, ",", {x2, 0, a}}]

Thanks,

Ion

• Domnule Nechita, the explanation is that Row is for formatting (display) only. Bine ai venit pe Mathematica.SE! – Szabolcs Mar 14 '12 at 21:59

Use Sequence to paste in the range in Integrate :

Remove[intCube]
intCube[n_, a_] :=
Module[{xvars, range, expression},
xvars = Table[Symbol["x" <> ToString[i]], {i, 1, n}];
range = Table[{xvars[[i]], 0, a}, {i, 1, n}];
(* range=Row[Riffle[range,","]]; *)
Print[range];
expression = Product[xvars[[i]], {i, 1, n}]; Print[expression];
Integrate[expression, Sequence @@ range]
];

{{x1,0,a},{x2,0,a}}

x1 x2

a^4/4

• Thanks mate, works like a charm. I also learned about @@, I am new to mathematica so these tips are useful to me – Ion Nechita Mar 14 '12 at 12:15
• @Ion Well, then you can go ahead and change expression using @@ – b.gates.you.know.what Mar 14 '12 at 12:17
• :) expression = Times @@ xvars; right ? – Ion Nechita Mar 14 '12 at 12:22


text:
# A Cryptic Riddle for Puzzlers

Happy as a Lark in a meadow,

Kind as a anonymous donor,

as Sunny and the midsummer's day,

Reliable as the sun rise that comes daily,

-Joe

Hint 1

You will have to do some cryptography.

Hint 2:

The cipher used is a simple substitution.

Hint 3

The cipher used is not a caesar cipher

• Why is this being downvoted again? Is it because of the rot13(ureevat)? – Cloudy7 Oct 10 '19 at 5:37

Sophia

which you find by

Taking the capital letters in the text ('HLKSRZ') and seing that as a Atbash ciphertext (i.e. $$A\leftrightarrow Z$$, $$B\leftrightarrow Y$$ $$\dots$$)

or

Make it Sophia-Lucy or Lucy-Sophia if you want to consider the red herring :) (If you take the italic letters in the plaintext, you get 'LUCY')

I've noticed that

there are four italicized letters which spell out "Lucy."

So, Joe wants to go to prom with

Lucy!

It's fairly simple, and I don't see any substitution here, unless I'm missing something.

• – WhatsUp Oct 8 '19 at 23:27
• @WhatsUp - Not my thoughts, exactly... :) – Voldemort's Wrath Oct 9 '19 at 0:04
• Sorry, I set that up. That is not the correct answer. There is some encryption involved. – Neo1009 Oct 9 '19 at 15:36
• @Neo1009 is it useful though, or just a red herring? – eru-cs Oct 9 '19 at 16:36
• @Neo1009 - Unfortunately, hints don't count as part of the "actual puzzle." – Voldemort's Wrath Oct 9 '19 at 22:00


text:
# Calculate delta h and delta e example

## Thermodynamics Calculator Shodor

Help understanding delta U (or H) of reaction Physics Forums. 12/06/2014В В· Thermodynamics is a branch of physics which deals with the energy and work of a system. For example, if we bring a hot The change in entropy (delta, 23/04/2009В В· With given reaction and thermodynamics quantities, you can calculate delta H easily. Make sure you write units along the calculation..

### 15 Thermodynamics Problems (Delta G Delta S Delta H)

Delta E Color Difference Algorithms. 15 Thermodynamics Problems (Delta G, E. II and III. 3) Calculate the entropy and-thermodynamics/15-thermodynamics-problems-delta-g-delta-s-delta-h, 9/09/2013В В· Delta H versus delta E Sehat Nauli. Loading Enthalpy - the four common ways to calculate delta H - Duration: 23:48. Scott Milam 20,656 views. 23:48..

What does negative delta S imply? Freezing or condensing water is a good example of a process where $\Delta S < 0$. = \Delta H^\circ -T\Delta S^\circ -RT \ln(P) 15/10/2008В В· Chemistry. What is delta H and delta E? and what is the formula of those 2? thanks?

### Thermodynamics Calculated Solution Calculate Delta H and

What Does Delta H Represent in Chemistry? Education. How do I calculate delta H with bond What is delta S in chemistry and what is the difference between delta H and How can one calculate the bond energy of, 5.02a Spontaneous vs Nonspontaneous Example of the the surroundings from Delta H. when you dealt with thermochemistry and learn to calculate the Delta.

### Understanding the Difference between Delta H and Delta S

Understanding the Difference between Delta H and Delta S. Isothermal Expansion. Isothermal Expansion of an ideal gas. You stay on the isotherm throughout the entire expansion which means that $$\Delta E = 0$$ and $$q = -w$$. Thermodynamics: Examples for chapter 3. 1. 2 is an ideal gas, calculate в€†H (i.e. per mole). To calculate change in.

9/09/2013В В· Delta H versus delta E Sehat Nauli. Loading Enthalpy - the four common ways to calculate delta H - Duration: 23:48. Scott Milam 20,656 views. 23:48. Use the data sheet to calculate Delta H, Delta S and Delta G at 25 degrees Celsius for each of the following reactions. In each case show that Delta G- Delta H-T

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text:
# Giải câu 7 trang 23 toán VNEN 9 tập 1

• 1 Đánh giá

Câu 7: Trang 23 sách VNEN 9 tập 1

Thực hiện phép tính:

a) - $\frac{1}{\sqrt{3} + 1}$ ; b) $\frac{\sqrt{2} - 1}{\sqrt{2} + 2}$ - $\frac{1}{1 + \sqrt{2}}$ + $\frac{\sqrt{2} + 1}{\sqrt{2}}$ ;

c) - 2 + $\frac{10 - x}{\sqrt{x} + 2}$ với x $\geq$ 0 ; d) $\frac{x\sqrt{x} - y\sqrt{y}}{\sqrt{x} - \sqrt{y}}$ với x $\geq$ 0, y $\geq$ 0 và x $\neq$ y.

Bài làm:

Giải câu a)

- $\frac{1}{\sqrt{3} + 1}$=$\frac{\sqrt{3} + 1}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$ - $\frac{\sqrt{3} - 1}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$ = $\frac{\sqrt{3} + 1 - \sqrt{3} + 1}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$ = $\frac{2}{3 - 1}$ = 1.

Giải câu b)

- $\frac{1}{1 + \sqrt{2}}$ + $\frac{\sqrt{2} + 1}{\sqrt{2}}$ = $\frac{\sqrt{2} - 1}{\sqrt{2}(1 + \sqrt{2})}$ - $\frac{1}{1 + \sqrt{2}}$ + $\frac{\sqrt{2} + 1}{\sqrt{2}}$ = $\frac{\sqrt{2} - 1}{\sqrt{2}(1 + \sqrt{2})}$ - $\frac{\sqrt{2}}{\sqrt{2}(1 + \sqrt{2})}$ + $\frac{(\sqrt{2} + 1)(1 + \sqrt{2})}{\sqrt{2}(1 + \sqrt{2})}$ = $\frac{\sqrt{2} - 1 - \sqrt{2} + 2 + 1 + 2\sqrt{2} }{\sqrt{2}(1 + \sqrt{2})}$ = $\frac{2 + 2\sqrt{2}}{\sqrt{2}(1 + \sqrt{2})}$ = $\sqrt{2}$.

Giải câu c)

- 2 + $\frac{10 - x}{\sqrt{x} + 2}$ = $\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{\sqrt{x} + 2}$ + $\frac{10 - x}{\sqrt{x} + 2}$ = $\frac{x - 4 + 10 - x}{\sqrt{x} + 2}$ = $\frac{6}{\sqrt{x} + 2}$

Giải câu d)

= $\frac{(\sqrt{x})^{3} - (\sqrt{y})^{3}}{\sqrt{x} - \sqrt{y}}$ = $\frac{(\sqrt{x} - \sqrt{y})((\sqrt{x})^{2} + \sqrt{x}\sqrt{y} + (\sqrt{y})^{2})}{\sqrt{x} - \sqrt{y}}$ = $(\sqrt{x})^{2}$ + $\sqrt{x}$$\sqrt{y}$ + $(\sqrt{y})^{2}$.

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Cập nhật: 07/09/2021