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# Is $dy/dx$ not a ratio?

In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.

I wonder if the author say this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?

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I suppose it can within non-standard analysis. – Raskolnikov Feb 9 '11 at 16:31
In the standard formulation of calculus, dy and dx are simply not numbers, so it doesn't make sense to ask this question. – Qiaochu Yuan Feb 9 '11 at 16:38
Although it sure behaves a lot like a ratio... for instance 1/(dy/dx) = dx/dy. – user7530 Mar 3 '11 at 16:53

Historically, when Leibniz conceived of the notation, $\frac{dy}{dx}$ was supposed to be a quotient: it was the quotient of the "infinitesimal change in $y$ produced by the change in $x$" divided by the "infinitesimal change in $x$".

However, the formulation of calculus with infinitesimals in the usual setting of the real numbers leads to a lot of problems. For one thing, infinitesimals can't exist in the usual setting of real numbers! Because the real numbers satisfy an important property, called the Archimedean Property: given any positive real number $\epsilon\gt 0$, no matter how small, and given any positive real number $M\gt 0$, no matter how big, there exists a natural number $n$ such that $n\epsilon\gt M$. But an "infinitesimal" $\xi$ is supposed to be so small that no matter how many times you add it to itself, it never gets to $1$, contradicting the Archimedean Property. Other problems: Leibniz defined the tangent to the graph of $y=f(x)$ at $x=a$ by saying "Take the point $(a,f(a))$; then add an infinitesimal amount to $a$, $a+dx$, and take the point $(a+dx,f(a+dx))$, and draw the line through those two points." But if they are two different points on the graph, then it's not a tangent, and if it's just one point, then you can't define the line because you just have one point. That's just two of the problems with infinitesimals. (See below where it says "However...", though).

So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a quotient, now it's a limit: $$\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}.$$ And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is not a quotient.

However, Leibniz's notation is very suggestive and very useful; even though derivatives are not really quotients, in many ways they behave as if they were quotients. So we have the Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du}\;\frac{du}{dx}$$ which looks very natural if you think of the derivatives as "fractions". You have the Inverse Function theorem, which tells you that $$\frac{dx}{dy} = \frac{1}{\quad\frac{dy}{dx}\quad},$$ again, almost "obvious" if you think of the derivatives as fractions. So, because the notation is so nice and so suggestive, we keep the notation even though the notation no longer represents an actual quotient, it now represents a single limit. In fact, Leibniz's notation is so good, so superior to the prime notation and to Newton's notation, that England fell behind all of Europe for centuries in mathematics and science because, due to the fight between Newton's and Leibniz's camp over who had invented Calculus and who stole it from whom (consensus is that they each discovered it independently), England's scientific establishment decided to ignore what was being done in Europe with Leibniz notation and stuck to Newton's... and got stuck in the mud in large part because of it.

(Differentials are part of this same issue: originally, $dy$ and $dx$ really did mean the same thing as those symbols do in $\frac{dy}{dx}$, but that leads to all sorts of logical problems, so they no longer mean the same thing, even though they behave as if they did).

So, even though we write $\frac{dy}{dx}$ as if it were a fraction, and many computations look like we are working with it like a fraction, it isn't really a fraction (it just plays one on television).

However... There is a way of getting around the logical difficulties with infinitesimals; this is called nonstandard analysis. It's pretty difficult to explain how one sets it up, but you can think of it as creating two classes of real numbers: the ones you are familiar with, that satisfy things like the Archimedean Property, the Supremum Property, and so on, and then you add another, separate class of reals numbers that includes infinitesimals and a bunch of other things. If you do that, then you can, if you are careful, define derivatives exactly like Leibniz, in terms of infinitesimals and actual quotients; if you do that, then all the rules of Calculus that make use of $\frac{dy}{dx}$ as if it were a fraction are justified because, in that setting, it is a fraction. Still, one has to be careful because you have to keep infinitesimals and regular real numbers separate and not let them get confused, or you can run into some serious problems.

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+1 Exceptionally good answer, very clear, very informative. – Orbling Feb 10 '11 at 3:09
As a physicist, I prefer Leibniz notation simply because it is dimensionally correct regardless of whether it is derived from the limit or from nonstandard analysis. With Newtonian notation, you cannot automatically tell what the units of $y'$ are. – rcollyer Mar 10 '11 at 16:34
@Kevin: Look at the history of math. Shortly after Newton and his students (Maclaurin, Taylor), all the developments in mathematics came from the Continent. It was the Bernoullis, Euler, who developed Calculus, not the British. It wasn't until Hamilton that they started coming back, and when they reformed math teaching in Oxford and Cambridge, they adopted the continental ideas and notation. – Arturo Magidin Mar 22 '11 at 1:42
Mathematics really did not have a firm hold in England. It was the Physics of Newton that was admired. Unlike in parts of the Continent, mathematics was not thought of as a serious calling. So the "best" people did other things. – André Nicolas Jun 20 '11 at 19:02
@BabakSorouh: Upvoting a bunch of answers of mine in a short time period, when you probably did not even have time to read them all, is not really very productive. And there is a pretty good chance that the votes will be undone by the software that checks for anomalous voting patterns in any case. Please, only up-vote that which you read and actually find worthy of the vote. – Arturo Magidin Jun 9 '12 at 4:57

Just to add some variety to the list of answers, I'm going to go against the grain here and say that you can, in an albeit silly way, interpret $dy/dx$ as a ratio of real numbers.

For every (differentiable) function $f$, we can define a function $df(x; dx)$ of two real variables $x$ and $dx$ via $$df(x; dx) = f'(x)\,dx.$$ Here, $dx$ is just a real number, and no more. (In particular, it is not a differential 1-form, nor an infinitesimal.) So, when $dx \neq 0$, we can write: $$\frac{df(x;dx)}{dx} = f'(x).$$

All of this, however, should come with a few remarks.

It is clear that these notations above do not constitute a definition of the derivative of $f$. Indeed, we needed to know what the derivative $f'$ meant before defining the function $df$. So in some sense, it's just a clever choice of notation.

But if it's just a trick of notation, why do I mention it at all? The reason is that in higher dimensions, the function $df(x;dx)$ actually becomes the focus of study, in part because it contains information about all the partial derivatives.

To be more concrete, for multivariable functions $f\colon R^n \to R$, we can define a function $df(x;dx)$ of two n-dimensional variables $x, dx \in R^n$ via $$df(x;dx) = df(x_1,\ldots,x_n; dx_1, \ldots, dx_n) = \frac{\partial f}{\partial x_1}dx_1 + \ldots + \frac{\partial f}{\partial x_n}dx_n.$$

Notice that this map $df$ is linear in the variable $dx$. That is, we can write: $$df(x;dx) = (\frac{\partial f}{\partial x_1}, \ldots, \frac{\partial f}{\partial x_n}) \begin{pmatrix} dx_1 \\ \vdots \\ dx_n \\ \end{pmatrix} = A(dx),$$ where $A$ is the $1\times n$ row matrix of partial derivatives.

In other words, the function $df(x;dx)$ can be thought of as a linear function of $dx$, whose matrix has variable coefficients (depending on $x$).

So for the $1$-dimensional case, what is really going on is a trick of dimension. That is, we have the variable $1\times1$ matrix ($f'(x)$) acting on the vector $dx \in R^1$ -- and it just so happens that vectors in $R^1$ can be identified with scalars, and so can be divided.

Finally, I should mention that, as long as we are thinking of $dx$ as a real number, mathematicians multiply and divide by $dx$ all the time -- it's just that they'll usually use another notation. The letter "$h$" is often used in this context, so we usually write $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$ rather than, say, $$f'(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}.$$ My guess is that the main aversion to writing $dx$ is that it conflicts with our notation for differential $1$-forms.

EDIT: Just to be even more technical, and at the risk of being confusing to some, we really shouldn't even be regarding $dx$ as an element of $R^n$, but rather as an element of the tangent space $T_xR^n$. Again, it just so happens that we have a canonical identification between $T_xR^n$ and $R^n$ which makes all of the above okay, but I like distinction between tangent space and euclidean space because it highlights the different roles played by $x \in R^n$ and $dx \in T_xR^n$.

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Cotangent space. Also, in case of multiple variables if you fix $dx^2,\ldots,dx^n = 0$ you can still divide by $dx^1$ and get the derivative. And nothing stops you from defining differential first and then defining derivatives as its coefficients. – Alexei Averchenko Feb 11 '11 at 2:30
Well, canonically differentials are members of contangent bundle, and $dx$ is in this case its basis. – Alexei Averchenko Feb 11 '11 at 5:54
Maybe I'm misunderstanding you, but I never made any reference to differentials in my post. My point is that $df(x;dx)$ can be likened to the pushforward map $f_*$. Of course one can also make an analogy with the actual differential 1-form $df$, but that's something separate. – Jesse Madnick Feb 11 '11 at 6:18
Sure the input is a vector, that's why these linearizations are called covectors, which are members of cotangent space. I can't see why you are bringing up pushforwards when there is a better description right there. – Alexei Averchenko Feb 11 '11 at 9:12
I just don't think pushforward is the best way to view differential of a function with codomain $\mathbb{R}$ (although it is perfectly correct), it's just a too complex idea that has more natural treatment. – Alexei Averchenko Feb 12 '11 at 4:25

It is best to think of "d/dx" as an operator which takes the derivative, with respect to x, of whatever expression follows.

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It is not a ratio, just as $dx$ is not a product.

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I wonder what motivated the downvote. I do find strange that students tend to confuse Leibniz's notation with a quotient, and not $dx$ (or even $\log$!) with a product: they are both indivisible notations... My answer above just makes this point. – Mariano Suárez-Alvarez Feb 10 '11 at 0:12
I think that the reason why this confusion arises in some students may be related to the way in which this notation is used for instance when calculating integrals. Even though as you say, they are indivisible, they are separated "formally" in any calculus course in order to aid in the computation of integrals. I suppose that if the letters in $\log$ where separated in a similar way, the students would probably make the same mistake of assuming it is a product. – Adrián Barquero Feb 10 '11 at 4:09
I once heard a story of a university applicant, who was asked at interview to find $dy/dx$, didn't understand the question, no matter how the interviewer phrased it. It was only after the interview wrote it out that the student promptly informed the interviewer that the two $d$'s cancelled and he was in fact mistaken. – jClark94 Jan 30 '12 at 19:54

My favorite "counterexample" to the derivative acting like a ratio: the implicit differentiation formula for two variables. We have $$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$$

The formula is almost what you would expect, except for that pesky minus sign.

See http://en.wikipedia.org/wiki/Implicit_differentiation#Formula_for_two_variables for the rigorous definition of this formula.

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Typically, the $\frac{dy}{dx}$ notation is used to denote the derivative, which is defined as the limit we all know and love (see Arturo Magidin's answer). However, when working with differentials, one can interpret $\frac{dy}{dx}$ as a genuine ratio of two fixed quantities.

Draw a graph of some smooth function $f$ and its tangent line at $x=a$. Starting from the point $(a, f(a))$, move $dx$ units right along the tangent line (not along the graph of $f$). Let $dy$ be the corresponding change in $y$.

So, we moved $dx$ units right, $dy$ units up, and stayed on the tangent line. Therefore the slope of the tangent line is exactly $\frac{dy}{dx}$. However, the slope of the tangent at $x=a$ is also given by $f'(a)$, hence the equation

$$\frac{dy}{dx} = f'(a)$$

holds when $dy$ and $dx$ are interpreted as fixed, finite changes in the two variables $x$ and $y$. In this context, we are not taking a limit on the left hand side of this equation, and $\frac{dy}{dx}$ is a genuine ratio of two fixed quantities. This is why we can then write $dy = f'(a) dx$.

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The notation $dy/dx$ - in elementary calculus - is simply that: notation to denote the derivative of, in this case, $y$ w.r.t. $x$. (In this case $f'(x)$ is another notation to express essentially the same thing, i.e. $df(x)/dx$ where $f(x)$ signifies the function $f$ w.r.t. the dependent variable $x$. According to what you've written above, $f(x)$ is the function which takes values in the target space $y$).

Furthermore, by definition, $dy/dx$ at a specific point $x_0$ within the domain $x$ is the real number $L$, if it exists. Otherwise, if no such number exists, then the function $f(x)$ does not have a derivative at the point in question, (i.e. in our case $x_0$).

For further information you can read the Wikipedia article: http://en.wikipedia.org/wiki/Derivative

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So glad that wikipedia finally added an entry for the derivative...  – The Chaz 2.0 Aug 10 '11 at 17:50
I think I laugh at almost every comment you post! :) – Steve D Nov 5 '11 at 2:54
@Steve: I wish there were a way to collect all the comments that I make (spread across multiple forums, social media outlets, etc) and let you upvote them for humor. Most of my audience scoffs at my simplicity. – The Chaz 2.0 Jan 27 '12 at 21:39

$\frac{dy}{dx}$ is not a ratio - it is a symbol used to represent a limit.

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# 2009-10-12 Diaspora

Last edit

Summary: Maybe I'll take another look at my friend's copy before buying anything.

Changed:

< with free form aspects à la [[Subjective Fudge of Zo]].
< I wonder whether that's what Diaspora is about at all.

to

> with free form qualities à la [[Subjective Fudge of Zo]].
> Too bad the skill list available from their website
> is a plan ol' skill list.

I’ve selectively read the Spirit of the Century rule book in anticipation of a one-shot. It’s called a pick-up game, but it’s not easy to pick up – it just seems to lend itself to little planning and episodic play. I discovered Diaspora at a friend’s place, and it seemed smaller.

I like small books! Hopefully the rules are quicker to read.

Since we’re also playing a Mongoose Traveller game, I’m interested. I also like the rant at the beginning of the 16 page thread on rpg.net.

I’m interested in replacing the skill list with free form qualities à la Subjective Fudge of Zo. Too bad the skill list available from their website is a plan ol’ skill list.

But unfortunately the price is not my friend:

 Subtotal $34.95 Shipping and Handling$12.89 Total $47.84 Maybe I’ll take another look at my friend’s copy before buying anything. ## Comments Last edit Summary: fixing links Changed: < -- AlexSchroeder 2009-10-16 11:06 UTC to > -- [[AlexSchroeder]] 2009-10-16 11:06 UTC Changed: < -- AlexSchroeder 2009-10-16 11:21 UTC to > -- [[AlexSchroeder]] 2009-10-16 11:21 UTC Changed: < -- AlexSchroeder 2009-10-16 18:51 UTC to > -- [[AlexSchroeder]] 2009-10-16 18:51 UTC Hey, I looked at my Lulu shopping cart again, and shipping and handling went up by a dollar:$13.97! Weird.

AlexSchroeder 2009-10-16 11:06 UTC

Buying it from Sphärenmeister in Germany is much better:

 Bezeichnung GP Diaspora - Hard SF FATE 28,95 € Versandmethode 1,80 € Gesamtbetrag ohne !MwSt. 28,74 €

Often small shops will not sell without VAT to Switzerland. This is cool. About \$43.

AlexSchroeder 2009-10-16 11:21 UTC

You probably know this already, but I read on an amazon.de info page that certain tax exemptions exist when importing from Germany to Switzerland. Sure enough, the package arrived without me having to pay anything extra.

– Marco 2009-10-16 14:38 UTC

Yup. No VAT if and only if 7.6% of the package does not exceed CHF 5.- – otherwise you’ll pay 7.6% + CHF 18.- service charge. A source of much annoyance!!

AlexSchroeder 2009-10-16 18:51 UTC

Please make sure you contribute only your own work, or work licensed under the GNU Free Documentation License. See Info for text formatting rules. You can edit the comment page if you need to fix typos. You can subscribe to new comments by email without leaving a comment.

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# Prove that it is possible to trisect 72 degrees

## Homework Statement

Prove that it is possible to trisect 72 degrees.

## Homework Equations

$$cos(3\theta)=4cos^3(\theta)-3cos(\theta)$$

## The Attempt at a Solution

well,72/3=24, So I need to see if cos24 is constructible, Which means that it is a root of a polynomial with a degree that is a power of 2.(?)I think.

I did this:$$cos(72)=\frac{\sqrt5-1}{4}$$
and $$Let \theta=24$$
so then letting $$cos\theta=x$$ in the equation above,
I get:
$$(\sqrt5)=(16x^3+12x+1)$$
Now I get a little confused about what I'm trying to do.
Here's what I need help with:
Firstly, am I on the right track here?
Secondly, Is the "degree of a power of 2" (above) correct?
What is it that I need to do with this polynomial from here?
I've confused myself.
CC

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HallsofIvy
Homework Helper

## Homework Statement

Prove that it is possible to trisect 72 degrees.

## Homework Equations

$$cos(3\theta)=4cos^3(\theta)-3cos(\theta)$$

## The Attempt at a Solution

well,72/3=24, So I need to see if cos24 is constructible, Which means that it is a root of a polynomial with a degree that is a power of 2.(?)I think.
Yes, that is (almost) correct. A number is "constructible" if and only if it is "algebraic of order a power of 2" which means it is a zero of a polynomial with integer coefficients of degree a power of two. The "with integer coefficients" is why I said "almost".

I did this:$$cos(72)=\frac{\sqrt5-1}{4}$$
and $$Let \theta=24$$
so then letting $$cos\theta=x$$ in the equation above,
I get:
$$(\sqrt5)=(16x^3+12x+1)$$
Not quite. In the first place, the right hand side should be
$$16x^3- 12x+ 1$$
In the second, you can't have $\sqrt{5}$- that's not an integer. What you want to say is that if x= cos(24) , then x satisfies
$$5= (16x^3- 12x+ 1)^2[/itex] which is a 6th degree equation. Now, you need to show that it can be reduced to either a 4th or 2nd degree equation- that it has either rational numbers or square roots as solutions. I'm not at all sure how you are going to do that! Now I get a little confused about what I'm trying to do. Here's what I need help with: Firstly, am I on the right track here? Secondly, Is the "degree of a power of 2" (above) correct? What is it that I need to do with this polynomial from here? I've confused myself. CC Hey, Thanks for looking at this. Now I'm not QUITE so cunfused. I'll fix my algebra and see where I can get. CC OK Please let me know of this is correct. I fixed my sign error(der) and squared the thing to get rid of the square root: [tex]5=256x^6-384x^4+32x^3+144x^2-24x+1$$
$$256x^6-384x^4+32x^3+144x^2-24x-4=0$$
Which factors like this:
$$4(4x^2+2x-1)(16x^4-8x^3-16x^2+8x+1)$$
The quadratic factor has square roots as the zeros and the 4th degree has cos24 as one of the roots. So x=cos24 satisfies that 4th degree polynomial factor with integer coefficients. 4 is a degree of a power of 2, so cos24 is constructible?
Am I any closer?
CC

HallsofIvy
Homework Helper
Wow, I am impressed! When I wrote my previous reply, I was tempted to add "Good luck with trying to factor that"!
Yes, if cos(24) satisfies the original equation (and you say it does) then it must satisfy one or the other of the factors equal to 0. You should be able to solve 4x2+ 2x-1= 0 to determine if it is that one but it really doesn't matter- cos(24) is either algebraic of order 4 or algebraic of order 2. In either case it is constructible.

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Hurkyl
Staff Emeritus
Gold Member
Some ("most", actually) irreducible 4-th degree polynomials have roots that are not constructible.

HallsofIvy
Homework Helper
It ocurred to me after I responded that you do need to show that the 4th degree polynomial is irreducible.

If the second degree polynomial happened to be reducible then it would follow that the root is either algebraic of order 1 or order 2, both of which are algebraic or order a power of 2.

If the 4th degree factor were separable, then it would either be factored into two quadratic factors (in which case x would be algebraic of order 1 or 2, both powers of 2) or factored into a cubic and a first degree factor. In that last case, where there is a first degree factor, then
$$(16x^4-8x^3-16x^2+8x+1)$$
has a rational root. By the "rational root theorem", any rational roots must have denominator that evenly divides 16: i.e. $\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$ or [/itex]\pm 16[/itex] and denominator that evenly divides 1: i.e. $\pm 1$. In other words, the only possible rational roots are $\pm 1$, $\pm \frac{1}{2}$, $\pm \frac{1}{4}$, $\pm\frac{1}{8|}$, or $\pm\frac{1}{16}$.

Do any of those satisfy the equation?

Now, you need to show that it can be reduced to either a 4th or 2nd degree equation- that it has either rational numbers or square roots as solutions. I'm not at all sure how you are going to do that!
I'm a little confused about this. The 4th degree polynomial that I got has no rational roots, so that means that cos24 isn't rational, but it's constructible? Yes?
And also, from the quadratic, can I conclude that cos72 is constructible? That one has$$\frac{\sqrt5-1}{4}=cos72$$ as a root.
And then since cos72 is constructible, so is a regular pentagon?
I thought that this might be true because 360/5=72.Is this the angle of interest? Or is there another method to show the pentagon is constructible?

I have attempted to prove that the hexagon is constructible and I keep getting a cubic. i know I'm wrong. That's another post tho.

BTW, on that factorization, I had gotten stuck because I had the wrong sign the first time. I scrawled about 3 pages trying to figure it out and then I busted out the TI89. Don't be too impressed.
CC

Some info on trisection

The ever-dubious wikipedia has some info on trisection -- full disclaimer, I wrote it -- at
< http://en.wikipedia.org/wiki/Angle_trisection >. The book by Ian Stewart I own, and is
available in updated editions.

Another reference that trisects 72 degrees (big .pdf file, search for "72") is
< http://www.math.uncc.edu/~droyster/courses/spring02/classnotes/Chapter03.pdf [Broken] >.
The construction is a little messy, but looks valid to me.

dino, not "dubi," -- too fast on the keyboard.

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Sometimes too much knowledge can distract us in the wrong direction. This problem actually doesn't require anything about the forms of equations for cos 24. Its a much simpler problem.

If you have a 72-degree angle, then construct a 60-degree angle, and subtract from 72, leaving a 12-degree angle. Double it, and you have the 24-degree angle, which is 1/3 of 72.


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# What is the balancing mass required to balance a mass of 5 kg, at a distance of 2 cm from the axis of rotation? (Assume that balancing mass is placed opposite to the unbalanced mass and 1 cm away from the axis of rotation)

1. 5 kg
2. 10 kg
3. 15 kg
4. 0.2 kg

Option 2 : 10 kg

## Balancing of Multiple Rotating Masses MCQ Question 1 Detailed Solution

Concept:

Balancing:

• It is the process of designing or modifying machinery so that the unbalance is reduced to an acceptable level and if possible is eliminated entirely.
• In rotation unbalancing, unbalance force ( Fun ) is constant in magnitude but changing in direction.
• Unbalanced rotating force ( Fun )  = m × r × ω2

Calculation:

Given:

m1r1ω12 = m2r2ω22

5 × 2 × ω2 = m × 1 × ω2

m = 10 kg

Hence the required mass at a distance of 1 cm will be 10 kg.

# A mass m1 attached to a shaft at radius r1, rotating with angular velocity ω rad/s, can be balanced by another single mass m2 which is attached to the opposite side of the shaft at radius r2, in the same plane, if

1. m1r2 = m2r1
2. m1r1 = m2r2
3. m1r1ω1 = m2r2ω2
4. m1r2ω1 = m2r1ω2

Option 2 : m1r1 = m2r2

## Balancing of Multiple Rotating Masses MCQ Question 2 Detailed Solution

Concept:

The shaft is rotating in a plane with angular velocity ω rad/sec.

The mass m1 is attached at a distance of r1 and mass m2 is attached at a distance of r2

Now in the same plane according to internal balancing we know that,

m1r1ω2 = m2r2ω

m1r1 = m2r2   (∵ Angular velocity ω is same for both the masses as both of them are attached to the same shaft or ω1 = ω= ω)

Here, the center of gravity lies on the same axis.

RA = RB =0 means no dynamic reactions at the supports. Therefore, the shaft is free from any dynamic bending stresses.

• There are two types of balancing i.e. External balancing and Internal balancing.
• Balancing is generally referred as external balancing.

# Two masses m are attached to opposite sides of a rigid rotating shaft in the vertical plane. Another pair of equal masses m1 is attached to the opposite sides of the shaft in the vertical plane as shown in figure. Consider m = 1 kg, e = 50 mm, e1 = 20 mm, b = 0.3 m, a = 2 m and a1 = 2.5 m. For the system to be dynamically balanced, m1 should be _______ kg.

## Balancing of Multiple Rotating Masses MCQ Question 3 Detailed Solution

Concept:

For dynamic equilibrium,

∑ F = 0 & ∑ M = 0 about any is valid.

Calculation:

Given:

m = 1 kg, e = 50 mm, e1 = 20 mm, b = 0.3 m, a = 2 m, a1 = 2.5 m.

By symmetry we have,

meω2 + m1e1ω2 = m1e1ω2 + meω2

Clearly ∑ F = 0

Now for moment,

We will balance all the moment due to forces except reaction, about O.

meω2 × (b) + m1e1ω2 × a1– meω2 × (a + b) = 0

m1e1ω2 × a1– meω2 × a = 0

$${m_1} = \frac{{mae}}{{{e_1}{a_1}}}$$

$${m_1} = \frac{{1\left( 2 \right)\left( {50} \right)}}{{\left( {2.5} \right)\left( {20} \right)}}$$

m1 = 2kg

# Three masses are connected to a rotating shaft supported on bearings A and B as shown in the figure. The system is in a space where the gravitational effect is absent. Neglect the mass of shaft and rods connecting the masses. For m1 = 10 kg, m2 = 5kg and m3 = 2.5 kg and for a shaft angular speed of 1000 radian/s, the magnitude of the bearing reaction (in N) at location B is ________

## Balancing of Multiple Rotating Masses MCQ Question 4 Detailed Solution

Concept:

For static balancing:

∑fX = ∑ mrw2cosθ and ∑fy = = ∑mrw2sinθ = 0

where, m = mass of element, r = distance of element from the fixed point, ω = Angular velocity of masses.

Calculation:

Given:

m1 = 10 kg, m2 = 5 kg, m3 = 2.5 kg, r1 = 0.1 m, r2 = 0.2 m, r3 = 0.4 m, ω = 1000 rad\sec

Now, we know that

∑fX = ∑ mrω2cosθ

∑fx = [10 × 0.1 × ω2 × cos 0°] – [5 × 0.2 × ω2 × cos 60°] – [2.5 × 0.4 × w2 × cos 60°] = 0

and

∑fy  = ∑mrω2sinθ

∑f= [0.1 × 10 × ω2 sin 0°] + [5 × 0.2 × ω2 sin 60°] – [2.5 × 0.4 × ω2 sin 60°] = 0

Resultant force is given by

$${F_{resultant}} = \sqrt {{{\left( {\sum {F_X}} \right)}^2} + {{\left( {\sum {F_y}} \right)}^2}} = 0$$

Since this is a balanced system so, net force on bearing is zero and so the magnitude of bearing reaction at point B is zero.

# What is the condition for Dynamic balancing of revolving masses?

1. Force polygon is closed
2. Couple polygon is closed
3. Neither force polygon nor couple polygon is closed
4. Both force and couple polygons are closed

Option 4 : Both force and couple polygons are closed

## Balancing of Multiple Rotating Masses MCQ Question 5 Detailed Solution

Concept:

Types of balancing:

DYNAMIC BALANCING: When several masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also form couples. A system of rotating masses is in dynamic balance when there does not exist any resultant centrifugal force as well as the resultant couple.

Dynamic balance:

$$\begin{array}{l} \sum F = 0 \Rightarrow \sum {m\mathop ω \nolimits^2 r = 0} \\ \sum M = 0 \Rightarrow \sum {m\mathop ω \nolimits^2 rl = 0} \end{array}$$

m = mass of the revolving object, ω = Angular velocity of the revolving mass, r = crank radius, l = distance of the revolving mass from the line of rotation

STATIC BALANCING: A system of rotating masses is said to be in static balance if the combined mass centre of the system lies on the axis of rotation.

Static balance: ΣF = 0

These are the cases of balancing:

1. Balancing of a single rotating mass by a single mass rotating in the same plane.

(Another rotating mass placed diametrically opposite in the same plane balances the unbalanced mass.)

2. Balancing of a single rotating mass by two masses rotating in different planes.

(Two masses placed in two different parallel planes balance the unbalanced mass.)

3. Balancing of different masses rotating in the same plane.

4. Balancing of different masses rotating in different planes.

# A system in dynamic balance implies that

1. the system is critically damped
2. there is no critical speed in the system
3. the system is also statically balanced
4. there will be absolutely no wear of bearings

Option 3 : the system is also statically balanced

## Balancing of Multiple Rotating Masses MCQ Question 6 Detailed Solution

Explanation:

If the rotor is statically balanced, it will not roll under the action of gravity, regardless of the angular position of the rotor. The requirement for static balance is that the centre of gravity of the system of masses is at the axis of rotation.

Static balance: ΣF = 0

A rotating system of mass is in dynamic balance when the rotation does not produce any resultant centrifugal force or couple.

Dynamic balance: ΣF = 0 and ΣM = 0

It can be said that if the rotor is dynamically balanced then it is also statically balanced. The converse is not true for all the rotors. A statically balanced rotor is not always dynamically balanced, the exception being the single plane rotors.

# In order to have a complete balance of the several revolving masses in different planes

1. the resultant force must be zero
2. the resultant couple must be zero
3. both the resultant force and couple must be zero
4. none of these

Option 3 : both the resultant force and couple must be zero

## Balancing of Multiple Rotating Masses MCQ Question 7 Detailed Solution

The unbalanced forces in the reference plane are m1r1ω2, m2r2ωand m3r3ω2 acting radially outwards. The unbalanced couples in the reference plane are m1r1ω2l1, m2r2ω2l2 and m3r3ω2lwhich may be represented by vectors parallel to the respective force vectors, i.e., parallel to the respective radii of m1, m2 and m3.

For complete balancing of the rotor, the resultant force and the resultant couple both should be zero. i.e., m1r1ω2 + m2r2ω+ m3r3ω= 0                         ________ (1)

and

m1r1l1ω2 + m2r2l2ω+ m3r3l3ω= 0          ----(2)

If the eq.(1) and (2) are not satisfied, then there are unbalanced forces and couples.

A mass placed in the reference plane may satisfy the force equation but the couple equation is satisfied only by two equal forces in different transverse planes. Thus, in general, two planes are needed to balance a system of rotating masses.

Therefore, in order to satisfy eq. (1) and (2) introduce two counter-masses mc1 and mc2 at radii rc1 and rc2 respectively. Then eq. (1) may be written as

m1r1ω2 + m2r2ω+ m3r3ω+ m c1r c1ω2 + m c2r c2ω= 0         ----(3)

m1r1 + m2r+ m3r3 + mc1rc1 + mc2rc2 = 0           ----(3.1)

Σmr + mc1rc1 + mc2rc2 = 0             ----(3.4)

Let the two counter masses be placed in transverse planes at axial locations O and Q, i.e., the counter mass mc1 be placed in the reference plane and the distance of the plane of mc2 be lc2 from the reference plane.

Equation (2) modifies to (taking moments about O)

m1r1l1ω+ m2r2l2ω+ m3r3l3ω+ mc2rc2lc2ω= 0            ----(4)

m1r1l1 + m2r2l2 + m3r3l3 + mc2rc2lc2 = 0             ----(4.1)

Σmrl + mc2rc2 = 0                ----(4.2)

# An unbalanced couple of magnitude 300 N-m is noticed on a shaft of length 200 cm. The dynamic reactions at the bearings are

1. 300 N and -300 N
2. 300 N and 300 N
3. 150 N and 150 N
4. 150 N and -150 N

Option 4 : 150 N and -150 N

## Balancing of Multiple Rotating Masses MCQ Question 8 Detailed Solution

Concept:

The dynamic reaction is due to the unbalanced couple, the static reaction is due to the weight of the shaft.

The dynamic reaction due to the couple on the shaft will be equal and opposite of each other.

This reaction is given by

$$R = \pm \frac{C}{L}$$

Where R is the reaction and C is the unbalanced couple and L is the length between the supports.

Calculation:

Given that the unbalanced couple magnitude is 300 N-m;

L = 200 cm = 2 m;

∴ Reactions will be $$R = \; \pm \frac{{300\;N - m}}{{2\;m}} = \; \pm \;150\;N$$

# Two masses are attached to opposite sides of a rigid rotating shaft in vertical plane. Another pair of equal mass m1 is attached to the opposite sides of shaft in the vertical plane as shown in figure. Consider m = 1 Kg, e = 50 mm, e1 = 20 mm, b = 0.3 m, a = 2 m and a1 = 2.5 m. For the system to be dynamically balanced m1 should be

1. 4 kg
2. 2 kg
3. 3 kg
4. 5 kg

Option 2 : 2 kg

## Balancing of Multiple Rotating Masses MCQ Question 9 Detailed Solution

Concept:

For dynamic equilibrium,

∑ F = 0 & ∑ M = 0 about any is valid.

Calculation:

Given:

m = 1 kg, e = 50 mm, e1 = 20 mm, b = 0.3 m, a = 2 m, a1 = 2.5 m.

By symmetry we have,

meω2 + m1e1ω2 = m1e1ω2 + meω2

Clearly ∑ F = 0

Now for moment,

We will balance all the moment due to forces except reaction, about point O

meω× (b) + m1e1ω× a– meω× (a + b) = 0

m1e1ω× a– meω× a = 0

$${m_1} = \frac{{mae}}{{{e_1}{a_1}}}$$

$$m_1~=~\frac{(1)(2)(50)}{(20)(2.5)}$$

m1 = 2 kg.

# In case of rotating masses, the magnitude of the balancing mass is ______ when the speed of the shaft is doubled.

1. Doubled
2. Halved
3. Unaffected

Option 3 : Unaffected

## Balancing of Multiple Rotating Masses MCQ Question 10 Detailed Solution

Explanation:

Unbalanced is the unequal distribution of the weight of a rotor about its rotating axis. When a rotor is unbalanced, it imparts vibratory forces to the rotor. Unbalance causes vibrations on the machine.

There are two types of balancing:

Static balancing:

If the combined mass centre of the system lies on the axis of rotation.

$$\begin{array}{l} \sum F = 0 \Rightarrow {m_1}{r_1}{\omega ^2} + {m_2}{r_2}{\omega ^2} + {m_3}{r_3}{\omega ^2} = 0\\ \Rightarrow {m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} = 0 \;\;\;eq(i)\end{array}$$

Dynamic Balancing:

When serval masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also form couples.

$$\begin{array}{l} \sum F = 0 \Rightarrow {m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} = 0\\ \sum C = 0 \Rightarrow {m_1}{r_1}{l_1} + {m_2}{r_2}{l_2} + {m_3}{r_3}{l_3} = 0 \;\;\;eq(ii)\end{array}$$

We can see that eq (i) and eq (ii) does not contain the term containing speed of the shaft, thus it is unaffected.


text:
Diagonals in diamons/rhombus

Rhombus ABCD has side length AB = 4 cm and a length of one diagonal of 6.4 cm. Calculate the length of the other diagonal.

Result

u2 =  4.8 cm

Solution:

$a=4 \ \\ u_{1}=6.4 \ \\ a^2=(u_{1}/2)^2 +(u_{2}/2)^2 \ \\ u_{2}=2 \cdot \ \sqrt{ a^2-(u_{1}/2)^2 }=2 \cdot \ \sqrt{ 4^2-(6.4/2)^2 }=\dfrac{ 24 }{ 5 }=4.8 \ \text{cm}$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

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Tips to related online calculators
Pythagorean theorem is the base for the right triangle calculator.

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text:
## Memory cell based on a $\varphi$ Josephson junction    [PDF]

E. Goldobin, H. Sickinger, M. Weides, N. Ruppelt, H. Kohlstedt, R. Kleiner, D. Koelle
The $\varphi$ Josephson junction has a doubly degenerate ground state with the Josephson phases $\pm\varphi$. We demonstrate the use of such a $\varphi$ Josephson junction as a memory cell (classical bit), where writing is done by applying a magnetic field and reading by applying a bias current. In the "store" state, the junction does not require any bias or magnetic field, but just needs to stay cooled for permanent storage of the logical bit. Straightforward integration with Rapid Single Flux Quantum logic is possible.
View original: http://arxiv.org/abs/1306.1683


text: Yes, by the theorem $\displaystyle 7^p+1\equiv8\pmod{p}$ and $\displaystyle 8\not\equiv 0\pmod{p}$ since p is odd.


text:
Create a GUI Piano [closed]

Challenge

Create a GUI Keyboard with as few characters as possible.

Example

Because this was an assignment in one of my courses, I can't show the source code. However, here is a screen shot of my keyboard.

In this example, my keys were of type JButton and I used a Midi Synthesizer to produce the sound (with the default ADSR envelope values).

Rules

• You are allowed to use standard external libraries.
• Be creative with your sound. You can use 8-bit, a sitar, etc.
• For simplicities sake, you may have five keys; black and white, from C to E (the first five keys on my keyboard).
• Most importantly... showcase your work!

NOTICE: Depending on what language you choose to work with, this may be a rather large task.

This is my first question on SE Code Golf. If anything is unclear, please ask for further details.

EDIT: The due date for this challenge will be 9/22/12. If you post an answer after this date, I will look over it regardless (and possibly +1 it).

• Restrictions on which language to use are not liked very much here. Consider dropping your restriction or name an important reason. – FUZxxl Sep 15 '12 at 18:08
• @FUZxxl As stated under the Example section, this was a term project for our Java class. It is still being used as a term project for that very class. But I suppose I'm just paranoid so I'll drop the restrictions. I think you meant which languages not to use... but whatever, I removed them. – Rob Sep 15 '12 at 19:02
• What are the minimal requirements to be considered a "GUI keyboard"? I infer from what's already present that it must display a GUI and produce some sound, but what restrictions are there on: a) the input mechanism; b) the sound envelope; c) the scale used; d) the accuracy of the tuning; e) the proportions of the keys? – Peter Taylor Sep 15 '12 at 22:02
• @MikeDtrick, that answers 0/5 of my questions. I'm not asking how your implementation worked: I'm asking how I can know whether my (hypothetical) implementation is a valid competitor, because there's no point shortening an entry by 20% if so doing takes it from being a valid entry to an invalid one. – Peter Taylor Sep 16 '12 at 8:03
• @MikeDtrick: For example, you could require that the buttons look exactly like the ones in your example, pixel by pixel. At the other extreme you could allow any arrangement of five GUI buttons of any type. – han Sep 16 '12 at 10:25

Mathematica 319 259 255

Edit: Keys now depress (as buttons) when clicked.

This will play the grand piano notes {"C","C#","D","D#","E"}, where "C" is middle C. z[n_] plays the note.

z@n_ := EmitSound@Sound[SoundNote[n, .3, 1]]; w = {10, 300}; b = {35, 180};
Graphics[Inset[Button["", z[#[[1]]], Background -> If[#[[2]] == w, None, Black],
ImageSize -> #〚2〛], #〚3〛] & /@ {{"C", w, {-.4, 0}}, {"D", w, {0, 0}}, {"E", w, {.4, 0}},
{"C#", b, {-.2, 0.31}}, {"D#", b, {.2, 0.31}}}, PlotRange -> 1]

The keyboard can be extended to 18-keys by using fewer than double the characters:

z@n_ := EmitSound@Sound@SoundNote[n, .3, 1];
w = {"C", "D", "E", "F", "G", "A", "B", "C5", "D5", "E5", "F5"};
b = {"C#", "D#", "", "F#", "G#", "A#", "", "C#5", "D#5"}; i = ImageSize; t = Thread;
l = List; s = Inset; m = Table; u = Button;
Graphics[Join[t[s[u @@@ t[l["", y /@ w, i -> {5, 350}]] /. y -> z, m[{90 k, 0}, {k, -5, 5}]]],
Delete[t[s[u @@@ t[l["", y /@ b, Background -> Black, i -> {28, 212}]] /.
y -> z, m[{90 k + 45, 220}, {k, -5, 3}]]], {{3}, {7}}]],
AspectRatio -> .45, PlotRange -> {{-500, 500}, {-610, 610}}, i -> {800, 430}]

• +1 I have no doubt in my mind that this will work... I just wish I could play on it. – Rob Sep 16 '12 at 0:08
• I left a .cdf version of the file in my DropBox at dropbox.com/sh/m3y0fs0v0nidqt5/UTv_0YGpz5 You can share this with others. There should be no licensing problems because it is being used for non-commercial, educational purposes. You will need to download the free Wolfram CDF player if you don't already have it. – DavidC Sep 16 '12 at 0:35
• David, I need w = {67, 300} to get your result; any idea why the difference? Also, may I edit this code to shorten it, if I am able? – Mr.Wizard Apr 6 '13 at 14:49
• Mr.Wizard. w = {67,300} works fine on v. 9 so if you want to change it, or for that matter, shorten any of the code, go right ahead. Adjusting the button size was hit or miss. Strange things happened for reasons I cannot explain. (For instance, adding more buttons affects the proportions of the original buttons.) – DavidC Apr 6 '13 at 16:14

Web page (840/796 characters)

>>> Start playing (Internet Explorer is not supported for multiple reasons; Google Chrome and Opera work best.)

I probably could make this a bit shorter, yet it is a good start. The lower score is after replacing all occurrences of   with the character itself and removing the keyword new, the latter change breaking Google Chrome compatibility.

Save this code as a text file with a name ending in .htm or .html and open it in Chrome or Opera (Safari might also work), or just open the solution's JSBin page to start playing. I reused the WAV file header from my solution to the Twinkle Twinkle Little Star code golf problem.

An important feature is that the sound diminishes as time passes. To observe this behavior, try holding down a key for a few seconds and listening to what happens.

Here is a more readable version of the code:

• +1 Works nice in Firefox 15, although I would have chosen a better sounding instrument. – DavidC Sep 17 '12 at 1:30

Groovy: 577 (703 with whitespaces)

The first 5 notes. Others could be added easily, it's somewhat dynamic.

Damn swing. Probably with a swing lib it would be better.

Plays through JFugue.

On groovy 2.0.2

import java.awt.event.*
class Note { def n; boolean s; def p() { new org.jfugue.Player().with {play n;close()}} }
notes=['C','C#','D','D#','E'].inject([]){ l,n -> l<< Note[n:n,s:n=~/#/]}
h=300
l=0
w=60
x=0
new groovy.swing.SwingBuilder().edt {
frame size:[notes.size()*30+30,h],
show:true,
defaultCloseOperation:javax.swing.JFrame.EXIT_ON_CLOSE,
{ l = layeredPane() }
notes.each { n ->
C=java.awt.Color
s=n.s
p=panel bounds:(s ? [x-15,0,w-30,h-100] : [x,0,w,h]),
background: s ? C.BLACK : C.WHITE,
border: lineBorder(1, color: C.BLACK)
if(!s)x+=w
}
}

R - 491 characters

I'm a little late but I just saw this post yesterday.

Works on a Mac, uses playRWave and packages tuneR and splancs.

a=array
x=c(7,2)
y=c(5,2)
z=c(1,1,3,3)
par(mar=rep(0,4))
plot(NA,xli=c(0,9),yli=c(0,3))
N=list(a(c(0,3,3,2,2,0,0,0,0,z,0),x),a(c(3,6,6,5,5,4,4,3,3,0,0,z,1,1,0),c(9,2)),a(c(6,6,7,7,9,9,6,0,z,0,0),x),a(c(2,4,4,2,2,z,1),y),a(c(5,7,7,5,5,z,1),y))
c=c(NA,NA,NA,1,1)
for(i in 1:5){polygon(N[[i]],c=c[i])}
h=c(261.63,293.66,329.63,277.18,311.13)
library(tuneR)
setWavPlayer("~/Library/Audio/playRwave")
repeat{P=data.frame(locator(1));play(sine(h[sapply(N,function(x)splancs::inout(P,x))],bit=16))}

Ungolfed:

par(mar=rep(0,4))
plot(NA,xlim=c(0,9),ylim=c(0,3)) #Create empty plot: due to fuzzy matching of arguments, xlim can be reduced to xli
N=list(array(c(0,3,3,2,2,0,0,0,0,1,1,3,3,0),dim=c(7,2)), #C polygon
array(c(3,6,6,5,5,4,4,3,3,0,0,1,1,3,3,1,1,0),dim=c(9,2)), #D polygon
array(c(6,6,7,7,9,9,6,0,1,1,3,3,0,0),dim=c(7,2)), #E polygon
array(c(2,4,4,2,2,1,1,3,3,1),dim=(5,2)), #Db polygon
array(c(5,7,7,5,5,1,1,3,3,1),dim=(5,2)))  #Eb polygon
c=c(NA,NA,NA,1,1) #Colors: by default 1 is "black"
for(i in 1:5){polygon(N[[i]],color=c[i])}
h=c(261.63,293.66,329.63,277.18,311.13) #Notes frequency in hertz: C4, D4, E4, Db4 and Eb4
library(tuneR)
setWavPlayer("~/Library/Audio/playRwave") #This can be change to other wav player I think
repeat{
P=data.frame(locator(1)) #Grab coordinates of selected point
H=h[sapply(N,function(x)splancs::inout(P,x))] #In which polygon does the selected point belong to, then map it to its ferquency
s=sine(H,bit=16) #By default create a 1sec note at the given frequency with 44100 sampling rate
play(s)
}


text:
let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume $f(z) = f(2z)$, prove that f is constant

1. $f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume that $f(z) = f(2z)$ for all $z \in \mathbb{C}$. Prove that f is constant...

Then we are supposed to use this result to solve the second question which is ...

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be analytic throughout $\mathbb{C}$ and satisfy $f(2z) = 2f(z)$ for all z. Prove that there exists $c \in \mathbb{C}$ such that $f(z) = cz$ for all z.

• do you mean $f$ is constant when you say $Z$ is constant? what have you tried? – user87543 Feb 7 '14 at 2:04
• ya sorry i changed that typo, I think the second part has something to do with cauchy-reimann equations, since thats what we've been doing in class. But in part 1 the function is just continuous, not analytic. So I'm not really sure. – user2253455 Feb 7 '14 at 2:10

Note that $f(z) = f({1 \over 2^n} z)$. Letting $n \to \infty$ gives $f(z) = f(0)$.

• Could you please explain to me how is it possible to take $f(z)=f(\frac{z}{2^n})$? – user114873 Apr 28 '17 at 9:31
• We are given that $f(z) = f ({1 \over 2} z )$, repeating gives $f(z) = f ({1 \over 2} z ) = f ({1 \over 2^2} z ) = f ({1 \over 2^3} z ) = \cdots$. – copper.hat Apr 28 '17 at 12:45

compute $$f'(z)=\frac{1}{2}f'(2z)2=f'(2z)$$ apply the answer to the first question to conclude $f'(z)=C$ for all $z$, so $f(z)=Cz+D$ and $$D=f(0)=2f(0)=2D \Rightarrow D=0$$

Working backward... $$f(z) = f(z/2) = f(z/4) = f(z/8) \cdots$$

Let $\epsilon >0$ be given. Then there is a $delta >0$ such that $|z|<\delta$ implies $|f(z)-f(0)|<\epsilon$. So for any we can chose an $n$ such that $z/2^n$ is as small as we please, so $|f(z)-f(0)|$ is less than any $\epsilon>0$. Hence

$$f(z) \equiv f(0)$$

For part (2), it is easy to see that $f(0)=2f(0)$ and hence $f(0)=0$. Let $g(z)=f(z)/z$. This has a removable singularity at $z=0$, so we can assume that $g$ is continuous.

$$g(2z)= f(2z)/(2z) = (2 f(z))/(2z) = f(z)/z= g(z)$$

Hence $g(z) = c$ a constant, and $f(z) = cz$

• I do not really understand why would most of the users post full solutions without missing any details when the OP has not showed any of his work.... you could have given some hint... It is your personal opinion anyways.... – user87543 Feb 7 '14 at 2:14
• Point taken. Here is the condition under which I try to give full answer: The OP is new, and OP has not accepted or voted on an answer and the question is more than 10 minutes old. Under these conditions, I feel that the OP should be encouraged and not feel unwelcome. Many first time users of the site do feel intimidated. – user44197 Feb 7 '14 at 2:22
• don't worry, the full solution doesn't make any sense to me either... i would have preferred some form of hint. – user2253455 Feb 7 '14 at 2:23
• Could you please explain to me how is it possible to take $f(z)=f(\frac{z}{2^n})?$ – user114873 Apr 28 '17 at 9:32


text:
Last edited by Dugor
Wednesday, May 13, 2020 | History

4 edition of Handbook of Gaussian basis sets found in the catalog.

# Handbook of Gaussian basis sets

## by Raymond Poirier

Written in English

Subjects:
• Molecular orbitals.,
• Gaussian basis sets (Quantum mechanics)

• Edition Notes

Bibliography: p. 28-33.

Classifications The Physical Object Statement Raymond Poirier, Roy Kari, Imre G. Csizmadia. Series Physical sciences data ;, 24 Contributions Kari, Roy., Csizmadia, I. G. LC Classifications QD461 .P63 1985 Pagination xii, 674 p. ; Number of Pages 674 Open Library OL3026491M ISBN 10 0444424938 LC Control Number 85006900

For the Pople's basis sets the following notaton is used: n-ij+G, or n-ijk+G when 1 diffuse s-type and p-type gaussian with the same exponents are added to a standard basis set on heavy atoms. The n-ij++G, or n-ijk++G are obtained by adding 1 diffuse s-type and p-type gaussian on heavy atoms and 1 diffuse s-type gaussian on hydrogens. 'This book is a guide to modeling and analyzing non-Gaussian and correlated data. There is clearly a need for such a book to help less experienced data scientists The data sets and models are well explained, and the limitations of each type of model on the various data sets is Cited by: 5.

Computational Quantum Chemistry removes much of the mystery of modern computer programs for molecular orbital calculations by showing how to develop Excel spreadsheets to perform model calculations and investigate the properties of basis sets. Using the book together with the CD-ROM provides a unique interactive learning tool. In addition, because of the integration of theory with Book Edition: 1.   STO-nG, kode basis set yang sering digunakan (T. Daniel Crawford, ). Himpunan basis diklasifikasikan menurut jumlah fungsi primitifnya dan cara pembentukan fungsi gaussian terluaskan. Himpunan basis minimal berisi satu fungsi basis untuk mewakili masing-masing (kulit dalam dan kulit valensinya) orbital atomik.

Physical Sciences Data, Volume Gaussian Basis Sets for Molecular Calculations provides information pertinent to the Gaussian basis sets, with emphasis on lithium, radon, and important ions. This book discusses the polarization functions prepared for lithium through radon for further improvement of the basis sets. Advanced Mathematics for Engineers Wolfgang Ertel translated by Elias Drotle and Richard Cubek October 1, and some important sections added such as radial basis functions, Gaussian processes and Exercise A permutation is a bijective mapping from a nite set onto itself. Applied.

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### Handbook of Gaussian basis sets by Raymond Poirier Download PDF EPUB FB2

Handbook of Gaussian Basis Sets: A Compendium for Ab-Initio Molecular Orbital Calculations (Physical Sciences Data) 1st Edition by R.

Poirier (Author), R. Kari (Author), I.G. Csizmadia Price: $Book Review. Handbook of gaussian basis sets: A compendium for ab‐initio molecular orbital calculations. By Raymond Poirier, Roy Kari, and Imre Csizmadia. Elsevier science publishers B.V., Amsterdem, Hans : Hans Ågren. Handbook of Gaussian basis sets: a compendium for ab-initio molecular orbital calculations. General. Guide to Choosing a Basis Set. Guide to the Appendices. References. Appendix A. Summary of Basis Sets Arranged by the Table Number. Appendix B. Summary of Basis Sets Arranged by Type and Size. Appendix C. Slater Exponents for Atoms He to Xe. Appendix D. Slater Exponents for STO-ENG Expansions. This chapter discusses different types of basis sets for electronic structure calculations with the main focus on Gaussian type basis sets for molecular calculations. After a brief overview of the basis set approximation in quantum chemistry, it summarizes and discusses the basic requirements of an ideal basis by: 9. Last updated on: 05 January [G16 Rev. C] Quick Links. Basis Sets; Density Functional (DFT) Methods; Solvents List SCRF. The values of the orbital exponents ($$\zeta s \text{ or } \alpha s$$) and the GTO-to-CGTO contraction coefficients needed to implement a particular basis of the kind described above have been tabulated in several journal articles and in computer data bases (in particular, in the data base contained in the book Handbook of Gaussian Basis Sets: A. Compendium for Ab initio Molecular Orbital. Last updated on: 22 August [G16 Rev. C] Quick Links. Basis Sets; Density Functional (DFT) Methods; Solvents List SCRF. Gaussian basis sets The STO-nG basis sets are rather unsatisfactory as they include only one contracted Gausssian (CG) per atomic orbital. Improved basis sets are obtained by including: •More than one CG per atomic orbital, e.g.: DZ (“doublezeta”),TZ (“triple zeta”),QZ (“quadruplezeta”). Contracted basis sets of double zeta (DZ) valence quality for H, He, first- and second-row atoms are constructed from fully-optimized Gaussian basis sets generated in this work. A single set of Gaussian polarization functions (d symmetry) optimized at the Mϕller-Plesset second-order (MP2) level was added to the DZ by: Basis sets2 Most methods require a basis set be specified; if no basis set keyword is included in the route section, then the STO-3G basis will be used. The following basis sets are stored internally in the Gaussian program (see references cited for full descriptions), listed below by their corresponding Gaussian keyword (with two exceptions):File Size: KB. Gaussian DFT Versus Plane-Wave DFT. Gaussian Basis Set. Parallel Efficient All-Electron. Core regions included in calculation. First row transition metals can readily be calculated. Ab Initio MD expensive. Pulay forces. Different basis sets for molecules and solids. PlaneWave Basis Set. Parallel Efficient Requires pseudopotentials to File Size: KB. Physical Sciences Data, Volume Gaussian Basis Sets for Molecular Calculations provides information pertinent to the Gaussian basis sets, with emphasis on lithium, radon, and important ions. This book discusses the polarization functions prepared for lithium through radon for further improvement of the basis : Anatomy of a Basis Set: H atom, STO-3G A basis set of Contracted Gaussian-Type Orbitals (CGTO’s) needs to specify the exponents (ζi’s) and the contraction coeﬃcients (ci’s). These are given below in the format used by the Gaussian program (exponents ﬁrst) H 0 S 3 0 File Size: KB. In the dual family sets of Faegri (27), which are the underlying Gaussian primitives used in the ANO-RCC (atomic natural orbital—relativistic and (semi-)core correlation) basis sets of Roos et al. (30), energy optimizations at the average-level Dirac–Hartree–Fock (DHF) level of theory were carried out using the lowest states arising from the 7s 2 6d 1 5f n − 1 configurations for all actinide elements (including Lr). Gaussian Basis Function. While the STO basis set was an improvement over hydrogenic orbitals in terms of computational efficiency, representing the STOs with Gaussian functions produced further improvements that were needed to accurately describe molecules. A Gaussian basis function has the form shown in Equation \ref{}. Note that in all. GAUSSIAN 09W TUTORIAL AN INTRODUCTION TO COMPUTATIONAL CHEMISTRY USING G09W AND AVOGADRO SOFTWARE Anna Tomberg [email protected] This is a quick tutorial that will help you to make your way through the ﬁrst steps of computational chemistry using Gaussian 09W software (G09).File Size: 1MB. Basis Function Regression. One trick you can use to adapt linear regression to nonlinear relationships between variables is to transform the data according to basis have seen one version of this before, in the PolynomialRegression pipeline used in Hyperparameters and Model Validation and Feature idea is to take our multidimensional linear model:$\$ y = a_0 + a_1.

We present a library of Gaussian basis sets that has been specifically optimized to perform accurate molecular calculations based on density functional theory.

It targets a Cited by: Gaussview/Gaussian Guide and Exercise Manual Introduction In this manual some of the principal features of the Gaussview and Gaussian programs are highlighted to enable the student to start working productively with both programs.

The examples/descriptions are inevitably brief and do not aim to be a comprehensive Size: 2MB. Medium basis sets based upon contractions of Gaussian primitives are developed for the third-row elements K through Zn. The basis functions generalize the G and G* sets Cited by: The start of a given basis set and AIMP is identified by the line where label" is defined below, in the input description to seward.

Then, comment lines, effective charge, and basis set follow, with the same structure that the all-electron Basis Set Library (see items 1. to 4. in Sec.) Next, the AIMP/ECP/PP is specified as follows.This is a pretty big topic, but the basic important qualities of an atom-centered basis set are: The number of separate, unlinked functions it uses at the valence level (the so-called *zeta* count), which allow the modeling of that many different electronic environments, practically speaking.


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# Projection of a Torus

(Difference between revisions)
Jump to: navigation, search
 Revision as of 11:04, 5 June 2009 (edit)← Previous diff Revision as of 11:14, 5 June 2009 (edit) (undo)Next diff → Line 6: Line 6: A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections. A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections. - |ImageDesc=The four-dimensional object is defined parametrically by $(x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v))$. A [[Stereographic Projection| stereographic projection]] is used to map this 4-d object into 3-d, using a projection point of $(0,0,0,\sqrt{2})$ for the first object in this page's main image. This projection is centered above the four-dimensional object, projecting the symmetric torus into three-dimensional space. The projection point is shifted to be closer to one part of the four-dimensional object than the other to create the second object in the main image, projecting an uneven object into three-dimensional space. + |ImageDesc=The four-dimensional object is defined [[Parametric Equations|parametrically]] by $(x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v))$. A [[Stereographic Projection| stereographic projection]] is used to map this 4-D object into 3-D, using a projection point of $(0,0,0,\sqrt{2})$ for the first object in this page's main image. This projection is centered above the four-dimensional object, projecting the symmetric torus into three-dimensional space. For the second object, the projection point is shifted to be closer to one part of the four-dimensional object than the other, creating an uneven object in 3-D. This object's unevenness is similar to the shadow of a symmetric object being asymmetric because of the light source's positioning. |AuthorName=Thomas F. Banchoff |AuthorName=Thomas F. Banchoff |AuthorDesc=Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967. |AuthorDesc=Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967.

## Revision as of 11:14, 5 June 2009

Projection of a Torus
A four-dimensional torus projected into three-dimensional space.

# Basic Description

It is impossible to visualize a complete four-dimensional object, since we have only ever lived in three-dimensional space. However, there are ways to capture parts of the four-dimensional object in three-dimensional space. A useful analogy is a world map. We can capture the essence of the three-dimensional globe on a two-dimensional map, but only by using a projection, which translates a three-dimensional object onto a two-dimensional surface at the expense of distorting the object in some way.

A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections.

# A More Mathematical Explanation

The four-dimensional object is defined parametrically by UNIQ4aa823cc42b822 [...]

The four-dimensional object is defined parametrically by $(x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v))$. A stereographic projection is used to map this 4-D object into 3-D, using a projection point of $(0,0,0,\sqrt{2})$ for the first object in this page's main image. This projection is centered above the four-dimensional object, projecting the symmetric torus into three-dimensional space. For the second object, the projection point is shifted to be closer to one part of the four-dimensional object than the other, creating an uneven object in 3-D. This object's unevenness is similar to the shadow of a symmetric object being asymmetric because of the light source's positioning.

# Teaching Materials

There are currently no teaching materials for this page. Add teaching materials.

# About the Creator of this Image

Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967.

If you are able, please consider adding to or editing this page!

Have questions about the image or the explanations on this page?
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.


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# How to link 'no result behaviour' to node?

When editing a view, i'm faced with a difficulty.
There's an image field, in the field settings i set the images (submitted when creating the node) to be linking to the node.
The problem is that in set up a default image in the no result behaviour setting.

I'd like this 'default image' to be linking to the node as well.

Any ideas on how to do this ?

An easy way would be to set up a default image for the field. This setting is on the field settings page for the field on the node.

OR

You could to use tokens in the view. Add the path field to the view (set it to not visible), make sure it is higher than the image field. Then you should be able to add an anchor tag around your image tag and use the token value. eg.

OR

You could to use tokens in the view like this. Add the path field to the view (set it to not visible), make sure it is higher than the image field. Under the 'rewrite results' section, check 'Output this field as a link' and add the link path token there.


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# fritzw/beamerthemeLUH

An unofficial LaTeX Beamer theme that is trying to follow the Coprorate Design Guidelines of Leibniz Universität Hannover.
TeX
Fetching latest commit…
Cannot retrieve the latest commit at this time.
Type Name Latest commit message Commit time
Failed to load latest commit information.
.gitignore
beamercolorthemeLUH.sty
beamerinnerthemeLUH.sty
beamerouterthemeLUH.sty
beamerthemeLUH.sty
example-0.png
example-1.png
example-2.png
example.pdf
example.tex

# LUH Beamer Theme

An unofficial LaTeX Beamer theme that is trying to follow the Coprorate Design Guidelines of Leibniz Universität Hannover. Here are some examples of how it might look:

## Installation

Simply clone the repository (or copy the files) to a location where LaTeX can find the .sty files.

• One way would be to copy them to your working directory.

• Or you could use the following commands to specify one or more directories where LaTeX should look for files:

\makeatletter
% The folder beamerthemeLUH is in the parent directoy.
\def\input@path{{../beamerthemeLUH/}{/some/other/path/}}
\makeatother

## Configuration

You should use the following commands to set up the logos:

\unilogo{\includegraphics[height=\LUHLogoHeight]{luh_logo_rgb}}
\logo{\includegraphics[height=\LUHLogoHeight]{department_logo}}
\titleimage{\includegraphics[width=.7\paperwidth]{luh_title_image}}

You will have to obtain the university logo and title image yourself (in PDF, PNG or JPEG format) because I'm not sure if it's okay to upload them here.

If you want to leave out any of the abovementioned design elements, simply pass an empty argument as in \titleimage{}.

For further customization please read the source code. If you find anything that can be improved, feel free to write an issue or send a pull request.

You can’t perform that action at this time.


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Seminar Calendar
for events the day of Wednesday, March 7, 2012.

.
events for the
events containing

Questions regarding events or the calendar should be directed to Tori Corkery.
    February 2012            March 2012             April 2012
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
1  2  3  4                1  2  3    1  2  3  4  5  6  7
5  6  7  8  9 10 11    4  5  6  7  8  9 10    8  9 10 11 12 13 14
12 13 14 15 16 17 18   11 12 13 14 15 16 17   15 16 17 18 19 20 21
19 20 21 22 23 24 25   18 19 20 21 22 23 24   22 23 24 25 26 27 28
26 27 28 29            25 26 27 28 29 30 31   29 30

Wednesday, March 7, 2012

GEAR Brown Bag
12:00 pm   in 331 Altgeld Hall,  Wednesday, March 7, 2012
 Del Edit Copy


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# Swirly bokeh like that of a Helios 44m 2/58?

I recently discovered the Helios 44m, a lense which produces some awesome swirly bokeh effect unlike anything else I've seen. I can't get one for myself, but I was wondering if there's a way to recreate this? I know the basics about DOF and I can somewhat achieve bokeh in Blender, but that kind of effect would require something else, probably some compositing? Or is it much more simple? Anyone out there capable of achieving something like that? It'd be lovely for some vintage renders.

• Do you have an image of exactly what you are going for? Depending on the image, I am sure there is a way to fake it. – Pythogen Jan 24 at 6:34
• I added a picture of the swirly bokeh. – kramkor Jan 24 at 13:18

This answer will achieve a similar effect, not quite the same or as good as what you are probably looking for. The problem is I couldn't find a way to get the bokeh dots to 'light' up or contrast against the other ones more - essentially I couldn't get them to look more like bokeh dots... maybe less motion blur? I am not sure. With that being said, this may work well enough for whatever project you are doing. Hopefully, this answer can serve as a good starting point for you to build off of or give someone else an idea to build off of as they form their answer.

Essentially I will just be using tiny refraction circles with traditional depth of field combined with motion blur to achieve the swirling effect. Here are a couple of example renders that I added the effect.

# How to achieve the swirly effect

We will follow these steps in making this swirly bokeh effect:

1. Make the small 'Bokeh Dots'
4. Compositing (Optional)

## Make the Small Bokeh Dots

1. Create a circle and fill in the face
2. Put the circle close to the camera in your scene as shown in the pic with two array modifiers. Look at the picture for guidance on how big the circles should be and how to set up the modifiers. I find that the circle should be just big enough to fit 70 by 40 (array modifier values) on a 16:9 ratio camera.
3. Add a Material to the Circle and set the surface node to a refraction node with a Roughness value of 0, IOR value of 1.2ish and set the color to as white as you can go.
4. Apply the Array Modifiers and then go into Edit Mode and press p and select separate by loose parts - separating them may take a while!
5. In Object Mode with all the circles selected, go to View->Set Origin->Origin to Geometry.
6. Open search by hitting F3 (or space if you go legacy hotkeys) and type Randomize and click Randomize Transform.
7. On the bottom left, change the randomize transform settings and click "Scale Even" and change the x scale value to a value around .5 (or whatever you like). You could also mess with the random transform values a bit, not a lot, but a bit.

And boom you have small bokeh dots! (I skipped this step in the render with the white balls and monkey heads because the white balls were the bokeh)

## Adding the Depth of Field & Motion Blur

Go to your camera and enable depth of field. Adjust the values to how you like. I find that distance values around .5 with F-Stop values of around 3 work well.

For the Motion blur

1. Press 0 to open your camera view, stay in camera view for the rest of the steps.
2. Select your camera and press i and click key frame rotation.
3. Go back one frame from the current, press r, type in -45, press enter, press i and click rotation.
4. Return to the current frame by going forward one frame. Then go foward one more frame and press r, type in 45 (this time its positive), press enter, press i and click rotation.
6. Go to the render panel and turn on motion blur. Set the shutter speed to around .05ish

## Optional Compositing

I noticed that the blur seemed a little more oil painty than the results so I attempted to make the render a little more oil painty using this post-processing trick. It is easier to set up than you think. For the color ramp, just select a point and keep pressing the plus button to add more points halfway in between. Then just change it to constant and duplicate it for all the RGB channels. This is the only post-processing that I could think of, maybe someone else will think of more.

• It's like a Monet filter 8*) – batFINGER Jan 25 at 11:02
• Oh wow, the one with Suzanne looks awesome, it looks exactly like what I was looking for. I don't know if someone will find a "simpler" way, but yours is pretty good. Thanks! – kramkor Jan 25 at 17:57

You can use a panaromic lens type and play with the ratio of the aperture:

Sunny Vondelpark from HDRIhaven

• I've noticed that this method works really well with HDRI maps, but I can't seem to make it work with models, I'll have to look into it more though. – kramkor Jan 26 at 2:05
• What do you mean by "make it work with my models"? Can you provide an example please? @kramkor – brockmann Jan 26 at 14:58


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💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

WZ

# We know from Example 1 that the region $\Re = \{ (x, y) \mid x \ge 1, 0 \le y \le \frac{1}{x} \}$ has infinite area. Show that by rotating $\Re$ about the x-axis we obtain a solid with finite volume.

## $\pi$

#### Topics

Integration Techniques

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Lucas this region We need to show that by rotating this region about axe access on with the battalion solid with finite volume. If I write way is the volume off? Is this true regional state irritated about X axis? And for each axe area of the section of this What solid if Iko too. So is a disc area off this section. Is he going to pie Times one over X square. This is the area. I was a section here. So the volume as Iko too integral from one to infinity. Off area with this section. Yaks. Now about that mission This's vehicle to the limit. A goes to infinity into girl from one, eh? Hi, moms. One over X squared? Yeah, they're asleep. We come to the limit. A toast to unity. And this is what? Negative. Hi. One over X from one too, eh? This is the limit. A toast to unity makes you a pie bond Over a minus one on one A goes to infinity One hour a goes to zero. So the volume off this solid if we goto Hi. It is a finite volume

WZ

#### Topics

Integration Techniques

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text:
Created by Created December 10, 2012 Last modified December 17, 2012 Tags

### Summary

A very simple model of a photodiode receiver using an
opamp as a transimpedance amplifier.

Simulate > DC Sweep > Run DC Sweep

### Description

A very simple model of a photodiode receiver using an opamp as a transimpedance amplifier.

More complex models take account of the non-linearity of the LED optical power o/p vs current characteristics and may have detailed modelling of rise and fall times and propagation delay.

D1 and D2 are separated by some distance.

Note that, although D1 provides a load that represents an LED,  in terms of voltage drop, resistance and junction capacitance  etc., the LED model has no optical parameters. The only  coupling to the receiver is via the behavioural current source  Ictr1 monitoring the current through D1.

Ictr1 is the (LED current)*(the attenuation between LED and photodiode)*(photodiode responsivity)

In this example responsivity=0.6 A/W, which is the typical responsivity of a silicon photodiode.

http://en.wikipedia.org/wiki/Photodiode

D2 has no optical effect. It is present purely to model the reverse leakage (dark current) and junction capacitance effects of a photodiode.

Note that the transimpedance:

V(out)/I(in) set by:

V(out) = -I(in)*Rfb

is a bit too high and the opamp output hits the -5V rail before the LED current reaches its maximum.

Cfb is to compensate for the photodiode capacitance.

Simulate > DC Sweep > Run DC Sweep


text:
# cirq.SimulatorStep¶

class cirq.SimulatorStep(state, measurements, qubit_map, dtype)[source]
__init__(state, measurements, qubit_map, dtype)[source]

Results of a step of the simulator.

qubit_map
A map from the Qubits in the Circuit to the the index
of this qubit for a canonical ordering. This canonical ordering
is used to define the state (see the state() method).
measurements
A dictionary from measurement gate key to measurement
results, ordered by the qubits that the measurement operates on.

Methods

 bloch_vector(index) Returns the bloch vector of a qubit. density_matrix(indices) Returns the density matrix of the wavefunction. dirac_notation(decimals) Returns the wavefunction as a string in Dirac notation. sample(qubits, repetitions) Samples from the wave function at this point in the computation. sample_measurement_ops(measurement_ops, …) Samples from the wave function at this point in the computation. set_state(state, numpy.ndarray]) Updates the state of the simulator to the given new state. state() Return the state (wave function) at this point in the computation.


text:
## Posts Tagged ‘circle’

### Classic bump deformation example mesh

Saturday, June 28th, 2014

This is code I end up re-writing every once in a while to demonstrate k-harmonic deformation on a simple 2d domain. The domain is just a square. This codes meshes the domain so that two inner circles show up in the edge set. Then I can select the vertices on and outside the outer circle and on and inside the inner circle to create a bump.

R = 1;
r = 0.15;
outer_sam = 200;
inner_sam = ceil(outer_sam/R*r);
inner_theta = linspace(0,2*pi,inner_sam+1)';
outer_theta = linspace(0,2*pi,outer_sam+1)';
P_inner = r*[cos(inner_theta(2:end)) sin(inner_theta(2:end))];
P_outer = R*[cos(outer_theta(2:end)) sin(outer_theta(2:end))];
E_inner = [1:size(P_inner,1);2:size(P_inner,1) 1]';
E_outer = [1:size(P_outer,1);2:size(P_outer,1) 1]';
P_bb = 1.1*[-R -R;-R R;R R;R -R];
E_bb = [1 2;2 3;3 4;4 1];
[P_bb,E_bb] = resample_polygon(P_bb,E_bb,2*pi*R/outer_sam);
P = [P_inner;P_outer;P_bb];
E = [E_inner;size(P_inner,1)+[E_outer;size(P_outer,1)+E_bb]];
[V,F] = triangle(P,E,[],'Flags',sprintf('-Yq32a%0.17f',(2*pi*R/outer_sam)^2),'Quiet');
tsurf(F,V)

### Re: Suggester shapes

Sunday, May 8th, 2011

Ken recently posted about shapes that suggest math problems. At the end of his post he leaves a riddle:

Find a shape whose area is 1/2 the area of a square and whose perimeter is the same as the perimeter of the square.

Find a shape whose area is 1/2 the area of a circle and whose perimeter is the same as the perimeter of the circle.

My idea for a solution for this new riddle was to intersect to equal-size circles. Notice as long as the circles are equal size if I take the shape that’s one of the circles minus the other circle, the perimeter will be the same as a full circle.

So then I just need to find how far apart the circles should be. I know that if the distance between the circles’ centers is zero then the left over shape will have area zero. And if the distance is twice their radii then then the left over area will be equal to a full circle. So if I want the left over area to be half the area of a circle I need the distance to be somewhere between zero and twice the radii.

I wrote out the algebra blindly hoping that the equations would collapse leaving me with a beautiful, succinct expression for the necessary distance. The problem turns out to be fairly intangible. Leaving me with a gnarly expression for the distance that is not easily solved. Here’s a picture and the expression:

I can approximate d numerically, which reveals d ≈ 0.8079455066. I’m not the first person to try this, this problem also makes an appearance on the Wolfram circle-circle intersection page.

This leaves me wondering if there is a simpler answer to my riddle. Maybe the solution doesn’t involve circle-circle intersection at all!

### Barycentric versus voronoi regional area in mass matrices

Friday, June 25th, 2010

In a recent project, we were minimizing an energy over a surface approximated by a triangle mesh. Our finite element method approach reduced our energy to solving a system of linear equations with coefficients from a stiffness matrix and mass matrix. In the lumped version of a mesh’s mass matrix, diagonal entries correspond to areas on the surface around each vertex. You want all the areas to sum to the total area of the surface and you would like the area for each vertex to some how reflect the influence of its incident triangles.

### Barycentric

One natural choice is a barycentric area. Take the barycenter (centroid, center of mass), which lies at the intersection of medians. Using those medians divide the triangle into three quadrilaterals each corresponding its incident point:

These quadrilaterals are by definition of equal area. The barycentric mass matrix entry for vertex i is the sum of its corresponding quadrilaterals from all incident triangles, or more simply: one-third of the total area of all neighboring triangles. Notice that the contribution of a triangle to its vertices is the same for all three, regardless of there placement around the triangle.

### Voronoi

Another, also natural, choice is voronoi area. Take the circumcenter, which lies at the center of the circumscribing circle of the triangle. It follows quickly that the circumcenter is equidistant from all vertices on the triangle. Connect this point to the midpoints along the edges of the triangle, again forming three quadrilaterals.

Note that this time the quadrilaterals are not equal. They are not even guaranteed to be positive. For obtuse triangles, the acute angled vertices will receive negative areas (assuming you demand their sums equal the area of the triangle). Here is a plot of voronoi areas as the angle of an isosceles triangle grows from 0 (acute) to π (obtuse). The x axis is the angle of the isosceles triangle and the y axis is area. The green line shows the area of corresponding to the vertex at the angle that’s growing, the blue corresponds to the other two areas.

Notice that at π/2 we might expect the other two areas to cross into negative but actually this happens later. So voronoi areas won’t produce negative coefficients until fairly obtuse.

Again, the voronoi mass matrix entry for vertex i is the sum of its corresponding quadrilaterals from all incident triangles. Since areas can be negative, this means negative entries could end up in the voronoi mass matrix.

### Connectivity dependence

The Voronoi area of a vertex is called so because the area represents its corresponding voronoi region: all the points on the surface for which this vertex is the closest vertex (assuming the mesh has “nice” triangles: non-obtuse). In this way the voronoi area depends only on the vertices positioning on the surface not the connectivity of the mesh.

The barycentric area, while having the advantage of always being positive, is heavily connectivity dependent as a vertex gets one-third of the area of any connected triangle.

For a simple example, here’re two meshes. One with regular positions and regular connectivity:

And one with regular positions and the same regular connectivity but a single edge at each highlighted vertex has been flipped:

And here are the barycentric (left) and voronoi (right) areas for the highlighted vertices on the regular connectivity mesh.

Now, notice how when the edge is flipped the barycentric area changes drastically, and actually changes for all vertices incident on the flipped edge. The voronoi however stays the same for this vertex and all vertices.

### Hybrid approach to vonoroi

To handle the problem of negative voronoi areas for some obtuse triangles, there exists a hybrid approach. For acute triangles, use voronoi as is. For obtuse triangles snap the circumcenter to the midpoint along the edge opposite the large angle and use the corresponding quads (I described this in an earlier post). Notice for a right angle triangle the circumcenter corresponds with this midpoint, so as a triangle becomes obtuse there isn’t a jump in area distribution. A keen eye will notice that I am doing this in the voronoi area picture above for the obtuse triangle.

### Areas of quadrilaterals within a triangle determined by its circumcenter (or perpendicular bisectors), using matlab

Thursday, February 11th, 2010

First, I need to locate the circumcenter of the triangle. The circumcenter is the center of the circle circumscribing the three points of the triangle. Thus, it is equidistance from all three points.

I do not need the actual world coordinates of the circumcenter, rather it is okay for me to just know the relative distance to each corner. The law of cosines tells me that the relative distance of the circumcenter from the line opposite point A (called a in the picture) is cos A (that is cosine of the angle at A). Likewise for B and C. Knowing these I can convert to barycentric coordinates by multiplying by the opposite edge lengths.

cosines = [ ...
(edge_norms(:,3).^2+edge_norms(:,2).^2-edge_norms(:,1).^2)./(2*edge_norms(:,2).*edge_norms(:,3)), ...
(edge_norms(:,1).^2+edge_norms(:,3).^2-edge_norms(:,2).^2)./(2*edge_norms(:,1).*edge_norms(:,3)), ...
(edge_norms(:,1).^2+edge_norms(:,2).^2-edge_norms(:,3).^2)./(2*edge_norms(:,1).*edge_norms(:,2))];
barycentric = cosines.*edge_norms;
normalized_barycentric = barycentric./[sum(barycentric')' sum(barycentric')' sum(barycentric')'];

A quality of the barycentric trilinear coordinates of a point in a triangle is that they specify the relative area of inner triangles form by connecting the corners to that point. In our case, the triangles made by connecting the circumcenter to each corner. Here, the aquamarine, red, and purple triangles corresponding to points A, B, and C.

Now I find the area of the triangle and then multiple against the normalized barycentric coordinates of the circumcenter to get the areas of each of these inner triangles:

areas = 0.25*sqrt( ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(-edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)));
partial_triangle_areas = normalized_barycentric.*[areas areas areas];

Finally the quadrilateral at each point is built from half of the neighboring inner triangles: the green, blue and yellow quads corresponding to A, B, and C.

Note, that the fact that these inner triangles are cut perfectly in half follows immediately from the fact the the circumcenter is equidistance from each corner. So the last step is just to add up the halves:

quads = [ (partial_triangle_areas(:,2)+ partial_triangle_areas(:,3))*0.5 ...
(partial_triangle_areas(:,1)+ partial_triangle_areas(:,3))*0.5 ...
(partial_triangle_areas(:,1)+ partial_triangle_areas(:,2))*0.5];

Note: I’ve suffered trying to use mathematic or maple to reduce these steps to a simple algebra formula (I must be mistyping something). As far as I can tell the above steps check out (of course, it only makes sense for cases where the circumcenter is within or on the triangle: i.e. the triangle is not obtuse). I have checked out my equations on maple with the following functions:

mycos:=(a,b,c)->(b^2+c^2-a^2)/(2*b*c);
barycentric:=(a,b,c)->a*mycos(a,b,c);
normalized_barycentric:=(a,b,c)->barycentric(a,b,c)/(barycentric(a,b,c)+barycentric(b,c,a)+barycentric(c,a,b));
area:=(a,b,c)->sqrt((a+b-c)*(a-b+c)*(-a+b+c)*(a+b+c))/4;
inner_triangle:=(a,b,c)->normalized_barycentric(a,b,c)*area(a,b,c);

Which then you can run:

To see a beautiful long algebra formula for the quadrilateral at point A.
Which in matlab corresponds to:

[-sqrt(-(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3))).* ...
(2.*edge_norms(:,2).^2.*edge_norms(:,3).^2  + edge_norms(:,1).^2.* ...
edge_norms(:,2).^2  - edge_norms(:,2).^4  + edge_norms(:,1).^2.* ...
edge_norms(:,3).^2  - edge_norms(:,3).^4)./ ...
(8.*(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3))), ...
-sqrt(-(edge_norms(:,2) + edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) - edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) + edge_norms(:,3) - edge_norms(:,1))).* ...
(2.*edge_norms(:,3).^2.*edge_norms(:,1).^2  + edge_norms(:,2).^2.* ...
edge_norms(:,3).^2  - edge_norms(:,3).^4  + edge_norms(:,2).^2.* ...
edge_norms(:,1).^2  - edge_norms(:,1).^4)./ ...
(8.*(edge_norms(:,2) + edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) - edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) + edge_norms(:,3) - edge_norms(:,1))), ...
-sqrt(-(edge_norms(:,3) + edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) - edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) + edge_norms(:,1) - edge_norms(:,2))).* ...
(2.*edge_norms(:,1).^2.*edge_norms(:,2).^2  + edge_norms(:,3).^2.* ...
edge_norms(:,1).^2  - edge_norms(:,1).^4  + edge_norms(:,3).^2.* ...
edge_norms(:,2).^2  - edge_norms(:,2).^4)./ ...
(8.*(edge_norms(:,3) + edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) - edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) + edge_norms(:,1) - edge_norms(:,2)))];

Note: It seems also that the wikipedia page for the circumscribed circle has a direct formula for the barycentric coordinates of the circumcenter. Though, wolfram’s mathworld has an ostensibly different formula. Upon a quick review Wolfram’s is at least easily verifiable to be the same as mine (recall these only need to be relative to each other).

### Average ratio of incenters to circumcenters, using matlab

Monday, February 8th, 2010

The incenter radius of a triangle is given by the formula:

2*A      with A = area
r = -------
(a+b+c)

the circumcenter radius is given by

abc
R = -----
(4A)

We want the ratio of the incenter to circumcenter so that’s

r         8A2
_ = ---------------
R   ((a+b+c)*(abc))

So then if you have a triangle mesh with a variable edge_norms which is an n by 3 matrix of the lengths of edges in your mesh, a way to get the incenter radii is, first applying Heron’s formula to find the areas of each triangle:

semi_perimeters = (edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3))*0.5;
areas = sqrt( ...
semi_perimeters.* ...
(semi_perimeters-edge_norms(:,1)).* ...
(semi_perimeters-edge_norms(:,2)).* ...
(semi_perimeters-edge_norms(:,3)));
ratios = 8*areas.*areas./(sum(edge_norms')'.*prod(edge_norms')')

Then just take the mean:

mean(ratios)

Note: I coded up the calculations of the incenters and circumcenters to check this out so I figure I might as well post that, too:

r=2*areas./(edge_norms(:,1)+edge_norms(:,2)+edge_norms(:,3))
R=(edge_norms(:,1).*edge_norms(:,2).*edge_norms(:,3))./(4*areas)


text:
# Lie algebras of the unit group of a finite dimensional algebra

Given a finite dimmensional associative algebra $$A$$, it can be proven that the unit group $$A^{\ast}:=\{a\in A \mid \exists\, b \in A \mbox{ such that } ab=ba=1_A\}$$ is always a Lie group (The unit group of a finite dimensional associative algebra is a Lie group?), but I have problems trying to identify its Lie algebra. I think that its Lie algebra $$(\mathfrak{a},[.,.])$$ should be isomorphic (as Lie algebras) to $$(A,[.,.]_c)$$, where $$[a,b]_c=ab-ba$$ (as the case $$A=M_n(\mathbb{K})$$). It is true?? I hope somebody can help me

• You should add the assumption of the linked question and answer that $\Bbb K$ is a complete valued field (as otherwise the concept of "Lie group" makes little sense to begin with). – Torsten Schoeneberg Feb 11 at 1:31
• Well, of course $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. – GaSa Feb 11 at 1:36
• Of course ... (Many people are investigating $p$-adic Lie groups, but of course that need not be your business.) Your conjecture sounds plausible, but what have you tried? And what construction of the Lie algebra of a Lie group are you using in this generality -- invariant vector fields, tangent space, invariant derivations ...? – Torsten Schoeneberg Feb 11 at 1:39
• I have tried considering $\mathfrak{a}$ as tangent space but the problem is choose a basis on $A$ (based in the case $A=M_n(\mathbb{K}$)). – GaSa Feb 11 at 1:43


text:
Linear Predictive Coding example in MATLAB

I have some data that is highly correlated and I wanted to see if I could try and encode it using linear predictive coding (LPC). Here is how I've been understanding the process:

Encoding

1. Generate predictive filter coefficients
2. Filter signal with these coefficients to obtain and error signal $$r[n]$$.
3. Check if $$r[n]$$ is white.
4. If it is, store the variance of $$r[n]$$ along with the coefficients and you are done.
5. If not, increase order until (hopefully) $$r[n]$$ becomes white.

Decoding

1. With the encoding variance, produce a white signal.
2. Pass that white signal through the inverse filter.
3. You produce a signal $$s[n]$$ that has the same autocorrelation sequence and power spectrum as your original.

I've completed all the encoding steps - computed my coefficients, checked if my residual signal was white (autocorrelation is a delta function) and then stored its variance. However, the trouble comes during the decoding process. I took the variance I encoded, produced white noise, and filtered it with the inverse filter. Not only was my signal nowhere near a representation of my old signal, but the two signals had very different autocorrelation sequences as well. I've posted my code and my outputs below. I mainly just want to know if I'm misunderstanding something about LPC or if I'm implementing it wildly incorrectly.

load('amp.mat') %this gets loaded as the variable y with timestamp t

order = 15;
[a,variance] = aryule(y,order);
error = filter(a,1,y);

%Now check to see if error is white

[acs, lags] = xcorr(error,'coeff');
plot(lags,acs), grid;

noise = sqrt(variance)*randn(length(y),1);

%compare noise to error signal
plot(t,error,t,noise)

%synthesize signal
yrecon = filter(1,a,noise);
%actual signal, as produced by our residual
yact = filter(1,a,error);

plot(t,yrecon,t,yact);
title('Signal vs synthesized signal')

%check autocorrelation sequence of each and plot
[yacs, ylags] = xcorr(y,'coeff');
[yracs, yrlags] = xcorr(yrecon, 'coeff');
subplot(2,1,1)
plot(yrlags,yracs), grid
title('Autocorrelation sequence of reconstructed signal')
subplot(2,1,2)
plot(ylags, yacs), grid
title('Autocorrelation sequence of signal'}

• wow it's exremely efficient to store only the variance of $r[n]$... ;-) consider a signal of length 1 G samples and you reduce it to a few prediction coefficients and a single variance ? something wrong ? – Fat32 Nov 25 '18 at 16:02
• I agree that something seems wayyyy off haha. I mean I've tried increasing the number of prediction coefficients, but it neither makes the error 'more white' nor does it seem to make the auto correlation sequences more similar. I apologize, but I'm just really out of my depth here, I've looked all over for examples of implementing LPC for compression but I can't find any worked out examples so I've been trying to do it myself. I just figured that if I can get the residual to be white by removing all correlations in the signal, I've finished the encoding process. I'm guessing that's not true? – compscinoob Nov 25 '18 at 19:23
• do you know what LPC is used for and how it does achieve its purpose ? – Fat32 Nov 25 '18 at 20:24
• LPC is mianly used for speech because it's designed to fit the model of our vocal system. Our speech is modeled as a filter that is excited by either an impulse train or white noise. I do not know how it achieves its purpose. Although it's used mainly for speech, I don't see why it can't be used for any general signal that's correlated and can be reduced to some residual white noise. Maybe I'm misunderstanding everything, but I would really appreciate it if you could tell me how I could correct my already existing implementation and/or my understanding of LPC. – compscinoob Nov 27 '18 at 14:10
• You are ok with the predictive part of the LPC. But it seems you have a problem with the coding part of it; that's where the compression actually takes place. You should consider the quantization stage carefully. I suggest you read Introduction to Data Compression by Khalid Sayood. For a full grasp of LPC and other data compression methods. – Fat32 Nov 27 '18 at 15:16


text:
# Change the order of triple integration

1. Nov 23, 2009

### coco87

1. The problem statement, all variables and given/known data
Rewrite $$\int_{0}^{2}\int_{0}^{y^3}\int_{0}^{y^2}dzdxdy$$ as an integral with order $$dydzdx$$.

2. Relevant equations
N/A

3. The attempt at a solution
Honestly, I got as far as sketching it:

and after sketching it, I'm lost...

I can't figure out how to set $$z$$ or $$y$$, but I'm fairly sure that $$x$$ is $$[0,8]$$.

could anyone possibly offer some help?

Thanks

2. Nov 23, 2009

### Staff: Mentor

Is this all you're given in the problem? Just change the order of integration? I'm concerned that you might be omitting some information in the problem.

BTW, I've always found it to be easier to include the variable in one of the limits of integration, when I'm dealing with iterated integrals.
$$\int_{y = 0}^{2}\int_{x = 0}^{y^3}\int_{z = 0}^{y^2}dz~dx~dy$$

3. Nov 23, 2009

### coco87

Mark44:

Thank you for your tip :)

I'm actually not leaving anything out at all. Many of these questions are vague (and rather annoying). I think I might have it, but am not sure:

$$\int_{0}^{8}\int_{0}^{\sqrt[3]{x}}\int_{0}^{\sqrt{z}}dydzdx$$. However, this seems way to easy to be true...

4. Nov 23, 2009

### Staff: Mentor

Well, you can check. both integrals represent the volume of the region, so both integrals should produce the same value.

5. Nov 23, 2009

### coco87

Mark44:

Hmm, how would I check without a function to integrate?

6. Nov 23, 2009

### Staff: Mentor

The function is 1.