text:
# Design for an analogue oscillator that doesn't drift in frequency?

VCOs (Voltage Controlled Oscillators) in analogue music synthesizers are notorious for drifting with temperature. What design solutions are there for an analogue audio VCO that keeps close (as in human hearing) to the frequency to which it's been tuned?

• See my answer to this stack exchange question Jul 3, 2012 at 13:50
• mctylr below makes a good point about the accuracy needed in multitrack recording Jul 3, 2012 at 20:06
• Just for interest - the "synthetic rock" circuit that I cite in my prior answer would hold a SSB signal in zero beat for long periods. Memory says possibly many 10's of minutes - will be in references. Zero beat means that the signmal is stable enough in frequency relative to a crystal derived transmitter that no audible 'beat' difference can be heard. As the oscillator will be at IF or RF frequencies the required stability is far better than 1 Hz across the audio pass band across the same period of time. Whether the SR scales down well is TBD but higher frequency plus a PLL would work. Jul 4, 2012 at 1:18

## 7 Answers

Use parts with lower temperature coefficients (NP0 or C0G for ceramic capacitors for examples). This is typically the most expensive option, but is simplest at a first-pass optimization.

Use a quality voltage regulator to supply the oscillator, one that is immune to temperature variability within the design's operating temperature range.

Minimize the dependency on variable components (capacitor, inductor, or resister) for tuning. Pad the variable components with fixed components to minimal the variables component values. For example replace a 100k$\Omega$ potentiometer with a 47k resisters on the two legs of a 10k$\Omega$ potentiometer as potentiometer may have a temperature coefficient of 1000ppm, while the fixed metal film 1% resistors may have tempco of 200-500ppm.

Use parts with complementary temperature coefficients of components (or additional components with a temperature variability that is well characterized like a thermistor e.g. +10 ohms per degree increase) that cancel out a change in values when the temperature changes. Eg. Oscillator drift correction circuit (mainly in reference to RF oscillator drift, but the principles are consistent)

You could consider a low frequency crystal oscillator such as the 32.768 kHz commonly referred to as a watch crystal as it is typically used in Real-Time Clock (RTC) circuits, as well as low-power microcontrollers. Using it in a VXO with a small tuneability, (aka "pull") ~10% I believe and a frequency divider you can generate a very stable audio oscillator that is tuneable over a narrow range.

The other is to make the oscillator circuit's environment thermal stable using a) insulation to minimize and slow thermal change, and if necessary b) temperature stabilized heat / cooling such as an oven-controlled crystal oscillator, OCXO.

If you look at quartz oscillator modules or "cans", pay attention to their output type, most are designed for digital timekeeping / clock generation and only output a digital signal, though sine-wave or clipped sine-wave modules or XO are available.

In response to necessary stability, that is application dependent. If you want to be able to match the oscillators frequency over time (such as in a multiple track recording studio, where tracks are layered and each track is recorded separately) then stability is important because while absolute accuracy of human hearing is moderate (no better than 1% I would guess), relative frequency mismatch is easily detectable to a much smaller degree (again I would guess approx 0.1 - 0.01%).

Analog oscillators are going to drift. For your purposes, a crystal is exact enough. You can try to tweak a analog oscillator using something derived from a crystal as a reference. But if you have the crystal there already, you might as well use it to make the desired frequencies directly.

Even a low end DSP clocked from a crystal can synthesize audio sine waves. It can even digitally produce each of the sines internally, then add them to make a composite output signal. This can be various harmonics with their own gains and phase shifts, or even arbitrary frequencies. There is a reason you don't see analog synthesizers anymore.

I go into detail about how to generate sine waves inside a processor in this answer.

• The technical/logical answer (go on then I'll give it a vote), but in the audio world analogue is still cool. Old analogue synths go for large sums along with valve(tube) amps, reel to reel machines and the like... Jul 3, 2012 at 15:57
• @OliGlaser: That may be true, but then it's about religion not electronics, and is off topic here. Jul 3, 2012 at 16:17
• true, true - there's plenty of snake oil sold to help perpetuate the religion too. It's interesting from my point of view to have connections in both "worlds" (i.e. EE/Music). I find it curious to see how much folk are willing to pay for e.g. a couple of hand wired transistors in a little metal box, but it's proof that the snake oil works a treat on most. Jul 3, 2012 at 18:59
• @Rob Kahn - to give an example of one of the dsPICs Olin mentions in the linked post, I found the dsPIC33FJ64GP802 to be very useful. It comes with a good quality onboard 16-bit DAC, which makes it very convenient for such a project - just add a differential opamp and you're done. I used one for a temporary function generator a while back, and was very impressed with the THD+N. Can also be used for audio in/out with the ADC and DMA (though the ADC is only 12-bit so external e.g. Wolfson ADC is preferable) Jul 3, 2012 at 20:31

Just to chime in, I repaired analog synthesizers for about 10 years and am well acquainted with the drifting issue. Many musicians like the analog oscillator precisely because of its imperfections, and in my humble opinion this is not so much "snake oil" as it is a matter of taste and preference. Let us not forget that perfect waveforms do not necessarily make music more enjoyable.

If you can download a moog MicroMoog service manual (available free on internet), in section 2-8 there is a description of their oscillator which was (to me) impressively stable. The oscillator is so stable due to the clever circuit described in section 2-3-3, which acts as a current source.

Basically the circuit is designed to maintain a constant temperature on the transistor array IC which drives the oscillator. I thought this worked really well: the instrument turned on, warmed up (in less than a minute) and then remained stable after that, unlike many other analog synths.

Best of luck with your project. If you come up with a good design, please post it. Chris Rowland

The solution in use since the 1970s has been to use a pair of matched PNP transistors, to form an exponential voltage-to-current converter. The input base-emitter voltage is exponentially related to the collector current. By having the second transistor configured, so that its current is in the opposite direction to the first, most of the temperature dependence is cancelled out.

Any remaining temperature dependence is dealt with by using a thermistor in thermal contact with the matched transistor pair, in the feedback path of an op-amp voltage summer at their input.

The apparently simple Thomas Henry VCO-1 design puts this into effect.

I used to work with a device that would generate high quality / low distortion sine waves and at the same time had TCXO stability. The trick the designers used was a VCO, which was controlled by the output of a PLL. The PLL in turn was fed with the analog oscillator signal and with a digital clock from a microcontroller. That microcontroller that was able to make about any frequency by counting (and finally dividing by 2 to make it a square wave).

Use the (now hard to find) CEM3340 IC.

By the way, for some users the instability of an analogue VCO is not an issue. It gives extra interest to the possible sound creations - with it as one component. Traditional non-electronic musical instruments also change in tone with temperature (and humidity).

A typical crystal oscillator is good to +/- 50 pm. As Olin Lathrop said, this is good enough. For example 440 Hz concert A would be 440 +/ 0.022 Hz. I don't think most people could here that.

However if you wish to calibrate one standard way is with a calibration table. Since part of the drift is due to temperature you measure the frequency at various temperatures and store that in a PROM. Then you use a temperature sensor and that table to calculate what to set the VCO input to.

• Did you mean "a typical crystal oscillator is good to +/- 50 ppm"? Aug 30, 2012 at 12:18

format: plain


text:
# Error estimate for sequence defining e

Is there an estimate for error of $e-\left(1+\frac1n\right)^n$ ?thanks

-

The power series for $\log(1+x)$ yields $$n\log\left(1+\frac1n\right)=1-\frac1{2n}+O\left(\frac1{n^2}\right)$$ applying $e^{x+y}=e^xe^y$ and the power series for $e^x$ gives $$\left(1+\frac1n\right)^n=e\left(1-\frac1{2n}+O\left(\frac1{n^2}\right)\right)$$ Therefore, $$e-\left(1+\frac1n\right)^n=\frac{e}{2n}+O\left(\frac1{n^2}\right)$$


text:
Share

For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7, (a – b) x + (a + b) y = 3a + b – 2 - CBSE Class 10 - Mathematics

ConceptAlgebraic Methods of Solving a Pair of Linear Equations Cross - Multiplication Method

Question

For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

Solution

2x + 3y -7 = 0

(a – b)x + (a + b)y - (3a +b –2) = 0

a_1/a_2 = 2/(a-b) = 1/2

b_1/b_2 = 3/(a+b)

c_1/c_2 = -7/-(3a+b-2) = 7/(3a+b-2)

For infinitely many solutions,

a_1/a_2 = b_1/b_2 = c_1/c_2

2/(a-b) = 7/(3a+b-2)

6a + 2b - 4 = 7a - 7b

a - 9b = -4 ... (i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a - 3b

a - 5b = 0 ... (ii)

Subtracting equation (i) from (ii), we get

4b = 4

b = 1

Putting this value in equation (ii), we get

a - 5 × 1 = 0

a = 5

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

Is there an error in this question or solution?

APPEARS IN

NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 3: Pair of Linear Equations in Two Variables
Ex. 3.50 | Q: 2.1 | Page no. 62

Video TutorialsVIEW ALL [2]

Solution For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7, (a – b) x + (a + b) y = 3a + b – 2 Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method.
S


text:
# Chu space

A Chu space over a set $\Sigma$ is a triple $(\mathcal{A},r,\mathcal{X})$ with $r:\mathcal{A}\times\mathcal{X}\rightarrow\Sigma$. $\mathcal{A}$ is called the carrier and $\mathcal{X}$ the cocarrier.

Although the definition is symmetrical, in practice asymmetric uses are common. In particular, often $\mathcal{X}$ is just taken to be a set of function from $\mathcal{A}$ to $\Sigma$, with $r(a,x)=x(a)$ (such a Chu space is called normal and is abbreviated $(\mathcal{A},\mathcal{X})$).

We define the perp of a Chu space $\mathcal{C}=(\mathcal{A},r,\mathcal{X})$ to be $\mathcal{C}^{\perp}=(\mathcal{X},r^{\smallsmile},\mathcal{A})$ where $r^{\smallsmile}(x,a)=r(a,x)$.

Define $\hat{r}$ and $\check{r}$ to be functions defining the rows and columns of $\mathcal{C}$ respectively, so that $\hat{r}(a):\mathcal{X}\rightarrow\Sigma$ and $\check{r}(x):\mathcal{A}\rightarrow\Sigma$ are given by $\hat{r}(a)(x)=\check{r}(x)(a)=r(a,x)$. Clearly the rows of $\mathcal{C}$ are the columns of $\mathcal{C}^{\perp}$.

Using these definitions, a Chu space can be represented using a matrix.

If $\hat{r}$ is injective then we call $\mathcal{C}$ separable and if $\check{r}$ is injective we call $\mathcal{C}$ extensional. A Chu space which is both separable and extensional is biextensional.

 Title Chu space Canonical name ChuSpace Date of creation 2013-03-22 13:04:51 Last modified on 2013-03-22 13:04:51 Owner Henry (455) Last modified by Henry (455) Numerical id 6 Author Henry (455) Entry type Definition Classification msc 03G99 Defines perp Defines carrier Defines cocarrier Defines normal Defines normal Chu space Defines separable Defines extensional Defines biextensional Defines row Defines column


text:
## Hybridization

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Andrea Raymundo 1B
Posts: 32
Joined: Wed Nov 15, 2017 3:04 am

### Hybridization

I am not sure if i heard right but when figuring out how many hybridization orbitals one has to use, is it the same as the number of regions of electron desnities??

NatalieSDis1A
Posts: 39
Joined: Fri Apr 06, 2018 11:05 am

### Re: Hybridization

The sum of the subscripts should equal the total number of regions of electron density.

Marisol Sanchez - 1E
Posts: 38
Joined: Fri Apr 06, 2018 11:05 am

### Re: Hybridization

Yes, this is what Professor Lavelle said during class.

jessicasam
Posts: 36
Joined: Wed Nov 15, 2017 3:02 am
Been upvoted: 1 time

### Re: Hybridization

Yes, that's true. The idea of a hybridized orbital is that it is a way for us to describe and understand how bonding can happen when it appears that there are not enough unpaired electrons to form all of the bonds in a molecular structure. If we see that a molecular structure has 4 areas of electron density, then it must have 4 hybridized orbitals.


text:
# College Math Teaching

## March 26, 2012

### Humiliated by Convex Splines

Update The writing in this article leaves something to be desired. I am going to clean this up a bit.

————————-
I was working on an analysis problem. One of the steps I needed to take was: given a set of points $p_1, p_2, p_3,...., p_i = (x_i,y_i)$ in the first quadrant of the standard Cartesian plane and given the fact that these points converged to the origin AND given the fact that these points met a convexity property in that for all $i, 0 < \frac{y_{i+2}-y{i+1}}{x_{i+2}-x_{i+1}} < \frac{y_{i+1}-y{i}}{x_{i+1}-x_{i}}$ could one then fit a convex $C^1$ curve though all of these points (and hence to the origin)? The answer turns out to be “no” but happily one can put an decreasing $C^1$ convex curve though an infinite subset of these. One standard way is to use the Beizer quadratic spline.

But I figured: “hey, we are talking about, what, a 4’th degree polynomial through two points that contains position and derivative information at the endpoints…how hard can that be?”

Then I tried it…..and found some more humility.

First I found out: it would be impossible to put a convex curve $f(x)$ through points $(x_0,y_0), (x_1, y_1)$ that had $p'(x_0) = m_0, p'(x_1) = m_1, m_0 < m_1$ unless certain conditions were met.

Here is why: suppose $\frac{f(x_1) - f(x_0)}{x_1 - x_0} = m \geq m_1$ The Mean Value theorem assures the existence of $x, x \in (x_0, x_1)$ where $f'(x) = m$ which is impossible due to convexity, unless $m = m_1$, in which case an application of the Mean Value Theorem to $f$ on $[x_0, x_f]$ (where $x_f$ is the first place that $f' = m_1$) shows that convexity is impossible.

So what about a convex polynomial that runs through the first and last point of a three point convex set and has the required derivatives at the first and last point (“knot”)? It turns out that such a problem has been the focus of much research and, in general, is impossible to do (reference).

But we can get a non-polynomial function (in particular a Frobenius type polynomial with fractional powers) that yields one sided differentiability
at the first and last point, which permits appropriate “piecing together” of splines to yield a $C^1$ convex curve.

The key will be to show that given three knots of “convex data” and a derivative condition at the first and last knot, one can always fit in a convex “Frobenius” polynomial through the first and last knot that meet the specified derivative condition at those knots.
The set up: the last knot will be taken to be the origin; the first knot will be given in terms of the sum of two line segments; the segment ending at the origin will have the specified slope at the origin and the second segment will have the slop of the “third knot”. The missing knot (the one that the function will not run through) can be thought of as being the sum of a segment whose slope is greater than the slope at the origin along with a segment ending at the third knot whose slope is the specified slope at the third knot. Denote the first knot by $(0, g(0))$ and $(1, g(1))$.

Theorem: Given real numbers $0 < m_{1} < m_{2}$ and $\rho$, $0 <\rho$ there is a convex Frobenius polynomial $g$ such that:

(i) $g(0)=0$

(ii)$g(1)=(1-\rho )m_{2}+\rho m_{1}$

(iii) $g^{\prime }(0)=m_{1}$

(iv) $g^{\prime }(1)=m_{2}$

Where the derivatives in question are the appropriate one-sided derivatives.

Proof. Define $g(x) = Dx + Ax^{1 + \alpha} + Bx^{1+ \alpha + \beta} + Cx^{1 + \alpha + \beta + \delta}$.
If we can find such a $g$ that meets conditions i, ii, iii and iv with $A,B,C, D$ positive, convexity follows.

Set $D=m_{1}$ to satisfy properties i and iii.

$m_{1}+A+B+C=(1-\rho )m_{2}+\rho m_{1}$

$A+B+C=(1-\rho )m_{2}+\rho m_{1}-m_{1}=(1-\rho )m_{2}-(1-\rho )m_{1}$

$A+B+C=(1-\rho )(m_{2}-m_{1})$

$m_{1}+(1+\alpha )A+(1+\alpha +\beta )B+(1+\alpha +\beta +\delta )C=m_{2}$

Or

$(1+\alpha )A+(1+\alpha +\beta )B+(1+\alpha +\beta +\delta )C=m_{2}-m_{1}$

So set $\Delta =m_{2}-m_{1}$

We have

$A+B+C=(1-\rho )\Delta$

$(1+\alpha )A+(1+\alpha +\beta )B+(1+\alpha +\beta +\delta )C=\Delta$

This leads to the augmented matrix:

$\left[ \begin{array}{cccc} 1 & 1 & 1 & (1-\rho )\Delta \\ (1+\alpha ) & (1+\alpha +\beta ) & (1+\alpha +\beta +\delta ) & \Delta\end{array}\right]$

Perform the following row operation: $R_{2}\rightarrow R_{2}-(1+\alpha )R_{1}$

$\left[ \begin{array}{cccc} 1 & 1 & 1 & (1-\rho )\Delta \\ 0 & \beta & \beta +\delta & \Delta -(1+\alpha )(1-\rho )\Delta \end{array} \right] =\left[ \begin{array}{cccc} 1 & 1 & 1 & (1-\rho )\Delta \\ 0 & \beta & \beta +\delta & \Delta (-\alpha +\rho +\alpha \rho ) \end{array} \right] =$

$\left[ \begin{array}{cccc} 1 & 1 & 1 & (1-\rho )\Delta \\ 0 & \beta & \beta +\delta & \Delta (\rho -\alpha (1-\rho )) \end{array} \right]$

Now perform $R_{2}\rightarrow \frac{1}{\beta }R_{2}$

$\left[ \begin{array}{cccc} 1 & 1 & 1 & (1-\rho )\Delta \\ 0 & 1 & 1+\frac{\delta }{\beta } & \Delta \frac{1}{\beta }(\rho -\alpha (1-\rho )) \end{array} \right]$

Now perform $R_{1}\rightarrow R_{1}-R_{2}$

$\left[ \begin{array}{cccc} 1 & 0 & -\frac{\delta }{\beta } & (1-\rho )\Delta -\Delta \frac{1}{\beta } (\rho -\alpha (1-\rho )) \\ 0 & 1 & 1+\frac{\delta }{\beta } & \Delta \frac{1}{\beta }(\rho -\alpha (1-\rho )) \end{array} \right]$

So our solution is :

$\left[ \begin{array}{c} A \\ B \\ C \end{array} \right] =\left[ \begin{array}{c} \Delta \frac{1}{\beta }(\beta (1-\rho )-\rho +\alpha (1-\rho ))+\frac{\delta }{\beta }u \\ \Delta \frac{1}{\beta }(\rho -\alpha (1-\rho ))-(1+\frac{\delta }{\beta })u \\ u \end{array} \right]$

Where $u$ is some real number $u\geq 0$

Note that $\Delta > 0$ and $0< \rho$ (as $u$ can be made as small as
desired)

Hence $\rho > \alpha (1-\rho )\rightarrow \frac{\rho }{1-\rho }> \alpha$

Now for the $A$ term:

$\beta (1-\rho )-\rho +\alpha (1-\rho )>0$
$\beta (1-\rho ) > \rho -\alpha (1-\rho )$

$\beta >\frac{\rho -\alpha (1-\rho )}{(1-\rho )}=\frac{\rho }{1-\rho } -\alpha >0$

Now choose $\delta >0$ and

$\Delta (\frac{(\rho -\alpha (1-\rho ))}{\beta +\delta })>u>0$

And convexity is guaranteed because the coefficients are all positive.

So now we can “piece together” as many of these splines as needed by matching the derivatives at the end points.

Next: now that we have something that works we should say something about how closely this “Frobenius polynomial” approximates the knot that it misses. That will be the subject for a subsequent post.


text:
# Scale invariance in sandpile model and forest fire model

I asked a similar question but the wrong way here. Because my intention was to ask about non thermodynamic system, i will be more specific:

• What is the relation between critical behaviour and the scale invariance in these two model (sandpile, forest fire)?

What i can't figure out is the meaning of "infinity correlation length" in these two model.

-
Dear emanuele, check e.g. the page and picture at en.wikipedia.org/wiki/… - The correlation length is what it always is - the thickness of the boundary between well-defined phases. On the picture, there are no such well-defined phases occupying a big volume so everything is the boundary and the corr. length is infinite. The sandpile model is interesting for creating "self-organized criticality": one gets scale invariance i.e. infinite corr. length even without fine-tuning any parameters, unlike in phase transitions in solids/liquids etc. –  Luboš Motl May 5 '12 at 15:28
Dear Lubos, do you mean that the phases, in the sand pile, are represented by the slopes at the different points of the sand pile? –  Emanuele Luzio May 5 '12 at 15:50
I think it's right. A feature of the model is, of course, that there are never qualitatively separated phases, but that's one of the "virtues" of the model. One doesn't have a sharp definition of a phase here because it doesn't exist, regardless of the parameters. The correlation length should have a well-defined definition but I wouldn't be able to define it. At the end, it's infinite, anyway, so there's no finite number to calculate. ;-) So the only thing that can really be meaningfully discussed is why the behavior is analogous to other critical points and CFTs. –  Luboš Motl May 5 '12 at 18:22
Thanx for the reply. Do you have references of these analogies? –  Emanuele Luzio May 5 '12 at 20:10

This is a tricky question and in my opinion some elder physicists are deliberately try to confuse students by using euphemisms when characterizing phase transitions.

Roughly speaking (there might be counter examples, please comment. I am interested of finding all of them):

1. First order phase transition:

• finite correlation length
• scales as e.g. $k^\alpha, \alpha = 2$ (short or finite range interaction) in fourier space (1D)
2. Second order phase transition:

• infinite correlation length
• scales as e.g. $k^\alpha, \alpha = 1$ (long or infite range interaction) in fourier space (1D)
• scale invariance
• (I think SOC goes here...)

Note that there is a continuous transition with exponent $\alpha$ that escapes my mind. Also $\alpha$ is dependent of dimension and probably something else (see e.g. anne tanguy et al. From individual to collective pinning: effect of long range interactions, PRE 1998). Also Daniel Fisher has a nice paper, Collective transport in random media: from superconductors to earthquakes.

Also I just stumbled upon: http://www.tcm.phy.cam.ac.uk/~bds10/phase/introduction.pdf which has a nice overview.

In general the purpose of these correlation lengths, roughness exponents and orders of phase transitions is just to find universality classes. The goal is to group the phenomena together and say "look, all these systems have properties of X,Y and Z. This simple model has the same properties. So by explaining the simple mode, I explain all these systems."

Simple characterization of phase transitions can be found at: http://link.aps.org/doi/10.1103/RevModPhys.76.663

-
What is that k-space alpha exponent nonsense in your answer? –  Ron Maimon Aug 31 '12 at 17:21
Could you be more spesific? Alpha = 2 means nearest neighbour interaction (= laplace, short range) and alpha = 1 means long range, infinity range (if I recall right, it means 1/r^2 interaction => infinite). Just fourier transform it and you get the exact formulation. The point is that k-space formulation is essentially saying short or long range interaction (and possibly something extra). This is what I meant by "elder physicists are deliberately try to confuse students by using euphemisms when characterizing phase transitions". –  Juha Sep 1 '12 at 20:06
Ok, -1. This is categorically false. Second order phase transition can scale with any $\alpha$ less than two, and even bigger than 2 (as in a Lifschitz point). Please erase this answer, as it is wrong. –  Ron Maimon Sep 2 '12 at 13:06
1. do you have a reference? 2. Read the second paragraph. –  Juha Sep 2 '12 at 18:07
I don't use references to back up what I am saying, I can give you a model. Please delete this answer, there are not just counterexamples, it's like saying "every function has a derivative with is cos(x). Are there counterexamples?" There are hardly any examples! You are possibly confusing a dynamic exponent with a critical exponent. The correlation functions in k space fall off as a nontrivial power, there is no "1" "2" dichotomy in phase transitions. You made this answer up, and you made it up wrong. –  Ron Maimon Sep 2 '12 at 20:43


text:
107 147
Assignments Done
100%
Successfully Done
In January 2023

# Answer to Question #111745 in Mechanics | Relativity for evonn

Question #111745
If kinetic energy of an object is been constant at all times, how does momentum of that object works?.....i think kinetic energy can change to another type of energy such as potential energy(ex:projected ball).that is the way energy is conserved.but here is asking about a motion which kinetic energy is always constant.then how momentum works of that object??i think there is no doubt about linearly motions,because if kinetic energy is constant ,momentum also constant. (mv*v) /2=k ,m is constant .v is also constant. if theres a direction change in linear motions v should get zero for a moment. its mean in there kinetic energy is not constant,thats why im saying there is no doubt about linear motions. because if k is constant then p is also constant.but in angular motions(circle) if kinetic energy is constant ,then linear momentum is not constant. because of the change of direction.but angular momentum can be constant.because we get it using the right hand law.am i right or not?
1
2020-04-24T10:52:14-0400

1.Your conclusion is correct but they don't take into account case when "m" is a change.

If "E=mv^2\/2" and "\\vec{p}=m\\vec{v}" then "|\\vec{p}|=\\sqrt{2mE}=\\sqrt{2k}\\sqrt{m}=const\\sqrt{m}" or

"m|\\vec{v}|=const\\sqrt{m}" then "|\\vec{v}|=\\frac{const}{\\sqrt{m}}" . With an increase in "m" ,"v" decreases and vice versa.

2.In angular motion (circle) with "|\\vec{v}|=const" angular momentum must be constant.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


text:
# How can I derive n-step off-policy temporal difference formula?

I was reading the book "Reinforcement Learning: An Introduction" by Sutton and Barto. In section 7.3, they write the formula for n-step off-policy TD as

$$V(S_t) = V(S_{t-1}) + \alpha \rho_{t:t+n-1}[G_{t:t+n} - V(S_{t-1})],$$

where $$V(S_{t})$$ is state value function of the state $$S$$ at time $$t$$ and $$G_{t:t+n} \doteq \sum_{i=t}^{t+n-1}\gamma^{i-t}R_{i+1} + \gamma^n V(S_{t+n})$$ and $$\rho_{t:t+n-1}$$ is the importance sampling ratio.

I tried to prove this equation for $$n = 1$$ using the incremental update of the value function. Now I end up with this formula: $$V(S_t) = \frac{1}{t} \sum_{j=1}^{t} \rho_{j}G_{j}$$ $$V(S_t)= \frac{1}{t}(\rho_{t}G_{t} + \sum_{j=1}^{t-1}\rho_{j}G_{j})$$ $$V(S_t) = \frac{1}{t}(\rho_t G_t + (t-1)V(S_{t-1}))$$ $$V(S_t)=V(S_{t-1}) + \frac{1}{t}(\rho_{t}G_{t} - V(S_{t-1}))$$ I know I'm wrong because this does not match with the above equation. But can anyone please show me where I am wrong?

• Do you remember which version of the book you were reading at the time? – nbro Nov 5 '20 at 22:40
• It was a draft version – Swakshar Deb Nov 7 '20 at 14:26


text:
# Generators and relation question

#### nhk

What is the minimum number of generators needed for Z2+Z2+Z2? Find a set of generators and relations for this group.

#### tonio

What is the minimum number of generators needed for Z2+Z2+Z2? Find a set of generators and relations for this group.

Three (3): $$\displaystyle \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2=\left\{a,b,c\;;\;a^2=b^2=c^2=1\,,\,\,[a,b]=[a,c]=[b,c]=1\right\}$$ (change the powers to coefficients if you prefer an additive notation).

With the "obvious" definitions , the above group is a three-dimensional vector space over the field $$\displaystyle \mathbb{F}_2:=\mathbb{Z}_2:=\mathbb{Z}/2\mathbb{Z}$$ .

Tonio

nhk

#### nhk

How would I prove that the minimum number of generators is 3? ( I think this is probable a stupid question, but my book does not explain anything about this generator stuff!!). i also do not understand to [a,b]=[a,c]=[b,c]=1 part. Thanks for your help.

#### Swlabr

How would I prove that the minimum number of generators is 3? ( I think this is probable a stupid question, but my book does not explain anything about this generator stuff!!). i also do not understand to [a,b]=[a,c]=[b,c]=1 part. Thanks for your help.
By definition, $$\displaystyle [g, h]:=g^{-1}h^{-1}gh$$. This is called the commutator. Note that two elements $$\displaystyle g$$ and $$\displaystyle h$$ commute if and only if $$\displaystyle [g, h] = 1$$.

So, what they are saying is that the three generators all commute. In general, if $$\displaystyle G_1 = <X_1 ; R_1>$$ and $$\displaystyle G_2 = <X_2 ; R_2>$$ then $$\displaystyle G_1 \times G_2 = <X_1, X_2 ; R_1, R_2, [X_1, X_2]>$$ because the direct product (although in your question it is the diect sum, but that is fine) means that the elements of each individual group commute with all the others. This holds if and only if the generators commute from the respective groups.

Now, as for the number of generators, note that every element has order two and your group is of order 8 and abelian. This means that it cannot be 2-generated. However, I will leave it to you to show that these facts also means that it cannot be 2-generated (a quick proof by contradiction will suffice).

nhk

#### tonio

By definition, $$\displaystyle [g, h]:=g^{-1}h^{-1}gh$$. This is called the commutator. Note that two elements $$\displaystyle g$$ and $$\displaystyle h$$ commute if and only if $$\displaystyle [g, h] = 1$$.

So, what they are saying is that the three generators all commute. In general, if $$\displaystyle G_1 = <X_1 ; R_1>$$ and $$\displaystyle G_2 = <X_2 ; R_2>$$ then $$\displaystyle G_1 \times G_2 = <X_1, X_2 ; R_1, R_2, [X_1, X_2]>$$ because the direct product (although in your question it is the diect sum, but that is fine) means that the elements of each individual group commute with all the others. This holds if and only if the generators commute from the respective groups.

Now, as for the number of generators, note that every element has order two and your group is of order 8 and abelian. This means that it cannot be 2-generated. However, I will leave it to you to show that these facts also means that it cannot be 2-generated (a quick proof by contradiction will suffice).

Hmmm...if the OP doesn't know the commutator notation I can't understand how she/he is trying to cope with presentations/generators/relations questions, which is usually studied after a basic course in group theory.

Anyway, and i shoty: as the hint in my first post "hinted", the fact that the given group is in fact an elementary finite abelian group makes it clear that its minimal number of generators = its dimension as avector space over some finite field...
Another easy proof could also use the Fratinni subgroup of G since the given group is a 2-group.

Tonio

nhk

#### Swlabr

Anyway, and i shoty: as the hint in my first post "hinted", the fact that the given group is in fact an elementary finite abelian group makes it clear that its minimal number of generators = its dimension as avector space over some finite field...
I decided not to mention that proof as I was unsure if they would have covered this. Needless to say, my grasp of when concepts are encountered is somewhat lacking...

That said, I have been reading a book recently called "Groups, Graphs and Trees". It's excellent, because it covers a lot of useful things, but in a really easy to read way because it's an undergrad book, it was taken from an undergrad course given by someone! It covers things like ends and introduces the Coxeter groups, and I was just flicking through it and I saw a wreath product! My point is, people cover weird things in the undergrads...

#### tonio

I decided not to mention that proof as I was unsure if they would have covered this. Needless to say, my grasp of when concepts are encountered is somewhat lacking...

That said, I have been reading a book recently called "Groups, Graphs and Trees". It's excellent, because it covers a lot of useful things, but in a really easy to read way because it's an undergrad book, it was taken from an undergrad course given by someone! It covers things like ends and introduces the Coxeter groups, and I was just flicking through it and I saw a wreath product! My point is, people cover weird things in the undergrads...

I have the book "Groups, Graphs and Trees", by John Meier. I don't know why you think it is an undergraduate one. It certainly covers some basic stuff, just like free groups and actions, but only shortly and it gets into deep waters right away.
This book could perhaps be covered in a seminar course for advanced undergraduates-graduates, but I doubt highly it could be an open course for unders in most schools I know of.

Tonio

#### Drexel28

MHF Hall of Honor
I decided not to mention that proof as I was unsure if they would have covered this. Needless to say, my grasp of when concepts are encountered is somewhat lacking...

That said, I have been reading a book recently called "Groups, Graphs and Trees". It's excellent, because it covers a lot of useful things, but in a really easy to read way because it's an undergrad book, it was taken from an undergrad course given by someone! It covers things like ends and introduces the Coxeter groups, and I was just flicking through it and I saw a wreath product! My point is, people cover weird things in the undergrads...
I have the book "Groups, Graphs and Trees", by John Meier. I don't know why you think it is an undergraduate one. It certainly covers some basic stuff, just like free groups and actions, but only shortly and it gets into deep waters right away.
This book could perhaps be covered in a seminar course for advanced undergraduates-graduates, but I doubt highly it could be an open course for unders in most schools I know of.

Tonio
Just to add in, I saw the commutator quite a bit as an undergrad (which I guess is right now) and we even covered minimally basic examples regarding generators and relations.

#### Swlabr

I have the book "Groups, Graphs and Trees", by John Meier. I don't know why you think it is an undergraduate one. It certainly covers some basic stuff, just like free groups and actions, but only shortly and it gets into deep waters right away.
This book could perhaps be covered in a seminar course for advanced undergraduates-graduates, but I doubt highly it could be an open course for unders in most schools I know of.

Tonio
The second last paragraph of the preface,

"Groups, Graphs and Trees was developed from notes used in two undergraduate course offerings at Lafayette College, and can certainly serve as a primary text for an advanced undergraduate course."

#### tonio

The second last paragraph of the preface,

"Groups, Graphs and Trees was developed from notes used in two undergraduate course offerings at Lafayette College, and can certainly serve as a primary text for an advanced undergraduate course."

"From two under. courses" seems to be the key, although I still believe this is just one example more of overstatement of his own work by an author.
Here's the list of under. courses at Lafayette College:
Lafayette:Math:Courses

Making a quick check reveals they don't even have any course in group theory, and the theory covered by Abstract Algebra I-II seems to be below the average of what's covered in other places.

So perhaps some teacher did more group theory in some course(s) and that's what the author meant, though it is highly doubtable that most of the material could be gathered from under. courses.

Tonio


text:
Sunday, April 18, 2021 ... //

Decline of education, specialized expertise, and the theoretical background. Stakeholders vs shareholders

Czech politicians have announced that back in 2014, an explosion of a Czech ammo depot near the Slovak border (the village of Vrbětice) was caused by Russian agents. They also wanted to do something with the explosives in Bulgaria. We may get the full intelligence report in days.

In fact, they were the very two same agents Petrov and Bashirov who have allegedly poisoned Skripal using Novičok and yesterday, the Czech Police promoted its photographs of the two men who are "wanted"! Well, it is possible that these "molodtsy" do this much work. After all, the tensions were high in 2014 and the Russian agents' working days should be similarly full as those of James Bond 007. But even if it is true that they contributed to the explosion, I find it obvious that the timing of the revival of this scandal (or a conspiracy theory?) is explained by the takeover of the power over the U.S. by the New Loony Left, their being mental cripples, and their fanatical desire to cripple the relationships with Russia (and others).

Even if the agents have caused it, it is a tragedy that this is being exposed again, in the time when America is becoming nasty against Russia again and Russia is understandably preparing to acquire Ukraine if the tensions run even higher. It is very bad that Rosatom – which is the most sensible competitor, as far as I can say – is being eliminated from the tender to extend Dukovany, the older among the 2 nuclear plants in Czechia (the "nuclear" part in both was made in Russia and it's worked great in those decades). Seconds ago, Russia expelled 20 Czech diplomats, a revenge for the expulsion of 18 Russian diplomats from Czechia. Bad. I am rather uncertain whether the Russian agents are innocent but I am almost sure that the timing now is caused by the rise of the Left in the U.S. – and about our government's role of a cheap slut directly controlled by the D.C. now.

But back to the title.

I finally watched the April 1st-2nd

Prague Café interview with Klaus Part I
Prague Café interview with Klaus Part II
which I previously hesitated to watch, despite the fact that in recent months, I have read, watched, or listened to every single one interview with expresident Václav Klaus, as he emerged as a leading warrior against Covidism (not only) in Czechia. Note that this particular program is called "Prague Café" because Žantovský wants to retake this nice intellectual phrase (and the coffee shops themselves) from the progressives who became associated with the phrase and who consequently reduced the credibility of the phrase dramatically.

Mr Petr Žantovský, a writer, publicist, and economist, was hosting the interview. Well, he came to the Klaus Institute to record the videos. He may look "more moderate" than Klaus from some perspectives but he's been close enough to Klaus for decades, anyway. They wrote a book together in 1998 etc.

In the first part, they discussed the increasingly superficial approach of the people to events and information, an evolution towards idiocracy that is being solidified by the increasingly low-quality media and other forces. All the focus is on the ephemeral, separate events. But people are no longer interested (and they are told not to be interested) in the underlying trends, the context, ideologies, and the umbrella theories within which the events may be explained. Klaus mentioned a would-be "cool" Czech journal from the communist era, "100+1 Foreign Curiosities" (Žantovský remembered the name), as a template of this misguided emphasis of the irrelevant separate factoids.

People are just led not to understand the trends and the theoretical underpinnings of their fields or any fields; they are encouraged not to understand what matters about the world. They can't correctly place any of the separate facts into the context, they can't compare the individual events with others, they can't sensibly interpret anything, and they don't even understand that the interpretation is the hard thing – depending on the main part of the actual knowledge. I couldn't agree more. Well, here I probably wrote something slightly different but I believe that Klaus' comments about these matters are equivalent. He says that everyone wants to talk about "facts" only. But there aren't any facts per se. Facts only exist when processed through theories and a methodology to evaluate them. All facts are theory-laden, a general fundamental principle that is being often contradicted explicitly. And because most of the people don't understand any theories, they don't properly understand any facts, either.

This general point has many dimensions. For example, we would fully agree with Richard Lindzen that contemporary "climatologists" are learning to do things incorrectly because they increasingly ignore the "theories of atmospheric physics", the essential theoretical background. It's really what matters for being a real competent scientist; having a computer model (plus some self-confidence and the views that "everyone can do everything, anyway") just cannot possibly replace it. Having isolated facts or pieces of information or computer programs is totally worthless if you need to make valid predictions or wise recommendations and you haven't mastered the "textbook theoretical material"!

This issue is slightly related to another topic they talked about, the education systems. Our, Central European, education systems had some standards and for centuries, thanks to folks like Marie Theresa and John Amos Comenius, they insisted on the students' comprehensive and full understanding of a field and the proper relationships between all subjects that make the field. Comenius already emphasized that the insights in a class must be connected with other insights in the same class and even in other classes! Instead, it's been normal in the West (even in 1968-1969 when Klaus visited the U.S. as a student; he says that there are basically no quality colleges in the U.S. anymore) to register for theoretical physics but also Icelandic medieval poetry (greetings to Jane Tolmie) and everything is fine. Clearly, the expertise obtained by spending the time with all these distractions is unavoidably shallower. Central Europe has mostly abandoned its Central European education philosophy and adopted the more superficial Western one.

Students can't solve exams that their equally old counterparts could solve in 1890 (or another moment many decades ago). This isn't just a hypothesis, this experiment has been tried by Klaus' older son when he was the principal of the prestigious PORG liberal high school. The students' inability to solve the old exams seems to be in agreement with the fact that they don't take the education (especially in the college) as seriously as Klaus' generation when he was a student (1958-1963, in this case). Students tend to dedicate "two days before the exam" to the material; Klaus and classmates actually discussed it in the pub etc. (so did we, not sure whether it was to the same extent).

Another aspect of the "end of the world" is all about the stakeholders vs shareholders debate, a topic that was hot among economists half a century ago. It should be even hotter today because the "stakeholders" are winning but the economists who discuss this important issue are no longer the leading voices in the field of economics.

That's also something that I've been deeply annoyed by – and I often discussed it in the context of the claims "Twitter has the right to censor conservatives because it is a private company". You know, I've been an actual stockholder of Twitter several times. The share was minor but I am sure that many shareholders have similar opinions about the censorship as I do, they have a much greater share, and they still can't actually control the company, either. It is totally obvious that actual profit-seeking shareholders wouldn't like to eliminate a substantial part of the GOP voters (consumers bringing profit to the company) by the censorship. Instead of shareholders (who own the company because they have invested some capital into it in the past), the key decisions are increasingly done by "stakeholders", various important but partial interest groups within the corporation who are employees, not owners. They may be important but they are not directly motivated to see the company thrive as a whole – only the shareholders have a clear motivation to do it. The interest of the stakeholders – various jobs at all levels, a loose generalization of labor unions – is unavoidably dominated by partial interests which can't be identified with the well-being of the company as a whole and that is why this "system" just can't be equally good for the company (and the society as a whole). To promote their interests, these stakeholders have increasingly exploited the moralist language. You need to be responsible in this way, another way.

I am sorry but it is extremely irresponsible and pathological to allow these (under healthy circumstances) irrelevant individuals and "bullshitters about responsibility" to decide about the most important matters in the companies. It's the end of capitalism, the end of the world. Companies led by stakeholders don't differ much from companies led by workers or labor unions – a system controlled in this way is undoubtedly a form of socialism (whose most important "proletarians" are much more lazy and smug than the hard-working and poor Soviet-style working-class-based proletarians, however). For example, when Audi banned an ad with a little girl eating a banana, it wasn't done by the real owners, I believe. Such decisions are being done by someone who doesn't actually own many shares of the companies – and the fact that such people (often straight SJW loons and even more often, not quite straight ones) control the companies means that we no longer live in capitalism. Can't the actual owners of the companies flush the SJWs and other "stakeholders" (who are sucking quite some blood from the company) down the toilet? What is a better places where all the world's SJWs should be flushed? In proper capitalism, they could but we don't live in proper capitalism anymore. So flushing stinky SJWs, just like any other excrements, down the toilet would probably be considered politically incorrect. The actual capitalists don't have sufficient freedom, legal strength, and/or testicles in the present conditions to deal with these harmful, parasitic "stakeholders". The precise choice of words is mine (LM), of course.

As another example of a similar phenomenon, Klaus mentioned the current president of the Federal Reserve, Jeremy Powell, who claimed that the powers of the Fed should be extended to... include the fight against inequality and the climate change! As Klaus correctly points out, this is the end of the world, too. The central banks were created to keep a stable currency, to prevent hyperinflation that the Czech readers may know from Remarque's novels (about Germany after the First World War), and they should do nothing else! But this toxic garbage – far left propaganda and pseudoscience such as the climate hysteria and fights against alleged inqualities – are spreading everywhere and their dishonest cheese carriers are increasingly stealing the power and money that they shouldn't possess and they wouldn't possess in a system that is both fair and efficient – in proper capitalism.

Second part

The second part begins with Žantovský's remarks about the divided society (the very meme that the society is divided is dividing it, it acts as a self-fulfilling prophesy) and about the cult of youth. The old generation should no longer occupy the space. My Oma is an old environmental swine. (LM: See also I am scared by innocent children. Incidentally, the song is from late 2019 and it's bizarre that to try to save a greater number of these Omas, Germany and others soon picked the opposite extreme, the Covid-19 lockdowns.)

Concerning the divided society, Klaus said that the implicit assertion that the "society is divided in a 50-50 way" is too optimistic because an apparent majority has been persuaded to follow the rosy future according to the only politically correct truth while the sensible people have been reduced to a self-evident minority. Increasingly drowning in the ocean of brainwashed superficial loons, the likes of your humble correspondent (and some of you, of course) have been turned into marginal characters!

The Sarajevo Assassination was mentioned (this term mostly means Klaus' forced resignation in 1997, while he was visiting Sarajevo) and Klaus bragged that his foes from 1997 are no longer young while he is still alive these 23+ years later. A simply clever usage of double standards. ;-)

Klaus insists that Václav Havel was the forerunner or template of the contemporary progressivist, overtly "apolitical" political philosophy or style; many friends tell Klaus that he seems biased or obsessed and he should already allow Havel to sleep in Hell. (The Klaus-Havel ideological conflict was surely the actual, deep political confrontation after 1989, and everything else was a detail. I surely see how Havel's progressivism was ahead of his time and how many theses of Havel perfectly agree with the current illnesses of the society but I still find it illegitimate for Klaus to automatically assume that Havel would help every single pathology that is taking place today.) Klaus says that Havel needs to be studied as a social phenomenon, there is no personal obsession here. On the other hand, Klaus rarely talks about the Sarajevo Assassination, and when (senior painter) Knížák and his spouse told him (in the luxurious home of the Knížáks) that the assassination was immoral, Klaus said "yes and no". For Klaus, the assassination was primarily a political confrontation as such, with some form (involving the human dimension and betrayal) – it was a branch of Klaus' confrontation with Havlism, he would probably agree.

Žantovský mentioned that he, Klaus, and our side are being labeled by the progressives. We are the misinformers and conspiracy theorists, among other things. Can't the outcome be that the public will be fed up with these labels and these constant unjustified vacuous insults will become counterproductive? Klaus says that these insults are a "way to simplify the argumentation". Some trivial sins are easier to understand for many people than actual deep mistakes. PM Babiš should have fired the ex-ex-minister of healthcare Col Prymula (Czechia already has the 4th minister of health in the Covid era) much earlier, for profound mistakes, but he only fired him after a trivial "ugly" visit to a Vyšehrad Restaurant, against his own lockdown rules. And that firing due to a trivial sin was no anomaly. Prymula, a high quality chess player, accidentally went to see a soccer match and Babiš fired him again (from the role of his personal aide) for being a spectator, another superficial trivial sin. Babiš doesn't decide about such ministers according to the merit of their work; but according to the moral appearances. Similarly, when labels are attached to Klaus, the attachers and audiences don't have to study what Klaus is actually saying. It's so comfy. People should think about this methodology – and how they're being had.

Klaus would be happy if the society were getting saturated with this propaganda but he seems to think it is just a wishful thinking. A pessimistic number that Klaus mentioned was that in a poll at the Parliamentary Letters – which have an overwhelmingly non-PC, anti-Brussels... audience (and in these issues, they are on Klaus' side) – 77% of readers supported a "full lockdown". It's a terrifyingly high number but one must understand that the readers of PL are generally pro-authoritarian and the lockdown seems rather natural in that mental context. It is clearly not "quite" a bunch of Klaus voters (and the support for the lockdown was actually well below 50% at iDNES.cz and maybe even Novinky.cz). But Klaus is rightfully frustrated. He complained that the changing trends towards shrinking Covid-19 "cases" didn't seem to make the people significantly more anti-lockdown. Again, the data themselves don't work. It matters how they're interpreted and distributed to assorted drawers.

Klaus looks at these questions (about the public that might be fed up) as an economist, through "economics of information". So we must distinguish the supply of information and demand for information; the "information" without extra words is "nothing". The Czech Public Television produces some supply of factoids about Covid, Klaus has never contributed to the demand, but others do. (Klaus is just showing a newly published book – hot from the printing presses – "Reason Against [Covid] Panic" that he is seeing for the first time; the Czech medical elite has co-authored the book.)

Žantovský discusses a topic that I saw in some Klaus' texts LOL that when it comes to the paradigm shifts, the world is turning away from Orwell – where the transformation of the people was "forced" – to Huxley where the people are changing themselves voluntarily. I know very well that Klaus totally agrees with it (Orwell was the Bible but Huxley turned more relevant) but I don't know whether Žantovský constructed the same ideas independently of Klaus ;-). Maybe Orwell was right for the old communism but Huxley is appropriate for its current replacement.

Klaus finally mentioned the aforementioned last high quality university that is left in the U.S., the Hillsdale College near Detroit. In a recent issue of their university journal, the president – whom Klaus knows, Klaus is sometimes invited there – was analyzing the three main dystopias, Orwell's 1984, Huxley's Brave New World, and Koestler's Darkness at Noon (and another, American one that Klaus doesn't know). The president has established a one-semester course about the dystopias (where he teaches as an instructor, too)! ;-) The Academic boss feels that it is a fundamental topic of the present, wants the students to think whether the dystopias are easy to bring, and so on. Klaus also promotes another book by his institute, The Self-Destruction of the West. It is a crucial ongoing process even though some would-be wise pundits keep on childishly warning about the threats coming from the East. These childish people can't see the tip of their noses and are completely missing any real content in the real world events.

At the very end, Žantovský asked – like other journalists – whether Klaus wants to return to the real, party-based politics. Klaus was disappointed that Žantovský joined the spectrum of journalists who always kill a one-hour interview by this superficial question "Will you be at the Castle again". That's sad, Klaus says. To show his respect, Žantovský agreed that this was an OK answer to his last question, and thanked President (Emeritus) Klaus for the interview (while saying it is a good tradition to keep the title "President" after the tenure).


text:
Tag Info

7

Yes. Such proofs are possible for El Gamal. It involves a zero knowledge proof of equality of a discrete log, together with the homomorphic property of El Gamal encryption. Recall that given $E(a)$ and $E(b)$, anyone can form $E(a/b)$ using the homomorphic property of El Gamal. Suppose $E(a/b)=(r,s)=(g^k,h^k a/b)$ (where $g$ is the generator and $h$ is ...

6

Well, whether it is a secure tweakable block cipher depends on how resistant (E,D) are to related key attacks; that's not a standard assumption for block ciphers. For example, this would not be a secure tweakable block cipher with 3DES; because every 8th bit is ignored, the attacker can effectively test the value of the 7 adjacent bits (except for the 7 ...

4

Encryption using a block cypher such as AES by passing plaintext blocks directly to the encryption function is known as Electronic Code Book mode (ECB) and is not CPA secure as (as you say in your question) it is entirely deterministic and two identical plaintext blocks will result in two identical ciphertext blocks. To prevent this an initialisation ...

2

The Caesar cipher (aka Shift cipher) has, as you said, a key space of size 26. To achieve perfect secrecy, it thus can have at most 26 plaintexts and ciphertexts. With a message space of one character (and every key only used once), it would fit the definition of perfect secrecy. For the usual use with messages longer than one character, or multiple ...

2

Not a complete answer, but since you mentioned "unmodified RSA" I feel it's relevant. Something stronger than vanilla RSA is necessary, even if it isn't semantic security. Example: What if you have a public key exponent of 3 and the symmetric key being encrypted is 16 bytes long? Using raw RSA, $m^e$ would be about $128 * 3 = 384$ bits long and thus ...

1

To be secure against a chosen-plaintext attack, an encryption scheme must be non-deterministic — that is, its output must include a random element, so that e.g. encrypting the same plaintext twice will result in two different ciphertexts. Indeed, if that was not the case, an attacker could easily win the IND-CPA game just by using the encryption ...

1

You are in a twist here: semantic security (equal to IND-CPA) can only be fulfilled by probabilistic encryption schemes. You need a deterministic encryption scheme for your drop-out tolerance. As it was pointed out previously, any homomorphic encryption allows you to proof in zero knowledge the equality of two ciphertexts: known: $c_0 = ... 1 Well, the obvious way to do this is: Before the protocol occurs, Alice runs the$Gen$procedure to create a public and a private key For her round, Alice sends her public key to Bob For his round, Bob selects a random symmetric key$\in \{0,1\}^n$, encrypts it with Alice's public key, and sends that encryption to Alice. Alice decrypts the message that Bob ... 1 To expand / generalize @poncho's reply, given a block cipher$(E,D)$with keylength$n$, you can make a new one$(E',D')$with key length$n+1$, which ignores the last bit of the key and just runs$(E,D)$. If$(E,D)$is a secure PRP, then so is$(E',D')$. But plugging$(E',D')\$ into the OP's construction does not yield a secure tweakable block cipher. To see ...

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


text:
# A riddle: What is the inertial frame of a clock? [closed]

Parts of the clock must be moving for the clock to tick.

-
Rest frame of the center of mass – Mitchell Porter Feb 5 at 9:37
Not even wrong. – user20622 Feb 5 at 9:57
I'll reopen this if you can elaborate more -- currently your question is rather vague. – Manishearth Feb 5 at 14:02
Not to mention we frown on riddle questions. The FAQ does say you should only ask practical, answerable questions based on problems you actually face, which excludes cases where you already know the answer and you're just teasing the community to find it (among other things). – David Zaslavsky Feb 5 at 20:58

## closed as not a real question by Manishearth♦Feb 5 at 14:02

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

But since the relative velocities and accelerations are ridiculously small compared to $c$ you would have to be an ubernerd to want to take them into account. As Mitchell says in his comment, you would just take the centre of mass of the clock to define its frame.


text:
# Math Help - generating function

1. ## generating function

If you have a three decks of cards, how many ways are there to pick a hand of m cards from n distinct cards in a single deck. Use a generating function.
I have Gn(x) = (1+x+x^2+x^3)^n but don't know where to go from here.

2. Hello,

Find the expansion of the generating function, and the answer you're looking for will be the coefficient in front of $x^m$

3. I have seen an example with two decks where the binomial theorem was used twice, is it possible to use it three times, if so how?

4. Originally Posted by CoraGB
If you have a three decks of cards, how many ways are there to pick a hand of m cards from n distinct cards in a single deck. Use a generating function.
I have Gn(x) = (1+x+x^2+x^3)^n but don't know where to go from here.

Hi CoraGB,

Break the problem into two parts: (1) Find the OPSGF (ordinary power series generating function) of the number of ways to choose the deck; (2) find the OPSGF of the number of hands in a deck.

(1) Is easy: it's just $f(x) = 3$.

(2) Let $a_m$ be the number of ways to choose a hand of m cards from a deck of size n, and let $g(x) = \sum a_m x^m$. Then $g(x) = (1+x)^n$.

Finally, the OPSGF of the number of ways to choose the deck and choose a hand is just

$f(x) \cdot g(x) = 3 (1+x)^n$.


text:
# Is every sequence of functions with uniformly bounded variation almost equicontinuous?

Let $$f_n: [0, 1] \to \mathbb R$$ be a sequence of functions.

Given a measurable subset $$E$$ of $$[0, 1]$$, we say that the sequence $$f_n$$ is equicontinuous on $$E$$ if for every $$x \in E$$, and $$\varepsilon > 0$$ there exists a $$\delta > 0$$ and all $$n \in \mathbb N$$ we have $$|f_n(x) - f_n(y)| < \varepsilon$$ for all $$y \in [0, 1]$$ such that $$|y - x| < \delta$$.

We say a sequence of functions $$f_n: [0, 1] \to \mathbb R$$ has uniformly bounded variation if there exists some $$M > 0$$ such that each $$f_n$$ has total variation less than or equal to $$M$$.

Question: Given any sequence of functions $$f_n: [0, 1] \to \mathbb R$$ of uniformly bounded variation, is it true that there exists a measurable subset $$E$$ of $$[0, 1]$$ with Lebesgue measure $$1$$ and a subsequence $$f_{n_k}$$ of lower density $$1$$ such that $$f_{n_k}$$ is equicontinuous on $$E$$?

Note: We define the lower density of a subsequence $$f_{n_k}$$ to be the number

$$\liminf_n\frac {\#\{k \, | \, k \in \mathbb N, \, n_k < n\}}{n}$$

where $$\#$$ denotes the cardinality of a finite set.

• I think if you take the pairs $(n,k)\in\mathbb{N}^2;0\leq k<n$ in lexicographic order and for each pair $(k,n)$ you define the function $f_{n,k}$ defined by $0$ in $[0,\frac{k}{n}]$, $1$ in $[\frac{k+1}{n},1]$ and interpolating linearly in $[\frac{k}{n},\frac{k+1}{n}]$ that gives a counterexample (you don't even need the subsequences to have lower density $1$) Jun 4 at 15:00
• Hmm, the points at which equicontinuity fail would depend a lot on the chosen subsequence in that case. Do you have a particular subsequence in mind? Jun 4 at 15:31
• Oh sorry, I was thinking of uniformly equicontinuous (it was the only type of equicontinuity I had seen). Now it makes sense why the subsequence needs to have lower density $1$ Jun 4 at 15:41
• Ah it does so happen that pointwise equicontinuity implies uniform equicontinuity (if I’m not mistaken), so it makes sense your counterexample would work. Jun 5 at 0:29
• That is true if we are talking about the whole interval $[0,1]$. With some subset $E$ of measure $1$ both conditions may not be equivalent, for example we can construct a family of functions pointwise equicontinuous in $(0,1)$ but not uniformly equicontinuous in $(0,1)$: the functions $f_n$ given by $0$ in $[0,\frac{n-1}{n}]$ and $nx-(n-1)$ in $[\frac{n-1}{n},1]$ Jun 5 at 0:36

Consider the pairs $$(n,k)\in\mathbb{N}^2$$ with $$0\leq k<2^n$$, they form a sequence in lexicographic order. Consider now the functions $$f_{n,k}$$ which are defined as $$0$$ in $$[0,\frac{k}{2^n}]$$, $$1$$ in $$[\frac{k+1}{2^n},1]$$, and $$2^nx-k$$ in $$[\frac{k}{2^n},\frac{k+1}{2^n}]$$, and suppose there is a set $$E\subseteq [0,1]$$ and a subsequence of $$f_{n,k}$$ as in the question.
If we define $$A_n=\bigcup\{(\frac{k}{2^n},\frac{k+1}{2^n});(n,k)\text{ is in the subsequence}\}$$, then the fact that the subsequence has density $$1$$ implies that when $$n$$ tends to $$\infty$$, the measure of $$A_n$$ tends to $$1$$. This implies that there is a point $$x\in E$$ which is in infinitely many of the $$A_n$$: for example, take a subsequence $$A_{n_i}$$ such that $$\sum_i(1-m(A_{n_i}))<1$$; then $$m(\cap_k A_{n_i})>0$$), so $$E\cap(\cap_k A_{n_i})\neq\emptyset$$.
Suppose then that $$x\in E\cap(\cap_i A_{n_i})$$ and we are given $$\varepsilon<\frac{1}{2}$$; there is no $$\delta$$ that satisfies the definition of equicontinuity at $$x$$: indeed, take $$i$$ such that $$2^{-n_i}<\delta$$, then there is some interval $$(\frac{k}{2^{n_i}},\frac{k+1}{2^{n_i}})$$ contained in $$(x-\delta,x+\delta)$$ and such that $$(n_i,k)$$ is in the subsequence. So the image of $$(x-\delta,x+\delta)$$ under the function $$f_{n_i,k}$$ is all $$[0,1]$$, thus it contains values at distance $$>\varepsilon$$ of $$f_{n_i,k}(x)$$.


text:
# fourier transform of cumulative function

Hi

I've encountered a test that uses the cumulative value of a finite time series to deterime the data set's stationarity.

I would like to know the characteristics of this test in frequency space, so would therefore need to know the fourier transform of this cummalative function.

I was assuming that the function could be writen as the product of the initial time series function, f, and a heavyside step function, H. With the defining function written as:

$g(t) = \sum^{t}_{n=1} f(n) H(t-n)$

I would like to know fourier transform of the function g(t). Is there a straightforward solution to this function or does the use of the discontinuous step function make a solution impossible.

-

## 1 Answer

Read

Discrete-Time Signal Processing (2nd Edition) (Prentice-Hall Signal Processing Series) [Hardcover] Alan V. Oppenheim (Author), Ronald W. Schafer (Author), John R. Buck (Author)

and all will be revealed.

-


text:
# Induction problem

dustwave
Hello, I have this problem in my calculus class where I have to prove a formula with induction.

the problem is:

( 1 * 3 * 5 * ... * (2n - 1) ) / ( 1 * 2 * 3 * ... * n) =< 2^n

=< = equal or lesser than

P(1) is easy to solve, and so is P(k), but I start having problems with P(k+1) to prove the formula.. can someone give me a hand? =)

thanks!

/Maximilian

Homework Helper
Yes, P(1) is "easy to solve": 1/1= 1< 21.
I'm not sure what you mean by saying P(k) is "easy to solve"- there's nothing to solve there!

You do, of course, assume that (1*2*...*(2k-1))/(1*2*...*k)<2k

Now, for P(k+1), you have to look at (1*2*..*(2k-1)(2(k+1)-1)/(1*2*...*k*(k+1))
2(k+1)-1= 2k+1 of course, so this is
(1*2*...*(2k-1)*(2k+1)/(1*2*...*k*(k+1)= {(1*2*...*(2k-1))/(1*2*...*k)}{(2k+1)/(k+1)}< 2k{(2k+1)/(k+1)}.

Looks to me like you need to prove that (2k+1)/(k+1)< 2 for all k! I would be inclined to write that as a "lemma" first and use induction to prove it.

tongos
(2n-1)!/{(2^(n-1)(n-1)!(n!)}=<2^n

{2^n(n!)}{2^(n-1)(n-1)!}>=(2n-1)!

(2n-1)!/(n!)(n-1)!=<2^(2n-1)

C(2n-1,n)=<2^(2n-1)
True?


text:
# Converting Degrees to Radians

Converting Degrees to Radians
Go back to  'Radian'

Let us analyze the following situation for a better understanding.

Drag the slider to change the angle of the car's position with reference to its starting point on the x-axis.

The angle's measurement in degrees is changing with the car's respective position as it moves forward.

What are the different units to represent this measurement? Let's explore this through the simulation and then read the lesson to learn about radian.

Do you know that radian was the SI supplementary unit for measuring angles before 1995? It was later changed to a derived unit. In this mini-lesson, we will explore the world of converting degrees to radians using interactive simulations. We will also cover the various concepts on the subject including converting degrees to radians formula, converting degrees to radians calculator, converting degrees to radians worksheet, degrees to radians chart, and degrees to radians equation.

Do not forget to try your hand at solving a few interesting practice questions at the end of the page.

## Explain Degrees

A degree, which is referred to as the degree of arc or arc degree, is the unit of measuring a plane angle.

It is denoted by the symbol $$(^\circ)$$.

A complete rotation is denoted by the angle measuring $$360^\circ$$ and the device used to measure an angle in degrees is known as the protractor.

The following figure shows different scales of measuring an angle in degrees:

Radian is another unit of measuring an angle which is made by wrapping a radius along the circle.

One Radian is $$\frac{180}{\pi}$$ degrees, which is equal to about 57.2958 degrees.

One complete counterclockwise revolution, in radians, is equal to $$2 \pi$$.

## How to Convert Degrees to Radians?

The amount of rotation from the starting side to the terminal side determines the measure of an angle.

One complete counterclockwise revolution in radians is equal to $$2 \pi$$.

Similarly, in degrees, one complete counterclockwise revolution is $$360^\circ$$.

Comparing both the values, we get,

\begin{align*} 360^\circ &= 2 \pi \\ 1^\circ &= \frac{360}{2 \pi} \\ 1^\circ &= \frac{180}{\pi} \end{align*}

Thus, to convert degrees to radians, we use the formula:

 Radians = Degrees $$\times \frac{\pi}{180}$$

Think Tank
• How would you experimentally calculate the value of $$\pi$$ using a given circle?

## Degrees to Radians Conversion Calculator

Check out this degrees to radians conversion calculator to see how degrees are converted to radian.

## Degrees to Radians Conversion Table

You can refer to the following table to see the radian value of some angles given in degrees:

$$0^\circ$$ $$0$$ rad
$$30^\circ$$ $$\frac{ \pi }{ 6}$$ rad
$$45^\circ$$ $$\frac{ \pi }{ 4}$$ rad
$$60^\circ$$ $$\frac{ \pi }{ 3}$$ rad
$$90^\circ$$ $$\frac{ \pi }{ 2}$$ rad
$$120^\circ$$ $$\frac{ 2 \pi }{ 3}$$ rad
$$135^\circ$$ $$\frac{ 3 \pi }{ 4}$$ rad
$$150^\circ$$ $$\frac{ 5 \pi }{ 6}$$ rad
$$180^\circ$$ $$\pi$$ rad
$$270^\circ$$ $$\frac{ 3 \pi }{ 2}$$ rad
$$360^\circ$$ $$2 \pi$$ rad

Challenging Question
• A certain line from the origin divides the third quadrant into two equal sections. What will be the anti-clockwise angle subtended between this line and the reference line in radians?

## Solved Examples

 Example 1

In a circle with center O, points A and B lie on the circle. $$\angle{AOB }$$ has a measure of $$60^\circ$$. What is the radian measure of $$\angle{AOB}$$?

Solution

We have,

$$\angle{AOB} = \frac{\pi}{3}$$

Degrees to radians conversion formula is given as,

Radians = Degrees $$\times \frac{\pi}{180}$$

Thus converting the given angle from degrees to radians, we get,

\begin{align*} \angle{AOB} \text { in radians} &= \angle{AOB} \text { in degrees} \times \frac{\pi}{180} \\ \angle{AOB} &= 60 \times \frac{\pi}{180} \\ &= \frac{\pi}{3} \end{align*}

Hence,

$$\angle{AOB} = \frac{\pi}{3} \ \text {rad}$$

 $$\therefore$$ $$\angle{AOB} = \frac{\pi}{3} \ \text {rad}$$
 Example 2

Katie bought a pizza from a nearby store. Her mother gave her a slice which measured $$\frac{1}{4}^ \text {th}$$ of the complete pizza.

Katie asked her mother about the angle that her slice of pizza made.

Can you help Katie find the measurement of the angle in degree and radian?

Solution

Here,

The value portion Katie got = $$\frac{1}{4}^ \text {th}$$ of the complete pizza.

Since the shape of a pizza is circular, and the angle for a complete circle = $$360^\circ$$, thus we can conclude that:

Angle for the complete pizza = $$360^\circ$$

The angle for the $$\frac{1}{4}^ \text {th}$$ slice = $$\frac{360^\circ}{4} = 90^\circ$$

We know,

The value of this angle in radians = value in degrees $$\times \frac{\pi}{180}$$

Thus,

The angle of Katie's slice in radians = $$90^\circ \times \frac{\pi}{180} = \frac{\pi}{2}$$

 $$\therefore$$ The angle of Katie's slice = $$90^\circ = \frac{\pi}{2}$$

## Interactive Questions

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

## Let's Summarize

The mini-lesson targeted the fascinating concept of converting degrees to radians. The math journey around converting degrees to radians starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## 1. Explain the radian of a circle?

Radian is another unit for measuring angles. It is the angle made by wrapping a radius along the circle.

One radian is $$\frac{180}{\pi}$$ degrees, which is equal to about 57.2958 degrees.

One complete counterclockwise revolution in radians is equal to $$2 \pi$$

More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus


text:
Combinatorics and Graph Theory

•  Jinting Liang, Michigan State University
•  Enriched toric $[\vec{D}]$-partitions
•  12/07/2022
•  3:00 PM - 3:50 PM
•  Online (virtual meeting) (Virtual Meeting Link)
•  Bruce E Sagan (bsagan@msu.edu)

In this talk I will discuss the theory of enriched toric $[\vec{D}]$-partitions. Whereas Stembridge's enriched $P$-partitions give rises to the peak algebra which is a subring of the ring of quasi-symmetric functions QSym, our enriched toric $[\vec{D}]$-partitions will generate the cyclic peak algebra which is a subring of cyclic quasi-symmetric functions cQSym. In the same manner as the peak set of linear permutations appears when considering enriched $P$-partitions, the cyclic peak set of cyclic permutations plays an important role in our theory.

Contact

Department of Mathematics
Michigan State University


text:
Services & ResourcesWolfram Forums
 MathGroup Archive
 2005 January February March April May June July August September October November December

Solving a weakly singular integral equation - Take 2.

• To: mathgroup at smc.vnet.net
• Subject: [mg54485] Solving a weakly singular integral equation - Take 2.
• From: Zaeem Burq <Z.Burq at ms.unimelb.edu.au>
• Date: Mon, 21 Feb 2005 03:44:55 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

Dear all,

I am still trying to solve a Linear Volterra type-2 integral equation by
using successive approximations method. I am mainly interested in the
behaviour of the solution near zero.

The unknown function is f[t]. Define

p[x_]:= Exp[- 0.5 * x^2]/Sqrt[2 Pi] (* Gaussian probability density
function*)

c[t_]:= Sqrt[2.1 t Log[Log[1/t]] ]

G[s_,t_]:= (c[t] - c[s]) p[(c[t]-c[s])/Sqrt[t-s]] / (t-s)^1.5  (* The
kernel *)

The integral equation is:

f[t] = c[t]p[c[t]/Sqrt[t]] / t^1.5 - \int_{0}^{t} G[s,t] f[s] ds

As you can see, the kernel has a $(t-s)^1.5$ in the denominator, i.e.,
singularities along the diagonal (s=t). These singularities are
appropriately killed by the function p, but Mathematica seems to have
trouble with them:

Since c is differentiable (except at zero), the factor c[t] - c[s] in the
numerator of the kernel is O(t-s) when t-s is small and t > 0. This means
that the denominator (t-s)^1.5 is only as bad as Sqrt[t-s]. This combined
with the exponential function, means that the kernel is actually at-least
Holder continuous of order -0.5. Even when t=0, the exponential function
kills the pole at zero. It is known from other sources, that this integral
equation has a unique solution.

I wrote down the following routine based on a paper in the Mathematica
Journal: n is the number of iterations of the approximation process:

\!$$Clear[c, approxsoln, K]\n n = 3\n 3\n \(p[x_] := \[ExponentialE]\^\(\(-x\^2$$/2\)\/\@$$2 \[Pi]$$;\)\n
$$c[t_] := Sqrt[2.1\ t\ Log[Log[1/t]]];$$\n
$$approxsoln[x_] = 0;$$\n
$$G[s_, t_] := \(\(\$$$$\((c[t] - c[s])$$\ p[$$c[t] - c[s]$$\/\@$$t - s$$]\)\
\)\/$$(t - s)$$\^1.5;\)\n
$$For[j = 1, j <= \ n, \(j++$$, \[IndentingNewLine]values =
Table[{t, $$\(\$$$$c[t] p[c[t]\/\@t]$$\)\/$$(t)$$\^1.5\  -
NIntegrate[G[s, t]*approxsoln[s], {s, 0, t}]}, {t,
0.0000000001,  .01,  .01\/10}]; \n
approxsoln[t_] =
InterpolatingPolynomial[values, t]; \[IndentingNewLine]Print[
j]];\)\[IndentingNewLine]
Plot[approxsoln[x], {x, 0,  .01}]\)

The PROBLEMS:

1. I have had to truncated the integral away from zero - i.e., s runs from
0 to t, but t runs from 0.0000000001 to 0.01. It should run from 0 to 0.1.
This is not entirely satisfactory, as I am mainly interested in behaviour
near zero.

2. As the code stands, Mathematica tells me that the numerical integrals
converge too slowly.

I'd appreciate any help.

Best, Zaeem.

________________________________

Zaeem Burq
PhD Stochastic Processes,
Dept. of Mathematics and Statistics,
Unimelb.

Room 201, Richard Berry Building
University of Melbourne,
Parkville, VIC 3052.

ph: 8344 4248.

http://www.ms.unimelb.edu.au/~zab

• Prev by Date: Re: Re: Re: Re: Re: Bug Report - Two numerical values for a same variable
• Next by Date: Re: Plotting symbolic representation of numerica function
• Previous by thread: Re: Printing numbers
• Next by thread: Re: Solving a weakly singular integral equation - Take 2.


text:
I am trying to understand a test called Hadamard Overlap Test, which consists of a destructive swap test (section IV of swap test and Hong-Ou-Mandel effect are equivalent) right after a Hadamard test. The circuit is from the Variational Quantum Linear Solver paper:

The authors claim that "conditioning the measurement on the ancilla qubit to yield the $$|\boldsymbol{0}\rangle$$ state, we can perform the depth-two Overlap circuit between registers S1 and S2 to get": \begin{aligned} P(0) &=\frac{1}{2}\left(\left\langle\mathbf{0}\left|U^{\dagger} V\right| \mathbf{0}\right\rangle\left\langle\mathbf{0}\left|V^{\dagger} U\right| \mathbf{0}\right\rangle\right.+\left\langle\mathbf{0}\left|U^{\dagger} A_{l^{\prime}} A_{l} V\right| \mathbf{0}\right\rangle\left\langle\mathbf{0}\left|V^{\dagger} A_{l}^{\dagger} A_{l^{\prime}}^{\dagger} U\right| \mathbf{0}\right\rangle \\ &\left.+\operatorname{Re}\left[\left\langle\mathbf{0}\left|U^{\dagger} A_{l} V\right| \mathbf{0}\right\rangle\left\langle\mathbf{0}\left|V^{\dagger} A_{l^{\prime}}^{\dagger} U\right| \mathbf{0}\right\rangle\right]\right) \end{aligned}

$$|\boldsymbol{0}\rangle$$ represents a vector of 0 qubits, those making up the registers $$S_1$$ or $$S_2$$. I can't understand how to derive this result. In particular, I have the following questions:

• What are we exactly measuring by $$P(0)$$? From the circuit, it seems like all the qubits are measured, and that is also the case when you do a swap test according to the reference above. I initially thought that $$P(0)$$ is the probability of getting 0 in the ancilla, but this should not be the case, because otherwise what is the effect of the CNOT and H performed? What I mean is that the gates inside the box are not controlled on the ancilla, so they cannot change the outcome of the ancilla.
• How do you get to this result? Below is my attempt and what I achieved so far

\begin{aligned} & |0 \rangle |\boldsymbol{0} \rangle |\boldsymbol{0} \rangle \xrightarrow{V, U} |0 \rangle V|\boldsymbol{0} \rangle U|\boldsymbol{0} \rangle \xrightarrow{H}\frac{1}{\sqrt{2}}\left(|0 \rangle +|1 \rangle \right)V|\boldsymbol{0} \rangle U|\boldsymbol{0} \rangle \xrightarrow{A_l, A_{l'}^{\dagger}}\\ &\frac{1}{\sqrt{2}}\left(|0 \rangle V|\boldsymbol{0} \rangle U|\boldsymbol{0} \rangle +|1 \rangle A_lV|\boldsymbol{0} \rangle A_{l'}^{\dagger}U|\boldsymbol{0} \rangle \right) \xrightarrow{H}\\ & \frac{1}{2}\left[|0 \rangle(V|\boldsymbol{0} \rangle U|\boldsymbol{0} \rangle +A_lV|\boldsymbol{0} \rangle A_{l'}^{\dagger}U|\boldsymbol{0} \rangle )+|1 \rangle (V|\boldsymbol{0} \rangle U|\boldsymbol{0} \rangle -A_lV|\boldsymbol{0} \rangle A_{l'}^{\dagger}U|\boldsymbol{0} \rangle )\right] \end{aligned} Now, one measures the ancilla, if it results in a 0 state, the system collapses in $$$$\frac{1}{2}(V|\boldsymbol{0} \rangle U|\boldsymbol{0} \rangle +A_lV|\boldsymbol{0} \rangle A_{l'}^{\dagger}U|\boldsymbol{0} \rangle)$$$$ How does one implement the the CNOTs and H now? What are the further steps that yields the results of the authors?

• It seems that the VQLS paper has not adequately explained their procedure. As you said, they are not measuring only the ancilla as in a standard hadamard test. Instead, they are measuring each qubit. The question that I have is how do they calculate the overlap after measuring the probabilities of each state? The Garcia-Escartin et al. paper that you reference has an equation for estimating the overlap given in (20), however, it is not clear how that may apply here. – thespaceman Apr 22 at 23:54
• Have you tried calculating the probability that that ancilla, and all the other measurement results, give 0? The reason I suggest this is that the boxed area is equivalent to measuring in the Bell basis, which kind of teleports one system to the other. That's how I see you getting the cross terms like $\langle 0|U^\dagger V|0\rangle$ (ever seen the Choi-Jamiolkowski isomorphism?). – DaftWullie Apr 23 at 7:00
• As I understand it, the hadamard overlap test calculates the product of two different expectation values. In equations (C3) - (C5) of the VQLS paper you can see that they accomplish this in an analogous way as the simple hadamard test. Given your comment, are you suggesting that $Re[\gamma_{ll'}]=P(00...0) - P(11...1)$? Where $P(00...0)$ means the $0$ state for all qubits. If not, what is your suggestion for calculating $Re[\gamma_{ll'}]$? – thespaceman Apr 26 at 17:31


text:
# Does having a finite number of generators, each having finite order, imply that the group is finite?

Suppose that I have a group $G$ with a finite number of generators $(a_1,a_2,\ldots,a_n)$, and the orders $|a_1|,|a_2|,\ldots,|a_n|$ are all finite. Does it follow that the order of the group $|G|$ is necessarily finite?

I'm not sure how to even approach proving/disproving this problem. To try disproving it, I tried finding an infinite group that had a finite number of generators each with finite order, but everything I tried didn't seem to work:

• $\mathbb{Z}$ has generators of infinite order
• $GL_n(\mathbb{R})$ the set of invertible $n\times n$ matrices, also has generators of infinite order
• The infinite cyclic group has a finite number of generators, but of infinite order

I don't really know many examples of groups (infinite or finite), so I got bummered here.

• You may be interested in looking up Dietzmanns's Lemma: "In any group a normal finite subset consisting of elements of finite order generates a finite group". books.google.co.uk/… Dec 24 '16 at 11:37
• Note that 'the' infinite cyclic group is isomorphic to $\Bbb{Z}$. Have you also considered subgroups of $\operatorname{GL}_n(\Bbb{R})$ generated by matrices of finite order? Dec 24 '16 at 11:49

Consider the free group $G$ on $a$ and $b$, mod the relation that $a^2 = b^2 = e$. Then $G$ admits sequences of arbitrary length, namely $a$, $ab$, $aba$, $abab$, $ababa$, $\ldots$.

• I actually thought of this group as a possible counterexample, but I'm completely new to group theory, so I'm not sure how to prove that all the sequences are distinct? Dec 24 '16 at 4:07
• @JoshuaLin This is isomorphic to the group of maps $f:\mathbb Z\rightarrow \mathbb Z$ of the form $f(x)=a\pm x$ for integer $a$. Yould take $a$ is the map $x\mapsto -x$ and $b$ is the map $x\mapsto 1-x$. This makes it easier to prove that the elements are distinct. Dec 24 '16 at 4:11
• It's also the infinite dihedral group. Dec 24 '16 at 4:14
• @TannerSwett Eric Wofsey's example provides a more concrete representation, with $g$ and $f$ generating the group and having the order indicated. As he mentions, "the explicit representation as permutations makes it clear that all the words you can form by alternating $g$ and $f$ really are distinct elements, since you can just compute what they are as functions". At any rate, if in any group we can have two elements of order two which generate it, and alternating sequences of these are distinct, then this certainly holds in the free group on these two elements, modding that they have order 2. Dec 24 '16 at 5:57
• An different alternative representation could be to let $a$ and $b$ be reflections in two lines in the plane whose mutual angle is an irrational multiple of $360^\circ$. That is, for example, in matrices, $a=({}^1\,{}_{-1})$ and $b=({}^{0.6}_{0.8}\,{}^{\;0.8}_{-0.6})$ in $GL(\mathbb Q,2)$. Dec 24 '16 at 11:45

Define $f:\mathbb{Z}\to\mathbb{Z}$ by $$f(n)=\begin{cases} n+1&\text{ if n is even} \\ n-1&\text{ if n is odd}\end{cases}$$ and $g:\mathbb{Z}\to\mathbb{Z}$ by $$g(n)=\begin{cases} n-1&\text{ if n is even} \\ n+1&\text{ if n is odd.}\end{cases}$$

Then $f$ and $g$ are both permutations of $\mathbb{Z}$, and have order $2$ in as elements of the group of all permutations of $\mathbb{Z}$. But notice that $$(g\circ f)(n)=\begin{cases} n+2&\text{ if n is even} \\ n-2&\text{ if n is odd}\end{cases}$$ so $g\circ f$ is a permutation of infinite order (as you iterate it on even and odd integers, you just keep adding or subtracting $2$, respectively). It follows that the group of permutations generated by $f$ and $g$ is infinite, even though $f$ and $g$ both have finite order.

(In fact, it turns out that this is the same as silvascientist's example. The explicit representation as permutations makes it clear that all the words you can form by alternating $g$ and $f$ really are distinct elements, since you can just compute what they are as functions.)

In plane Euclidean geometry, the group generated two reflections in two parallel lines provides an easy example. The product of the two reflections is a nonzero translation, which has infinite order.

• nice example with geometry. Dec 24 '16 at 15:21

I. The dihedral group $D_n$ is a group of order $2n$ generated by two elements $a_n,b_n$ of order $2.$ Then $a=(a_1,a_2,a_3,\dots)$ and $b=(b_1,b_2,b_3,\dots)$ are elements of order $2$ in the direct product $\prod_{n=1}^\infty D_n,$ and they generate a subgroup $D=\langle a,b\rangle$ which is infinite because each $D_n$ is a homomorphic image of $D$ via the projection map.

II. Any permutation can be expressed as a product of two involutions [W. R. Scott, Group Theory, Prentice–Hall, 1964, Exercise 10.1.17]. In particular, given a permutation $\pi$ of infinite order, we can write $\pi=\alpha\beta$ for some involutions $\alpha$ and $\beta,$ and then $G=\langle\alpha,\beta\rangle$ is an infinite group generated by two elements of order $2.$ For a concrete example where $\pi$ is an infinite cycle in $\operatorname{Sym}(\mathbb N),$ take $$\alpha=(1\ 2)(3\ 4)(5\ 6)(7\ 8)\cdots,$$ $$\beta=(1)(2\ 3)(4\ 5)(6\ 7)\cdots,$$ $$\pi=\alpha\beta=(\cdot\ \cdot\ \cdot\ 7\ 5\ 3\ 1\ 2\ 4\ 6\ 8\ \cdot\ \cdot\ \cdot).$$

III. Any countable group can be embedded in a $2$-generator group with generators of prescribed orders $p\ge3$ and $q\ge2$ [F. Levin, Factor groups of the modular group, J. London Math. Soc. 43 (1968), 195–203]. Hence, any countable group can be embedded in a group generated by three elements of order $2.$

IV. See Burnside's problem for examples of finitely generated infinite groups in which each element has finite order.


text:
# Generating titles

One of Blot’s principles is ﬂexibility. I want to be able to publishing a tiny snippet of text to my blog without Blot complaining. Blot will work around me in the background to generate appropriate metadata (such as publish date or permalink) if none is speciﬁed.

For instance, this is the pseudo code for how Blot generates a title from a ﬁle:

If one of the ﬁrst 3 nodes of the entry is a heading: Title is the text of the earliest greatest heading.

Else if there is a node with text in the entry: Title is the ﬁrst sentence of the ﬁrst node with text.

Else Title is derived from the ﬁle’s name. For instance /word_doc.txt becomes Word doc’ and /2015/why-hello-there.txt becomes Why hello there’

This is complex requires some ﬂexibility. But the beneﬁt is that Blot will not complain if you do not specify a title for a blog post.

Everything is ﬂexible.


text:
Tufts University  |  School of Arts and Sciences  |  School of Engineering  |  Find People  |

# People

Robert Lemke Oliver
Assistant Professor

Contact Info:

Tufts University
Department of Mathematics
503 Boston Avenue
Bromfield-Pearson
Room 116
Medford, MA 02155

Email @tufts.edu:
robert.lemke_oliver
Phone: 617-627-0436
Personal site

Expertise:
Number theory

Research:
My research is in analytic number theory, but I am often considering problems of an arithmetic flavor. Analytic number theory essentially began with Riemann's seminal 1860 memoir, in which he showed that the apparent randomness and noise in the distribution of prime numbers is controlled by the zeros of a single complex function, the so-called zeta function. Riemann's memoir outlined a program that led to a proof of the "prime number theorem," and it is also the source of what we now call the Riemann hypothesis, one of the seven Millennium prize problems and arguably the most important unsolved problem in mathematics. Riemann's zeta function is the prototype of a class of functions called L-functions, which together can capture finer, or sometimes just different, information about the primes. For example, studying L-functions lets us prove statements like 25 percent of primes end in the digit 1; more generally, whenever primes are grouped into classes (e.g., by their last digit), L-functions are often the "right" tool to prove that each class gets its fair share. I am fascinated by these sorts of equidistribution problems, as well as their natural outgrowth.

Another central focus of my research is problems related to "elliptic curves." An elliptic curve is simply a curve in the xy-plane cut out by the equation $$y^2=x^3+ax+b$$, for some coefficients a and b. While this is a very simple equation (e.g., it can be studied in first-year calculus), all of the standard analytic tools go out the window when we ask only for the rational solutions to the equation. In general, whenever we ask about only the rational solutions to a polynomial equation, this is called a Diophantine problem. It turns out that elliptic curves are an interesting boundary case within the class of Diophantine problems: they sometimes admit infinitely many rational solutions, and sometimes only finitely many. However, it turns out that even if there are infinitely many solutions, there is a finite set of solutions from which all others can be generated; in particular, the set of rational points has the structure of a finitely-generated abelian group. In principle (though not always in practice), for any given elliptic curve, we can at least determine the number of generators needed, and often the actual generating set as well. But how does the number of generators vary as we vary the coefficients defining the elliptic curve in some way? How many generators are needed for a "typical" elliptic curve? Are there elliptic curves that need more than 100 generators, or more than your favorite arbitrarily large number? These are questions of active study.

The bridge between elliptic curves and prime numbers is through L-functions and the theory of modular forms. A modular form is a complex function satisfying certain dazzling symmetry properties that almost no function should be able to satisfy. There is a natural recipe, reliant on these symmetries, that produces from a modular form an L-function; that is, modular forms are a natural source of L-functions, which in turn naturally encode information about the primes. Remarkably, there is also a (not at all obvious!) way of starting from an elliptic curve and producing a modular form which "knows" everything about the elliptic curve. This is the so-called modularity theorem, one of the crowning achievements of 20th century number theory, and whose proof resolved Fermat's Last Theorem. Thus, modular forms are attached to elliptic curves, but they also arise naturally in combinatorics (e.g., the partition generating function is a modular form), in studying quadratic forms, and in many other arenas I know less about. My current research in the area is focused on the application of modular forms to elliptic curves, but I have previously thought about and remain interested in these other applications as well.