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# Mutually Exclusive Events

If two events have no elements in common (Their intersection is the empty set.), the events are called mutually exclusive.   Thus, $P\left(A\cap B\right)=0$ .  This means that the probability of event $A$ and event $B$ happening is zero.  They cannot both happen.

Example 1:

A pair of dice is rolled.  The events of rolling a $5$ and rolling a double have NO outcomes in common so the two events are mutually exclusive.
A pair of dice is rolled.  The events of rolling a $4$ and rolling a double have the outcome $\left(2,2\right)$ in common so the two events are not mutually exclusive.

Example 2:

From a group of $6$ freshmen and $5$ sophomores, $3$ students are to be selected at random to form a committee.  What is the probability that at least $2$ freshmen are selected?

The committee will have at least $2$ freshmen if either $2$ freshmen and $1$ sophomore are selected (event $A$ )or $3$ freshmen are selected (event $B$ ).   Since the two events are mutually exclusive

$\begin{array}{l}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)\hfill \\ \begin{array}{l}P\left(A\right)=\frac{{}_{6}{}^{}C{}_{2}\cdot {}_{5}{}^{}C{}_{1}}{{}_{11}{}^{}C{}_{3}}=\frac{15\cdot 5}{165}=\frac{75}{165}=\frac{5}{11}\\ \end{array}\hfill \\ \begin{array}{l}P\left(B\right)=\frac{{}_{6}{}^{}C{}_{3}}{{}_{11}{}^{}C{}_{3}}=\frac{20}{165}=\frac{4}{33}\\ \end{array}\hfill \\ P\left(A\cup B\right)=\frac{5}{11}+\frac{4}{33}=\frac{15}{33}+\frac{4}{33}=\frac{19}{33}\hfill \end{array}$

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# Problem with addition and macro argument

I need to use a formula like this:

``````y = 1.8pt * x + 1.5pt
``````

inside a macro, where `x` is a scalar macro argument. My idea:

``````\kern \advance #1\dimexpr1.8pt\relax by \dimexpr1.5pt\relax
``````

This doesn't really work, the `\advance` command seems wrongly formatted.

• This is a duplicate of something that is really not easy to find on this site – percusse Sep 27 '13 at 14:28

`\kern` requires a `<dimen>` after it; it can also be a `<skip>`, that is coerced to a `<dimen>` by removing the `plus` and `minus` parts.

A `<dimen>`, according to the TeXbook, is basically either an internal dimen register (for instance `\parindent`, `\dimen2` or any control sequence defined with `\newdimen`) or an explicit specification `<decimal number><unit of measure>` (I'll gloss over the finer details).

Also `<factor><dimen register>` is allowed. So `\kern 0.5\parindent` is a legal specification.

You can't follow `\kern` with the instructions to get a length, so an assignment instruction such as `\advance` is illegal.

However, e-TeX introduced a new possibility: a `<dimen>` can also be an expression computed with `\dimexpr`. The syntax is extended so that `\dimexpr<dimen expression>` can replace an internal dimension in the rules above.

Thus `\kern 0.3\dimexpr 1.7pt+\baselineskip\relax` (just a silly example) is legal when e-TeX is used, which is the case in all modern TeX distributions when running (pdf)LaTeX, XeLaTeX or LuaLaTeX.

`\dimexpr` calls can be nested; the `\relax` at the end is optional and marks the end of the expression, so that TeX won't continue to search for terms belonging to the current expression when it finds `\relax` (that will vanish).

Therefore you're looking for

``````\kern \dimexpr 1.5pt + #1\dimexpr 1.8pt\relax\relax
``````

The alternative way without `\dimexpr` is

``````\dimen0=1.8pt
\dimen0=#1\dimen0
\kern\dimen0
``````

using the scratch register `\dimen0`, or with the “official” LaTeX commands `\setlength` and `\addtolength`. For such a computation, it's not necessary to allocate a new register, but it doesn't hurt.

What's the difference between the two approaches? The `\dimexpr` way is certainly more flexible; but remind that division with `\divide` truncates, while `\dimexpr(dimen expression)/<integer>\relax` rounds. This might be a factor in some computation (but this is more relevant in `\numexpr`, because dimensions are actually integer multiples of scaled points, so the rounding happens at the scaled point level). Indeed e-TeX provides `\numexpr`, `\dimexpr`, `\glueexpr` and `\muexpr` to operate on numbers, dimensions, glues and muglues.

By the way, `\advance` must be followed by an internal register name (`\count`, `\dimen`, `\skip` or `\muskip` type), so `\advance\dimexpr...` is illegal also with e-TeX, which would balk with

``````! You can't use `\dimexpr' after \advance.
``````
• Thanks for the long explanation, I know too little about "registers" and such. Normally working with general purpose statically typed programming languages, things like R and LaTeX will clearly remain a mystery to me till I die. – Emit Taste Sep 28 '13 at 13:00

The following works:

``````\kern \dimexpr 1.5pt + #1\dimexpr 1.8pt\relax\relax
``````

Old school:

``````\setlength(\y}{1.8pt}
\setlength{\y}{#1\y}


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ACTA issues

An improvement on the bound for $C_2(n)$

Rajeev Gupta

Acta Sci. Math. (Szeged) 83:1-2(2017), 263-269
88/2015

 Abstract. The inequality $C_2(n)\leq2 K^\C_G$, where $K_G^\C$ is the complex Grothendieck constant and $C_2(n)=\sup\big \{\|p(\boldsymbol T)\|:\|p\|_{\D ^n,\infty }\leq1, \|\boldsymbol T\|_{\infty } \leq1 \big\},$ for each $n\in\N,$ is due to Varopoulos. Here the supremum is taken over all commuting $n$-tuples $\boldsymbol T:=(T_1,\ldots,T_n)$ of contractions and all complex polynomials $p$ in $n$ variables of degree at most $2$ and of supremum norm at most $1$ over the polydisc. We show that $C_2(n)\leq\frac {3\sqrt{3}}{4} K^\C_G$ for each $n\in\N.$ DOI: 10.14232/actasm-015-088-4 AMS Subject Classification (1991): 47A13 Keyword(s): complex Grothendieck constant, von Neumann inequality, Varopoulos-Kaijser polynomial Received December 18, 2015, and in final form February 17, 2016. (Registered under 88/2015.)


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Products

# grating equation

### Transmission Gratings | Edmund Optics

Transmission gratings offer low polarization sensitivity as opposed to reflection gratings because incident light is not reflected by a mirror coating such as aluminum. Since light is transmitted through the grating, transmission gratings can be used in compact, in-line configurations. They are ideal for use in monochromators and spectrometers.

### Packaging, characterization and calibration of fiber Bragg

Packaging, characterization and calibration of fiber Bragg grating temperature sensors Conference Paper (PDF Available) · August 2012 with 408 Reads DOI:

### Fiber Bragg Gratings: Theory, Fabrication, and

Here we offer a short explanation of FBGs provided as excerpts from the SPIE Tutorial Text, Fiber Bragg Gratings: Theory, Fabrication, and Applications. Bragg gratings are one of the most useful, reliable, versatile, practical, and attractive passive devices in the fields of optical fiber communications and fiber optic sensors.

### Part 2: The Grating - B&W Tek

Based on equation 2-3, it is clear that the maximum spectral range of a spectrometer is determined by the detector length (L D), the groove density (1/d) and the focal length (F). The minimum wavelength difference that can be resolved by the diffraction grating is given by. where N is the total number of grooves on the diffraction grating. This

### Diffraction Grating Physics - Newport Corporation

The Diffraction Grating Equation A typical diffraction grating consists of a substrate, usually of an “optical material”, with a large number of parallel grooves ruled or replicated in its surface and overcoated with a reflecting material such as aluminum.

### A Guide to Fiber Bragg Grating Sensors - InTech - Open

A Guide to Fiber Bragg Grating Sensors 3 principle is used until today for the study and determination of crystal structure, particularly in thin film research. Sir Bragg, together with his father, won the Nobel Prize for Physics in 1915 for an important step in the development of X-ray crystallography.

### Part 5: Spectral Resolution - B&W Tek

Measuring Spectral Resolution. It is important to understand that the observed signal (S o) is not solely dependent on the spectral resolution of the spectrometer (R) but it is also dependent on the linewidth of the signal (S r). As a result, the observed resolution is the convolution of the two sources, Equation 5-1

### (PDF) Encapsulated fiber Bragg grating sensor for strain

The Several studies have been reported for producing experimental procedure provided robustness to the sensor and encapsulated fiber Bragg grating for simultaneous the matrix equation approach has the potential for strain and measurement of temperature and strain in order to prevent the temperature independent measurements.

### Diffraction grating - Wikipedia

In optics, a diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions. The emerging coloration is a form of structural coloration. The directions of these beams depend on the spacing of the grating and the wavelength of the light so that the grating acts as the dispersive element.

### - Tutorials

(1) known as the grating equation states that a diffraction grating with spacing will deflect light at discrete angles (), dependent upon the value λ, where is the order of principal maxima. The diffracted angle, , is the output angle as measured from the surface normal of the diffraction is easily observed from Eq. 1 that for a given order , different wavelengths of

### Di raction Grating Equation with Example Problems

Di↵raction Grating Equation with Example Problems1 1 Grating Equation In Figure 1, parallel rays of monochromatic radiation, from a single beam in the form of rays 1 and 2, are incident on a (blazed) di↵raction grating at an angle i relative to the grating normal. These rays are then di↵racted at an angle r.

### Grating Physics - BGU

The quality and spacing of the grooves are crucial to the performance of the grating, but the basic grating equation may be derived by supposing a section through the grating surface, normal to the ruling direction as a sawtooth pattern, shown below.

### Encapsulated fiber Bragg grating sensor for strain and

Encapsulated fiber Bragg grating sensor for strain and temperature measurements Leandro de Marchi Pintos, Ivo de Lourenço Junior, Jean Carlos Cardozo da Silva

### Grating Equations : SHIMADZU CORPORATION

SHIMADZU DIFFRACTION GRATINGS " Grating Equations" As shown in and is the angle between the incident light and the normal to the grating (the incident angle) and ß is the angle between the diffracted light and the normal to the grating (the diffraction angle), then, they satisfy the following relationship: as shown in , in case of transmission grating

### Diffraction Gratings Tutorial - Thorlabs

Diffraction Gratings Tutorial. Diffraction gratings, either transmissive or reflective, can separate different wavelengths of light using a repetitive structure embedded within the grating. The structure affects the amplitude and/or phase of the incident wave, causing interference in the output wave.

### Coupling Equation - an overview | ScienceDirect Topics

Coupling of light into the waveguide is governed by the grating equation, n eff = n c sin θ + mλ0 /Λ, such that a change in effective index, n eff, will cause a shift in the coupling angle, θ, where m is the waveguide mode, λ 0 is the incident wavelength, Λ is the grating period, and n c is the refractive index of the cover material above

SHIMADZU DIFFRACTION GRATINGS " Resolution" If there are two spectrum lines corresponding to two neighboring wavelengths λ and λ+Δλ, the resolution is a measure of the extent to which Δλ can be reduced while still being able to distinguish between the general, the grating's diffracted light has a limited width defined by the diffraction limit.

### US7554734B1 - Polarization independent grating - Google

A grating has a high diffraction efficiency into the minus first diffracted order in transmission, for both TE and TM polarizations. The incident angle may optionally be chosen so that the minus first diffracted order in reflection would be retroreflected back to the incident beam. The grating may be formed from various materials and/or layers, where the thicknesses of the individual layers

### Spectroscopic Properties of Gratings | Spectrogon

The Grating Equation A beam of light which falls on a grating will be diffracted into one or several beams. The directions of these beams depend on the wavelength and direction of the incident beam, and on the groove frequency of the grating. The grating equation is a good starting point when describing the properties of gratings.

### Diffraction gratings - Scholarpedia

Figure 1 represents a diffraction grating. The periodic profile $$\mathcal{P}$$ of period d along the x-axis separates air from a grating material which is generally a metal or a y-axis is perpendicular to the average profile plane and the z-axis is the axis of invariance of the suppose that the incident light can be described by a sum of monochromatic radiations of


text:
# Plot - Andrews curves¶

from __future__ import annotations

from gemseo.api import configure_logger
from gemseo.post.dataset.andrews_curves import AndrewsCurves

configure_logger()

<RootLogger root (INFO)>

iris = load_dataset("IrisDataset")

## Plot Andrews Curves¶

We can use the AndrewsCurves plot which can be viewed as a smooth version of the parallel coordinates. Each sample is represented by a curve and if there is structure in data, it may be visible in the plot.

AndrewsCurves(iris, "specy").execute(save=False, show=True)

[<Figure size 640x480 with 1 Axes>]

Total running time of the script: ( 0 minutes 0.320 seconds)

Gallery generated by Sphinx-Gallery


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Propagation delay in asynchronous counter

Are the ripple counters useful in real life?

I know, for example, if we set an up counter with 3 FFs, CBA. Followings are the block and the waveform diagram:

After the count 001, it will appear 000 (the 2nd CLK period), which makes the operation false. In fact, this situation is inevitable in real life because propagation delay is inherent for all flip flops.

As a result, I wonder how can we implement asynchronous counters?

• CD4060 was the "goto" counter when simple clock generation to high orders of 2 were needed in the 70's and still sold today in high volume. – Tony Stewart Sunnyskyguy EE75 Apr 8 '18 at 5:26

It is simply a matter of can you wait for the final output stage to settle before taking a count reading. In which case you build in the sample delay with hardware or software.

When you gate/stop the count to measure each stage in parallel then synchronous counters are better. There is still a minimum wait time before sampling but it is much more predictable, as the wait time applies to the entire counter.

If it is just a 'pre-scaler' used to divide a fixed high frequency to a lower frequency and it runs continuously then there is no delay issue. In a way the count of any one stage is unknown and has a "don't care" policy.

• Do you mean that asynchronous counters are not used when we want to examine the circuit at certain stages, but rather to wait until the exact moment on the count? – W.Joe Apr 8 '18 at 6:21
• Correct. By their nature the count of anyone stage is unknown unless you halt the count and wait before sampling. Many watch's and Real-time-clocks use a 32.768 KHZ crystal that goes through an 11 stage binary divider to end up with a accurate 1 HZ clock pulse. No one cares what count anyone stage has, as it is about dividing down the 32.768 KHZ only. – Sparky256 Apr 8 '18 at 6:31
• The problem with asynchronous counter is the "ripple". Say take an example of 4-bit asynchronous counter counting up. The transition from 0111 --> 1000 goes through or ripple through 3 intermediate states because of accumulaton of propogation delays of each preciding Flip-Flop. These are glitches in the asynchronous counters.

If you plug-in some LEDs at the 4-bit outputs of the counter, you wont notice this glitch at all as these ripples are fast enough and you see that its counting correctly. But if you are feeding this count to some other digital logic for sampling, these glitches may cause unwanted errors.

• A solution to the above problem is strobing. It is basically like latching the count output.

The idea is to sample the count, only after it is settled to a stable value. In the above circuit, the count transition is happening at positive edge of clock. What we do is, we sample the count only when clock is low. The clock is like 'enable' signal of the strobing circuit. The one condition here is that, by the time clock transits from high to low, all ripples should have been passed by. i.e, the accumulated propagation delay should be less than clock's "high time". It make sure that correct count is sampled every time.

• The problem with asynchronous counter is the limitation of clock frequency/limted speed, due to accumulated delays of each Flip-Flop. The counter should satisfy: $$\Sigma d_i < T_{clk}$$

Due to these design complexities, we prefer synchronous counters, where timing is predictable.

Ripple counters are useful in real life. Frequency dividers are an obvious one.

A 16 stage ripple counter will use 1 + 1/2 + 1/4 + ... ~=2 units of power.

A synchronous counter would use 16 units of power.

Generally, synchronous logic solves one set of problems at the expense of

• more Power
• more gates
• lower speed
• new race conditions introduced by synch
• increased emi from continuous clocking and spectrally narrow transitions

A better question might be why do people think synchronous counters are so especially useful? Why is synchronising the entire counter/system seen as the correct solution to the timing of some critical outputs?

Philosophically, the orthodoxy of synchronous logic may have resulted in limited skill and techniques and strategies for making complex unclocked logic systems work, and explain why there are so few asynchronous clockless cpus, even for very simple instruction sets. The Greenarrays F18 is the only one that springs to mind, but able to achieve PIC like power consumption at hundreds of MHz


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## Friday, September 8, 2017

### Field Weakening, Part 1

Previously on Motor Diddlers, we learned how to optimize the torque produced by an IPM operating in the current-limited regime. We can't stay in this regime forever; at higher RPM's, all motors enter a voltage-limited operating regime.

What's surprising is how quickly IPM's become voltage-limited. Using the data from the last post, we can plot the maximum achievable current versus speed:

The "base speed" of the motor is very low, about 800 rad/s (2500 RPM). Furthermore, we quickly run out of volts past base speed; the motor cannot turn much faster than 1500 rad/s (5000 RPM) if we constrain ourselves to the MTPA trajectory. This is a result of the much higher inductance of an IPM; at high electrical frequencies this inductive impedance limits how much current we can put into the stator.

Fortunately, we can use this higher inductance to our advantage. Recall that $$V_q=R_s I_q+\omega L_d I_d+\omega\lambda$$; this means that if $$I_d$$ is negative, we can cancel out the $$\omega\lambda$$ term (what would conventionally be called "back EMF" in a surface PM machine) in the expression for $$V_q$$. On a surface PM machine, this current would be wasted; however, if $$L_d < L_q$$, the $$(L_d-L_q)I_d I_q$$ term in the torque equation will generated positive torque! Furthermore, the $$L_d/\lambda$$ ratio is often an order of magnitude higher for IPM's, which greatly reduces the amount of d-axis current required to cancel out the flux linkage. This is, in a nutshell, why IPM's field weaken much better than their surface PM counterparts.

It is fairly straightforward to compute the optimal field-weakening trajectory:

Starting from the motor equations $$\begin{array}{lcl} \tau=\frac{3}{2}n_p(\lambda I_q+(L_d-L_q)I_d I_q)\\ V_d=R_s I_d-\omega L_q I_q\\ V_q=R_s I_q+\omega L_d I_d+\omega\lambda\\ V_s=\sqrt{V_d^2+V_q^2}\\ I_s=\sqrt{I_d^2+I_q^2} \end{array}$$ we note that in the voltage limited operating regime, maximum torque must be achieved when the voltage vector is on the boundary of the allowable area, in which case the optimization problem becomes an equality: $$\begin{cases} \frac{3}{2}n_p(\lambda I_q+(L_d-L_q)I_d I_q) = \tau_0\\ (R_s I_d-\omega L_q I_q)^2 + (R_s I_q+\omega L_d I_d+\omega\lambda)^2 = V_0^2 \end{cases}$$ If we ignore saturation, this system is the intersection of a constant-torque hyperbola (which is independent of speed) and a series of shrinking ellipses.

This system is polynomial (and in fact only a quartic) and can be solved in many ways. For example, one element of the reduced Groebner basis of the ideal generated by the two equations is the very long $$(L_d-L_q)^2(R_s^2+L_q^2w^2)I_q^4 + (L_q^2\lambda^2\omega^2+R_s^2\lambda^2+(L_d-L_q)^2(2R_s\tau_0\omega-V_0^2))I_q^2- 2\lambda\tau_0(R_s^2+L_d L_q\omega^2)+ R_s^2\tau_0^2+L_d^2\tau_0^2\omega^2$$ where we have included the factor of $$3/2n_p$$ in $$\tau_0$$ for brevity's sake. The roots of this polynomial are easily found with a variety of numeric or analytic methods; in the case of multiple real solutions the correct one is the $$(I_d,I_q)$$ vector with the shortest length.
Unfortunately, the system is no longer polynomial if the motor saturates (that is, $$L_d$$ and $$L_q$$ are functions $$l_d(I_d,I_q)$$, $$l_q(I_d,I_q)$$ of the axis currents). In order to solve the system in this case, we will have to turn to more complicated (and less reliable) numeric methods, the nature of which will be the subject of the next post in this series.


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## Proposition: Addition of Integers

Any integer $$x\in\mathbb Z$$ is by the corresponding proposition an equivalence class
$x:=[a,b],$
where $$a$$ and $$b\neq 0$$ denote natural numbers representing the equivalence class $$x$$. Given two integers $$x:=[a,b]$$, $$y:=[c,d]$$, $$a,b,c,d\in \mathbb N$$, the addition of integers is defined by using the addition of the natural numbers $$a+c$$ and $$b+d$$:

$\begin{array}{rcl} x+y&:=&[a+c,b+d], \end{array}$

where $$[a+c,b+d]$$ is also an integer, called the sum of the integers $$x$$ and $$y$$. The sum exists and is well-defined, i.e. it does not depend on the specific representatives $$[a,b]$$ and $$[c,d]$$ of $$x$$ and $$y$$.

| | | | | created: 2014-09-21 14:25:51 | modified: 2017-08-17 23:09:45 | by: bookofproofs | references: [696]

## 7.Proposition: Existence of Inverse Integers With Respect to Addition

(none)

### Bibliography (further reading)

[696] Kramer Jürg, von Pippich, Anna-Maria: “Von den natürlichen Zahlen zu den Quaternionen”, Springer-Spektrum, 2013


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Heat Transfer Video Lessons

Concept

# Problem: An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 degrees Celsius and rejects heat to a room at a temperature of 19.2 degrees Celsius. Suppose that liquid water with a mass of 76.3kg at 0.0 degrees Celsius is converted to ice at the same temperature. Take the heat of fusion for water to be Lf = 3.34x105 J/kg.How much energy E must be supplied to the device?Express your answer in joules.

###### FREE Expert Solution

For us to get the amount of energy that must be supplied to the device, we first need to get the energy rejected to the room.

To get the quantity of heat rejected to the room, we use the formula:

$\overline{)\frac{{\mathbf{Q}}_{\mathbf{c}}}{{\mathbf{Q}}_{\mathbf{h}}}{\mathbf{=}}\frac{{\mathbf{T}}_{\mathbf{c}}}{{\mathbf{T}}_{\mathbf{h}}}}$

85% (186 ratings)
###### Problem Details

An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 degrees Celsius and rejects heat to a room at a temperature of 19.2 degrees Celsius. Suppose that liquid water with a mass of 76.3kg at 0.0 degrees Celsius is converted to ice at the same temperature. Take the heat of fusion for water to be Lf = 3.34x105 J/kg.

How much energy E must be supplied to the device?


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# Exponential distribution Q-Q plot homework question

For an exponential QQ plot, we fix the theoretical distribution to have $λ=1$. In order to compare the plotted points with the line $y=x$, what kind of rescaling should be applied to the data and why?

“Hint“ given with this problem: If $X$ follows an exponential distribution with parameter $λ$, then $λX$ follows an exponential distribution with parameter 1.

I don’t know how to use this hint in answering this question. This is a homework problem. So far, I have considered the graph of $\ln(x)$ but I am not sure what to do with that and the hint.

• Is this homework? Please add the self-study tag and read our policy on that meta.stackexchange.com/questions/10811/… – Momo Jul 17 '14 at 23:32
• Maybe it's just me, but I feel like there is some information missing. Is the exponential qq plot a qq plot with an exponential distribution with $\lambda=1$ as the distribution from which the theoretical quantiles are derived? And you have some data $x$ that are supposedly from this distribution and you want to check that? Why would you need to rescale $x$ anyway, what is the overall goal? – Momo Jul 17 '14 at 23:54
• I'm still confused. What are the plotted points, $x$ and $y$? – Momo Jul 17 '14 at 23:57
• Yes, exactly as you say. The theoretical distribution is exponential with lamda 1. The rescaling idea is similar to the idea of rescaling data in order to "normalize" it. For example, if a normal qq plot shows positively skewed data you would rescale the data by taking the log or square root. I hope this makes it more clear. I am a bit confused as well, as I am just learning this stuff, but I would appreciate the input. – user162381 Jul 18 '14 at 0:24
• here, you would not want to "normalize" the data, but instead to make it fit to the exponential distribution, so that the qq plot would lie on the y=x line. – user162381 Jul 18 '14 at 0:25

Time to put this one to bed, I think

For an exponential qq plot, we fix the theoretical distribution to have λ=1.

Sure, because in a QQ plot we care about how far it is from exponential, not what the parameter value is. But in any case, you can see approximately what it is from the slope of the line the points should lie along in the QQ plot.

In order to compare the plotted points with the line y=x, what kind of rescaling should be applied to the data and why?

While you could rescale the data by dividing by their mean (yielding data with a mean of 1), I think that you should not do so.

If you do this with exponential data, the result will be data that no longer has an exponential distribution. In small samples the effect may be strong. Further, if the data aren't exponential, that adjustment may be badly impacted by large outliers.

It may do okay with very large samples, but there's really no need.

“Hint“ given with this problem: If X follows an exponential distribution with parameter λ, then λX follows an exponential distribution with parameter 1.

Yeah, as I thought -- my first guess is they probably want you to estimate $\lambda$ by $\hat{\lambda}=1/\bar{x}$, and then scale to unit mean -- but as I already explained, I think that's actually not a great idea, because it alters the very distribution you're checking for.

Better I think to plot the the data, and if you want a line to compare to, put some moderately robust estimate of the trend as a line on the graph ... without putting too much stock in the position of the line (it's more a guide for the eye, to help you spot where the thing curves).

So far, I have considered the graph of ln (x) but not sure what to do with that and the hint.

Okay, that's another way to go. If you do it that way, consider $Z\sim \text{Exp}(1)$, and $X=Z/\lambda$. Then $\ln(X)=\ln(Z)-\ln\lambda$, implying that a plot of quantiles of $\ln(X)$ should be parallel to expected quantiles of $\ln(Z)$, and the distance to the $y=x$ line tells you about $\lambda$. But to be honest I don't think this is the intent of the question.


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# Arithmetic Progression proof for sum of an integer

Question:-

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that resulting sum is the square of an integer.

MyApproach:

Let d be the common difference and d^4 be the fourth power.I am considering the four consecutive terms be (a+2d),(a+d),(a-d),(a-2d).What do the question mean by telling that the common difference is added to the four consecutive terms ?What is the product of the terms equals to ?

• First, you haven't identified four consecutive terms: you skipped right over the middle term $a$. So it should be adjusted to (for example) $a - d, a, a + d, a + 2d$. The product of four consecutive terms is $(a + 2d)(a + d)a(a - 2d)$. – user296602 Feb 11 '18 at 16:11

If the common difference is $2d$,

$$(a-3d)(a-d)(a+d)(a+3d)+(2d)^4$$

$$=(a^2-d^2)(a^2-9d^2)+(2d)^4$$

$$=(a^2)^2-2a^2(5d^2)+(5d^2)^2$$

$$=?$$

• Is it (a^2-(5d^2))^2? Is this the answer that is asked. – user517784 Feb 11 '18 at 16:14
• @CalculusProgrammer, $$(x-y)^2=?$$ – lab bhattacharjee Feb 11 '18 at 16:15
• Yes thanks for showing the steps. – user517784 Feb 11 '18 at 16:18


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zbMATH — the first resource for mathematics

Regularity of the free boundary for the porous medium equation. (English) Zbl 0910.35145
This paper deals with the initial value problem for the porous medium equation $$u_t=\Delta u^m$$ with compactly supported initial data, written for pressure $$f=mu^{m-1}$$. The authors prove that, under rather general assumptions on the initial data, the free boundary is a smooth surface locally in time. First, a model linear degenerate equation is studied on the half-space $$x\geq 0$$. The basic idea is to establish Schauder type coercive estimates for solutions of the model equation. Then the result is extended to a certain class of quasilinear degenerate evolution equations. Then the proof of regularity of the free boundary is given. Using a global change of coordinates, the authors transform the free boundary problem to a fixed boundary problem for a degenerate quasilinear equation, which can be solved in appropriately defined Hölder spaces.

MSC:
 35R35 Free boundary problems for PDEs 35K60 Nonlinear initial, boundary and initial-boundary value problems for linear parabolic equations 35B65 Smoothness and regularity of solutions to PDEs
Full Text:
References:
 [1] Sigurd Angenent, Analyticity of the interface of the porous media equation after the waiting time, Proc. Amer. Math. Soc. 102 (1988), no. 2, 329 – 336. · Zbl 0653.35040 [2] D. G. Aronson, Regularity propeties of flows through porous media, SIAM J. Appl. Math. 17 (1969), 461 – 467. · Zbl 0187.03401 [3] D. G. Aronson, Regularity properties of flows through porous media: A counterexample., SIAM J. Appl. Math. 19 (1970), 299 – 307. · Zbl 0255.76099 [4] D. G. Aronson, Regularity properties of flows through porous media: The interface., Arch. Rational Mech. Anal. 37 (1970), 1 – 10. · Zbl 0202.37901 [5] D. G. Aronson, L. A. Caffarelli, and Juan Luis Vázquez, Interfaces with a corner point in one-dimensional porous medium flow, Comm. Pure Appl. Math. 38 (1985), no. 4, 375 – 404. · Zbl 0544.35058 [6] D. G. Aronson and J. L. Vázquez, Eventual \?^{\infty }-regularity and concavity for flows in one-dimensional porous media, Arch. Rational Mech. Anal. 99 (1987), no. 4, 329 – 348. · Zbl 0642.76108 [7] Luis A. Caffarelli, Interior a priori estimates for solutions of fully nonlinear equations, Ann. of Math. (2) 130 (1989), no. 1, 189 – 213. · Zbl 0692.35017 [8] Luis A. Caffarelli and Avner Friedman, Regularity of the free boundary for the one-dimensional flow of gas in a porous medium, Amer. J. Math. 101 (1979), no. 6, 1193 – 1218. · Zbl 0439.76084 [9] Luis A. Caffarelli and Avner Friedman, Regularity of the free boundary of a gas flow in an \?-dimensional porous medium, Indiana Univ. Math. J. 29 (1980), no. 3, 361 – 391. · Zbl 0439.76085 [10] L. A. Caffarelli, J. L. Vázquez, and N. I. Wolanski, Lipschitz continuity of solutions and interfaces of the \?-dimensional porous medium equation, Indiana Univ. Math. J. 36 (1987), no. 2, 373 – 401. · Zbl 0644.35058 [11] Luis A. Caffarelli and NoemíI. Wolanski, \?^{1,\?} regularity of the free boundary for the \?-dimensional porous media equation, Comm. Pure Appl. Math. 43 (1990), no. 7, 885 – 902. · Zbl 0728.76103 [12] K. Höllig, H.O. Kreiss, $$C^{\infty }$$ regularity for the porous medium equation, Univ. of Winsconsin Madison, Computer Scienced Dept., Technical report # 600. · Zbl 0597.35107 [13] Barry F. Knerr, The porous medium equation in one dimension, Trans. Amer. Math. Soc. 234 (1977), no. 2, 381 – 415. · Zbl 0365.35030 [14] J. J. Kohn and L. Nirenberg, Degenerate elliptic-parabolic equations of second order, Comm. Pure Appl. Math. 20 (1967), 797 – 872. · Zbl 0153.14503 [15] M. V. Safonov, The classical solution of the elliptic Bellman equation, Dokl. Akad. Nauk SSSR 278 (1984), no. 4, 810 – 813 (Russian). [16] M. V. Safonov, Classical solution of second-order nonlinear elliptic equations, Izv. Akad. Nauk SSSR Ser. Mat. 52 (1988), no. 6, 1272 – 1287, 1328 (Russian); English transl., Math. USSR-Izv. 33 (1989), no. 3, 597 – 612. [17] Lihe Wang, On the regularity theory of fully nonlinear parabolic equations. I, Comm. Pure Appl. Math. 45 (1992), no. 1, 27 – 76. · Zbl 0832.35025 [18] Lihe Wang, On the regularity theory of fully nonlinear parabolic equations. II, Comm. Pure Appl. Math. 45 (1992), no. 2, 141 – 178. · Zbl 0774.35042
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Question

# A particle is projected from point $$A$$ with velocity $$\displaystyle u\sqrt{2}$$   at angle $$45$$ to horizontal as shown in diagrame. It strikes the plane $$BC$$ at right angle. What is the velocity of particle at the time of impact?

A
3u2
B
u2
C
2u3
D
u

Solution

## The correct option is D $$\displaystyle \frac {2u}{\sqrt{3}}$$The horizontal component of the velocity remains the same=$$\sqrt{2}u cos45^{\circ}=u$$Let the vertical velocity of the particle when it strikes the plane be $$v$$ downwards. Since it hits the plane perpendicularly, the final velocity forms an angle of $$60^{\circ}$$ with the vertical.Thus $$\dfrac{u}{v}=tan60^{\circ}=\sqrt{3}$$$$\implies v=\dfrac{u}{\sqrt{3}}$$Thus the magnitude of the velocity when it hits the plane is $$\sqrt{v_x^2+v_y^2}=\sqrt{u^2+(\dfrac{u}{\sqrt{3}})^2}=\dfrac{2u}{\sqrt{3}}$$Physics

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## Basic College Mathematics (10th Edition)

Move the decimal one place to the right to make 2.2 a whole number. Move the decimal one place to the right in 8.24 to avoid changing the quotient. $2.\underrightarrow2\bigr)\overline{8.\underrightarrow24}$
$8.24\div2.2$ = $(8.24\times10)\div(2.2\times10)$ = $82.4\div22$


text:
# Local Lorentz Frames

1. May 25, 2013

### WannabeNewton

Hey guys. I'm having doubts about something in a GR text I read (Gravitation - T.Padmanabhan). In general, for Lorentzian 4-manifolds, we call a basis of smooth vector fields $(e_{\mu})^{a}$ (that is, a set of smooth vector fields that define a basis for the tangent space at each event of space-time) a tetrad if $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ i.e. $(e_{0})^{a}(e_{i})_{a} = 0,(e_{0})^{a}(e_0)_{a} = -1, (e_{i})^{a}(e_{j})_{a} = \delta_{ij}$ so the basis of smooth vector fields are also orthonormal. In Padmanabhan's text, when defining Fermi-Walker transport, he says (I tweaked the notation a bit) "Consider an observer who is moving along some worldline (which is not necessarily a geodesic) with the 4-velocity $u^{a}$. Let $(e_{\mu})^{a}$ be a tetrad of basis vectors transported along the trajectory of this observer...The first condition we will impose is that the basis should satisfy the standard relations, $(e_{\mu})^{a}(e_{\nu})_{a} = g_{\mu\nu}$."

This seems to go against the very definition of a tetrad (as an orthonormal basis for the tangent space at each point on the observer's worldline) because the way it is defined in this text, the basis vectors need not be orthonormal. What gives? He later says that Fermi-Walker transport ensures that the tetrad remains orthonormal from the initial erected tetrad (which is of course true in a general context) but he doesn't even define it to be orthonormal to start with so what's up with that?!

My next question involves the use of the term "local Lorentz frame / local inertial frame" in GR. Here is the issue: consider an observer moving along an arbitrary worldline and let $p$ be an event on the observer's worldline. Now by a procedure much like the Gram-Schmidt scheme, we can always find a basis $(e_{\mu})^{a}|_p$ for $T_p M$ such that $g_{\mu\nu}|_p = \eta_{\mu\nu}$ with respect to this basis i.e. we can always find a point-wise tetrad $(e_{\mu})^{a}|_p$ for $T_p M$. Now I've seen some sources (including Padmanabhan) call this point-wise tetrad a local Lorentz frame / locally inertial reference frame for the observer at $p$ which I find incorrect because from what I've learned, the concept of a locally inertial reference frame at $p$ requires that the frame be derived from a coordinate system $\{x^{\mu}\}$, which we call locally inertial coordinates, that the observer must setup in a neighborhood $U$ of $p$ (and when I say the frame is derived from $\{x^{\mu}\}$ I mean $(e_{\mu})^{a}|_p = (\partial_{\mu})^{a}|_p$) such that $\partial_{\alpha}g_{\mu\nu}|_p = 0$ (this further ensures that the metric tensor is approximately the Minkowski metric for any $q\in U$ to first order).

This physically ensures that the locally inertial reference frame at $p$ is that of a freely falling observer (because it is precisely the freely falling observers who are locally inertial) i.e. $u^{a}\nabla_{a}u^{b}|_p = (e_{0})^{a}\nabla_{a}(e_{0})^{b}|_p = (\partial_{0})^{a}\nabla_{a}(\partial_{0})^{b}|_p = \Gamma^{b}_{ac}|_p (\partial_{0})^{c}(\partial_{0})^{a}|_p = 0$
where I have used the fact that in any reference frame of an observer at an event $p$ on his/her worldline, the 4-velocity $u^{a}|_p = (e_0)^a|_p$.

So you can see why I am confused when Padmanabhan, and others, call the tetrad $(e_{\mu})^{a}|_p$ at an event $p$ on an arbitrary observer's worldline (i.e. a point-wise tetrad which is not necessarily derived from a locally inertial coordinate system about $p$) a local Lorentz frame / locally inertial frame since no restriction is placed on the observer being in free fall. What am I missing here? Thanks in advance for any help.

Last edited: May 25, 2013
2. May 26, 2013

### ForMyThunder

It's most likely a typo: should be eta and not g. This, $g(e_\mu,e_\nu)=(e_\mu)^a(e_\nu)_a=g_{\mu\nu}$, is the definition of the components of $g$, the $g_{\mu\nu}$.

The book "Gravitation" by Misner, Thorne and Wheeler gives the definition for local Lorentz frame as you have described; with the Christoffel symbols of the coordinate system at the point zero. I don't have a copy of Padmanabhan's text, so I am unable to comment further.

3. May 26, 2013

### WannabeNewton

Yeah I was hoping it was just a typo and not an error in my understanding. Thanks for clearing that up. So the "standard relations" amongst the tetrad alluded to by Padmanabhan's text should be $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ (which is, for example, how Wald defines a tetrad) and not $g_{\mu\nu}$ correct?

As for the local Lorentz frame part, look at the very first page of this: http://web.mit.edu/edbert/GR/gr2.pdf
They say that "Orthonormal bases correspond to locally inertial frames" which is definitely not as strong as the definition of a locally inertial frame in Carrol, MTW etc. and physically it makes no sense to simply claim any observer who erects an orthonormal frame (tetrad) at each event on his/her worldline is locally inertial but according to that document, such a frame is locally inertial which would make the observer a freely falling one thus giving rise to the issues I described in the OP. Thanks again for the response Thunder.

4. May 26, 2013

### ForMyThunder

Yes.

What I get from reading the document, and I may be wrong in this, is that it is saying "given any point $p\in M$ and any orthonormal basis $e_\mu\in T_p M$, there is an inertial coordinate system (that is, such that the Christoffel symbols are 0) such that the coordinate frame coincides with the orthonormal basis at the point $p$". This is done using normal coordinates http://en.wikipedia.org/wiki/Normal_coordinates.

5. May 26, 2013

### WannabeNewton

Yes geodesic normal coordinates are how I learnt about the construction of locally inertial coordinate systems and the associated locally inertial reference frames as well but in the document I linked, in the very first paragraph where I drew the quote from post #3, I did not see anything about the requirement that a local inertial frame at an event be derived from a locally inertial coordinate system about that event, which is what I was criticizing. I'm wondering if it links you to a different document? The section is titled "Orthonormal Bases, Tetrads, and Commutators".

Regardless, are we in agreement that simply specifying a tetrad at a given event on an observer's worldline is not enough to claim that the tetrad corresponds to a locally inertial frame, and that we further require that the tetrad be derived from a locally inertial coordinate system in order to claim that it does indeed correspond to a locally inertial frame at the given event? This way only freely falling observers may erect locally inertial frames, in accordance with the equivalence principle, and not any arbitrary observer.

6. May 26, 2013

### ForMyThunder

I believe the term "correspond" is causing the confusion. To me, this just means that there is some sort of mapping from one set to the other, in this case from the set of orthonormal bases at p to the set of locally inertial reference frames. It doesn't exactly give a definition; in his first set of notes, he says "there exist locally inertial reference frames, corresponding to locally ﬂat coordinates carried by freely
falling observers". He never actually defines them rigorously, just in a way to give the reader an intuitive idea about them (which I hate, being a mathematician at heart).

Does this at all answer your question? I have a very bad habit of going off on tangents.

7. May 26, 2013

### WannabeNewton

I would say so yes, thanks; I absolutely hate the non-rigorous accounts as much as you do and this is why I try to stick to Wald as much as possible and rely heavily on it as my primary source. But to sum it up, how do you stand with regards to my last statement:

8. May 26, 2013

### Bill_K

No, absolutely not! The relationship is correct as given.

The quantities that specify a tetrad eμa may be regarded in several ways. They are an example of hybrid quantities, quantities that have indices of two different types. For fixed a, eμa represents a contravariant vector in the usual sense. Just as the gμν may be used to raise and lower tensor indices, ηab may be used to raise and lower the internal Lorentz indices, Va = ηabVb.

And eμa may be used to convert indices from one type to the other. We call this latter process strangulation. Given a vector Vμ, its strangled components in the Lorentz basis are Va = eμaVμ. Conversely, the vector may be unstrangled, Vμ = eμaVa. Consistency is guaranteed by the orthonormality relation eμaeμb = δab, and completeness relation, eμaeνa = δμν .

gμν, ηab and eμa may all be regarded as different components of the same object. In this sense, eμa therefore is just a different way of expressing the metric information, and its differentiation leads to curvature information, the Riemann tensor. In this process in place of the Christoffel symbols we use instead the Ricci rotation coefficients, e[μ,ν]a.

Last edited: May 26, 2013
9. May 26, 2013

### George Jones

Staff Emeritus
Given any event $p$ and any unit timelike element $\bf{u}$ of $T_p$, there always exists a timelike geodesic through $p$ that, at $p$, has tangent $\bf{u}$.

Last edited: May 26, 2013
10. May 26, 2013

### WannabeNewton

Bill I'm confused because Wald and Padmanabhan give the exact same equation just Wald has the Minkowski metric on the right and the other has the general metric tensor on the right. Why is it ok to say $(e_{\mu})^{a}(e_{\nu})_{a} = g_{\mu\nu}$ and still claim the tetrad is that of an orthonormal frame (here $\mu$ labels the basis vectors and $a$ labels the abstract index of a single basis vector).

George, yes that is true but Padmanabhan makes explicit reference to an arbitrary accelerated observer in space-time and talks about local Lorentz frames at events on this observer's worldline, which I cannot agree with (page 183 second paragraph since you have the book) since an arbitrary observer cannot set up locally inertial coordinates at events on his/her worldline and this obstructs him/her from erecting a locally inertial frame / local Lorentz frame.

Thanks for the responses guys.

11. May 26, 2013

### Ben Niehoff

No. A frame is a collection of vectors at a point. A "tetrad" field (I hate that terminology...far too 4d-centric; I call them orthonormal frame fields) is a collection of vector fields that are orthonormal at every point. But there is no requirement that each vector field be integrable (i.e., that it can be written $\partial / \partial x^\mu$ in some coordinates).

Even so, at each point, the vector fields comprise an orthonormal frame, which is the same thing as a local Lorentz frame. The vectors at $p$ can always be written, at $p$ only, as the coordinate vector fields of some geodesic normal coordinates. But this does not hold in a neighborhood of $p$. And that's fine.

Anyway, I see this mistake a lot in this forum, so I urge you to learn it correctly the first time: A frame is a set of basis vectors at a point. A frame field is a collection of vector fields such that they form a basis at every point. Frame fields can be orthonormal or otherwise. A coordinate system is not a frame; a coordinate system is a way to label points by mapping to $R^n$. But a coordinate system does allow you to define a coordinate frame, which is the collection of partial derivatives $\partial / \partial x^\mu$. Other kinds of frames may or may not be coordinate frames, depending.

12. May 26, 2013

### WannabeNewton

I didn't say the coordinate system was the frame though, I said the local Lorentz / locally inertial frame is derived from the locally inertial coordinate system (i.e. the basis vectors constituting the frame at a given event coincide with the coordinate vector fields of the chart at said event). This is, for example, how Carroll defines it on page 74 at the top. And when I said tetrad, I meant the tetrad evaluated at the event in question, so sorry for not clarifying that.

Can I say local Lorentz frames the same as instantaneously co-moving locally inertial frames? I.e. a locally inertial observer at that event co-moving with the arbitrary observer at that event?

EDIT: Thanks for the reply Ben!

Last edited: May 26, 2013
13. May 26, 2013

### George Jones

Staff Emeritus
Yes, but the tetrad for an accelerated observer at p is also the tetrad for a an inertial observer at p This is all that Padmanabhan means, and he is not incorrect. At a different q on the accelerated observer's worldline, there will a different comoving inertial observer.

14. May 26, 2013

### WannabeNewton

Ok thanks this clears it up. So we are basically taking a co-moving locally inertial observer at that event and taking his local Lorentz frame and passing it on to the original accelerated observer at that specific event? So this is basically the same thing as instantaneously co-moving inertial references frames for accelerated observers in SR-sort of like instantaneously co-moving locally inertial observers :p?

15. May 26, 2013

### George Jones

Staff Emeritus
Yes.

16. May 26, 2013

### WannabeNewton

So this is what he (Padmanabhan) means when he says the local Lorentz frames are different at each event. Cool, thanks. And when he says "treating Lorentz transformations as..." is he talking about the infinitesimal Lorentz transformation to go from one co-moving locally inertial observer to the very next one at the following event with the boost being the velocity of the very next co-moving observer, at the following event, with respect to the first co-moving observer to whom the Lorentz transformation is being applied?

17. May 26, 2013

### George Jones

Staff Emeritus
By definition, the components of the metric $g$ with respect to any basis $\left\{e_\mu\right\}$ (which may or may not be orthonormal) are $g_{\mu\nu} = g\left( e_\mu ,e\nu \right)$.

18. May 26, 2013

### Bill_K

I think you've got the orthonormality condition and the completeness condition reversed. To get the norm of a vector you must contract on μ, whereas you're contracting on a. Orthonormality says that eμa eμb = δab.

Also, eμa eνa = ημν is not even a covariant statement. There is no such thing as ημν, unless you're restricting the discussion to Minkowski coordinates, which we are not!

19. May 26, 2013

### George Jones

Staff Emeritus
I will have to look at at what Padmanabhan says, but, right now, I am on my way out to brunch with my family, then grocery shopping, then...

20. May 26, 2013

### WannabeNewton

You are certainly correct George but perhaps if I give the reference in Wald, it might direct you better to where my confusions are coming from. On page 49 section 3.4b, titled "Orthonormal Basis (Tetrad) Methods", he says "Thus, we may wish to introduce a 'nonholonomic' i.e. noncoordinate, orthonormal basis of smooth vector fields $(e_{\mu})^{a}$ satisfying $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$..." but you are certainly correct that $g(e_{\mu},e_{\nu}) = g_{\mu\nu}$. How should I differentiate between the two different expressions given?

EDIT: Ah ok, well thanks for all the help! Have fun at brunch, I look forward to hearing from ya.

EDIT 2: Sorry Bill, I literally just saw your post. I gave the reference in Wald where I got that from so maybe you could take a look and see if/where I'm interpreting it incorrectly? I'll read the rest of your post in the meantime, thanks! Carroll gives a similar thing on page 112 expression 3.66 (he calls it the 'canonical form').

Last edited: May 26, 2013
21. May 26, 2013

### Fredrik

Staff Emeritus
For any set $\{e_\mu\}_{\mu=0}^3$ of vector fields, and any coordinate system x defined on a a relevant subset of spacetime, we have
$$g(e_\mu,e_\nu) =g\left((e_\mu)^\rho\partial_\rho,(e_\nu)^\sigma\partial_\sigma\right) =g_{\rho\sigma}(e_\mu)^\rho(e_\nu)^\sigma =\left(g\otimes e_\mu\otimes e_\nu\right)\left(\partial_\rho,\partial_\sigma,dx^\rho,dx^\sigma\right)$$ In abstract index notation, the right-hand side is written as
$$g_{ab}(e_\mu)^a(e_\nu)^b = (e_\mu)^a (e_\nu)_a.$$ So if the $e_\mu$ are the basis vectors associated with a coordinate systems, or if we have simply defined the notation $g_{\mu\nu}$ by $g_{\mu\nu}=g(e_\mu,e_\nu)$, then the condition $(e_\mu)^a (e_\nu)_a=g_{\mu\nu}$ doesn't tell us anything at all about the $e_\mu$. It's just a condition satisfied by any four vector fields.

The condition $(e_\mu)^a (e_\nu)_a=\eta_{\mu\nu}$ on the other hand, is the definition of what it means for $\{e_\mu\}_{\mu=0}^3$ to be an orthonormal set. The condition can also be written as $g(e_\mu,e_\nu)=\eta_{\mu\nu}$, since we always have $(e_\mu)^a (e_\nu)_a=g(e_\mu,e_\nu)$.

Edit: The proof of $g(e_\mu,e_\nu)=(e_\mu)^a (e_\nu)_a$ above would have been clearer if I had done this:
$$g(e_\mu,e_\nu) =g\left((e_\mu)^\rho\partial_\rho,(e_\nu)^\sigma\partial_\sigma\right) =g_{\rho\sigma}(e_\mu)^\rho(e_\nu)^\sigma =(e_\mu)^\rho(e_\nu)_\rho=(e_\mu)^a (e_\nu)_a.$$ The last step follows from the calculation I did in my explanation of abstract index notation below.

Last edited: May 26, 2013
22. May 26, 2013

### dextercioby

It's confusing for me: the Greek indices are world indices and the Latin ones Minkowskian ?

23. May 26, 2013

### Bill_K

OMG Fredrik, this is absolutely false!!

Orthonormality is eμ0 eμ0 = 1, eμ0eμ1 = 0, and so on, ten equations. Do you see, I hope, that each equation is summed on μ?? These are ten vector dot products, each one of them equal to either 0 or 1.

The condition you're confusing this with is eμ0 eν0 + eμ1 eν1 + eμ2 eν2 + eμ3 eν3 = ημν (whatever ημν is supposed to mean in a general coordinate system!) Do you see that this set of equations is summed on the Lorentz index a and is totally different from the other one??

Last edited: May 26, 2013
24. May 26, 2013

### Fredrik

Staff Emeritus
I don't know what that means, but the convention used by Wald is to e.g. write a tensor of type (1,2) as $T^a{}_{bc}$. The latin indices just tell us what type of tensor we're dealing with. The components of this tensor in a coordinate system would be written as $T^\mu{}_{\nu\rho}$.

Also, when you see something like $\omega_a X^a$, it should be interpreted as what would be written as
$$(\omega\otimes X)(\partial_\mu,dx^\mu)=\omega(\partial_\mu)X(dx^\mu) =\omega(\partial_\mu)dx^\mu(X) =\omega(\partial_\mu)X(x^\mu) =\omega_\mu X^\mu$$ in the index-free notation. So we have $\omega_a X^a=\omega_\mu X^\mu$, but it takes a small calculation to see it. (Note that I used the notation X both for a tangent vector and the corresponding "dual dual vector").

Last edited: May 26, 2013
25. May 26, 2013

### WannabeNewton

As per Wald's notation, the Greek indices label which basis vector we are talking about and the Latin ones denote each basis vector using the abstract index notation. So $(e_0)^{a}$ is the time-like basis vector etc. and he says the orthonormal basis of smooth vector fields $(e_{\mu})^{a}$ satisfy $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ which I interpreted as saying $(e_{0})^{a}(e_{0})_{a} = -1, (e_{i})^{a}(e_{j}) _{a} = \delta_{ij}, (e_{0})^{a}(e_{i})_{a} = 0$ where $i,j=1,2,3$. I thought this was the orthonormality condition but I may be wrong. I mean this is basically just the statement that $g_{ab}(e_{\mu})^{a}(e_{\nu})^{b} = \eta_{\mu\nu}$ which is how I thought tetrads were always defined (so that the associated orthonormal frame at each event can be physically thought of as a clock for the "time axis" and three meter sticks held along mutually orthogonal "spatial axes"). This is also how Carroll defines it in appendix J, expression J.1. which is also what Fredrik wrote (hi Fredrik, how's it goin pal xP)


text:
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## Kryptos

Cryptography article on Nova

## Euler, born 15th April 1707

My favourite sum: $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{\pi^2}{6}$ To Euler, The master of infinity, who summed with impunity! PS: A nice article on infinity, explaining concepts up to Cantor’s work. via http://www.mathed.org via amazon.

## Coincidences or just bad luck

How many times have you tried to search for a particular item and find that the only thing missing is the one that you want? Two years back, I was searching for a 1995 article in the Rocky Mountain Journal … Continue reading

Posted in Fun Stuff, Number Theory, Probability | 1 Comment

## Recurring Decimals

Came across a funny little problem last week. Explain the following: For primes $p \ge 7$ , the decimal expansion of $\frac{1}{p}$ is non-terminating. Is there a formula for the number of repeating digits? $\frac{1}{7} = 0.\overline{142857}$ 6 digits [tex]\frac{1}{11} … Continue reading

Posted in Number Theory, Problems | 2 Comments

## Video Lectures

When you are based in a small Asian country, you get precious little opportunities to hear good lectures from distinguished lecturers. Luckily, there is an increasing trend of video-casts available on certain websites. The silly thing is that even when … Continue reading

Posted in General, Number Theory | 1 Comment


text:
# Euclid's Division Lemma

Euclid's Division Lemma
Go back to  'Integers'

In this mini-lesson, we will learn about the Euclid division lemma by understanding the Euclid division algorithm, the division using Euclid division lemma, and how to apply them while solving problems. We will also discover interesting facts around them.

Paul wants to plant some saplings in his backyard.

He has 48 sapling plants with him.

He is making 9 rows such that he can plant 5 saplings in each row.

If he plants the saplings in the above way, he can plant a total of $$5 \times 9 = 45$$ sapling plants.

The sapling plants remaining with him would be $$48 - 45 = 3$$

We can write the above scenario mathematically as:

The above scenario shows the way of representing the division of positive integers with the help of Euclid's division lemma.

Now let's explore Euclid's division lemma in this lesson.

## Lesson Plan

 1 What Is Euclid’s Division Lemma? 2 Thinking Out of the Box! 3 Solved Examples on Euclid's Division Lemma 4 Important Notes on Euclid's Division Lemma 5 Interactive Questions on Euclid's Division Lemma

## What Is Euclid’s Division Lemma?

Euclid’s Division Lemma (lemma is similar to a theorem) says that given two positive integers, $$a$$ and $$b$$, there exists unique integers, $$q$$ and $$r$$, such that:

$\boxed{a = bq + r,\;0 \leqslant r < b}$

The integer $$q$$ is the quotient and the integer $$r$$ is the remainder.

The quotient and the remainder are unique.

In simple words, Euclid's Division Lemma statement is that if we divide an integer by another integer (non-zero integer), we will get a unique integer as qoutient and a unique integer as remainder

Here are some examples of the application of this lemma:

$23 = 2 \times 11 + 1 \\[0.2cm] 54 = 7 \times 7 + 5 \\[0.2cm] 63 = 9 \times 7 + 0$

## How to Divide Integers Using Euclid’s Lemma?

Euclid’s lemma or Euclid's Division Lemma statement says that for given two positive integers, $$a$$ and $$b$$, there exists unique integers $$q$$ and $$r$$ such that:

$a = bq + r,\;0 \leqslant r < b$

So, let's take $$a = 9$$ and $$b = 1$$

$9 = 1 \times 9 + 0$

Here, $$q = 9$$ and $$r = 0$$, we can clearly see that $$0 \leqslant r < 1$$

Now, let's take $$a = 9$$ and $$b = 2$$

$9 = 2 \times 4 + 1$

Here, $$q = 4$$ and $$r = 1$$, we can clearly see that $$0 \leqslant r < 2$$

Now, let's take $$a = 9$$ and $$b = 3$$

$9 = 3 \times 3 + 0$

Here, $$q = 3$$ and $$r = 3$$, we can clearly see that $$0 \leqslant r < 3$$

Now, let's take $$a = 9$$ and $$b = 4$$

$9 = 4 \times 2 + 1$

Here, $$q = 2$$ and $$r = 1$$, we can clearly see that $$0 \leqslant r < 4$$

We can observe that when we divide two integers using the Euclid lemma division algorithm, we can always get a non-negative integral remainder which is always less than the divisor.

## What Is the Use of Euclid's Division Lemma?

Euclid's division lemma has many uses.

Let's learn about two of the most important uses of this algorithm.

### Calculating HCF of Two Numbers

We use Euclid's division lemma to find the HCF of large numbers which is typically difficult to calculate using basic HCF calculation techniques.

The HCF of two numbers can be calculated with the help of Euclid's division lemma by following these steps.

Let's take two numbers for which we need to find the HCF.

$$c$$ and $$d$$ are two numbers such that $$c > d$$.

Step 1:- Apply Euclid’s division lemma to $$c$$ and $$d$$. We can find whole numbers, $$q$$ and $$r$$ such that $$c = dq + r,\;0 \leqslant r < d$$.

Step 2:- If $$r = 0$$, $$d$$ is the HCF of $$c$$ and $$d$$. If $$r \neq 0$$, apply the division lemma to $$d$$ and $$r$$.

Step 3:- Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

### Example

Use Euclid’s algorithm to find the HCF of 4052 and 12576.

### Solution

We can observe that 12576 > 4052, hence $$c = 12576$$ and $$d = 4052$$.

According to Euclid's division lemma, we have:

$12576 = 4052 \times 3 + 420$

Here the remainder $$420 \neq 0$$, we again apply Euclid's division lemma to 4052 and 420

$4052 = 420 \times 9 + 272$

Let's repeat the same process until we get the remainder 0

Since the remainder is now zero, we can stop the process.

The divisor at this stage is 4; the HCF of 12576 and 4052 is 4

Explore the HCF calculator below to find the HCF of any two numbers using Euclid's division lemma.

Finding Properties of Numbers

We can understand the various properties of odd and even numbers using Euclid's division lemma.

### Example

Show that every positive even integer is of the form $$2q$$ and that every positive odd integer is of the form $$2q + 1$$, where $$q$$ is some integer.

Solution

Assume $$a$$ is a positive integer and $$b = 2$$.

By Euclid’s algorithm, we can write:

$a = 2q + r$

Where $$q \geq 0$$ and $$r$$ can take either $$0$$ or $$1$$ by the rule $$0 \leqslant r < b$$

Thus, $$a = 2q$$ or $$a = 2q + 1$$.

If $$a$$ is of the form $$2q$$, then $$a$$ is an even integer, and if $$a$$ is of the form $$2q + 1$$, then it is an odd integer.

For example, if we substitute $$q = 1, 2$$, and $$3$$ then $$a = 2q$$ is even

$$a = 2 \times 1 = 2$$
$$a = 2 \times 2 = 4$$
$$a = 2 \times 3 = 6$$

And, if we substitute $$q = 1, 2$$, and $$3$$ then $$a = 2q + 1$$ is odd

$$a = 2 \times 1 + 1= 3$$
$$a = 2 \times 2 + 1= 5$$
$$a = 2 \times 3 + 1= 7$$

Think Tank
1. Does Euclid’s Division Lemma hold if $$a$$ is a negative integer?
2. Find any positive even integer of the form $$6q$$, $$6q + 2$$ or $$6q + 4$$ where $$q$$ is some integer.

## Solved Examples on Euclid's Division Lemma

 Example 1

Paul knows that $$n$$ is an odd integer.

He wants to prove that $$n^2 - 1$$ is a multiple of 8

Can you help him?

Solution

By Euclid’s Division Lemma, $$n$$ can be written as:

$n = 2k + 1,\;k \in \mathbb{Z}$

Thus,

\begin{align}&{n^2} - 1 \;= {\left( {2k + 1} \right)^2} - 1\\& \qquad\quad= \left( {4{k^2} + 4k + 1} \right) - 1\\& \qquad\quad= 4{k^2} + 4k\\&\qquad\quad = 4k\left( {k + 1} \right) & \end{align}

Note that the product $$k(k+1)$$ will always be even.

This is because either $$k$$ is even, or $$k + 1$$ is even.

Thus,

$$4k(k + 1)$$, must be a multiple of 8.

 $\therefore n^2 - 1 \text{ is a multiple of 8 }$
 Example 2

Paul is emptying two water tanks that are filled with water using a jug.

The water tanks have capacities of 420 liters and 130 liters.

What is the maximum capacity of the jug he should use so that no water remains in the tanks?

Solution

Given:

The capacities of tanks are 420 liters and 130 liters.

To find the maximum capacity of the jug, we need to find the HCF of 420 and 130.

Let's apply Euclid's division lemma to calculate the HCF of these numbers.

$420 = 130 \times 3 + 30 \\[0.2cm] 130 = 30 \times 4 + 10 \\[0.2cm] 30 = 10 \times 3 + 0$

The HCF of 420 and 130 is 10.

 $$\therefore$$ Maximum capacity of the jug is 10 liters.
 Example 3

The cube of a positive integer is divided by 9.

What can be the possible remainders?

Solution

We have to use Euclid's Division Lemma to find the remainders.

If we can express the cube of any positive integer in the form $$9q + r$$, then we can solve the problem easily.

We will then have to find what values $$r$$ can take.

Let $$n$$ be any positive integer.

Note that when we divide $$n$$ by 3, we can get three possible remainders: 0, 1, or 2.

Thus, $$n$$ can be written as:

\begin{align}&n = 3k\\&\quad{\rm{or}}\\&n = 3k + 1\\&\quad{\rm{or}}\\&n = 3k + 2\end{align}

Let's consider each case separately.

Case 1: In this case, we have:

${n^3} = {\left( {3k} \right)^3} = 27{k^3} = 9\left( {3{k^3}} \right)$

Thus, the cube of $$n$$ will be perfectly divisible by 9 in this case (the remainder will be 0).

Case 2: In this case, we have:

\begin{align}&{n^3} = {\left( {3k + 1} \right)^3}\\&\quad = {\left( {3k} \right)^3} + {\left( 1 \right)^3} + 3\left( {3k} \right)\left( 1 \right)\left( {3k + 1} \right)\\&\quad= 27{k^3} + 1 + 9k\left( {3k + 1} \right)\\&\quad = 9k\left( {3{k^2} + 3k + 1} \right) + 1\\&\quad = 9\left\{ {k\left( {3{k^2} + 3k + 1} \right)} \right\} + 1 & \end{align}

It is clear that on dividing the cube of $$n$$ by 9, in this case, we will be left with a remainder of 1.

Case 3: In this case, we have:

\begin{align}&{n^3} = {\left( {3k + 2} \right)^3}\\&\quad= {\left( {3k} \right)^3} + {\left( 2 \right)^3} + 3\left( {3k} \right)\left( 2 \right)\left( {3k + 2} \right)\\&\quad = 27{k^3} + 8 + 18k\left( {k + 2} \right)\\&\quad= 9k\left( {3{k^2} + 2k + 4} \right) + 8\\&\quad = 9\left\{ {k\left( {3{k^2} + 2k + 4} \right)} \right\} + 8 & \end{align}

In this case, on dividing the cube of $$n$$ by 9, we will be left with a remainder of 8.

Thus, we conclude that dividing the cube of a positive integer by 9 can yield three possible remainders:
$\boxed{{\text{Three Possible Remainders: }}0,1,8}$

In other words, the cube of any integer can be expressed as follows:

\begin{align}&{n^3} = 9k\\&\quad\;{\rm{or}}\\&{n^3} = 9k + 1\\&\quad\;{\rm{or}}\\&{n^3} = 9k + 8\end{align}

You can verify this result by taking some cubes and checking the remainders you obtain upon their division by 9.

 $\therefore \text{Possible remainders are 0, 1, 8}$

Important Notes
1. When you divide one integer by another non-zero integer, you are left with a quotient and a remainder.
2. The remainder is always less than the divisor.
3. Euclid's lemma is used to calculate the HCF between large numbers.

## Interactive Questions on Euclid's Division Lemma

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 Euclid's division lemma. The math journey around Euclid's division lemma started with what a student already knew about division and went on to creatively crafting the fresh concept of Euclids division lemma 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.

## About 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.

## FAQs on Euclid's Division Lemma

### 1. What is lemma and algorithm?

A lemma is a proven statement used for proving another statement.

An algorithm is a series of well-defined steps that give a procedure for solving a type of problem.

### 2. What is a lemma in Greek?

The word "lemma" is derived from an ancient Greek word which means "a thing that is received."

### 3. Is lemma a proof?

Yes, a lemma is an already proven statement.

In mathematics, lemmas are used to prove other results.

This is why lemma is sometimes called a "helping theorem."

### 4. How do you use Euclid’s division algorithm?

Euclid's division lemma is used to find the HCF between two large numbers or it can be used to prove the properties of a number.

### 5. What is the difference between lemma and corollary?

A lemma is a proven statement used for proving another statement.

A corollary is a theorem that can be deduced from a previous, more notable statement. A corollary has lesser importance.

### 6. How do you find the HCF using Euclid's Division Lemma?

You can find the HCF between two numbers using Euclid's division lemma by following these steps.

### 7. What is Euclid's formula?

If $$a, b, c$$ are Pythagorean triplets $$a^2 + b^2 = c^2$$, then Euclid's formula suggests that.

$$a = m^2 - n^2$$
$$b = 2mn$$
$$c = m^2 + n^2$$

Where $$m$$ and $$n$$ are positive integers.

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text:
# maximum number of edges to be removed to possess a property

I am working on a problem. We know that on squaring a cycle, degree of every vertex is 4. For squares of cycles, we know if we delete any arbitrary edge then still eccentricity is same for all vertices. And if we delete 3 edges incident on a single vertex, it loses its property that eccentricity is same for all vertices, because eccentricity is not same for those graphs having a vertex of degree1. How to find out the maximum number of edges to be deleted so that eccentricity is still same. Any idea or suggestion will be helpful. Thanks for your help.

The square $G$ of a cycle graph of length $n$ has vertices $0,1,\ldots, n-1$ and an edge between $a$ and $b$ iff $a-b\equiv -2, -1, 1, \text{ or }2\pmod n$. Note that this implies that the claim that $\rho(a)=4$ for all vertices holds only if $n>4$. Therefore, in what follows, we shall assume that $n>4$. Let $F(n)$ be the minimum number of edges that can be removed such that the same-eccentricity-property of $G$ is destroyed.

The distance in $G$ from vertex $0$ to vertex $k$ is quite obviously $d(0,k)=\min\{\lceil\frac k2\rceil,\lceil\frac {n-k}2\rceil\}$ and hence the eccentricity of $0$ is $\epsilon(0)=\left\lceil\frac{\lfloor \frac {n}2\rfloor}2\right\rceil=\lfloor\frac{n+2}4\rfloor$ (and by symmetry for all vertices). We observe the following about shortest paths from $0$ to $k$:

• All steps go clockwise or all go counterclockwise. This is because two consecutive steps in opposite direction either cancel or can be replaced by a single step.
• If $\lceil \frac k2\rceil=\lceil \frac {n-k}2\rceil$, shortest paths of both clockwise and anticlockwise orientation exist.

Lemma 0. If $n\not\equiv 1\pmod 4$, then $F(n)>1$.

Proof: Let $G^\star$ be any graph obtained from $G$ by removing one edge. Since each edge in $G$ is part of a triangle, $d^\star(v,w)\le d(v,w)+1$. Therefore, eccentricity can only change if the distance to the farthest vertices grows. For $v=0$, those are as follows:

• If $n=4m-2$: $w=2m$ with a path of length $m$ clockwise and counterclockwise
• If $n=4m$: $w=2m$ with as path of length $m$ clockwise and counterclockwise; $w=2m\pm1$ with a path beginning and a path ending with an edge of length $1$
• If $n=4m-1$: $w=2m$ with a path of length $m$ clockwise (all steps of length $2$) and several paths counterclockwise (with one step of length $1$); $w=2m+1$ similarly.

In all cases, there are two edge-disjoint paths of the required length, hence the distance does not change when removing a single edge. $_\square$

Lemma 1. If $n\equiv 1\pmod 4$, then $F(n)=1$.

Proof: Assume $n=4m+1$ (so that $\epsilon(v)=m$) and let $G'$ be the graph $G$ with edge $(0, 2)$ removed. Clearly, $d'(1,v)=d(1,v)$ for all $v$ and hence $\epsilon'(1)=\epsilon(1)$. The only path $0\stackrel 2\to2\stackrel 2\to\ldots\stackrel 2\to2m$ of length $m$ from $0$ to $2m$ does not exist in $G'$, hence $$\epsilon'(0)\ge d'(0,2m)>m=\epsilon(1).$$ Removing a single edge can destroy the same-eccentricity-property. $_\square$

Lemma 2. If $n\equiv 3\pmod 4$, then $F(n)=2$.

Proof: Let $n=4m-1$ (so that $\epsilon(v)=m$ for all $v$) and let $G''$ be $G$ with edges $(0,1)$ and $(0,2)$ removed. Then $$\epsilon''(0)\ge d''(0,2m-2)\ge \min\{d(n-1,2m-2),d(n-2,2m-2)\}+1=m+1$$ because any path from $0$ must have $n-1$ or $n-2$ as next node. On the other hand, $d''(1,v)=d(1,v)$ unless all shortest $G$-paths from $1$ to $v$ contain either $(0,1)$ or $(0,2)$. The latter is never the case and the former only if $v=0$ (but that case does not matter as $d''(1,0)=2\le m$) as otherwise the shortest path continues $1\stackrel 1\to 0\stackrel 2\to n-2$ and can be replaced with $1\stackrel 2\to n-1\stackrel1\to n-2$. Hence $\epsilon''(1)=\epsilon(1)=m$. $_\square$

Lemma 3. If $n\equiv 0\pmod 4$, then $F(n)=2$.

Proof: Let $n=4m$ (so that $\epsilon(v)=m$) and again let $G''$ be $G$ with edges $(0,1)$ and $(0,2)$ removed. Then $$\epsilon''(0)\ge d''(0,2m-1)\ge \min\{d(n-1,2m-1),d(n-2,2m-1)\}+1=m+1$$ whereas just as above $\epsilon''(1)=\epsilon(1)$. $_\square$

Lemma 4. If $n\equiv 2\pmod 4$, then $F(n)=3$.

Proof: Let $n=4m-2$ (so $\epsilon(v)=m$) with $m\ge 2$ and let $G^\star$ be any graph obtained by removing two edges from $G$. We will show that $d^\star(0,v)\le m$ for $1\le v\le 2m-1$. Then by symmetry we will have $\epsilon^\star(v)=m$ for all $v$.

For $v=2m-1$, note that there are four edge-disjoint paths of length $m$: $0\stackrel 1\to 1\stackrel2\to\ldots\stackrel2\to 2m-1$, $0\stackrel 2\to 2\stackrel2\to\ldots\stackrel2\to 2m-2\stackrel1\to 2m-1$, ans similarly in negative direction. At most two of these can be destroyed by the edge removal, hence $d^\star(0,v)=m$.

For $v=2m-2$, we have edge-disjoint paths $0\stackrel2\to n-2\stackrel2\to\ldots\stackrel2 2m-2$, $0\stackrel2\to2\stackrel2\to\ldots\stackrel2\to 2m-2$ and $0\stackrel 1\to1\stackrel2\to\ldots\stackrel1\to 2m-2$ of lengths $\le m$. Again, one must survive the edge removal, $d^\star(0,v)\le m$.

For $2<v<2m-2$ odd, we have two edge-disjoint paths $0\stackrel 1\to 1\stackrel 2\to\ldots\stackrel2\to v$ and $0\stackrel 2\to 2\stackrel 2\to\ldots\stackrel1\to v$ of length $<m$. Unless each of these paths contains exactly one of the removed edges, we are done. If $(0,1)$ is a removed edge, then $0\stackrel 1n-1\to 2\stackrel 1\to 1\stackrel 2\to\ldots\stackrel2\to v$ has length $m$ (note that $(1,2)\ne(v-1,v)$ and we are done. Similarly, we are done if $(v-1,v)$ is a removed edge. Therefore, we may assume that in each of the two paths, an edge $(k,k+2)$ is removed and that edges $(k,k+1)$ and $(k+1,k+2)$ are both present in $G^\star$ so that we obtain a path of length $\le m$.

For $2\le v<2m-2$ even, we have two edge-disjoint paths $0\stackrel 1\to1\stackrel 2\to\ldots\stackrel1\to v$ and $0\stackrel 22\stackrel 2\to\ldots\stackrel2\to v$ of length $<m$. Again we may assume that exactly one edge of each path is missing. Again, if $(0,1)$ is removed, we can replace it with $0\stackrel 1n-1\stackrel 21$ andsimilarly if $(v-1,v)$ is missing. Again we conclude that in each of the two paths, an edge $(k,k+2)$ is removed and that edges $(k,k+1)$ and $(k+1,k+2)$ are both present in $G^\star$ so that we obtain a path of length $\le m$.

For the remaining case $v=1$, note that we have three edge-disjoint paths $0\stackrel 1\to 1$, $0\stackrel 2\to 2\stackrel 1\to 1$ and $0\stackrel 1\to n -1\stackrel 2\to 1$ of length $\le 2\le m$, at least one of which is present in $G^\star$. Together with your own result about three edges, the claim of the lemma follows. $_\square$

In summary, we have therefore

Theorem. Let $n>4$. Then $$F(n)=\begin{cases}2 & \text{if }n\equiv 0\pmod 4\\1 & \text{if }n\equiv 1\pmod 4\\3 & \text{if }n\equiv 2\pmod 4\\2 & \text{if }n\equiv 3\pmod 4\\\end{cases}$$

Remark: Note the degenerate cases $F(4)=1$ and $F(3)=1$.

• Didn't the problem ask for the maximum number of edges to be removed so that the same-eccentricity-property of $G$ is still preserved? – Jared May 12 '13 at 6:25

Here is what I can say with certainty

Let $n\ge 5$. The maximum number of edges that can be removed from a squared cycle so that the eccentricity of each vertex remains unchanged is at least $n-2-(n+2)\operatorname{mod}4$.

To show this is true, I will give an explicit set of removed edges in each case. Label consecutive vertices of the cycle $1,2,3,\ldots, n$ and call an edge 'outer' if it connects adjacent vertices in the cycle, and 'inner' if it connects vertices a distance of two from each other. Splitting into four cases, depending on the residue class of $n$ modulo $4$, we can remove the following edges:

• $n\equiv 0\operatorname{mod}4$: $n-6$ outer edges and $2$ inner edges. Remove all outer edges except those connecting $(1,2),(2,3),(3,4),(\frac{n}{2}+1,\frac{n}{2}+2),(\frac{n}{2}+2,\frac{n}{2}+3),$ and $(\frac{n}{2}+3,\frac{n}{2}+4)$. Remove inner edges connecting $(1,3)$ and $(\frac{n}{2}+2,\frac{n}{2}+4).$
• $n\equiv 1\operatorname{mod}4$: $n-5$ outer edges and $0$ inner edges. Remove all outer edges except those connecting $(1,2),(2,3),(3,4),(\frac{n+3}{2},\frac{n+3}{2}+1),$ and $(\frac{n+3}{2}+1,\frac{n+3}{2}+2)$. Remove no inner edges.
• $n\equiv 2\operatorname{mod}4$: $n-4$ outer edges and $2$ inner edges. Remove all outer edges except those connecting $(1,2),(2,3),(\frac{n}{2}+1,\frac{n}{2}+2),$ and $(\frac{n}{2}+2,\frac{n}{2}+3)$. Remove inner edges connecting $(1,3)$ and $(\frac{n}{2}+1,\frac{n}{2}+3)$.
• $n\equiv 3\operatorname{mod}4$: $n-5$ outer edges and $2$ inner edges. Remove all outer edges except those connecting $(1,2),(2,3),(3,4),(\frac{n+3}{2},\frac{n+3}{2}+1),$ and $(\frac{n+3}{2}+1,\frac{n+3}{2}+2)$. Remove inner edges connecting $(1,3)$ and $(\frac{n+3}{2},\frac{n+3}{2}+2)$.

This should all match up with the conjecture. The above is proved case-by-case by induction. Notice that the remaining graphs in all cases have two gaps of outer edges. In each gap, we can add $2$ vertices and $2$ edges so that the eccentricity of each vertex is raised by $1$ (note that the eccentricity of a squared cycle with $n+4$ edges is $1$ greater than the eccentricity of a squared cycle with $n$ edges).

The hard part is getting rid of the words 'at least' in the above claim. It can be shown that when $n\equiv 1\operatorname{mod}4$, we cannot remove any inner edges, and we need at least $5$ outer edges to preserve eccentricity. This establishes the result for $n\equiv 1\operatorname{mod}4$.

When $n\equiv 0\operatorname{mod}4$, we can remove at most $2$ inner edges, but I can't seem to show that $6$ outer edges are necessary.

The other cases are even trickier. I have reason to believe my result is true for all cases, but a formal proof eludes me.


text:
[Theorem 3.6.5, Tohoku].

Proposition 20.21.7 (Grothendieck). Let $X$ be a Noetherian topological space. If $\dim (X) \leq d$, then $H^ p(X, \mathcal{F}) = 0$ for all $p > d$ and any abelian sheaf $\mathcal{F}$ on $X$.

Proof. We prove this lemma by induction on $d$. So fix $d$ and assume the lemma holds for all Noetherian topological spaces of dimension $< d$.

Let $\mathcal{F}$ be an abelian sheaf on $X$. Suppose $U \subset X$ is an open. Let $Z \subset X$ denote the closed complement. Denote $j : U \to X$ and $i : Z \to X$ the inclusion maps. Then there is a short exact sequence

$0 \to j_{!}j^*\mathcal{F} \to \mathcal{F} \to i_*i^*\mathcal{F} \to 0$

see Modules, Lemma 17.7.1. Note that $j_!j^*\mathcal{F}$ is supported on the topological closure $Z'$ of $U$, i.e., it is of the form $i'_*\mathcal{F}'$ for some abelian sheaf $\mathcal{F}'$ on $Z'$, where $i' : Z' \to X$ is the inclusion.

We can use this to reduce to the case where $X$ is irreducible. Namely, according to Topology, Lemma 5.9.2 $X$ has finitely many irreducible components. If $X$ has more than one irreducible component, then let $Z \subset X$ be an irreducible component of $X$ and set $U = X \setminus Z$. By the above, and the long exact sequence of cohomology, it suffices to prove the vanishing of $H^ p(X, i_*i^*\mathcal{F})$ and $H^ p(X, i'_*\mathcal{F}')$ for $p > d$. By Lemma 20.21.1 it suffices to prove $H^ p(Z, i^*\mathcal{F})$ and $H^ p(Z', \mathcal{F}')$ vanish for $p > d$. Since $Z'$ and $Z$ have fewer irreducible components we indeed reduce to the case of an irreducible $X$.

If $d = 0$ and $X = \{ *\}$, then every sheaf is constant and higher cohomology groups vanish (for example by Lemma 20.21.2).

Suppose $X$ is irreducible of dimension $d$. By Lemma 20.21.4 we reduce to the case where $\mathcal{F} = j_!\underline{\mathbf{Z}}_ U$ for some open $U \subset X$. In this case we look at the short exact sequence

$0 \to j_!(\underline{\mathbf{Z}}_ U) \to \underline{\mathbf{Z}}_ X \to i_*\underline{\mathbf{Z}}_ Z \to 0$

where $Z = X \setminus U$. By Lemma 20.21.2 we have the vanishing of $H^ p(X, \underline{\mathbf{Z}}_ X)$ for all $p \geq 1$. By induction we have $H^ p(X, i_*\underline{\mathbf{Z}}_ Z) = H^ p(Z, \underline{\mathbf{Z}}_ Z) = 0$ for $p \geq d$. Hence we win by the long exact cohomology sequence. $\square$

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