text:
# Intuition behind the “large but finite search space” of the proof of the four colour theorem?

I know the four colour theorem was solved by a computer checking a large number of cases. What I don't understand is why there are only a large but finite number of cases. It seems like there should be infinitely many planar graphs... What is the intuition behind cutting this down to a finite number of configurations?

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The question in the title and the question in the body are different: one askes for the intuition about the theorem and the other for the intuition about the (current) proofs. People are answering the body... (I would love an intuitive explanation for the theorem itself, though!) –  Mariano Suárez-Alvarez Sep 14 '10 at 15:26
I think the wikipedia article I linked to does an adequate job of explaining the content of the theorem. –  Seamus Sep 15 '10 at 10:40
sure but I meant, an intuitive explanation of why one would expect the theorem to be true, not of what the statement means. –  Mariano Suárez-Alvarez Sep 15 '10 at 11:45
In the comments to my answer, Seamus clarifies that he gets how the proof works and would like a heuristic argument as to why there should be a finite unavoidable set of reducible configurations. I don't know. Any ideas? –  David Speyer Sep 15 '10 at 14:11

I'm not sure whether or not you understand how the proof of the $4$ color map theorem worked. I'll sketch that, and then make some philosophical comments.

Let's start by proving the $6$ color map theorem. Without loss of generality, we may assume that all the faces of our graph are triangular, because adding edges just makes a graph harder to color. (We are coloring vertices.) Let $V$, $E$ and $F$ be the number of vertices, edges and faces. Every edge is in two faces, and every face has three edges, so $F=(2/3)E$. We have $V-E+F=2$ by Euler's relation, so $V=E/3 +2$. Since the sum of all vertex degrees is $2E$, the average degree of a vertex is $2E/V = (2E)/(E/3 + 2) < 6$. So the average degree is less than $6$ and there must be a vertex $v$ of degree $\leq 5$. Delete $v$, $6$-color the rest of the graph (by induction), and add $v$ back. Since the vertex has degree $\leq 5$, there is some color which does not border $v$, color $v$ that color.

To prove the $5$ color map theorem requires one new idea. Find the vertex $v$ as before, color the rest of the graph and add it back. But we have a problem if the $v$ has degree $5$ and its neighbors all have different colors. The idea to get around this is to consider the subgraphs $G_{ij}$ consisting of vertices of colors $i$ and $j$, for different pairs $(i,j)$. Because $G_{ij}$ and $G_{k \ell}$ can not cross, for $i$, $j$, $k$ and $\ell$ distinct, some of these graphs must be disconnected. By considering the finitely possible many arrangements of these graphs, one sees that, in each case, one can take some component of $G_{ij}$ and switch the colors $i$ and $j$ on that component in order to make $v$ have only $4$ distinct colors of neighbor. See, for example, Wikipedia for details.

The $4$ color map theorem requires the recoloring idea of the previous paragraph, plus one more idea. Instead of showing that there is always a vertex of degree $3$, $4$ or $5$, one shows that one of a finite list of larger configurations is unavoidable. This list tends to have 600-2000 members, depending on exactly which copy of the proof you read. The idea to showing that this list is unavoidable is an averaging argument like the one I used; the technical term to search for is "discharging". For each of these hundreds of configurations, one removes it from the graph, colors the rest, and analyzes the possible positions of the graphs $G_{ij}$. In each case one can recolor some of the components of the $G_{ij}$ in such a way that, when the configuration is put back in, the whole graph will be colorable.

As I understand it, computers are used in two ways. The less interesting one is that there is a routine which verifies that the discharging argument is correct, finds the possible positions of the $G_{ij}$ and checks that they can be recolored.

The more interesting one is that, when the routine in the previous paragraph fails, the computer perturbs the discharging argument, finds a new resulting list of configurations and tries again. It is basically a genetic process, where the computer tweaks the discharging algorithm until it finds a list of configurations which works.

The leap of faith which Appel and Haken took was to believe that there was some sufficiently complex discharging algorithm, and some list of configurations for which this procedure would work, and they just had to search long enough. Since then, this search has been implemented many times, with the details of the search algorithm different in each case, and many different solutions have been found.

To my mind, the whole thing reminds me of parallel evolution. There are lots of ways to make a wing, and if you leave an organism for long enough in an environment that rewards flight then it will develop one. But all of those ways are so complex that a human intellegence couldn't engineer them. Similarly, there are lots of discharging algorithms that prove the $4$ color map theorem, but they are all too bizarre for a human mathematician to find directly.

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"There are lots of ways to make a wing, and if you leave an organism for long enough in an environment that rewards flight then it will develop one." Is that a theorem of evolutionary biology? If so, I'd like to see the proof. (More plainly, I think you are probably exaggerating a bit for dramatic effect, n'est-ce pas?) –  Pete L. Clark Sep 14 '10 at 22:05
Yup, I am exaggerating. Also, engineers have been known to design wings. But this really is the vague intuition I have for why these searches work. –  David Speyer Sep 14 '10 at 23:00
The approach above is the usual one for proving the 5-color theorem of Heawood, however, relatively recently there is a new approach which does not require the use of Euler's Polyhedral formula but depends directly on the Jordan Curve Theorem: Thomassen, Carsten (1994), "Every planar graph is 5-choosable", Journal of Combinatorial Theory, Series B 64: 180–181 . –  Joseph Malkevitch Sep 15 '10 at 2:18
Nitpicking aside, this is a very nice answer. –  Pete L. Clark Sep 15 '10 at 4:05
"one of a finite list of larger configurations is unavoidable. This list tends to have 600-2000 members" This is exactly the point I don't get, the point I wanted an intuitive explanation of. –  Seamus Sep 15 '10 at 10:44

Perhaps this expository account of an "improved" proof of the 4-color theorem will help:

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From "Kempe's" proof we know that the size of the unavoidable set of reducible configurations cannot be 5. Then for any proof of the 4CT of this kind "What is the exact minimum size of the unavoidable set of reducible configurations?" or "Is the number of 633 reducible configurations in the RSST proof (1996) mininum?"

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×

# T's in the plane

You can fit uncountably many lines in the plane (e.g. by having them parallel). You can also fit uncountably many circles in the plane (e.g. by arranging them concentrically).

Can you fit uncountably many capital letter "T"s in the plane? Give an example or prove that you can't.

Note by Maggie Miller
8 months, 3 weeks ago

Sort by:

For every length L, we can think of the plane divided up in to square cells of side length L. Since there are countablly many cells in the grid, as least one cell must contain parts of uncountably many Ts. But there is no dense way to stack Ts, unlike lines and circles. I don't have a quick geometric argument as to why not, but the greatest number of parts of distinct Ts I can get arbitrarily close to each other is 10---that's a lot, but it's countable · 8 months, 3 weeks ago

I think that's a really good start (and intuitively the whole answer). Try taking little boxes that contain the intersection point (like where the two lines cross) of each T instead of boxes around whole Ts. If you can argue that you only need countably many boxes and that you can only fit countably many intersection point in each box, then you're done. :) · 8 months, 3 weeks ago

format: plain


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# 1.4: Arc Length

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

We can find the arc length of a curve by cutting it up into tiny pieces and adding up the length of each of the pieces. If the pieces are small and the curve is differentiable then each piece will be approximately linear.

We can use the distance formula to find the length of each piece:

$L = \sqrt{ \left(\Delta{x}\right)^2+ \left(\Delta{y}\right)^2}.$

Multiplying and dividing by $$\Delta{t}$$ gives

$L = \sqrt{ \left(\dfrac{\Delta{x}}{\Delta{t}}\right)^2+ \left(\dfrac{\Delta{y}}{\Delta{t}}\right)^2} \, \Delta {t}.$

Adding up all the lengths and taking the limit as $$\Delta{t}$$ approaches 0 gives the formula

$L = \int _a^b {\sqrt{ \left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} \, dt}.$

Example $$\PageIndex{1}$$

Find the arc length of the curve defined parametrically by $$x(t) = t^2 + 4t$$ and $$y(t) = 1 - t^2,$$ from $$0 < t < 2$$.

Solution

We calculate the derivatives:

$x' = 2t + 4 \;\;\; \text{and}\;\;\; y' = -2t. \nonumber$

Hence, the integrand to integrate is

$\sqrt{ \left(\dfrac{dx}{dt}\right)^2+ \left(\dfrac{dy}{dt}\right)^2} = \sqrt{ 8\,t^2 + 16\, t + 16} \nonumber$

and the full integral to solve, with limits,

$\int_0^2 \sqrt{ 8\,t^2 + 16\, t + 16} \, dt \nonumber$

is difficult (but not impossible) to do by hand. Either by hand or by computer we get

$L \approx 12.74. \nonumber$

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.


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# Armstrong number in Python

Written by

Himani Kohli

## Armstrong Number Program in Python

In this program where we are going to learn whether an N-digit integer is an Armstrong number or not.

A number is said to be an Armstrong number if it is equal to the sum of the cubes of its own digit.

Example: 153 is an Armstrong number, 153=1*1*1 + 5*5*5 + 3*3*3

Algorithm:

1. Input from the user.
2. Take the length of the number.
3. Assign a = n
4. Initialize arm = 0.
5. Using for loop, find whether a given number is Armstrong or not.
6. Using if condition, if arm !=n then print it is not an Armstrong.
7. Else print it is an Armstrong number.
8. Exit

Code:

n=input("Enter a number :")
l=len(n)
n=int(n)
a=n
arm=0
for i in range(l+1):
b=a%10
a=a/10
arm=arm+(b**l)
if (arm != n):
print("It is not an Armstrong no.")
else:
    print("It is an Armstrong no.")

Output:

Enter a number :153
It is an Armstrong no.


text:
# CAEN V488 TDC

The 12 bit ADC conversion time is 5.7 s. A sliding scale technique is used to reduce differential nonlinearity by injecting an offset. This reduces the dynamic range to 0 3840. ADC values between 3841 and 4095 indicate an error.

The TDC measures the time interval between the falling ECL inputs to the start/stop and the channel.

You can change the dynamic range of the TDC by setting bits using the library function c775setFSR.

 Address Pin Setting SW 1 2 SW 2 2 SW 3 0 SW 4 0

# initialize TDC

-> c775Init(0x610000)
Initialized TDC ID 0 at address 0x90610000


text:
## The Annals of Probability

### Mutually catalytic branching in the plane: Finite measure states

#### Abstract

We study a pair of populations in $\mathbb{R}^{2}$ which undergo diffusion and branching. The system is interactive in that the branching rate of each type is proportional to the local density of the other type. For a diffusion rate sufficiently large compared with the branching rate, the model is constructed as the unique pair of finite measure-valued processes which satisfy a martingale problem involving the collision local time of the solutions. The processes are shown to have densities at fixed times which live on disjoint sets and explode as they approach the interface of the two populations. In the long-term limit, global extinction of one type is shown. The process constructed is a rescaled limit of the corresponding $\mathbb{Z}^{2}$-lattice model studied by D. A. Dawson and E. A. Perkins and resolves the large scale mass-time-space behavior of that model.

#### Article information

Source
Ann. Probab., Volume 30, Number 4 (2002), 1681-1762.

Dates
First available in Project Euclid: 10 December 2002

Permanent link to this document
https://projecteuclid.org/euclid.aop/1039548370

Digital Object Identifier
doi:10.1214/aop/1039548370

Mathematical Reviews number (MathSciNet)
MR1944004

Zentralblatt MATH identifier
1017.60098

#### Citation

Dawson, Donald A.; Etheridge, Alison M.; Fleischmann, Klaus; Mytnik, Leonid; Perkins, Edwin A.; Xiong, Jie. Mutually catalytic branching in the plane: Finite measure states. Ann. Probab. 30 (2002), no. 4, 1681--1762. doi:10.1214/aop/1039548370. https://projecteuclid.org/euclid.aop/1039548370

#### References

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• and XIONG, J. (2002). Mutually cataly tic branching in the plane: Infinite measure states. Electron. J. Probab. 7.
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• (2002). Mutually cataly tic branching in the plane: Uniqueness. Ann. Inst. H. Poincaré Probab. Statist. To appear.
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• [25] FELLER, W. (1971). An Introduction to Probability Theory and Its Applications 2, 2nd ed. Wiley, New York.
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• KNOXVILLE, TENNESSEE 37996-1300 E-MAIL: jxiong@math.utk.edu


text:
# Does TRACON airspace override Class B airspace in terms of control authority when controlling approach or departure procedures?

This is a simple schematic description of Class Airspace.

My understanding is that Class B has strict rules, specific lower and upper altitude limits and requires clearance to enter. However, I have also come across the Standard Operating Procedures Manual for the N90 TRACON (ORDER 7110.1D), which designate the sectors of local airspace differently. Those also define lower and upper altitude limits. Clearly, the N90 TRACON overlaps with the Class B airspace around the JFK, LGA, TEB and EWR airports.

My confusion has to do primarily with this overlap. I was thinking that Class B airspace has to do strictly with terminal procedures (approaches/departures), that is, all take-off and landing procedures take place within this airspace.

However, I have seen SID and STAR procedures running over fixes located close to the airports, which define much higher altitudes than the Class B upper limits (around JFK and the above airports, the upper limit is 7000ft for Class B airspace, while lower limits vary).

Take, for example, the LENDY SIX ARRIVAL, of JFK airport. I read over the JENNO fix information: "Expect FL230".

What is this supposed to mean: "Expect FL230"? Will an approaching aircraft enter the procedure at a Flight Level of 23000ft or be at that flight level whenever approaching the airport at that fix point? How does Class B relate to these operations? I thought that approach operations take place within Class B but I am confused because the altitudes I see in SID/STAR procedures or the lower/upper altitude limits in the various airspace sectors of Standard Operating Procedure (SOP) plans* are quite different. It appears as if flights that approach for arrival, for example, are controlled before entering Class B (i.e. at higher altitudes) when inside a TRACON, e.g. N90.

*(I cannot link to an SOP Manual because they are intended for internal distribution, so I cannot find something official online, however, the information is pretty much like that document).

Class B is a legal restriction for pilots. ATC has to give them clearance to enter it. It's there to protect the traffic around a busy airport from aircraft that are not being controlled by ATC.

The TRACON airspace is just for controllers. Pilots aren't expected to know sector boundaries. They don't have to worry about that at all. Each controller will tell the pilot the frequency to contact the next controller whose airspace they are about to enter. Few TRACONs make their sector boundaries public and they change often.

A pilot flying under VFR does not necessarily even need to talk to controllers outside of A, B, C or D airspace, regardless of what ATC sector he might be in.

As long as you have the correct type of transponder, it is entirely legal to fly VFR through the layer of air above the class B below 18,000 feet without contacting atc. (This is class E airspace). Whatever controller whose sector you're flying through will keep any IFR traffic away from you, but they don't need to give you vectors or anything. You have to stay away from clouds. It's entirely the pilot's responsibility to see and avoid things.*

SIDs and STARs are only used by IFR traffic that is being controlled by ATC, but not necessarily by TRACON the whole time. Some STARs start hundreds of miles from the airport. When they begin the arrival they will be controlled by en-route controllers, who will tell them when they need to switch to TRACON and what frequency to use.

* I'm leaving out flight following for simplicity.

I may not understand your question correctly but I think you are a little confused about Airspace here. There is no TRACON airspace. TRACON is defined here:

A TRACON is an ATC facility usually associated with Class B airspace. Because the Class B airspace is so busy they usually have their own radar controllers instead of calling the tower directly or using ATC Center. The documents you referring too probably reference ATC procedures for incoming traffic rather than restrictions on pilots. For example which controller is responsible for which airspace.

Like you said Class B airspace is typically airspace over crowded airports and is more restricted than most types of airspace. When approaching this facility you will often be directed to "Approach" which is the actual TRACON. You may be handed off to different controllers as you transit the class B airspace.

• There is no TRACON airspace—well, there is, that's why OP is confused. There is regulatory airspace (Class A/B/C/D/E/G), and there is airspace owned by a certain facility (of which the majority by volume will be Class E, in all likelihood, but could also be any of the other airspace classes). From a pilot's point of view, nothing changes from a regulatory standpoint and very little changes from a services-provided standpoint when one changes from ARTCC-controlled airspace to TRACON-controlled, or vice versa. Mar 25, 2021 at 14:30


text:
In mathematics, interval is a concept relating to the sequence and set-membership of one or more numbers.

## Algebra

In elementary algebra, an interval is a set that contains every real number between two indicated numbers, and possibly the two numbers themselves. Interval notation is the notation in which permitted values for a variable are expressed as ranging over a certain interval; "5 < x < 9" is an example of the application of interval notation. In conventional interval notation, parentheses ( () ) indicate exclusion while square brackets ( [] ) indicate inclusion. For example, the interval "(10,20)" indicates the set of all real numbers between 10 and 20 but does not include 10 or 20, the first and last numbers of the interval, respectively. On the other hand, the interval "[10,20]" includes both every number between 10 and 20 as well as 10 and 20. Other possibilities are listed below.

## Higher mathematics

In higher mathematics, a formal definition is the following: An interval is a subset of a totally ordered set with the property that whenever and are in and then is in .

As mentioned above, a particularly important case is when , the set of real numbers.

Intervals of are of the following eleven different types (where a and b are real numbers, with a < b):

1. itself, the set of all real numbers
2. the empty set

In each case where they appear above, a and b are known as endpoints of the interval. Note that a square bracket [ or ] indicates that the endpoint is included in the interval, while a round bracket ( or ) indicates that it is not. For more information about the notation used above, see Naive set theory.

Intervals of type (1), (5), (7), (9) and (11) are called open intervals (because they are open sets) and intervals (2), (6), (8), (9), (10) and (11) closed intervals (because they are closed sets). Intervals (3) and (4) are sometimes called half-closed (or, not surprisingly, half-open) intervals. Notice that intervals (9) and (11) are both open and closed, which is not the same thing as being half-open and half-closed.

Intervals (1), (2), (3), (4), (10) and (11) are called bounded intervals and intervals (5), (6), (7), (8) and (9) unbounded intervals. Interval (10) is also known as a singleton.

The length of the bounded intervals (1), (2), (3), (4) is b-a in each case. The total length of a sequence of intervals is the sum of the lengths of the intervals. No allowance is made for the intersection of the intervals. For instance, the total length of the sequence {(1,2),(1.5,2.5)} is 1+1=2, despite the fact that the union of the sequence is an interval of length 1.5.

Intervals play an important role in the theory of integration, because they are the simplest sets whose "size" or "measure" or "length" is easy to define (see above). The concept of measure can then be extended to more complicated sets, leading to the Borel measure and eventually to the Lebesgue measure.

Intervals are precisely the connected subsets of . They are also precisely the convex subsets of . Since a continuous image of a connected set is connected, it follows that if is a continuous function and I is an interval, then its image is also an interval. This is one formulation of the intermediate value theorem.

## Intervals in partial orders

In order theory, one usually considers partially ordered sets. However, the above notations and definitions can immediately be applied to this general case as well. Of special interest in this general setting are intervals of the form [a,b].

For a partially ordered set (P, ≤) and two elements a and b of P, one defines the set

[a, b] = { x | axb }

One may choose to restrict this definition to pairs of elements with the property that ab. Alternatively, the intervals without this condition will just coincide with the empty set, which in the former case would not be considered as an interval.

## Interval arithmetic

Interval arithmetic, also called interval mathematics, interval analysis, and interval computation, was introduced by mathematicians in the 1950s and 1960s as an approach to putting bounds on rounding errors in mathematical computation. Where classical arithmetic defines operations on individual numbers, interval arithmetic defines a set of operations on intervals:

T · S = { x | there is some y in T, and some z in S, such that x = y · z }.

The basic operations of interval arithmetic are, for two intervals [a,b] and [c,d] that are subsets of the real line (-∞, ∞),

• [a,b] + [c,d] = [a+c, b+d]
• [a,b] - [c,d] = [a-d, b-c]
• [a,b] * [c,d] = [min (ac, ad, bc, bd), max (ac, ad, bc, bd)]
• [a,b] / [c,d] = [min (a/c, a/d, b/c, b/d), max (a/c, a/d, b/c, b/d)]

Division by an interval containing zero is not defined under the basic interval arithmetic.

The addition and multiplication operations are commutative, associative and sub-distributive: the set X ( Y + Z ) is a subset of XY + XZ.

## Relational operations

Relational operations on intervals can be defined in tri-state logic {true, false, uncertain}:

• T · S is true if for any x in T, and any y in S, x · y is true
• T · S is false if for any x in T, and any y in S, x · y is false
• otherwise T · S is uncertain

Often intervals are considered as estimations of some individual numbers. In that case for both arithmetic and relational interval operations the following is true: if x in T and y in S, then the result of T · S contains x · y.

## Alternative notation

An alternative way of writing intervals, commonly seen in France and some other European countries, is shown below:

• ]a,b[ = { x | a < x < b }
• [a,b] = { x | axb }
• [a,b[ = { x | ax < b }
• ]a,b] = { x | a < xb }

Another notation in use in some countries (e.g. Czech Republic) uses parentheses for an open interval and angle brackets for an closed interval, with numbers separated by semicolon.


text:
Chemistry / Sigma and Pi Bonds
• Solutions 19

# Solution: How many sigma and pi bonds are in the molecule NH 2CHCCH2? ...

###### Question

How many sigma and pi bonds are in the molecule NH 2CHCCH2? You need to draw the Lewis structure to see all the bonds.

1. 6 sigma and 2 pi

2. 3 sigma and 2 pi

3. 6 sigma and 4 pi

4. 8 sigma and 0 pi

5. 8 sigma and 2 pi

6. 8 sigma and 3 pi


text:
# Input.GetAxis returns wrong sign only for 1 and -1

I'm using an Xbox 360 controller on Windows 10 with Unity, and the maximum values have the opposite sign as the rest of that side. For example, if I tilt the stick up, I get values from 0.1 to 0.99, but then I get a -1 when the stick is at maximum. Have I configured it to be this way by mistake? I have reproduced the issue in a clean project and using another controller.

void Update () {
float pitch = Input.GetAxis("Vertical");
Debug.Log (pitch);
}

I have discovered that if the axis is inverted (i.e. not as shown in the image) then returned values have expected signs. I will add this workaround as an answer for anyone else who may encounter this problem, but I will refrain from accepting it to allow for answers addressing the root cause.

• That's really odd. Might be a Unity bug. – Draco18s no longer trusts SE Dec 31 '17 at 1:27
• Hm! I've never seen this before. Are you able to reproduce the issue with different controllers, or in a fresh project? – DMGregory Dec 31 '17 at 3:56
• @DMGregory Yes, I have reproduced it with a different controller (but I only have one USB wireless dongle) and in a fresh project. :/ – Tashus Dec 31 '17 at 23:10
• Have you checked the joystick calibration in the control panel? For example: maketecheasier.com/calibrate-game-controller-windows10 – Ed Marty Jan 1 '18 at 3:02
• @EdMarty Thanks for the suggestion, but calibration didn't change it. The issue isn't present in control panel. :/ – Tashus Jan 1 '18 at 22:23

I had previously discovered a workaround that gave me reasonable behavior but did not reveal the underlying problem. I have revisited the project and have determined the root cause.

Two axes were named "Horizontal".

It seems that the InputManager by default has two axes named "Horizontal" and two axes named "Vertical." Deleting one of the sets completely solved the issue.

The previous solution:

Setting the axis to inverted causes the returned values to behave in a sensible way. Now if I tilt the stick up, I get values from -0.1 to -0.99 and -1 when the stick is at maximum, and I get values from 0.1 to 0.99 and 1 when I tilt the stick down. I can just as easily write my script expecting an inverted axis, so this allows me to continue. Unfortunately I still don't know what is causing the odd behavior when the axis isn't inverted.

• when inverted, what happens as the stick moves to maximum down? – Stephan Oct 30 '18 at 15:45
• @Stephan Edited. Unfortunately I have not worked on this project for a while, so I am not sure whether I can still reproduce this behavior. – Tashus Oct 30 '18 at 15:51
• Should be easy enough to test. I'll spin one up later this afternoon, and get back to you – Stephan Oct 30 '18 at 15:54


text:
# Problem with citation multiple authors

I need to cite a paper with multiple authors:

@article{DBLP:journals/tit/HammonsKCSS94,
author    = {A. Roger Jr. Hammons  and
P. Vijay Kumar and
A. Robert Calderbank and
Neil J. A. Sloane and
Patrick Sol{\'e}},
title     = {The Z$_{\mbox{4}}$-linearity of Kerdock, Preparata, Goethals,
and related codes},
journal   = {IEEE Transactions on Information Theory},
volume    = {40},
number    = {2},
year      = {1994},
pages     = {301-319},
ee        = {http://dx.doi.org/10.1109/18.312154},
bibsource = {DBLP, http://dblp.uni-trier.de}
}

But when I made the citation it appears as [HKC+94] and I want to be [HKCSS94]. can anyone help me?

-
Which bibliography style, and which document class, do you use? Separately, do use a citation management package such as natbib? – Mico Mar 2 '13 at 0:06
I am new using latex, i use report as document class,alpha as bibliography style and i dont use natbib – Dimitri Mar 2 '13 at 0:08
Thanks. The appearance of the entries in the References section is determined by the bibliography style -- the argument of the command \bibliographystyle. To have a chance to help you, we thus need to know which bibliography style you employ. – Mico Mar 2 '13 at 0:11
Without switching to biblatex it isn't possible. – Marco Daniel Mar 2 '13 at 9:09
@Dimitri: I provided an answer. – Marco Daniel Mar 3 '13 at 9:35

## 1 Answer

As written in the comments, without using biblatex it isn't possible. By using the package biblatex you have the options minalphanames/maxalphanames. The option maxalphanames requires a number and it represents the number of authors which will be counted for the label of the style alphabetic. In your case it should be 5.

Here a MWE with the correct compilation order using arara

% arara: pdflatex
% arara: biber
% arara: pdflatex
% arara: pdflatex
\listfiles
\RequirePackage{filecontents}
\begin{filecontents*}{\jobname.bib}
@article{DBLP:journals/tit/HammonsKCSS94,
author    = {A. Roger Jr. Hammons  and
P. Vijay Kumar and
A. Robert Calderbank and
Neil J. A. Sloane and
Patrick Sol{\'e}},
title     = {The Z$_{\mbox{4}}$-linearity of Kerdock, Preparata, Goethals,
and related codes},
journal   = {IEEE Transactions on Information Theory},
volume    = {40},
number    = {2},
year      = {1994},
pages     = {301-319},
ee        = {http://dx.doi.org/10.1109/18.312154},
bibsource = {DBLP, http://dblp.uni-trier.de}
}
}
\end{filecontents*}

\documentclass[12pt]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage[style=alphabetic,maxnames=99,maxalphanames=5]{biblatex}
\addbibresource{\jobname.bib}

\begin{document}
\verb+\cite{DBLP:journals/tit/HammonsKCSS94}+\cite{DBLP:journals/tit/HammonsKCSS94}
\printbibliography
\end{document}

-
Thanks for the answer, i already try with biblatex, but when i do, the references don´t appear...i don´t know why, can you give any idea? – Dimitri Mar 3 '13 at 18:11
Did you compile with pdflatex+biber+pdflatex? Do you get any errors? – Marco Daniel Mar 3 '13 at 18:40
I fix it, i think that was a problem with the compilation, but now is ok. Thanks again for the answer, that really help me, and besides you were the only one who really try to help me with my question – Dimitri Mar 3 '13 at 18:52


text:
# Finding the positive value K

Say we have some positive value K. How can we know what K is if the function $f(x)=K tan(x)$ intersects its derivative perpendicularly on a graph?

So I know how to do this using some form of a software but don't know how to do it theoretically, or by paper and pencil, to make sure I am getting the same value. How I tried to approach the problem: $f(x)=K tan(x)$. $f(x)'=K sec^2(x)$. Then we have $-K/sec^2(x)$ since we are trying to find the derivative perpendicularly. Then to find K, I set the the two equal and that is where I am stuck. Any help with shown steps would be appreciated.

You are fine with $f'(x)=K\sec^2 x$. When you say "then we have $-K/\sec^2 x$" it is not clear what you mean. To have a perpendicular intersection of two curves, you need their derivatives to be negative reciprocals, so now you need to find $f''(x)$. Having done that, you require that $f'(x)f''(x)=-1$ for some $x$ and $K$.
• It would be where $f(x)$ and $f'(x)$ make a 90 degree angle. But since the slope of $f(x)$ is $f'(x)$ and the slope of $f'(x)$ is $f''(x)$ we get the condition I gave at the end. – Ross Millikan Oct 9 '13 at 22:19


text:
# Show that questions involving orthonormal basis and inner product space

The question is:

Show that if $$V$$ is an inner product space and {$$v_1, . . . , v_n$$} is an orthonormal basis, then $$v =\sum\limits_{i=1}^{n} \langle v, v_i\rangle v_i$$, for all $$v ∈ V$$ .

To be perfectly honest, I'm unsure of where to begin with this question. Any help would be appreciated, thank you.

• Begin with: $v=\sum_i a_i v_i$ for some $a_i$ since $v_i$ form a basis. Now how would you compute $a_i$? – Michal Adamaszek Nov 25 '19 at 11:47
• Regarding the original version of the question: the Gram-Schmidt process is for constructing an orthonormal basis. Since we already are given an orthonormal basis, I can't see why constructing one would be useful. – Ben Grossmann Nov 25 '19 at 11:49

Every vector $$v \in V$$ can be written uniquely as: $$v = \sum_{j=1}^{n}\alpha_{j}v_{j},$$ once $$\{v_{1},...,v_{n}\}$$ is an orthonormal basis. Thus, using the linearity of the inner product, we get, for each $$k=1,...,n$$: $$\langle v, v_{k}\rangle = \langle \sum_{j=1}^{n}\alpha_{j}v_{j}, v_{k}\rangle = \sum_{j=1}^{n}\alpha_{j}\langle v_{j},v_{k}\rangle = \alpha_{k},$$ because $$\langle v_{j},v_{k}\rangle = \delta_{jk}$$. Thus, for each $$k=1,...,n$$, $$\alpha_{k}=\langle v,v_{k}\rangle$$ and: $$v = \sum_{j=1}^{n}\alpha_{j}v_{j} = \sum_{j=1}^{n}\langle v, v_{j}\rangle v_{j}$$
One approach is as follows: it suffices to note that if $$\langle v,v_i\rangle = \langle w,v_i \rangle$$ for all $$i = 1,\dots,n$$, then $$v$$ and $$w$$ must be the same vector. Apply this fact to $$w = \sum_{i=1}^n \langle v,v_i \rangle v_i$$.