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суббота, 3 июля 2010 г.
суббота, 12 июня 2010 г.
Glossary
An algorithm is a detailed set of instructions for solving a problem. One example of an algorithm is the method of long division. Algorithms are named after the mathematician al-Khwarazmi.
The numerals 1,2,3,4,5,6,7,8,9, and 0 that we use today are often called Arabic numerals because al-Khwarazmi wrote a popular book about them. This number system probably comes from India.
Calculus is a branch of mathematics that involves rates of change, the slopes of curves, length, area, and volume. One important calculus technique is the careful study of very small changes. Sometimes we think of these changes as "infinitely small." Calculus is also concerned with limits and "infinitely large" numbers.
The coordinate plane is sometimes called the Cartesian plane in honor of René Descartes.
The distance around the outside of a shape is called its circumference. The word "circumference" is usually used for circles; for polygons like squares and triangles, we use the word "perimeter." Sometimes "circumference" is used for three-dimensional objects, like spheres. In this case, the circumference is the distance around the object at its widest point. You might think of this as the distance around the equator of a sphere.
Conic sections are the curves formed when a plane slices through a double cone.
Conic sections include hyperbolas, parabolas, and ellipses. They also include circles, because a circle is a special kind of ellipse. Lines and points can also be made by the intersection of a plane and a double cone, but they usually do not count as conic sections.
The coordinate plane is a plane that contains two lines called axes.
Coordinates based on distance from the axes can be used to locate points in the plane.
The diameter of a shape is the greatest possible distance between two points on that shape.
A directrix is a special line. A conic section such as a parabola can be defined by its distance from the directrix and a point called the focus.
Ellipses are one kind of conic section. They look like ovals, but they have a more precise definition. An ellipse can be described by the equation x2/a2 + y2/b2 = 1.
An ellipse can also be described using foci. An ellipse has two focus points, and the sum of the distances from a point on the ellipse to each focus is always constant. This fact can be written as the equation r1 + r2 = 2a.
A circle is a special kind of ellipse with only one focus.
The equator is the circle around the Earth exactly between the North Pole and the South Pole.
Fermat's Last Theorem states that the equation xn + yn = zn has no solutions when all of the variables are integers (numbers in the set . . . -2, -1, 0, 1, 2, 3, . . .), n is greater than 2, and x, y, and z are not all 0. Pierre de Fermat first stated this theorem, in the margin of one of his books, along with a note that the margin was not big enough to hold the proof. This theorem was not published until after his death. For hundreds of years, nobody could find a correct proof of the theorem. Andrew Wiles finally published a proof in 1995.
A focus is a special point. For example, the center of a circle can be considered its focus. A conic section such as a parabola can be defined by its distance from the focus and a line called the directrix.
The plural of "focus" is "focuses" or "foci."
Hyperbolas are one kind of conic section. A hyperbola can be described by the equation x2/a2 - y2/b2 = 1.
A hyperbola can also be described using foci. A hyperbola has two focus points. The difference of the distances from a point on the ellipse to each focus is always constant. This fact can be written as the equation r2 - r1 = 2a.
The latitude of a point on the Earth is its distance around the Earth from the equator. Since the Earth is a sphere, latitude is measured in degrees. Circles of points on the earth that all have the same latitude are called lines of latitude.
Nicole Oresme used "latitude" as a more general term for a type of coordinate.
The longitude of a point on the Earth is its distance around the Earth from the Prime Meridian. Since the Earth is a sphere, longitude is measured in degrees. Circles of points on the earth that all have the same longitude are called lines of longitude.
Nicole Oresme used "longitude" as a more general term for a type of coordinate.
A matrix is made up of numbers arranged in a rectangular shape of rows and columns.
Parabolas are one kind of conic section. A parabola that opens upward can be described by the equation y = ax2 + bx + c.
A parabola can also be described using a focus and a directrix. The distance from any point on the parabola to the focus is the same as the distance to the directrix.
Plane loci are curves that are defined by their distances from other objects in the plane. The conic sections are plane loci.
In polar coordinates, points are located by their distance from the origin, often labeled r, and their angle from the positive x-axis, often labeled with the Greek letter θ or theta.
The Prime Meridian is a line of longitude that travels from the North Pole to the South Pole through Greenwich, England.
The slope of a line is a number that tells how much it is slanted compared to the x-axis. A line with the equation y = mx + b has slope m. If we draw a tangent line to a curve at a particular point, then we can also define the slope of the curve at that point, by saying that the line and the curve have the same slope at the point where they touch.
A tangent line to a curve only touches the curve at one point.
There is also a function called the tangent function that is important in trigonometry.
Trigonometry is the study of angles.
The Witch of Agnesi is a curve. It can be described by the equation y = a3/(x2 + a2).
This curve is named after Maria Gaetana Agnesi, who included it in a calculus textbook she published in 1748. Agnesi called her curve the versiera, which means "turning" in Italian. When her book was translated into English, the translator confused the word "versiera" with the word "avversiera," which means witch, so he called the curve a witch.
пятница, 11 июня 2010 г.
среда, 9 июня 2010 г.
Basic Mathematical Symbols
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can be read ``for all a and b in the set of real numbers, a less than b implies a cubed is less than b cubed'' (or ``for all real numbers a and b, if a is less than b, then a cubed is less than b cubed'').
Written: | Spoken: | |
---|---|---|
![]() | = | for all (or ``for every'') |
![]() | = | there exists |
![]() | = | such that |
![]() | = | is orthogonal to (perpendicular to) |
![]() | = | intersection |
![]() | = | union |
![]() | = | is contained in |
![]() | = | is an element of |
![]() | = | is not an element of |
![]() | = | implies |
![]() | = | if and only if (or ``is equivalent to'') |
![]() | = | if and only if (same as above) |
![]() | = | less than |
![]() | = | greater than |
![]() | = | less than or equal to |
![]() | = | greater than or equal to |
![]() | = | the set of real numbers |
![]() | = | the set of natural numbers |
![]() | = | the set of integers |
![]() | = | the set of rational numbers |
![]() | = | the set of complex numbers |
Diophantine Equation
Version 1.1
Introduction
The following diophantine equation was a topic problem from #math. The origin of the problem is unknown to me(landen). To explore ideas and cut down on mistakes I used the software package for number theory, PARI/GP, which is free for noncommercial use. This package is convenient for calculations in algebraic extensions of Q and Z.
2a + 3b = 5c (1)The problem can be split into four cases: a is odd with a ≥ 3, a=2, a=1, and a is even with a ≥ 4. Two of the cases can be solved by just trying all possibilities (mod 3) and (mod 8). The remaining cases are a=1, and a is even with a ≥ 4, which take more work.
Proof: Case 1, a is odd with a ≥ 3
2a + 3b = 52n+1 and then (mod 8)
2a = 0 (mod 8); 3b = 1 or 3 (mod 8); 52n+1 = 5 (mod 8)
So (1) fails (mod 8) for this case. This case gives no solutions.
Q.E.D.
Proof: Case 2, a = 2
Q.E.D.
Proof: Case 3
2 + 3b = 5c has only one solution, b = 1 and c = 1 (2)Assume b = 2m+1 and c = 2n+1 with n > 0. They must be odd by the argument in Case 1. Multiplying the equation by 3 and rearranging terms we get:
32(m+1) - 15 52n = -6 (3)In this form we see that any solutions to (2) must also be among solutions to:
x2 - 15 y2 = -6 (4)Of course (4) will have many more solutions than (3) but solutions to (4) are easy to study in Z[w], where w is sqrt(15). The only solutions to (4) which have a chance of solving (3) are the ones in which y is a power of 5. One way to show that (3) has no solutions is to show that when y is divisible by 3 it is also divisible by 11. This property of the solutions was first conjectured after using PARI/GP to experiment. It was then easy to prove that conjecture by considering the solutions to (4) (mod 5) and (mod 11). From the theory of binary quadratic forms we know that all the solutions to (4) can be written as:
s(k) = s0uk = x(k) + y(k)w (5)By factoring y(k) when it is a multiple of 5, it was conjected that y(k) is also divisible by 11 and thus is not a power of 5. When s(k) is computed (mod 5) and (mod 11) the results are periodic with a period of ten. Notice that
k is in Z
w = sqrt(11)
s0 = 3-w a solution of (4) and u = 4+w, the fundamental unit of Z[w].
norm(x(k) + y(k)w) = x(k)2 - 15 y(k)2 = -6
PARI/GP can compute solutions using (5) directly. Here are solutions for a few values of k:
k = -12; s(k) = (194888982963 - 50320119025*w); norm(s(k)) = -6
k = -11; s(k) = (24754146477 - 6391493137*w); norm(s(k)) = -6
k = -10; s(k) = (3144188853 - 811826071*w); norm(s(k)) = -6
k = -9; s(k) = (399364347 - 103115431*w); norm(s(k)) = -6
k = -8; s(k) = (50725923 - 13097377*w); norm(s(k)) = -6
k = -7; s(k) = (6443037 - 1663585*w); norm(s(k)) = -6
k = -6; s(k) = (818373 - 211303*w); norm(s(k)) = -6
k = -5; s(k) = (103947 - 26839*w); norm(s(k)) = -6
k = -4; s(k) = (13203 - 3409*w); norm(s(k)) = -6
k = -3; s(k) = (1677 - 433*w); norm(s(k)) = -6
k = -2; s(k) = (213 - 55*w); norm(s(k)) = -6
k = -1; s(k) = (27 - 7*w); norm(s(k)) = -6
k = 0; s(k) = (3 - w); norm(s(k)) = -6
k = 1; s(k) = (-3 - w); norm(s(k)) = -6
k = 2; s(k) = (-27 - 7*w); norm(s(k)) = -6
k = 3; s(k) = (-213 - 55*w); norm(s(k)) = -6
k = 4; s(k) = (-1677 - 433*w); norm(s(k)) = -6
k = 5; s(k) = (-13203 - 3409*w); norm(s(k)) = -6
k = 6; s(k) = (-103947 - 26839*w); norm(s(k)) = -6
k = 7; s(k) = (-818373 - 211303*w); norm(s(k)) = -6
k = 8; s(k) = (-6443037 - 1663585*w); norm(s(k)) = -6
k = 9; s(k) = (-50725923 - 13097377*w); norm(s(k)) = -6
k = 10; s(k) = (-399364347 - 103115431*w); norm(s(k)) = -6
k = 11; s(k) = (-3144188853 - 811826071*w); norm(s(k)) = -6
k = 12; s(k) = (-24754146477 - 6391493137*w); norm(s(k)) = -6
k = 13; s(k) = (-194888982963 - 50320119025*w); norm(s(k)) = -6
u5 = 15124 + 3905*wy(3) is 55 which is divisible by 5 and 11. This means that divisiblity by 5 and 11 occur together at y(3 + k*5). Thus y is never a power of 5.
u5 = 1 (mod 5) Proves solutions repeat with a period of 5 (mod 5).
u5 = -1 (mod 11) Proves solutions repeat with a period of 10 (mod 11).
But every 5th term (mod 11), the coefficient of w (mod 11) in u5 = 0, so is divisible by 11 also.
Q.E.D.
As a check for errors, PARI/GP calculated the solutions (mod 5) and (mod 11). Here are some results:
** means coefficient of w, y(k) is divisible by 5 and 11 both
k = -7; s(k) = 2 + 0*w (mod 5); s(k) = 7 + 0*w (mod 11) **
k = -6; s(k) = 3 + 2*w (mod 5); s(k) = 6 + 7*w (mod 11)
k = -5; s(k) = 2 + w (mod 5); s(k) = 8 + w (mod 11)
k = -4; s(k) = 3 + w (mod 5); s(k) = 3 + w (mod 11)
k = -3; s(k) = 2 + 2*w (mod 5); s(k) = 5 + 7*w (mod 11)
k = -2; s(k) = 3 + 0*w (mod 5); s(k) = 4 + 0*w (mod 11) **
k = -1; s(k) = 2 + 3*w (mod 5); s(k) = 5 + 4*w (mod 11)
k = 0; s(k) = 3 + 4*w (mod 5); s(k) = 3 + 10*w (mod 11)
k = 1; s(k) = 2 + 4*w (mod 5); s(k) = 8 + 10*w (mod 11)
k = 2; s(k) = 3 + 3*w (mod 5); s(k) = 6 + 4*w (mod 11)
k = 3; s(k) = 2 + 0*w (mod 5); s(k) = 7 + 0*w (mod 11) **
k = 4; s(k) = 3 + 2*w (mod 5); s(k) = 6 + 7*w (mod 11)
k = 5; s(k) = 2 + w (mod 5); s(k) = 8 + w (mod 11)
k = 6; s(k) = 3 + w (mod 5); s(k) = 3 + w (mod 11)
k = 7; s(k) = 2 + 2*w (mod 5); s(k) = 5 + 7*w (mod 11)
k = 8; s(k) = 3 + 0*w (mod 5); s(k) = 4 + 0*w (mod 11) **
Proof: Case 4
a is even with a ≥ 4 has only the solution a=4,b=2,c=2Assume a is even with a ≥ 4. Then b = 2m and c = 2n by considering equation (1) (mod 8). Rearranging (1) we have:
2a = 52n - 32m = (5n - 3m)(5n + 3m)By looking at the sum and difference of the two factors on the right, we see that they have 2 as their greatest common divisor. This means that
(5n - 3m) = 2, and therefore n=1 and m=1 by Case 3, so (a,b,c) = 4,2,2)Q.E.D.
http://efnet-math.org/math_tech/dioph2.htm
Positive Solutions to linear diophantine equation
Let a,b,m,n be integers > 0, let integer k >= 0. define g = gcd(a,b)
Theorem: am + bn = ab/g + g + kg has a solution m,n, given a,b,k. (eqn 1)
Proof: n must have a value such that ab/g + g + kg -bn is divisible by a. This can be accomplished by selecting n to be in the set {1,2,... a/g}. One of these numbers is sufficient because
ab/g + g + kg has only a/g possible residues (mod a/g).
We need to show that m>=1 is possible. The maximum value of bn is ba/g. So
am >= ab/g + g + kg -ab/g = g + kg (eqn 2)
m >= g/a + kg/a (eqn 3)
g/a > 0 is so m >= 1. Any k>0 can only help this.
Notice that if the +g term is ommited from (eqn 1), then (eqn 3) becomes m >= kg/a (eqn 4), thus m=0 cannot be ruled out. So (eqn 1) is the best we can do. Q.E.D
Special case, There are non-negative solutions starting at n=(a-1)(b-1)
x>=0,y>=0,g=1
applying the theorem with m=x+1, n=y+1;
a(x+1)+b(y+1)=ab+1+k
ax+by = (a-1)(b-1) + k
This is a common form of the problem.
LATIN TERMS USED IN MATHEMATICS
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a fortiori
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a posteriori
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a priori
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e.g.
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exempli gratia
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id est
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i.e.
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ipso facto
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n.b.
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nota bene
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per impossibile
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QED
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QEF
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quod erat demonstrandum
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quod erat faciendum
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OTHER COMMON LATIN TERMS
ab initio
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accessit
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ad hoc
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ad hominem
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ad nauseam
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alma mater
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alumnus/alumna
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a.d.
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anno domini
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bona fide
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carpe diem
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cf.
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circa
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confer
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cum laude
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de facto
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de jure
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dixi
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emeritus
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ergo
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erratum/errata
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et al.
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et cetera
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ex post facto
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ibid.
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ibidem
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in re
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inter alia
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in toto
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in vacuo
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magna cum laude
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modus operandi
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mutatis mutandis
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non sequitur
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per se
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post hoc, ergo propter hoc
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post scriptum
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prima facie
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pro forma
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qua
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quod vide
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q.v.
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sine qua non
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summa cum laude
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tabula rasa
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verbatim
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videlicet
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http://www.mathacademy.com/pr/prime/articles/latin/index.asp