Jun 611:50 AM Using technology and a matrix approach we can verify our solution. Planes have a pretty special property. 2. 1 0 Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. 4. Colloquially, curves that do not touch each other or intersect and keep a fixed minimum distance are said to be parallel. Learn more about this Silicon Valley suburb, America's richest neighborhood. CS 506 Half Plane Intersection, Duality and Arrangements Spring 2020 Note: These lecture notes are based on the textbook “Computational Geometry” by Berg et al.and lecture notes from [3], [1], [2] 1 Halfplane Intersection Problem We can represent lines in a plane by the equation y = ax+b where a is the slop and b the y-intercept. The sum of two measures of two acute angles is greater than 90 degrees. Three points must be coplanar. For lines, rays, and line segments, intersect means to meet or cross. yes, three planes can intersect in one point. (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3). Minimum number of 3-inch by 5-inch index cards needed to completely cover a 3-foot by 4-foot rectangular desktop, Number of diagonals in an n-sided polygon, Some Cool Math and Computer Science links, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Sometimes. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. It is very easy to find a system of three equations in three unknowns for each one of the eight different relative positions, except for one: case 6 (where the three Planes intersect in a line). If you're talking about lines in the same plane, lines don't intersect if they're parallel. General solution to system of differential equation question...? Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Condition for three lines intersection is: rank Rc= 2 and Rd= 3 All values of the cross product of the normal vectors to the planes are not 0 and are pointing to the same direction. Let $\ell'$ be a line different from but parallel to $\ell$. The intersection of the three planes is a point. The planes will then form a triangular "tube" and pairwise will intersect at three lines. Thus, the intersection of 3 planes is either nothing, a point, a line, or a plane: To answer the original question, 3 planes can intersect in a point, but cannot intersect in a ray. 3. How many multiples of 8 are between 100 and 175? Finally we substituted these values into one of the plane equations to find the . There are infinitely many planes through $\ell'$, but only one of them intersects $\ell$, and only two of them are parallel to one of the first two planes. Each plan intersects at a point. 3 Plane Intersection. Jun 611:50 AM Using technology and a matrix approach we can verify our solution. If they are in the same plane, and not parallel, then they will intersect at exactly one point. true. Given 3 unique planes, they intersect at exactly one point! Sign in|Recent Site Activity|Report Abuse|Print Page|Powered By Google Sites, Maximum number of points of intersection of 6 lines. A projective plane can be thought of as an ordinary plane equipped with additional "points at infinity" where parallel lines intersect. Huh? he might have made a mistake when writing out the test or got it out of a faulty text or you might have misread. Intersection of Three Planes. true. Three planes can mutually intersect but not have all three intersect. In the figure below, rays BA and BC meet at endpoint B, so their intersection forms angle ∠ABC. true. Just two planes are parallel, and the 3rd plane cuts each in a line. Find the equation of the plane that contains the point (1;3;0) and the line given by x = 3 + 2t, y = 4t, z = 7 t. Lots of options to start. In geometry, parallel lines are lines in a plane which do not meet; that is, two straight lines in a plane that do not intersect at any point are said to be parallel. What is A? Now, if we draw a 3rd line, that can intersect the other two lines in at most 2 points as shown below. How many five-digit numbers greater than 70,000 are mountain numbers? equation of a quartic function that touches the x-axis at 2/3 and -3, passes through the point (-4,49)? You can also rotate it around to see it from different directions, and zoom in or out. Integer coordinates enclosed by 2 squares. The intersection of the three planes is a line. In three dimensions, no. The text is taking an intersection of three planes to be a point that is common to all of them. How many times during the day do the hands of a clock overlap? The point can be located by starting from the origin then following the red arrows on the grid. Not for a geometric purpose, without breaking the line in the sketch. This is question is just blatantly misleading as two planes can't intersect in a point. three planes can intersect as a point or as a line. Penny Note that there is no point that lies on all three planes. (g) If a point lies outside a line, then exactly one plane contains both the line and the point … Increase Brain Power, Focus Music, Reduce Anxiety, Binaural and Isochronic Beats - Duration: 3:16:57. Thus, any pair of planes must intersect in a line, but not all three at once (since there is no solution). Each plane cuts the other two in a line and they form a prismatic surface. Or three planes can, like the pages in the spine of a book, can intersect in one single line. three noncollinear points determine a plane. Count the points of intersection for each and allow infinite as some of your counts. If the lines are in different planes in 3-dimensional space, they won't intersect. We can use a matrix approach or an elimination approach to isolate each variable. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. The new app allows you to explore the concepts of solving 3 equations by allowing you to see one plane at a time, two at a time, or all three, and the intersection point. When two lines, rays, or line segments intersect, they have one common point. It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. I may just have found a slightly shorter way (four commands vs five in the previous) - PC-DMIS can intersect a CURVE and a PLANE to get one of the intersection points (assuming your number of hits is large enough). Yes, look at … Think about what a plane is: an infinite sheet through three... See full answer below. The system has one solution. In one of the positions, the three Planes share only one point (they intersect at that point). You can edit the visual size of a plane, but it is still only cosmetic. Justify your answer. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Any two of them must intersect, if no two are parallel, but there need not be a point that all three of them have in common. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. This lines are parallel but don't all a same plane. (b) Give an example of three planes in R^3 that intersect in pairs but have no common point of intersection. Probability that a number created randomly from 3 digits is even. The second and third planes are coincident and the first is cuting them, therefore the three planes intersect in a line. There is nothing to make these three lines intersect in a point. The East wall, South wall and floor are three planes that intersect at one point, the point on the floor in the South-East corner of the room. Leave a comment * Sometimes. (e) A line contains at least two points (Postulate 1). (a) Give an example of three planes in R^3 that have a common line of intersection. The East wall and the South wall are pieces of two planes that do intersect. In America's richest town, $500k a year is below average. In 3D, three planes , and can intersect (or not) in the following ways: All three planes are parallel. Four points may be coplanar or noncoplanar. The intersection is the line that forms the South-East corner of the room. Thus any two distinct lines in a projective plane intersect in one and only one point. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Given 3 unique planes, they intersect at exactly one point! two planes intersect in a line. false. Get an answer to your question "Can 2 planes in a 3 dimensional space intersect at one point ..." in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Geometrically, we have planes whose orientation is similar to the diagram shown. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. (1) (2) (3) point of intersection 3 4 z. value. the lines intersect at a point. Precalculus 3-D Cartesian Coordinate System Lines in Space. Homework Statement The three lines intersect in the point (1; 1; 1): (1 - t; 1 + 2*t; 1 + t), (u; 2*u - 1; 3*u - 2), and (v - 1; 2*v - 3; 3 - v). Justify your answer. Join Yahoo Answers and get 100 points today. 2 Answers bp Dec 14, 2016 We can use a matrix approach or an elimination approach to isolate each variable. true. 3. We know a point on the line is (1;3… (g) If a point lies outside a line, then exactly one plane contains both the line and the point … Similarly, if we draw a 4th line, that can intersect the other 3 lines in at most 3 points and so on. Planes have a pretty special property. Two points must be collinear. Sometimes. Renaissance artists, in developing the techniques of drawing in perspective, laid the groundwork for this mathematical topic. a line and a plane can intersect in a point. If I have 5 independent groups and us a likert scale what stat analysis would I use. Two planes can intersect on exactly one point? Get your answers by asking now. Three lines intersect at one point. The number 23AB3 is exactly divisible by 99. Sometimes. Examples Example 3 Determine the intersection of the three planes: 4x y — z — 9m + 5y — z — was the question about three planes? (e) A line contains at least two points (Postulate 1). 1 0 Justify your answer. (c) Give an example of three planes in R^3 that intersect in a single point. If two planes intersect, then their intersection is a line (Postulate 6). For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). A line and a plane intersect at exactly one point. If we found in nitely many solutions, the lines are the same. For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). The intersection of the first two is a line$\ell$. Three planes may all intersect each other at exactly one point. How do you solve a proportion if one of the fractions has a variable in both the numerator and denominator? Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Using that, we only need to create one line to find the other point. Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! Calculator won't calculate sin divided by anything, shows error? Other geometric figures. (1) (2) (3) point of intersection 3 4 In four dimensions or higher, yes. In how many days will Jessica read 270 pages of a book? Music for body and spirit - Meditation music Recommended for you 3:16:57 Doesn't matter, planes have no geometric size. When two distinct planes meet, they intersect at a line. If we found no solution, then the lines don’t intersect. Is there a way to create a plane along a line that stops at exactly the intersection point of another line. The first and second are coincident and the third is parallel to them. Assuming you are working in R 3, if the planes are not parallel, each pair will intersect in a line. two planes intersect at exactly one point. Two planes contain the same point. Here are the ways three planes can associate with each other. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Some geometric figures intersect at more than one point. These planes do not intersect. The measure of an angle is greater than the measure of its complement. This means that, instead of using the actual lines of intersection of the planes, we used the two projected lines of intersection on the x, y plane to find the x and y coordinates of the intersection of the three planes. Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Any two straight lines can intersect each other in one point. true. Only lines intersect at a point. How many 5 digit numbers are perfect squares? Precalculus . If two planes intersect, then their intersection is a line (Postulate 6). was the question about three planes? Science Anatomy & Physiology ... Can two planes intersect in exactly one point? no they cant... only a line can work for this.. all the time... Can two two-dimensional planes intersect in a point? It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. One Comment on “Intersection of 3 planes at a point: 3D interactive graph” Leesa Johnson says: 24 Jan 2017 at 9:13 pm [Comment permalink] Nice explanation for me to understand the interaction of 3d planes at a point using graphical representation and also useful for the math students. Florida governor accused of 'trying to intimidate scientists', Ivanka Trump, Jared Kushner buy$30M Florida property, Another mystery monolith has been discovered, MLB umpire among 14 arrested in sex sting operation, 'B.A.P.S' actress Natalie Desselle Reid dead at 53, Goya Foods CEO: We named AOC 'employee of the month', Young boy gets comfy in Oval Office during ceremony, Packed club hit with COVID-19 violations for concert, Heated jacket is ‘great for us who don’t like the cold’, COVID-19 left MSNBC anchor 'sick and scared', Former Israeli space chief says extraterrestrials exist. Question 97302: can 3 planes intersect in exactly one point? 3 Plane Intersection. Planes intersect along a line. three planes can intersect as a point or as a line. This commonly occurs when there is one straight plane and two other planes intersect it at acute or obtuse angles. Still have questions? (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3).


# If sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)

1. Sep 23, 2007

### cks

if sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)

I know how to prove but i have difficult in choosing the signs.

sinh(y) = x = [exp(y)-exp(-y)]/2

there are two equations I can find

exp(2y) - 2x exp(y) -1 =0 OR exp(-2y) + 2xexp(-y) -1 =0

exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)

if I select the signs of +

exp(y) = x + sqrt(x^2 +1) & exp(-y) = -x + sqrt(x^2 + 1)

then by substituting to cosh(y) = [exp(y) + exp(-y)] / 2

then, I can find the answer of
cosh(y)=sqrt(1+x^2)

But, what should I say to justify my actions of selecting the +ve sign.

2. Sep 23, 2007

### neutrino

Why not just substitute $$x = \frac{\left(e^y - e^{-y}\right)}{2}$$ into the expression $$\sqrt{1+x^2}$$ and simplify?

3. Sep 23, 2007

### cks

Thanks, but I'd like to find cosh(y) without prior knowledge that it's equal to sqrt(1+x^2)

4. Sep 23, 2007

### robphy

For real y, exp(y)>0.

5. Sep 23, 2007

### arildno

Utilize the identity:
$$Cosh^{2}(y)-Sinh^{2}(y)=1$$
Along with the requirement that Cosh(y) is a strictly positive function.
(Alternatively, that it is an even function with Cosh(0)=1)

Last edited: Sep 23, 2007
6. Sep 23, 2007

### cks

still don't quite understand

7. Sep 24, 2007

### Gib Z

arildno pretty much gave you the identity that you need. If you need to prove it, Just replace cosh y and sinh y with their exponential definitions and simplify. Then isolate cosh y on the left hand side and it is clear. Basically when we take the final square root, one could normally argue you could have either the positive or the negative root. However cosh is always positive.

8. Sep 24, 2007

### robphy

There are different ways to approach the title of this thread, depending on what you are given to start with and what constraints are imposed... which should include your working definitions of sinh and cosh.

Rather than merely proving an identity, it appears that the OP wishes to obtain the title by using the definitions of sinh and cosh starting from their definitions using the sum and difference of exponential functions. In addition, one might taking the point of view that one doesn't know at this stage any properties of cosh except for its working definition. So, it might be "cheating" to use any other properties of cosh which you didn't derive already.

cks, is this correct?

(If I'm not mistaken, my suggestion "For real y, exp(y)>0" provides a reason to choose the positive sign.)

9. Sep 25, 2007

### cks

Oh, I see thank you. Let me rephrase what you all said

exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)

exp(y) is always > 0

so if we select exp(y) = x - sqrt(x^2 +1)

and most importantly , sqrt(x^2 +1) > x

so exp(y) < 0 which is false.

As a result, we have to select exp(y) = x + sqrt(x^2 +1)

The similar argument can be applied to exp(-y)

Thank you all of you, really appreciate that.


# Math Help - Simultaneous Equations with way too many unknowns..

1. ## Simultaneous Equations with way too many unknowns..

Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.

'Find the values of m such that these equations have no solutions':

3x - my = 4
x + y = 12

'Find the values of m and a such that these equations have infinite solution sets':

4x - my = a
2x + y = 4

So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot

2. Originally Posted by Fnus
Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.

'Find the values of m such that these equations have no solutions':

3x - my = 4
x + y = 12

'Find the values of m and a such that these equations have infinite solution sets':

4x - my = a
2x + y = 4

So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.

So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.

Similarly a = 8.

-Dan

3. there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.

Question 1:
'Find the values of m such that these equations have no solutions':

3x - my = 4
x + y = 12

Question 2:
'Find the values of m and a such that these equations have infinite solution sets':

4x - my = a
2x + y = 4

Does that change anything? <.<

5. $x+y=12\Rightarrow y=-x+12$ so the slope is -1.

$3x-my=4\Rightarrow y=\frac{3}{m}x-\frac{4}{m}$

now solve $\frac{3}{m}=-1$ you should get $m=-3$

6. for the second one we have

$4x-my-a=0$
and
$2(2x+y)=2(4)\Rightarrow 4x+2y-8=0$

from here it should be obvious what values to make a and m.

7. Hello, Fnus!

Find the values of $m$ such that these equations have no solutions:
. . $\begin{array}{cc}(1)\\(2)\end{array}
\begin{array}{cc}3x - my \\ x + y \end{array}
\begin{array}{cc} = \\ = \end{array}
\begin{array}{cc}4 \\ 12\end{array}$

If you are familiar with determinants, there is a simple solution.

A system has no solution if its determinant equal zero.

The determinant is: . $\begin{vmatrix} 3 & \text{-}m \\ 1 & 1\end{vmatrix} \:=\:(3)(1) - (\text{-}m)(1) \:=\:3 + m$

Therefore: . $3 + m \:=\;0\quad\Rightarrow\quad\boxed{m = -3}$

Otherwise, we can try to solve the system.

We have equation (1): . $3x - my \:=\:4$
. . .Multiply (2) by $m:\;\;mx + my \:=\:12m$

. . . . . . . . . . . . Add: . $mx + 3x \:=\:12m + 4$

. . . . . . . . . . .Factor: . $(m + 3)x \:=\:12m + 4$

. . . . . . . . . . . Then: . . . . . . $x\:=\:\frac{12m + 4}{m + 3}$

We see that $x$ is undefined if $m = -3.$

Therefore, the system has no solutions for $m = -3.$

8. Originally Posted by topsquark
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.

So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.

Similarly a = 8.

-Dan
Originally Posted by putnam120
there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.
But if one equation is merely a multiple of the other, they represent the same line, so any point (x,y) on the line is a solution. So we have an infinite number of solutions.

-Dan


# Relationship between O and o notation

In big-O notation, $f(x) = O(g(x))$ as $x\rightarrow \pm\infty$ if

$$\exists C, \delta>0: \forall |x| \geq \delta: |f(x)| < C |g(x)|$$

and, for the case I'm more interested in here, $f(x) = O(g(x))$ as $x\rightarrow 0$ if

$$\exists C, \delta>0: \forall |x| \leq \delta: |f(x)| < C |g(x)|$$

In little-o notation, $f(x) = o(g(x))$ means

$$\lim_{|x|\rightarrow0}\frac{f(x)}{|g(x)|} = 0$$

I've come across big-O before, but this is the first time I've seen little-o. I have experience in applied math, but very little in pure math, and my first reaction (for $x\rightarrow 0$) was "they mean the same thing". Then I thought "but little-o is a more precise upper bound".

What is the relationship between these two notations, and what does the distinction mean practically?

• Well, if $f(x)$ is $o(g(x))$ it is also $O(g(x))$ but maybe not the other way around. It is not "more precise", but rather stronger: intuitively, if $f(x)$ is $o(g(x))$ then $g(x)$ grows "much faster" than $f(x)$, whereas $O(g(x))$ implies only that $f(x)$ is eventually bounded above (by some constant multiple of) $g(x)$ -- indeed, the growth rates could be comparable. – The_Sympathizer Sep 20 '13 at 11:54
• An example -- though note here I'm using the O-notation for growth near infinity, not near 0: If $f(x) = x$ and $g(x) = x^2$, then $f(x)$ is $o(g(x))$ AND $O(g(x))$, but if $f(x) = x^2$ then $f(x)$ is only $O(g(x))$ and NOT $o(g(x))$. – The_Sympathizer Sep 20 '13 at 11:56
• An example for near 0: $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x^2}$. Then $f(x)$ is $o(g(x))$ (as $x \rightarrow 0$) while if $f(x) = \frac{1}{x^2}$ then $f(x)$ is only $O(g(x))$ and not $o(g(x))$. – The_Sympathizer Sep 20 '13 at 11:58

In the briefest possible terms, $f \in O(g)$ is like saying $f \leqslant g$ asymptotically, and $f \in o(g)$ is like saying $f < g$ asymptotically.

As an example, if $f(x) = x$ and $g(x) = 2x$,we have that that $f \in O(g)$, but $f \notin o(g)$ and $g \notin o(f)$. If $f(x) = x^2$ and $g(x) = x$, then $f \in o(g)$ and $f \in O(g)$.

As an aside, be very aware of context. In various probabilistic situations, we define $O$-notation with a limit as $x\rightarrow 0$. In most algorithmic applications, we take limits as $x \rightarrow \infty$. Always make sure it's clear what you're working with before using either notation. $O$-notation is probably the most abused thing in all of mathematics.

• Also note: Notation $f \in O(g)$ is quite rare, compared to notation $f = O(g)$. – GEdgar Sep 20 '13 at 13:42
• Indeed. I didn't want to start listing abuses of $O$ notation because I don't think I'd ever stop. It's more common to say things like $2x = O(x)$. – ymbirtt Sep 20 '13 at 14:13
• $x^2$ is not in $O(x)$, at least when $x \to \infty$. Typo? – ftfish Sep 20 '13 at 14:44
• From OP's definitions, we're sending $x \rightarrow 0$, so I'm remaining consistent with that. – ymbirtt Sep 20 '13 at 15:08
• @GEdgar thanks, discussion of this helps me to get to grips with it. I've always been a little bit uncomfortable with the use of $=$, and thinking about it in terms of sets is more consistent. – TooTone Sep 21 '13 at 9:07

Little $o$ means lim sup of absolute value $=0$, big $O$ means lim sup of absolute value $< \infty$.

• Thanks, but I thought big O could be used for $\rightarrow 0$ as well as $\rightarrow \infty$? The former was what I was more interested in because it then has a similar meaning to little o. (My post was a little ambiguous and I see you changed $<$ in my post to $\ge$ -- I've changed that back and added some more which hopefully makes it clearer.) – TooTone Sep 20 '13 at 14:31
• You are referring to my correction? Okay, I am sorry then the same remains true. It should be indicated though wether you look at the behaviour at $0$ or at $\infty$. A continuous variable transformation, e.g. $x \mapsto x^{-1}$ exchanges between the setting and my statement remains valid;) – Marc Palm Sep 20 '13 at 14:39
• Yes you're right I should have indicated whether I was looking at limit towards zero or infinity. I can see how a reciprocal transformation might make the situations equivalent; however I still don't quite understand your answer. – TooTone Sep 20 '13 at 19:52
• my answer remains valid no matter to where the limit goes. – Marc Palm Sep 21 '13 at 14:57
• I think what I don't understand is, in simple terms, your use of "sup" means. – TooTone Sep 21 '13 at 18:01

check this out :

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

:-) it's not just Big O notation in the page

Looking at how $O(1)$ and $o(1)$ are sufficient to understand the general case.

Suppose $f \in O(1)$. Then

$$\lim_{x \to +\infty} f(x) < +\infty$$

if the limit exists. Or more generally, if the limit does not exist, then

$$\sup_{x \to +\infty} f(x) < +\infty$$

Now suppose $f \in o(1)$. Then

$$\lim_{x \to \infty} f(x) = 0$$


# Fixed Point Theorems

Theorem 1. Let $B=\{x\in \mathbb R^n :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^n$ . Any continuous function $f:B\rightarrow B$ has a fixed point.

Theorem 2. Let $X$ be a finite dimensional normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.

Theorem 3. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.

Theorem 4. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, closed, and bounded set. Then given any compact mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.

For some authors Theorem 1 is Brouwer's fixed-point theorem. For others Brouwer's fixed-point theorem is Theorem 2. Actually there is no difference because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball.

My problem is with theorems 3 and 4. For some authors Theorem 3 is Schauder's fixed-point theorem, for others Schauder's fixed-point theorem is Theorem 4.

Are Theorem 3 and Theorem 4 are equivalent? If not, are Theorems 1 and 2 special cases of Theorem 4?

-
"because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball" is not quite true, it could be lower-dimensional. Theorems 1 and 2 are special cases, because in that case, a continuous $f \colon K \to K$ is compact. – Daniel Fischer Jul 13 '13 at 15:39
What is the nonl tag for? – draks ... Jul 13 '13 at 20:08
@draks... My guess is a foreshortened nonlinear-something that never really got off the ground. – Sharkos Jul 13 '13 at 22:56

Your statement of Theorem 4 is missing an assumption on $K$, such as being convex, or at least homeomorphic to such a set (convex, closed, bounded). Without such an assumption, rotation of a circle gives a counterexample. Also, I think that in Theorem 4 you want the normed space to be complete, i.e., a Banach space.

Theorem 3 is contained in Theorem 4, because on a compact set every continuous map is compact. Theorem 4 cannot be easily obtained from Theorem 3 (I think) because if we tried to simply replace $K$ with $\overline{f(K)}$ (which is compact), we can't apply Theorem 3 because $\overline{f(K)}$ is not known to be convex.

Both 3 and 4 were stated and proved by Schauder in his 1930 paper Der Fixpunktsatz in Funktionalraümen, which is in open access. Here is Theorem 3:

Satz I. Die stetige Funktionaloperation $F(x)$ bilde die konvexe, abgeschlossene und kompakte Menge $H$ auf sich selbst ab. Dann ist ein Fixpunkt $x_0$, vorhanden, d.h. es gilt $F(x_0)=x_0$.

And this is Theorem 4 (in slightly less general version: the image of $F$ is assumed compact instead of relatively compact; possibly because the latter concept wasn't in use).

Satz II. In einem "B"-Raume sei eine konvexe und abgeschlossene Menge $H$ gegeben. Die stetige Funktionaloperation $F(x)$ bilde $H$ auf sich selbst ab. Ferner sei die Menge $F(H)\subset H$ kompakt. Dann ist ein Fixpunkt vorhanden.

("B"-Raume is what is now called a Banach space.) So, it is correct to call both Theorem 3 and Theorem 4 "Schauder's fixed-point theorem".

And yes, Theorems 1 and 2 follow by specialization of Theorem 3 or 4 to finite dimensions.

-


# Math Help - ship problem

1. ## ship problem

A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.

Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km

iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h

but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz

2. At t=0, they are both 10 kms from their intersection.

Think Pythagoras. Let D=square of distance between ships.

$D(t)=(10-30t)^{2}+(10-40t)^{2}$

$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$

$D'(t)=5000t-1400$

$5000t-1400=0$

$t=\frac{7}{25}$

3. Originally Posted by galactus
At t=0, they are both 10 kms from their intersection.

Think Pythagoras. Let D=square of distance between ships.

$D(t)=(10-30t)^{2}+(10-40t)^{2}$

$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$

$D'(t)=5000t-1400$

$5000t-1400=0$

$t=\frac{7}{25}$
thankz...but i was wondering if i did the first part right that i showed?(shown below) because im not too sure if i was meant to include negative signs with any of the answers? can u please just check that for me thankz

Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km

iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h

4. Originally Posted by dopi
A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.
Hello, dopi,

with your problem you use automatically a coordinate system: The pos. x-axis points to East and the pos. y-axis points to North.

At the time t = 0 the ship P is at P(-10, 0) and the ship Q is at Q(0, 10). The movement of both ships is described by a straight line. The speed is described by a vector:
$\overrightarrow{v_P}=(30, 0)$
$\overrightarrow{v_Q}=(0, -40)$

Thus P is moving along the line:
$\overrightarrow{x_P}=(-10, 0)+t*(30, 0)$
and Q is moving along the line:
$\overrightarrow{x_Q}=(0, 10)+t*(0, -40)$

The distance between the ships is: $\vec{d}=\overrightarrow{x_P}-\overrightarrow{x_Q}$

Originally Posted by dopi
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km

iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
I don't understand why you want to calculate the relative coordinates. It isn't necessary with your problem.

Originally Posted by dopi
but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
This was already done by galactus.

Greetings

EB

5. Hello, dopi!

You seem to be using a coordinate system and vectors
. . and I don't know what you plan to do with them.

galactus gave you the solution . . . so why struggle with your approach?

I'll make some diagrams to accompany his excellent solution.

Originally Posted by galactus

At t=0, they are both 10 kms from their intersection.
Code:
                        * Q
|
|
| 10
|
|
|
P * - - - - - - - - o
10

$t$ hours later, $P$ has moved $30t$ km east to point $A$
. . and $Q$ has moved $40t$ km south to point $B.$
Code:
                        *
|
| 40t
|
o B
|
| 10-40t
* - - - o - - - - *
30t  A  10-30t
And we want to minimize the distance $\overline{AB}.$

Originally Posted by galactus

Think Pythagoras. Let D = square of distance between ships.

$D(t) \:= \10-30t)^{2} + (10-40t)^{2}" alt="D(t) \:= \10-30t)^{2} + (10-40t)^{2}" />

$D'(t) \:= \:2(10-30t)(-30) + 2(10=40t)(-40)$

$D'(t) \:= \:5000t - 1400$

$5000t -1400\:=\:0$

$t \:= \:\frac{7}{25}$


# Math Help - Tricky Double Integral with polar coordinates

1. ## Tricky Double Integral with polar coordinates

Evaluate the integral:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$

where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Tried drawing the region $D$ and it looks like I have to use polar coordinates.

Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.

I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.

Thanks in advance, any pointers would be greatly appreciated.

2. Originally Posted by craig
Evaluate the integral:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$

where $D$ is the region $D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Tried drawing the region $D$ and it looks like I have to use polar coordinates.

Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.

I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.

Thanks in advance, any pointers would be greatly appreciated.
In general to convert, you do the following:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$

Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:

$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$

Hence:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$

Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.

Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$

Hence, in this particular situation:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

3. Hi

The area of integration is the part of the disk which is above the red line

And $dx dy$ becomes $r dr d\theta$

4. Originally Posted by craig
Evaluate the integral:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$

where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Tried drawing the region $D$ and it looks like I have to use polar coordinates.

Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.

I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.

Thanks in advance, any pointers would be greatly appreciated.
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html

If you draw the xy domain you can see we want the area to the left of the line but inside the circle,

If we convert this to polar co-ordinates this means

$\frac{ \pi }{4} \le \theta \le \frac{ 5 \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis. And $\frac{ 5 \pi }{4}$ comes from the angle between the line in the first quadrent to the line in the 3rd quadrent.

For r, we are going from the origin to the radius of the circle. So the interval of r must be

$0 \le r \le \sqrt{3}$

We can now compute this integral,

$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$

$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$

$= \iint_Q rcos^2 \theta dr d \theta$

$= \int_0^{ \sqrt{3} } rdr \int_{ \frac{ \pi }{4} }^{ \frac{ 5 \pi }{4} } cos^2 \theta d \theta$

Which can be easily computed using double angle formulas

5. Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

Originally Posted by Mush
In general to convert, you do the following:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$

Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:

$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$

Hence:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.

Originally Posted by Mush
Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.

Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$
I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess

Originally Posted by Mush
Hence, in this particular situation:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

Anyway on with the integral:

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =

$\frac{3}{4}\bigg[\theta + \frac{1}{2}\sin{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ =

$\frac{3}{4}\bigg(\frac{5\pi}{4} + \frac{1}{2}\sin{\frac{5\pi}{2}} -\frac{pi}{4} - \frac{1}{2}\sin{\frac{\pi}{2}} \bigg)$

$\frac{3}{4}(\pi + 0)$ = $\frac{3\pi}{4}$

Thanks a lot for the help!

6. Originally Posted by running-gag
Hi

The area of integration is the part of the disk which is above the red line

And $dx dy$ becomes $r dr d\theta$
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!

Originally Posted by AllanCuz
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html

If you draw the xy domain you can see we want the area under the line but inside the circle,

If we convert this to polar co-ordinates this means

$0 \le \theta \le \frac{ \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis.

For r, we are going from the origin to the radius of the circle. So the interval of r must be

$0 \le r \le \sqrt{3}$

We can now compute this integral,

$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$

$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$

$= \iint_Q rcos^2 \theta dr d \theta$

$= \int_0^{ \sqrt{3} } rdr \int_0^{ \frac{ \pi }{4} } cos^2 \theta d \theta$

Which can be easily computed using double angle formulas
Thanks a lot for the reply, really helped.

7. Originally Posted by craig
Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.

I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess

** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

Anyway on with the integral:

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =

$\frac{3}{4}\bigg[1 + \cos{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ = $\frac{3}{4}(\pi)$ = $\frac{3\pi}{4}$

Thanks a lot for the help!
I don't think you're performed that integration properly, have you?

The integral of $1 + \cos(2\theta)$ is $\theta + \frac{1}{2}\sin(2\theta)$.

8. Originally Posted by craig
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!

Thanks a lot for the reply, really helped.
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P

9. Originally Posted by Mush
I don't think you're performed that integration properly, have you?
Yes & no. I did the integration properly, just forgot to update the latex code...oops.

Edited my above working now, sorry about that.

10. Originally Posted by AllanCuz
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P
Haha thought that was what you meant, thanks again for the reply.

11. Originally Posted by craig
Yes & no. I did the integration properly, just forgot to update the latex code...oops.

Edited my above working now, sorry about that.
Indeed. Good show.


# Elementary question about a fake-proof and greatest common divisors

I have a question to an excercise - for which I have a wrong solution - and I wanted to ask you to help me understand my thinking error. The excercise was as follows:

Let $$a, b, n \in \mathbb{N}$$. Show that $$\gcd(a,b) = \gcd(a,b+na)$$.

My solution was (in a nutshell):

• Let $$d := \gcd(a,b)$$.
• Then I showed that $$d$$ is a common divisor of both $$a$$ and $$b + na$$
• Then I showed, that any common divisor $$c$$ of $$a$$ and $$b+na$$ is smaller-or-equal than $$d$$
• After that, I concluded that by definition $$d = \gcd(a,b+na)$$, and with the definition of $$d$$ I would have $$\gcd(a,b) = \gcd(a,b+na)$$, so my proof was completed.

However, my tutor did not accept the proof. He showed me the 'correct' proof, in which I woould have $$d := \gcd(a,b)$$ and $$e := \gcd(a,b+na)$$ and then demonstrate $$d \leq e \leq d$$.

Anyway, I do understand his proof. But I still do not understand where my thinking error is. After all, if my starting line would be, let's say, $$d := 9$$ and I would have been able to show the steps afterwards, then I would be able to conclude $$9 = \gcd(a,b+na)$$, no?

EDIT Wow, you guys answered fast! Thank you very much. I will ask my tutor on thursday again. For completeness sake, I will formulate my complete proof (my original proof is in german, so maybe there will be something lost in translation, the [Reference] are references to our script).

Let $$a,b,n \in \mathbb{N}$$. Let $$d := \gcd(a,b)$$. By definition it means $$d \mid a$$ and $$d \mid b$$, so with [Reference] the following holds true: $$d \mid (b + na)$$. Therefor, $$d$$ is a common divisor of both $$a$$ and $$b + na$$. We will show now, that $$d$$ is the greatest common divisor.

Let $$c$$ be any common divisor of $$a$$ and $$b + na$$. Therefor $$c$$ divides $$a$$, and as such $$c \mid na$$ und therefor $$c \mid ((b + na) - na) = b$$ (again because of [Reference]). So $$c$$ divides $$b$$, and therefor $$c$$ is a common divisor of both $$a$$ and $$b$$. Since $$d$$ was the greatest common divisor of $$a$$ and $$b$$, we have $$c \leq d$$.

By definition of the greatest common divisor we get $$d = \gcd(a,b+na)$$.

End of Proof. My tutor wrote, that I only showed $$\gcd(a,b+na) \leq \gcd(a,b)$$ and also need to show the other direction. Then he showed me the 'correct'/'ideal' proof today. But I will ask him again on thursday!

• You need to be more specific with your proof. From the outline there should be no objection, so either your tutor is wrong or you made an error. We can't determine which. – Matt Samuel Oct 30 '18 at 2:36
• Possibly the confusion arises because you used two different proofs to show that neither gcd exceeds the other, but you can just repeat the same proof replacing $b$ and $n$ with $b+na$ and $-n$. – Neil Oct 30 '18 at 9:24
• This has been the way I have understood things: The following is true $\forall k \in \mathbb N$, for any pair of integers $(n,j)$: $$n\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{n\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ $$j\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{j\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ – Adam L Nov 3 '18 at 17:14
• The above two equalities can be summed only in integer multiples of $n$ and $j$,ie $\forall a,b \in \mathbb Z$ $$(an+bj)\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{(an+bj)\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ – Adam L Nov 3 '18 at 17:14

The outline of the proof that you have explained is correct.

More than likely the problem was with the detail in one of the intermediate steps.

I would check each step carefully again and there is always room for skipping a point in writing a proof.

You want to know where the error is in your proof. Given no details no one can give the answer. All we can do is give only our opinion, so do not treat this as an answer.

The sketch given is correct, but it is merely a restatement of what is required to be proved. Possible that both you and your tutor are correct, but tutor did not understand the proof and misjudged it. But given that you expect people to be able to analyze your "proof" without providing details I am inclined to believe you are wrong. (IT IS AN OPINION!)

• Yes, I also did assume that I was wrong. This is why I provided only the outline sketch since I thought my thinking error would have been with my 'proof-strategy'. I translated now my complete proof and my tutors response as an edit to the original question. – DaGoddyMan Oct 30 '18 at 3:18

Ok well firstly, if your tutor has not mentioned to you something by the name of the Euclidean Algorithm, then he or she isn't being very forthcoming about their level of expertise in the subject it's as simple as that. But follow the link I provided there, and apply the method to both greatest common divisor expressions, and this might be able to help you reach a better level of understanding on the subject, the process is clearly explained in the link.

Also important to note that you shouldn't worry about studying a modern day definition of an algorithm,don't start to think this has anything to do with programming or a need to learn a programming language, this algorithm was invented over 2300 years ago, there was no necessity to update java at that point.

Secondly if your tutor's final step is to demonstrate that $$d \leq e \leq d$$, then it is in fact their solution that is completely bogus and circular, since this statement is equivalent to the original equality that we are trying to prove!

I recommend purchasing a hard cover book in Analytic Number Theory, one that provides the solutions for every exercise it contains, so that if you really are unable to find a proof you are confident with, you can look at the one the book has provided.

I currently use one that I have found to be priceless, "Problems in Analytic Number Theory" Second Edition, written by M.Ram Murty. But do ask your supervisors at your college what they recommend.

• Proving that $d=e$ by proving that $d\leq e\leq d$ is a standard approach and there is nothing bogus in it. – Ister Oct 31 '18 at 7:40
• It's the same question. If you asked me to prove dragons are real and I replied "I shall do so by proving dragons are real, hence proving dragons are real" would you be satisfied? – Adam L Oct 31 '18 at 18:49
• No, it's entirely different thing. You have a single statement to prove ($d=e$) but instead you prove two separate statements usually following different approaches ($d\leq e$ is one proved statement and $e\leq d$ is another one). Only now you apply a not so obvious implication that if $d\leq e\leq d$ holds then $d=e$ so if we've proved the former then the latter is truth indeed. As I said - this method is pretty common, especially when calculating limes but also in some other cases. – Ister Nov 2 '18 at 20:54
• $d=e$ is just an abbreviation for $d \leq e \leq d$. $e \leq d$ means $e$ is less than or equal to $d$. $d \leq e$ means $d$ is less and or equal to $e$. $d=e$ is equivalent to the statement $e \leq d$ and $d \leq e$, since $e \lt d=\lnot(e \gt d)$ – Adam L Nov 3 '18 at 9:09
• I totally disagree that it's just an abbreviation. You define $\leq$ as equal or less (you need to have at least weak order in order to write something like that at all) so equality precedes weak inequality. Of course proving that $d\leq e \leq d \iff d=e$ is trivial but still it's a property of weak inequality. That's why it is worth and can be used as a proof method to separately show each of the weak inequalities and then conclude that it means equality in the end. – Ister Nov 5 '18 at 9:51

Your proof is indeed correct. It can be reorganized more symmetrically as follows.

Notice that if $$\ c\mid a\$$ then $$\ c\mid b \!+\! na \iff c\mid b.\,$$ Therefore $$\,a,\, b\!+\!na\,$$ and $$\,a,\, b\,$$ have exactly the same set $$\,\cal C\,$$ of common divisors $$\,c,\,$$ so they have the same greatest common divisor $$(= \max \,\cal C)$$

• @Ister Please read more carefully. The answer says nothing at all about "invalidating OP's proof". Nor is there any such mistake. Analyzing (e.g. simplifying) proofs is an important part of teaching. I often point out such simplifcations arising by exploiting innate symmetry, e.g. yesterday using reflection (negation) symmetry to simplify sum computation, and many others. – Bill Dubuque Oct 31 '18 at 16:15
• I must have had bad day, sorry. My comment removed. – Ister Nov 2 '18 at 20:55

In my opinion your tutor is plainly wrong, too much focused on the proof he's accustomed to. It's often a problem. Let me say I faced this issue all the time since I tended to think out of the box and find my own, correct solutions, different to the canonical one.

You proved by definition, which is a fully valid proof. Thus technically you don't have to prove anything else.

Yet, since your tutor apparently don't understand this basics, you have to be smarter than him and be able to counterargument him (i.e. explain why his statement is plainly wrong).

My tutor wrote, that I only showed $$gcd(a,b+na)\leq gcd(a,b)$$ and also need to show the other direction.

In the first part of your proof by showing that $$d$$ is a divisor of both $$a$$ and $$b$$ you've proved that $$gcd(a,b) = d \leq gcd(a,b+na)$$. By definition of $$gcd$$ (name even represents that!) for any given divisor $$z$$ of both $$x$$ and $$y$$ we have $$z\leq gcd(x,y)$$. By substituting $$x=a$$ and $$y=b+na$$ you have the statement that your tutor claims to be missing.

Let me emphasise it once more - you don't need that explanation in your proof, which is perfectly valid. This is only to show that your tutor's claim is wrong.

As an anecdote...

In the course of probabilistic we once had some simple homework task from combinatorics. I don't remember details now but it's actually irrelevant. I had done my calculation, verified them and as a result got a nice looking solution.

On the next lesson I was presenting the solution on the blackboard. Imagine my surprise when I heard my solution is wrong. Someone else had done the task on the blackboard the right way and indeed the results didn't look alike at all. If I remember correctly there were even different powers used in it! Of course, the new solution was correct and in accordance to the key, yet I was sure mine was correct as well (despite all odds and apparent difference) so I stood up for it and asked to point out where had I made an error. No-one, including the tutor could find anything. Eventually the tutor said it might be that the two polynomials are equivalent and if I was able to prove that equivalence I will have the whole course passed with maximum grade.

Needless to say it took me 3 months of work, some horrible calculations and some peculiar theorem from number theory but I did it. The two results are equivalent.

End of Proof. My tutor wrote, that I only showed $$\gcd(a,b+na) \leq gcd(a,b)$$ and also need to show the other direction.

I am pretty sure, that $$\gcd(a,b+na) \geq gcd(a,b)$$ follows from $$d:=\gcd(a,b)$$ and $$d\mid(b+na)$$ (plus the definition of $$\gcd$$).

• Can whoever downvoted this please explain, what my error is. If it was due to my bad formatting, I fixed it. – Nathrat Oct 30 '18 at 11:26
• I didn't downvote but your answer may not be clear. You are trying to give a counterargument to tutor's wrong quoted statement. You should though be more firm in your statement (do the math and make sure you are right) plus add a clear explanation that it means OP is correct in his proof. Note - you are right, but gap in your proof might be to difficult to fill. Try making it more complete (or check my answer as it is actually filling this gap. – Ister Oct 31 '18 at 8:01


# Creating an integral from 2 functions

Let $$R$$ be the region in $$\mathbb{R}^2$$ below the line $$y = x + 2$$ and above the parabola $$y = x^2$$. Check the integral of these $$2$$ functions in terms of $$dx\cdot dy$$ and then $$dy\cdot dx$$ ' I am having an issue figuring out what the integrals will range from. I have: $$G =\{(y,x) : -1 < x < 2 \text{ and } x^2 < y < (x + 2)\}\to dy.dx$$ $$H =\{(x,y) : 0 < y < 4 \text{ and } y -2 < x < \sqrt{y}\} \to dx.dy$$

However when I create the integrals in terms of $$dx\cdot dy$$ and $$dy\cdot dx$$ they differ? Any help please, did i get the range of the $$G$$ and $$H$$ wrong? Its a parabola cut with a line

For $$dydx$$ the integral over $$1$$ is given by $$\int_{-1}^2\int_{x^2}^{x+2}dydx=\int_{-1}^2-x^2+x+2dx=[\frac13x^3+\frac12x^2+2x]_{-1}^2=\frac92$$ with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $$0\le y\le1$$ we have that $$-\sqrt{y}\le x\le\sqrt{y}$$ and when $$1\le y\le4$$ we have $$y-2\le x\le\sqrt{y}$$. So the integral over the function $$1$$ is $$\int_0^1\int_{-\sqrt{y}}^\sqrt{y}dxdy+\int_1^4\int_{y-2}^\sqrt{y}dxdy=\int_0^12\sqrt{y}\,dy+\int_1^4\sqrt{y}-y+2\,dy$$ $$=[\frac43y^{\frac32}]_0^1+[\frac23y^{\frac32}-\frac12y^2+2y]_1^4=\frac92$$ So the two regions are now equal.

• Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated! – Shaun Weinberg Mar 31 at 10:36

For the region $$H$$, the lower limit of $$x$$ shouldn't be $$y-2$$.

It should be the maximum of $$-\sqrt{y}$$ and $$y-2$$. In fact, when $$0 \le y \le 1$$, the lower limit is $$-\sqrt{y}$$.

That is

$$H =\{(x,y): 0 \le y \le 4 , \max(-\sqrt{y}, y-2) < x < \sqrt{y} \}$$

That is from the picture below, the left limit of the region consists of the green color and blue color part.

• I originally wrote an answer for the duplicate, so I copied it here. (+1) btw – Peter Foreman Mar 30 at 19:13
• I did the same thing. ;) – Siong Thye Goh Mar 30 at 19:14


# Limits problem involving greatest integer function and an unknown function.

Given that $$\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$$, find $$\lim_{x \to 0} \lfloor f(x) \rfloor$$ and find if $$\lim_{x \to 0} \lfloor \frac{f(x)}{x} \rfloor$$ exists. My math teacher says that since the denominator in the first limit is non negative and the limit itself is positive, he says that $$\lim_{x \to 0} f(x) = 0^{+}$$ and thus $$\lim_{x \to 0} \lfloor f(x) \rfloor = 0$$. I find this acceptable but my friend assumes $$f(x) = 2x^{2} + \infty^{-}x^{3}$$ and claims that $$\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$$ but that $$\lim_{x \to 0} \lfloor f(x) \rfloor$$ does not exist as $$\lim_{x \to 0^{+}} f(x) = 0^{+}$$ and $$\lim_{x \to 0^{-}} f(x) = 0^{-}$$. So who's right and who's wrong? If either of the two are wrong please explain why?

• If $f(x)=2x^2$, does it satisfy any of either of your claims? – Andrew Chin Nov 8 at 15:49
• Your math teacher is correct. – Paramanand Singh Nov 8 at 15:58
• Thanks for answering Andrew Chew! If $f(x) = 2x^{2}$ then quite obviously, $\lim_{x \to 0} frac{f(x)}{x^{2}} = 2$ and $\lim_{x \to 0} \lfloor f(x) \rfloor = 0$. But the problem is that $f(x)$ can be $2x^{2}$ or it can be any other expression that satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. We need to find what value $\lim_{x \to 0} \lfloor f(x) \rfloor$ equals irrespective of the function $f(x)$, as long as it satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. – K Darshan Nov 8 at 17:17
• Paramanand Singh, I appreciate your effort but can you give the reason as to why my friend was wrong? – K Darshan Nov 8 at 17:20
• Oh what I would give if tomorrow I'd wake up and find everyone has stopped plugging $\infty$ into expressions and treating it as though it were a number! – fleablood Nov 9 at 1:34

The problem is, simply put, that an expression like $$\infty^- \cdot x^3$$ doesn't make sense; you won't find an actual function (meaning a function which only uses numbers and not $$\infty$$) which behaves like that. Even if you take a function like $$f(x) = 2x^2 + (-10000) \cdot x^3$$, as you approach zero, eventually the $$x^3$$ term will not matter anymore: it will be much smaller than the $$2x^2$$ term, if only the $$x$$ you insert is "small enough", so close enough to zero. That means that for $$x$$ small enough, this $$f(x)$$ will still be greater than $$0$$.

So yeah, your teacher is correct. In general, you should always be very skeptical when people use infinity like that. Without proper care, infinity doesn't actually make a whole lot of sense :)

Elaborating on my comment the counter-example given by your friend does not make any sense. You can't write expressions like $$f(x) =2x^2+\infty^{-}x^3$$. The usage of symbol $$\infty$$ is always specified by specific definitions and typical examples are expressions like $$x\to\infty$$ and $$\lim_{x\to 0}1/x^2=\infty$$.

On the other hand the correct argument goes like this. Since $$f(x) /x^2\to 2$$ the expression $$f(x) /x^2$$ is positive as $$x\to 0$$ and hence $$f(x) >0$$ as $$x\to 0$$. Further $$f(x) =x^2\cdot \dfrac {f(x)} {x^2}\to 0\cdot 2=0$$ and hence $$f(x) <1$$ as $$x\to 0$$. It follows that $$\lfloor f(x) \rfloor =0$$ as $$x\to 0$$ and thus $$\lim_{x\to 0}\lfloor f(x) \rfloor =0$$.

The expression $$f(x) /x= x(f(x) /x^2)$$ also tends to $$0$$ but is positive if $$x\to 0^{+}$$ and negative if $$x\to 0^{-}$$ and hence $$\lfloor f(x) /x\rfloor =0$$ if $$x\to 0^{+}$$ and $$\lfloor f(x) /x \rfloor =-1$$ as $$x\to 0^{-}$$ so that the limit $$\lim_{x\to 0}\left\lfloor \dfrac{f(x)} {x} \right\rfloor$$ does not exist.

I'm not entirely sure I understand your teacher's argument (I sure as heck don't get your friend's), but I think it is valid.

$$1 > \epsilon > 0$$ and there is a $$\delta$$ so that $$|x|< \delta$$ implies $$|\frac {f(x)}{x^2} - 2| < \epsilon$$ so $$2-\epsilon \frac {f(x)}{x^2} < 2-\epsilon$$. Thus $$1 < \frac {f(x)}{x^2} < 3$$ and $$f(x) > 0$$ and $$x < \min(\delta,\frac 12)$$ then

$$0 < x^2 < \frac {f(x)} < 3x^2 < \frac 34$$ so $$0 and $$\lfloor f(x)\rfloor = 0$$ for all $$x < \min(\delta, \frac 12)$$.

Which means $$\lim_{x\to 0}\lfloor f(x)\rfloor = 0$$.

Which I think is your teacher's argument.

I can't see why your friend assumes $$f(x) = 2x^2 + \infty - x^3$$, which doesn't even make sense. (What is $$f(5)$$? Is it $$\infty -25$$? What's that?) so I can't tell you why he is wrong other than that what he says makes no sense.

• I'm guessing here but you probably didn't understand what $0^+$, $\infty^-$, etc. actually mean. Generally a number like $a^{+}$ is a number just greater than $a$, something like $a+0.000...1$ and $a^{-}$ is a number just less than $a$, something like $a-0.000...1$. What I meant by $\infty^{-}$ is that it's a number that is very large but not $\infty$. However, on second thoughts, I believe that a number like that doesn't exist. – K Darshan Nov 9 at 13:27
• Infinity is NOT a number. $\infty^{-}$ is oxymoronic and self-contradictory. $f(x) = 2x^2 +\infty^{-}x^3$ is meaningless and illdefined (what number is $f(5) = 50+150\infty^-$?). Your friend is more than wrong. Your friend is spouting uninterpretable gibberish. – fleablood Nov 10 at 16:10


# Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?

We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this. To be specific, it gives us 3 diophantine equations

$$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$

and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.

So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.

Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$.

• You could find the primitive element of $\mathbb{Q}(\sqrt[3]{3},\sqrt[3]{2})$, find its minimal polynomial and show that it is of degree larger than $3$. – Michael Burr Sep 16 '17 at 1:22
• Algebraic number theory might be able to give a solution by considering discriminants of rings of integers - but I'm a bit rusty on that. – Daniel Schepler Sep 16 '17 at 1:27
• @MichaelBurr Thanks! I think we can even solve the general problem with that idea and computers. However, I don't know how to do this by hands. It seems that $\sqrt[3]{2}+\sqrt[3]{3}$ is a degree 9 element, but I cannot convince it without any help of computer programs. – Seewoo Lee Sep 16 '17 at 1:34
• $a$ cannot be divisible by $2$ (first equation). Therefore $b$ is divisible by $2$ (second equation). Therefore, $c$ is divisible by $2^2$ (third equation). Hence $b$ is divisible by $2^3$ (second equation). Keep repeating and you get that $b$ is divisible by $2^k$ for any $k$. For other pair of primes it is the same business. – Hellen Sep 16 '17 at 1:34
• What is $\Bbb{Q}\left( \sqrt[3]{2} \right)$? – gen-z ready to perish Sep 16 '17 at 4:03

Here’s another proof, making light use of $p$-adic theory:

First, note that Eisenstein tells us that both $\Bbb Q(\sqrt[3]3\,)$ and $\Bbb Q(\sqrt[3]2\,)$ are cubic extensions of the rationals, and thus either they are the same field or their intersection is $\Bbb Q$. Therefore, a question equivalent to the stated one is, “Why is $\sqrt[3]2\notin\Bbb Q(\sqrt[3]3\,)$?” I’ll answer the new question.

Note that $\Bbb Q(\sqrt[3]3\,)$ may be embedded into $\Bbb Q_2$, the field of $2$-adic numbers, because $X^3-3\equiv(X+1)(X^2+X+1)\pmod2$, product of two relatively prime polynomials over $\Bbb F_2$, the residue field of $\Bbb Q_2$ (more properly the residue field of $\Bbb Z_2$). So Strong Hensel’s Lemma says that $X^3-3$ has a linear factor in $\Bbb Q_2[X]$, in other words, there’s a cube root of $3$ in $\Bbb Q_2$. But if there were a cube root of $2$ in $\Bbb Q(\sqrt[3]3\,)$, there’d be such a root in $\Bbb Q_2$, and there isn’t since it’d have to have (additive) valuation $v_2(\sqrt[3]2\,)=1/3$. No such thing, the value group of $\Bbb Q_2$ is $\Bbb Z$.

I really like all of the 3 answers, and I think I found one more solution which uses Trace, not Norm. (I don't know why I didn't try to use trace map before.) So I'm going to write it down here.

Suppose $$K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{3})$$. Consider a trace map $$\mathrm{Tr}_{K/\mathbb{Q}}:K\to \mathbb{Q}, \quad \alpha \mapsto \sum_{i=1}^{n}\sigma_{i}(\alpha)$$ where $$\sigma_{1}(\alpha)=\alpha, \dots, \sigma_{n}(\alpha)$$ is roots of minimal polynomial of $$\alpha$$. (We can consider trace map as a trace of a linear map $$m_{\alpha}:K\to K, x\mapsto \alpha x$$.) This is a $$\mathbb{Q}$$-linear map and we can check that $$\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{2})=\sqrt[3]{2}+\sqrt[3]{2}w+\sqrt[3]{2}w^{2}=\sqrt[3]{2}(1+w+w^{2})=0$$ where $$w=e^{2\pi i /3}$$ and $$\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{3})=0$$, too. If we assume $$\sqrt[3]{3}=a+b\sqrt[3]{2}+c\sqrt[3]{4}\,\,\,\,(a, b, c\in \mathbb{Q})$$ as above, by taking trace on both sides we get $$0=3a$$ and $$a=0$$. By multiplying $$\sqrt[3]{2}$$ and $$\sqrt[3]{4}$$ on both sides, we get $$b=c=0$$ and contradiction. I think this methods can be used to prove for higher power cases, such as $$\sqrt[n]{3}\not\in \mathbb{Q}(\sqrt[n]{2})$$ for $$n\geq 2$$.

• Great idea! I used it to prove that roots appear in root extensions only in "obvious" ways. – Orest Bucicovschi Sep 16 '17 at 6:53
• Could you please detail how do you get $b=c=0$? I am not sure how to prove $Tr_{K/\Bbb Q}(\sqrt[3]3\sqrt[3]2)=0$ – PerelMan 2 days ago

Let $\omega := e^{2 \pi i/3}$, and consider the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of $x^3 - 2$, and the unique automorphism $\sigma$ such that $\sigma(\sqrt[3]{2}) = \omega \sqrt[3]{2}$, $\sigma(\omega \sqrt[3]{2}) = \omega^2 \sqrt[3]{2}$, $\sigma(\omega^2 \sqrt[3]{2}) = \sqrt[3]{2}$. Suppose that $\sqrt[3]{3} \in \mathbb{Q}[\sqrt[3]{2}]$ is equal to $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ for $a, b, c \in \mathbb{Q}$. Then $\sigma(\sqrt[3]{3})$ is also a root of $x^3 - 3 = 0$, which implies it's equal to either $\sqrt[3]{3}$, $\omega \sqrt[3]{3}$, or $\omega^2 \sqrt[3]{3}$.

However, in the first case, $$\sigma(\sqrt[3]{3}) = \sqrt[3]{3} \Rightarrow a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4} = a + b \sqrt[3]{2} + c \sqrt[3]{4}.$$ Rewriting $\omega^2 = - \omega - 1$ and using the linear independence of the basis $\{ 1, \omega, \sqrt[3]{2}, \omega \sqrt[3]{2}, \sqrt[3]{4}, \omega \sqrt[3]{4} \}$ of the splitting field, we get $b = c = 0$, which gives a contradiction since $\sqrt[3]{3}$ is irrational.

Similarly, in the second case, $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{2}$, which is impossible since $\sqrt[3]{3/2}$ is irrational. And in the third case, $\sigma(\sqrt[3]{3}) = \omega^2 \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{4}$, which is impossible since $\sqrt[3]{4/3}$ is irrational.

HINT:

If $p$ a rational number, not a cube, and $a$, $b$, $c$ rational numbers so that

$$(a + b \sqrt[3]p + c \sqrt[3]{p}^2)^3$$ is rational, then at most one of the numbers $a$, $b$, $c$ is nonzero.

To prove this, it's enough to show that if $(b,c) \ne (0,0)$ then $a=0$. Consider $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. We have $\omega^2 + \omega + 1=0$, and $\omega^3 = 1$.

From $$(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$$ we get $$(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$$ Since $(b,c)\ne (0,0)$ we conclude $$a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2 = \sqrt[3]{q} \omega^{\pm 1}$$ and the conjugate equality $$a + b \sqrt[3]{p}\omega^2 + c \sqrt[3]{p}^2 \omega = \sqrt[3]{q} \omega^{\mp 1}$$

To these two equalities we add $$a + b \sqrt[3]p + c \sqrt[3]{p}^2= \sqrt[3]{q}$$ and we get $3 a = 0$, so $a = 0$.

$\bf{Added:}$

Details:

Assume $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2 \ne a + b \sqrt[3]{p} + c \sqrt[3]{p}^3$. We get $$b\omega + c \sqrt[3]{p}\omega^2 = b + c \sqrt[3]{p}$$ and with $\omega^2 =-1-\omega$ we get $(b-c\sqrt[3]{p}) \omega = b + 2 c \sqrt[3]{p}$, and so $\omega = \frac{b + 2 c \sqrt[3]{p}}{b-c\sqrt[3]{p}}$, contradiction ( since, say, $\omega$ is not real). So $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2$ must be one of the other two roots of the equation $x^3 = q$.

$\bf{Added 2:}$ We follow the beautiful idea of @See-Woo Lee: to use traces.

$\bf{Fact:}$ Let $\alpha$ a real radical, that is $\alpha \in \mathbb{R}$, $\alpha^m\in \mathbb{Q}$ for some $m$. Assume moreover that $\alpha$ is irrational. Then $\operatorname{trace}^L_{\mathbb{Q}}\alpha= 0$ for any $L$ finite algebraic extension of $\mathbb{Q}$ containing $\alpha$.

Proof: We may assume $\alpha >0$. Let $m$ smallest so that $\alpha^m \in \mathbb{Q}$. Clearly $m>1$. Let's prove that the polynomial $X^m - \alpha^m$ is irreducible over $\mathbb{Q}$. It factors over $\mathbb{C}$ as $\prod_{i=0}^m(X- \alpha \omega^i)$. If several of these linear factors produced a polynomial with rational coefficients, we would have the free term $$\prod_{i \in I} (- \alpha \omega^i) \in \mathbb{Q}$$ Taking absolute values we would have $\alpha^k \in \mathbb{Q}$ for some $1\le k < m$, not possible.

Denote by $K = \mathbb{Q}(\alpha)$. Then we have $$\operatorname{trace}^K_{\mathbb{Q}}(\alpha) = \sum_{i=0}^{m-1} \alpha \omega^i = 0$$. Hence for every $L\supset K$ we get $$\operatorname{trace}^L_{\mathbb{Q}} \alpha = [L: K] \operatorname{trace}^K_{\mathbb{Q}}\alpha = 0$$

$\bf{Main\ result:}$ Let $\alpha_i$ real radicals. Assume moreover that the ratios $\frac{\alpha_i}{\alpha_j}$ for $i\ne j$ are irrational. The the $\alpha_i$'s are linearly independent over $\mathbb{Q}$.

Proof: Let $a_i\in \mathbb{Q}$ so that $$\sum a_i \alpha_i = 0$$ Fix $i$, $1\le i \le k$. We have $$a_i = -\sum_{k \ne i} a_k \frac{\alpha_k}{\alpha_i}$$

Let a finite extension $L$ of $\mathbb{Q}$ containing all the $\alpha_i$ Taking traces on both sides we get $$d \cdot a_i = \sum_{k \ne i} a_k \operatorname{trace}^L_{\mathbb{Q}} \left(\frac{\alpha_k}{\alpha_i}\right )=0$$, since $\frac{\alpha_k}{\alpha_i}$ is a real irrational radical. So all the $a_i$ are $0$.

In words: incommensurable real radicals are linearly independent over $\mathbb{Q}$.

• This looks great, but I don't get how to get from $(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$ to $(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$. Any tips? – Robert Lewis Sep 16 '17 at 3:32
• @Robert Lewis: Thanks. Added some details. – Orest Bucicovschi Sep 16 '17 at 4:06

Somehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\mathbf Q (\omega)$, where $\omega$ is a primitive cubic root of $1$. Since $\mathbf Q(\sqrt[3] 2)$ and $\mathbf Q(\sqrt[3]3)$ have degree $3$ over $\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\sqrt[3]2)/k$ and $k(\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\in (k^{*})^{3}$. Norming down to $\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$.


The $${Y_{1}^{0}}^{*}Y_{1}^{0}$$ and $${Y_{1}^{1}}^{*}Y_{1}^{1}$$ functions are plotted above. The more important results from this analysis include (1) the recognition of an $$\hat{L}^2$$ operator and (2) the fact that the Spherical Harmonics act as an eigenbasis for the given vector space. As one can imagine, this is a powerful tool. Forgot password? When this Hermitian operator is applied to a function, the signs of all variables within the function flip. (ℓ+m)!Pℓm(cosθ)eimϕ.Y^m_{\ell} (\theta, \phi) = \sqrt{\frac{2\ell + 1}{4\pi} \frac{(\ell - m)! 4Algebraic theory of spherical harmonics. Plots of the real parts of the first few spherical harmonics, where distance from origin gives the value of the spherical harmonic as a function of the spherical angles, https://brilliant.org/wiki/spherical-harmonics/. 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A conducting sphere of radius RRR with a layer of charge QQQ distributed on its surface has the electric potential everywhere in space: V={14πϵ0QR2r3sinθcosθcosϕ,  r>R14πϵ0Qr2R3sinθcosθcosϕ,  rR \\ Spherical Harmonics and Linear Representations of Lie Groups 1.1 Introduction, Spherical Harmonics on the Circle In this chapter, we discuss spherical harmonics and take a glimpse at the linear representa-tion of Lie groups. \hspace{15mm} \ell & \hspace{15mm} m&\hspace{15mm} Y^m_{\ell} (\theta, \phi) \\ \hline For each fixed nnn and ℓ\ellℓ there are 2ℓ+12\ell + 12ℓ+1 solutions corresponding to the 2ℓ+12\ell + 12ℓ+1 choices of mmm at fixed ℓ.\ell.ℓ. So the solution can thus far be written in the form. When we consider the fact that these functions are also often normalized, we can write the classic relationship between eigenfunctions of a quantum mechanical operator using a piecewise function: the Kronecker delta. Spherical Harmonics are considered the higher-dimensional analogs of these Fourier combinations, and are incredibly useful in applications involving frequency domains. \hspace{15mm} 2&\hspace{15mm} -2&\hspace{15mm} \sqrt{\frac{15}{32\pi}} \sin^2 \theta e^{-2i\phi} \\ The first is determining our $$P_{l}(x)$$ function. \hspace{15mm} 2&\hspace{15mm} 1&\hspace{15mm} -\sqrt{\frac{15}{8\pi}} \sin \theta \cos \theta e^{i \phi} \\ The spherical harmonics. As the general function shows above, for the spherical harmonic where $$l = m = 0$$, the bracketed term turns into a simple constant. As such, this integral will be zero always, no matter what specific $$l$$ and $$k$$ are used. 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Information contact us at [ email protected ] or check out our page...


# Exercise 11.3.2

Suppose that we hash a string of $r$ characters into $m$ slots by treating it as a radix-128 number and then using the division method. We can easily represent the number $m$ as a 32-bit computer word, but the string of $r$ characters, treated as a radix-128 number, takes many words. How can we apply the division method to compute the hash value of the string without using more than a constant number of words of storage outside the string itself?

Yes, this follows pretty easily from the laws of modulo arithmetic. We need to observe that if the string is $s = \langle a_n, \ldots, a_1, a_0 \rangle$, then its hash $h(s)$ is going to be:

\begin{aligned} h(s) &= \left( \sum_{i=0}^{n}{a_i \cdot {128}^{i}} \right) \bmod m \\ &= \sum_{i=0}^{n}{ \Big( \left( a_i \cdot {128}^{i} \right) \bmod m \Big) } \\ &= \sum_{i=0}^{n}{ \Big( ( a_i \bmod m ) ( {128}^{i} \bmod m ) \Big) } \\ \end{aligned}

We can easily compute $a_i \bmod m$ without extra memory. To compute ${128}^{i} \bmod m$ without extra memory, we just need to observe that $k^i \bmod m = k(k^{i-1} \bmod m) \bmod m$, that is, we can compute the module for each power incrementally, without ever needing unbound memory.

### Python code

K = 128

def consthash(digits, m):
result = 0
power = 1

for d in reversed(digits):
result += ((d % m) * power) % m
result %= m
power = (power * K) % m

return result


# Thread: Differentiate using implicit differentiation

1. ## Differentiate using implicit differentiation

find dy/dx (3xy)^(1/2)=3x-2y

Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.

2. Originally Posted by math34
find dy/dx (3xy)^(1/2)=3x-2y

Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.

$\sqrt{3xy} = 3x - 2y$

So
$\frac{1}{2} \frac{1}{\sqrt{3xy}} ( 3y + 3xy') = 3 - 2y'$

$\frac{3y}{2\sqrt{3xy}} + \frac{3x}{2\sqrt{3xy}}y' = 3 - 2y'$

$\frac{3x}{2\sqrt{3xy}}y' + 2y' = 3 - \frac{3y}{2\sqrt{3xy}}$

$\left ( \frac{3x}{2\sqrt{3xy}} + 2 \right )y' = 3 - \frac{3y}{2\sqrt{3xy}}$

$y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2}$

$y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2} \cdot \frac{2 \sqrt{3xy}}{2 \sqrt{3xy}}$

$y' = \frac{6\sqrt{3xy} - 3y}{3x + 4\sqrt{3xy}}$ <-- To get your given answer, multiply top and bottom by $\sqrt{3xy}$ again.

We should rationalize this, but it is easier to use the original equation:
$\sqrt{3xy} = 3x - 2y$

$y' = \frac{6(3x - 2y) - 3y}{3x + 4(3x-2y)}$

$y' = \frac{18x - 12y - 3y}{3x + 12x - 8y}$

$y' = \frac{18x - 15y}{15x - 8y}$

-Dan

3. Hello, math34!

Find $\frac{dy}{dx}:\;\;(3xy)^{\frac{1}{2}} \:=\:3x-2y$

Would it be easier to first square both sides to get rid of the fractional exponent,
then differentiate implicitly? . . . absolutely!

Answer: (6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy . . . this can't be right!

Square: . $3xy \:=\:9x^2 - 12xy + 4y^2$

We have: . $3xy' + 3y \:=\:18x - 12xy' - 12y + 8yy'$

Then: . $15xy' - 8yy' \:=\:18x - 15y$

Factor: . $(15x - 8y)y' \:=\:18x - 15y$

Therefore: . $\boxed{y' \:=\:\frac{18x - 15y}{15x - 8y}}$ . . . which is Dan's answer.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

It can be done head-on . . .

Differentiate implicitly: . $\frac{1}{2}(3xy)^{-\frac{1}{2}}(3y + 3xy') \:=\:3 - 2y'$

We have: . $3y + 3xy' \:=\:2\sqrt{3xy}(3 - 2y')$

We are told that: . $\sqrt{3xy} \:=\:3x - 2y$

. . Hence, we have: . $3y + 3xy' \:=\:2(3x - 2y)(3 - 2y')$

Expand: . $3y + 3xy' \:=\:18x - 12xy' - 12y + 8yy'$

Simplify: . $15xy' - 8yy' \:=\:18x - 15y$

Factor: . $(15x - 8y)y' \:=\:18x - 15y$

Therefore: . $\boxed{y' \:=\:\frac{18x-15y}{15x-8y}}$

4. ## Thanks

I got the same answer as you and Dan but, still am clueless about how the given solution was obtained. I think this is a case of the experts making a mistake, while the layman suffers!!!

Thanks guys

5. I am not happy about the idea of squaring the equation, and here's why:

Say we have:
$y = x$

Square both sides:
$y^2 = x^2$

Now take the derivative:
$2yy' = 2x$

$y' = \frac{x}{y}$

Now, this certainly is a differential equation for the original function, but we also have another solution: y = -x. In this case the differential equation has more solutions than what we started with.

Just something to watch out for.

-Dan


# A sine wave equation?

1
Date created: Sat, Jul 17, 2021 3:51 PM
Date updated: Sun, Jun 26, 2022 6:06 PM

Content

Video answer: How to write a sine wave equation by looking at a graph | trigonometry

## Top best answers to the question «A sine wave equation»

Sine Wave… A general form of a sinusoidal wave is y(x,t)=Asin(kx−ωt+ϕ) y ( x , t ) = A sin ( kx − ω t + ϕ ) , where A is the amplitude of the wave, ω is the wave's angular frequency, k is the wavenumber, and ϕ is the phase of the sine wave given in radians.

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### 👋 What is a sine wave equation?

Sine Wave… A general form of a sinusoidal wave is y(x,t)=Asin(kx−ωt+ϕ) y ( x , t ) = A sin ( kx − ω t + ϕ ) , where A is the amplitude of the wave, ω is the wave's angular frequency, k is the wavenumber, and ϕ is the phase of the sine wave given in radians.

Video answer: Determining the equation for a sinewave from a plot

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