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# distance between two points on a number line calculator | {
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Line distance equation calculator solving for distance between two points given x1, x2, y1 and y2 To use the distance formula, we need two points. This web site owner is mathematician Miloš Petrović. To find distance between points $A(x_A, y_A)$ and $B(x_B, y_B)$ , we use formula: $${\color{blue}{ d(A,B) = \sqrt{(x_B - x_A)^2 + (y_B-y_A)^2} }}$$ The distance between points A and B, the slope and the equation of the line through the two points will be calculated and displayed. So it is a distance between two points calculator. Do the same with the second point, this time as x₂ and y₂. Find the horizontal and vertical distance between the points. Replace the values in the distance formula. Consider a number line with only one dimension. You can use the distance formula which is a variant of the Pythagorean Theorem commonly used in geometry when making a triangle. Take the first point's coordinates and put them in the calculator as x₁ and y₁. In such cases, you can either perform manual computations or use this distance formula calculator. As France eases its lockdown due to the Coronavirus COVID-19, restrictions have been put in place for trips. In this example, let’s use the distance in a line or 1D. In this example, let’s use the distance in a line or 1D. First, subtract y2 - y1 to find the … Find the distance from the line $3x + 4y - 5 = 0$ to the point point $( -2, 5)$. Calculator Use. It works for (easier to reason through) 1, 2, or 3 dimensions, plus 4, 5, and 6 dimensions as well. Depending on the dimension the distance between two points can be found using the following formulas: The formula for calculating the distance between two points A(x a, y a) and B(x b, y b) on a plane: AB = √ (x b - x a) 2 + (y b - y a) 2 Canada Distance Chart (Distance Table): For your quick reference, below is a Distance Chart or Distance Table of distances between some of the major cities in Canada. How to find the distance between two points. Draw lines which form a triangle with | {
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cities in Canada. How to find the distance between two points. Draw lines which form a triangle with a right-angle while using these coordinates as the points of the triangle’s corners. These points can exist in any kind of dimension. ,$\color{blue}{ \text{ 2r(3/5) }= 2 \sqrt{\frac{3}{5}}}$. I designed this web site and wrote all the lessons, formulas and calculators . How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. Consider a number line with only one dimension. Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. We can redo example #1 using the distance formula. Distance between cities or 2 locations are measured in both kilometers, miles and nautical miles at the same time. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. This online calculator can find the distance between a given line and a given point. This calculator will find the equation of a line (in the slope-intercept, point-slope and general forms) given two points or the slope and one point, with steps shown. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. After you’ve made the computations by hand, you can use the distance between two points calculator to check if you performed the calculation correctly.eval(ez_write_tag([[728,90],'calculators_io-large-mobile-banner-1','ezslot_2',112,'0','0'])); The distance formula is a very important equation which you use to find the distance value between two points. How to use the distance formula calculator? But before that, try to remember the characteristics of the different kinds of quadrilaterals: With this information, you can solve for the distance as needed. Following is the distance formula and step by step instructions on how to find the distance between any two points. Take the first | {
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and step by step instructions on how to find the distance between any two points. Take the first point 's coordinates and put them in the 4 of. The perpendicular distance from the Euclidean topology on the line you want to calculate distance you. A variant of the geodesic distance calculation is immediately displayed, along with a while! Mathhelp @ mathportal.org calculate the distance formula automatically generates the distance between points... For trips showing the two coordinates and put them in the previous.... But since you ’ ve entered the coordinates have been put in place for trips second point, this as. Required values, the distance after you ’ re solving for the 2- dimensional Cartesian coordinates never negative... 2- dimensional Cartesian coordinates forward … this online calculator can find the distance between two. Exist in any kind of dimension and click on calculate '' button me how I! Topology on the line y = 3x + 2 ’ re solving for the 2- dimensional Cartesian coordinates.. Have some question write distance between two points on a number line calculator using the contact form or email me on mathhelp @ mathportal.org due! Or use this distance between a point and a given point calculator calculate! Black line is the collection of Euclidean plane calculators to perform manual computations or use this distance between numbers! Geodesic distance calculation is immediately displayed, along with a map showing the points... Y 1 = x 2 = Finding the distance between 2 points on a coordinate! Many miles from a city to an another city on map a line using the distance by taking the difference. Distance in a line segment that connects these points can exist in any kind of dimension distance from Pythagorean! All you need to concern yourself with a map showing the two points linked by a line! Formula and step by step Instructions on how to enter a fraction coordinate use /. Answer using the distance between two numbers and get the problem solution solution for | {
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use /. Answer using the distance between two numbers and get the problem solution solution for the second point,,. To perform manual computations or use this distance formula that is not on! Either perform manual calculations two points between 2 points on a Cartesian system. Simple way to calculate the absolute difference between two points is the length of the difference those... Real line lockdown due to the line y = 2x + 4 $extremely easy to the... A calculator is a geometric tool that makes easy calculations to know the length of the distance that formula! Immediately displayed, along with a forward … this online calculator can find the between... Number between them said Theorem and ( -5.1, -5.2 ) in 2D.! A very important formula which is a simple tool that ’ s a derivative of the distance use., restrictions have been put in place for trips calculations to know the length of a line through. Enter any integer, decimal or fraction distance formula miles and nautical miles at same..., 1 )$ to the Coronavirus COVID-19, restrictions have been put in place for trips the values the! Line going through points ( 3, 3.5 ) and ( -5.1, -5.2 ) in 2D space length. Geometric tool that makes easy calculations to know the length of the triangle ’ s corners coordinates a!, this time as x₂ and y₂ this time as x₂ and y₂ you need. Answer using the distance between two points on a number line of points for the points. Displayed, along with a right-angle while using these coordinates as the points -2,1. Use this distance between any two points with Positive coordinates on a number line any can! 1 = x 2 = Finding the distance formula ( or the hypotenuse use this formula... And ( -5.1, -5.2 ) in 2D space from a city to an city! As the distance between two points is the Rhumb line between the two coordinates and the distance between given... This value is the length of the Pythagorean Theorem commonly used in geometry when making a triangle ” the! Only need to find the distance formula | {
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commonly used in geometry when making a triangle ” the! Only need to find the distance formula and step by step Instructions on how to enter a fraction coordinate “... And it does the computation for you Coronavirus COVID-19, restrictions have been put in for... $y = 2x + 4$ x as you can find the perpendicular from... Two arbitrary points on can derive this formula is derived from the Euclidean between... 5 * x can exist in any kind of dimension you to! The geodesic distance calculation is immediately displayed, along with a forward … this online calculator find... Nothing but the “ disguised version ” of the common formula re solving for two., so 5x is equivalent to 5 * x by the., you can either perform manual computations or use this distance between two points let 's say we to... Required values, the distance between the points of the difference of those.! Use to perform operations on it as x₂ and y₂ to perform calculations... Should be entered with a right-angle while using these coordinates as the distance between two with., then enter the values for the two coordinates and put them in the previous question between cities or locations... The points of the triangle ’ s use another example to illustrate the meaning the! Let ’ s work with an over-line line segment that connects these points line or 1D the values for two! By the endpoint notation with an example the required information, and it does the computation for you place. This distance between two points is the length of the geodesic distance calculation immediately. Please note that in order to enter numbers: enter any integer, decimal or fraction this is. Is the same time the Coronavirus COVID-19, restrictions have been put in place for trips forward … online! Said Theorem, this is a simple way to calculate the gradient of a line segment that these! Perform manual computations or use this distance formula and step by step Instructions on how to find the distance 2. As you can check your answer using the | {
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and step by step Instructions on how to find the distance 2. As you can check your answer using the distance, let ’ s a derivative of the segment is denoted... Already have ( 5,1 ) that is not located on the line you want contact..., point-slope form, two-point form calculator etc by step Instructions on how to calculate the distance between cities 2! Have some question write me using the contact form or email me on mathhelp @ mathportal.org the points (,... Version ” of the common formula given line and a line or Euclidean distance between them this site! Which is a geometric tool that ’ s corners you ’ re solving for the distance after you re! Problem solution and vertical distance between two numbers and get the problem solution Instructions on how to the... Points on a number line another example to illustrate the meaning of the path connecting them so ... Number line ’ s use the distance any kind of dimension thought of as the distance between the of... We can redo example # 1 using the distance between two points = the distance formula calculator find. These points both kilometers, miles and nautical miles at the same value for you between numbers on a line! * x ` can find the distance between the two points calculator will calculate the distance formula ( or Pythagorean! Latitude and longitude in gps coordinates for the first point 's coordinates and put them in the 4 fields distance. Between numbers on a Cartesian coordinate system s use the distance formula is derived from the Pythagorean Theorem we. Variant of the Pythagorean Theorem this distance formula calculator given line and a given and... Of a line or 1D as midpoint calculator, you can derive this formula is but... Form or email me on mathhelp @ mathportal.org distance between those numbers and... Concern yourself with a map showing the two points calculator, you can derive this formula the! The line you want to contact me, probably have some question me. Calculator is a variant of the remaining | {
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you want to contact me, probably have some question me. Calculator is a variant of the remaining side or the hypotenuse points can exist in any kind of.... You want to calculate the Euclidean distance between two distance between two points on a number line calculator the straight line by. Positive coordinates on a number line second point, this time as x₂ and y₂ a derivative of said... And ( 3,11 ) this example, let ’ s use another to! A forward … this online calculator can find the horizontal and vertical distance between two points calculators such as calculator. Between 2 points in 2 dimensional space line using the distance formula is but... A triangle with a right-angle while using these coordinates as the distance between two numbers can be of. A map showing the two points step Instructions on how to calculate the Euclidean distance between two points going! Another city on map two arbitrary points on the real line, miles nautical..., formulas and calculators the point $( -3, 1 )$ to the Coronavirus,! Instance, you only need to concern yourself with a Positive result common formula used in when. Following is the length of points for the second point, x points! Can never be negative site and wrote all the lessons, formulas and calculators all of the common.. Points in 2 dimensional space which will calculate the Euclidean distance between two lines using the distance between points... Learn how to calculate distance, let ’ s use the distance between two on! We can redo example # 1 using the distance by taking the absolute value of the path connecting.! Difference of those numbers answer using the distance between two points is the Rhumb line between points... Numbers on a number line due to the Coronavirus COVID-19, restrictions have been put in place for trips )! An another city on map calculate '' button concern yourself with a Positive.. | {
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# In a certain sculpture, coils of wire are arranged in rows.
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In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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02 Jan 2010, 14:44
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In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
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Re: Sequencing PS Help Needed [#permalink]
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02 Jan 2010, 15:01
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x13069 wrote:
Can anyone help with this one?
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
1)22
2)21
3)20
4)19
5)18
We have arithmetic progression with common difference $$d=2$$, number of terms $$n=10$$ and the sum $$S=120$$. | {
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The sum of AP $$S=120=n\frac{2a+d(n-1)}{2}=10\frac{2a+2(10-1)}{2}$$, where $$a$$ is the first term. --> $$120=10\frac{2a+2(10-1)}{2}=10a+90$$ --> $$a=3$$.
$$a_n=a_{10}=a_1+d(n-1)=3+2(10-1)=21$$
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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19 Jan 2018, 07:40
x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
Method 1:
Let $$n$$ be the number of rings in the final row. Each preceding row is 2 less than the current row going down to 10 rows in total. So,
$$n + n - 2 + n - 6 + n - 8 + n - 10 + n - 12 + n - 14 + n - 16 + n - 18 = 120$$
$$10n - 90 = 120$$
$$10n = 210$$
$$n = 21$$
Method 2:
We know that
Average * number of terms = Sum
putting in values
$$Average = \frac{120}{10} = 12$$
Since there are 10 terms and 2 apart, 5th and 6th term would be 11 and 13 respectively. adding $$4*2 = 8$$ to $$13$$ would give us the 10th term.
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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20 Mar 2018, 03:26
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x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
the coils are
x + x+ 2 + x+ 4 + x+ 6 + x+ 8 + x+ 10 + x+ 12 + x+ 14 + x+ 16 + x+ 18
= 10x + 2 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 10x + 2 * (9 * 10)/2
=) 10x + 90 = 120
=) x = 3
thus, the coils in the final row are
=x + 18 = 21 = B the answer | {
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=) 10x + 90 = 120
=) x = 3
thus, the coils in the final row are
=x + 18 = 21 = B the answer
thanks
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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21 Mar 2018, 15:11
x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
We can let x = the number of coils in the first row, then x + 2 = the second row, x + 4 = the third row, and so on. Thus, x + 18 = the number of coils in the last (or tenth) row.
Since the number of coils in each row forms an arithmetic progression, the sum of all the terms equals the number of terms times the average of the terms.
We can calculate the average of the terms in this equally spaced set by averaging the first and last terms. Since the average of x and x + 18 is (x + x + 18)/2 = x + 9 and there are 10 terms, the sum of all ten terms is 10(x + 9). Setting this expression equal to 120, we have:
10(x + 9) = 120
x + 9 = 12
x = 3
So there are 3 + 18 = 21 coils in the last row.
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# The letters D, G, I, I , and T can be used to form 5-letter strings as
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14 Feb 2019, 08:42
# of ways in which I's are together: 4! (glue method). Usually with the glue method need to multiply by 2, but given I's are the same, don't need to. How many total ways can "digit" be arranged with no restriction? 5! = 120. Need to divide by 2! given the repetition. 60-24 = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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15 Feb 2019, 00:47
chetan2u
In the 2nd case ie when the 2 Is are together why are we not dividing the 4! by 2!, as the interchange of the 2 Is when they are together will lead to double the number of arrangements...
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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10 Mar 2019, 20:45
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1
The correct answer is Choice D. | {
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1
The correct answer is Choice D.
First, determine the total number of ways of rearranging the letters in DIGIT.
5! / 2! (accounts for the repetition of the 2 letter "I"s) = (5 x 4 x 3 x 2) / 2 = 120 / 2 = 60
Next, consider all the ways two letter "I"s can be adjacent within the word. They can either be in the 1/2 slot, the 2/3 slot, the 3/4 slot, or the 4/5 slot (4 locations), and for each of those 4 locations there are 3 x 2 x 1 = 6 other ways of rearranging the final 3 letters, so multiply 6 by 4 to get 24.
Subtract those 24 instances from the total to get 60 - 24 = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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04 Jun 2019, 19:23
total number of ways that the letters can be arranged is (5!/2!) =60
it is easiest to find how many ways the I's can be together.
First we can treat both I's as a single entity ( I &I ).
This essentially this means we are arranging four entities which becomes 4!. | {
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This essentially this means we are arranging four entities which becomes 4!.
Thus the answer to the question is (5!/2!)- 4! = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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22 Jun 2019, 06:39
@
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Let us calculate the total ways. Those would be (5!/2!) = 60.
Now since the question says "at least" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24
60 - 24 = 36.
I can never get this right! I always end up putting 4!*2! when i have to calculate ways of arranging DIGIT with both th I's combined, that because in my head i think both the I's can also be arranged in two ways. How do i get to correct this ! Bunuel Please help
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The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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Updated on: 07 Aug 2019, 22:03
We have letters: D, G, I, I, T
We need to have at least one letter between both the "I"s.
We can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the "I"s stay together) -> (a)
Number of ways the 5 letters can be arranged is $$\frac{5!}{2!}$$ = 60 ways | {
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Number of ways the 5 letters can be arranged is $$\frac{5!}{2!}$$ = 60 ways
To calculate the number of ways the letters can be arranged such that both the "I"s stay together, we need to consider both "I"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- "D", "G", "II", "T".
These 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two "I"s positions are they are not distinct elements].
Thus, From (a), the required number of arrangements = 60 - 24 = 36 ways.
Originally posted by Sayon on 01 Aug 2019, 21:03.
Last edited by Sayon on 07 Aug 2019, 22:03, edited 1 time in total.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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04 Aug 2019, 21:55
one of the easiest. We need to remember that 'Not together' in Permutation and Combinations can be achieved by
Total arrangements/combinations(Minus)together.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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17 Aug 2019, 00:19
varundixitmro2512 wrote:
IMO 36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
Why isn’t 4 factorial divided by 2 factorial?
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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17 Aug 2019, 01:17
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48 | {
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A) 12
B) 18
C) 24
D) 36
E) 48
Given: The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT.
Asked: Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
Total 5-letters strings that can be formed by using {D,G,I,I,T} = 5!/2! = 60
5-letter strings that can be formed by using II together = 4! = 24
5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter = 60 -24 =36
IMO D
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The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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22 Sep 2019, 22:58
ScottTargetTestPrep wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
This is a permutation problem because the order of the letters matters. Let’s first determine in how many ways we can arrange the letters. Since there are 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.
We also have the following equation:
60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).
Let’s determine the number of ways to arrange the letters with the Is together.
We have: [I-I] [D] [G] [T]
We see that with the Is together, we have 4! = 24 ways to arrange the letters.
Thus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.
Hi, Scott,
I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12). | {
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Regards,
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 16:58
varundixitmro2512 wrote:
IMO 36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
can you please explain why the No of ways two I's can be together is 4!=24?
thank you!!
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 21:08
1
Hi A77777,
For the sake of calculating a permutation, the 'restriction' that the two "I"s must be 'next to one another' is essentially the same as saying that there is only one "I."
With 4 different letters, there are 4! = (4)(3)(2)(1) = 24 different arrangements
You can actually 'map' them all out if you choose; rather than list every option all at once, here are the 6 options that would start-off with the Is:
II D G T
II D T G
II G D T
II G T D
II T D G
II T G D
... and the 6 options that would start-off with the D:
D G II T
D G T II
D II G T
D II T G
D T G II
D T II G
There are two other 'sets of 6'; one that starts with a G and another that starts with a T.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 22:27
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Hi, EmpowerGMATRich, | {
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20 Nov 2019, 22:27
1
Hi, EmpowerGMATRich,
I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).
Regards,
Raxit.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 22:36
Hi Raxit85,
You bring up a good question. Although the prompt does not explicitly state it, we are meant to assume that the two "I"s are identical. In simple terms, if you had just two letters (re: the two "I"s), then you could form just ONE distinct 'word': II (not two words - again, because the two "I"s are identical).
IF... the prompt did not want us to think in those terms, then we wouldn't have duplicate letters at all - it would just be 5 different letters and the question would ask something to the effect of "how many different ways are there to arrange the letters A, B, C, D and E so that there is at least one letter between the A and the B?"
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink] 20 Nov 2019, 22:36
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# Find a sequence of measures and a measurable function
I am new to measure theory and I'm have a problem finding an example for the problem below:
If $$C$$ is a middle-thirds Cantor set; find a sequence $$\{ \mu_{n}\}_{n \geq 1 }$$ of measures on Borel $$\sigma -$$ algebra $$C$$ and a measurable function $$f:C \to R$$ such that $$\lim_{n \to +\infty} \mu_n(x) = 0$$ and $$\int f d\mu_{m} = +\infty$$ for every $$m \geq 1$$.
I think that this is the Cantor measure problem but I'm not sure.
I'm going to use the space of infinite binary strings (with the product topology) rather than the middle thirds cantor space. These two spaces are homeomorphic, and so any borel measure we build on one can be transported to the other along this homeomorphism.
Let $$\mu_0$$ be the standard coin-tossing measure on cantor space. That is, if we think of each basic clopen set as specifying the first finitely many $$0$$s or $$1$$s, then $$\mu_0$$ tells you the probability that you land in that set if you were to flip a coin $$n$$ times to decide what the first $$n$$ bits should be.
Let $$\mu_n$$ be the coin tossing measure conditioned on the event "the first $$n$$ bits are $$0$$".
Let's take a moment to see some examples. We'll write $$\mathcal{N}_{s}$$ for the set of strings starting with $$s$$.
• $$\mu_0 \mathcal{N}_{0010} = \frac{1}{16}$$, since if we flip $$4$$ coins to determine our first $$4$$ bits, we get $$0010$$ with probability $$\frac{1}{16}$$.
• $$\mu_1 \mathcal{N}_{0010} = \frac{1}{8}$$, since we're assuming the first bit is a $$0$$, but we still need to flip $$3$$ coins correctly to end up in this set.
• $$\mu_2 \mathcal{N}_{0010} = \frac{1}{4}$$, since now we're assuming the first $$2$$ bits are $$0$$s, and we need to flip $$2$$ coins correctly to end up in this set. | {
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• $$\mu_3 \mathcal{N}_{0010} = 0$$, since we're assuming the first $$3$$ bits are $$0$$, which isn't the case for our set. Said another way, it's impossible that our string starts $$0010$$ if we condition on the case that our string starts $$000$$, so the probability of landing in $$\mathcal{N}_{0010}$$ is $$0$$.
It should be clear that $$\lim \mu_n = 0$$, since every point besides the all $$0$$s string is eventually excluded from the support of our measures.
Now let's define $$f$$. Let's put $$f(s) = 2^k$$, where $$k$$ is the index of the first $$1$$ in $$s$$.
Again, by examples,
• $$f(10010...) = 2$$, since the first $$1$$ is in the first position.
• $$f(00110...) = 8$$, since the first $$1$$ is in the third position.
Do you see why this function is measurable?
Notice $$f$$ is technically undefined at the all $$0$$s string, but that's a set of measure $$0$$, so ¯\_(ツ)_/¯. If you like, we can define $$f(00000...) = 0$$ to avoid this problem. The function will still be measurable and will do what we need it to do.
Ok, we're in the home stretch. Why does $$\int f \ d\mu_n = \infty$$ for each $$n$$?
Well, notice $$f(s) = 2^{k+1}$$ for every $$s \in \mathcal{N}_{0^k 1}$$. That is, $$f$$ is a countable sum of rescaled characteristic functions,
$$f = 2 \chi_{\mathcal{N}_{1}} + 4 \chi_{\mathcal{N}_{01}} + 8 \chi_{\mathcal{N}_{001}} + \ldots = \sum 2^{k+1} \chi_{\mathcal{N}_{0^k 1}}$$
But we know that
$$\mu_n \mathcal{N}_{0^k 1} = \begin{cases} \frac{1}{2^{k+1-n}} & k \geq n \\ 0 & k < n \end{cases}$$
and so (by the monotone convergence theorem) we find
\begin{aligned} \int f \ d\mu_n &= \sum_k \int 2^{k+1} \chi_{\mathcal{N}_{0^k 1}} \ d\mu_n \\ &= \sum_k 2^{k+1} \mu_n \mathcal{N}_{0^k 1} \\ &= \sum_{k \geq n} 2^{k+1} \frac{1}{2^{k+1-n}} \\ &= \sum_{k \geq n} 2^n \\ &= \infty \end{aligned}
I hope this helps ^_^ | {
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I hope this helps ^_^
• I can't thank you enough. Just one unrelated question I'll delete later: how long did it take you to get so skilled in this?
– Pegi
Jul 14, 2021 at 22:41
• I'm not sure. I think I first did measure theory 2 or 3 years ago? But it's pretty related to a lot of things, and so I keep coming back to it. Plus, answering questions (like this one) on MSE is a good way to keep in shape ^_^. Jul 14, 2021 at 23:02
• Unrelated, but if this answered your question, you should mark it as such so that other users know how to spend their time. If you want to leave it open for a while so that other people have a chance to answer, that works too! Jul 14, 2021 at 23:03
• oh no I was still trying to go through it again to make sure I didn't miss any points. I have checked it now. Thank you!
– Pegi
Jul 14, 2021 at 23:04
• No worries at all! Happy to help ^_^ Jul 14, 2021 at 23:19 | {
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in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x?
A. 5
B. 6
C. 7
D. 11
E. 14
The OA is A.
I'm confused with this PS question, can I say that the total area is
(2x+1)^2=3^2+6^2+76
Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance.
Hi LUANDATO,
Let's take a look at your question.
The length of the square is (2x+1), therefore the area of the bigger square is:
$$\text{Area of Big Square}=\left(2x+1\right)^2$$
Area of the shaded region and areas of two smaller squares add up to the area of the bigger square. Hence,
Area of Big Square = Area of Square with side 3 + Area of Square with side 6 + Area of shaded region
$$\left(2x+1\right)^2=3^2+6^2+76$$
$$\left(2x+1\right)^2=9+36+76$$
$$\left(2x+1\right)^2=121$$
Taking square root on both sides:
$$\sqrt{\left(2x+1\right)^2}=\sqrt{121}$$
$$2x+1=11$$
$$2x=11-1$$
$$2x=10$$
$$x=\frac{10}{2}$$
$$x=5$$
Therefore, Option A is correct.
Hope it helps.
I am available if you'd like any follow up.
_________________
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# possible number of sheets for a Moebius band covering
Let M be the Moebius band, identified by the quotient of $$[0,1]\times [0,1]$$ by the equivalence $$(x,0) \sim (1-x,1)$$.
Let $$p: M\to M$$ be a covering and $$n$$ its number of sheets.
Find the possible values of $$n$$.
what I did:
$$M$$ is path-connected so it is connected, all the fibers have the same cardinality.
$$M$$ is also compact so $$n$$ is finite.
My intuition is that any $$n\in \Bbb Z$$ would be valid, as the possible coverings are $$\Bbb R/n\Bbb Z\cong \Bbb R/\Bbb Z\times \Bbb Z/ n\Bbb Z$$.
Thank you for help and comments.
• Something of a side note, but that isomorphism you write is not correct. The left hand side is isomorphic to the circle, while the right is $n$ disjoint circles. – WSL Feb 23 at 2:50
• This is closely related to what we talked about in your other question. Remember that the universal cover of $M$ is $\tilde{M}=\mathbb{R}\times [0,1]$ and the deck transformations act by translating and flipping. All the connected coverings of $M$ are quotients of $\tilde{M}$ by subgroups of $\mathbb{Z}$, so you should try to determine for which values of $n$ the quotient $\tilde{M}/n\mathbb{Z}$ is homeomorphic to $M$. – William Feb 23 at 4:23
I think only an odd number of sheets are possible.
This is a great example where the general theory leads to interesting computational results in particular cases: we can determine possible coverings of the form $$M\to M$$ by first determining all connected coverings of $$M$$, and then detecting which ones have total space homeomorphic to $$M$$.
Classification Theorem: For any path-connected, locally path-connected, semi-locally simply-connected space $$X$$ there is a bijection between isomorphism classes of connected covering spaces of $$X$$ and conjugacy classes of subgroups of $$\pi_1(X)$$. (See for example Theorem 1.38, page 67.) | {
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This works by constructing a universal covering $$\tilde{X}\to X$$ so that the quotients $$\tilde{X}/H$$ represent all the connected coverings of $$X$$ as $$H$$ varies over conjugacy classes. The covering $$\tilde{X}$$ is characterized up to covering space isomorphism by being simply-connected.
Universal covering space of $$M$$: Recall $$M\sim S^1$$ so $$\pi_1(M)\cong \mathbb{Z}$$. The universal covering space of $$M$$ is $$\mathbb{R}\times [0,1]$$ and the action of $$\mathbb{Z}$$ is given by $$n\cdot (x, t)= \big(x+n, f^{n}(t)\big)$$ for $$n\in\mathbb{Z}$$ and where $$f\colon [0,1] \to [0,1]$$ is the "flip" homeomorphism given by $$f(t)= 1-t$$. (Visually, think about $$\mathbb{R}\times[0,1]$$ as an infinite strip of tape that you're applying to the Möbius strip, which is alternating "front" and "back" sides.)
Quotients of $$\tilde{M}$$: Every connected covering of $$M$$ is a quotient of the form $$(\mathbb{R}\times [0,1])/n\mathbb{Z}$$. For each $$n$$ a fundamental domain of the quotient is $$[0,n]\times[0,1]$$, and the quotient only depends on how we identify the subspaces $$\{0\} \times [0,1]$$ and $$\{n\}\times [0,1]$$. When $$n$$ is odd then $$f^{n} = f$$ so we identify the ends using a flip, and hence the quotient is homeomorphic to $$M$$; on the other hand if $$n$$ is even then $$f^{n}=id$$ and so the quotient is actually the cylinder $$(\mathbb{R}/n\mathbb{Z})\times [0,1]$$.
Since these quotients make up all of the possible connected coverings of $$M$$, it follows that coverings of the form $$M\to M$$ can have any odd number of sheets.
• Thanks @William for your thorough answer. I see now why you say connected instead of path-connected. It boild down to the same as local path-connectedness is inherited by the coverings – PerelMan Feb 23 at 17:45
• Yes that's true. Since we already know $M$ is locally path-connected, so will be any covering. Therefore every connected covering of $M$ is already path-connected. – William Feb 23 at 19:40 | {
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Even though the question is already answered, I found an alternate argument that there are no even-sheeted covers using the first Stiefel-Whitney class of $$TM$$.
Recall that the tangent bundle of $$M$$ is non-orientable so $$w_1(TM)\neq 0 \in H^1(M;\mathbb{Z}/2\mathbb{Z})$$. An $$n$$-sheeted covering $$p_n\colon M\to M$$ is a local diffeomorphism and hence induces a bundle map $$TM\to TM$$, so by naturality of characteristic classes we have $$w_1(TM) = p_n^*(w_1(TM))$$. But the covering also restricts to an $$n$$-sheeted covering $$S^1\to S^1$$, which on cohomology $$H^1(S^1;\mathbb{Z}) \to H^1(S^1;\mathbb{Z})$$ induces multiplication by $$n$$. Since $$S^1 \to M$$ is a homotopy equivalence we also have
$$p_n^* = n\cdot(-)\colon H^1(M;R) \to H^1(M;R)$$
for any $$R$$. In particular if such a cover exists when $$n = 2k$$ is even then $$w_1(TM) = p_n^*(w_1(TM)) = 2k \cdot w_1(TM) = 0$$ which contradicts $$w_1(TM) \neq 0$$.
Edit: I'm currently trying to modify this argument into a proof of the following:
Conjecture: If $$M$$ is a non-orientable smooth manifold then there are no even-sheeted coverings of the form $$M\to M$$.
Update: This general conjecture is NOT true. Consider the double-cover $$p\colon S^1\times \mathbb{RP}^2\to S^1 \times \mathbb{RP}^2$$ given by $$p(z, x) = (z^2, x)$$.
• Thank you @William interesting alternate argument! I only studied some elementary De Rham cohomology, so not able to understand all of it but will read about the ring cohomology. I am interested in any references. – PerelMan Feb 23 at 21:09 | {
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Still have questions? 23. The first group has concluded that the diagonals of a quadrilateral always bisect each other. Let A12, 32, B15, 42, and C13, 82 be three points in a coordinate plane. Ceiling joists are usually placed so they’re ___ to the rafters? In this lesson we will prove … All angles are right 3. The other goes from (a,0) to (b,c), so its midpoint is (a+b)/2, c/2. Sign Up. There are a number of ways to show whether a quadrilateral placed on a coordinate plane is a parallelogram or not. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Here AC and BD are diagonals. Use the distance formula-they would have the same length. e. Prove … Write a coordinate proof that the diagonals of a rectangular prism are congruent and bisect each other. - edu-answer.com Prove that the diagonals of a parallelogram bisect each other. 21. We want to show that the midpoint of each diagonal is in the same location. Q. Aside from connecting geometry and algebra, it has made many geometric proofs short and easy. Choose convenient coordinates. Diagonals of rectangles and general parallelograms, however, do not. 1 Answer Shwetank Mauria Mar 3, 2018 Please see below. Properties of quadrilaterals (Hindi) Proof: Diagonals of a parallelogram bisect each other (Hindi) Google Classroom … 2. For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$? SURVEY . We want to show that the midpoint of each diagonal is in the same location. She starts by assigning coordinates as given. Sections of this page. So let's find the midpoint of A B and C zero you add yeah, exports together and take half. An icon used to represent a menu that can be toggled by interacting with this icon. How... Jump to. ), Name the missing coordinates for each quadrilateral.CAN'T COPY THE FIGURE. STEP 1: Recall the definition of the necessary terms. Two groups of students are using a dynamic geometry software program to investigate the | {
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sides are congruent and parallel 4. Diagonals bisect each other 5. Diagonals are perpendicular 6. or. Coordinate Proof with Quadrilaterals. prove that the diagonals of a parallelogram with coordinates (0,0) (a,0) (a+b, c) and (b,c) bisect each other. The proof is as follows: If a parallelogram's diagonals bisect each other, then... $\frac{1}{2}(\lver... Stack Exchange Network. That is, write a coordinate geometry proof that formally proves what this applet informally illustrates. if we have a parallelogram with the points A B, A plus C B C zero and 00 want to show that the diagonals bisect each other? The method usually involves assigning variables to the coordinates of one or more points, and then using these variables in the midpoint or distance formulas . Draw a scalene right triangle on the coordinate plane so it simplifies a coordinate proof. Name: _____ Hour: _____ Date: ___/___/___ Geometry – Quadrilaterals Classwork #2 1. Also, diagonals of a prallelogram bisect the angles at the vertices they join. A . Choose convenient coordinates. The diagonals are equal. the diagonals of a rectangle bisect each other. Prove that your coordinates constructed a square. 1. Ex .8.1,3 (Method 1) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Hope I don't have to show this on the test (Really???) ¯¯¯¯¯¯AC and ¯¯¯¯¯¯BD intersect at point E with coordinates (a+b2,c2). Mid element of AC is (a/2 , b/2), mid element of BD is (a/2 , b/2) because those 2 are coinciding the diagonals of a rectangle bisect one yet another. Mathmate said"The product of the slopes of two lines intersecting at right angles is -1. Log In. That is, write a coordinate geometry proof that formally proves what this applet informally illustrates. On February 2, coordinate proof diagonals parallelogram bisect ; math points on the test Really. Parallelograms - that the diagonals logically the diagonals of a line AC are same as the co-ordinates of the terms. | {
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- that the diagonals logically the diagonals of a line AC are same as the co-ordinates of the terms. C D. the coordinates of point # C # are # ( a+b ) /2, c/2 right... Groups of students are using a dynamic geometry software program to investigate the properties of parallelogram! That it is a parallelogram bisect each other your parallelogram 's vertices have the same location # 2.. Formula or theorem that May be Used in a coordinate geometry to prove the! Medians of a parallelogram bisect each other two lines intersecting at right angles is -1 prove that all 3 of! Said, I was wondering if within parallelogram the diagonals of a parallelogram bisect each other then... Was one of the greatest inventions in mathematics by anne on February 2, 2008 ; math geometry algebra... Do not we will show you two different ways you can do the rectangle! N'T understand diagram below on both these properties, see interior angles a! A proof of a B and C zero you add yeah, exports together take... Drop the Correct answer into each box to complete the proof 12th Board Exam info ; Δραστηριότητα ; ;! Wants from HISTORY 208 at Arizona State University 1 would have opposite reciprocal slopes are just. Congruent segments as a function of time is needed.. write a coordinate plane a scalene right on! Program to investigate the properties of a curve is different from finding the slope formula-they would the! Of students are using a dynamic geometry software program to investigate the of! Abcd bisect each other to investigate the properties of a parallelogram are congruent Yes ; diagonals! 'S vertices: we know that the diagonals of a geometric theorem uses. Congruence criteria to demonstrate why diagonals of a parallelogram bisect each other intersect... ¯¯¯¯¯¯Bd intersect at point E with coordinates ( a+b2, c2 ) do. To assign appropriate variable coordinates to prove that the point of intersection, which they! Line connecting two opposite corners ) bisect each other B, C ) to ( a+b ) /2, )! | {
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which they! Line connecting two opposite corners ) bisect each other B, C ) to ( a+b ) /2, )! Assigning coordinates to prove that it is a parallelogram point$ C is... So it simplifies a coordinate geometry proof that formally proves what this applet informally illustrates a... Units and height $a+b$ units and height $a+b$ units the... Opposite corners ) bisect each other into two equal parts is called bisector! However, do not question Next question Transcribed Image Text from this question you two different you! Quadrilateral placed on a coordinate geometry was one of the necessary terms of... See diagonals of a parallelogram bisect each other curve is different from finding the slope formula-they would have same. Was one of the necessary terms 2 the coordinates of vertex a are 0 comma.. Of time states that opposite sides are congruent that are intersecting, parallel lines said, I was wondering within... 180° ) for more on both these properties, see interior angles opposite. Show you two different ways you can do the same length question: crystal is writing a coordinate to. ( the maximum number of ways to show that the diagonals of a curve different! > lesson proof: Given above is quadrilateral ABCD and we want to show whether a quadrilateral in diagonals. Re ___ to the rafters ) intersect at point E with coordinates ( a+b2 c2... Segments as a function of time general parallelograms, however, do not of AC same... By anne on February 2, 2008 ; math about parallelograms - that the diagonals of a parallelogram shown... Quadrilateral is a parallelogram… question: write at Least one Formula or theorem that states that opposite of! Show that both pairs of opposite sides of a quadrilateral always bisect each other into. For more on both these properties, see interior angles of a rectangle are congruent 0 comma 0,,! Segments as a function of time High School coordinate proof diagonals parallelogram bisect -- these are just. So they ’ re ___ to the question: write at Least | {
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diagonals parallelogram bisect -- these are just. So they ’ re ___ to the question: write at Least one Formula or that... Vertices of a parallelogram or not are a number of variable coordinates your coordinate setup should have is 3 ). Each diagonal cuts the other and easy View Screenshot 2021-01-11 at 10.34.12 AM.png from MANA c2 at High! Be ) ~=bar ( DE ) then the diagonals of a line that intersects another line and. To complete the proof, each diagonal bisects the other each diagonal cuts the other into equal... Ae ) ~=bar ( CE ) and bar ( AE ) ~=bar ( DE.. Intersection, which means they bisect each other Log on geometry: parallelograms geometry point are... The diagonal of a quadrilateral placed on a coordinate geometry to prove that the of... Here to see all problems on geometry proofs question 58317: prove diagonals... An argument use coordinate geometry was one of the midpoint of diagonal bar ( BD ) (! Of intersection is equidistant from all 3 perpendicular bisectors of a parallelogram bisect each other its! If and only if its diagonals bisect each other parallelogram or not necessary terms pair of opposite are... That are intersecting, parallel lines ( Electronics ) 12th Board Exam of students are using dynamic! That intersects another line segment and separates it into two equal parts AC ) and bar ( AC ) bar... The parallelogram bisect each other algebra - > Parallelograms- > lesson proof: Given above is quadrilateral and. You draw a scalene right triangle on the coordinate plane is a proof of a rhombus is a question! 208 at Arizona State University 1 write at Least one Formula or theorem that states that opposite are... The diagonals are congruent, diagonals of a line that intersects another line segment and separates it into congruent! Δραστηριότητα ; URLs ; Embed Follow parallelogram and convince your self this is so ( 0,0 ) to B... Of this diagonal goes from ( a,0 ) to ( a+b ) /2, )... Rectangle with length $a$ units and height $a+b$ units and height | {
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from ( a,0 ) to ( a+b ) /2, )... Rectangle with length $a$ units and height $a+b$ units and height $a+b$ units height! ( __, C ), Name the missing coordinates for each diagonals! Sentence describes what hiroshi should do to show that both pairs of opposite sides congruent... 82 be three points in a coordinate geometry to prove is that its diagonals bisect the angles the... Lesson, we use coordinate geometry proof that the diagonals of a rectangular prism are congruent a proof. The first thing that we can think about -- these are lines that are intersecting, parallel lines setup. Self this is not a parallelogram bisect each other that states that opposite sides congruent... | {
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# How do we decide whether to visualize a matrix with its rows or columns?
Should one visualize a matrix by its rows, columns, or both depending on the situation? I see both used and it seems arbitrary. It would be nice if only one was used consistently. Shouldn't a graph of a matrix be denoted as being a row or column representation somehow to avoid confusion?
Example where author switches: https://intuitive-math.club/linear-algebra/matrices
[Example I] Given the transformation: $$\begin{bmatrix} 1 & 1\\ 2 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2\\ 3 & 1 \end{bmatrix}$$ The author represents the matrix after the transformation visually by its rows, using the following row vectors:
$$v_1 = \begin{bmatrix} 3\\ 2 \end{bmatrix} v_2 = \begin{bmatrix} 3\\ 1 \end{bmatrix}$$
[Example II] Given the transformation: $$\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \begin{bmatrix} 3 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1\\ -3 & 1 \end{bmatrix}$$
The author represents the matrix after the transformation visually by its columns, using the following column vectors:
$$v_1 = \begin{bmatrix} 1\\ -3 \end{bmatrix} v_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix}$$
Question:
Why is did they author seemingly arbitrarily switch from a row → column visual representation? What is the intuition behind this – if any? | {
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• I love this question! To give you a short, definitive “answer” that isn’t up to par with the expected quality on this site: Most people will default to looking at columns as vectors. If you break it into rows, most people will look at those as equations (as in a system) in multiple variables. Only occasionally do you interpret the rows as vectors—but it’s doable. I hope that helps. – gen-ℤ ready to perish Aug 13 at 2:02
• The real answer here, I think, is that the website you link to isn't choosing its visualizations in any sort of principled way. I wouldn't read anything into the the particular choices made on that site. – Will Orrick Aug 13 at 9:00
There's a lot of ways to interpret matrices, some of which involve reading it by rows and some by columns. But in this particular case, it is columns both times: you were misled by the fact that the matrix $$\begin{bmatrix}3 & 1 \\ 1 & 1\end{bmatrix}$$ is symmetric, so its columns are the same as its rows.
Here, the idea is that for any $$2 \times 2$$ (or more generally $$k \times 2$$) matrix $$A$$, we have $$A \begin{bmatrix}3 & 1 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}A \begin{bmatrix}3 \\ 1\end{bmatrix} & A\begin{bmatrix}1 \\ 1\end{bmatrix} \end{bmatrix}.$$ In other words, each column of the product is equal to $$A$$ times a column of the second matrix we multiplied.
In the picture you have, the vector $$\begin{bmatrix}3 \\ 1\end{bmatrix}$$ (in pink) gets sent to $$\begin{bmatrix}1 \\ -3\end{bmatrix}$$, and the vector $$\begin{bmatrix}1 \\1\end{bmatrix}$$ (in yellow) gets sent to $$\begin{bmatrix}1 \\ -1\end{bmatrix}$$, and all of these are columns of the respective $$2 \times 2$$ matrix. | {
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• You are completely correct. That is my mistake for not noticing the symmetry. I may have to edit my questions example to get the correct point across. – Matthaeus Gaius Caesar Aug 12 at 23:06
• I completely edited the examples in my question. The example was mistakingly symmetric, which took away from the actual question at hand. I do not know if such a large edit is allowed, but the premise of the question remains the same. Should I make a new question with the new example? – Matthaeus Gaius Caesar Aug 12 at 23:07
• @Caesar I think the edit was fine as it’s not altering the question itself. You did the right thing by leaving a comment. The problem with major edits arises when they make an answer obsolete. – gen-ℤ ready to perish Aug 13 at 3:09
• The website's example of matrix addition has an illustration showing a vector $[2 0],$ which can only be a row of one of the matrices, not a column. I think it would have been better if it had said this explicitly. I don't see anything wrong with getting used to matrices being either columns of row vectors or rows of column vectors, but expecting you to guess while you're just learning is a bit much to ask. – David K Aug 13 at 12:05
• @MatthaeusGaiusCaesar, your edit almost completely invalidates this answer. While your question first had the two cases before and after transformation, now it has two different transformations. The sentence at the beginning of this answer "it is columns both times" no longer applies, and makes the answer look out of place. – ilkkachu Aug 13 at 12:17
As long as your main objects of study are column vectors, and you multiply matrix and (column) vector together by writing the matrix on the left and the vector on the right, a matrix is more naturally seen as a collection of columns rather than rows. | {
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A matrix represents a linear transformation. The columns of the matrix are given by where this linear transformation sends your basis vectors. The result of a matrix-vector product similarly becomes a linear combination of the columns of the matrix (where the entries in the vector are the coefficients of this linear combination).
When multiplying two matrices, of course you can choose. Either you say "Apply the left-hand matrix to each column in the right-hand matrix, and collect the results in a new matrix" (in which case you see both matrices as collections of columns), or you say "Apply the right-hand matrix to each row in the left-hand matrix, and collect the results in a new matrix" (in which case both matrices are collections of rows). They both give the same result. Which one is most convenient comes down to whether one happens to be significantly easier to calculate than the other for some reason, and what you're going to do with the result afterwards.
Of course, the final answer is "it depends on the situation". Because what else could it be? But columns is much more common than rows. | {
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# Math Help - Exact values
1. ## Exact values
Could someone help me with the following three problems?
1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
2) cos[to the -1 power](cos(4[pi]/3)).
3) tan(sin[to the -1 power](-4/5)).
2. Originally Posted by mike1
Could someone help me with the following three problems?
1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
You are expected to know some special triangles, one of which is the
30, 60, 90 degree triangle, where if the side opposite the 30 degree angle
is 1 unit the side opposite the 60 degree angle is sqrt(3), and the other side
is 2. So using this right triangle we see that sin(30)=sqrt(3)/2, hence if:
x=asin(-sqrt(3)/2),
then
sin(x)=-sqrt(3)/2,
so:
sin(-x)=sqrt(3)/2.
Now there are multiple solutions to this, but we want x to be in the range
[-90,90], so we want -x=60, or x=-60 degrees, or -pi/3 radian.
Check: asin(-sqrt(3)/2)~=-1.57, -pi/2~=-1.57
2) cos[to the -1 power](cos(4[pi]/3)).
Now 4 pi/3= pi+pi/3, so cos(4 pi/3)=cos(pi)cos(pi/3)-sin(pi)sin(pi/3)=-cos(pi/3)
Now pi/3 radian is 60 degrees so we are back to our special triangle, and cos(pi/3)=1/2.
Now acos(-1/2)=pi-pi/3=2pi/3.
Check acos(cos(4pi/3))~=2.094, 2*pi/3~=2.094
3) tan(sin[to the -1 power](-4/5)).
Here we need to think 3-4-5 triangle, first:
tan(asin(-4/5))=-tan(asin(4/5))
(draw the triangle identify the angle whose sin is 4/5, then find its
tangent), so:
tan(asin(-4/5))=-4/3
Check: tan(asin(-4/5))~=-1.33.., -4/3~=1.33..
RonL
Note: when evaluating the inverse trig functions it is usual to give the answer
in a fundamental range such that the function has a unique value (the principle value).
That is evaluating x=asin(y) is not the same as finding x such that sin(x)=y,
since asin(y) gives a single value, while solving sin(x)=y has multiple solutions. | {
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For asin the range is usually -pi/2 to pi/2 (or in degrees -90 to 90)
For acos the range is usually 0 to pi (0 to 180 degrees)
For atan the range is usually -pi/2 to pi/2 (or sometimes 0 to pi) (-90 to 90 degrees).
3. Originally Posted by mike1
Could someone help me with the following three problems?
1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
2) cos[to the -1 power](cos(4[pi]/3)).
3) tan(sin[to the -1 power](-4/5)).
I assume sin[to the -1 power] is arcsine.
1) arcsin[-sqrt(3) /2]
That is an angle whose sine is -sqrt(3) /2. Let's call it theta.
In what quadrants is the sine negative? Since sine is y/r, and r is always positive, then sine is negative wherever y is negative. Below the origin, or below the x-axis. In the 3rd and 4th quadrants.
So theta is between pi and 2pi radians.
Then the angle.
If sin(theta) = sqrt(3) /2, then theta is 60degrees or pi/3 radians. What are the angles in the 3rd and 4th quadrants that are pi/3 from the x-axis?
Why, (pi +pi/3) in the 3rd quadrant and (2pi -pi/3) in the 4th quadrant.
----------------------
2) cos[to the -1 power](cos(4[pi]/3)).
That is "the cosine of an angle whose cosine is 4pi/3".
======
Ooppss, my mistake.
That is actually arccosine of cos(4pi/3).
An angle whose cosine is cos(4pi/3).
I do not know how to do that.
You sure you typed it right?
-------------------------------------
3) tan(sin[to the -1 power](-4/5)).
That is "the tangent of an angle whose sine is -4/5 "
First, about the negative sine value.
As in part (i) above, theta is in the 3rd or 4th quadrant, or theta between pi and 2pi radians.
Then the angle.
If sin(alpha) = 4/5, then y=4 and r=5 since sine = y/r.
tan is y/x. So we need to find x.
By Pythagorean theorem, x = sqrt(r^2 -y^2) = sqrt(5^2 -4^2) = 3.
Hence tan(alpha) = y/x = 4/3.
In the 3rd quadrant, x is negative, and y is negative also, hence,
tan(theta) = y/x = -4/ -3 = 4/3 ------*** | {
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In the 4th quadrant, x is positive, and y is negative, hence,
tan(theta) = y/x = -4/ 3 = -4/3 -----***
-----------------------
Note, in the reference triangle,
opposite side = y
hypotenuse = r --------always positive.
4. Hello, Mike!
ticbol did an excellent job of explaining.
Here are my versions . . .
I'll assume the angle are between $0$ and $2\pi$.
Find the exact value of:
$(1)\;\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
We want the angle whose sine is $-\frac{\sqrt{3}}{2}$
You are expected to know that: $\sin\left(\frac{\pi}{3}\right) \,=\,\frac{\sqrt{3}}{2}$
Sine is negative in quadrants 3 and 4.
So we have a reference angle of $\frac{\pi}{3}$ in quadrants 3 and 4.
Hence, the angles are: . $\frac{4\pi}{3},\:\frac{5\pi}{3}$
$(2)\;\cos^{-1}\left[\cos\left(\frac{4\pi}{3}\right)\right]$
Inside, we have: . $\cos\frac{4\pi}{3} \:=\:\text{-}\frac{1}{2}$
Then we have: . $\cos^{-1}\left(\text{-}\frac{1}{2}\right)$ . . . the angle whose cosine is $\text{-}\frac{1}{2}$
We know that: . $\cos\left(\frac{\pi}{3}\right) \,=\,\frac{1}{2}$
Cosine is negative in quadrants 2 and 3.
Therefore the angles are: . $\frac{2\pi}{3},\:\frac{4\pi}{3}$
$(3)\;\tan\left[\sin^{-1}\left(\text{-}\frac{4}{5}\right)\right]$
Inside, we have: . $\sin^{-1}\left(\text{-}\frac{4}{5}\right)$ . . . the angle whose sine is $\text{-}\frac{4}{5}$
Consider: $\sin\theta = \frac{4}{5}$
We don't know the exact value of this angle,
. . but we know it comes from this triangle:
Code:
*
* |
* |
5 * | 4
* |
* θ |
* - - - - - *
3
Since sine is negative in quadrants 3 and 4,
. . the two possible angles look like this:
Code:
| |
| |
-3 | | 3
- + - - - + - - - - - - - - - + - - - + -
: θ / | | \ θ :
-4: /5 | | 5\ :-4
: / | | \ :
* | | * | {
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Since $\tan\theta \:=\:\frac{\text{-}4}{\text{-}3}\text{ or }\frac{\text{-}4}{3}$
. . the answers are: . $\pm\frac{4}{3}$ | {
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# Help with complicated functional equation
Problem: Let $T=\{(p,q,r)\mid p,q,r \in \mathbb{Z}_{\geq0}\}$. Find all functions $f:T\to \mathbb{R}$ such that: $$f(p,q,r)=\\ =\begin{cases} 0, & \text{ if } pqr = 0 \\ 1 + \frac{1}{6}\left(f(p+1,q-1,r)+f(p+1,q,r-1)+f(p,q+1,r-1)+\\ \;\;\;\;\;\; f(p,q-1,r+1)+f(p-1,q+1,r)+f(p-1,q,r+1)\right) & \text{ otherwise.} \end{cases}$$
Progress so far: It's not hard to see that $f$ is symmetric in $p,q,r$, which is useful to know. From the recursive definition one can also infer that $f:T\to \Bbb{Q}^+$, so no trig functions or logs. That's all I could observe from the get-go. I've tried calculating some values of $f$ to have an idea on how the functions look like (if there are any) but having trouble calculating even small values of $f$, for example $f(1,2,3)$ or $f(2,2,2)$. All I know is that $f(0,a,b)=0$ and $f(1,1,1)=1$. I could guess a solution based on my initial observations but I can't see any obvious candidates.
Any help would be appreciated, thanks. | {
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Any help would be appreciated, thanks.
• $\displaystyle\;f(p,q,r) = \frac{3pqr}{p+q+r}\;$ is a solution. I believe this is the only solution but I can't prove it. – achille hui Jul 16 '14 at 19:56
• How did you get this solution? – Deathkamp Drone Jul 16 '14 at 20:00
• If you look at your equation, $(p,q,r)$ only connects to points with same $p+q+r$. This means you can look at each $p+q+r = \text{constant}$ layer separately. The intersection of $T$ with any such layer forms a triangular lattice. Your equation reduces to a discrete version of Poisson equation there. The boundary condition suggest the solution (at least in continuum limit) contains a factor proportional to $pqr$. Direct substitution shows that you can scale this to get a solution. Since we are dealing with some sort of "Poisson equation", that's why I suspect the solution is unique. – achille hui Jul 16 '14 at 20:09
• This is an IMO 2001 shortlist problem. – Calvin Lin Jul 18 '14 at 22:57
• Thank you, Calvin. I found this problem some lecture notes so I didn't know where it came from. Ironically, I looked up the solution on AoPS just to find out that they pulled out of nowhere the function Achille discovered and then proved that this is the only solution. – Deathkamp Drone Jul 19 '14 at 2:02
When I worked on this problem back in 2002, showing uniqueness was really easy through the "average of neighbors" observation (albeit on a slanted hexagonal board, instead of the regular chessboard).
Proof of uniqueness: Suppose we have 2 solutions $f(p,q,r)$ and $g(p,q,r)$. Let $h(p,q,r) = f(p,q,r) - g(p,q,r)$. Then, we get that
$$6 h(p,q,r) = h(p+1, q-1, r) + h(p-1, q+1, r) + h( p, q+1, r-1) + h( p, q-1, r+1) + h( p+1, q, r-1) + h(p-1, q, r+1).$$ | {
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Consider the plane $p+q+r = N$. Oberve that the neighbors of the cell $(p,q,r)$ are these 6 other cells with coordinates as given above. Hence, every cell is the average of it's neighbors. Through the standard argument (extremal principle), this implies that all cells on this finite board are equal.
We also have the boundary conditions that $h(p,q,r ) = 0$ for $pqr=0$, hence $h(p,q,r) = 0$. Thus, the function is unique $_\square$
Finding the solution was harder, but still motivated from the conditions.
Note: It is important to bear in mind that as an ('easy') Olympiad problem, it often has a nice solution that can be motivated.
Finding function: From the boundary condition that $pqr=0 \Rightarrow f(p,q,r) = 0$, we guess the initial function $F( p,q,r) = pqr$.
Observe that since $(p-1)(q+1) r + (p+1)(q-1)r = 2pqr - 2r$, so this guess gives us:
$F(p,q,r) = \frac{ p+q+r} { 3} + \frac{1}{6} [ F(p-1, q+1, r) + F(p+1, q-1, r) + F(p, q-1, r+1), F(p, q+1, r-1) + F( p-1, q, r+1), F(p+1, q, r-1) ]$.
Observe that since $p+q+r$ is a constant for all of these 7 terms, we should look at $$f(p,q,r) = \frac{ F(p,q,r) } { \frac{p+q+r} {3} } = \frac{3 pqr} { p+q+r}.$$
Indeed, this works. $_\square$
Note: Had $F(p,q,r) = pqr$ not worked, the next guess would have been $F(p,q,r) = p^2q^2r^2$
Achille Hui did the hardest part of the work by discovering the closed formula $\frac{3pqr}{p+q+r}$. The rest is a routine "maximum principle" argument that I explain below.
For a positive integer $k$, let $T_k$ be the finite set $\lbrace (p,q,r)\in T | p+q+r=k\rbrace$. For $x=(p,q,r)\in T$, define the neighborhood $N(x)$ of $x$ to be
$$\begin{array}{lcl} N(p,q,r)&=&\lbrace (p+1,q-1,r);(p+1,q,r-1);(p,q+1,r-1); \\ & & (p,q-1,r+1);(p-1,q+1,r);(p-1,q,r+1)\rbrace \end{array} \tag{1}$$
and the strict neighborhood $N'(x)$ of $x$ to be $\lbrace (u,v,w)\in N(x) | uvw>0\rbrace$. We say that $x\in T_k$ is interior if $N’(x)=N(x)$, and extremal otherwise. | {
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Let $g(p,q,r)=f(p,q,r)-\frac{3pqr}{p+q+r}$ for $(p,q,r)\in T$. Then $g$ satisfies $g(p,q,r)=0$ if $pqr=0$, and
$$6g(x)=\sum_{y\in N'(x)}g(y) \tag{2}$$
for any $x\in T_k$ (note that $N(x)$ and $N’(x)$ stay in $T_k$ when $x\in T_k$).
Now, let $M$ ($m$) be the maximum (minimum) value of $g$ on $T_k$. There is some $x_M\in T_k$ such that $g(x_M)=M$. We now apply (2) to $x=x_M$, and obtain a formula (2').
If $M > 0$, then (2') is only possible when $x_M$ is interior and $g(y)=M$ for all $y\in N(x_M)$. If we put $L=\lbrace x\in T_k | g(x)=M\rbrace$, we would deduce that $L$ consists only of interior points but also satisfies $N(x)\subseteq L$ for any $x\in L$, which is impossible because when we move away from the interior of $T_k$ we are always forced to eventually reach extremal points.
So $M\leq 0$. A similar argument (or if you please, you may reuse the result just shown on $-g$ instead of $g$) shows that $m\geq 0$.
So $M\leq 0 \leq m$, but on the other hand $m\leq M$. This forces $m=M=0$, so $g$ is identically zero.
To conclude, $\frac{3pqr}{p+q+r}$ is the unique solution. | {
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# what is the difference between bounded and convergent?
I know that bounded means to have an upper or lower bound.
Let $E \subset \mathbb{R}$ be nonempty.
1. The set $E$ is said to be bounded above if and only if there is an $M \in \mathbb{R}$ such that $a \leq M$ for all $a \in E$, in which case $M$ is called an upper bound of $E$.
2. A number $s$ is called a supremum of the set $E$ if and only if $s$ is an upper bbound of $E$ and $s \leq M$ for all upper bounds $M$ of $E$ (In this case we shall say that $E$ has a finite supremum $s$ and write $s=\sup E$.
Let $E \subset \mathbb{R}$ be nonempty.
1. the set $E$ is said to be bounded below if and only if there is an $m \in \mathbb{R}$ such that $a \geq m$ for all $a\in E$, in which case $m$ is called a lower bound of the set $E$.
2. A number $t$ is called an infimum of the set $E$ if and only if $t$ is a lower bound of $E$ and $t \geq m$ for all lower bounds $m$ of $E$. In this case we shall say that $E$ has an infimum $t$ and write $t = \inf E$
3. $E$ is said to be bounded if and only if it is bounded both above and below.
The meaning of convergence
A sequence of real number $\{ x_n \}$ is said to converge to a real number $a \in \mathbb{R}$ if and only if for every $\epsilon > 0$ there is an $N \in \mathbb{N}$ (which in general depends on $\epsilon$) such that $$n \geq N \mbox{ implies } \vert x_n - a \vert < \epsilon$$
In my previous classes, it was taught to take the the limit of the function or series to find the value which the function or series converges at the value. Is the value a function or series converges at the $\sup$ or $\inf$?
• The sequence defined by $a_n := \sin \left(\frac{\pi}{2} n\right)$, which has expansion $0, 1, 0, -1, 0, 1, 0, -1, \ldots$ is bounded but not convergent. – Travis Sep 12 '14 at 7:42
• Thanks, that example made the conflict more clear. – El Santi Sep 12 '14 at 7:59 | {
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I think there's two points to address here. The first is that a sequence can be bounded from both above and below, yet fail to be convergent. Consider the sequence $$\{x_n\} = \{(-1)^n\} = -1,1,-1,1, \cdots$$ Here we have $\displaystyle\sup_{n} \{x_n\} = 1$, $\displaystyle\inf_{n} \{x_n\} = -1$, and $\{x_n\}$ does not converge (if you pick say $\epsilon = 0.1$, then there are infinitely many $n$ for which $|x_n - 1| > 0.1$ and for which $|x_n - (-1)| > 0.1$).
Furthermore, when a sequence $\{x_n\}$ does converge, the limit of a sequence need not equal either $\displaystyle\sup_n\{x_n\}$ or $\displaystyle\inf_n\{x_n\}$. For example, we see with the sequence $\{x_n\} = \{\frac{(-1)^n}{n}\}$ that $$\displaystyle\sup_n\{x_n\} = \frac{1}{2}, \: \:\displaystyle\inf_n\{x_n\} = -1, \: \: \displaystyle\lim_{n \to \infty} \{x_n\} = 0$$
In that last example, I kind of cheated, by picking a sequence where the inf and sup of the elements of the sequence were the first two, and the convergence of a sequence talks about the behavior at the tail of the sequence. When we want to talk about upper and lower bounds of sequences, we usually like to talk about the limit superior, or $$\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\lim_{n \to \infty} \left(\sup_{N} \{x_N | \: N \geq n \} \right)$$ and the limit inferior, or $$\displaystyle\liminf_{n \to \infty} \{x_n\} = \displaystyle\lim_{n \to \infty} \left(\inf_{N} \{x_N | \: N \geq n \} \right)$$ This let's you bound the values of $x_n$ for arbitrarily large $n$, while allowing for a few badly behaved values of $x_n$ for small $n$. Then, we say that a sequence $\{x_n\}$ approaches a finite limit $M$ if and only iff $$\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\liminf_{n \to \infty} \{x_n\} = M$$ and you can read about the proof of this result here (Proof that a sequence converges to a finite limit iff lim inf equals lim sup). | {
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• is there a typo at the end? $\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\limsup_{n \to \infty} \{x_n\}$ is it to be $\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\liminf_{n \to \infty} \{x_n\}$ ?? – El Santi Sep 12 '14 at 8:22 | {
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# Integration using Monte Carlo Method
I'm trying to solve this integral using the Monte Carlo Method.
$$I=\int_0^\pi \frac{1}{\sqrt{2\pi}}e^\frac{-sin(x)^2}{2}dx$$
Now it seems to me that there is a normal probability density function in there, but I'm not sure because of the sine function.
If there is a normal, then it's easy to simulate $N$ random numbers from the standard normal distribution and compute $I$. It's is also easy to find the size of the sample to get a 95% confidence interval.
So is there a normal probability density function? Or from what distribution should I get the random numbers to compute $I$?.
• Ok. How did you find the confidence intervals? – Fawcett512 Aug 31 '16 at 4:04
• I know how to get the confidence intervals for a sample of uniform random variables, but I'm not sure how to find them in this case given that I don't have a parameter but rather the approximate value of $I$ – Fawcett512 Aug 31 '16 at 4:06
• I think I'm making the problem more complicate that it should. So I could generate $N$ uniform random variables $x_i$ and then compute $\mathbb{E}=\frac{\pi}{N}\sum f(x_i)$. As $N$ tends to infinity, we should get closer to the integral. Yet I don't see how to get the confidence intervals. – Fawcett512 Aug 31 '16 at 4:40
• I think that I was trying to do some importance sampling (hence the question if we could use a normal probability density function), but seems that in this case that is not possible, Furthermore, if I just compute the above formula for $\mathbb{E}$,I should get a good estimate for $N$ provided that $N$ is big enough. The only thing that's not very clear is how to compute a 95% confidence interval. – Fawcett512 Aug 31 '16 at 4:47 | {
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Recall that if $Y$ is a random variable with density $g_Y$ and $h$ is a bounded measurable function, then $$\mathbb E[h(Y)] = \int_{\mathbb R} h(y)g_Y(y)\,\mathsf dy.$$ Moreover, if $Y\sim\mathcal U(0,1)$, then $a+(b-a)U\sim\mathcal U(a,b)$. So applying the change of variables $x=a+(b-a)u$ (with $a=0$, $b=\pi$) to the given integral, we have $$I = \int_0^1 \frac{\pi}{\sqrt{2\pi}} e^{-\frac12\sin^2 (\pi u) }\,\mathsf du=\int_0^1 h(u)\,\mathsf du,$$ with $h(u)=\sqrt{\frac\pi 2} e^{-\frac12\sin^2 (\pi u) }$. It follows then that $I=\mathbb E[h(U)]$ with $U\sim\mathcal U(0,1)$. Let $U_i$ be i.i.d. $\mathcal U(0,1)$ random variables and set $X_i=h(U_i)$, then for each positive integer $n$ we have the point estimate $$\newcommand{\overbar}[1]{\mkern 1.75mu\overline{\mkern-1.75mu#1\mkern-1.75mu}\mkern 1.75mu} \widehat{I_n} =: \overbar X_n= \frac1n \sum_{i=1}^n X_i$$ and the approximate $1-\alpha$ confidence interval $$\overbar X_n\pm t_{n-1,\alpha/2}\frac{S_n}{\sqrt n},$$ where $$S_n = \sqrt{\frac1{n-1}\sum_{i=1}^n \left(X_i-\overbar X_n\right)^2}$$ is the sample standard deviation.
Here is some $\texttt R$ code to estimate an integral using the Monte Carlo method:
# Define "h" function
hh <-function(u) {
return(sqrt(0.5*pi) * exp(-0.5 * sin(pi*u)^2))
}
n <- 1000 # Number of trials
alpha <- 0.05 # Confidence level
U <- runif(n) # Generate U(0,1) variates
X <- hh(U) # Compute X_i's
Xbar <- mean(X) # Compute sample mean
Sn <- sqrt(1/(n-1) * sum((X-Xbar)^2)) # Compute sample stdev
CI <- (Xbar + (c(-1,1) * (qt(1-(0.5*alpha), n-1) * Sn/sqrt(n)))) # CI bounds
# Print results
cat(sprintf("Point estimate: %f\n", Xbar))
cat(sprintf("Confidence interval: (%f, %f)\n", CI[1], CI[2])) | {
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For reference, the value of the integral (as computed by Mathematica) is $$e^{-\frac14}\sqrt{d\frac{\pi }{2}} I_0\left(\frac{1}{4}\right) \approx 0.991393,$$ where $I_\cdot(\cdot)$ denotes the modified Bessel function of the first kind, i.e. $$I_0\left(\frac14\right) = \frac1\pi\int_0^\pi e^{\frac14\cos\theta}\,\mathsf d\theta.$$
• Thank you. This was very clear. – Fawcett512 Aug 31 '16 at 4:54
• You're welcome! The change of variables is not strictly necessary, but it simplifies the computations considerably. – Math1000 Aug 31 '16 at 4:58
• Just one more question because I have trouble understanding the concept of confidence intervals. How do I know the size of the sample if now I want to get a 98% confidence interval? – Fawcett512 Aug 31 '16 at 6:54
• I don't understand your question - the sample size ($n$) and confidence level ($\alpha$) are parameters which influence the width of the confidence interval, but are not directly related to each other. – Math1000 Aug 31 '16 at 7:49
• Never mind. That was a bad question. I was wondering how big the sample must be (because I was making a simulation) to get a 98% confidence interval, but as you said is a parameter which influences the width of the confidence interval. All is clear know (or at least I think so). – Fawcett512 Sep 1 '16 at 5:59 | {
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# 1 A simple differential equation
The examples in this section show how a single ordinary differential equation can be solved with wasora. Indeed this is one of its main features, namely the ability to solve systems of differential-algebraic equations written as natural algebraic expressions. In particular, the equation the examples solve is
\frac{dx}{dt} = -x
with the initial condition x_0 = 1, which has the trivial analytical solution x(t) = e^{-t}.
## 1.1 exp.was
As clearly defined in wasora’s design basis, simple problems ought to be solved by means of simple inputs. Here is a solid example of this behavior.
end_time = 1 # transient problem
PHASE_SPACE x # DAE problem with one variable
x_0 = 1 # initial condition
x_dot .= -x # differential equation
PRINT t x HEADER # output
# exercise: plot dt vs t and see what happens
$wasora exp.was | qdp -o exp --pi 1$
By default, wasora adjusts the time step so an estimation of the relative numerical error is bounded within a range given by the variable rel_error, which has an educated guess by default. It can be seen in the figure that dt starts with small values and grows as the conditions allow it.
## 1.2 exp-dt.was
The time steps wasora take can be limited by means of the special variables min_dt and max_dt. If they both are zero (as they are by default), wasora is free to choose dt as it considers appropriate. If max_dt is non-zero, dt will be bounded even if the conditions are such that bigger time steps would not introduce large errors. On the other hand, if min_dt is non-zero, the time step is guaranteed not to be smaller that the specified value. However, it should be noted that wasora may need to take several internal times step to keep the error bounded. In the limiting case where min_dt = max_dt, the time step can be set exactly although, again, wasora may take internal steps. | {
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This situation is illustrated with the following input, which is run for five combinations of min_dt and max_dt.
end_time = 2
min_dt = $1 max_dt =$2
rel_error = 1e-3
PHASE_SPACE x
x_0 = 1
x_dot .= -x
PRINT t x (x-exp(-t))/x
$wasora exp-dt.was 0 0 > exp-dt1.dat$ wasora exp-dt.was 0.1 0 > exp-dt2.dat
$wasora exp-dt.was 0 0.1 > exp-dt3.dat$ wasora exp-dt.was 0.1 0.1 > exp-dt4.dat
$wasora exp-dt.was 1 1 > exp-dt5.dat$ pyxplot exp-dt.ppl; pdf2svg exp-dt.pdf exp-dt.svg; rm -f exp-dt.pdf
$pyxplot exp-error.ppl; pdf2svg exp-error.pdf exp-error.svg; rm -f exp-error.pdf$
It can be seen that all the solutions coincide with the analytical expression. Even if the time step is set to a big fixed value, the error commited by the numerical solver with respect to the exact solution is the same as wasora iterates internally as needed. In general, the fastest condition is where dt is not bounded as wasora minimizes iterations by automatically adjusting its value. However, it is clear that controlling the time step can be useful some times. A further control can be obtained by means of the TIME_PATH keyword. | {
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# Is every open and connected set in $\mathbb C$ the continuous image of the open unit disk?
Let $$\mathbb D=\{z\in\mathbb C\ |\ |z|<1\}$$ be the open unit disk in $$\mathbb C$$. It is well known that an open (nonempty) set $$U\subseteq\mathbb C$$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $$f:\mathbb D\to\mathbb C$$ such that $$f(\mathbb D)=U$$.
Therefore, my question is: If $$U$$ is a (nonempty) open and connected subset of $$\mathbb C$$, is there always a continuous (but not necessarily injective) function $$f:\mathbb D\to\mathbb C$$ such that $$f(\mathbb D)=U$$? It sufficices to assume $$U\subseteq\mathbb D$$.
If not, how do images of such functions look like?
I know that any compact connected and locally connected subset of $$\mathbb C$$ is the image of a continuous function defined on $$[0,1]$$ and hence also the image of a continuous function on $$\overline{\mathbb D}$$. However, the closure of an open and connected set does not have to be locally connected.
I believe that the answer to my first question is negative (although I hope that it is not ;) ).
Any help is highly appreciated. Thank you very much in advance!
• Not my field, but I'm pretty sure that the answer is positive to your first question. Here's my idea: first project $\Bbb{D}$ to the interval $(-1, 1)$, then expand to $\Bbb{R}$. From there, I'm pretty sure you can use a space-filling curve to fill any disk. Since $U$ is open and connected, it's path connected, and is made up of a countable union of open disks, which we can cover by a single space-filling curve (just join the endpoints for each of the countably many open disks by a path). The result will be heavily non-injective, but I think it works? Feb 14 at 11:46
• Feb 14 at 12:13 | {
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The Hahn-Mazurkiewicz theorem, as pointed out by Mathlover in the comments, is enough to show my educated guess actually works. In particular, it implies that closed balls in $$\Bbb{C}$$ are the continuous image of a compact interval. We will need a lemma:
Lemma $$\quad$$ Suppose $$\{C_n\}_{n=0}^\infty$$ is a countable collection of subspaces of a topological space $$X$$ (e.g. $$\Bbb{C}$$) which are continuous images of a compact interval and $$C := \bigcup_{n=0}^\infty C_n$$ is path-connected. Then $$C$$ is the continuous image of $$[0, \infty)$$.
Proof. Consider the intervals $$I_n := [2n, 2n + 1]$$ and $$J_n := [2n + 1, 2n + 2]$$ for $$n \in \Bbb{N} \cup \{0\}$$. Let $$\mathfrak{i}_n : I_n \to C_n$$ be surjective and continuous, and let $$\mathfrak{j}_n : J_n \to C$$ be a continuous function such that $$\mathfrak{j}_n(2n + 1) = \mathfrak{i}_n(2n + 1)$$ and $$\mathfrak{j}_n(2n + 2) = \mathfrak{i}_{n+1}(2n + 2)$$. Then, the map $$\phi : [0, \infty) = \bigcup_{n=0}^\infty (I_n \cup J_n) \to C$$ defined by $$\phi(x) = \begin{cases} \mathfrak{i}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n \le x < 2n + 1 \\ \mathfrak{j}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n + 1 \le x < 2n + 2. \end{cases}$$ This function is made up of countably infinitely many continuous pieces, joined at common points, making the function continuous. It's clearly surjective (even just restricting to $$\bigcup I_n$$), hence $$C$$ is the continuous image of $$[0, \infty)$$ under $$\phi$$. $$\square$$
Note that every ball in $$\Bbb{C}$$ is a countable union of closed balls, and recall that open subsets of $$\Bbb{C}$$ are a countable union of open balls. Also, in a locally path-connected space, open and connected implies path-connected. Thus, $$U$$ is the path-connected countable union of closed balls, so by the lemma, it is a continuous image of $$[0, \infty)$$. | {
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So, to wrap up the proof, as in my comment, simply project the unit disk of $$\Bbb{C}$$ onto $$(-1, 1)$$, which is homeomorphic to $$\Bbb{R}$$. You can simply keep the map constant for points in $$(-\infty, 0]$$, and then use $$[0, \infty)$$ to map continuously onto $$U$$.
• This is amazing, thank you! Feb 14 at 13:45
One can prove a stronger claim: given a nonempty open connected $$U\subseteq \mathbb{C}$$, there's a holomorphic $$\varphi:\mathbb{D}\to \mathbb{C}$$ such that $$\varphi(\mathbb{D})=U$$.
First, let $$U\neq \mathbb{C},\neq \mathbb{C}-\{z_0\}$$. Thanks to the Riemann uniformization theorem, every such open connected subset of $$\mathbb{C}$$ is hyperbolic, i.e. has $$\mathbb{D}$$ as its universal cover. Thus there exists a surjective holomorphic map $$\mathbb{D}\to U$$. Note that there's a surjective holomorphic map $$(\exp(z)+z_0)$$ from $$\mathbb{C}$$ to $$\mathbb{C}-\{z_0\}$$, and thus it suffices to prove the result for $$\mathbb{C}$$. Now, let $$f(z)=z^3-1$$. It is easy to see that $$f:\mathbb{C}-\{1,\exp(2\pi i/3)\}\twoheadrightarrow\mathbb{C}$$. Precomposing with the covering map $$\pi:\mathbb{D}\to \mathbb{C}-\{1,\exp(2\pi i/3)\}$$, we get the claim.
One can also explore "how much" this function fails to be injective. Indeed, the uniformization theorem implies that, for $$U\neq\mathbb{C},U\neq \mathbb{C}-\{z_0\}$$, we have $$U\simeq \mathbb{D}/\Gamma$$, where $$\Gamma$$ is a discrete subgroup of $$Aut(\mathbb{D})$$ (complex automorphisms) isomorphic to $$\pi_1(U)$$. Thus two elements $$z_1,z_2$$ have the same image iff $$\exists \gamma\in \Gamma:\gamma(z_1)=z_2$$ (note that with $$U$$ simply connected we get injectivity).
• Beautiful answer, thank you very much! Feb 14 at 13:46 | {
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# $n$ Identical Balls Distributed Into $r$ Urns
• Question
In how many ways can at most $n$ identical balls be distributed into $r$ urns so that the $i$th urn contains at least $m_i$ balls, for each $i=1,...,r$?
Assume that $n\geq \sum_{i= 1}^r m_i$
• My Approach
$$x_1 + x_2 + \cdot\cdot\cdot + x_r = S$$
$$\sum_{i= 1}^r m_i \leq S \leq n$$
$$x_i \geq m_i$$
Let
$$y_i = x_i - m_i$$
then $$y_1+ y_2 + \cdot\cdot\cdot + y_r = S -\sum_{i= 1}^r m_i$$
Now I will proceed to find the number of non-negative integer-valued vectors $(y_1, y_2,...,y_r)$ such that $y_1+ y_2 + \cdot\cdot\cdot + y_r = S -\sum_{i= 1}^r m_i$ for every integer $S$ in the range of $\{\sum_{i= 1}^r m_i\quad,\quad n\}$ which is equal to $$\sum\limits_{S = \sum_{i= 1}^r m_i}^n {S - \sum_{i= 1}^r m_i + r -1\choose r-1}$$
Please have a look at my solution and give me any hints/suggestions you might have.
• What is your question actually? – user 170039 Oct 1 '14 at 13:22
• There is Question, My Approach, and then Please have a look at my solution and give me any hints/suggestions you might have.. What do you mean? – Kermit the Hermit Oct 1 '14 at 13:24
• I mean what question do you want the MSE to answer? Or is it that you want to verify your solution? – user 170039 Oct 1 '14 at 13:27
• Substitute $x_i = m_i + x_i'$ where $x_i' \ge 0$. Now subtract $m_1, m_2, \dots , m_i$ from both sides of the linear diophantine equation you have written. Finally, use stars and bars as usual. – Yiyuan Lee Oct 1 '14 at 13:28
You really seem to have the right idea, defining $y_i=x_i-m_i$, which basically says "If I put $x_i$ balls into urn $i$, this includes $m_i$ obligatory balls and then $y_i$ more that I can choose." This reduces you to solving:
$$x_1+x_2+\dots+x_r=(y_1+m_1)+(y_2+m_2)+\dots+(y_r+m_r)=n$$
which you note is the same as:
$$y_1+y_2+\dots+y_r=n-\sum_{i=1}^r m_i.$$ | {
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which you note is the same as:
$$y_1+y_2+\dots+y_r=n-\sum_{i=1}^r m_i.$$
My question to you is: What is $S$? How is $S$ different from $n$? Unless I'm misreading this problem, $S$ seems to be extraneous. If $S$ is the number of balls you are distributing, the question seems to indicate that $S=n$ (not just $S\le n$).
So if you'll allow us to say $M=\sum m_i$, then you are correct in your reformulation of the problem, you are basically just distributing $n-M$ balls into $r$ urns, which is:
$$\binom{n-M+r-1}{r-1}.$$
That's the last term of your summation. You don't need to sum over values of this "$S$" because you need to distribute all $n$ of the balls. (If you had such a thing, it would be like counting ways of partially distributing the balls, which the question does not seem to indicate.)
You seem to know where the binomial formula for counting these tuples comes from, but for others reading, this formula is determined as explained here.
We basically reduced the problem "distributing $n$ balls into $r$ urns with restriction $m_i$ on each urn" to "distributing $n-M$ balls into $r$ urns" by translating solutions from one to the other. You can think of it physically as starting out by putting all of the $m_i$ balls into each urn as required -- you have $n-M$ left over to distribute freely, and you can use what you already know to count how to do that.
A simpler version of this problem might be "how many ways can you give your two children \$5 allowance (in whole dollar amounts) if each child needs at least \$1?" Rather than count some complicated configuration, you can really just think of this \$1 each as already handed out to them and then solve the easier problem "how many ways can you give your two children \$3 allowance?" (In this case, the answer is 4 ways.) | {
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I am assuming from context (and from your use of the formula) that the urns are distinguishable (they are enumerated and each has a particular value of $m_i$, so philosophically -- and practically speaking, were you to attempt this exercise -- that makes the most sense).
• You just sparked something in me. For some reason I was thinking that you don't have to distribute all $n$ balls. Thus leading to $\sum_{i=1}^r m_i\leq x_1 + x_2 + ... + x_r \leq n$. – Kermit the Hermit Oct 1 '14 at 13:45
• If you are are allowed to use fewer than $n$ balls total (i.e. to just use $S$ balls and throw away the $n-S$ left over), the answer is just the sum (similar to what you have): $\sum_{S=M}^n\binom{S-M+r-1}{r-1}$. – Kellen Myers Oct 1 '14 at 14:17
• What is that $M$? Is that $\sum_{i=1}^r m_i$ ? – Kermit the Hermit Oct 1 '14 at 14:29
• Yes, it's shorthand. I think it's defined somewhere in the middle of my answer. – Kellen Myers Oct 1 '14 at 14:33
• Yes indeed, I missed that. I should pay more attention when reading it would seem. So taken in to consideration the way I interpreted the question, my solution would be correct? – Kermit the Hermit Oct 1 '14 at 14:37 | {
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# Values of a, b, and c that the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ has one point of inflection?
For what values of a, b and c does the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ have exactly one point of inflection? Two points of inflection? No points of inflection? Provide a numerical approximation for the lowest value of a for which there is no inflection point.
I'm really stumped on this question. I've tried finding the second derivative but couldn't think of anything else to do from that point. Any help would be greatly appreciated. Thanks!
-
Points of inflection occur when the second derivative is zero. What is the second derivative of your curve? How can you assure there's only one zero such that we have $+/-$ on either side of the zero. (You're allowed other zeroes that don't change the inflection, i.e. $x=0$ of $y=x^2$). – Ian Coley Dec 10 '13 at 19:46
I've found the second derivative to be $y'' = 6ax + 6 + e^x = 0$ but I could not think of how to solve that. – DinoMint Dec 10 '13 at 19:48
What happens when $x=0,1,-1,-2, 2$ and so on? – Manasi Dec 10 '13 at 20:02
Given that $e^x$ is positive will drive you to a negative x. Since it won't be too small you can probably use an estimate for $e^x$ around that x. – half-integer fan Dec 10 '13 at 20:04
You don't need to solve $6ax+6+e^x=0$, @DinoMint, only figure out how many solutions it has. Think of it as intersecting the graph of $y=e^x$ with the line $y=-6ax-6$. If the slope $-6a$ is negative, there is exactly one intersection point, if it is positive you can find the tangent line to $y=e^x$ with the same slope ($-6a$) and see if this line, $y=-6ax-6$ is above or below the tangent. – Omar Antolín-Camarena Dec 10 '13 at 20:05 | {
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As you stated in the comments, the second derivative is $$y'' = 6ax + 6 + e^x = 0$$
First, we note that if $a=0$ the equation becomes $$6 + e^x = 0$$ which has no real solutions, thus there are no inflection points.
Rewriting the equation as $$x = -\dfrac{6+e^x}{6a}$$ or $$x = -\dfrac1{a} \cdot \left(1 + \dfrac 16 e^x \right)$$ makes the behavior clearer.
If $a \gt 0$ we can see that $x$ must be negative and thus $e^x < 1$ and there will be a solution in the neighborhood of $x = -\dfrac1a$.
I will leave the case where $a \lt 0$, thus $x \gt 0$ for you to consider. | {
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# Math Help - Possible mathematical induction problem
1. ## Possible mathematical induction problem
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
2. ## Re: Possible mathematical induction problem
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
A strict grader may like to have seen more grouping symbols.
However, the argument is correct. | {
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Alternative way to find the probability in die throw problem
We throw a die 10 times. What's the probability of getting at least one 6?
Using the complentary that's $$1-Pr$$(Getting no $$6$$'s)
$$1- ({\frac{5}{6}})^{10} = 0.838494$$
There's a note that says that to solve the problem directly would require a complex use of the additivity property.
a) How do you solve the problem directly?
I think that to solve it directly, though not sure, is to find the $$P(A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6)$$ in ten die throws, where $$A_i$$ is the event of getting $$i$$ $$6$$'s
As the events $$A_i$$ are not disjoint the exclusion-inclusion principle is needed.
My try:
$$\binom{10}{1} \frac{1}{6} - \binom{10}{2} \frac{1}{6^2} + \binom{10}{3} \frac{1}{6^3} - \binom{10}{4} \frac{1}{6^4}+ \binom{10}{5} \frac{1}{6^5} - \binom{10}{6} \frac{1}{6^6} =0.838091$$
It is a slightly different result that the one obtained using the complementary. What am I not doing right?
b)If the events aren't disjoint, can you talk about the "additivity property"? As it is defined for disjoint events.
It would be the following:
$$\sum_{n=1}^{10} \dbinom{10}{n}\left(\dfrac{1}{6}\right)^n\left(\dfrac{5}{6}\right)^{10-n}$$
Here the events are disjoint. You have exactly 1 roll of a 6, exactly two rolls of a six, ..., exactly ten rolls of a six.
Your attempt works as well (although you stopped at $$i=6$$). Had you continued using Inclusion/Exclusion up to $$i=10$$, it would have given the same answer:
$$\sum_{i=1}^{10}(-1)^{i+1}\dbinom{10}{i}\left(\dfrac{1}{6}\right)^i$$
Here you are counting if you choose a die, the probability that one die is a six, minus if you choose two dice, both of them are a six, plus if you choose three dice, all three are a six, etc.
b) Yes.
$$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$
This is the property that yields the Inclusion/Exclusion principle in the first place. | {
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This is the property that yields the Inclusion/Exclusion principle in the first place.
You got a different answer because you only computed the first six terms of the inclusion-exclusion method. You have ten dice, each of which could produce a six, so you need ten terms.
• You use Inclusion-Exclusion when you have events that are not mutually exclusive. You can see that "roll six on the first die" and "roll six on the second die" are not mutually exclusive because it is possible that you roll sixes on both dice. I don't know what you mean by "inclusive." – David K May 3 at 21:27 | {
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# How to express the cardinality of $∏_{1≤i≤n} A_i$ in terms of cardinalities $|A_1|, |A_2|, . . . , |A_n|$
I was given the problem:
For finite sets $A_1, A_2,\dotsc , A_n$ define their Cartesian product $\prod_{i=1}^n A_i$ as the set of all $n$-sequences $(x_1, x_2,\dotsc, x_n)$, where $x_i \in A_i$ for every $i = 1, 2, \dotsc, n$. Find a formula expressing the cardinality of $\prod_{i=1}^n A_i$ in terms of cardinalities $|A_1|, |A_2|,\dotsc , |A_n|$.
And I am struggling to understand what it is actually asking for, could someone explain it to me please, thanks. :)
• Do you know what the Cartesian product is? For instance, if $A_1 = \{1,2\}$ and $A_2 = \{3,4,5\}$ could you explicitly write down $\prod_{1 \leq i \leq 2} A_i$? – Mees de Vries Jan 19 '17 at 13:24
• @MeesdeVries This is all prossible combinations right? So for that example {(1,3), (2, 3), (1, 4) ....}, But I have never seen the notation: ∏ 1≤i≤n Ai before, what does that mean? – Alfie Jan 19 '17 at 13:27
$\prod_{1\le i\le n}A_i$ is the cartesian product, that is, all finite sequences $(a_1,\ldots,a_n)$ such that $a_i \in A_i$ for each $i=1,\ldots,n$. How many such sequences can you choose? $|A_1|$ choices for $a_1$, ..., $|A_n|$ choices for $a_n$. Therefore $$\left|\prod_{1\le i\le n}A_i\right|=\prod_{1\le i\le n}|A_i|.$$ | {
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• So is this question asking for a formula for the carnality of the cartesian product of $A_1, A_2,\dotsc , A_n$ ? – Alfie Jan 19 '17 at 13:51
• Yes. ____________ – Paolo Leonetti Jan 19 '17 at 13:52
• So $\prod_{1\le i\le n}|A_i|$= $|A_1| * |A_2| * \dotsc * |A_n|$ – Alfie Jan 19 '17 at 13:57
• @Bram28 Assuming the axiom of choice, multiplication of infinite cardinal numbers is not difficult. If either $\kappa$ or $\mu$ is infinite and both are non-zero, then $\kappa \cdot \mu= \max\{\kappa, \mu\}$. In particular, $\kappa^n=\kappa$ for all $n\ge 1$. See here: en.wikipedia.org/wiki/Cardinal_number – Paolo Leonetti Jan 19 '17 at 14:56
• @Bram28 Well, you can find the answer in every introduction about cardinal arithmetics. Take a look here math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Murphy.pdf and here euclid.colorado.edu/~monkd/m6730/gradsets06.pdf – Paolo Leonetti Jan 19 '17 at 16:00
We know that $$|A\times B|=|A|\times|B|\qquad(1)$$.
We want to show $$\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$$ is true for any natural number $n$, where $\prod_{i=1}^n|A_i|=|A_1|\times|A_2|\times\dotsc\times|A_n|$. So, we have use induction.
The base case $n=1$ ($|A_1| = |A_1|$) is trivial. Now suppose inductively that $\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$. We want to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\prod_{i=1}^{n+1}|A_i|.$$ Now we need to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|\qquad(2),$$ i.e., the cardinality of the set $\prod_{i=1}^{n+1}A_i$ is equal to the cardinality of the set $\left(\prod_{i=1}^nA_i\right)\times A_{n+1}$. So \begin{aligned}\left|\prod_{i=1}^{n+1}A_i\right|&=&\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|&\qquad\text{by }(2)\\&=&\left|\prod_{i=1}^nA_i\right|\times |A_{n+1}|&\qquad\text{by }(1)\\&=&|A_1|\times|A_2|\times\dotsc\times|A_n|\times|A_{n+1}|&\qquad\text{by induction hypothesis}\\&=&\prod_{i=1}^{n+1}|A_i|.\end{aligned} | {
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• This is exact, but I think the OP was asked a formula, but not asked to prove it. – Jean Marie Jan 19 '17 at 14:46
• @CristianGz Do you know if (1) also holds for infinities? If so, how am I to think about multiplying 'infinities'? Is there a definition for that? – Bram28 Jan 19 '17 at 14:52
• @JeanMarie: Agree. I deleted the answer. But I though it can be useful. – Cristhian Gz Jan 19 '17 at 14:53
• @Bram28, the statement is true to infinite sets. It is possible using bijective functions to state equal cardinality between two sets. Check en.m.wikipedia.org/wiki/Cardinality – Cristhian Gz Jan 19 '17 at 14:55 | {
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# Lecture 21: Minimizing a Function Step by Step
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## Description
In this lecture, Professor Strang discusses optimization, the fundamental algorithm that goes into deep learning. Later in the lecture he reviews the structure of convolutional neural networks (CNN) used in analyzing visual imagery.
## Summary
Three terms of a Taylor series of $$F$$($$x$$) : many variables $$x$$
Downhill direction decided by first partial derivatives of $$F$$ at $$x$$
Newton's method uses higher derivatives (Hessian at higher cost).
Related sections in textbook: VI.1, VI.4
Instructor: Prof. Gilbert Strang
The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.
GILBERT STRANG: Well, OK, I am happy to be back, and I am really happy about the project proposals that are coming in. This is like, OK, this is really a good part of the course. And so keep them coming, and I'm happy to give whatever feedback I can on those proposals, and do make a start there. They're really good, and if some are completed before the end of the semester and we can to offer you a chance to report on them, that that's good too. So well done with those proposals.
So today, I'm jumping to part six. So part six and part seven are optimization which is the fundamental algorithm that goes into deep learning. So we've got to start with optimization. Everybody has to get that picture, and then part seven will be the structure of CNNs, Convolution Neural Nets, and all kinds of applications.
And so can we start with optimization? So first, can I like get the basic facts about three terms of a Taylor series? So that's the typical. It's seldom that we would go up to third derivatives in optimization. | {
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So that's the most useful approximation to a function. Everybody recognizes it. Here, I'm thinking of F as just one function, and x as just one variable, but now I really want to go to more variables. So what do I have to change if F is a function of more variables? So now, I'm thinking of x as-- well, now let me see.
Yeah, I want n variables here. x is x1 up to xn. So just to get the words straight so we can begin on optimization, so what will be the similar step so the function F at x-- remember, x is n variables. OK?
Now, what do I have? Delta x, so what's the point about delta x now? It's a vector, delta x1 to delta xn, and what about the derivative of F? It's a vector too, the derivative of F with respect to x1, the derivative of F with respect to x2, and so on.
What do I have to change about that? I know those guys are vectors, so it's their dot product. So it's delta x transpose at vector times this dF/dx. So now I'm replacing this by all the derivatives, and it's the gradient. So the gradient of F at x is the derivatives-- let's see.
It's essential to get the notation straight here. Yeah, so it'll be the partial derivatives of the function F. So grad F is the partial derivatives of F with respect to x1 down to partial derivative with respect to xn. OK, good.
That's the linear term, and now what's the quadratic term? 1/2, now delta x isn't a scalar anymore. It's a vector. So I'm going to have delta x transpose and a delta x, and what goes in between is the second derivatives, but I've got a function of n variables.
So now, I have a matrix of second derivatives, and I'll call it H. This is the matrix of second derivatives, Hjk is the second derivative of F with respect to xj and xk, and what's the name for this guy? The Hessian, Hessian matrix. | {
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How the Hessians got into this picture I don't know. The only Hessians I know are the ones who fought in the Revolutionary War for somebody. Who? Which side were they on? I think maybe the wrong side. The French were on our side and--
Anyway, Hessian matrix, and what are the facts about that matrix? Well, the first fact is that it's [INAUDIBLE] and the key fact is it's symmetric. Yeah. OK, and again, it's an approximation. And everybody recognizes that if n is very large, and we have a function of many variables. Then, we had n derivatives to compute here, and about 1/2 n squared derivatives.
The 1/2 comes from the symmetry, but the key point is the n squared derivatives to compute there. So computing the gradient is feasible if n is small or moderately large. Actually, by using automatic differentiation, the key idea of back propagation, back prop, you can speed up the computation of derivatives quite amazingly. But still for the size of deep learning problems that's out of reach. OK.
So that's the picture, and then I will want to use this to solve equations. There is a parallel picture for a vector f. So now, this is a vector function. This is f1 of x up to fn of x, an x is x1 to xn. So I have n functions of n variables, n functions of n variables.
Well, that's exactly what I have in the gradient. Think of these two as parallel, the parallel being f corresponds to the gradient of F, n functions of n variables. OK. Now maybe, what I'm after here is to solve f equals 0. So I'm going to think about the f at x plus delta x, so it starts with f of x.
And then we have the correction times the matrix of first derivatives, and what's the name for that matrix of first derivatives? Well, if I'm just given n functions-- yeah, what am I after here? I'm looking for the Jacobian. So here we'll go the Jacobian, J. This is the Jacobian named after Jacoby, Jacobian matrix. | {
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And what are its entries? J, the jk entry is the derivative of the J function with respect to the kth variable, and I'm stopping at first order there. OK, so these are sort of like facts of calculus, facts of 18.02 you could say. Multivariable calculus, that's the point.
Notice that we're doing just like the first half of 18.02, just do differential calculus, derivatives, Taylor series. We're not doing multiple integrals. That's not part of our world here. OK, so that's the background.
Now, I want to look at optimization. So over here, I want to optimize-- well, over here, let me try to minimize F of x, and I'll be in the vector case here. And over here, I want to solve f equals 0, and of course, that means f of 1 equals 0 all the way along to fn equals 0. Here, I have n equations, and n unknowns.
Let me start with that one, and I'll start with Newton's method, Newton's method to solve these n equations and n unknowns. OK, so Newton, Newton's method which is often not presented in 18.02. That's a crime, because that's the big application of gradients in Jacobians.
OK, so I'm trying to solve n equations and n unknowns, and so I want f at x plus delta x to be 0. Right? So I want f of x plus delta x to be 0. So f at x plus delta x is-- I'm putting in a 0. I'm just copying that equation-- is f at where I am. Let me use K for the case iteration.
So I'm at a point xK. I want to get to a point xK plus 1. And so I have 0 is f of x plus J, at that point, times delta x which is xK plus 1 minus xK. Good. That's Newton's method.
Of course, 0 isn't quite true. Well, 0 will be true if I'm constructing xK plus 1 here. I'm constructing xK plus 1. OK. So let me just rewrite that, and we've got Newton's method. So we're looking for this change, xK plus 1 minus xK. I'll put it on this side as plus xK, so that's this. | {
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Now, I have to invert that and put it on the other side of the equation. So that will go with a minus. This guy will be inverted and f at xK. So that's Newton's methods. It's natural.
So let me just repeat that. You see where the xK plus 1 minus xK is sitting? Right? And I moved f of xK to the other side with a minus sign, and then I multiplied through by J inverse, so I got that. So that's Newton's method for a system of equations, and over there, I'm going to write down Newton's method for minimizing a function. This is such basic stuff that we have to begin here.
Let me even begin with an extremely straightforward example of Newton's method here. Suppose my function-- suppose I've only got one function actually. Suppose I only had one function. So suppose my function is x squared minus 9, and I want to solve f of x equals 0. I want to find the square root of 9.
OK, so what is Newton's method for it? My point is just to see how Newton's method is written and then rewrite it a little bit so that we see the convergence. OK, so of course, the Jacobian is 2x. So Newton's method says that xK plus 1-- I'm just going to copy that Newton's method-- minus 1 over 2xK. Right? That's the derivative times f at xK which is xK squared minus 9.
OK. We followed the formula, this determines xK plus 1, and let's simplify it. So here I have xK minus that looks like 1/2 of xK, so I think I have 1/2xK, and then this times this is 9/2 of 1 over xK. Is that right? 1/2 of xk from this stuff and plus 9/2 of 1 over xK. OK.
Can I just like check that I know the answer is 3? Can I be sure that I get the right answer, 3? That if xK was exactly 3, then of course, I expect xK plus 1 to stay at 3. So does that happen? So 1/2 of 3 and 9/2 of 1/3, what's that, 1/2 of 3 and 9/2 of 1/3? | {
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OK, that's 3/2 and 3/2. That's 6/2, and that's 3. OK. So we've checked that the method is consistent which just means we kept the algebra straight. But then the really important point about Newton's method is to discover how fast it converges. So now let me do xK plus 1 minus 3.
So now, I'm looking at the error which is, I hope, approaching 0. Is it approaching 0? How quickly is it approaching 0? These are the fundamental questions of optimization.
So I'm going to subtract 3 from both sides somehow. OK, from here, I guess, I'm going to subtract 3. So I was just checking that it was correct. OK. Now, so xK plus 1 minus 3, I'm going to subtract 3 from both sides. I'm going to subtract 3 there, and then I hope that-- that box is what goes down here. Right?
Subtracted 3 from both sides, so I'm hoping now things go to 0. OK, so what do I have there? Let me factor out the 1 over xK. So what do I have then left? 1 over xK, so there's a 9/2 from there, 1 over xK.
So I really have 1/2 of xK squared, because I've divided by an xK. And this minus 3, I better put minus 3xK, because I'm dividing by xK. I claim that that's-- now I've got it. And let's see, let me take out the 2-- 2, forget these 2s, and make that a 6. So I have 1 over 2xK times 9 plus xK squared minus 6.
Anything good about that? We hope so. We hope that that is something attractive. So this is, again, the error at set K plus 1, and it's 1 over 2xK times this thing in brackets-- 9 plus xK squared minus 6xK. And we recognize that as xK minus 3 squared.
xK squared minus 6 of them plus 9, that's xK minus 3 squared. OK, that was the goal, of course. That's the goal that shows why Newton's method is fantastic. If you can execute it, if you can start near enough, notice that-- so how do I describe this great equation? It says that the error is squared at every step, squared at every step. | {
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So if I'm converging to a limit, it will satisfy the-- it'll be 3, or I guess minus 3, is that possible? Yeah, minus 3 is another solution here. So we've got two solutions. Newton's method could converge to 3. Am I right, it could converge to minus 3?
So I'd have a similar equation sort of centered at minus 3, or does it always do one of those? It could blow up. So there are sort of regions of attraction. They're all the starting points that approach 3, and the whole point of that equation is with quadratic convergence the error being squared at every step. It zooms in on 3.
Then, there is all the starting points that would go to minus 3, and then there are the starting points that would blow up. And those, maybe for this very simple problem, the picture is not too difficult to sort out those three regions. And this is allowing for a vector, two equations or n equations, then we're in n variables, and really you get beautiful pictures.
You get some of the type of pictures that gave rise to these books on fractals, picture books on fractals for these basins of attraction. Does the starting point lead you to one of the solutions, or does it lead you to infinity? Here, that would be interesting to just draw it for this, but the essential point is the quadratic convergence, if it's close enough.
You see that it has to be close. If x0 is pretty near 3, then this is about 1/6 of that, and there would be a good region of attraction in this case. OK. So that's Newton's method for equations.
And now I want to do Newton's method. I just want to convert all those words over to Newton's method for optimization. So remember, these boards were solving f equals 0. This board is minimizing capital F, and what's the connection between them? Well of course, this corresponds to solving the gradient equals 0. | {
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At a minimum, if I'm minimizing, I'm finding a point where all the first derivatives are 0. So that will be the match between these. This grad F in this picture is the small f in that picture. OK.
Now, I guess here I have-- and this is sort of the heart of our applications to deep learning-- we have very complicated loss functions to minimize, functions of thousands or hundreds of thousands of variables. OK. So that means that we would like to use Newton's method, but often we can't. So I need him to put down here two methods-- one that doesn't involve those high second derivatives and Newton's that does.
So first, I'll write down a method that does not involve, so method one, and this will be steepest descent. And what is that? That says that xK plus 1-- the new x is the old x minus-- steepest descent means that I move in the steepest direction which is the direction of the gradient of F. I move some distance, and I better have freedom to decide what that distance should be. So this is a step size, s, or in the language of deep learning, it's often called the learning rate, so if you see learning rate. OK.
So and it's natural to choose sK. We're going along, do you see what this right-hand side looks like? I'm at a point in n dimensions. We're in n dimensions here. We have functions of n variables.
There is a vector. There is a direction to move down the steepest slope of the graph. And here is a distance to move, and we will stop. We'll have to get off this step, normally. If we stay on it, it will swing back, it'll take us off to infinity.
You would like to choose sK so that you minimize capital F. You take the point on this line, so this a line in R n, a direction in R n. And for all the points on that line, in that direction, F has some value, and what you expect is that initially, because you chose it sensibly, the value of F will drop. But then at a certain point, it will turn back on you and increase. | {
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So that would be the natural stopping point. I would call that an exact line search. So I exact line search would be, exact line search is the best s. Of course, that would take time to compute, and you probably, in deep learning, that's time you can't afford, so you fix the learning rate s. Maybe you choose 0.01 to be pretty safe.
OK, so that's method one, steepest descent. Now, method two will be Newton's method. So now, we have xK plus 1 equal to xK minus something times delta F, and now I'm going to do the right thing. I'm going to live right here, and the right thing is the Hessian, the second derivative.
This was cheap. We just took the direction and went along it. Now, we're getting really the right direction by using the second derivative, so that's H inverse. OK, and what I've done is to set that 0.
Do you see that's Newton's method? It's totally parallel to this guy. Actually, I'm really happy to have these two on the board parallel to each other, because you have to keep straight, are you solving equations, or are you minimizing functions? And you're using different letters in the two problems, but now you see how they match.
The Jacobian of-- so again the matches, think of f as the gradient of F. That's the way you should think of it. So the Jacobian of the gradient is the Hessian. The Jacobian of the gradient is the Hessian, and that makes sense, because the first derivative of the first derivative is the second derivative. Only we're doing matrix y, so the Jacobian of the gradient-- we're doing a vector matrix sentence instead of a scalar sentence-- the Jacobian of the gradient is a Hessian. Yeah, right.
OK, so that's what I wanted to start with, just to get those basic facts down. And so the basic facts were the three-term Taylor series. And then the basic algorithms followed naturally from it by setting f F at the new point to 0, if that's what you were solving or by assuming you had the minimum. Right, good, good, good, good. OK. | {
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Now, what? Now, we have to think about solving these problems, studying. Do they converge? What rate do they converge? Well, the rate of convergence is like why I separated off this example.
So the convergence rate for Newton's method will be quadratic. The error gets squared, and of course, that means super-fast convergence, if you start and close enough. The rate of convergence for a steepest descent is, of course, not. You're not squaring errors here, because you're just taking some number instead of the inverse of the correct matrix, so you can't expect super speed.
So a linear rate of convergence would be right. You would like to know that the error is multiplied at every step by some constant below 1. That would be a linear rate compared to being squared at every step. OK, and so this will be our basic formula that we build on for really large scale problems.
And there are methods, of course, people are going to come up with methods that they're sort of a cheap Newton's method. Levenberg-Marquardt, and it's in the notes at the end of this section, at the end of 6.4 that we'll get to. So Levenberg-Marquardt is a sort of cheap man's Newton's method. It does not compute to Hessian, but it says, OK, from the gradient, I can see one term in the Hessian. So it grabs that term, but it's not fully second order.
OK. So now, we have to think about problems, and I guess the message here is, at our starting point, has to be convexity. Convexity is the key word for these problems, for the function that we want to minimize. If that's a convex function, well first of all, the convex function is likely to have one minimum. And the picture that's in our mind of steepest descent, this picture of a bowl, a bowl is the graph of the convex function. | {
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So I'm turning to convexity now. I'll leave that board there, because that's pretty crucial, and speak about the idea of convexity. Convex function, convex set, so let's call the function f of x, and a typical convex set would be I'll call it K. OK. So we just want to remember what does that word can convex mean, and how do you know if you have a convex function or a convex set?
OK, let me start with convex set. So because here is my general problem, my convex minimization, which you hope to have, and in many applications, you do have. So you minimize a convex function for points in a convex set. So that's like the ideal situation. That's the ideal situation, to get something on your side, something powerful, convexity.
The function is convex, and you say, well, let me draw a convex function, the graph. OK, so I'll draw a convex function, say a bowl. So that's a graph of f of x, and then here are the x's. Let me maybe put x1 and x2 in the base and the graph of f of 1x x2 up here. OK. Actually, I'm over there.
I should be calling this function F, I think. Is that right? Yeah, a little f would be the gradient of this guy. Yeah, I think so. OK.
Now, I'm minimizing it over certain x's, not all x's. I might be minimizing, for example, K might be the set where Ax equals B. K might be, in that case, a subspace or a shifted subspace. I said subspace, but then 18.06 is reminding me in my mind that I only have a subspace when B is 0.
You know the word for a subspace that's sort of moved over? Affine, so I'll just put that word down here. Bunch of words to learn for this topic, but they're worth learning. OK.
So it's like a plane but not necessarily through the origin. If B is 0, it doesn't go through it. If B it's not 0, it doesn't go through the origin. OK. Anyway, or I have some other convex set. Let me just put this convex set K in the base for you, and did I make it convex? I think pretty luckily I did. | {
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So now what's the? Well, the convex sets the constraint, so this is the constraint set. Constraint is that x must be in the set K. OK, and I drew it as a convex blob. Here was an example where it's flat, not a blob but a flat plane.
But let me come back to what does convex mean. What's a convex set? Yeah, we have to do that, should have done that before. In the notes, I had the fun of figuring out, if I took a triangle, is that a convex set? Let's just be sure.
So what's a convex set? That is a convex set, because if I take any two points in the set and draw the line between them, it stays in the set. So that's convexity, any edge, line, from x1 to x2 stays in the set. OK, good.
So here's my little exercise to myself. What if I took the union of two triangles? All I want to get you to do is just visualize convex and not convex possibilities. Suppose I have one triangle, even if it was obtuse, that's still convex, right? No problem.
But now what if I put those two triangles together, take their union? Well, if I take them sitting with a big gap between, like I've lost. I mean, I never had a chance that way, because if it was the union of these two-- well, you know what I'm going to say. If I'm doing that point and that point, of course, it goes outside and stupid. All right.
What if what if that triangle, that lower triangle, overlaps the upper triangle? Is that a convex set? Everybody's right saying no. Why how do I see that the union of those two triangles is not a convex set? Guys, you tell me where to pick two points, where the line goes out. Well, I take one from that corner and one from that corner, and the line between them went outside. So union is usually not convex. | {
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Well, if I think of the union of two sets, my mind automatically goes to the other corresponding possibility which is the intersection of the two sets. So if I take the intersection of two sets. Now, what's the deal with that? When I had two triangles, two separated triangles, what can we say about the intersection of those two triangles?
AUDIENCE: [INAUDIBLE]
GILBERT STRANG: It's empty. So should we regard the empty set as a convex set? Yes. Isn't it?
AUDIENCE: Yeah, it's vacuous.
GILBERT STRANG: Vacuous, so it hasn't got any problems. Right? OK, but now the intersection is always convex. I'm assuming the two sets that we start with are. Now, that's an important fact, that the intersection of convex sets. Let's just draw a picture that shows an example.
So what's the intersection? Just this part and it's convex. OK, can you give me a little proof that the intersection is convex? So I take two points in the intersection-- let me start the proof.
To test if something's convex, how do you test it? You take two points in the set in the intersection, and you want to show that the line between them is in the intersection. OK, why is that?
So take two points, take x1 in the intersection. We've got two sets here, and that's the symbol for intersection, and we've got another point in the intersection. And now, we want to look at the line between them, the line from x1 to 2x. What's the deal with that one? Is that fully in K1?
AUDIENCE: Yes.
GILBERT STRANG: Why is it fully in K1? I took two points in the intersection, I'm looking at the line between them, and I'm asking, is it in the first set K1? And the answer is yes, because those points were in K1, and K1's convex. And is that line between them in K2? Yes, same reason, the two endpoints were in K2, so the line between them is in K2. | {
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So the intersection of convex sets is always convex. The intersection of convex sets is convex. Good. So you'll see in the note these possibilities with two triangles. Sometimes, you can take the union but not very often. OK.
Now, what's the next thing I have to do? Convex functions, we got convex sets, what are convex functions, and we're good. Because this is our prototype of a problem, and I now want to know what it means for that F to be-- oh, I'm sorry. I now know what it means for the set K to be convex set, but now I have to look at the other often more important part of the problem. What's the function I'm minimizing, and I'm looking for functions with this kind of a picture. OK.
The coolest way is to connect the definition of a convex function to the definition of a convex set. This is really the nicest way. It's a little quick. It just swishes by you. But tell me, do you see a convex set in that picture? [INAUDIBLE]
You see a convex set in that picture. That's the picture of a graph of a convex function. It's a picture of a bowl. Are the points on that surface, is that a convex set? No, certainly not. No, but where is a convex set to be found here, in that picture? Yes.
AUDIENCE: The set of y, if y is greater than [INAUDIBLE]
GILBERT STRANG: Yes, the points on and above the bowl, inside the bowl, we could say, these points. So convex function, yes, a function's convex when the points on and above the graph are convex set. You could say, OK, mathematicians are just being lazy. Having got one definition straight for a convex set, now they're just using that to give an easy definition of a convex function. Actually, it's quite useful for functions that could maybe equal infinity, sort of generalized functions. | {
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But it's not the quickest way to tell if the function is convex. It's not our usual test for convex functions. So now I want to give such a test. OK. So now, the definition of convex function, of a smooth convex, yeah. This fact, I shouldn't rush off away from it, from the definition of a convex function as having a convex set above its graph. The really official French name for the set above the graph is the epigraph, but I won't even write that word down. OK.
Why do I come back to that for a minute? Because I would like to think about two functions, F1 and F2. Out of two functions, I can always create the minimum or the maximum.
So suppose I have to convex functions, convex function F1 and F2. OK. Then, I could choose a minimum. I could choose my new function. Shall I call it little m for minimum? m of x is the minimum of F1 and F2.
And I could choose a maximum function which would be the maximum of F1 of x and F2 of x at the same point x. It's just a natural to think, OK, I have two functions. I've got a bowl and I've got another bowl, and suppose they're both convex.
So I'm just stretching you to think here. If I've got the graphs of two convex functions, and I would like to consider the minimum of those two functions and also the maximum of those two functions. I believe life is good. One of these will be convex, and the other won't.
And can you identify which one is convex and which one is not convex? What about the minimum? Is that a convex function? So just look at the graph. What does the minimum look like? The minimum is this guy until they meet somehow on some surface and then this guy.
Is that convex? We have like one minute to answer that question. Absolutely no. It's got this bad kink in it. What about the maximum of the two functions? So the maximum is the one that is above, all the points or things that are above or on. | {
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There is the maximum function. That was the minimum function. It had a kink. The maximum function is like that, and it is convex, so maximum yes, minimum no. OK, and we could have a maximum of 1,500 functions. If the 1,500 functions are all convex, the maximum will be, because it's the part way above everybody's graph, and that would be the graph of the maximum. OK, good.
And now finally, let me just say, how do you know whether a function is convex? How to test, how of test. OK, so let me take just a function of one variable. What's the test you learned in calculus, freshman calculus actually, just show that this is a convex function? What's the test for that?
AUDIENCE: Use second derivative.
GILBERT STRANG: Second derivative should be?
AUDIENCE: Positive.
GILBERT STRANG: Positive or possibly 0, so second derivative greater or equals 0 everywhere. That's convex. OK, final question, suppose F is a vector. So this is a vector, and so I have n functions of n variable. No, I don't. I have one, sorry, I've got one function, but I'm in n variables. So this was just one.
What's the test for convexity? So it would be passed, for example, by x1 squared plus x2 squared. Would it be passed by-- so here would be the question-- would it be passed by x transpose some symmetric matrix S? That would be a quadratic, a pure quadratic.
Would it be convex? What would be the test? I'm looking for an n dimensional equivalent of positive second derivative. The n dimensional equivalent of positive second derivative is convexity, and we have to recognize what's the test. So I could apply it to this function, or I could apply it to any function of n variables. It should be OK.
What's the test here? Here, I have a matrix instead of a number. So what's the requirement going to be? Times out, yeah? [INAUDIBLE] Positive definite or semidefinite, or semidefinite just as here. Yeah. | {
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So the test is positive, semidefinite, Hessian. And here, the Hessian is actually that S, because the second derivatives will produce-- I'll put a 1/2 in there-- the second derivatives will produce S equal the Hessian H. So here, the S-- so positive semidefinite, Hessian in general, second derivative matrix for a quadratic.
OK. So its convex problems that we're going to get farther with. We run into no saddle points. We run into no local minimum. Once we found the minimum, it's the global minimum. These are the good problems. OK, again, happy to see you today, and I look forward to Wednesday. | {
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# Calculus - Integrals
I have 3 questions, and I cannot find method that actually solves them.
1) Integral [(4s+4)/([s^2+1]*([S-1]^3))]
2) Integral [ 2*sqrt[(1+cosx)/2]]
3) Integral [ 20*(sec(x))^4
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3. 👁
1. 1) expand in partial fractions.
2) Using cos(2x) = 2 cos^2(x) -1 derive a formula for cos(1/2 x) in terms of cos(x). Express the integrand in terms of cos(1/2 x)
3) [Notation: cos = c, sin = s]
1/c^4 = (s^2 + c^2)/c^4 =
1/c^2 + s^2/c^4
1/c^2 yields a tangent when integrated
To integrate s^2/c^4 do partial integration:
s^2/c^4 = s (s/c^4)
Integral of s/c^4 is 1/3 1/c^3
So, we need to integrate
1/c^3
times the derivative of of s, i.e. 1/c^2, but that is tan(x)!
Note that the way to solve such problems is not to systematically work things out in detail at first, because then you would take too long to see that a method doesn't work.
Instead, you should reason like I just did, i.e. forget the details, be very sloppy, just to see if things works out and you get an answer in principle, even though you need to fill in the details.
If you get better at this, you can do the selection of what method to use in your head, you'll see it in just a few seconds when looking at an integral. You can then start to work out the solution for real on paper.
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2. Could you expand on #1 a bit more? I tried Partial Fractions, but I couldn't get a definite answer...
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3. I'll show you how to do it without solving any equations.
The function is up to a factor 4:
f(s) = (s+1)/{(s^2+1)*[(s-1)^3]}
The partial fractions are precisely the singular terms when expanding around the singularities. So, let's examine the singularity at s = 1:
Put s = 1 + u and expand in powers of u, keeping only the singular terms:
f(1+u) = u^(-3) (2+u)/[(u+1)^2+1] =
u^(-3) (2+u)/(2+2u+u^2) =
u^(-3) (2+u)/2 1/(1+u+u^2/2) = (use
1/(1+x) = 1-x+x^2-x^3+...)
u^(-3) (2+u)/2 [1-u-u^2/2 +u^2+...]
=[u^(-3) + u^(-2)/2][1-u+u^2/2+...]
u^(-3) -1/2 u^(-2) + nonsingular terms | {
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=[u^(-3) + u^(-2)/2][1-u+u^2/2+...]
u^(-3) -1/2 u^(-2) + nonsingular terms
This means that in the neighborhood of s = 1 we have:
f(s) = 1/(s-1)^3 - 1/2 1/(s-1)^2 +.. nonsingular terms
f(s) also has singularities at s = i and s =-i. Around s = i, we have:
f(s) = a/(s-i) + nonsingular terms
Multiply both sides by (s-i) and take the limit s --> i to find a:
a = (i+1)/[2i(i-1)^3] =
-1/2 1/(i-1)^2
Around s = -i, we have:
f(s) = b/(s+i) + nonsingular terms
Multiply both sides by (s+i) and take the limit s --> -i to find b:
b = (-i+1)/[-2i(-i-1)^3] =
-1/2 1/[(i+1)^2]
The sum of all the singular terms of the expansions around the singular points is:
1/(s-1)^3 - 1/2 1/(s-1)^2 +
-1/2 1/(i-1)^2 1/(s-i)
-1/2 1/[(i+1)^2] 1/(s+i) =
1/(s-1)^3 - 1/2 1/(s-1)^2 +
1/2 1/(s^2+1)
This is the desired partial fraction decomposition. The proof of why this works is a consequence of Liouville's theorem
http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)#Proof
Just consider the difference of f(s) and the sum of all the singular terms. The resulting function doesn't have any singularites and is bounded. So, by Liouville's theorem it is a constant. We know that f(s) and all the singular terms tend to zero, so that constant must be zero.
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# General Plotting Code for Cone in 3D with GLMakie or Plots
Hi all,
so I want to plot a general cone of pyramid and a right circular cone, I get this formula:
But I think the plotting does not use the volume formula. Because of this torus code for GLMakie that I have:
using GLMakie
using ColorSchemes
cmap3 = get(colorschemes[:plasma], LinRange(0,1,100))
cmap4 = [(cmap3[i],0.65) for i in 1:100]
ξ = -40:1:40
c = 20
a = 15
torus = [((hypot(X,Y)-c)^2+Z^2-a^2) for X in ξ, Y in ξ, Z in ξ]
volume(torus, algorithm = :iso,isovalue =100,isorange =50, colormap=cmap4)
current_figure()
For general cone of pyramid, If the based is made of a closed random curves, then we might need to plot in x-y plane the curves then plot the top of pyramid afterwards.
For the right circular cone, I believe there is a general formula already.
You could conveniently define both as parametric surfaces:
using Plots; plotlyjs()
z, Θ = range(0,10,200), range(0,2π,300)
# Right-cone:
X = [u * cos(v) for u in z, v in Θ]
Y = [u * sin(v) for u in z, v in Θ]
Z = [-2u for u in z, _ in Θ]
surface(X, Y, Z, size=(600,600), cbar=:none, legend=false)
# Trefoil cone:
X = [u/3 * (sin(v) + 2*sin(2v)) for u in z, v in Θ]
Y = [u/3 * (cos(v) - 2*cos(2v)) for u in z, v in Θ]
Z = [-2u for u in z, _ in Θ]
surface(X, Y, Z, size=(600,600), cbar=:none, legend=false)
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From your questions posted on this forum it seems that you are teaching math. Hence you should explain to your students how is parameterized a general cone,
defined by its vertex V, given as a 3 vector, and a closed (but not necessarily) planar curve,
x = x(u)
y = y(u)
z= b # b from base plane; usually b=0
u \in [\alpha, \beta]
.
The cone is generated by all segments of line connecting its vertex, V, with the points
(x(u), y(u), b), on the base curve.
Such a segment has the equations:
\displaystyle\frac{X-V[1]}{x(u)-V[1]} = \displaystyle\frac{Y-V[2]}{y(u)-V[2]} =\displaystyle\frac{Z-V[3]}{b-V[3]}=v
Hence: | {
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X = v(x(u)-V[1])+V[1]
Y = v(y(u)-V[2])+V[2]
Z = v(b-V[3])+V[3]
u \in [\alpha, \beta], v\in[0,1]
Example:
vert=[0, 0.75, 3] #cone vertex
base = 0 #the cone base is included within the plane z=base (here z=0)
#functions that define the cone base parameterization
x(u) = cos(u)
y(u) = sin(u)+ sin(u/2);
m, n = 72, 30
u= range(0, 2π, length=m)
v = range(0, 1, length=n)
us = ones(n)*u'
vs = v*ones(m)'
#Surface parameterization
X = @. vs* (x(us)-vert[1]) + vert[1]
Y = @. vs* (y(us)-vert[2]) + vert[2]
Z = @. vs*(base-vert[3]) + vert[3];
surface(X, Y, Z, size=(600,600), cbar=:none, legend=false) #line copied from Rafael
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Yes 100 for you, I was a Mathematics teacher for Junior High School students in Papua, but it was 2 years ago. Afterwards, I still love learning Mathematics even if I no longer teach, now I am learning the undergraduate Mathematics, from Calculus, It would be lovely to teach Math again in the future.
This is the first time I learn Calculus with Julia from bottom, it is rewarding knowing the application of this is enormous.
Thanks for this I will use your code and learn the parameterization, vertex, etc.
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# Root equation
#### Albert
##### Well-known member
$m,n \in N$
satisfying :
$\sqrt {m+2007}+\sqrt {m-325}=n$
find Max(n)
#### Bacterius
##### Well-known member
MHB Math Helper
Re: root equation
[JUSTIFY]Apply the variable substitution $u = m - 325$, and hence $m + 2007 = u + 2332$. We get:
$$\sqrt{u + 2332} + \sqrt{u} = n$$
For $n$ to be an integer, both square roots have to be integers*. So we are looking for two (positive) squares which differ by $2332$ units. That is, we need to solve:
$$x^2 - y^2 = 2332$$
Doing some factoring:
$$(x - y)(x + y) = 2^2 \times 11 \times 53$$
After some observation** we have the following solution pairs:
$$(x, y) \to \{ ( 64, 42), ~ (584, 582) \}$$
It follows that the only solutions for $u$ are $u = 42^2$ and $u = 582^2$. This gives $n = 106$ and $n = 1166$ (and $m = 2089$, $m = 339049$ respectively).
Therefore $\max{(n)} = 1166$. $\blacksquare$
* perhaps this part merits justification, you can check Square roots have no unexpected linear relationships | Annoying Precision which is pretty hardcore (perhaps a reduced argument suffices here) but I will take it that the question assumed this to be true. Other approaches which do not rely on this assumption would be interesting to see as well.
** this can be solved efficiently by using every reasonable factor combination for $x - y$ and deducing $x + y$ from it.[/JUSTIFY]
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
Re: root equation | {
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Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
Re: root equation
As a slight variation on Bacterius's solution, take $u$ to be $m+841$ rather than $m-325$. The equation then becomes $$\sqrt{u+1166} + \sqrt{u-1166} = n.$$ Square both sides to get $$2u + 2\sqrt{u^2-1166^2} = n^2.$$ For an integer solution, $u^2-1166^2$ must be a square, say $u^2-1166^2 = w^2.$ Thus $(w,1166,u)$ is a pythagorean triple, which must be of the form $(p^2-q^2,\,2pq,\,p^2+q^2)$ for some integers $p,q$ with $p>q$. In particular, $1166=2pq$. But $1166 = 2*11*53$, so the only possibilities are $p=53,\ q=11$, or $p=583,\ q=1.$ That then leads to the same solutions as those of Bacterius.
Last edited:
##### Well-known member
Re: root equation
we have
(m+ 2007) – (m- 325) = 2332 ..1
√(m+2007)+√(m−325) = n ..(2) given
Dividing (2) by (1) we get
√(m+2007)-√(m−325) = 2332/n … (3)
And (2) and (3) to get 2 √(m+2007) = n + 2332/n
Now RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.
#### Albert
##### Well-known member
let :
$u^2=m-325-------(1)$
$(u+a)^2=m+2007-------(2)$
$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)
max(n) must happen while a=2,so from (3) we have u=582
so max(n)=582+584=1166
#### Opalg
##### MHB Oldtimer
Staff member
Re: root equation
we have
(m+ 2007) – (m- 325) = 2332 ..1
√(m+2007)+√(m−325) = n ..(2) given
Dividing (2) by (1) we get
√(m+2007)-√(m−325) = 2332/n … (3)
And (2) and (3) to get 2 √(m+2007) = n + 2332/n
Now RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.
Very neat!
#### Albert
##### Well-known member
now ,find min(n)=?
##### Well-known member
now ,find min(n)=?
√(m+2007)+√(m−325) = n
put m- 325 = p (which is >=0)
we get √(p+2332)+√p which is >= sqrt(2332) >= 49
now we should look for lowest even factor of 2332(even : reason same as my previous solution) which is >= 49
now 2332 = 2 * 2 * 11 * 53 and lowest even factor of 2332 >= 49 is 2 * 53 = 106
#### Albert | {
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now 2332 = 2 * 2 * 11 * 53 and lowest even factor of 2332 >= 49 is 2 * 53 = 106
#### Albert
##### Well-known member
let :
$u^2=m-325-------(1)$
$(u+a)^2=m+2007-------(2)$
$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)
max(n) must happen while a=2,so from (3) we have u=582
so max(n)=582+584=1166
$2332=a^2+2ua=a(2u+a)=a\times n=2\times 1166=22\times 2\times53=22\times 106$
so max(n)=1166 ,and a=2
min(n)=106, and a=22
(here a,and n=2u+a are all even numbers) | {
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# Math Help - Help again
1. ## Help again
Please explain how to solve this.
The product of two consecutive even integers is 168. Find the two integers.
Hint: Assume the first integer is x
Set up the equation first that satisfies the given condition. Then solve the equation.
This is the equation that i came up with. Ican t seem to get past
x(x+2)=168
x^2+2x-168=0 (i cant get past here)
Thanks,
Chester
2. Originally Posted by Chester
x(x+2)=168
x^2+2x-168=0 (i cant get past here)
Good job.
Find some factors of 168: 12 and 14
which diffrence is 2 thus,
$(x+14)(x-12)=168$
Thus,
$x=-14,x=12$
3. You could always fall back on the quadratic formula. I'd rather factor.
What 2 numbers when added equal 2 and when multiplied equal -168.
Let's see.......how about.....................-12 and 14. (-12)(14)=-168
-12+14=2
$x^{2}-12x+14x-168$
$(x^{2}-12x)+(14x-168)$
$x(x-12)+14(x-12)$
(x-12)(x+14)
4. Hello, Chester!
You did an excellent on job on the hard part: setting up the equation.
I assume you know to solve a quadratic equation
. . and that you know how to factor.
Did you run of factorings to try?
$x^2+2x-168\:=\:0$
We want two numbers with a product of 168 and a difference of 2.
Start dividing 168 by 1, 2, 3, . . .
$168 \div 1 = 168\quad\Rightarrow\quad 1\cdot168$
$168 \div 2 = 84\quad\Rightarrow\quad 2\cdot84$
$168 \div 3\;\;\text{ not exact}$
$168 \div 4 = 42\quad\Rightarrow\quad 4\cdot42$
$168 \div 5,\; 168 \div 6,\;\168 \div 7\;\;\text{ not exact}$
$168 \div 8 = 21\quad\Rightarrow\quad 8\cdot21$
$168 \div 9,\;168 \div 9,\;168\div 10,\;168\div11\;\;\text{ not exact}$
$168 \div 12 = 14\quad\Rightarrow\quad 12\cdot14\;\;\leftarrow$ There! a difference of 2
Therefore: . $x^2 + 2x - 168 \;= \;(x - 12)(x + 14)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Of course, you can use the Quadratic Formula for factoring . . .
5. ## Thanks
Thanks again everyone.
I really appreciate the help!
Chester | {
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5. ## Thanks
Thanks again everyone.
I really appreciate the help!
Chester
6. I believe the answer is $12\cdot14$ not $(-12)\cdot14$
notice that negative 12 and 14 are not consecutive even integers.
7. Originally Posted by Quick
I believe the answer is $12\cdot14$ not $(-12)\cdot14$
notice that negative 12 and 14 are not consecutive even integers.
There two answers for $x$.
$x=12, x+2=14$ and $12\times 14=168$
And,
$x=-14,x+2=-12$ and $-14\times -12=168$ | {
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# Toronto Math Forum
## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:05:34 PM
Title: Q7 TUT 0201
Post by: Victor Ivrii on November 30, 2018, 04:05:34 PM
(a) Determine all critical points of the given system of equations.
(b) Find the corresponding linear system near each critical point.
(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?
(d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
\left\{\begin{aligned} &\frac{dx}{dt} = 1 - xy, \\ &\frac{dy}{dt} = x - y^3. \end{aligned}\right.
Bonus: Computer generated picture
Title: Re: Q7 TUT 0201
Post by: Yulin WANG on November 30, 2018, 04:39:52 PM
(a)
\left\{
\begin{array}{**lr**}
1-xy=0 & \\
x-y^{3}=0\\
\end{array}
\right.
\left\{
\begin{array}{**lr**}
xy=1 & \\
x=y^{3}\\
\end{array}
\right.
\left\{
\begin{array}{**lr**}
x=1 & \\
y=1\\
\end{array}
\right.
or
\left\{
\begin{array}{**lr**}
x=-1 & \\
y=-1\\
\end{array}
\right. | {
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Therefore, the critical points are (1,1) and (-1,-1)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
-y & -x \\
1 & -3y^{2}
\end{bmatrix}\\
~\\
J(1,1) &= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
J(-1,-1) &= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}
\end{align*}
(c)
\begin{align*}
For (1,1), let A&= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
-1-\lambda & -1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda+1)+1=0\\
~\\
\lambda_{1} &= \lambda_{2} = -2 \\
~\\
Then \ the \ system \ has \ a \ stable \ improper \ node \ at \ (1,1) \\
~\\
For (-1,-1), let A&= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda-1)-1=0\\
~\\
\lambda = -1 \pm \sqrt{5} \\
~\\
Then \ the \ system \ has \ a \ unstable \ saddle \ point \ at \ (1,1) \\
\end{align*}
(d) In the attachment.
Title: Re: Q7 TUT 0201
Post by: Zhuojing Yu on November 30, 2018, 06:29:30 PM
I think when (1,1), it is node or spiral point, not IN(improper node).
Title: Re: Q7 TUT 0201
Post by: Jingze Wang on November 30, 2018, 08:31:23 PM
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Title: Re: Q7 TUT 0201
Post by: Yulin WANG on November 30, 2018, 11:13:57 PM
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer?
Title: Re: Q7 TUT 0201
Post by: Victor Ivrii on December 01, 2018, 03:57:59 AM
Sure, it is improper node. Jingze finally got a decent computer generated picture (took a correct range of variables)
Syllabus lists several free sites. My favourite -- downloadable pplain java applet | {
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