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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2800 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 196 Kudos [?]: 1162 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : | {
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on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [1] , given: 186 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... | {
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http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 | {
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Re: Inequalities trick [#permalink] 11 Nov 2014, 08:52
1
KUDOS
For statement I
(1) (1-2x)(1+x)<0
Once you get this into the req form
2 (x-1/2) (x+1) > 0
sol +++++ -1 ------ 1/2 +++++
As Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there.
So sol for this x> 1/2 and x<-1
Integers greater than 1/2 and less than -1 , thus |x| may be >= 1. Unsure
Thus Insufficient.
For Statement II
(2) (1-x)(1+2x)<0
Once you get this into the req form
2(x+1/2) (x-1) > 0
sol +++++ -1/2 ------ 1 +++++
So sol for this x>1 and x< -1/2
Integers greater than 1 and less than -1/2 , thus |x| may be >= 1. Unsure
Thus Insufficient.
Combining Both the statements
x<-1 and x>1
Thus Integers for this range will give |x| > 1
Thus Sufficient.
Hope this helps. I am not very confident of my solution though. Its my first solution gmatclub
mayankpant wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - + | {
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If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Hi
Can you please explain this question to me using the graph. I am missing the point when graph is being used here?
If x is an integer, is |x|>1?
(1) (1-2x)(1+x)<0
(2) (1-x)(1+2x)<0
For me ,the first equations roots are -1 and 1/2. Now I am struggling to get to the correct sign using the graph method here.
Same for second equation: roots are 1 and -1/2 but struggling for the sign.
THanks
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Re: Inequalities trick [#permalink] 16 Mar 2010, 09:46
Can u plz explainn the backgoround of this & then the explanation.
Thanks
Re: Inequalities trick [#permalink] 16 Mar 2010, 09:46
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# Why do we care about the fact that a series converges, but not what it converges to?
In the 10 tests of convergence/divergent (that I know), them being,
by defn, integral test, div test, comparison test, limit comparison test, gs test, alternating series test, p-series test, root test, ratio test
the only ones that can tell me what the series converges to is by defn and the gs test.
The other ones can tell me if the series converges or diverges.
But I've always wondered, why do we care whether it converges, if we cannot exactly figure out what it converges to? Why is it helpful to know only the behaviour and not the actual value? | {
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• Of course we care about both, but in practice it's generally a lot easier to decide convergence than to actually compute the sum. A good convergence result often comes with information about the speed of convergence...so that we can use it to compute the series numerically, which is often the best we can manage. – lulu Apr 20 '17 at 16:58
• There are ways (in specific cases) to prove that if a sequence converges, then it must converge to $L$. For example, some recurrences, if they converge, converge to a fixed point. Also, for many of the tests, you can get estimates on the remaining error. – Michael Burr Apr 20 '17 at 16:59
• You can numerically compute the sum and see it goes to a certain approximate value, but what if it diverges slowly? The most dangerous and famous example is the harmonic series, which diverges but really slowly. So, knowing if it converges in the first place is essential. – AspiringMathematician Apr 20 '17 at 17:10
• Well, a professor of mine once said "Before you go digging around in the mud, you want to know that there's a potato in there". In context it depends on what you want to do. If the goal is not to actually solve an equation but just to understand a behavior--- say, you want to know if a function is bounded or not. If it is bounded/unbounded you can say such and such will happen. We just want to know the behavior, we don't care what the bound itself is nescessarily. Except when we do. – fleablood Apr 20 '17 at 17:15
Here are some ideas:
• Tests like the integral test actually include bounds on the error. For example, if the integral test applies to $\sum a_n$, then you could compute $\sum_{n=1}^Na_n$ and use the integral test on the remainder to bound the remainder.
• The alternating series test also comes with a bound on the error, so you know how close your partial sum is to the right answer (if you were working on a computer, for example). | {
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• The ratio test essentially tells you that the the terms (eventually) act like a geometric series. Therefore, if you can get a handle on how close something is to being a geometric series, you can use the geometric series computation to bound the remainder.
• If you have a recurrence like $x_n=ax_{n-1}+b$, then you can prove that if the sequence converges, then it must converge to a fixed point, i.e., a point that satisfies $L=aL+b$, $L=\frac{b}{1-a}$. However, to use this, one must first justify why the sequence converges at all.
• :o I didn't know about this "bounds on error" thing... thanks! – K Split X Apr 20 '17 at 17:08
If the series $\sum_{n=0}^\infty a_n$ converges, then I already know what it converges to: the actual value is $\sum_{n=0}^\infty a_n$.
There are lots of ways to represent numeric values, and "the sum of this series" is one of them. Often, it is a very useful way to write the actual value. Sometimes, it's even the best way we have to do so.
I don't want to replace any of the answers given so far and rather stress just one point: To sometimes consider the questions of convergence independently from the computation of the limit is an important educational step, in order to build up a more abstract point of view on convergence.
• I guess, there are many reasons to discuss convergence of series as one of the first topics in an analysis course. Analysis of series is a bit more abstract, than analysis of sequences, but simple enough to present elementary proofs for the important theorems. Students see how exchange of limits can lead to wrong results. (This can be also archived with sequences, but harmonic series as counterexample is simpler to remember.) They can to practice how to estimate and handle an abstract term like a sum with unknown entries.
Why is it important to consider convergence and computation of limit points separate ? | {
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Why is it important to consider convergence and computation of limit points separate ?
• Some spaces are constructed to be the completion of certain smaller spaces. For example the real numbers can be constructed as the set of all limit points of rational Cauchy-convergent sequences (with identifying some limit points as being equal). Or sometimes you build function spaces in this way, for example the set of limits of smooth functions with respect to some Sobolev norm. When working in these spaces, it is often important to handle convergence of sequences without using a closed form for the limit point.
• Many numerical simulations use some kind of series or sequences to approximate complicated terms, like solutions of differential equations. In these situations the computer will compute sufficiently many terms to estimate the exact limit. It is of great interest to prove theoretically, that these series converge to a unique limit. In many cases the limit of the sequence exist only as an abstract object, being a solution of a certain differential equation. It is common the consider the general convergence to some point (~ Stability) independently from conditions to ensure that an existing unique limit is indeed the correct solution (~ Consistency). Therefore I consider it to be an important lesson to see how convergence and computing the limit point can be treated somehow independently from each. It can happen surprisingly easy, that numerical simulations converge to a false solutions if one of both condition is not satisfied, see for example the plots one the first slides here: Slides 3-5. | {
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A rhombus is a quadrilateral with all sides of equal length. Next: Rectangle→ Chapter 3 Class 8 Understanding Quadrilaterals; Concept wise; Rhombus, Rectangle, Square. Square. The main difference between Rhomboid and Rhombus is that the Rhomboid is a parallelogram in which adjacent sides are of unequal lengths and angles are oblique and Rhombus is a quadrilateral in which all sides have the same length. Another interesting thing is that the diagonals (dashed lines) meet in the middle at a right angle. What is the difference between Rhombus and Rectangle? Yes, because a square is just a rhombus where the angles are all right angles. Main Difference . The difference between a kite and a rhombus is that a kite does not always have four equal sides or two pairs of parallel sides like a rhombus. Is Rhombus a Square? Keep in mind that the question "Is a rhombus a square?" 1 Descriptions; 2 Rhombus vs Parallelogram; 3 Comparison Chart; Descriptions A rhombus . (c) All angles are equal to 90 degrees. Venn diagrams are schematic diagrams used in logic and in the branch of mathematics known as … Rectangle Square Example 7 Example 8 Ex 3.4, 5 Ex 3.4, 6 … The three special parallelograms — rhombus, rectangle, and square — are so-called because they’re special cases of the parallelogram. that's the main difference!! Here are the differences between a square and a rhombus. As nouns the difference between rhombus and square is that rhombus is while square is any simple object with four nearly straight and nearly equal sides meeting at nearly right angles. Square… ADVERTISEMENT. Rhomboid. Trigonometry. Think about what happens when you "tilt" a square into a rhombus, keeping the horizontal lines of the square horizontal, but rotating the vertical lines a little bit clockwise. The other opposite sides need not be parallel. Please find the attached file for the answer. In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the | {
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geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. On the other hand, if you bisect the diagonals of a rectangle, they bisect each other at equal length but … To calculate the perimeter, you have 4 x length of one side or 4A, where A is the length of the side. If there are 32.6 million Americans in this age group, find the tot … al U.S. … Among these, many people get confused with rhombus and parallelograms and wonder if they are similar or if the terms are used interchangeably. Therefore, the difference between a square and a rhombus is that {eq}\displaystyle{ \rm \boxed{ \text{1) A square has all sides equal and all angles equal, but a rhombus … The table below offers a clear distinction between them. The plural is rhombi or rhombuses, and, rarely, rhombbi or rhombbuses (with a double b). On a square, both of the diagonals equal each other, but on a rhombus … I seems clear that the orientation of the figure has a lot to do with the name applied to it. • Considering the diagonals; – The diagonals of the rhombus bisect each other at right angles, and the triangles formed are equilateral. Also opposite sides are parallel and opposite angles are equal. Trapezoid A quadrangle which has only one of the two opposite sides as parallel is called a Trapezoid. WHAT IS DIFFERENCE BETWEEN SQUARE AND RHOMBUS WHAT IS DIFFERENCE BETWEEN SQUARE AND RHOMBUS JUSTIFY YOUR ANSWER pls ans both u will get 28 pt I(t) = 27, 992 * (1.017) ^ t covert to log Americans who are 65 years of age or older make up 13.1% of the total population. To distinguish a rectangle from rhombus following property should be kept in mind: 2. All sides of the rectangle are at right angle while this is not the case with rhombus shape. In other words … 3. While it may be a little bit confusing since a square is also a rectangle, there is one difference that makes it easy to distinguish one from the other. Thus a rhombus is not a square unless the angles | {
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makes it easy to distinguish one from the other. Thus a rhombus is not a square unless the angles are all right angles. Compare the new area to the original area. Its properties are (a) All sides are equal. The rhombus has a square as a special case, and is a special case of a kite and parallelogram. What’s the difference between a square and a rectangle? rhombus . The most common types of quadrilaterals are a square, rectangle, rhombus, parallelogram, trapezium, and kite. Rhombus Rhombus and Kite You are here. Rhombuses, rectangles, and squares 2. If both pair of opposite sides of a … All the angles of a square are equal … As nouns the difference between rhomboid and rhombus is that rhomboid is a parallelogram which is neither a rhombus nor a rectangle while rhombus is . 3. – The … English (fish) (wikipedia rhombus) Noun … The top right one, the original “diamond” is actually a square. 10 thoughts on “ Rhombus vs Diamond ” howardat58 February 11, 2016 at 2:57 am. As with all quadrilaterals, the sum … in the above quadrilateral family tree works just like. In this article, let us discuss the difference between rhombus and parallelogram in detail. Square A square is a parallelogram with right angles and equal sides. Rhombus and square have all sides equal, to distinguish a rhombus from square following property should be kept in Each angle of square has to be 90 degrees unlike rhombus. A square must have 4 right angles. (d) The diagonals are equal. Difference Between Rhombus and Parallelogram. (b) Opposite sides are equal and parallel. Unlike rhombus square needs to have all angles equal to 90 degree. A rhombus, also known as “rhom” or “diamond,” is an equilateral quadrilateral, a term that refers to a figure with four parallel sides (the lines will never intersect even if they … Solved Problems : Home >> Parallelogram >> Difference & Similarity between Square & Parallelogram >> Difference & Similarity between Square & Parallelogram. The Rhombus. To distinguish a | {
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>> Difference & Similarity between Square & Parallelogram. The Rhombus. To distinguish a rectangle from rhombus following property should be kept in mind: 2. Unlike rhombus only opposite sides of rectangle are equal. Rhombus vs … Harlon Moss. Area of rhombus = (1/2) x (d₁ x d₂) Here d₁ and d₂ are diagonals. A square is a quadrilateral with all sides equal in length and all interior angles right angles. This is true for all squares, so ALL SQUARES ARE RHOMBI (the plural of rhombus). but square is a quadilateral having all 4 sides equal but having all angles of its 4 sides as 90 degree . (In addition, the square is a special case or type of both the rectangle and the rhombus.) means Is every rhombus also always a square? As a adjective square is shaped like a (the polygon). Trigonometry Ratios. A kite is one type of four sided figure known as a quadrilateral. A kite is a four-sided shape that has two sets of adjacent sides that have equal lengths. It is more common to call this shape a rhombus, but some people call it a rhomb or even a diamond. Each angle of rectangle has to be 90 degrees unlike rhombus. Views: 2. A rhombus, on the other hand, does not have any rules about its angles, so there are many many, examples of a rhombus that are not also squares. In contrast, a rhombus is a shape having four equal sides but in a diamond shape. A rhombus is a four-sided shape where all sides have equal length (marked "s"). All rhombuses and squares are also kites. A rhombus is a quadrilateral with all sides equal in length. To distinguish a rectangle from rhombus following property should be kept in mind: 2. Same is the case with a rhombus … In general, a rhombus has the following special properties • All four sides are equal in length. Other Names. The rhombus is often called a diamond, after the … Each angle of square has to be 90 degrees unlike rhombus. Updated: February 27, 2019. As you can see in diagram … The shaded region represents the result of the operation. The | {
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27, 2019. As you can see in diagram … The shaded region represents the result of the operation. The different quadrilaterals are square, rectangle, rhombus, parallelogram, kite, trapezium which have some similar characteristics. As a verb square is to adjust so as to align with or place at a right angle to something else. Rhomboid vs. Rhombus. Rhombus and square have all sides equal, to distinguish a rhombus from square following property should be kept in Each angle of square has to be 90 degrees unlike rhombus. Sum of interior angles will be 360 degree Difference between square and rhombus: A square also fits the definition of a rectangle (all angles are 90°), and a rhombus (all sides are equal length). Each angle of rectangle has to be 90 degrees unlike rhombus. There are many shapes which give the impression of being associated to 1 one other, nevertheless you probably have a take a look at it, there are only some variations between them. 5 Min Read. As a adjective rhomboid is resembling, or shaped like a rhombus or rhomboid. Similarly, the area is A = b x h. Main Difference Between Parallelogram vs Rhombus . • Area of any can be calculated using the formula base ×height. The square can be considered as a special case of the rhombus, where the internal angles are right angles. Each angle of rectangle has to be 90 degrees unlike rhombus. Unlike rhombus only opposite sides of rectangle are equal. A square, however, has four equal interior angles. Rhombus and Parallelograms are different although they both have four sides and four vertices and look almost similar. (AB=DC=AD=BC) • The diagonals of the rhombus bisect each other at right angles; diagonals are perpendicular to each other, in addition to the following properties of a … Area of Parallelogram … Note that a square is also a rhombus because it has 4 equal sides as well. Rhombus, Rectangle, Square; Rhombus and Kite. Opposite sides of a rhombus are parallel. Rectangle and rhombus are special cases of the | {
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and Kite. Opposite sides of a rhombus are parallel. Rectangle and rhombus are special cases of the parallelograms. rhombus is a quadilateral having all 4 sides equal but not having all angles 90 degree. Learn difference between square & rhombus topic of maths in details explained by subject experts on vedantu.com. The difference is a rhombus is made up of 2 acute and 2 obtuse angles and a square is made up of 4 right angles. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. Rhombus Square Three dimensional object Trapezium Triangle Quadrilateral. Data-Handling. Answer: Rhombus is a four-sided quadrilateral with all its four sides equal in lengths.Rhombus is also referred to as a slanting square. You can come back to reference it in the future if you have any … Published: 14 Jan, 2021. • Rhombus and rectangle are quadrilaterals. CONTINUE READING BELOW. ADVERTISEMENT. Arithmetic Mean Frequency Distribution Table Graphs Median Mode Range . The major difference between rectangle and rhombus is that, the opposite sides of the rectangle are equal. Contents. The properties of square and rhombus are similar, the only distinguishing property is that square has all the angles equal to 90 0, and rhombus does not.. Rhombus is also called an equilateral quadrilateral. This article will explore the difference between a rhombus and a parallelogram. Square… Opposite angles are congruent in a rhombus. A square however is a rhombus … Diagram 2. 3. A Rhombus is a quadrilateral: A Square is also a quadrilateral: Opposite sides of Rhombus are parallel to each other : Opposite sides of Square are also parallel to each other: All sides of Rhombus are of equal length: All side of Square are also of equal length: Adjacent Angles of Rhombus are supplementary: Adjacent angles of Square are also supplementary : Diagonals of Rhombus … Rhomboid is a related term of rhombus. … Both a square and a rectangle may have their four angles at | {
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Rhomboid is a related term of rhombus. … Both a square and a rectangle may have their four angles at 90°, have parallel opposite sides with the same length, and have their diagonals equal in length, but only a square … Answer: No, a rhombus is not a square. ! Last updated at Sept. 25, 2018 by Teachoo Subscribe to our Youtube Channel - https://you.tube/teachoo. But these shapes are different from each other because of their few properties. The three-level hierarchy you see with. The name "rhombus" comes from the Greek word rhombos: a piece of wood whirled on a string to … Properties of rhombus: It has four equal sides ; Both diagonals are intersecting each other at right angle. 3 difference between Rhombus & square 2 See answers sufiyaanwar200451 sufiyaanwar200451 Answer: Here is four diffence b/w rhombus and square hope it help u. Step-by-step explanation: Purplehidie Purplehidie Answer: The sides of a square are perpendicular to each other whereas the sides of a rhombus are not perpendicular to each other. Unlike rhombus only opposite sides of rectangle are equal. Main Difference. Parallelograms are different although they both have four sides all have the length... H. Main difference between square and a parallelogram with right angles and equal sides ; both are... Rectangle and rhombus are special cases of the figure has a lot to do with the name to... A shape having four equal sides each angle of square has to be 90 unlike... Or place at a right angle almost similar special properties • all four sides and vertices! Represents the result of the rectangle and rhombus is a quadrilateral with all four. Rectangle→ Chapter 3 Class 8 Understanding Quadrilaterals ; Concept wise ; rhombus rectangle... Sides that have equal lengths referred to as a special case or type four. Angle to something else its sides are equal in lengths.Rhombus is also referred to as a adjective is. Object Trapezium Triangle quadrilateral equal sides to calculate the perimeter, you have x. | {
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in explained. Square… answer: rhombus is not the case with rhombus and kite plural rhombus... Since equilateral means that all of its 4 sides equal but having all angles 90 degree these, people! Comparison Chart ; Descriptions a rhombus or rhomboid a quadilateral having all 4 equal. Have four sides all have the same length after the … Learn difference between parallelogram vs rhombus. its are! The following special properties • all four sides all have the same.! Rectangle are equal and d₂ are diagonals the rectangle are equal having all angles are all right angles so. For all squares, so all squares, so all squares are (... | {
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difference between rhombus and square 2021 | {
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# Math Help - Jacobi Symbol
1. ## Jacobi Symbol
Evaluate the Jacobi symbol ((n−1)(n+1)/n) for any odd natural number n.
Trying out some numbers, I THINK it alternates between 1 and -1, but how can we PROVE it formally?
Any help is appreciated!
[also under discussion in math link forum]
2. $\left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$
For $n$ odd, $\left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.
For $n$ even, take $n=2k$, then $\left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
3. Originally Posted by chiph588@
$\left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$
For $n$ odd, $\left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.
For $n$ even, take $n=2k$, then $\left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
Thank you!
But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
4. Originally Posted by kingwinner
Thank you!
But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
Oops! You're right! | {
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5. But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)
Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?
Thanks!
6. Originally Posted by kingwinner
But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)
Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?
Thanks!
$\left(\frac{a}{n}\right)
= \begin{cases}
\;\;\,0\mbox{ if } \gcd(a,n) \ne 1
\\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$
So sub in $0$ and $1$ and see what you get.
7. Originally Posted by chiph588@
$\left(\frac{a}{n}\right)
= \begin{cases}
\;\;\,0\mbox{ if } \gcd(a,n) \ne 1
\\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$
So sub in $0$ and $1$ and see what you get.
gcd(0,1)=1, so that rule says that (0/1)=+1 OR -1, but how do we know whether it is +1 or -1?
8. Jacobi symbol
Apparently the answer is $0$. I haven't had too much exposure to the Jacobi symbol so I can't really tell you why. Check out the site for yourself though.
9. $\left( \frac{0}{1} \right)=1$ because the set of prime factors of $1$ is empty. if $n > 1$ is an odd integer, then $\left( \frac{0}{n} \right)=0.$
10. Originally Posted by NonCommAlg
$\left( \frac{0}{1} \right)=1$ because the set of prime factors of $1$ is empty. if $n > 1$ is an odd integer, then $\left( \frac{0}{n} \right)=0.$
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??
Also, is it true that, by definition, (a/1)=1 for any integer a?
Can someone clarify this? Thank you!
11. Originally Posted by kingwinner
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??
Also, is it true that, by definition, (a/1)=1 for any integer a? | {
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Also, is it true that, by definition, (a/1)=1 for any integer a?
Can someone clarify this? Thank you!
12. Originally Posted by NonCommAlg
Is there any real reason why we define (a/1)=1 for any integer a?
Why not define it to be 0? why not -1?
13. Originally Posted by kingwinner
Is there any real reason why we define (a/1)=1 for any integer a?
Why not define it to be 0? why not -1? | {
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# Number of elements vs cardinality vs size
I have been wondered the definition of cardinality and number of elements. One mathematician told me that one can't said that the cardinality or size of the set $\{1\}$ is one, it should be said that the number of elements in the set is one. I guess that his opinion is that there is no term size in mathematics. Is these true? On the other hand, if we have an infinite set like $\mathbb Z$, can we say that the number of elements of $\mathbb Z$ is infinite or is the term "number of elements" used only in finite sets?
-
I have no idea what the mathematician was thinking. The cardinality of $\{1\}$ is indeed $1$. Of course it’s also true that $1$ is the number of elements in the set. I would probably not use the expression number of elements when the set is infinite, but I’d not complain if someone said ‘$\Bbb R$ has $2^\omega$ elements’ instead of ‘the cardinality of $\Bbb R$ is $2^\omega$’. – Brian M. Scott Jun 15 '12 at 10:51
For finite sets there is absolutely no problem; a set has cardinality or size $n$ if it is in bijection with $\{ 1, 2, ... n \}$, and it has cardinality or size $0$ if it is empty. (The empty set is unique!) It is silly to prevent yourself from being able to say this. – Qiaochu Yuan Jun 15 '12 at 12:39
Converting my comment to an answer:
I have no idea what the mathematician was thinking: the cardinality of the set $\{1\}$ is indeed $1$, and that’s a perfectly normal way to express the fact. Of course it’s also true that $1$ is the number of elements in the set.
I would probably not myself use the expression number of elements when speaking of an infinite set, but ‘$\Bbb R$ has $2^\omega$ elements’ is a perfectly fine synonym of ‘the cardinality of $\Bbb R$ is $2^ω$’. | {
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-
Okay. I guess the mathematician wanted that I won't be his graduate student as he asked me to modify my bachelor thesis so heavily. That was one point where I made a "mistake" and got a bad grade. And is the word "size" a synonym of the cardinality? – mathenthusiast Jun 15 '12 at 15:52
@mathenthusiast: I wouldn’t say so: it’s more general, as it can also refer to length, area, volume, etc. – Brian M. Scott Jun 15 '12 at 21:38
The question you need to ask yourself is "what is the meaning of the notion number?"
Natural numbers can be used to measure size, length, etc. while rational numbers measure ratio between two integers, real numbers measure length... what do complex numbers measure?
In mathematics we allow ourselves to define new ways of measuring things, and usually we refer to the values of these measures as numbers. Indeed there is no real difference between the length of a $100$ meters running track, and a $100$ meters long hot dog...
When measuring the size of a set we came up with a clever way to discuss infinite sets. This way is "cardinality", and the cardinal of a set represents in a good sense the "number of elements in the set". So it has a perfectly good meaning to say that a set has $\aleph_0$ many elements, or that a set is of size $\aleph_1$.
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# Integration by Partial Fractions $\int \sec^3xdx$
I have looked at multiple ways to do partial fractions to integrate $$\sec^3x$$, but there is a part where I keep getting stuck when I see the partial fractions get split up.
$$\int \frac{\cos x}{\cos^4 x} = \int \frac{1}{(1-y^2)^2}$$
The next step I have seen is shown like this on Wikipedia:
$$\frac{1}{(1-u^2)^2}=\frac{1/4}{1-u}+\frac{1/4}{(1-u)^2}+\frac{1/4}{1+u}+\frac{1/4}{(1+u)^2}.$$
Where does the $$\frac{1}{4}$$ come from in the numerator?
And for the denominator, why are there two additional positive fraction decompositions $$1+u$$ and $$(1+u)^2$$?
• See this page for the general theorem in use. – Bernard May 2 at 12:35
• – cgiovanardi May 2 at 14:54
And for the denominator, why are there two additional positive fraction decompositions
Because $$1-u^2 = (1-u)(1+u).$$ It's the "difference of squares" factorization. And so, squaring both sides, we also have $$(1-u^2)^2 = (1-u)^2(1+u)^2$$.
In partial fractions decomposition, all irreducible factors must appear as denominators, to the highest power that appears, as well as to all lower powers. So that means we need a denominator for $$(1-u)$$ and $$(1-u)^2,$$ but also for $$(1+u)$$ and $$(1+u)^2.$$
Where does the $$\frac{1}{4}$$ come from in the numerator?
For partial fraction decomposition, the numerators must be chosen so that if you were to give every term common denominator and combine the numerators, you would get the starting fraction back. It takes a bit of algebra to find what numerators do the job, but that is where the $$\dfrac{1}{4}$$ comes from.
But let's do the algebra. The normal way to find these numerators is to introduce variables for them and solve. Like so:
$$\dfrac{1}{(1-u^2)^2} = \dfrac{1}{(1+u)^2(1-u)^2} = \dfrac{A}{1+u} + \dfrac{B}{(1+u)^2} + \dfrac{C}{1-u} + \dfrac{D}{(1-u)^2}$$
here the "variable" numerators are called $$A, B, C,$$ and $$D.$$
We solve for them by clearing denominators: | {
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We solve for them by clearing denominators:
$$1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2$$
Then in general you may have to expand all those binomials and match like terms of the resulting polynomials in $$u$$. But a shortcut can be to sub in the values of the roots in $$u$$. For example, at $$u=1,$$ the above equation becomes
$$1 = A(1-1)^2(1+1) + B(1-1)^2 + C(1-1)(1+1)^2+ D(1+1)^2 = D(1+1)^2.$$
So we have $$D=1/4.$$
Similarly, by next subbing $$u=-1,$$
$$1 = B(1-(-1))^2$$
So we have $$B=1/4.$$
To figure out that $$A$$ and $$C$$ are also $$1/4$$, we can either sub in some other, non-root values of $$u$$, or bite the bullet and do the algebra. (But if this were a partial fractions problem with no powers of irreducibles, we'd be done).
I'll do the algebra, I guess. Expand, and gather like terms in $$u$$, treating the coefficients $$A, B, C,$$ and $$D$$ as variables.
$$1 = A(1-u-u^2+u^3) + B(1-2u+u^2) + C(1+u-u^2-u^3) + D(1+2u+u^2) \\ = (A+B+C+D) + (-A-2B+C+2D)u + (-A+B-C+D)u^2 + (A-C)u^3.$$
Now for two polynomials in $$u$$ to be equal for all values of $$u$$, every coefficient must be equal. On the left-hand side we have the polynomial $$1$$, which has a constant coefficient of $$1$$ and all higher coefficients are $$0$$.
So this gives us the following equations:
$$1 = A+B+C+D\\ 0 = -A-2B+C+2D\\ 0 = -A+B-C+D\\ 0 = A-C.$$
We could solve this system of four linear equations in four unknowns, using Gaussian elimination or substitution or matrix methods. But first let's remember that we already know that $$B=D=1/4.$$ So we don't have four unknowns any more, our only unknowns left are $$A$$ and $$C$$. By the fourth equation, $$A=C$$, so our only unknown is really $$A$$. The first equation becomes
$$1 = A+B+C+D = A + \dfrac{1}{4} + A + \dfrac{1}{4}.$$
So $$2A=1-\dfrac{2}{4} = \dfrac{1}{2}.$$ And so $$A=B=C=D=\dfrac{1}{4}.$$ | {
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So $$2A=1-\dfrac{2}{4} = \dfrac{1}{2}.$$ And so $$A=B=C=D=\dfrac{1}{4}.$$
• Oh, seeing the way you wrote that makes sense (going from $1-u^2$ to $(1-u^2)^2)$, just adding ^2 to the factors – Evan Kim May 2 at 12:38
This is actually a problem of partial fractions.
When dividing one fraction to partial fractions, all possible cases of denominator should be considered. In this case, the denominator can be factorised to $$(1-u)^2(1+u)^2$$, so there are actually in total 8 possible fractions, with denominators being $$1-u$$, $$(1-u)^2$$, etc. Setting all the numerators to A, B, C, D, etc. for different denominator, layout the expression and compare coefficients, the result should then be that 1 fraction equal to 4 partial fractions.
Note that $$(1-y^2)^2=(y-1)^2(y+1)^2$$. On the other hand, every rational function of the form$$\frac{p(x)}{(x-r_1)^{m_1}\times\cdots\times(x-r_k)^{m_k}},$$where $$\deg p(x), can be written as a linear combination of rational functions of the type $$\frac1{(x-r_j)^N}$$, with $$N\in\{1,2,\ldots,m_j\}$$. The rest is a matter of computations. | {
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# Applying the chain rule and product rule
AP Calc: FUN‑3 (EU)
## Video transcript | {
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- [Instructor] What we're going to do in this video is try to find the derivative with respect to X of X squared sin of X. All of that to the third power. And what's going to be interesting is that there's multiple ways to tackle it. And I encourage you to pause the video and see if you could work through it on your own. So there's actually multiple techniques. One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to X of something to the third power. So, if I take the derivative it would be the derivative with respect to that something. So that would be three times that something squared times the derivative with respect to X of that something. Where the something, in this case, is X squared sin of X. X squared sin of X. This is just an application of the chain rule. Now, the second part, what would that be? The second part here do this in another color. In orange. Well here, I would apply the product rule. I have a product of two expressions. So I would take the derivative of, let me write this down. So this is gonna be the product rule. Product rule. I would take the derivative of the first expression. So, X, derivative of X squared is two X. Let me write a little bit to the right. This is gonna be two X times the second expression sin of X. Plus the first expression X squared times the derivative of the second one. Cosin of X. That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front. Which actually, let me just rewrite that. So all of this I could rewrite as let's see, this would be three times if I have the product raised to the second power I could take each of them to the second power and then take their products. So X squared squared is X to the fourth. And then sin of X squared is sin squared of X. And then all of that is being multiplied by that. And, if we want, we can algebraically simplify. We | {
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And then all of that is being multiplied by that. And, if we want, we can algebraically simplify. We can distribute everything out. In which case, what would we get? Well let's see. Three times two is six. X to the fourth times X is X to the fifth. Sin squared of X times sin of X is sin of X to the third power. And then, let's see, three so plus three, X to the fourth times X squared is X to the sixth power. And then I have sin squared of X sin squared of X cosin of X. So there you have it, that's one strategy chain rule first, and then product rule. What would be another strategy pause the video and try to think of it. Well, we could just algebraically use our exponent properties first. In which case, this is going to be equal to the derivative with respect to X of if I'm taking X squared times sin of X to the third power instead I could say X to the third power which is going to be X to the sixth. And then sin of X to the third power. Sin of X to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have if I'm doing the product things to some exponent, well that's the same thing if each of them raised to the exponent and then the product of the two. Now how would we tackle this? Well, I here, I would do the product rule first. So let's do that. So let's do the product rule. So, we're gonna take the derivative of the first expression. So, derivative of X to the sixth is six X to the fifth. Times the second expression. Sin to the third of X. Or, sin of X to the third power. Plus the first X to the sixth times the derivative of the second and I'm just gonna write that D DX of sin of X to the third power. Now, to evaluate this right over here it does definitely make sense to use the chain rule. Use the chain rule. And so what is this going to be? Well I have the derivative of something to the third power. So that's going to be three times that something squared times the derivative of that something. So in this | {
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going to be three times that something squared times the derivative of that something. So in this case, the something is sin of X. And the derivative of sin of X is cosin of X. Then I have all of this business over here. I have six X to the fifth. Sin to the third or sin of X to the third power. Plus X to the sixth. And if I were to just simplify this a little bit in fact, you see it very clearly. These two things are equivalent. This this term is exactly equivalent to this term the way it's written. And then this is exactly, if you multiply three times X to the sixth sin of X squared cosin of X. So, the nice thing about math if we're doing things that make logical sense we should get to the same endpoint. But the point here is that there's multiple strategies. You could use a chain rule first and then the product rule. Or you could use a product rule first, and then the chain rule. In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster. But sometimes these two are pretty close. But sometimes it'll be more clear than not which one is preferable. You really want to minimize the amount of hairiness, the amount the number of steps the chances for careless mistakes you might have. | {
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# Number of rectangles with odd area.
We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?
I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.
• But where was I wrong? – Omer Jun 14 '18 at 19:29
• @Omer: In counting the number of possible starting vertices (121). Not all of them are valid for all of the sub-cases you count. So, you go wrong at the very first step. Compare your thinking to how Hagen von Eitzen counted the cases, to correct the error. – Nominal Animal Jun 14 '18 at 19:55
• "and than assume that each possible odd width and height could be realized with this vertex." - no, s/he just assumed that 5 different odd x pairs and 5 different odd y pairs included the x and y of this vertex. If you start at (0,0) then in the x direction you can go to (1,0), (3,0), (5,0), (7,0), (9,0) but if you start at (5,0) you can go to (0,0), (2,0), (4,0), (6,0), (8,0), (10,0), that's 6 choices! – immibis Jun 15 '18 at 3:55
The reasoning is ok, except for this:
There are 5 possibilities to each | {
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The reasoning is ok, except for this:
There are 5 possibilities to each
Sadly, no. If the chosen vertex is at $(1,1)$ (grid starting at $(0,0)$) you have 6 possibilities for each direction.
In general, you have 6 possibilities if the coordinate is odd, 5 if it's even.
Fix: Because there $36$ all-even-coordinates vertices, $25$ all-odd, and $121-36-25=60$ mixed vertices, the correct counting is
$$36 \times 5^2+ 25 \times 6^2 + 60 \times5 \times 6=3600$$
Dividing by $4$ you get the $900$ rectangles.
I assume you're looking for rectangles with sides parallel to those of the square and vertices that are lattice points. A rectangle with integer sides has odd area iff the sides are both odd. For each pair of odd integers $x, y$ with $1 \le x,y \le 9$, there are $(11-x)(11-y)$ possible positions for a rectangle of size $x \times y$. Thus the number of rectangles is $$\sum_{x \in \{1,3,5,7,9\}} \sum_{y \in \{1,3,5,7,9\}} (11-x)(11-y) = 900$$ | {
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Fundamental group of this space
Based on this question: What is the homology groups of the torus with a sphere inside?
I'm trying to find the fundamental group of this space using the Seifert–van Kampen theorem. If $U$ is the torus and $V$ is the sphere, then $U\cap V$ is the circle, thus we have the following fundamental groups:
$\pi_1(U)=\mathbb Z\times\mathbb Z$
$\pi_1(V)=0$
$\pi_1(U\cap V)=\mathbb Z$
If we use the group presentation notation we have:
$\pi_1(U)=\langle\alpha,\beta\mid \alpha\beta=\beta\alpha\rangle$
$\pi_1(V)=\langle\emptyset\mid\emptyset\rangle$
$\pi_1(U\cap V)=\langle\gamma\mid\emptyset\rangle$
Thus using the Seifert–van Kampen theorem:
$\pi_1(X)=\langle\alpha,\beta\mid\alpha\beta=\beta\alpha,\beta\rangle$
Note that I added $\beta$ above because when we turn around the generator of $S^1$ which is $U\cap V$, we span one of the generators of the torus which is $U$.
Thus the fundamental group of this space is $\mathbb Z\times \{0\}$ which is $\mathbb Z$ itself.
My approach is correct?
Thanks a lot!
-
It looks like that one generator of fundamental group of torus vanishes since you can collapse a close path around the equator. – Sigur Jan 25 '13 at 2:02
@Sigur yes, you're right. – user42912 Jan 25 '13 at 2:10
@Sigur but what can you say about my proof? – user42912 Jan 25 '13 at 2:11
The approach is correct, but you can use a simpler method. The space you care about is homotopic to the wedge sum of two spheres and one circle. Thus the fundamental group is Z. To see why this space is homotopic to the space I said, you just check first that you can contract the upper half-sphere, thus the lower half-sphere becomes to a sphere, and the torus becomes a sphere with its south and north poles glued together. A sphere with its south and north poles glued together is homotopic to a sphere with a line connect its south and north poles. After these deformations you have a space which is what I said above. | {
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-
Thank you very much you helped not only with this question but with the question I linked about homology groups of this space. – user42912 Jan 25 '13 at 11:16
one question, do you know where can I find a book with these constructions (except Hatcher's book)? these constructions aren't so obvious at the first glance. – user42912 Jan 25 '13 at 11:18
@user42912 I was about to introduce you the Hatcher's book. You just have to learn the first two chapters you will get the idea. There are a lot of deformations of CW cplx. – lee Jan 25 '13 at 11:40
Nitpicking, but I suppose you could be more explicit about the relation $\beta=\gamma$, which comes from the inclusion $\iota:U\cap V\to U$, $\iota_*(\gamma)=\beta$. So you would have
$$\pi_1(X)=\langle\alpha,\beta,\gamma|\alpha\beta=\beta\alpha,\beta=\gamma\rangle$$
As always, it depends how familiar your audience (and you!) are with the subject.
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# Show that the set of all finite subsets of $\mathbb{N}$ is countable.
Show that the set of all finite subsets of $\mathbb{N}$ is countable.
I'm not sure how to do this problem. I keep trying to think of an explicit formula for 1-1 correspondence like adding all the elements in each subset and sending that sum to itself in the natural numbers, but that wouldn't be 1-1 because, for example, the set {1,2,3} would send to 6 and so would the set {2,4}. Multiplying all the elements in each subset and sending that product to itself in the natural numbers wouldn't work either since, for example, {2,3} would send to 5 and so would the set {1,5}. Any ideas?
• $2\times3\not=5$ :) – Ben Millwood Sep 21 '12 at 17:53
• Set of Finite Subsets of Countable Set is Countable at ProofWiki. See also this question for a generalization of your question ($\mathbb N$ is replaced by arbitrary infinite set $X$ and it is shown that set of all finite subset of $X$ has the same cardinality as $X$). – Martin Sleziak Sep 22 '12 at 6:50
• You're right, an explicit bijection would be a neat proof. Hint: think like a computer scientist. – Colonel Panic Jul 17 '15 at 10:05
Consider an enumeration of the (positive) prime numbers by $n\mapsto p_n$. Then if you're given the set $\{n_1,...,n_k\}$, map that to $$p_{n_1}\cdots p_{n_k}.$$ For example, if we've enumerated the primes increasingly ($p_1=2$, $p_2=3$, and so on), then $$\{1,3,4\}\mapsto 70=2\cdot 5\cdot 7=p_1\cdot p_3\cdot p_4,$$ and $$\{1,3\}\mapsto 10=2\cdot 5=p_1\cdot p_3.$$ | {
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You'll need the fact that there are countably infinitely many prime numbers, and uniqueness of prime factorization. From there, it shouldn't be hard to prove this is an injection from the set of finite subsets of $\Bbb N$ into $\Bbb N$, so there are at most countably many such subsets. To show that there are at least (and so exactly) countably many finite subsets of $\Bbb N$, you need only find an injection from $\Bbb N$ into the set of finite subsets of $\Bbb N$, which should be easy.
Added: The described map "should" send $\emptyset$ to $1,$ the "empty product."
• So if I was given a set {1,3} and a set {1,3,4}, you're saying to map that to 1,3 and 1,2^2,3? Am I understanding correctly? – user39794 Sep 21 '12 at 17:28
• Not quite. I'll include an example to clarify. – Cameron Buie Sep 21 '12 at 17:31
• It's ok I think I get it now from reading the other answer! :) Thanks though! – user39794 Sep 21 '12 at 17:32
• @Diego: Yes. However, the given map is not a bijection with $\Bbb N.$ For example, there is no finite subset of $\Bbb N$ that is mapped to $4.$ Rather, the map is an injection into $\Bbb N.$ – Cameron Buie Feb 12 '16 at 12:34
• @Diego: Your reasoning is a bit faulty. After all, $\{1\}$ is a subset of $\Bbb N,$ too, and we certainly can't conclude that there exists a bijection between $\{1\}$ and $\Bbb N$! In order to conclude that a subset of $\Bbb N$ is in bijection with $\Bbb N,$ we have to know that it is an infinite subset. The last part is aimed at proving that the range of the map is, indeed, infinite. – Cameron Buie Feb 12 '16 at 19:33
Define three new digits: \begin{align} \{ &= 10 \\ \} &= 11\\ , &= 12 \end{align} Any finite subset of $\mathbb{N}$ can be written in terms of $0,1,\cdots,9$ and these three characters, and so any expression of a subset of $\mathbb{N}$ is just an integer written base-$13$. For instance, $$\{1,2\} = 10 \cdot 13^4 + 1 \cdot 13^3 + 12 \cdot 13^2 + 2 \cdot 13^1 + 11 \cdot 13^0 = 289873$$ | {
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So for each finite $S \subseteq \mathbb{N}$, take the least $n_S \in \mathbb{N}$ whose base-$13$ expansion gives an expression for $S$. This defines an injection into $\mathbb{N}$.
More generally, any set whose elements can be written as finite strings from some finite (or, in fact, countable) alphabet, is countable.
• +1 Really nice answer. – Fly by Night Sep 2 '13 at 18:07
• The user @TimothySwan tried to edit your question to add a comment, I have copied his comment here since the user can't comment himself. He commented: However, this proof is not correct since there must be an injection from $\mathbb N$. – Alice Ryhl Sep 22 '14 at 14:34
• Thanks @Darksonn. (For the benefit of anyone else reading, that's not true; every infinite set has an injection from $\mathbb{N}$ into it!) – Clive Newstead Sep 22 '14 at 20:32
• (+1 a while ago) Not that it will be relevant to most users, but depending on the axioms one is working with, it may be possible for an infinite set to have no injection from $\Bbb N.$ – Cameron Buie May 10 '15 at 9:22
• @TTL: You're missing 'and these three new characters', namely open curly bracket, close curly bracket, and comma. – Clive Newstead Nov 15 '19 at 22:23
Alternatively, just use base $2$, so send $\{a_1,...,a_n\}$ to $2^{a_1}+\cdots+2^{a_n}$. (Assuming your $\mathbb N$ includes $0$, this is $1-1$ and onto.) | {
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• +1 Put in other way: any finite subset of positive integers can be represented as a binary string (with a finite number of ones), which in turn can be interpreted as a binary number which correspond to a positive integer. – leonbloy Mar 17 '14 at 20:09
• In my opinion, this is the best possible proof because the bijection is very, very explicit. – Martin Brandenburg Jul 17 '15 at 8:31
• Any base worked, right?(the base being an integer) – Diego Feb 12 '16 at 4:42
• Any base gives you a $1-1$ function. Base $2$ is onto. @Diego – Thomas Andrews Feb 12 '16 at 11:52
• @MartinBrandenburg: Since the set {a1,...an} isn't an ordered one, we have several base 2 numbers mapping to the same set. How's a bijection best constructed here, then? The accepted answer is also just an injection, no? – Nikolaj-K Mar 8 '17 at 14:57
The other answers give some sort of formula, like you were trying to do. But, the simplest way to see that the set of all finite subsets of $\mathbb{N}$ is countable is probably the following.
If you can list out the elements of a set, with one coming first, then the next, and so on, then that shows the set is countable. There is an easy pattern to see here. Just start out with the least elements.
$$\emptyset, \{1\}, \{2\}, \{1, 2\}, \{3\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}, \{4\}, \ldots$$
In other words, first comes $\{1\}$, then comes $\{2\}$. Each time you introduce a new integer, $n$, list all the subsets of $[n] = \{1, 2, \ldots, n\}$ that contain $n$ (the ones that don't contain $n$ have already showed up). Therefore, all subsets of $[n]$ show up in the first $2^{n}$ elements of this sequence. | {
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• In other words: List the elements of $P(\emptyset)$. List the elements of $P(\{1\}) \setminus P(\emptyset)$. List the elements of $P(\{1,2\}) \setminus P(\{1\})$. List the elements of $P(\{1,2,3\}) \setminus P(\{1,2\})$. And so on. – Martin Brandenburg Jul 17 '15 at 8:29
• The lazy approach is a 'surjective' listing, $PL([0,1]),PL([0,2]),PL([0,3]),\dots,PL([0,n]),\dots$ where $PL$ is powerset listing (with commas) and $[0,n]$ is the initial segment. – CopyPasteIt Aug 18 '17 at 19:26
• how do you actually show this is surjective and injective? – Pinocchio Sep 5 '18 at 16:18
• For a proof specification see this answer: Neatest proof that set of finite subsets is countable?. – CopyPasteIt Jun 2 at 14:21
I'm surprised nobody else has said this, so I worry that I've missed the point of the question. But I would argue as follows: Let $N_i$ be the family of all subsets of $\Bbb N$ with exactly $i$ elements. $N_0$ is finite, and $N_1$ is countable. $N_2$ is countable also, since it's a subset of $\Bbb N\times \Bbb N$. There's an easy induction argument that $N_i$ is countable for all $i>0$, since it's contained in the product $\Bbb N\times N_{i-1}$ of two countable sets.
Then, since the set you want, $\bigcup N_i$, is a countable union of countable sets, it is countable. | {
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• Well, a countable union of countable sets needn't be countable. It depends on your axioms. – Cameron Buie Sep 29 '12 at 22:09
• What axioms would those be? – MJD Sep 29 '12 at 22:28
• Set theoretic axioms--which you need at least some of before you can talk about sets, countability, unions, etc. For example, there are models of ZF in which $\Bbb R$ is a countable union of countable sets. If one takes countable choice (or stronger) as an axiom, then countable unions of countable sets are countable (that's a strictly weaker statement than countable choice). Now, it's worth noting that countable unions of enumerated sets (countable sets with a given bijection with $\Bbb N$) are countable, with no need for choice principles. – Cameron Buie Sep 29 '12 at 23:07
Let $p_1,p_2,\ldots$ the sequence of primes. Then map each finite subset $\{a_1,\ldots,a_n\} \subset \mathbb N$ to $p_1^{a_1}\cdots p_n^{a_n}$ where $a_1 < \ldots < a_n$. This gives an injective map from the set of finite subsets of $\mathbb N$ to $\mathbb N$.
Alternatively, notice that for every $s$ there are only finitely many finite subsets of $\mathbb N$ whose sum is $\leq s$. Let $A_s$ be the set of those subsets, then $\mathbb N = \bigcup_{s} A_s$ is a countable union of finite sets, hence countable.
• You beat me to the post ... slightly different, but the same important idea - You can show that the set $C_N$ of subsets of $\mathbb N$ whose members sum to $N$ is finite for each $N\in\mathbb N$. Then each finite subset of $\mathbb N$ has a finite sum $N$ and hence is in one of the $C_N$, and the union of the $C_N$ is a countable union of finite sets, hence is countable. – Mark Bennet Sep 21 '12 at 18:41
• @Mark: Not that it's probably relevant to the OP, but a countable union of finite sets needn't be countable in general without sufficient Choice. Here, of course, we're dealing with well-ordered finite sets, so it works out. – Cameron Buie Jul 31 '14 at 10:48 | {
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If $S$ is the set of all finite subsets of $\mathbb{N}$, define $f:S\rightarrow \mathbb{Q}$ by
$f(A)=.x_1x_2x_3x_4\cdots$ where $x_{i}=\begin{cases}1 &\mbox{, if }i\in A\\0 &\mbox{, if } i\not\in A \end{cases}$.
Then $f$ is clearly injective, so $S$ is countable since $\mathbb{Q}$ is countable and any subset of a countable set is countable. | {
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# Show that for any real constants $a$ and $b$, where $b > 0$, $(n + a)^b = \Theta(n^b)$
I'm currently studying growth of function chapter in Introduction to Algorithm. In exercise 3.1-2 the question is:
Show that for any real constants $$a$$ and $$b$$, where $$b>0$$,
$$(n + a)^b = \Theta(n^b)$$.
I understand I need to find constants $$c_1$$, $$c_2$$, $$n_0 > 0$$ where
$$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ for all $$n \geq n_0$$.
But I don't understand what should I do after above step. Looking at the solution, it shows
$$n + a \leq n + |a|$$
$$n + a \leq 2n$$, when $$|a| \leq n$$
and
$$n + a \geq n - |a|$$
$$n + a \geq \frac{n}{2}$$ , when $$|a| \leq \frac{n}{2}$$
How did they come up with $$n + a \leq n + |a|$$ and $$n + a \geq n - |a|$$ just from $$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ ? Are they derived from $$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ ? Or are they just a general knowledge that can help solve this problem?
Thank you!
UPDATE: I have updated the question to be more specific.
• $(n+a)^b = n^b + c_1n^{b-1}a+... = \theta(n^b)$ – Mr. Sigma. Dec 19 '18 at 13:40
• @Mr.Sigma.: $b$ can be any positive real number, say 0.5. – rici Dec 19 '18 at 13:54
• O'Luck: $\mid a\mid$ is the absolute value of $a$; either $a$ or $-a$. If $a$ is positive, $n+ |a| = n + a$; otherwise $a$ is negative and $n+a < n < n + |a|$. What problem do you have with that? – rici Dec 19 '18 at 14:00
• "Where did they get those inequalities?" Imagine that $n$ is very large. – John L. Dec 19 '18 at 14:12
• Please do not post equations, text, and the like as images since these are incompatible with the site's search engine. Thank you. – dkaeae Dec 19 '18 at 14:20
I understand I need to find constants $$c_1$$, $$c_2$$, $$n_0 > 0$$ where
$$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ for all $$n \geq n_0$$.
But I don't understand what should I do after above step.
The approach you would like to use here is "going backwards from the target". | {
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The approach you would like to use here is "going backwards from the target".
Suppose we do have $$c_1n^b \leq (n+a)^b$$ for some constant $$c_1\gt0$$ when $$n$$ is large enough. What can we understand or deduce from it? What is needed to make it true?
Since the exponent is the same, we can use the law of same exponent, $$(x/y)^m = x^m/y^m$$ to simplify that inequality so that we will have the appearances of our variable $$n$$ closer to each other.
$$c_1\leq \frac{(n+a)^b}{n^b}=\left(\frac{n+a}n\right)^b$$
Raising both sides to the power of $$\frac1b$$, we have
$$(c_1)^{\frac1b}\leq \left(\frac{n+a}n\right)^{b\,\frac1b}=\frac{n+a}n = 1+\frac an$$
(Another way to obtain $$(c_1)^{\frac1b}\leq\frac{n+a}n$$ is to raise both sides of $$c_1n^b \leq (n+a)^b$$ to the power of $$\frac 1b$$.)
Since $$c_1$$ is a positive constant, $$(c_1)^{\frac1b}$$ is also a positive constant. To make the above inequality hold, we would like to make $$\left|\frac an\right|$$ small enough. For example, we can require $$\left|\frac an\right|\le \frac12$$. That is why you see the following condition.
$$|a|\le\frac12 n$$
What is nice here is that we can reverse the above argument to obtain a wanted constant $$c_1$$. Note that it is just as fine if we choose a different condition such as $$|a|\le\frac13 n$$ or $$|a|\le\frac1{2019}n$$ or infinitely many others.
If we start from $$(n+a)^b \leq c_2n^b$$, we could arrive at $$|a|\le n$$.
Are they derived from $$0\le c_1n^b\le(n+a)^b\le c_2n^b$$ ? Or are they just a general knowledge that can help solve this problem?
Yes, they are derived as you have just seen. You could say that they are general knowledge that helps solve this problem. You could also say that we just discovered some specific facts that help solve this problem. | {
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• Wow thank you so much @apass-jack for your answer. This type of step-by-step answer is what I'm looking for. I guess I'm pretty weak at math. Before I accept your answer, I want to ask a few questions. In $(c_1)^{\frac1b}\leq \left(\frac{n+a}n\right)^{b\,\frac1b}=\frac{n+a}n = 1-\frac an$ , isn't it supposed to be $1+\frac an$ at the end? – O'Luck Dec 19 '18 at 23:12
• Thanks for pointing out my typo. Updated. – John L. Dec 19 '18 at 23:35
• Thank you. Next question is: "To make the above inequality hold, we would like to make $\left|\frac an\right|$ small enough". Why? If we choose big value for $\left|\frac an\right|$ , let say $\left|\frac an\right| > 1$ the inequality is still satisfied right? – O'Luck Dec 20 '18 at 0:01
• Reason one: when $n$ goes to (positive) infinity, $|\frac an|$ goes to 0. So you cannot make it as big as you want. In fact, $|\frac an| \le |a|$ if we assume $n$ is a positive integer. Reason one is enough. Reason two: even if you could magically, suppose $a$ is negative, $\frac an$ is negative, too, which is bad for $1+\frac an$. – John L. Dec 20 '18 at 0:12
• umm, I mean $|\frac an|$ as a whole not only the $n$. So, with $(c_1)^{\frac1b}\leq 1 + \frac an$ if $|\frac an| > 1$ let say $|\frac an| = 2$ then $(c_1)^{\frac1b}\leq 1 + 2$ , the inequation is still satisfied, am i right? – O'Luck Dec 20 '18 at 1:10
If n is large then we have n/2 ≤ n + a ≤ 2n.
We therefore have $$(1/2)^b n^b ≤ n^b ≤ 2^b n^b$$.
There you have your two constants: $$(1/2)^b$$ and $$2^b$$. | {
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# Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die.
Bob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.
I started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.
Obviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem.
PS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.
• The probability asked for in the title $P(\text{D20 beats 3 D6s})$ is the opposite of the probability asked for in the details $P(\text{3 D6s beats D20})$. – N. Shales May 11 '17 at 1:57
Let the total on the three six-sided dice be $X,$ $3\leq X\leq18.$ Let the number on the twenty-sided die be $Y,$ $1\leq Y\leq20.$
Given any particular value of $X,$ the probability that the twenty-sided die will roll higher is $$P(Y>X \mid X) = \frac{20-X}{20} = 1 - \frac1{20}X.$$
The overall probability that the twenty-sided die will roll higher than the total on the other three dice is \begin{align} P(Y>X) &= \sum_{n=3}^{18} P(Y > X \mid X)P(X=n) \\ &= \sum_{n=3}^{18} \left(1 - \frac1{20}X\right)P(X=n). \end{align} | {
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The last line of that set of equations is just the expected value of $1 - \frac1{20}X.$ That is, \begin{align} P(Y>X) &= \mathbb E\left[1 - \frac1{20}X\right] \\ &= 1 - \frac1{20} \mathbb E[X] \\ &= 1 - \frac1{20} \left(\frac{21}{2}\right) \\ &= \frac{19}{40}. \end{align}
If the question is actually the one posed in the original question body rather than in the original title, namely the probability that $X > Y,$ then we simply observe that for any given value of $X,$ $$P(Y < X \mid X) = \frac{X-1}{20} = \frac{1}{20}X - \frac{1}{20}.$$
The rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \mid X).$ We find that \begin{align} P(Y<X) &= \mathbb E\left[\frac1{20}X - \frac1{20}\right] \\ &= \frac1{20} \mathbb E[X] - \frac1{20}\\ &= \frac1{20} \left(\frac{21}{2}\right) - \frac1{20} \\ &= \frac{19}{40}. \end{align}
This should not be surprising, because it also follows from $P(Y>X)=\frac{19}{40}$ and the "obvious" fact that $P(Y=X)=\frac1{20}.$
• Please see my answer for a simple proof that the two probabilities are equal (and hence both equal to $19/40$). – Barry Cipra May 11 '17 at 13:50
• I don't know if this counts as a shortcut, but to find the expected value of X faster, you could find the expected value of one dice and just multiply it by three instead of summing all the numbers from 3-18. I feel like that just works better in my head. – Henry Weng May 11 '17 at 14:47
Whatever Bob rolls with the $6$-sided dice has a $1$ in $20$ chance of being matched, for a tie, by Bill's roll of the $20$-sided die. Whatever sum, $S=a+b+c$, Bob rolls, if you turn his dice over, the sum is $(7-a)+(7-b)+(7-c)=21-S$. Similarly, whatever number $T$ Bill rolls, if you turn the $20$-sided die over, the number is $21-T$. Thus for each outcome in which Bob wins, there is an equally likely outcome in which Bill wins, and vice versa. Hence the probability of winning for each of them is the same, namely | {
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$${1\over2}\left(1-{1\over20}\right)={19\over40}$$
Remark: It's not literally necessary that the "complementary" number for each side of a die be the opposite face, just that there be a complemenary number somewhere. For $6$-sided dice, having opposite faces sum to $7$ is fairly standard; I believe it's also standard for $20$-sided dice to have opposite faces sum to $21$.
Also, I'd like to credit David K's answer with motivating this one. When I saw from his analysis that the two probabilities were equal, I decided there ought to be simple reason why. As luck would have it, I found one.
• The connection to my answer is even deeper than the coincidence that $P(Y>X)=P(Y<X)$: the usual easy method to calculate $E[X]$ uses the same "turn the dice over" argument (or something equivalent). But the really nice observation here is that the exact same symmetry works for both dice. – David K May 11 '17 at 13:57
• @DavidK, thanks. The symmetry principle also suffices, for example, to show that it's a fair game to roll $13$ regular dice against $7$ dodecahedral dice. In that case, however, computing the probability of a tie is not so easy. (Or is it?) – Barry Cipra May 11 '17 at 14:10
Your best bet is to keep track of two things for both sides: how likely this particular result is, and how likely anything less than this result is. Given these we can multiply them together relatively easily. | {
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$$\begin{array}{r|rr|rr|r} x & 3\text{d}6 = x & 3\text{d}6 < x & 1\text{d}20 = x & 1\text{d}20 < x & 1\text{d}20 < x \cap 3\text{d}6 = x\\ \hline 1 & 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 & 2 & 2 \\ 4 & 3 & 1 & 1 & 3 & 9 \\ 5 & 6 & 4 & 1 & 4 & 24 \\ 6 & 10 & 10 & 1 & 5 & 50 \\ 7 & 15 & 20 & 1 & 6 & 90 \\ 8 & 21 & 35 & 1 & 7 & 147 \\ 9 & 25 & 56 & 1 & 8 & 200 \\ 10 & 27 & 81 & 1 & 9 & 243 \\ 11 & 27 & 108 & 1 & 10 & 270 \\ 12 & 25 & 135 & 1 & 11 & 275 \\ 13 & 21 & 160 & 1 & 12 & 252 \\ 14 & 15 & 181 & 1 & 13 & 195 \\ 15 & 10 & 196 & 1 & 14 & 140 \\ 16 & 6 & 206 & 1 & 15 & 90 \\ 17 & 3 & 212 & 1 & 16 & 48 \\ 18 & 1 & 215 & 1 & 17 & 17 \\ 19 & 0 & 216 & 1 & 18 & 0 \\ 20 & 0 & 216 & 1 & 19 & 0 \\ \hline \text{total} & 216 & & 20 & & 2052 \end{array}$$
Total up everything in the rightmost column, divide by the totals for the $3\text{d}6 = x$ and $1\text{d}20 = x$ columns, and you win: Bob wins $\frac{2052}{4320} = \frac{19}{40} = 0.475$ of the time.
To get ties, or where Bill wins, change what pairs you multiply: both $=$ ones for ties, and $3\text{d}6 < x$ and $1\text{d}20 = x$ for Bill's wins.
This particular one is interesting: because both distributions are symmetrical, and they have the same mean, Bill and Bob will both win the same proportion of the time.
It occurs to me that there's another part to this question: how do we efficiently calculate result probabilities for combinations of dice?
The answer to that is an operation called convolution, which I'll present here in discrete form.
Given two functions $f(x)$ and $g(x)$, the convolution is $$(f * g)(x) = \sum_{k = -\infty}^\infty f(k)g(x-k)$$
This can be interpreted in probability theory as the following: we have two random variables $f$ and $g$, with probability functions $f(x)$ and $g(x)$. the probability function for $f + g$ -- adding the two results together -- is equal to $(f * g)(x)$. | {
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Obviously with those infinities in there, we have to fiddle with it a little to actually get anything done. In our case, because we're dealing with dice, our functions have what's called limited support: they're only non-zero in a small area, so we only need to cover that small area.
Let's do a specific example. Say I want to calculate the probability that I'll get a $7$ on $3\text{d}6$. $3\text{d}6$ is the same as $1\text{d}6$ + $2\text{d}6$, so I can convolve these two. I'll call their functions $f$ and $g$ respectively.
$f$ here has limited support: the only values it is non-zero for are $1$ through $6$. This allows us to change the limits of our summation to the bounds of $f$'s support.
\begin{align} (f * g)(7) &= \sum_{k=1}^6 f(k)g(7 - k)\\ &= f(1)g(6) + f(2)g(5) + f(3)g(4) + f(4)g(3) + f(5)g(2) + f(6)g(1) \\ &= \frac{1}{6}\cdot\frac{5}{36} + \frac{1}{6}\cdot\frac{4}{36} + \frac{1}{6}\cdot\frac{3}{36} + \frac{1}{6}\cdot\frac{2}{36} + \frac{1}{6}\cdot\frac{1}{36} + \frac{1}{6}\cdot\frac{0}{36} \\ &= \frac{15}{216} \end{align}
Using convolution, then, we can calculate the probabilities of the summed results of multiple dice, without necessarily considering every single simple event: to calculate, say, the probability distribution of $5\text{d}6$, we can take the distribution of $4\text{d}6$ and the distribution of $1\text{d}6$ and convolve them. And to get $4\text{d}6$'s distribution we can convolve $3\text{d}6$ and $1\text{d}6$, etc. So instead of counting out $7776$ possibilities, we instead handle $21\cdot6 + 16\cdot6 + 11\cdot6 + 6\cdot6 = 324$ total multiplications and a similar number of additions. | {
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• I didn't read your answer or the problem, but your LateX table skills are impressive – user2879934 May 11 '17 at 0:53
• That last paragraph is an important short cut. It means that you really only need to work out the cases that result in a tie. – amd May 11 '17 at 0:55
• I literally learned how to make the table to write the answer, and beat Excel until morale impro^W^W it built most of the rows for me. – Dan Uznanski May 11 '17 at 0:56
• I was initially skeptical that this is really "Your best bet", given the other answers, but it does appear to be a more general solution that is purely computational. – 6005 May 12 '17 at 5:34
Call the probability of scoring a $k$ with the D20 $q_k=q=1/20$ and scoring a $k$ with the 3 D6'S $p_k$ then we are looking for the probability that the D20 has the greater score
$$p_3\sum_{k=4}^{20}q_k+p_4\sum_{k=5}^{20}q_k+\ldots +p_{18}\sum_{k=19}^{20}q_k$$
Or simply
$$q(17p_3+16p_4+\cdots +2p_{18})$$
but by symmetry for every way to score $k$ with 3 D6s there is a way to score $21-k$, simply by subtracting the score on each die from $7$. The probability of rolling a total of $18$ with $3$ D6s is therefore the same as rolling a total of $3$ ($p_3=p_{18}$) and rolling a total of $17$ is the same probability as rolling $4$ ($p_4=p_{17}$) etc. So
$$\text{required probability}=19q(p_3+p_4+\cdots +p_{10})=19q(p_{11}+p_{12}+\cdots +p_{18})$$
But of course
$$p_3+p_4+\cdots +p_{18}=1$$
So
$$2(p_3+p_4+\cdots +p_{10})=1=2(p_{11}+p_{12}+\cdots +p_{18})$$ $$\implies p_3+p_4+\cdots +p_{10}=p_{11}+p_{12}+\cdots +p_{18}=\frac{1}{2}$$
Giving
$$\text{required probability}=19\cdot\frac{1}{20}\cdot\frac{1}{2}=\frac{19}{40}\tag{Answer}$$
Note that, since the probability of the players getting the same score is $(1/20)\cdot 1=1/20$, then the probability that the D20 loses is the same as it winning i.e. $19/40$. | {
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# Math Help - Different forms of same integral?
1. ## Different forms of same integral?
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral
$\int \frac{1}{3x}dx$
Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$
However, is it not true that I can also make this of the form
$\int \frac{f'(x)}{f(x)}dx$
via
$\frac{1}{3}\int \frac{3}{3x}dx$
which via a simple u sub yields
$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$
Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
$y=\frac{1}{3}\ln{x}+k$ and $\frac{1}{3}\ln{3x}+j$
as explicit functions of x?
Ta muchly.
2. Originally Posted by Xei
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral
$\int \frac{1}{3x}dx$
Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$
However, is it not true that I can also make this of the form
$\int \frac{f'(x)}{f(x)}dx$
via
$\frac{1}{3}\int \frac{3}{3x}dx$
which via a simple u sub yields
$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$
Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
Mr F says: They are equivalent. j + ln 3 is just as arbitrary as k .....
$y=\frac{1}{3}\ln{x}+k$ and $\frac{1}{3}\ln{3x}+j$
as explicit functions of x?
Mr F says: They already are explicit functions of x. Do you mean how to express them as explicit functions of y ....?
Ta muchly.
..
3. Originally Posted by Xei
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral
$\int \frac{1}{3x}dx$
Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$ | {
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Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$
However, is it not true that I can also make this of the form
$\int \frac{f'(x)}{f(x)}dx$
via
$\frac{1}{3}\int \frac{3}{3x}dx$
which via a simple u sub yields
$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$
Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
Since j and k are just arbitrary constants one can be set equal to the other.
By the way, just so that you are aware...
$\int \frac{1}{x}~dx = ln|x| + C$
-Dan
4. Yeah it's all right, I just wasn't bothered with the mod bit.
I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.
Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?
Mr F: Yeah, y please. As simple as it'll go.
5. Hello, Xei!
You are basically correct . . .
$\int \frac{1}{3x}\,dx \;=\;\frac{1}{3}\int \frac{1}{x}\,dx\;=\;\frac{1}{3}\ln{x}+K$ .[1]
$\frac{1}{3}\int\frac{3}{3x}\,dx \;=\;\frac{1}{3}\ln{3x}+J$ .[2]
Is it the case that both are valid, but the constants $K$ and $J$ are different? .
. . . yes
Is $K\:=\:J+\ln3$? .
. . . um, not quite
From [2], we have: . $\frac{1}{3}\ln(3x) + J \;=\;\frac{1}{3}\left[\ln(3) + \ln(x)\right] + J \;=\;\frac{1}{3}\ln(3) + \frac{1}{3}\ln(x) + J$
. . $= \;\frac{1}{3}\ln(x) + \underbrace{\frac{1}{3}\ln(3) + J}_{\text{a constant}} \;=\;\frac{1}{3}\ln(x) + K$
6. Originally Posted by Xei
Yeah it's all right, I just wasn't bothered with the mod bit.
I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing. | {
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Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?
Mr F: Yeah, y please. As simple as it'll go.
$y=\frac{1}{3}\ln{|x|}+k \Rightarrow 3y - 3k = \ln |x| \Rightarrow e^{3y - 3k} = x$
$\Rightarrow x = e^{3y}\, e{-3k} = A e^{3y}$. | {
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# [SOLVED]Volume of a pyramid
#### dwsmith
##### Well-known member
I am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.
Let $$L$$ be on the x axis and $$W$$ be on the y axis.
In the x-z plane, we have the line $$z = -\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\frac{h}{W/2}y + h$$.
My cross sections has width $$\Delta z$$. So I want to find the volume $$\int_0^hA(z)dz$$.
How can I do this?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Hint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\left(1-\frac{z}{h}\right)^2$ where $A=A(0)$. Then compute the integral $\int_0^h A(z)\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.
#### dwsmith
##### Well-known member
Hint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\left(1-\frac{z}{h}\right)^2$ where $A=A(0)$. Then compute the integral $\int_0^h A(z)\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.
So is A a plane? Where did $$\left(1-\frac{z}{h}\right)^2$$ come from?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
So is A a plane?
No, in my notations $A=A(0)$ is a number, the area of the pyramid's base.
Where did $$\left(1-\frac{z}{h}\right)^2$$ come from?
The length of the cross-section at height $z$ is $L\left(1-\frac{z}{h}\right)$, and the width at height $z$ is $W\left(1-\frac{z}{h}\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.
By the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.
#### dwsmith | {
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#### dwsmith
##### Well-known member
No, in my notations $A=A(0)$ is a number, the area of the pyramid's base.
The length of the cross-section at height $z$ is $L\left(1-\frac{z}{h}\right)$, and the width at height $z$ is $W\left(1-\frac{z}{h}\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.
By the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.
Why would we find the area as $$2x\cdot 2y$$ instead of $$x\cdot y$$?
#### Prove It
##### Well-known member
MHB Math Helper
I am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.
Let $$L$$ be on the x axis and $$W$$ be on the y axis.
In the x-z plane, we have the line $$z = -\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\frac{h}{W/2}y + h$$.
My cross sections has width $$\Delta z$$. So I want to find the volume $$\int_0^hA(z)dz$$.
How can I do this?
If it's a right pyramid, isn't the volume just \displaystyle \begin{align*} V = \frac{1}{3}A\,h = \frac{1}{3}L\,W\,h \end{align*}?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Why would we find the area as $$2x\cdot 2y$$ instead of $$x\cdot y$$?
Because the section of the cone using the $xz$ plane is a triangle with sides
\begin{align}
\end{align}
If you find the $x$ corresponding to a given $z$ from (2), then the line at height $z$ crosses the triangle from $-x$ to $x$, i.e., the length of the segment the triangle cuts on the line is $2x$. But again, this is easier to see from similar triangles. The overall intuition is that the length of the cross-section decreases linearly from $L$ at $z=0$ to $0$ at $z=h$. There is a single linear function that does this, and it is $L\left(1-\frac{z}{h}\right)$.
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- - - Updated - - -
If it's a right pyramid, isn't the volume just \displaystyle \begin{align*} V = \frac{1}{3}A\,h = \frac{1}{3}L\,W\,h \end{align*}?
We are trying to prove it.
#### MarkFL
Staff member
I am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.
Let $$L$$ be on the x axis and $$W$$ be on the y axis.
In the x-z plane, we have the line $$z = -\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\frac{h}{W/2}y + h$$.
My cross sections has width $$\Delta z$$. So I want to find the volume $$\int_0^hA(z)dz$$.
How can I do this?
You may want to read >>>this thread<<< for a tutorial on how to work this type of problem.
#### dwsmith
##### Well-known member
You may want to read >>>this thread<<< for a tutorial on how to work this type of problem.
I was reading a calculus book and had most of it figured out but the 2x 2y piece Makarov cleared up. | {
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# Math Help - Binomial series #2 stuck!
1. ## Binomial series #2 stuck!
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$
$= 5\left[1 + \frac{x}{10})^-4\right]$
$= 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$
But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
2. Originally Posted by mollymcf2009
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$
$= 5\left[1 + \frac{x}{10})^-4\right]$
$= 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$
But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
3. Originally Posted by Chris L T521
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
Sorry...
Here is the whole question:
Use the binomial series to expand the function as a power series.
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$
4. Binomial series are given by: $(1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$ ${\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$
for any $k \in \mathbb{R}$ and if $|y| < 1$. Here, $y = \tfrac{x}{10}$ and $k = -4$ .
So: $g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$
$g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $+ \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$
_______________
Let's look at the general term: $\frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$
Just looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative. | {
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So, if we factor out the negative signs: $\frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$
etc.
5. Originally Posted by o_O
Binomial series are given by: $(1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$ ${\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$
for any $k \in \mathbb{R}$ and if $|y| < 1$. Here, $y = \tfrac{x}{10}$ and $k = -4$ .
So: $g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$
$g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $+ \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$
_______________
Let's look at the general term: $\frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$
Just looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.
So, if we factor out the negative signs: $\frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$
etc.
Ok, I understand everything except where the 3!/3! got there.
Thanks!!!!
6. Focus on the expression: $\frac{4 \cdot 5 \cdot 6 \cdots (n+3)}{n!}$
We can simplify it by multiplying by "1": $\frac{{\color{red} 1 \cdot 2 \cdot 3} \cdot 5 \cdot 6 \cdots (n+3)}{n! \cdot {\color{red}3!}} = \frac{(n+3)!}{n! \cdot 3!}$ $= \frac{(n+3)(n+2)(n+1)n!}{n! \cdot 3!} = \frac{(n+3)(n+2)(n+1)}{ 6}$
So you can see why I multiplied both top and bottom by $3!$ so that I can simplify that long product into a simple factorial. | {
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# Convergence of the following sequence of functions.
For $$n \ge 1$$, let $$g_n(x) = \sin^2 \left (x + \frac 1 n \right ), x \in [0,\infty)$$ and $$f_n(x) = \int_{0}^{x} g_n (t)\ \mathrm {dt}.$$ Then
$$(1)$$ $$\{f_n \}$$ converges pointwise to a function $$f$$ on $$[0,\infty)$$, but does not converge uniformly on $$[0, \infty)$$.
$$(2)$$ $$\{f_n \}$$ does not converge pointwise to any function $$f$$ on $$[0, \infty)$$.
$$(3)$$ $$\{f_n \}$$ converges uniformly on $$[0,1]$$.
$$(4)$$ $$\{f_n \}$$ converges uniformly on $$[0, \infty)$$.
I have found that $$f_n (x) = \frac 1 2 \left (x - \sin x \cos \left (x + \frac 2 n \right ) \right)$$ which converges to the function $$f$$ on $$[0, \infty)$$ defined by $$f(x) = \frac 1 4 ( 2x - \sin {2x}), x \in [0, \infty)$$. But I am not quite sure about whether this convergence is uniform or not. Please help me in this regard.
Thank you very much.
• Yeah you are right. I have edited my body. Oct 12, 2018 at 19:05
• What is the definition of uniform convergence, and how is it different from point-wise convergence. Oct 12, 2018 at 19:28
hint
$$\sin^2(t+\frac 1n)=\frac{1-\cos(2t+\frac 2n)}{2}$$
$$f_n(x)=\int_0^x\sin^2(t+\frac 1n)dt=\frac{1}{2}\Big[t-\frac{\sin(2t+\frac 2n)}{2}\Bigr]_0^x$$
$$=\frac x2-\frac 14\sin(2x+\frac 2n)+\frac 14\sin(\frac 2n)$$
The pointwise limit is $$f(x)=\frac x2-\frac{\sin(2x)}{4}$$
For the uniform convergence, use MVT to get
$$|\sin(2x+\frac 2n)-\sin(2x)|\le \frac 2n$$
and
$$|f_n(x)-f(x)|\le \frac 1n.$$
The convergence is uniform at $$[0,\infty)$$.
• I have done that. First see my body and then answer. Oct 12, 2018 at 19:09
• @Dbchatto67 Your $f_n(x)$ is not correct. Oct 12, 2018 at 19:10
• Have you evaluated your integral between the limits $0$ and $x$? Oct 12, 2018 at 19:11
• In place of $t$ we have $t + \frac 1 n$. Oct 12, 2018 at 19:11
• Now calculate and simplify. Oct 12, 2018 at 19:13 | {
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# Moment equation from dynamics differs with equations from physics & statics. Where did I go wrong?
In my dynamics class, we're being asked to solve the following problem:
My Attempt:
Since I'm given the initial velocity, final velocity, and distance, I solved for the acceleration of the plane using kinematics:
$$a = \frac{v_f^2 - v_i^2}{2d} = \frac{(55.6 m/s)^2 - (16.7 m/s)^2}{2(425m)} = 3.31 m/s^2$$
Now this is where my approach starts to differ from the formulas recommended by the class. I decided to set the moment about point A equal to zero, because the plane is not rotating as it moves down the runway. I seem to remember from both statics and physics that if a body is not rotating about a given point, you can simply set the moment around that point to be zero.
Let N be the reaction force at B. Combining the above assumption with Newton's second law in the x direction, I get:
$$\Sigma F_x: R = ma = (140000kg)(3.31 m/s^2) = 4.63 *10^5 N$$
$$\Sigma M_A: -(15m)N + (2.4m)mg - (1.8m)R = 0$$
Solving for N gives me:
$$N = 1.64*10^5 N$$
According to the solution guide, this is incorrect.
Solution Guide's Explanation:
The guide uses a formula which was introduced to us in the textbook, which is as follows.
For some point P fixed on a rigid body with center of mass G, the moment about point P is given by:
$$\Sigma M_P = I_G\alpha + ma_Gd$$
where:
$$I_G$$ is the moment of inertia of the rigid body about G
$$\alpha$$ is the angular acceleration of the rigid body about G
$$a_G$$ is the acceleration of G
$$d$$ is the moment-arm distance, from P to G, of $$ma_G$$
The book cleverly chose a point C on the plane through which both R and A (the reaction force at wheel A) pass. See below:
Using the moment equation above and setting $$\alpha = 0$$ (because the plane isn't rotating) gives:
$$\Sigma M_C = ma_Gd = (15m)N - (3m-1.8m)mg$$
Finally, using $$a_G = 3.3 m/s^2$$ and solving for N, they got:
$$N = 2.57*10^5 N$$
Why I'm Confused: | {
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$$N = 2.57*10^5 N$$
Why I'm Confused:
From the physics and statics classes I've taken in the past, I've always been taught that $$\Sigma M = I\alpha$$; there was never that extra "$$+ mad$$" term thrown onto the end. That term is basically saying that an accelerating body with no rotation can still have a moment about a point. By contrast, in my physics and statics classes, I recall using the fact that if a body is stationary (no angular acceleration), we can set the moment about any point on the body equal to zero to help us solve.
My Guess at Where the Inconsistency Lies:
Here's my guess at where the inconsistency lies: in my statics classes, we assumed that the rigid bodies we were analyzing had not only zero angular acceleration, but also zero linear acceleration. (It's statics, after all!) In dynamics problems such as this one, however, there is a nonzero linear acceleration which must be accounted for. The "+mad" term must come from the fact that for any point object, the moment about some fixed point O is given by:
$$\Sigma M_O = \vec{r} \times \vec{F} = \vec{r} \times m\vec{a}$$
As I'm writing this, I think the intuition totally makes sense. It seems the formula used by the solution guide accounts for both rotation of the rigid body about its center, and linear acceleration of the body with respect to some point P outside the center of mass of the body.
I'll still post this though, in case anyone would like to correct me, add anything, or use this for their own reference.
• Your image font is not readable. Would you add the text of the image to your question. Thanks. Apr 30, 2020 at 0:42
• This is actually for once such a nice homework question, so much information and own effort instead of just a bad photo of a problem sheet. +1, thanks OP Apr 30, 2020 at 8:00 | {
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There is a slight mistake in your approach. It seems you have forgotten to add the contribution the moment due to inertial force acting at the center of mass $$G$$ to the moment equation. The moment equation should be : $$\sum M_A = -(15 m) N + (3 m) 140*10^3*a + (2.4 m) 140*10^3*g - (1.8 m) 140*10^3*a = 0$$ Evaluating for $$N$$ results in : $$N = 2.57 * 10^5 N$$ | {
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# Count the number of non-decreasing numbers with d digits, lower than X
A number is said to be made up of non-decreasing digits if all the digits to the left of any digit is less than or equal to that digit.
For example, the four-digit number 1234 is composed of digits that are non-decreasing. Some other four-digit numbers that are composed of non-decreasing digits are 0011, 1111, 1112, 1122, 2223.
Notice that leading zeroes are required: 0000, 0001, 0002 are all valid four-digit numbers with non-decreasing digits.
I arrived at the conclusion that there are in total $$\sum_{l=0}^9 \sum_{k=0}^l \sum_{j=0}^k ... \sum_{a=0}^b 1 = \binom{9+d}{d}$$ ways to calculate the non decreasing numbers, for d digits, without constrains.
My question is, suppose I want to find the number of non-decreasing numbers with D digits, under a given limit (in the same order of magnitude). For example, how many 4 digit numbers are non-decreasing and under 5000?
I realize that for this case, I need this summations, but I can't neither figure out how to solve them, nor how to generalize them $$\sum_{l=0}^9 \sum_{k=0}^l \sum_{j=0}^k\sum_{i=0}^{min(5,j)} 1$$
where I limit this first number to 5.
Any help?
• When I look at this post, I see eight-digit numbers such as $12341234$ rather than four-digit numbers such as $1234$. – N. F. Taussig Nov 17 '15 at 10:01
• Fixed the examples, posted this late at night. Thanks @N.F.Taussig – José Pedro Marques Nov 17 '15 at 10:37 | {
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We can represent a number with non-decreasing digits by placing vertical bars in a string of nine ones. The number of ones to the left of the vertical bar represents the digit. For instance, $$| 1 1 1 | | 1 1 1 1 1 1 |$$ represents the number $0339$, while $$1 1 | 1 1 | 1 1 | 1 1 | 1$$ represents the number $2468$. Since a particular number is determined by which four of the thirteen symbols (four vertical bars and nine ones) are vertical bars, the number of non-decreasing numbers with four digits (including leading zeros) is $$\binom{13}{4} = \binom{13}{9}$$ as you found. To determine how many four-digit numbers (including those with leading zeros) are less than $5000$, we must subtract from these those non-decreasing four-digit numbers that are at least $5000$. Notice that in those non-decreasing four-digit numbers that are at least $5000$, the first vertical bar must be placed to the right of at least five ones. For instance, $$1 1 1 1 1 | | 1 | 1 | 1 1$$ represents the number $5567$. Since a particular non-decreasing four-digit number that is at least $5000$ is determined by which four of the last eight symbols (four vertical bars and four ones) are vertical bars, the number of non-decreasing four-digit numbers that are at least $5000$ is $$\binom{8}{4}$$ Hence, there are $$\binom{13}{4} - \binom{8}{4}$$ non-decreasing four-digit numbers, including those with leading zeros, that are less than $5000$. | {
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Alternate Method: A non-decreasing number is completely characterized by the number of times each digit appears. Let $x_k$ represent the number of times the digit $k$ appears in the non-decreasing four-digit number, including those with leading zeros. Since there are four digits, $$x_0 + x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 4$$ A particular number is determined by which nine of the thirteen symbols (nine addition signs and four ones) is an addition sign. For instance, $$1 + + + 1 1 + + + + + + 1$$ represents the number $0339$, while $$+ + 1 + + 1 + + 1 + + 1 +$$ represents the number $2468$. Thus, the total number of non-decreasing four-digit numbers is the number of ways nine addition signs can be placed in a row of four ones, which is $$\binom{4 + 9}{9} = \binom{13}{9}$$ From these, we subtract those non-decreasing four-digit numbers that are at least $5000$. The digits of non-decreasing four-digit numbers that are at least $5000$ satisfy the equation $$x_5 + x_6 + x_7 + x_8 + x_9 = 4$$ A particular number is determined by which four of the eight symbols (four addition signs and four ones) is an addition sign. Hence, there are $$\binom{4 + 4}{4} = \binom{8}{4}$$ non-decreasing four-digit numbers that are at least $5000$. Hence, the number of non-decreasing four-digit numbers that are less than $5000$, including those with leading zeros, is $$\binom{13}{9} - \binom{8}{4}$$ | {
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# Thread: The centroid of a triangle with coordinates
1. ## The centroid of a triangle with coordinates
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.
I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
2. Originally Posted by darksoulzero
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.
I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
Read Centroid - AoPSWiki
3. ## centroid of triangle
Hi darksoulzero,
Your solution is confusing.Here is a suggested method.
connect M midpoint of DE and C (centroid) with extended lenght.F lies on this line. FC = 2CM. Slope diagram of C and M = 2/1/2.Slope diagram of FC is twice that or 4/1. Working from point C one point left and 4 points up gives F (2,8)
bjh | {
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bjh
4. If ABC is a triangle with $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ then its centroid is
$G\left( {\tfrac{{{x_1} + {x_2} + {x_3}}}{3},\tfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
5. Hello, darksoulzero!
$\text{Triangle }D{E}F\text{ has vertices: }D(1,3)\text{ and }E(6,1)\text{, and centroid at }C(3,4).$
$\text{Determine the coordinates of point }F.$
Code:
1
F o-----+
\ :
\ :4
\ :
\ :
\:0.5
(3,4)o---+
D C\ :
o \ :2
(1,3) * \:
M o
(3.5,2) * E
o
(6,1)
We have vertices $D(1,3)$ and $E(6,1)$, and centroid $C(3,4).$
The midpoint of $DE$ is: $M(3\tfrac{1}{2},\,2).$
The median to side $DE$ starts at $\,M$, passes through $\,C,$
. . and extends to $\,F$, where: . $FC \,=\,2\!\cdot\!CM.$
Going from $\,M$ to $\,C$, we move up 2 and left $\frac{1}{2}$
Hence, going from $\,C$ to $\,F$, we move up 4 and left 1.
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# F(x) symmetric about the line x=2
## Main Question or Discussion Point
Why is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?
## Answers and Replies
ShayanJ
Gold Member
Change the variable to y=x-2 and see whether the resulting function is even or odd or none!
You mean x = x-2 right? It doesn't stay the same on doing that. So it should not be symmetrical about x = 2, but it is. And I am not able to see how that happens.
ShayanJ
Gold Member
$f(x)=a(x-1)(x-2)(x-3) \rightarrow f(y)=a(y+1)y(y-1)$
$f(-y)=a(-y+1)(-y)(-y-1)=-a(y-1)y(y+1)=-f(y)$
So f(y) is odd which means f(x) is not symmetric around x=2 but is not that much asymmetric because we have f(2-x)=-f(x-2).
Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1
And this function is said to be symmetrical about the line x = 2. But I am unable to see how?
ShayanJ
Gold Member
Its the same trick. Just do the tranformation $x\rightarrow x+2$ and check whether the resulting function is even. If its even, then the original function is symmetric around x=2.
They are not the same, i.e. after changing x to x+2 in f(x), f(x) ≠ f(-x). But what I am reading, it says that f(x) is symmetrical about x=2 and I am still wondering that how would I go about proving it?
ShayanJ
Gold Member
You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.
PeroK
Homework Helper
Gold Member
Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1
And this function is said to be symmetrical about the line x = 2. But I am unable to see how?
Another way to look at it is as follows:
Imagine starting at x = 2 and moving the same distance, d, to the right and left (d > 0). So, to the right we have 2 + d and to the left we have 2 - d. | {
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Now, if f is symmetrical about x = 2, then f(2-d) = f(2+d) for all d. You could try that approach.
You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.
How did you reduce it down to that? Can you show me? I am unable to get to that point.
ShayanJ
Gold Member
$f(x+2)=a[ \frac 1 4 (x+2)^4-2(x+2)^3+\frac{11}{2} (x+2)^2-6(x+2)]+1=\\ a(x+2)[ \frac 1 4 (x+2)^3-2(x+2)^2+\frac{11}{2} (x+2)-6]+1=\\ a(x+2)(\frac 1 4 x^3+\frac 3 2 x^2+3x+2-2x^2-8x-8+\frac{11}{2}x+5)+1=\\ \frac 1 2 a(x+2)(\frac 1 2 x^3-x^2+x-2)+1=\frac 1 2 a (\frac 1 2 x^4+x^3-x^3-2x^2+x^2+2x-2x-4)+1=\\ \frac 1 2 a (\frac 1 2 x^4-x^2-4)+1$
And what is the reason that we transformed x to x+2?
ShayanJ
Gold Member
Simple. When we say a function is even, we mean its symmetric around x=0. So if a function is symmetric around $x=a \neq 0$, it means if we move the origin to x=a, the resulting function would be even.
Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?
ShayanJ
Gold Member
Check here!
But we aren't shifting the curve as in your link but instead we are shifting the coordinate axes. The curve stays where it was. The axes are what shift.
PeroK
Homework Helper
Gold Member
Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?
This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.
If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.
But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.
In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made. | {
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This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.
If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.
But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.
In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.
Wait. So you are saying that f(x) should change to a new variable so that it becomes f(z), right? But that would again mean f(x-2) and not f(x+2).. got me confused
ShayanJ
Gold Member
$z=x-2 \Rightarrow x=z+2$ so $f(x)=f(z+2)$!(Forget the confusing $x\rightarrow x+2$!)
It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?
For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.
ShayanJ
Gold Member
It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)? | {
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For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.
Let's start from the beginning.
At first we have a coordinate system which we call xy. Now I define a function y=f(x). Then I move the origin to x=a and name the new coordinate system zy. But this doesn't change the function, only the coordinate system has moved. But if I insist that the function f has the same form in terms of both x and z, then this means that the function has changed which isn't right.(Imagine y=x^2. z=x-a so x=z+a. But if I say that y=z^2, this function would have its minimum at z=0 so x=a which means the function has changed!) So I should have y=g(z). But g should be related to f somehow that we actually get the same function. So let's see what's the relationship. At x=0, f gives f(0), so at z=-a, g should give f(0). Then at x=a, f gives f(a), so at z=0, g should give f(a). Now we have two relationships g(-a)=f(0) and g(0)=f(a) and so we can deduce that g(z)=f(z+a)=f(x). | {
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# Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$
We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s>1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).
I was curious to see till what extent I can push the denominator so that it will still diverges.
So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).
No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$.
However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s>1$.
(By the same argument as previous).
This doesn't make sense to me though.
If this were to converge, then I should be able to find a $s_1 > 1$ such that
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$
Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$?
(or)
How should I make sense of (or) interpret this result?
(I am assuming that my convergence and divergence conclusions are right). | {
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(I am assuming that my convergence and divergence conclusions are right).
• They are right. You don't seem to have completed one of your sentences: "if this were to converge, then I should be able to find a s_1 > 1 such that"...? – Qiaochu Yuan Nov 11 '10 at 21:24
• There seems to be some bug with this. It doesn't display the entire thing. – user17762 Nov 11 '10 at 21:25
• @Sivaram: Yes even i tried editing it, and it still has not worked – anonymous Nov 11 '10 at 21:27
• There seems to be a bug/feature! when dealing with <. Probably confusion with html tags. Sivaram, please use \log n instead of just logn. – Aryabhata Nov 11 '10 at 21:32
• @ Chandru and Moron: Thanks for editing it. Now it looks fine. – user17762 Nov 11 '10 at 21:33
## 6 Answers
Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s > 1$, we can see that by comparing with the corresponding integral.
As to your other question, if
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$
does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.
What happens is that the first "few" terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$, we have that $\log n > n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.
• Sorry. I meant $\exists s_1 > 1$ such that $\displaystyle \sum_{2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{2}^{\infty} \frac{1}{n(\log n)^{s}}$ – user17762 Nov 11 '10 at 21:56 | {
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This is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the "boundary of convergence" in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic "A course in pure mathematics". For further work see his "Orders of infinity" and see especially this very interesting paper by G. Fisher on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered diagonalization before Cantor).
Note in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\rm\ s > 1\$ and divergent if $\rm\: s\le 1\$ where $\rm\: n_0,\ a\$ are any numbers large enough to ensure positivity of $\rm\ log_k x := \log \log \cdots \log x\:,\$ iterated $\rm\:k\:$ times.
$$\rm \sum_{n_0}^\infty \frac{1}{n \log n\ \log_2 n\ \cdots\ \log_{k-1} n\ (\log_k n)^s }$$
$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s }$$
• Off-topic question: What text is the excerpt from? Thank you! – Ayesha Jan 29 '14 at 1:04
• @Ayesha I say that in the text above it: Hardy's .... – Bill Dubuque Jan 29 '14 at 1:09
• Sorry, sorry, I somehow glanced over that. Thank you again for responding to my rather asinine question. – Ayesha Jan 29 '14 at 1:43
Du Bois-Reymond was interested in this question, positing some sort of series "between" all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging "ever more slowly" and divergent series diverging "ever more slowly", you can find a convergent and a divergent series "in between". Here the order relation is the one given by the (conclusive) ratio test.
Bill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series | {
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$\displaystyle \frac{1}{n(\log n)^2}, \frac{1}{n\log n (\log \log n)^2}, \ldots$
He then diagonalizes to obtain his $\omega$ code (Elias omega code, if you want to look it up), which corresponds to
$\displaystyle \frac{1}{n\log n\log \log n \cdots (\log^* n)^2},$
where $\log^* n$ is the number of times you need to apply $\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\log^* n$ to mean the $\log n \log\log n \cdots$ part. We can continue the iteration like this:
$\displaystyle \frac{1}{n\log n\log\log n \cdots \log^*n (\log \log^* n)^2}, \frac{1}{n\log n\log\log n \cdots \log^*n \log \log^* n (\log\log \log^* n)^2}, \cdots$
and so on. We can diagonalize again to obtain an $\omega + \omega$ code, and so on at least for $n \omega + m$. We an probably continue even to $\omega^2$ and beyond.
The corresponding divergent series are $\displaystyle \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \ldots, \frac{1}{n\log n\log\log n \cdots \log^* n}, \frac{1}{n\log n\log\log n \cdots \log^* n \log\log^* n}, \ldots$
One can ask whether there is a scale for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).
• Indeed, it deserves to be better known that Du Bois Reymond - not Cantor - is the true discoverer of diagonalization - in this context, i.e. asymptotic growth rates of functions. – Bill Dubuque Nov 11 '10 at 22:53
• Thanks Bill and Yuval. I was looking for series which diverges slower and slower and your answer helps me on that note. – user17762 Nov 11 '10 at 23:09 | {
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Another test that applies to series of positive decreasing terms (and in this particular one in a rather elegant fashion) is the following: $$\sum_{n=1}^\infty a_n<\infty \quad\Longleftrightarrow\quad\sum_{k=1}^\infty 2^ka_{2^k}<\infty.$$ In our case $$\sum_{k=1}^\infty 2^ka_{2^k}=\sum_{k=1}^\infty 2^k\frac{1}{2^k(\log 2^k)^s}= \frac{1}{(\log 2)^s}\sum_{k=1}^\infty \frac{1}{k^s},$$ and thus $$\sum_{n=2}^\infty \frac{1}{n(\log n)^s}\quad\Longleftrightarrow\quad \sum_{k=1}^\infty \frac{1}{k^s}\quad\Longleftrightarrow\quad s>1.$$
• Monotonicity (as is in this case) is required? – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:40
• @Martín-BlasPérezPinilla: I do not know the most general version of this criterion, but the version I know requires $a_n$ to be non-negative and decreasing. – Yiorgos S. Smyrlis Jan 31 '14 at 19:46
• Yes, now I remember I've seen in Baby Rudin. – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:56
By the integral comparison test the given series is convergent if and only if the integral $$\int_2^\infty\frac{dx}{x\log^s(x)}$$ is convergent and notice that $\log'(x)=\frac1x$ so
• if $s>1$ $$\int_2^\infty\frac{dx}{x\log^s(x)}=\frac1{1-s}\log^{1-s}(x)\Big|_2^\infty<+\infty$$ so the series is convergent
• if $s=1$ $$\int_2^\infty\frac{dx}{x\log(x)}=\log(\log(x))\Big|_2^\infty=+\infty$$ so the series is divergent.
• The exponent $s$ is just on the log term, not the entire term – user2566092 Jan 31 '14 at 19:38
• Ah I'll edit my answer! – user63181 Jan 31 '14 at 19:41
• @user2566092 I edited my answer. – user63181 Jan 31 '14 at 19:50
• What about $s < 1$? – mavavilj Sep 19 '15 at 17:25 | {
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Convergence does not change if you switch to base-2 logs. There is a theorem that says that if $a_n$ is a decreasing positive sequence, then $\sum_n a_n$ converges if and only if $\sum_k 2^k a_{2^k}$ converges. If you apply this theorem's transformation to your sequence, you get $\sum_k 1/k^s$, which obviously converges if $s > 1$ and diverges if $s \leq 1$. If you'd like a proof of that theorem (called the Cauchy condensation test), I can provide it. | {
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What's new
# Difference between probability density function and inverse cumulative distribution function?
#### bpdulog
##### Active Member
What is the difference between probability density function and inverse cumulative distribution function, if any? They both look the same
#### ShaktiRathore
##### Well-Known Member
Subscriber
Hi,
No they are not the same.
The inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.
Whereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.
thanks
#### brian.field
##### Well-Known Member
Subscriber
David also covers this extensively in his Study Notes for Miller. Have you read them?
#### bpdulog | {
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#### bpdulog
##### Active Member
Hi,
No they are not the same.
The inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.
Whereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.
thanks
If I understand what are you saying, in the inverse distribution the 1.645 represents 95% and less. And in the probability density function the 1.645 represents 95% only?
#### ShaktiRathore
##### Well-Known Member
Subscriber
Hi
in the inverse distribution the 1.645(quantile) represents 95%(cumulative probability). And in the probability density function P(x) ,when x=1.645 ,P(x) shall give the probability density at x,and we integrate the P(x) from -inf to x=1.645 to get the cumulative probability of 95%.
thanks
#### bpdulog
##### Active Member
Ah ok! I think I get it now. It just 2 different ways to display a distribution of data. The top is the PDF and the PPF is basically the CDF with the axes transposed.
#### David Harper CFA FRM | {
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#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Yuu definitely want to understand @ShaktiRathore 's above relationship. Miller's Chapter 2 does a decent job. We don't use the continuous pdf too much. The pdf integrates to the CDF, and we're arguably more interested in the relationships around the CDF, as Shakti illustrates. For example, if probability (p) = 5.0%, then:
• p = 5% = F(q), where F(.) is the cumulative distribution function, so if we are given the probability of 5% because we want a 95% confident normal VaR, then we use the inverse CDF to retrieve the quantile:
• q = F^(-1)(p) = F^(-1)(5%) = -1.645, where F^(-1) is the inverse CDF. "Inverse" implies that F[F^(-1)(p)] = p; e.g., F[F^(-1)(5%)] = F[-1.645] = 5%, just like also F^(-1)[F(q)] = q. This is why Dowd says "VaR is just a quantile;" i.e., because -1.645 is the quantile (q) retrieved by using the inverse CDF given a probability (p) of 5%. This is an essential FRM building block. I hope that adds something.
#### theapplecrispguy
##### New Member
In Question 300.2, the final price of the bond (i.e. the value of "x") is calculated using the inverse CDF such that 5% of the distribution is less than or equal to "x". This price is calculated to be $1.842 using p=5%. My interpretation of this is that 5% of the time the bond will be priced below$1.842 and the other 95% of the time the bond will be priced above $1.842. Is this correct? The question then states that the 95% VAR is given by$5.00 (the face value of the bond) minus $1.842 (q(.05)) equals$3.158. I don't understand conceptually how the formula for 95% VAR was determined to be $5.00 - q(.05). Can you please explain what the 95% VAR ($3.158) represents in the context of the area under the cumulative distribution graph?
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##### David Harper CFA FRM
Staff member
Subscriber
Hi @theapplecrispguy FYI, the source Q&A (with follow on discussion) is here at https://www.bionicturtle.com/forum/threads/p1-t2-300-probability-functions-miller.6728/ but briefly:
• Re: "My interpretation of this is that 5% of the time the bond will be priced below $1.842 and the other 95% of the time the bond will be priced above$1.842. Is this correct?" Yes, exactly correct. The integral (anti-derivative) of the given f(x) is the CDF which is F(x) = x^3/125. When we have the CDF, it means that cumulative p = x^3/125; i.e., p is the probability that the random variable will be less than or equal to x. Because 1.842^3/125 = 5.0%, under this distribution, there is a 5.0% probability that X will be less than $1.842; and, as you say, 95.0% of the time X will be greater than$1.842 (but less than $5.00, as the function domain is$0.00 < X < 5.00). As the question is testing the associated Miller, this is really the essential point of the question ....
• Because the VaR part here is really simplistic. VaR is a loss relative to something. This question gives the assumption that the expected future value is $5.00 so that the VaR can be computed as a simple "worst expected loss" of$5.00 - $1.842 =$3.158; ie, because there is a 5.0% probability of the final price falling below $1.842, there is a 5.0% probability of a loss worse than$3.158. If we were not given this assumption, the proper approaches are classically either: | {
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• Compute the loss relative to the expected future value as the proper mean of the future distribution. In this case, as stated in the question, this relative VaR (it is called because it is relative to the expected future value) would be $3.75 -$1.842 = $1.908. I just didn't want to make the question too difficult; or, • Compute the loss relative to the current value, which is properly termed the absolute VaR (relative to the initial value). I don't think we have information to do that, but it would be given by discounted present value [$5.00 par] - $1.842 = absolute VaR. This shows why a good question needs to be specific about the VaR it looks for, VaR is a worst expected loss over a horizon (choice #1) with a confidence level (choice #2) relative to something (definitional #3). I hope that helps! #### theapplecrispguy ##### New Member Thank you David for the explanation which is very helpful. If the question had asked for the 5% VAR would the answer have been the same? #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber Hi @theapplecrispguy Yes, but only because occasionally (or rarely) the 95.0% VaR will be represented by an author as the 0.050 VaR, treating them as the same idea. In any realistic use case, we are referring to the losses that happen in the worst 5.0% of times, the losing tail. It would not be realistic/sensible to speak of the losses that occur in the worst 95.0% of times (although mathematically it works of course). But, again, the idea is that we have a cumulative distribution function here, given by F(x) = p = x^3/125. Solving for x, we have x = (p*125)^(1/3). So for example, • at p = 5%, x = (5%*125)^(1/3) =$1.84
• at p = 10%, x = (10%*125)^(1/3) = $2.32 • at p = 20%, x = (20%*125)^(1/3) =$2.92
• at p = 50%, x = (50%*125)^(1/3) = $3.97; i.e., the median • at p = 95%, x = (95%*125)^(1/3) =$4.92 | {
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• at p = 50%, x = (50%*125)^(1/3) = $3.97; i.e., the median • at p = 95%, x = (95%*125)^(1/3) =$4.92
It's not a distribution of losses, it's a distribution of future values. We always want to be mindful of the difference, sometimes the distribution is losses. As this is future values, the risk tail is "down" at p = 5%, it is not "up" at p = 95%, so that either a 95% or 0.05 VaR would be realistically referring to the p = 5%. At the same time, there are valid usages of VaR for lower confidence levels, so we could have a 90% VaR (given by the loss of $5.00 minus$2.32) or a even an 80% VaR (given by the loss of $5.00 minus$2.92). I hope that helps! | {
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Last edited:
#### theapplecrispguy
##### New Member
Thank you. Clear and much appreciated! | {
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# what happens to i^(square root of 2)?
I always thought what be in store for us if take irrational number to power of a complex number? we already have natural number, integers, rational numbers, real numbers and complex numbers. Will it create a class of its own?
Note by Shiva Kumar
9 months, 2 weeks ago
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yes sir but when the value of k is increased in polar form of (i)^(square root of 2), we do not get the solutions that superimpose the previous solutions. so we instead get infinite no of solutions. it form a circle with magnitude equal to one and center as origin in Argand Plane after k becomes infinite. so does that mean (i)^(any irrational number) forms a circle with magnitude of value one and center at origin.and get same set of solutions. does this mean (i)^any irrational number has same set of values.
- 9 months, 2 weeks ago | {
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- 9 months, 2 weeks ago
The basic way to define powers of complex numbers is via the formula $z^w \; = \; e^{w\log z}$ and so the whole business revolves around the definition of the logarithm of complex numbers. We need $\log z \; = \; \ln |z| + i\mathrm{Arg}\,z$ and here is the real problem. There is no way of defining the argument function continuously (let along differentiably) on the whole complex plane.
Since we want the argument (and hence the logarithm) to be differentiable, it has to be defined on an open set (so that we can consider its derivative at all points), and complex functions are considered on open connected domains. The standard way of doing this for the argument is to cut the plane. This means considering the domain formed by the complex numbers with a straight line out from $$0$$ to infinity removed. The argument can be defined uniquely (and differentiably) on any such domain, and so can the natural logarithm of the modulus (since we have removed $$0$$).
For example, we could consider the cut plane $$\mathbb{C} \backslash (-\infty,0]$$ consisting of the complex numbers except for the nonpositive reals. This means that $$|z| > 0$$ and $$-\pi < \mathrm{Arg}\,z < \pi$$ for all $$z$$ in the cut plane, and we can define $$\log z$$ uniquely (and differentiably) now. This choice is called using the principal branch of the argument. In this case we would have $$\ln i = \tfrac12\pi i$$, and so $$i^\sqrt{2} = e^{\frac{\pi i}{\sqrt{2}}}$$.
A different cut would be to exclude the nonnegative reals $$[0,\infty)$$, and decide that $$2\pi < \mathrm{Arg}\,z < 4\pi$$ for all $$z$$ in this cut plane. This would make $$\ln i = \tfrac52\pi$$, giving a totally different value for $$i^{\sqrt{2}}$$.
This is what @Pi Han Goh means about this function being multivalued. There is no way to have a single function that works everywhere. | {
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The other option is to allow all these values to work at the same time! This means we have to leave the ordinary complex plane, and work with a Riemann surface. Imagine an infinite number of complex planes stacked one on top of another. Cut them all along the positive real axis, and stick the "first quadrant edge" of each sheet to the "fourth quadrant edge" of the sheet below. The resulting helicoidal shape allows you to consider complex numbers with positive moduli and all possible arguments. The number $$i$$ would lie in one sheet, the number $$ie^{2\pi i}$$ would lie in the sheet above, the number $$ie^{-6\pi i}$$ would lie in three sheets below, and so one. You could then define a logarithm on the whole Riemann surface at one go...
- 9 months, 1 week ago
Ah crap. I can't answer this question either. My first though is Equidistribution theorem, but I'm not entirely confident.
Summoning the great @Mark Hennings again!
- 9 months, 1 week ago
No, $$i^{\sqrt 2}$$ is just a multivalued complex numbers.
Let $$x = i^{\sqrt 2}$$, then $$\ln x = \sqrt 2 \cdot i = \sqrt 2 ( 0 + 1i )$$.
With $$0 + 1i = \cos \left (\frac\pi 2 + 2\pi k \right) + i \sin \left ( \frac\pi 2 + 2\pi k \right)$$, where $$k$$ is any integer.
Can you finish it off from here?
- 9 months, 2 weeks ago | {
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# $\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges?
Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
Thought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys
• What tests do you know for convergence? – Qiaochu Yuan Oct 20 '10 at 21:32
• Look at the absolute value of the ratio of a term in the series to the subsequent term. You should already know what the possible results imply... – Brandon Carter Oct 20 '10 at 23:35
Another way is
If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$
We have that
$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$
$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} < 2$
Thus $S_n < 2$
thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.
I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is obviously a number, so he doesn't have to prove that the series for e converges.
Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the ratio test:
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.
Therefore, the series converges. | {
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Therefore, the series converges.
• Just out of the curiosity, I plugged the series into Wolfram|Alpha. It gives the very nice result of sinh(1). If anyone knows how to prove that, it would be wonderful to know it. By the way, how can I add an hyperlink in the comment section? – Robert Smith Oct 21 '10 at 0:18
• $\sinh(x)=\frac{\exp(x)-\exp(-x)}{2}$. Also, [this](http://functions.wolfram.com/ElementaryFunctions/Sinh/) gives this. – J. M. is a poor mathematician Oct 21 '10 at 1:21
• @Robert: It is the Taylor series for sinh evaluated at 1. – GEdgar Jun 16 '12 at 16:28
• Thanks. I didn't know that. – Robert Smith Jun 16 '12 at 18:06
The series you have is
$1 + \frac{1}{3!} + \frac{1}{5!} \ldots$
If you add the even factorial terms, you get an upper bound i.e.,
$1 + \frac{1}{3!} + \frac{1}{5!} \ldots < \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$
This can be written more compactly as
$\sum_{n=0}^\infty \frac{1}{(2n+1)!} < \sum_{n=0}^\infty \frac{1}{n!} = e^1$
Therefore the series converges.
• I don't see the point of doing this when proving that the series for e converges is exactly as hard. – Qiaochu Yuan Oct 20 '10 at 22:23
• In short, $e=\cosh\;1+\;sinh\;1$. :) – J. M. is a poor mathematician Oct 21 '10 at 0:19
• @ Qiaochu Yuan. You make a valid point. I guess the Ratio test(which Robert Smith has mentioned) is the rigorous answer :). – svenkatr Oct 21 '10 at 3:53
We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots$$
Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
• The question was «does the series converge?» Your answer begins by asserting convergence of a series of equivalent difficulty :) – Mariano Suárez-Álvarez Oct 21 '10 at 14:12 | {
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Can you bound the series from above by one that you know converges? The factorials grow very fast, so you should be able to.
To elaborate on the first answer given to this question by Ross Millikan.
$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$
$$< 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } \frac{1}{(2n+1)!} < \frac{1}{4^n} \textrm{ for } n \ge1.$$
Hence by the comparison test the series converges.
Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that
$$R_N < \frac{1}{(2n+3)!} \left( 1 + \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$
METHOD I
We may simply resort to the Basel problem and get the inequality: $$0<\sum_{k=0}^{\infty}\frac{1}{(1+2k)!}\leq\sum_{k=0}^{\infty}\frac{1}{(1+k)^2}=\frac{\pi^2}{6}$$
METHOD II
According to Taylor's expansion we have that:
$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!}$$
For $x=1$ we get that the value of the series is $\sinh(1)$. The series converges.
Q.E.D.
This is how sometimes we can extract the exact closed form of some series :
$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$
The radius of convergence is $\infty$
$f \in C^{\infty}(\mathbb{R})$
$f''(x) = f(x)$
solving our equation, we get $f(x) = ae^{\lambda_1x}+be^{\lambda_2x}$
$\lambda^2-1=0$
$\lambda \in \{-1,1\}$
$f(x) = ae^{-x}+be^{x}$
$f(0)= 0 \implies a+b=0 \implies a=-b \implies f(x) = a(e^{x}- e^{-x})$
$f'(0) = 1 \implies a=\frac{1}{2}$
$f(x) = \frac{e^{x}-e^{-x}}{2}$
so our series $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$ does converge to $\sinh(1)$ | {
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so our series $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$ does converge to $\sinh(1)$
• This only proes that if the series converges, then its sum is $\sinh(1)$. – José Carlos Santos Jul 26 '18 at 20:24
• Yes that is true, but I wanted to say, regarding to other answers, that I would find the limit. So I didn't repeat any other done process here, that is namely true. Anyway that is the same, effectively the sum itself implies the convergence of that series to $\sinh(x)$. So you had not always to prove the convergence of series, since it does have a closed form. the closed form itself implies the convergence, that is so understandable. – Ахмед Jul 26 '18 at 20:26
• your implication is not true for closed form and finite values. that does mean, the sum does converge for all $x\in \mathbb{R}$ to $\sinh(x)$ which is finite. – Ахмед Jul 26 '18 at 20:33 | {
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# Statement Proof.
1. Jul 11, 2007
### linuxux
My question is this:
Theorem 1.4.13 part (ii) says: If $$A_n$$ is a countable set for each $$n \in \mathbf{N}$$, then $$\cup^{\infty}_{n=1} A_n$$ is countable.
I can't use induction to prove the validity of the theorem, but the question does say how does arranging $$\mathbf{N}$$ into a 2-d array:
1 3 6 10 15 ...
2 5 9 14 ...
4 8 13 ...
7 12 ...
11 ...
lead to the proof of part (ii) of Theorem 1.4.13?
so obviously it has something to do with the (x,y) co-ordinate system of the array, but I nt sure how it leads to the proof.
Last edited: Jul 11, 2007
2. Jul 11, 2007
### CRGreathouse
You can just show a function $f: A\to\mathbb{N}$.
$$f(0)=A_1(0)$$
$$f(1)=A_1(1)$$
$$f(2)=A_2(0)$$
$$f(3)=A_1(2)$$
$$f(4)=A_2(1)$$
$$f(5)=A_3(0)$$
$$f(6)=A_1(3)$$
. . .
3. Jul 11, 2007
### linuxux
but what is the significance of putting it into a 2-d array? i could just as easily say: 1, 2, 3, 4, ... and call it a 1-d array.
4. Jul 11, 2007
### CRGreathouse
$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...
$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...
$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...
$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...
$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...
... ... ... ...
5. Jul 11, 2007
### mathwonk
this proof is due to cantor, 100 years ago. just start couting from the upper left corner, counting by sw/ne diagonals.
6. Jul 12, 2007
### matt grime
To demonstrate a bijection between N (the indices you're inserting) and the countable union of N (the locations in the doubly infinite array). I.E the whole point of the question.
7. Jul 14, 2007
### linuxux
but how is this a 1-1 and onto function? how is it that $$A_n$$ is a function itself? the cantor theorem is the next section in my text so i can't apply that theory now. i understand that f(n) is a function, but why also $$A_n$$?
Last edited: Jul 14, 2007
8. Jul 14, 2007
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Last edited: Jul 14, 2007
8. Jul 14, 2007
### CompuChip
Here's a different explanation (sometimes I see that the understanding of the principle depends on the way it is explained, so maybe this will help you - maybe it will make it less clear to you, then please ignore this )
Suppose we put two copies of the natural numbers next to each other. We want to show that, doing this, the total number we get is still countable (that there are "as many" elements in this string of numbers as there are natural numbers, by assigning a natural number to each of the elements).
Clearly, just counting them in a line won't work. Then we assign each natural number to itself in the first copy, but when we're through with that one, we don't have any natural numbers left for the second copy. So what we do is: assign 1 to 1 from the first copy, 2 to 1 from the second copy, 3 to 2 from the first copy, 4 to 2 from the second copy, etc. - so
$$2n - 1 \mapsto n \text{ from the first copy of } \textbf{N}, 2n \mapsto n \text{ from the second copy of } \textbf{N}.$$.
Now consider the Cartesian product $$\textbf N \times \textbf N$$, consisting of all pairs of natural numbers (a, b). How do you prove they are countable. One way would be to start by assigning n to (n, 1), but when we're done, we won't have any natural numbers left to assign to (n, 2) and (n, 3), etc. So here's the trick: we organize the pairs like
(1,1) (1,2) (1,3) ...
(2,1) (2,2) (2,3)
(3,1) (2,3) (3,3)
....
and now we can map them like
1 2 6 7 ...
3 5 8 ...
4 9 10
without running out of numbers too soon.
In the same way, you can prove that $\textbf{N}^k$ is countable for any natural number k, as well as $\frac{1}{2} \textbf{N}$ (all halfintegers 0, 1/2, 1, 3/2, ...) and the integers $\textbf{Z}$ (it's basically two copies of the naturals, like in my first example, with a zero added) and even that the rationals $\textbf{Q}$ are countable. And, also the question from your initial post can be solved like this.
9. Jul 14, 2007 | {
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9. Jul 14, 2007
### linuxux
thanks, i understand the example you showed. and to think, this is just an introductory text and i'm self-studying!...
Last edited: Jul 14, 2007
10. Jul 14, 2007
### linuxux
(1,1) (1,2) (1,3) ...
(2,1) (2,2) (2,3)
(3,1) (2,3) (3,3)
this part i understand. each (a,b) has a particular n mapped to it, thus (a,b)~N. so i got that.
but, looking at this:
$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...
$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...
$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...
$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...
$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...
i understand this to mean that the number of sets of $$A_n$$ is countable, not that once the union is performed on the sets the elements of the union will be countable. I had a previous problem where i had to show that the union of $$A_1$$ and $$A_2$$ was countable, i did this by defining a function that chose the minimum element first in $$A_1$$ and then in $$A_2$$, then choosing the next smallest element in both, and so on, thus i was able to show $$A_1\cup{}A_2$$~$$\mathbb{N}$$. So in that exercise i addressed the elements in the sets, not the sets themselves. how does the section in red address the elements in the sets? is that even necessary?
**********
wait a minute: i know each $$A_n$$ is countable, so each $$A_n(a\in\mathbb{N})$$ addresses the elements in $$A_n$$. now i see.
Last edited: Jul 14, 2007
11. Jul 15, 2007
### matt grime
By construction: do you write down the same number twice or fail to fill in a box?
It isn't a function.
12. Jul 15, 2007
### linuxux
wouldn't it qualify as a function since $$A_n$$ is countable thus $$A_n \mapsto \mathbb{N}$$?
i thought $$\mapsto$$ represents some type of function. does it not?
i also have another question, what happens if a particular $$A_n$$ is finite?
Last edited: Jul 15, 2007
13. Jul 15, 2007
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Last edited: Jul 15, 2007
13. Jul 15, 2007
### CompuChip
Yeah, so for each n we can assign a natural number to each element from $$A_n$$, and in the end, have neither elements of $$A_n$$ nor natural numbers "left" (that is, there is a bijection). The claim is, that the union is countable, so there is a bijection from the naturals to the union. What the 2-D array tries to visualize, is the following: since each $$A_n$$ is countable, we can denote the elements by $$A_n(k)$$ (with k = 1, 2, 3, ...). Now you can order the elements in the union like this:
$$A_1(1) \, A_2(1) \, A_3(1) \, \cdots$$
$$A_1(2) \, A_2(2) \, A_3(2) \, \cdots$$
$$A_1(3) \, A_2(3) \, A_3(3) \, \cdots$$
and use a diagonal counting argument, like before.
Another way is this: each element from the union is indexed by two numbers (n, k) (assuming the union is disjoint, as we implicitly did before). So it's really trivial to write down a bijection from this union to $$\textbf{N} \times \textbf{N}$$. As shown in my earlier post, the latter is countable (there is a bijection to $$\textbf{N}$$, so by composing the bijections you get one from the union to $$\textbf{N}$$. Actually, it also uses the diagonal counting argument (to show that $$\textbf{N} \times \textbf{N}$$ is countable) but perhaps this approach is more intuitive (and once you've proven this result about $$\textbf{N} \times \textbf{N} \times \cdots \times \textbf{N}$$ you can use it and prove countability of other things, like this union of $$A_n$$, without explicitly invoking a diagonal counting argument).
Last edited: Jul 15, 2007
14. Jul 15, 2007
### CompuChip
Does it matter? If you have a set in the union, that is not countably infinite, but finite, will the union become bigger (so big, it will become uncountable?) | {
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If you want to do it officially, consider this: we said a set is countable if we can assign the numbers 1, 2, 3, ... to it. But of course, we can use another offset, and assign 2, 3, 4, ... to it. Or 154842, 154843, 154844, ...
So you could of course first count off the finite sets in the union (in order, no need for diagonals here), and then the infinite ones with the diagonal argument, but using a certain offset.
15. Jul 15, 2007
### linuxux
thanks a million man!
16. Jul 15, 2007
### CompuChip
Most welcome, this stuff is fun and moreover, it's important basics.
Later you will encounter more diagonal arguments, for example: if you want to prove that the real numbers are uncountable, or if you get to work with converging sequences.
17. Jul 16, 2007
### matt grime
You said A_n was a function, not that |-> was a function. Don't be sloppy. | {
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