# 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?

• 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).

• 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 ?

• 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.


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 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 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 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 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 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 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. 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difference between rhombus and square 2021


# 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!

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?

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?


# 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?

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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:

$$1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2$$

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 $$A=B=C=D=\dfrac{1}{4}.$$

• 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

- [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 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 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.


# 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

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 $$\sum_{x \in \{1,3,5,7,9\}} \sum_{y \in \{1,3,5,7,9\}} (11-x)(11-y) = 900$$


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!

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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 $$\{1,3\}\mapsto 10=2\cdot 5=p_1\cdot p_3.$$

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, $$\{1,2\} = 10 \cdot 13^4 + 1 \cdot 13^3 + 12 \cdot 13^2 + 2 \cdot 13^1 + 11 \cdot 13^0 = 289873$$

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.)

• +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.

• 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.

• 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

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.


# 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".

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.

• 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 $$2^b$$.


# 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}

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

$${1\over2}\left(1-{1\over20}\right)={19\over40}$$

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.

$$\begin{array}{r|rr|rr|r} x & 3\text{d}6 = x & 3\text{d}6 < x & 1\text{d}20 = x & 1\text{d}20 < x & 1\text{d}20 < x \cap 3\text{d}6 = x\\ \hline 1 & 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 & 2 & 2 \\ 4 & 3 & 1 & 1 & 3 & 9 \\ 5 & 6 & 4 & 1 & 4 & 24 \\ 6 & 10 & 10 & 1 & 5 & 50 \\ 7 & 15 & 20 & 1 & 6 & 90 \\ 8 & 21 & 35 & 1 & 7 & 147 \\ 9 & 25 & 56 & 1 & 8 & 200 \\ 10 & 27 & 81 & 1 & 9 & 243 \\ 11 & 27 & 108 & 1 & 10 & 270 \\ 12 & 25 & 135 & 1 & 11 & 275 \\ 13 & 21 & 160 & 1 & 12 & 252 \\ 14 & 15 & 181 & 1 & 13 & 195 \\ 15 & 10 & 196 & 1 & 14 & 140 \\ 16 & 6 & 206 & 1 & 15 & 90 \\ 17 & 3 & 212 & 1 & 16 & 48 \\ 18 & 1 & 215 & 1 & 17 & 17 \\ 19 & 0 & 216 & 1 & 18 & 0 \\ 20 & 0 & 216 & 1 & 19 & 0 \\ \hline \text{total} & 216 & & 20 & & 2052 \end{array}$$

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 $(f * g)(x)$.

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.

• 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. $19/40$.


# 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

$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$

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.

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.
$y=\frac{1}{3}\ln{|x|}+k \Rightarrow 3y - 3k = \ln |x| \Rightarrow e^{3y - 3k} = x$

$\Rightarrow x = e^{3y}\, e{-3k} = A e^{3y}$.


# [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

##### 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 $L\left(1-\frac{z}{h}\right)$.

- - - 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.


# 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.

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.


# 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


# 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:

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

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 : $$N = 2.57 * 10^5 N$$


# 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

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 $5000$.
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 $$\binom{13}{9} - \binom{8}{4}$$