# properties of matrix addition

To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Numerical and Algebraic Expressions. 1. Addition: There is addition law for matrix addition. 4. Matrix Vector Multiplication 13:39. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. Commutative Property Of Addition 2. The determinant of a matrix is zero if each element of the matrix is equal to zero. The inverse of a 2 x 2 matrix. All-zero Property. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. This is an immediate consequence of the fact that the commutative property applies to sums of scalars, and therefore to the element-by-element sums that are performed when carrying out matrix addition. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. Yes, it is! Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) There are a few properties of multiplication of real numbers that generalize to matrices. Examples . The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! Proof. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. Matrix Multiplication Properties 9:02. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. Properties of Matrix Addition and Scalar Multiplication. Properties of matrix multiplication. Matrix multiplication shares some properties with usual multiplication. 12. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants. Instructor. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. 13. Use properties of linear transformations to solve problems. A scalar is a number, not a matrix. If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. A B _____ Commutative property of addition 2. In a triangular matrix, the determinant is equal to the product of the diagonal elements. Then we have the following properties. The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. Laplace’s Formula and the Adjugate Matrix. Mathematical systems satisfying these four conditions are known as Abelian groups. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m×nmatrices. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m × n matrix and B and C be n × m matrices, then Equality of matrices 14. Andrew Ng. You should only add the element of one matrix to … EduRev, the Education Revolution! Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + − 1 A = 0. Taught By. A. Best Videos, Notes & Tests for your Most Important Exams. Try the Course for Free. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! 1. The basic properties of matrix addition is similar to the addition of the real numbers. 18. Transcript. There often is no multiplicative inverse of a matrix, even if the matrix is a square matrix. The determinant of a 2 x 2 matrix. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. What is the Identity Property of Matrix Addition? 2. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Learning Objectives. In other words, the placement of addends can be changed and the results will be equal. The first element of row one is occupied by the number 1 … Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. What is a Variable? Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? 8. det A = 0 exactly when A is singular. Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. Find the composite of transformations and the inverse of a transformation. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. Properties of Transpose of a Matrix. This property is known as reflection property of determinants. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. Reflection Property. Properties of scalar multiplication. This tutorial introduces you to the Identity Property of Matrix Addition. To find the transpose of a matrix, we change the rows into columns and columns into rows. The inverse of 3 x 3 matrix with determinants and adjugate . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Matrix Multiplication - General Case. Let A, B, and C be three matrices. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. We state them now. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix A. 11. Matrices rarely commute even if AB and BA are both defined. There are 10 important properties of determinants that are widely used. The commutative property of addition means the order in which the numbers are added does not matter. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Question 3 : then find the additive inverse of A. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the n×k zero-matrix. A+B = B+A 2. Matrix Matrix Multiplication 11:09. Properties involving Multiplication. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. Addition and Scalar Multiplication 6:53. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. 16. General properties. Inverse and Transpose 11:12. Important Properties of Determinants. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. The matrix O is called the zero matrix and serves as the additiveidentity for the set of m×nmatrices. If you built a random matrix and took its determinant, how likely would it be that you got zero? A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of matrix addition. Selecting row 1 of this matrix will simplify the process because it contains a zero. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Matrix addition is associative; Subtraction. However, unlike the commutative property, the associative property can also apply to matrix … 17. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. Let A, B, C be m ×n matrices and p and q be two non-zero scalars (numbers). Property 1 completes the argument. Given the matrix D we select any row or column. Let A, B, and C be mxn matrices. the identity matrix. We have 1. This project was created with Explain Everything™ Interactive Whiteboard for iPad. We can also say that the determinant of the matrix and its transpose are equal. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. So if n is different from m, the two zero-matrices are different. ... although it is associative and is distributive over matrix addition. Proposition (commutative property) Matrix addition is commutative, that is, for any matrices and and such that the above additions are meaningfully defined. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. In this lesson, we will look at this property and some other important idea associated with identity matrices. 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# Find the sum of a sequence [duplicate]

Based on: $\frac{1}{n*(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ where n is element of N find the sum of the following:

$\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+ ... +\frac{1}{38*39}+\frac{1}{39*40}$

How should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one.

Here are the options: a)$\frac{31}{40}$ b)$\frac{33}{40}$ c)$\frac{37}{40}$ d)$\frac{39}{40}$

## marked as duplicate by Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 14 '17 at 11:56

• It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. – Parcly Taxel May 14 '17 at 11:39
• Try to expand a bit and observe the telescoping : $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} -\frac{1}{4})$$ – Zubzub May 14 '17 at 11:44

$\frac{1}{1*2}+\frac{1}{2*3}+ ... +\frac{1}{38*39}+\frac{1}{39*40}=\\ (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{38}-\frac{1}{39})+(\frac{1}{39}-\frac{1}{40})=\\ \frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})-\frac{1}{3}+...+\frac{1}{38}+(-\frac{1}{39}+\frac{1}{39})-\frac{1}{40}=\\ \frac{1}{1}-\frac{1}{40}$

Can you see that all terms except for the first and last term cancel?

Since each and every term will be canceled out except $1$ and $-\dfrac{1}{40}$

So, the answer is $$1-\frac{1}{40}=\frac{39}{40}$$


# Is this induction procedure correct? ($2^n<n!$)

I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I’ve been able to solve some of the form $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$.

Now, I was presented this, for $n \ge 4$:

$$2^n<n!$$

I tried to do it with similar logic as the one suggested there. This is what I did:

Prove it for $n = 4$:
$$2^4 = 16$$
$$4! = 1\cdot2\cdot3\cdot4 = 24$$
$$16 < 24$$
Assume the following:
$$2^n<n!$$
We want to prove the following for $n+1$:
$$2^{n+1}<(n+1)!$$
This is how I proved it:

• So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
• By our assumption, we know that $2^n\cdot2 < n!\cdot2$
• This is because I just multiplied by $2$ on both sides.
• Then we’ll be finished if we can show that $n! \cdot 2 < (n+1)!$
• Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$
• Since both sides have $n!$, I can cancel them out
• Now I have $2<(n+1)!$
• This is clearly true, since $n \ge 4$

Even though the procedure seems to be right, I wonder:

• In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more “careful”?
• Is this whole procedure valid at all? I ask because, well, I don’t really know if it would be accepted in a test.
• Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.

#### Solutions Collecting From Web of "Is this induction procedure correct? ($2^n<n!$)"

Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.

For example, the base case could be re-written as follows:

When $n = 4$, we have $2^4 = 16 < 24 = 4!$

Next, the inductive hypothesis and the subsequent manipulations:

Suppose that for $n \geq 4$ we have $2^n < n!$

Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$.

Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D.

Note: I am not making a judgment about whether your write-up or the one I have included here is “better.” I’m only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).


Shall remainder always be positive?

My cousin in grade 10, was told by his teacher that remainders are never negative. In a specific example,

$$-48\mod{5} = 2$$

I kinda agree.

But my grandpa insists that

$$-48 \mod{5} = -3$$

Which is true? Why?

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$2$ and $-3$ are just two names for the same element in $\mathbb{Z}_5$, i.e. $2 \equiv -3 \mod 5$ – mm-aops May 19 '14 at 13:04
There are various conventions in use, e.g. see the links in this answer. – Bill Dubuque May 19 '14 at 13:08
Python and Ruby say it's 2; C, C#, and Java say it's -3. You could also say it's -8, 7, 12, etc. All are correct. The question is, which do you want? – Tim S. May 19 '14 at 14:00
Who says that remainder and modulus are the same thing? I was taught differently, and the people who developed the Ada programming language apparently thought so, as well. – O. R. Mapper May 19 '14 at 17:36
@John: C++ specification explicitly says (§5.6/4) that / must round towards zero and (a/b)*b + a%b must be equal to a. Which has just one solution, namely that a%b has the same sign as a (unless it is 0). Nothing implementation-defined here. – Jan Hudec May 20 '14 at 21:56

The first one is saying that $-48$ is $2$ more than a multiple of $5$. This is true. The second one is saying that $-48$ is $3$ less than a multiple of $5$. This is also true.

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Who says short answers are short? – Awal Garg May 20 '14 at 11:42
@AwalGarg I do. Axiomatically. – Cruncher May 20 '14 at 14:10
But what about the division? Would the division with module 2 be -48/5 = 10 remainder 2 ? – Pieter B May 21 '14 at 10:09
@PieterB -48/5 = -10 remainder 2 if you like, but it could also equal -9 remainder -3. – Jack M May 21 '14 at 10:14

The teacher is within his/her authority to define remainders to be numbers $r$, $0\leq r < d$ where $d$ is the divisor. This is just a systematic choice so that all students can apply the same rule and arrive at the same anwer, but the rule is only a convention, not a "truth." The teacher isn't wrong to define remainders that way, but it would be wrong for the teacher to insist that there is no other way to define a remainder. And, anyhow, what your grandpa says is also perfectly true :)

I imagine that ordinary long division is taught with this remainder rule because it makes converting fractions to decimals smoother when students do it later. Another reason is probably that mixed fraction notation (as far as I know) makes no allowance for negatives in the fraction. What I mean is that $1+\frac23=2-\frac13$, but the mixed fraction $1\frac23$ is not usually written as $2\frac{-1}{3}$, although one could make an argument that it makes just as much sense.

As far as modular arithmetic is concerned, you really want to have the flexibility to switch between these numbers, and insisting on doing computations with the positive version all the time would be hamstringing yourself.

Consider the problem of computing $1445^{99}\pmod{1446}$. It should not be necessary to compute powers and remainders of powers of $1445$ when you can just note that $1445=-1\pmod{1446}$, and then $1445^{99}=(-1)^{99}=-1=1445\pmod{1446}$

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But $(-48) / 5$ as a mixed numeral is $-9 \frac 35$ meaning $-(9 + \frac 35) = -9 - \frac 35$, which lines up more with Grandpa's way than with Teacher's way. – aschepler May 20 '14 at 19:48

The answer depends on whether you want to talk about modular arithmetic or remainders. These two perspectives are closely related, but different. In modular arithmetic, $2\equiv -3 \pmod 5$, so both answers are correct. This is the perspective most answers here have taken. In the division algorithm, though, where remainders are defined, in order to guarantee uniqueness, you need a specific range for the remainder -- when dividing integer $a$ by integer $d$, you get $a=qd+r$, and the integers $q$ and $r$ are unique if $0\leq r<d$ (note that this also places a restriction that $d$ be positive). There are other ways you could set up the condition, but you need some similar range in order to guarantee uniqueness, and this range is the simplest and most common, so in this sense, a negative remainder doesn't work.

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Excellent answer. The point is that we often want uniqueness of $q$ and $r$, which is guaranteed by requiring $0\leq r < d$. Without this condition, there are $infinitely$ many possible "remainders". – ChocolateAndCheese May 19 '14 at 19:57
@Chocolate: And it's worth noting that other normalizations are possible: sometimes you want $d/2 \leq r < d/2$. Other settings want $0 \leq |r| < d$, but for $rd \geq 0$. Or maybe $rn \geq 0$ or even $rnd \geq 0$. – Hurkyl May 20 '14 at 8:59
We indeed often want uniqueness of $q$ and $r$. But long division algorithm is only practical if $q$ is rounded towards zero (truncated) and that leads to negative (non-positive) remainder for negative numerator. But Euclidean division defines remainder as positive (non-negative). So either definition is sometimes needed with division as well. – Jan Hudec May 21 '14 at 4:57

When -48 is divided by 5 the division algorithm tells us that there is a "unique" reminder r satisfying $0\leq r<5$. In that case there is only one possibility, namely $r=2$.

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The division algorithm is defined on natural numbers, isn't it? Or I may be mistaken – Cheeku May 19 '14 at 13:30
I think it's usually defined for any integer numerator and any positive integer denominator. – poolpt May 19 '14 at 15:32
Negative denominator is not a problem. But long division needs negative remainders when numerator is negative. – Jan Hudec May 21 '14 at 4:52

Here is a different perspective which is more technical but is also reflected e.g. in Hardy and Wright.

On the whole it is best to regard the expression $"-48 \mod 5"$ on its own as representing the set of numbers $a:a\equiv -48 \mod 5$, so it isn't equal to a number at all, but to a set.

[Technically it is a coset in $\mathbb Z$ of the ideal generated by $5$, and consists of the numbers $-48 + 5 b$, where $b$ is an arbitrary integer].

Quite often it is useful to work with numbers rather than sets, and we choose a representative element of the set to work with. There are different ways in which this can be done (least positive, smallest absolute value etc). In this case $2$ and $-3$ are members of the set and could be used to represent it.

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Maybe the teacher was thinking about Euclidean division, which states:

Given two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b

We indeed have the remainder ("r") being positive in this case, but if we talk strictly about congruences the grandpa's expression as well as the teacher's are both true.

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For integers $a,b,c$ the following statements are equivalent:

1) $a\equiv b\text{ mod }c$

2) $a+c\mathbb{Z}=b+c\mathbb{Z}$

3) $c\mid a-b$

Note that $5\mid-48-2=-50$ and $5\mid-48-\left(-3\right)=-45$.

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By convention, for a division $\frac{a}{b}=c$, if we are looking for a whole integer for $c$ (with no further stipulations) , the rounding method is towards zero. If $a<0$ and $b>0$, this means that in the equation $\frac{a}{b}=c+\frac{r}{b}$, $r$ must be $\leq 0$.

Also remainders and modulus are two different things.

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This question touches parts of my old message Algebraic abstractions related to (big) integers in C++ on the boost developers mailing list.

Why should "modulo" be identical to "remainder"? Having a "remainder" function such that "r=remainder(a,m)" satisfies "0 <= r < m" is something convenient to have in a programming language. The quoted message presents some evidence that a "sremainder" function such that "s=sremainder(a,m)" satisfies "-m/2 <= s < m/2" would also be a good idea.

The meaning of modulo and modular arithmetic is not directly addressed by these considerations.

-

For variety, I want to point out that when doing division with remainder (as opposed to, e.g., modular arithmetic), there is sometimes utility in having "improper" results; e.g. saying $17 / 5$ is $2$ with remainder $7$.

An example where this would be useful is if you just need a value that is close to the correct quotient, and you can compute something close (and the appropriate remainder) relatively more easily.

-


# Finding $\int \sec^2 x \tan x \, dx$, I get $\frac12\sec^2x+C$, but an online calculator gets $\frac12\tan^2x+C$.

I tried to find a generic antiderivative for

$$\displaystyle \int \sec^2x \tan x \mathop{dx}$$

but I think there is something wrong with my solution because it doesn't match what I got through an online calculator.

What am I doing wrong?

Below is my solution.

We will use substitution:

$$u = \sec x \qquad du = \sec x \tan x \, dx$$

We substitute and apply the power rule:

$$\int (\sec x) (\sec x \tan x \, dx) = \int u \, du = \frac{1}{2} u^2 + C = \frac{\sec^2x}{2} + C$$

The solution I found with the online calculator is:

$$\frac{\tan ^2 x}{2} + C$$

The steps in the online solution make sense also, so I'm not sure what's going on.

The one thing I have some doubts about is whether I derived the $$du$$ from $$u = \sec x$$ correctly. But it seems okay to me. I used implicit differentiation with $$x$$.

• $\sec^2{(x)}=1+\tan^2{(x)}=\tan^2{(x)}+C$ hence these functions differ by a constant... – Peter Foreman Jul 23 at 18:23
• What happens if you try $u=\tan \theta?$ – Chris Leary Jul 23 at 18:25

$$\frac{\sec^{2}(x)}{2} + C = \frac{1+\tan^{2}(x)}{2} + C = \frac{1}{2} + \frac{\tan^{2}(x)}{2} + C$$

$$\frac{1}{2}$$ is just another constant, in indefinite integral constant doesn't really matter unless you're asked for the integrand original function so you can just kind of "combine" $$\frac{1}{2}$$ into C, you can also differentiate the answer to know that there's nothing wrong at all with your answer

The difference between your answer and the answer given is that they differ by a constant value.

You can see this by using the identity

$$1+\tan^2(x) = \sec^2(x)$$

Hence your answer can be converted to the given answer by subtracting $$-1/2$$, which is a constant.

As mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution

$$u = \tan x$$

• I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! – Cameron Williams Jul 23 at 18:29
• Thanks for editing the expressions – StackUpPhysics Jul 23 at 18:29

Try differentiating both of them and see that nothing has gone wrong at all.

Let $$u = \tan x$$.

Then $$du = \sec^{2} x dx$$.

Hence, the integral becomes

$$\int u du = \frac{u^{2}}{2} + C = \frac{\tan^{2}x}{2} + C.$$


# Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant?

Why is the difference between these two functions a constant?

$$f(x)=\frac{2x^2-x}{x^2-x+1}$$ $$g(x)=\frac{x-2}{x^2-x+1}$$

Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.

Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?

Thanks!

BACKGROUND: The background of this question is that I tried to find this integral:

$$\int\frac{x dx}{(x^2-x+1)^2}$$

As a solution I found:

$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$

Whereas my calculusbook gave as the solution:

$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$

I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.

• Did you actually graph both functions? That would tell a lot May 3, 2018 at 14:26
• @imranfat desmos.com/calculator/njhbjs54rv. Looks strangely like $2$ to me. Did you actually graph both functions? May 3, 2018 at 14:31
• @Oldboy Actually, it is. May 3, 2018 at 14:31
• It doesn't matter that the numerators have different degrees. What matters is that their difference is a multiple of the denominator.
– user856
May 3, 2018 at 16:14
• Fascinating background: at face value, I would wager that a lot of instructors would mark your answer wrong without paying much attention to how you got to it - and that most students wouldn't know to ask for a review. It is not obvious that they are both valid. I'd love to hear from educators here how would they approach this. May 4, 2018 at 0:23

Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?

• Not sure who is responsible for that "from review". This actually provides more of an answer to the question asked than the majority of the posts (though there are others that do better). May 3, 2018 at 18:14
• It absolutely does provide an answer to the question, and what's more it's a good answer. It gives a simpler example of the phenomenon which the OP requests intuition for, and the intuition is easier to obtain in this simpler example. May 3, 2018 at 18:27
• @JamesMartin: Indeed. How about: Would you be surprised that the difference of $\frac{1234}{5}$ and $\frac{234}{5}$ is $200$? May 4, 2018 at 11:26
• Yes, this is still surprising. It is roughly equally as surprising as the difference in the original question (to me anyways). May 4, 2018 at 21:21
• @Nova Often repeating a question with slightly different values IS the best answer. Countless times I have done this with students with great success, because it can relate a situation that the student already has intuition for to the one that they are stumped on. Maybe this answer doesn’t work for some, but that doesn’t inherently make it a bad answer. May 6, 2018 at 7:30

It is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x^n$.

Now consider $$\frac{p(x)}{p(x)}$$ This is clearly the constant $1$ (except at zeroes of $p(x)$).

Now separate the leading term: $$\frac{a_n x^n}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$

and re-write to create the difference:

$$\frac{a_n x^n}{p(x)} - \frac{a_n x^n - p(x)}{p(x)}$$

Obviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial.

Similarly, you could split $p(x)$ in many other ways.

I'm not sure anyone is speaking to your observation that the two numerators have different degrees.

Let's flip this around the other way:

\begin{align*} \frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1} \\ &= \frac{2x^2 -x}{x^2-x+1} \text{.} \end{align*} That is, we started with a thing having a linear numerator and added a constant to it. But when we brought the constant to have a common denominator, it picked up a degree two factor. Then the addition was forced to produce a degree two sum.

To sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators. So constants effectively have "degree two in the numerator" in your example.

• +1 for this > "Constants effectively have "degree two in the numerator" in your example." May 4, 2018 at 14:20
• This deserves a lot more upvotes for the "degree two in the numerator" note. It reconciles intuition to the raw result. May 5, 2018 at 15:09
• A further +1 for "To sum up". May 10, 2018 at 5:34

Look at it in reverse:

Take a polynomial fraction and add it to a constant. The result will be a polynomial fraction, with the same denominator and a different polynomial as the numerator.

This belongs to a specific set of questions "you cannot really answer to your students" if you are a teacher.

The difference is $2=\frac{2(x^2-x+1)}{x^2-x+1}$. This is why this seems weird (but true). Even this seems weird: $$\frac{15}{7}-\frac{1}{7}=2$$
The numerators differ by 14 (not 2) but the denominators are equal. The best is to try and explain this to yourself why this perfectly fine.

Note that in general given a rational function

$$f(x)=\frac{p(x)}{q(x)}\implies g(x)=f(x)+c=\frac{p(x)+c\cdot q(x)}{q(x)}$$

and $\deg(p(x)+c\cdot q(x))\le \max\{\deg(p(x)),\deg(q(x))\}$.

• Why? Where is "the difference"? May 5, 2018 at 5:59
• @RolazaroAzeveires It is just a generalized way to see that two rational function can differ for a constant c even if denominators are equal and the numerators differ in degree, that's exactly the point of the OP.
– user
May 5, 2018 at 6:44
• I think it is just a matter of preference, from my point of view this observation suffices to answer the OP completely and maybe also the upvoters have the same idea. There are many other answers here with concrete examples and discussion so the asker have a lot of points on view on that and you also are free to add your own answer according to your best interpretation. Thanks anyway for your advice and suggestions on that. Bye!
– user
May 6, 2018 at 11:13

$$(2x^2-x)-(x-2)=2x^2-2x+2$$ Hence$$f(x)-g(x)=2\bigg(\frac{x^2-x+1}{x^2-x+1}\bigg)=2(1)=2$$

In effect, this pair of equations is a very specific case where the numerator and denominator end up lining up, and thus you get a constant for all values.

• What kind of an answer is this? He already knows that. May 3, 2018 at 14:31

Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.

When the numerators differ in degree, the difference between the numerators is a nontrivial polynomial that goes to infinity as $x\to\pm\infty$.

However, nobody says that this polynomial cannot grow at the same rate as the (equal) denominators, in which case the ratio will be bounded (and in particular possibly constant). There's nothing in your observation "the denominators are equal and the numerators differ in degree" that link the denominator to either the numerators or their difference, so you have no reason to think the difference in numerators should dominate the growth of the entire fraction.

• Eactly, i already wanted to provide an answer before seeing this, when $x->\infty$ the first function tends to $2$, the second tends to $0$, which are both constants and give out a cte difference, this doesn't apply when the numerator is bigger than the denom, if the denom is 1 the difference would be diverging but once you divide it by a polynomial of order 2, things change to converge. May 4, 2018 at 14:14

Maybe instead of integrating you need just differentiating.

$$\frac{df}{dx}=\frac{(2x^2-x)(2x-1)-(x^2-x+1)(4x-1)}{denom^2}\\ =\frac{x^2-4x+1}{denom^2}$$

In the other hand:

$$\frac{dg}{dx}=\frac{(x-2)(2x-1)-(x^2-x+1)}{denom^2}\\ \ =\frac{x^2-4x+1}{denom^2}=\frac{df}{dx}$$

See that both functions have same derivates, which means they differ by the same value from $x$ to $x'$ , this only means they have a same growth rate and they have same difference all along the range of x axis.

• This is something OP himself did. He was solving an integral and found an alternate answer $g$. Since $f$ and $g$ are solutions to the same integral, they defer by a constant just like you demonstrate here. OP is asking for an intuition why the two rational functions defer by only a constant. May 6, 2018 at 6:03
• What do you intend to do with a primitive, why at all one think to ascend to an integral? one function may be integrated in an infinitely many ways but what's the point ? look in the op's post that he already found two functions completely different, you are missing the point. The intuition is clear, you shouldn't compare the numerator unconsiderably of the rest of the function, you should take it all into account, for example $f$ and $g$ are mutually excluding eachother in a derivate of the form $f+g$, but they are mutual-inclusive in $f/g$. Appearence is deceiving. May 6, 2018 at 12:46

You can construct a similar puzzle from the (simpler) observation that

$$\frac{x}{x+1} + \frac{1}{x+1} = 1 .$$

Translating, that's the same as $$\frac{u-1}{u} + \frac{1}{u} = 1$$ which isn't surprising at all.

• Yes, but in my example the degrees of the numerators are different, which caught me off guard May 3, 2018 at 14:34
• The degrees are different in my example too. But I agree that yours is at first surprising. Note: this kind of rational function identity is useful sometimes when calculus instructors want to make up problems that look harder than they are. May 3, 2018 at 14:35

Another way of stating what everyone has said:

$2x^2 - x= 2(x^2-x+1) + x - 2$

$\Rightarrow 2x^2-x = x-2 \mod x^2-x+1$

Because if $P\ and\ Q\ are\ polynomials$ then, $$\deg \left( P\pm Q \right)\le max\left( \deg \left( P \right),\deg \left( Q \right) \right)$$ where $deg$ is the degree of the polynomial. See this link

Here's another view on the problem:

Let's look at the fraction with the higher degree: $$f(x)=\frac{2x^2-x}{x^2-x+1}$$ We note that the degree of the numerator is the same as the one of the denominator.

Now whenever the numerator's degree is at least as large as the denominator's degree, we can do polynomial division. The result is a polynomial whose degree is the difference between the numerator's and the denominator's degree, and a rest whose degree is less than the denominator's degree.

Now since numerator and denominator have the same degree, the difference of the degrees is zero. But a polynomial of degree zero is a constant. And the division rest will be of lesser degree.

Let's do the polynomial division explicitly: $$\begin{array}{llccl} &(2x^2-\ \ x) & / & (x^2-x+1) &=& 2\\ -&\underline{(2x^2-2x+2)}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ x \ - 2 \end{array} %\begin{array}{rrrrcccl} %&(2x^2 & -x) & & / & (x^2-x+1) & = & 2\\ %-&(2x^2 & -2x & +2)\\ %\hline % & & x & -2 %\end{array}$$ You surely will recognize the rest as the numerator of $g(x)$

So, whenever you have a rational function whose numerator and denominator, after cancelling out, have the same non-zero degree, it is not only possible, but guaranteed that there exists another rational function with lesser degree in the numerator which differs from the original rational function only by a constant.

• BTW, does anyone know how to make the horizontal line end at the end of the polynomial above? May 4, 2018 at 5:51
• @farruhota: That first option destroys the alignment between the terms, and therefore is worse than the too-long line. That second option defeats the whole purpose of writing the polynomial division in its standard form. May 4, 2018 at 6:48
• How about this: $\begin{array}{ll} &(2x^2-x) & / (x^2-x+1) = 2\\ -&\underline{(2x^2-2x+2)}\\ & x-2 \end{array}$ May 4, 2018 at 7:04
• @farruhota: Still doesn't correctly align the terms. May 4, 2018 at 7:10
• Oh well, the rest should be adjusted: $\begin{array}{ll} &(2x^2-\ \ x) & / (x^2-x+1) = 2\\ -&\underline{(2x^2-2x+2)}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ - 2 \end{array}$ May 4, 2018 at 7:16

I feel that it's only natural that intuition would fail you in cases like these. If you wrote the two functions out as $$f(x)=\frac{2\left(x^2-x+1\right)+\left(x-2\right)}{x^2-x+1}$$ $$g(x)=\frac{x-2}{x^2-x+1}$$ it would be obvious that the difference between them was 2. This is the first thing you could trip over, the degree of the numerator don't actually differ by 1, they are the same.

I guess if you made it a rule to divide whenever you have a rational function where the degree of the numerator is greater than or equal to the degree of the denominator until it isn't, you would get consistent results. (Kinda like how you learn waaay back how $\frac{13}{7}$ is an improper fraction, because it's actually bigger than a whole number, and that you should write it as $1\frac{6}{7}$. a function $f(x)=\frac{P(x)}{Q(x)}$ where the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$ is called an improper rational function.)

But, and this might be another thing that's bugging you about this question, you can't just do that! (As my algebra teacher used to exclaim in exasperation.) You introduce a new constraint: $x^2-x+1\neq0$. Which isn't relevant if x is real, but does change things up a bit if it s a complex number.

This is probably what you get when you graph the two functions $f$ and $g$.

But if you allow for complex values of $x$, it can also look like this:

(Sans the vertical line at x=0.5, I don't know why that is happening, probably a limitation in the graphing engine on desmos.)

Not so obvious that the difference between the two is 2 in this case...

Sorry, I just realized this probably doesn't answer your question on how you can reconciliate this apparent discrepancy with your intuition, but I thought explaining why it may not be that clear-cut would help at least a little bit.

Playing with the parameters used in the graph might help further your understanding, here's a link to the interactive graph, with a little more in depth graphical interpretation of why you might doubt what you logically "know" to be true.

As for the equations I used in place of the given ones... You'll just have to trust my algebra (which you shouldn't in critical situations, but I think this is safe enough hehe). I did them on a legal pad, but I'll type them up if anyone insists.

Since the denominator is the same , you subtract one numerator from the other. In this special case you find the resultant numerator is twice the denominator. Feels like that was much simpler than the rest of your background calculus.

It must be very clear to you that $11/3$ and $2/3$ defer by an integer. Why is this so? Because in the world $$modulo 3$"$, 11 and 2 are but the same. In the same vein, we can say that in the world of $$modulo $x^2-x+1"$, $2x^2-x$ and $x-2$ are but the same because $2x^2-x=2(x^2-x+1)+x-2$. When we are in such a world, the same polynomial may appear in different forms with different "degrees". So in a world such as the ideal generated by $x^2-x+1$ (ie $<x^2-x+1>$ ), degree is no longer retained. One's intuition may fail here because one is not familiar with such a world where degree is not so well behaved.
• The question is not about constants; it is about degree. $\frac{x^3+1}{x^2+2}$ and $\frac{-2x+1}{x^2+2}$ also defer by a constant (polynomial) May 6, 2018 at 6:11


# Thread: [SOLVED] Finding Co-Ordinates of a Rectangle

1. ## [SOLVED] Finding Co-Ordinates of a Rectangle

Here's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:

Is my method correct, considering I did get the right answer?

But is there another simpler method to do this which would save time during an exam.

2. The diagonals bisect one another. The midpoint of $\displaystyle \overline{AC}$ is ?

3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?

4. Originally Posted by unstopabl3
Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
But the y-coordinate of $\displaystyle B~\&~D$ is 6.
You are given the x-coordinate of $\displaystyle D$, so $\displaystyle Dh,6)$

5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.

I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.

I am not after the answer, I am looking for an alternate method.

6. Originally Posted by unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.

I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.

I am not after the answer, I am looking for an alternate method.
Hi unstopabl3,

I really don't see what method you used in your first post, but here's how I'd do it.

Plato already told you that the midpoint of BD is M(6, 6).

This means the y-coordinates of B and D are also 6.

This distance from B to D is 20 (found using distance formula)

Each individual segment of the diagonals measure 10 since they are bisected.

Using the distance formula, it is easy to determine the x-coordinates of B and D.

M(6, 6) -----> D(h, 6) = 10

$\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$

7. Hello, unstopabl3!

Code:
              |           C(12,14)
|           o
|       *    *
|   *         *
*              *
B o - + - - - - - - - o D
*  |           *
------*-+-------*------------
*|   *
o
A|(0,-2)
|

The diagram shows a rectangle $\displaystyle ABCD.$
We have: .$\displaystyle A(0,-2),\;C(12,14)$
The diagonal $\displaystyle BD$ is parallel to the $\displaystyle x$-axis.

$\displaystyle (i)$ Explain why the $\displaystyle y$-coordinate of $\displaystyle D$ is 6.
The diagonals of a rectangle bisect each other.
. . Hence, the midpoint of $\displaystyle AC$ is the midpoint of $\displaystyle BD.$

The midpoint of AC is: .$\displaystyle \left(\tfrac{0+12}{2},\;\tfrac{-2+14}{2}\right) \:=\:(6,6)$

Therefore, $\displaystyle B$ and $\displaystyle D$ have a $\displaystyle y$-coordinate of 6.

The $\displaystyle x$-coordinate of $\displaystyle D$ is $\displaystyle h.$
$\displaystyle (ii)$ Express the gradients of $\displaystyle AD$ and $\displaystyle CD$ in terms of h,.
We have: .$\displaystyle \begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$

$\displaystyle m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$

$\displaystyle m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$

$\displaystyle (iii)$ Calculate the $\displaystyle x$-coordinates of $\displaystyle D$ and $\displaystyle B.$

Since $\displaystyle m_{AD} \perp m_{CD}$ we have: .$\displaystyle \frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$

Hence: .$\displaystyle (h+4)(h-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$

Therefore: .$\displaystyle D(16,6),\;B(-4,6)$

8. Originally Posted by masters
Hi unstopabl3,

I really don't see what method you used in your first post, but here's how I'd do it.

$\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!

This distance from B to D is 20 (found using distance formula)
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?

Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!

9. Originally Posted by unstopabl3
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!

How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?

Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).

10. Co ordinates of B and D are $\displaystyle (x_1 , 6) and (x_2, 6)$
Diagonal AC = BD
AC = 20 = BD.
Distance $\displaystyle BD^2 = (x_1 - x_2)^2$

So (x_1 - x_2) = 20...........(1)

Mid point point of AC = mid point of BD

$\displaystyle \frac{x_1+x_2}{2} = 6$

(x_1 + x_2) = 12......(2)
Solve Eq (1) ans (2) to find the coordinates of B and D.

11. Thanks for the responses guys!


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A system of linear equations is called homogeneous if the constants $b_1, b_2, \dots, b_m$ are all zero. By … Solve age word problems with a system of equations. Determining the value of k for which the system has no solutions. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. Section 7-1 : Linear Systems with Two Variables. Answer: x = .5; y = 1.67. Substitution Method. Systems of Linear Equations. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\left( {2 - x} \right) = 3x + 2$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}$$. Systems of linear equations can be used to model real-world problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. A large pizza at Palanzio’s Pizzeria costs $6.80 plus$0.90 for each topping. Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. If you're seeing this message, it means we're having trouble loading external resources on our website. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. 1. At the first store, he bought some t-shirts and spent half of his money. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. So far, we’ve basically just played around with the equation for a line, which is . A solution of the system (*) is a sequence of numbers $s_1, s_2, \dots, s_n$ such that the substitution $x_1=s_1, x_2=s_2, \dots, x_n=s_n$ satisfies all the $m$ equations in the system (*). Solving using Matrices by Elimination. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. So a System of Equations could have many equations and many variables. For example, the sets in the image below are systems of linear equations. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you're seeing this message, it means we're having trouble loading external resources on our website. Systems of Linear Equations and Problem Solving. Wouldn’t it be cl… Solving Systems of Linear Equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. 9,000 equations in 567 variables, 4. etc. But let’s say we have the following situation. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. The same rules apply. Solve the following system of equations by elimination. Problem 1 Two of the following systems of equations have solution (1;3). HIDE SOLUTIONS. If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. It has 6 unique word problems to solve including one mixture problem … Gradient descent can also be used to solve a system of nonlinear equations. Just select one of the options below to start upgrading. Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. Systems of linear equations are a common and applicable subset of systems of equations. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. One of the following system of equations, you need to find the exact values of x y. 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# Average number of selections before duplicate picked

I have a dataset of 1296 unique codes which can be numbered 1 through 1296. If numbers are selected at random, one at a time, with replacement. On average, how many iterations will it take to select a number that has already been selected?

Experimentally, (looping through the list of 1296 codes and creating a subset of selected codes using Python) it averages out at 45.875 times (this number includes the duplicate) but I would like to verify it with a calculation so any help would be appreciated.

This question has some similarities but I am unable to perform a calculation based on the answer:

Question with similarities

• This is an example of the generalized birthday problem where you have $1296$ "days" instead of $365$. The number that gives a $50\%$ chance of a match in $d$ "days" is about $\sqrt {2d\ln 2}$, which for you is about $42.39$ – Ross Millikan Aug 19 '17 at 1:15
• @RossMillikan instead of asking for where it switches from being below a $50\%$ chance to being above a $50\%$ chance, isn't the OP asking for the expected number of draws until a match occurs? Will those two necessarily be the same? I'm getting a different result (fixed minor typo in equation) getting a result closer to $\approx 45.788$ – JMoravitz Aug 19 '17 at 1:20
• @JMoravitz: I'll reopen it. I don't know an easy approach to the expected number version, but would expect it to be close to the $50\%$ probability number. No, they won't be the same because there is the long tail to high numbers. – Ross Millikan Aug 19 '17 at 2:10
• @RossMillikan well, the pdf is straightforward (and is included in one of your two links in some form) and we can apply the definition of expected value. I don't know of a clean way to simplify the sum by hand, but computers can calculate it easily enough for us. – JMoravitz Aug 19 '17 at 2:12

It is impossible to have gotten a duplicate on the first draw. It is impossible to have not gotten a duplicate by the 1297'th draw by pigeon-hole principle.

To have gotten your first duplicate on the $k$'th draw, you need the first $k-1$ draws to all be distinct and the $k$'th to be a duplicate.

The first draw will always be distinct. The second will be distinct from the first with probability $\frac{1295}{1296}$. The third will be distinct from the first two with probability $\frac{1294}{1296}$ and so on... the $(n)$'th will be distinct from the earlier $n-1$ with probability $\frac{1296-n+1}{1296}$. Multiplying these, we get for $n$ draws to all be distinct, this will occur with probability $\frac{1296\frac{n}{~}}{1296^n}$ where $x\frac{n}{~}$ represents a falling factorial $x\frac{n}{~}=\underbrace{x(x-1)(x-2)\cdots (x-n+1)}_{n~\text{terms in the product}}=\frac{x!}{(x-n)!}$.

Next, supposing $k-1$ distinct values have all been taken, for the $k$'th to duplicate one of the earlier results, this will occur with probability $\frac{k-1}{1296}$

We have then the probability distribution function for $X$, the number of draws until the first duplicate:

$$Pr(X=k)=\frac{(k-1)1296\frac{k-1}{~}}{1296^k}$$

Applying the definition of expected value for a pdf: $E[X]=\sum\limits_{k\in\Delta} kPr(X=k)$ we have then the expected value is

$$\sum\limits_{k=2}^{1297}\frac{k(k-1)1296\frac{k-1}{~}}{1296^k}\approx 45.7889$$

• As a minor aside, I was having difficulty with the link. Trying to write it as [linkname](actuallinkgoeshere) with parenthesis appearing in the link part of it was being chopped off, and replacing parenthesis with brackets was causing the calculation to time out. Tinyurl's aren't permenant... does anyone have a suggestion other than just hiding it with a spoiler tag like I have? – JMoravitz Aug 19 '17 at 2:35

With $n$ objects the expected time until the first repeat is exactly $$\mathbb{E}(T)=\int_0^\infty \left(1+{x\over n}\right)^ne^{-x}\,dx,$$ and approximately equal to $\sqrt{n\pi/2}.$

You can find a derivation of this formula at my answer here: Variance of time to find first duplicate

For $n=1296$ the exact formula gives $\mathbb{E}(T)\approx 45.78885405,$ while the approximation gives $\sqrt{1296\pi/2}\approx 45.11930893.$


# Find relationship between $a, b, c, f, g, h$

Given that - $$a=a_1 a_2$$ $$b=b_1 b_2$$ $$c=c_1 c_2$$ $$h=a_2 b_1 + b_2 a_1$$ $$g=a_1 c_2 + a_2 c_1$$ $$f=b_1 c_2 + b_2 c_1$$

Find the relationship between $a, b, c, f, g, h$

My Attempt: I could not see how I could exploit the symmetry of the equations to directly get an answer, so I tried to solve them as 6 simultaneous equations by - $$a_2=\frac{a}{a_1}$$ $$b_2=\frac{b}{b_1}$$ $$c_2=\frac{c}{c_1}$$ and put these values into the remaining 3 equations to obtain - $$a_1 b_1 h = a {b_1}^2 + b {a_1}^2$$ $$a_1 c_1 g = a {c_1}^2 + c {a_1}^2$$ $$c_1 b_1 f = c {b_1}^2 + b {c_1}^2$$ Then, I treated the last 2 equations as quadratics in $a_1$ and $b_1$; found their values using the quadratic formula and input those into the first of the last 3 equations above.

Then I found the value of $c_1$ using that and then the value of the remaining - $a_1, a_2, b_1, b_2, c_2$.

Then, when I finally input these values into any of the equations I got a tautology (which, as I now realize - too late - was doomed to happen from the very beginning, due to my approach).

============================================================

So, How should I go about finding the relation between $a, b, c, f, g, h$?

OR,

Equally - How do I eliminate $a_1, a_2, b_1, b_2, c_1, c_2$ from the equations?

I just need a hint on how I could exploit the symmetry of the equations.

• What does "Find the relationship between a,b,c,f,g,h" mean exactly? One could argue that the initial six equalities already gives a relationship among them. Are you looking for a polynomial expression in a,b,c,f,g,h (and not involving the subscripted variables) that equals 0? – Greg Martin May 19 '17 at 3:36
• Hint: consider $af^2+bg^2+ch^2$. – Greg Martin May 19 '17 at 3:38
• @GregMartin Yes, you're right I need a single expression in a, b, c, f, g, h without the subscripted variables. As an example of such an expression - $\frac{c^{2}}{2a}=\frac{f^{3}}{5g^{2}} +h tan(\frac{b}{a})$ The expression could include logs and trig functions (although I don't think they will be necessary) – Quantum Sphinx May 19 '17 at 3:39
• @GregMartin Yes, that is the kind of expression I need – Quantum Sphinx May 19 '17 at 3:43

Notice that $a,b,c,h,g,f$ are represented by staggered multiplication of $a_1,a_2,b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, and then we could rearrange how they are combined together. So symmetry is the key here.
$$h\cdot g \cdot f=(a_2 b_1 + b_2 a_1)(a_1 c_2 + a_2 c_1)(b_1 c_2 + b_2 c_1)$$ $$=a_2b_1a_1c_2b_1c_2 + a_2b_1a_1c_2b_2c_1+a_2b_1a_2c_1b_1c_2 + a_2b_1a_2c_1b_2c_1$$$$+b_2a_1a_1c_2b_1c_2+b_2a_1a_1c_2b_2c_1+b_2a_1a_2c_1b_1c_2+b_2a_1a_2c_1b_2c_1$$
$$=ab_1^2c_2^2+abc+a_2^2b_1^2c+a_2^2c_1^2b+a_1^2c_2^2b+a_1^2b_2^2c+abc+b_2^2c_1^2a$$ $$=2abc + a(b_1^2c_2^2+b_2^2c_1^2)+b(a_1^2c_2^2+a_2^2c_1^2) + c(a_2^2b_1^2+a_1^2b_2^2)$$ $$=2abc+a((b_1c_2 + b_2c_1)^2-2b_1c_2b_2c_1)+b((a_1c_2+a_2c_1)^2-2a_1c_2a_2c_1)+c((a_2b_1+a_1b_2)^2-2a_2b_1a_1b_2)$$ Thus $$hgf=2abc + a(f^2-2bc)+b(g^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2-bg^2-ch^2=0$$


# Find the Numbers

Status
Not open for further replies.

##### Full Member
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?

Set up:

Let x = large number
Let y = small number

x + y = 53...Equation A
3y = x + 19....Equation B

x + y = 53

y = 53 - x...Plug into B.

3(53 - x) = x + 19

159 - 3x = x + 19

-3x - x = 19 - 159

-4x = -140

x = -140/-4

x = 35...Plug into A or B.

I will use A.

35 + y = 53

y = 53 - 35

y = 18.

The numbers are 18 and 35.

Yes?

#### JeffM

##### Elite Member
Do the numbers satisfy both equation?

$$\displaystyle 35 + 18 = 53.$$ Checks.

$$\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks.

In algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math.

#### Subhotosh Khan

##### Super Moderator
Staff member
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?

Set up:

Let x = large number
Let y = small number

x + y = 53...Equation A
3y = x + 19....Equation B

x + y = 53

y = 53 - x...Plug into B.

3(53 - x) = x + 19

159 - 3x = x + 19

-3x - x = 19 - 159

-4x = -140

x = -140/-4

x = 35...Plug into A or B.

I will use A.

35 + y = 53

y = 53 - 35

y = 18.

The numbers are 18 and 35.

Yes?
When possible check your work. Most of the time that is a part of the process of solution.

There is a shorter way to accomplish the algebra/arithmetic part.

You have two equations,

x + y = 53...Equation A
3y = x + 19....Equation B

rewrite B to collect all the unknowns to LHS

x + y = 53...Equation A
3y - x = 19....Equation B'

Add A & B' (to eliminate 'x' from the equations) and get equation C

3y + y = 72....Equation C

4y = 72

y = 18

Use this value in equation 'A'

x + 18 = 53...Equation A

x = 53- 18 = 35

##### Full Member
When possible check your work. Most of the time that is a part of the process of solution.

There is a shorter way to accomplish the algebra/arithmetic part.

You have two equations,

x + y = 53...Equation A
3y = x + 19....Equation B

rewrite B to collect all the unknowns to LHS

x + y = 53...Equation A
3y - x = 19....Equation B'

Add A & B' (to eliminate 'x' from the equations) and get equation C

3y + y = 72....Equation C

4y = 72

y = 18

Use this value in equation 'A'

x + 18 = 53...Equation A

x = 53- 18 = 35

What is wrong with my method?

#### Dr.Peterson

##### Elite Member
Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better".

##### Full Member
Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better".
There are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool.

#### Dr.Peterson

##### Elite Member
Correct.

In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!

But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem.

##### Full Member
Correct.

In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!

But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem.
We can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y).

#### Jomo

##### Elite Member
I know that you can check these problems. Just admit that you like posting here.