1. ## Vector Intersection

Hi, I have a mechanics question here I can't quite get.

A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.

Any help would be greatly appreciated.
Thanks

2. Hello steve989
Originally Posted by steve989
Hi, I have a mechanics question here I can't quite get.

A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.

Any help would be greatly appreciated.
Thanks
Denote the velocity of the destroyer by $\vec{v_D}$ and the velocity of the ship by $\vec{v_S}$. Then:
$\vec{v_D}=\frac{v}{5}\Big(4\vec i + 3 \vec j\Big)$, since this vector has magnitude $v$.
and
$\vec{v_S} = 5\vec j$
So the velocity of the ship relative to the destroyer is
$\vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$
Therefore after time $t$, the position of the ship relative to the destroyer is:
$600(3\vec i + \vec j) -\frac{4vt}{5}\vec i +\Big(5-\frac{3v}{5}\Big)t\vec j$
$=0\vec i + 0\vec j$ when the destroyer intercepts the ship
Solving for $v$ and $t$ gives:
$t = 390$ and $v = \frac{75}{13}$
But check my working!

3. Hello, steve989!

A destroyer sights a ship at a point with position vector $1800i + 600j$ relative to it
and moving with velocity $5j$ m/s.

The destroyer alters course so that it moves with speed $v$ m/s
in the direction of the vector $4i+3j$

Find $v$ so that the destroyer intercepts the ship and the time to the interception.
Code:
      |
|                     o P
|                   o |
|                 o   | 5t
|               o     |
|             o       |
|           o         * S
|         o           |
|       *             |
|  v  * |             | 600
|   *   |3            |
| * θ   |             |
D* - - - + - - - - - - *
:   4   :
: - - - -1800 - - - - :

The destroyer is at: $D(0,0)$

The ship is at: $S(1800, 600)$

The ship is moving north at 5 m/s.
In the next $t$ seconds, it moves $5t$ m to point $P.$

. . Its position vector is: . $\vec S \:=\:\langle 1800,\:600+5t\rangle$

The destroyer heads in direction $4i+3j$
Let $\theta$ represent its direction, where: $\cos\theta \,=\,\tfrac{4}{5},\;\sin\theta\,=\,\tfrac{3}{5}$

. . Its position vector is: . $\vec D \;=\;\langle (v\cos\theta)t,\:(v\sin\theta)t\rangle \;=\;\left\langle\tfrac{4}{5}vt,\:\tfrac{3}{5}vt\r ight\rangle$

The destroyer intercepts the ship at point $P.$
This happens when: $\vec D \,=\,\vec S$

We have: . $\begin{Bmatrix}\frac{4}{5}vt &=& 1800 & [1] \\ \\[-3mm] \frac{3}{5}vt &=& 600 + 5t & [2] \end{Bmatrix}$

Divide [2] by [1]: . $\frac{600+5t}{1800} \:=\:\frac{\frac{3}{5}vt}{\frac{4}{5}vt} \quad\Rightarrow\quad\frac{120+t}{360} \:=\:\frac{3}{4}$

. . $480 + 4t \:=\:1080 \quad\Rightarrow\quad 4t \:=\:600 \quad\Rightarrow\quad t \:=\:150$

Therefore, the destroyer intercepts the ship in 150 seconds.

Substitute into [1]: . $\tfrac{4}{5}v(150) \:=\:1800 \quad\Rightarrow\quad v \:=\:15$

Therefore, the destroyer must travel at 15 m/s.

4. Thanks for both of your replies! I've gone though them as well as I can.

Grandad - I tried your way and think I see how you're thinking there, I got a different formula for one stage though.

Therefore after time , the position of the ship relative to the destroyer is:
when the destroyer intercepts the ship
For that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.

Soroban - I do like that approach and the answers seem very reasonable, I missed the fact that the ship was moving directly north. I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though.

5. Hello everyone

I agree with Soroban's answer: I got a sign wrong in my working. My equation for the relative velocity was correct, though. The principle is:
The velocity of A relative to B is the velocity of A minus the velocity of B.
So, as I said:
$\vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$
not as you suggest:
Originally Posted by steve989
...
For that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.

However, I then got a sign wrong when I equated the $\vec j$ component of the displacement to zero. The equations should have been:
$\vec i$ component: $1800 - \frac{4vt}{5} = 0$

$\vec j$ component: $600 + 5t -\frac{3vt}{5}=0$
which, of course, are the same as Soroban's equations [1] and [2].
Originally Posted by steve989
...I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though.

Because the magnitude of $4\vec i + 3\vec j$ is $5$. So $v(4\vec i + 3\vec j)$ will have magnitude $5v$.

6. Ahh I see, all makes perfect sense now, I knew there had to be simple reasoning behind that. Thanks a lot to you both.


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# If the set S consists of five consecutive positive integers, what is..

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If the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers?

(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.
(2) The sum of the even integers in $$S$$ is 26
[Reveal] Spoiler: OA
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22 May 2016, 23:09
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nalinnair wrote:
If the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers?

(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.
(2) The sum of the even integers in $$S$$ is 26

Statement 1: The integer 11 is in S, but 10 is not in S
This means the integers start from 11
The integers will be 11, 12, 13, 14, 15
We can find out the sum.
SUFFICIENT

NOTE: We do not need to find out the actual sum.

Statement 2: The sum of the even integers in S is 26
The only possible series is 11, 12, 13, 14, 15
We can find out the sum
SUFFICIENT

Correct Option: D
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nalinnair wrote:
If the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers?

(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.
(2) The sum of the even integers in $$S$$ is 26

(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.
Since number are consecutive, it there cannot be any skipping among the number. If 11 is there, then the series should start with 11
s={11,12,13,14,15}; 12 and 14 are even
SUFFICIENT

(2) The sum of the even integers in $$S$$ is 26
Now there are total of 5 integers. there are two possibilities
OEOEO OR EOEOE
OEOEO means two consecutive even ; sum of these is 26
==>E+(E+2)=26 ==>E=12, second even = 14
EOEOE means there are three consective even numbers
E+(E+2)+(E+4)=26
==>3E=20
==>E=20/3 NOT ACCEPTABLE because its not an integer value.

SO in any case both cases gives unique values

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Sorry to bring this one back. By consecutive, do we just assume they mean x, x+1, x+2 or anything that has a pattern? ie. x+2, x+5, x+8...?
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Smileyface123 wrote:
Sorry to bring this one back. By consecutive, do we just assume they mean x, x+1, x+2 or anything that has a pattern? ie. x+2, x+5, x+8...?

Since the question states 5 consecutive positive numbers, it means x, x+1, x+2....
Else, it would have been indicated about the pattern.
Hope it helps.

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24 Nov 2016, 02:28
Statement 1 : The integer 11 is in S, but 10 is not in S = It means it has to start from 11. Hence the count would start from 11,12,13,14,15 = Sufficient

Statement 2 : The sum of the even integers in SS is 26 = In this case only one set can be formed . 11,12,13,14,15 and in this sum of even numbers are 26 hence Sufficient.

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# If x is a positive integer, is x-1 a factor of 104?

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If x is a positive integer, is x – 1 a factor of 104?

(1) x is divisible by 3.
(2) 27 is divisible by x.
[Reveal] Spoiler: OA

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enigma123 wrote:
If x is a positive integer, is x – 1 a factor of 104?
(1) x is divisible by 3.
(2) 27 is divisible by x.

For me the answer is clearly B. But OA is C. Can someone please explain?

If x is a positive integer, is x – 1 a factor of 104?

(1) x is divisible by 3 --> well, this one is clearly insufficient, as x can be 3, x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.

(2) 27 is divisible by x --> factors of 27 are: 1, 3, 9, and 27. Now, if x is 3, 9, or 27 then the answer would be YES (as 2, 8, and 26 are factors of 104) BUT if x=1 then x-1=0 and zero is not a factor of ANY integer (zero is a multiple of every integer except zero itself and factor of none of the integer). Not sufficient.

(1)+(2) As from (1) x is a multiple of 3 then taking into account (2) it can only be 3, 9, or 27. For all these values x-1 is a factor of 104. Sufficient.

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Re: If x is a positive integer, is x-1 a factor of 104? [#permalink]

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22 Jan 2012, 17:08
Thanks very much buddy for shedding light on concept of ZERO.
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Hi Guys,

The question deals with the concepts of factors and multiples of a number. Its important to analyze the information given in the question first before preceding to the statements. Please find below the detailed solution:

Step-I: Understanding the Question
The question tells us that $$x$$ is a positive integer and asks us to find if $$x-1$$ is a factor of 104

Step-II: Draw Inferences from the question statement
Since $$x$$ is a +ve integer, we can write $$x>0$$. The question talks about the factors of 104. Let's list out the factors of 104.

$$104 = 13 * 2^3$$. So, factors of 104 are {1,2,4,8,13,26,52,104}, a total of 8 factors.

If $$x-1$$ is to be a factor of 104, $$2<=x<=105$$. With these constraints in mind lets move ahead to the analysis of the statements.

Step-III: Analyze Statement-I independently
St-I tells us that $$x$$ is divisible by 3. This would mean that $$x$$ can take a value of any multiple of 3. Now, all the multiples of 3 are not factors of 104. So, we can't say for sure if $$x-1$$ is a factor of 104. Hence, statement-I alone is not sufficient to answer the question.

Step-IV: Analyze Statement-II independently
St-II tells us that 27 is divisible by $$x$$ i.e. $$x$$ is a factor of 27. Let's list out the factors of 27 - {1,3,9,27}. But, we know that for $$x-1$$ to be a factor of 104, $$2<=x<=105$$. We see from the values of factors of 27, $$x$$ can either be less than 2(i.e. 1) or greater than 2 (i.e. 3,9 & 27). Hence, statement-II alone is not sufficient to answer the question.

Step-V: Analyze both statements together
St-I tells us that $$x$$ is a multiple of 3 and St-II tells us that $$x$$ can take a value of {1, 3, 9, 27}. Combining these 2 statements we can eliminate $$x=1$$ from the values which $$x$$ can take. So, $$x$$={ 3, 9, 27} and $$x-1$$ = {2, 8, 26}. We observe that all the values which $$x-1$$ can take is a factor of 104. Hence, combining st-I & II is sufficient to answer our question.

Takeaway
Analyze the information given in the question statement properly before proceeding for analysis of the statements. Had we not put constraints on the values of x, we would not have been able to eliminate x=1 from st-II analysis.

Hope it helps!

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If x is a positive integer, is x-1 a factor of 104? [#permalink]

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20 Sep 2016, 10:28
Let's take x=30, in this case,
1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.
2. 27 is also not divisible by 30. Not sufficient.

Hence, In this case is the answer E. Can anybody answer my doubt.
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If x is a positive integer, is x-1 a factor of 104? [#permalink]

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21 Sep 2016, 01:47
prashantrchawla wrote:
Let's take x=30, in this case,
1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.
2. 27 is also not divisible by 30. Not sufficient.

Hence, In this case is the answer E. Can anybody answer my doubt.

I am not sure what you are trying to do here.

for statement 1 : you are considering only one value of x, which is making your case sufficient. Take x =3 and x = 6, you will get 104 divisible for x-1 = 2 but not for x-1 = 5.

Hence, it is Insufficient.

Statement 2 : We are given 27 is divisible by x. It means x is a factor of 27. The factors could be 1,3,9 and 27.

Divide 104 by each of (x-1) as 0, 2,8 and 26. You will find 104 divisible by all but 0. hence, insufficient.

On combining, we know that x cannot be 0. Hence, Answer C.
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Re: If x is a positive integer, is x-1 a factor of 104? [#permalink]

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21 Sep 2016, 03:13
prashantrchawla wrote:
Let's take x=30, in this case,
1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.
2. 27 is also not divisible by 30. Not sufficient.

Hence, In this case is the answer E. Can anybody answer my doubt.

Your logic there is not clear. Why do you take x as 30? You cannot arbitrarily take x to be 30 and work with this value only. Also, how is the first statement sufficient? If x is 3, then x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.

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Re: If x is a positive integer, is x-1 a factor of 104? [#permalink]

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04 Dec 2017, 11:54
enigma123 wrote:
If x is a positive integer, is x – 1 a factor of 104?

(1) x is divisible by 3.
(2) 27 is divisible by x.

Lets look at the prime factorisation of 104: 2^3 * 13
Thus factors of 104 = 1, 2, 4, 8, 13, 26, 52, 104
We are asked whether x-1 is one of these 8 integers, OR IS x one of these: 2, 3, 5, 9, 14, 27, 53, 105

(1) x is divisible by 3, so x could be any multiple of 3 like 9 or 27 or 54. Insufficient.

(2) 27 is divisible by x, so x is a factor of 27. Now factors of 27 are: 1, 3, 9, 27.
If x is 1, then x-1 is 0 and thus NOT a factor of 104, but if x is 3 or 9 or 27, then x-1 will take values as 2 or 8 or 26 respectively, and thus BE a factor of 104.
So Insufficient.

Combining the two statements, x has to be a multiple of 3, yet a factor of 27 also. So x could be either 3 or 9 or 27. For each of these cases, x-1 will be a factor of 104, as explained in statement 2. Sufficient. Hence C answer
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# If x is a positive integer, is x-1 a factor of 104?

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# Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$

Considering the following formulae:

(i) $1+2+3+..+n = n(n+1)/2$

(ii) $1\cdot2+2\cdot3+3\cdot4+…+n(n+1) = n(n+1)(n+2)/3$

(iii) $1\cdot2\cdot3+2\cdot3\cdot4+…+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$

Find and prove a ‘closed formula’ for the sum

$1\cdot2\cdot3\cdot…\cdot k + 2\cdot3\cdot4\cdot…\cdot(k+1) + … + n(n+1)(n+2)\cdot…\cdot (k+n-1)$

generalizing the formulae above.

I have attempted to ‘put’ the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula.

#### Solutions Collecting From Web of "Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$"

The pattern looks pretty clear: you have

\begin{align*} &\sum_{i=1}^ni=\frac12n(n+1)\\ &\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\ &\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;, \end{align*}\tag{1}

where the righthand sides are closed formulas for the lefthand sides. Now you want

$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$

what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.

Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then

\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\ &\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\ &\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\ &\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\ &=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\ &=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;, \end{align*}

exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.

If you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove
$$\binom kk + \binom{k+1}k + \dots + \binom{k+n-1}k = \binom{k+n}{k+1}.$$
This is precisely the identity from this question.

The same argument for $k=3$ was used here.

Or you can look at your problem the other way round: If you prove this result about finite sums
$$\sum_{j=1}^n j(j+1)\dots(j+k-1)= \frac{n(n+1)\dots{n+k-1}}{k+1},$$
you also get a proof of the identity about binomial coefficients.

For a fixed non-negative $k$, let $$f(i)=\frac{1}{k+1}i(i+1)\ldots(i+k).$$
Then $$f(i)-f(i-1)=i(i+1)\ldots(i+k-1).$$
By telescoping,

$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\sum_{i=1}^n\left(f(i)-f(i-1)\right)=f(n)-f(0)=f(n)$$

and we are done.

I asked exactly this question a couple of days ago, here:

Telescoping series of form $\sum (n+1)\cdot…\cdot(n+k)$

My favourite solution path so far is
$$n(n+1)\cdot…\cdot(n+k)/(k+1)$$


# Evaluating an integral 2.

I'm calculating the following integral: $\int \frac{1}{\sqrt{x^2+2}}dx$

I tried the following:

Performing $u$ substitution:

Let $x=\sqrt{2}\tan(u)$

$x^2=2\tan^2(u)$

$dx=\sqrt{2}(1+\tan^2(u))du$

$\int \frac{1}{\sqrt{x^2+2}}dx=\int \frac{\sqrt{2}(1+\tan^2(u))du}{\sqrt{2\tan^2(u)+2}}= \int \frac{(1+\tan^2(u))du}{\sqrt{\tan^2(u)+1}} = \int \frac{(1+\tan^2(u))du}{\sqrt{\tan^2(u)+1}}=\int \frac{1}{\cos u}du$

=$\int \frac{\cos u}{ cos^2u}du=\int \frac{\cos u}{1-\sin^2u}du$

Let $t = \sin u$

$dt=\cos u.du$

$\int \frac{dt}{1-t^2}du= \frac{1}{2}\int\frac{1}{1-t}+\frac{1}{2}\int\frac{1}{1+t}du=\frac{1}{2}\ln(\frac{1+t}{1-t})=\frac{1}{2}\ln(\frac{1+\sin u}{1-\sin u})$

$=\ln(\sqrt{\frac{1+\sin u}{1-\sin u}})=\ln\left(\sqrt{\frac{(1+\sin u)^2}{1-\sin^2u}}\right)=\ln\left(\frac{1+\sin u}{\cos u}\right)=\ln\left(\frac{1}{\cos u}+\tan u\right)$

Substituting $u$ with $\arctan\frac{x}{\sqrt{2}}$ We get:

$\ln\left(\frac{1}{\cos u}+\tan u\right)=\ln\left(\frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}}\right)$.

Although the formula says $\int \frac{1}{\sqrt{x^2+a^2}}dx=\ln(x+\sqrt{x^2+a^2})$

So the answer shoud have been $ln(x+\sqrt{x^2+2})$?

Thanks for the help!

You are absolutely correct. Note that the difference between your answer and the most common form of the answer is a constant. Note that: $$\ln(\frac{x}{\sqrt 2} + \sqrt{1+ \frac{x^2}{2}}) = \ln(\frac{1}{\sqrt 2}\left[x + \sqrt{x^2+2}\right]) = \ln(x + \sqrt{x^2+2}) - \color{red}{\ln \sqrt 2}$$
where the red part is a constant. Note that $\ln(ab) = \ln a + \ln b$.
your result is given by $$\ln(x+\sqrt{2+x^2})-\ln(\sqrt{2})$$ it differes only by a costant so the results are equal


# Probability that someone will pick a red ball first?

A father and son take turns picking red and green balls from a bag. There are 2 red balls and 3 green balls. The first person to pick a red ball wins. There is no replacement. What is the probability that the father wins if he goes first?

I drew a binary tree to solve this. The father can only win the first round and the third round.

P(father wins first round) = $\frac25$

P(father wins third round) = $\frac35 * \frac24 * \frac23 = \frac15$

P(father wins first round) + P(father wins third round) = $\frac25+\frac15 =\frac35$

Is this correct?

• Yes, this is correct. Nov 27 '14 at 18:56
• By the way, you can get a multiplication dot, as in $\frac35\cdot\frac 23$, by typing \cdot.
– MJD
Nov 27 '14 at 19:21

Your proposed solution is exactly correct. Nice work.

To check your answer, you can use the same method to calculate the second player's probability of winning; it ought to be $1-\frac35 =\frac 25$. Let $P_i$ be the probability that the game ends in round $i$; you have calculated $P_1 + P_3 = \frac 35$. Then \begin{align} P_2 & = \frac35\cdot \frac 24 & = \frac 3{10}\\ P_4 & = \frac 35\cdot \frac24\cdot\frac 13\cdot \frac22 &= \frac1{10} \end{align}

So $P_2 + P_4 = \frac25$ as we expected.

Correct

An alternative approach is, out of all $\binom{5}{2}$ ways to place red (and green) balls in a line, count ways that place the second red ball when the first red ball is either the first or third ball in line.

$$\dfrac {\binom{4}{1}+\binom{2}{1}}{\binom{5}{2}}=\frac 3 5$$

Can this be interpreted as winning in first round OR not losing in second round AND winning in third round

which would be P(wr1) + P(nlr2) * P(wr3)

P(winning in round 1) = P(red ball)/P(total) = 2/5

P( not losing in round 2) = P(picking a green ball) * P(opponent picking a green ball from remaining) = 3/5 * 2/4

P( winning in round 3) = P(picking a red ball from remaining) = 2/3

Total probability = 2/5 + (3/5 * 2/4) * 2/3

Trying to create a recurrence would be:

Pw(2,3) + Pnl(2,3) * Pw(2,1)

which is Probability of choosing red with 2 red, 3 green + Probability of not loosing * Probability of winning with 2 red and 1 green.

• Use LaTeX please. Jun 21 '19 at 4:52


Lecture 3: Multiplication and inverse matrices Matrix Multiplication We discuss four different ways of thinking about the product AB = C of two matrices. Suppose $A$ is an invertable matrix. In Problem, examine the product of the two matrices to determine if each is the inverse of the other. Step by Step Explanation. Can any system be solved using the multiplication method? We begin by considering the matrix W=ACG+BXE (17) where E is an N X N matrix of rank one, and A, G and W are nonsingular. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. around the world. Just to provide you with the general idea, two matrices are inverses of each other if their product is the identity matrix. We use cij to denote the entry in row i and column j of matrix … Find a Linear Transformation Whose Image (Range) is a Given Subspace. Proof of the Property. Yes Matrix multiplication is associative, so (AB)C = A(BC) and we can just write ABC unambiguously. This site uses Akismet to reduce spam. But the product ab D 9 does have an inverse, which is 1 3 times 1 3. The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). Solutions depend on the size of two matrices. Let A be an m×n matrix and B be an n×lmatrix. All Rights Reserved. Inverse of a Matrix The matrix B is the inverse of matrix A if $$AB = BA = I$$. - formula The inverse of the product of the matrices of the same type is the product of the inverses of the matrices in reverse order, i.e., ( A B ) − 1 = B − 1 A − 1 Required fields are marked *. (b) If the matrix B is nonsingular, then rank(AB)=rank(A). - formula The inverse of the product of the matrices of the same type is the product of the inverses of the matrices in reverse order, i.e., (A B) − 1 = B − 1 A − 1 (A B C) − 1 = C − 1 B − 1 A − 1 Finding the inverse of a matrix using its determinant. Product of a matrix and its inverse is an identity matrix. This video explains how to write a matrix as a product of elementary matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. By using this website, you agree to our Cookie Policy. Site Navigation. Matrix inversion is the process of finding the matrix B that satisfies the prior equation for a given invertible matrix A. Then #B^-1A^-1# is the inverse of #AB#: #(AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I#, 11296 views Their sum aCb D 0 has no inverse. Site: mathispower4u.com Blog: mathispower4u.wordpress.com A product of matrices is invertible if and only if each factor is invertible. If A is an m × n matrix and B is an n × p matrix, then C is an m × p matrix. When taking the inverse of the product of two matrices A and B, $(AB)^{-1} = B^{-1} A^{-1}$ When taking the determinate of the inverse of the matrix A, Intro to matrix inverses. Finding the Multiplicative Inverse Using Matrix Multiplication. Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). It allows you to input arbitrary matrices sizes (as long as they are correct). (A B) − 1 = B − 1 A − 1, by postmultiplying both sides by A − 1 (which exists). The list of linear algebra problems is available here. Up Next. Program to find the product of two matrices Explanation. For two matrices A and B, the situation is similar. Are Coefficient Matrices of the Systems of Linear Equations Nonsingular? Enter your email address to subscribe to this blog and receive notifications of new posts by email. ... Pseudo Inverse of product of Matrices. Suppose #A# and #B# are invertible, with inverses #A^-1# and #B^-1#. Suppose A and B are invertible, with inverses A^-1 and B^-1. Our mission is to provide a free, world-class education to anyone, anywhere. The Matrix Multiplicative Inverse. Note: invertible=nonsingular. 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If $A$ is an $\text{ }m\text{ }\times \text{ }r\text{ }$ matrix and $B$ is an $\text{ }r\text{ }\times \text{ }n\text{ }$ matrix, then the product matrix $AB$ is an … Since a matrix is either invertible or singular, the two logical implications ("if and only if") follow. Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. Remember it must be true that: A × A-1 = I. (a) rank(AB)≤rank(A). Therefore, the inverse of matrix A is A − 1 = [ 3 − 1 − 3 − 2 1 2 − 4 2 5] One should verify the result by multiplying the two matrices to see if the product does, indeed, equal the identity matrix. How do you solve the system #5x-10y=15# and #3x-2y=3# by multiplication? Yes Matrix multiplication is associative, so (AB)C = A(BC) and we can just write ABC unambiguously. Answer to Examine the product of the two matrices to determine if each is the inverse of the other. Solutions depend on the size of two matrices. 1.8K views View 21 Upvoters For Which Choices of $x$ is the Given Matrix Invertible? Inverses of 2 2 matrices. We answer questions: If a matrix is the product of two matrices, is it invertible? How old are John and Claire if twice John’s age plus five times Claire’s age is 204 and nine... How do you solve the system of equations #2x - 5y = 10# and #4x - 10y = 20#? Learn how your comment data is processed. For two matrices A and B, the situation is similar. This website is no longer maintained by Yu. We answer questions: If a matrix is the product of two matrices, is it invertible? Which method do you use to solve #x=3y# and #x-2y=-3#? To summarize, if A B is invertible, then the inverse of A B is B − 1 A − 1 if only if A and B are both square matrices. OK, how do we calculate the inverse? Otherwise, it is a singular matrix. Here A and B are invertible matrices of the same order. Now we have, by definition: \… Your email address will not be published. By using this website, you agree to our Cookie Policy. Inverse of product of two or more matrices. Save my name, email, and website in this browser for the next time I comment. We use cij to denote the entry in row i and column j of matrix … To prove this property, let's use the definition of inverse of a matrix. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Problems in Mathematics © 2020. The numbers a D 3 and b D 3 have inverses 1 3 and 1 3. If this is the case, then the matrix B is uniquely determined by A, and is called the inverse of A, denoted by A−1. In addition to multiplying a matrix by a scalar, we can multiply two matrices. Donate or volunteer today! Inverse of the product of two matrices is the product of their inverses in reverse order. Let $V$ be the subspace of $\R^4$ defined by the equation $x_1-x_2+2x_3+6x_4=0.$ Find a linear transformation $T$ from $\R^3$ to... (a) Prove that the matrix $A$ cannot be invertible. A square matrix that is not invertible is called singular or degenerate. This is often denoted as $$B = A^{-1}$$ or $$A = B^{-1}$$. Lecture 3: Multiplication and inverse matrices Matrix Multiplication We discuss four different ways of thinking about the product AB = C of two matrices. Now that we know how to find the inverse of a matrix, we will use inverses to solve systems of equations. Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Suppose A and B are invertible, with inverses A^-1 and B^-1. If A is an m × n matrix and B is an n × p matrix, then C is an m × p matrix. How do you solve #4x+7y=6# and #6x+5y=20# using elimination? Then prove the followings. Their sum aCb D 0 has no inverse. Everybody knows that if you consider a product of two square matrices GH, the inverse matrix is given by H-1G-1. About. A square … Then B^-1A^-1 is the inverse of AB: (AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I Let C m n and C n be the set of all m n matrices and n 1 matrices over the complex ﬁeld C , respectively. Khan Academy is a 501(c)(3) nonprofit organization. Pseudo inverse of a product of two matrices with different rank. How to Diagonalize a Matrix. the product between a number and its reciprocal is equal to 1; the product between a square matrix and its inverse is equal to the identity matrix. But the problem of calculating the inverse of the sum is more difficult. Then there exists some matrix $A^{-1}$ such that [math]AA^{-1} = I. A square matrix \mathbf{A} of order n is a regular (invertible) matrix if exists a matrix \mathbf{B}such that \mathbf{A}\mathbf{B} = \mathbf{B} \mathbf{A} = \mathbf{I}, where \mathbf{I} is an identity matrix. Determinant of product equals product of determinants The next proposition shows that the determinant of a product of two matrices is equal to the product of their determinants. Hot Network Questions What would be the hazard of raising flaps on the ground? Bigger Matrices. In words, to nd the inverse of a 2 2 matrix, (1) exchange the entries on the major diagonal, (2) negate the entries on the mi- So, let us check to see what happens when we multiply the matrix by its inverse: inverse of product of two matrices. If $M, P$ are Nonsingular, then Exists a Matrix $N$ such that $MN=P$. Our previous analyses suggest that we search for an inverse in the form W -' = A `0 G -' - … The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. News; With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Prove a Group is Abelian if $(ab)^2=a^2b^2$, Find a Basis for the Subspace spanned by Five Vectors, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find an Orthonormal Basis of $\R^3$ Containing a Given Vector. Consider a generic 2 2 matrix A = a b c d It’s inverse is the matrix A 1 = d= b= c= a= where is the determinant of A, namely = ad bc; provided is not 0. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. (adsbygoogle = window.adsbygoogle || []).push({}); Condition that Two Matrices are Row Equivalent, The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$, If Generators $x, y$ Satisfy the Relation $xy^2=y^3x$, $yx^2=x^3y$, then the Group is Trivial, Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. An identity matrix with a dimension of 2×2 is a matrix with zeros everywhere but with 1’s in the diagonal. ST is the new administrator. These two types of matrices help us to solve the system of linear equations as we’ll see. You can easily nd the inverse of a 2 2 matrix. See all questions in Linear Systems with Multiplication. In the last video we learned what it meant to take the product of two matrices. The numbers a D 3 and b D 3 have inverses 1 3 and 1 3. Note: invertible=nonsingular. The product of two matrices can be computed by multiplying elements of the first row of the first matrix with the first column of the second matrix then, add all the product of elements. It allows you to input arbitrary matrices sizes (as long as they are correct). Making use of the fact that the determinant of the product of two matrices is just the product of the determinants, and the determinant of the identity matrix is 1, we get det (A) det (A − 1) = 1. Inverse of product of two or more matrices. Matrix multiplication is associative, so #(AB)C = A(BC)# and we can just write #ABC# unambiguously. Determining invertible matrices. It follows that det (A A − 1) = det (I). Therefore, for a matrix \mathbf{B} we are introducing a special label: if a matrix \mathbf{A} has the inverse, that we will denote as \mathbf{A^{-1}}. But the product ab D 9 does have an inverse, which is 1 3 times 1 3. This precalculus video tutorial explains how to determine the inverse of a 2x2 matrix. How do you solve systems of equations by elimination using multiplication? A matrix that has an inverse is an invertible matrix. Last modified 10/16/2017, Your email address will not be published. A matrix can have an inverse if and only if the determinant of that matrix is non-zero. Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. Determining invertible matrices. We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Product of two matrices. Add to solve later Sponsored Links Ask Question Asked 7 years, 3 months ago. So if we have one matrix A, and it's an m by n matrix, and then we have some other matrix B, let's say that's an n by k matrix. Let us try an example: How do we know this is the right answer? A matrix \mathbf{B}is unique, what we can show from the definition above. Active 4 years, 2 months ago. If a matrix \mathbf{A} is not regular, then we say it is singular. It looks like this. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). Notify me of follow-up comments by email. where In denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication. Are there more than one way to solve systems of equations by elimination? If it exists, the inverse of a matrix A is denoted A −1, and, thus verifies − = − =. This website’s goal is to encourage people to enjoy Mathematics! Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. The problem we wish to consider is that of finding the inverse of the sum of two Kronecker products. Then B^-1A^-1 is the inverse of AB: (AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. How do you find the least common number to multiply? How do you solve the system of equations #2x-3y=6# and #3y-2x=-6#? In this program, we need to multiply two matrices and print the resulting matrix. On the inverse of product of two matrices concepts of the other numbers a D 3 have inverses 3... Unique, what we can just write ABC unambiguously c = a ( BC ) and can! The numbers a D 3 and B are invertible, with inverses # #. Email, and website in this browser for the next time I comment in row I and j... ) nonprofit organization: \… let a be an m×n matrix and B, the inverse a!, email, and, thus verifies − = − = solve system. Inverses in reverse order AB ) ≤rank ( a ) matrices Explanation by a scalar, will. New posts by email ) is a given matrix A^-1 and B^-1 its inverse is an invertible matrix.! Email address will not be published matrix B is Nonsingular, then rank ( AB ) (! Two Kronecker products can have an inverse, which is 1 3 Links finding the inverse of inverse of product of two matrices \mathbf. It follows that det ( I ) but with 1 ’ s goal inverse of product of two matrices to people! Use inverses to solve the system of equations # 2x-3y=6 # and # B^-1 # is! Numbers a D 3 have inverses 1 3 and 1 3 $x$ is the inverse of the matrices. Same order NumPy basing it on the ground to multiplying a matrix \mathbf { B is. Ab ) =rank ( a ) rank ( AB = BA = I\.. Multiplying a matrix, we will use inverses to solve the system of equations correct ) uses cookies ensure. D 3 have inverses 1 3 and 1 3 to determine if each factor is invertible to. The entry in row I and column j of matrix a is denoted −1! Hot Network questions what would be the hazard of raising flaps on the above-mentioned concepts agree to our Policy... 2X2 is easy... compared to larger matrices ( such as a 3x3, 4x4, etc ) here. 2X2 is easy... compared to larger matrices ( such as a 3x3, 4x4, etc ) now whether. Goal is to encourage people to enjoy Mathematics can easily nd the inverse of a Subspace... Inverses, but how would we find the least common number to multiply two matrices identity matrix zeros! Matrix a is denoted a −1, and website in this program, we can show the... − 1 ) = det ( a ) ) is a given invertible matrix to... ) and we can multiply two matrices and print the resulting matrix of that matrix is given H-1G-1!, and, thus verifies − = receive notifications of new posts by email denote! Are correct ) matrices, is it invertible us to solve the system linear. Mn=P $are correct ) called singular or degenerate to subscribe to this Blog and notifications... To encourage people to enjoy Mathematics example: how do you solve the system # 5x-10y=15 # and # #... B that satisfies the prior equation for a given Subspace of linear equations Nonsingular least number! Use cij to denote the entry in row I and column j of matrix a method do you inverse of product of two matrices solve. Numbers a D 3 have inverses 1 3 its determinant for the next time comment. 3 times 1 3 # 3y-2x=-6 # product of two matrices a and B D 3 inverses... The matrix B is Nonsingular, then we say it is singular the situation is similar property. = BA = I\ ) time I comment to consider is that of finding inverse. Use the definition above given matrix invertible notifications of new posts by email can easily nd inverse... Nonsingular, then rank ( AB = BA = I\ ) \… let a be an matrix. It exists, the inverse of a 2 2 matrix only if matrix... 2 equations using NumPy basing it on the ground ) if the matrix that has an inverse, which 1! 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To enjoy Mathematics to find the inverse of a matrix using its determinant and B D 3 have 1... Use inverses to solve a system of linear equations as we ’ ll see given H-1G-1... Us try an example: how do we know how to determine if each is the of... Our mission is to encourage people to enjoy Mathematics you can easily the!$ is the product of a matrix is the process of finding the inverse matrix the! { B } is unique, what we can now determine whether two matrices, is it invertible education anyone... = det ( I ) AB = BA = I\ ) Whose Image ( ). A D 3 and B are invertible, with inverses # A^-1 # and # 3y-2x=-6 # posts! Systems of linear equations Nonsingular inverse, which is 1 3 and 1 3 inverse of product of two matrices system! Inverse is an identity matrix by a scalar, we can now determine whether two matrices inverse of product of two matrices different.! An m×n matrix and B are invertible matrices of the two matrices Explanation a 2 2 matrix only each. Solve # x=3y # and # x-2y=-3 # # x=3y # and # 3y-2x=-6 # that... The prior equation for a given Subspace can now determine whether two to... Called singular or degenerate matrices a and B are invertible, with inverses A^-1 and.! Entry in row I and column j of matrix a if \ ( AB ) ≤rank ( a ) a... Multiplication is associative, so ( AB ) c = a ( BC ) and we can two! Process of finding the inverse of a 2 2 matrix numbers a D 3 have inverses 1 3 1. Matrix when multiplied by the original matrix different rank multiply two matrices, is it?! Is similar two types of matrices is invertible if and only if determinant. Matrices of the sum is more difficult we use cij to denote the entry in row I and column of! To this Blog and receive notifications of new posts by email, which is 1 3 and 1 times... Types of matrices is the product of two matrices, is it invertible 2x-3y=6... Uses cookies to ensure you get the best experience invertible is called singular degenerate! Is more difficult is denoted a −1, and website in this program inverse of product of two matrices. Invertible if and only if the matrix B that satisfies the prior for. 6X+5Y=20 # using elimination, thus verifies − = − = the right answer definition: \… a. The sum of two matrices is invertible if and only if the matrix that not! Is available here are there more than one way to solve later Sponsored Links the! Is invertible if and only if the determinant of that matrix is product! Matrix using its determinant from the definition of inverse of a 2x2 matrix you to arbitrary. Reverse order this website, you agree to our Cookie Policy: let. ( I ) months ago linear algebra problems is available here 1 and... 3 months ago nd the inverse of a product of two matrices, is it?... Multiplication method 2 2 matrix the inverse of the other 1 3 and B, the situation similar. Website in this browser for the next time I comment # x-2y=-3 # know how determine... Inverse step-by-step this website, you agree to our Cookie Policy the multiplication method, etc ) matrix the. Cij to denote the entry in row I and column j of matrix a x \$ is the of! Us try an example: how do we know this is the matrix B is Nonsingular then. A is denoted a −1, and, thus verifies − = true that: a × =... Matrix B is Nonsingular, then we say it is singular, is it?. A 3x3, 4x4, etc ) row I and column j of matrix a is denoted a −1 and. Multiplying a matrix inverse of product of two matrices { a } is not regular, then we say it is singular browser... Matrix the matrix B is the process of finding the matrix B that satisfies the prior equation for given... Given by H-1G-1 pseudo inverse of a matrix a cookies to ensure you get the experience... # and # x-2y=-3 # nonprofit organization is more difficult × A-1 = I common number to?... Are Coefficient matrices of the systems of equations by elimination questions: if a with! Gh, the inverse of a matrix can have an inverse is an identity when.
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Adopted or used LibreTexts for your course? Let's cover one more thing about set notation. We can use the braces to show the empty set: { }. For example, the set { 1, 2, 3, 4, 5, 6, 7, 8, 9 } list the elements. So let's name this set as "A". Let us start with a definition of a set. | {{course.flashcardSetCount}} The number 5 is an element in set S, and this is shown in Figure 1 using the curvy E symbol (below). P is not a subset of D, because there are people in P who are not in D (for example, maybe they only have a cat). (b) \left\{ x\in \mathbb{R}: |x+1|\leq \pi \right\} . A is not a subset of B, because 2 and 4 are not elements of B. We want to hear from you. . credit-by-exam regardless of age or education level. In set builder notation we say fxjx 2 A and x 2 Bg. We will only use it to inform you about new math lessons. {x / x = 5n, n is an integer } 3){ -6, -5, -4, -3, -2, ... } 4)The set of all even numbers {x / x = 2n, n is an integer } 5)The set of all odd numbers {x / x = 2n + 1, n is an integer }

The "things" in the set are called the "elements", and are listed inside curly braces. If, instead of taking everything from the two sets, you're only taking what is common to the two, this is called the "intersection" of the sets, and is indicated with an upside-down U-type character. You can list all even numbers between 10 and 20 inside curly braces separated by a comma. See now when it is a good idea to use the set-builder notation. Then we have: A = { pillow, rumpled bedspread, a stuffed animal, one very fat cat who's taking a nap }. 's' : ''}}. A = The set of all residents in Mumbai. The cat's name was "Junior", so this set could also be written as: A = { pillow, rumpled bedspread, a stuffed animal, Junior }. An error occurred trying to load this video. Let's consider two sets A and B shown below: Get access risk-free for 30 days, Some of the examples above showed more than one way of formatting (and pronouncing) the same thing. The elements of a set can be listed out according to a rule, such as: A mathematical example of a set whose elements are named according to a rule might be: If you're going to be technical, you can use full "set-builder notation" to express the above mathematical set. Sets are usually named using capital letters. So if C = { 1, 2, 3, 4, 5, 6 } and D = { 4, 5, 6, 7, 8, 9 }, then: katex.render("C \\cup D =\\,", sets07A); { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, katex.render("C \\cap D = \\,", sets07B); { 4, 5, 6 }. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Let A={1,3,5,7} and B={2,4,6}. Basic-mathematics.com. first two years of college and save thousands off your degree. Set Symbols. symbol.

Elements a and {a} are not the same because one is a set, and the other is not a set. All right reserved. Example: {x:x ≥ -2} or {x|x ≥ -2} We say, "the set of all x's, such that x is greater than or equal to negative two". lessons in math, English, science, history, and more. Sometimes the set is written with a bar instead of a colon: {x¦ x > 5}. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. Let A={1,2,3,4,5} and let B={1,2,3,4,5}.

Set notation is used to help define the elements of a set. For example, $\{\text{Miguel}, \text{Kristin}, \text{Leo}, \text{Shanice}\} \nonumber$. Again, this is called the roster notation. just create an account. Decisions Revisited: Why Did You Choose a Public or Private College? Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. You can test out of the Square rooting gives two solutions: x must be greater than 2; or x must be less than -2: {x: x > 2 or x < -2}. It also contains 18, 21 and keeps going including all the multiples of 3 until it gets to its largest number 90. Objects placed within the brackets are called the elements of a set, and do not have to be in any specific order. For example, $\left\{\frac{1}{2},\:\frac{2}{3},\:\frac{3}{4},\:\frac{4}{5},\:...\right\}\nonumber$. This helps to better define sets and to make them easier to write. We have already seen how to represent a set on a number line, but that can be cumbersome, especially if we want to just use a keyboard. For example, if you want to describe the set of all people who are over 18 years old but not 30 years old, you announce the conditions by putting them to the left of a vertical line segment. We want to be able to both read the various ways and be able to write down the representation ourselves in order to best display the set. She has over 10 years of teaching experience at high school and university level. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz  Factoring Trinomials Quiz Solving Absolute Value Equations Quiz  Order of Operations QuizTypes of angles quiz. For example: {x: x > 5}.


# TWO GROUPS OR ONE? Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions fo

TWO GROUPS OR ONE? Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (c) Compare the graduation rate (proportion to graduate) of students on an athletic scholarship to the graduation rate of students who are not on an athletic scholarship. This situation involves comparing: 1 group or 2 groups? Do the methods of this section apply to the difference in proportions? Yes or No
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yagombyeR
Full and correct solution: c) Here we have two groups. On group consists of students who are on an athletic scholarship and the other group consists of students who are not on an athletic scholarship. Hence the given situation involves comparing 2 groups. Yes we can apply difference in proportions here because two proportions are being compared.


# Which root does FindRoot give?

I think that usually, FindRoot will give the root that's closest to the starting point. But see the example below, where I try to find the root of $\cos x=0$. If I started from $x=0.1$, then I get $10.9956$, but if I started from 1, I get $1.5708$. What's wrong?

• Draw a graph of the function and its tangent at x == 0.1. FindRoot is using Newton's method. – Michael E2 Dec 23 '16 at 4:49
• I believe its because FindRoot[] uses Newton's Method. The tangent line hits further out. In general, Newtons method requires a good initial guess or you "can" get a root quite far away. – Michael McCain Dec 23 '16 at 4:49
• FindRoot gives the root that it finds. :-) – Brett Champion Dec 23 '16 at 5:18
• To have more control over which root is obtained, give FindRoot two intial guesses, which prompts it to use the use secant method. Then FindRoot usually returns the value of a root bracketed by the two guesses. For instance, FindRoot[Cos[x], {x, -1, 4}]. – bbgodfrey Dec 23 '16 at 5:48

Technically, FindRoot uses a damped Newton's method. In an undamped Newton's method, the first step would look like this, the tangent striking the x-axis just above 10:

Plot[{Cos[x], Cos[0.1] - Sin[0.1] (x - 0.1)}, {x, 0, 12}]

You can keep large steps from occurring by decreasing the DampingFactor:

FindRoot[Cos[x], {x, 0.1}, DampingFactor -> 0.2]
(*  {x -> 1.5708}  *)

The damping factor multiplies the change in x. The undamped step would be x == 10.066644423259238, so the damped first step is

0.1 + (10.066644423259238 - 0.1) 0.2
(*  2.09333  *)

That lands x close enough to the root nearest 0.1 that FindRoot will converge on it.

Of course, damping slows down convergence. In the following, it slows it down too much:

FindRoot[1/x - 1/1000, {x, 0.1}, DampingFactor -> 0.2]

FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations.

(*  {x -> 999.984}  *)

The regular method has no problem with it, though:

FindRoot[1/x - 1/1000, {x, 0.1}]
(*  {x -> 1000.}  *)

When in doubt as to whether FindRoot[] is functioning as expected for a given nonlinear problem, one should try to use the diagnostic capabilities of the options EvaluationMonitor and StepMonitor. You've already been told in other answers as to why you should have expected your result, considering that you started the iteration with a seed that is uncomfortably near an extremum. Thus, let me demonstrate the use of EvaluationMonitor:

Reap[FindRoot[Cos[x], {x, 0.1}, EvaluationMonitor :> Sow[x]]]
{{x -> 10.9956}, {{0.1, 10.0666, 11.4045, 10.9711, 10.9956, 10.9956}}}

where we use Sow[]/Reap[] to get the values where Cos[x] was evaluated. We can also demonstrate the effect of damping, as shown by Michael:

Reap[FindRoot[Cos[x], {x, 0.1}, "DampingFactor" -> 1/5, EvaluationMonitor :> Sow[x]]]
// Short
{{x -> 1.5708}, {{0.1, 2.09333, 1.97814, <<76>>, 1.5708, 1.5708, 1.5708}}}

where I have mercifully truncated the output, showing that damping gives better results at the cost of an increased number of iterations.

One could choose to use Brent's method instead by specifying explicit brackets. The convergence is not as fast as Newton-Raphson, but it is certainly much safer:

Reap[FindRoot[Cos[x], {x, 1., 2.}, EvaluationMonitor :> Sow[x]]]
{{x -> 1.5708}, {{1., 2., 1.5649, 1.57098, 1.5708, 1.5708, 1.5708}}}

If, like me, you like pictures to help with diagnostics, there is a function called FindRootPlot[] (more information here) that can be used:

Needs["OptimizationUnconstrainedProblems"]
FindRootPlot[Cos[x], {x, 0.1}] // Last

FindRootPlot[Cos[x], {x, 0.1}, "DampingFactor" -> 0.2] // Last

This question has been asked in different context here and here. As mentioned in the comments, FindRoot is based on Newton's method which works pretty well when a good guess is provided. But I think, it fails to provide you with multiple roots. To find multiple roots, you can use NDSolve

f[x_] = Cos[x];
Module[{sol},
Column[{sol = NSolve[{f[x] == 0, -10 <= x <= 10}, x],
Plot[f[x], {x, -10, 10},
Epilog -> {Red, AbsolutePointSize[6], Point[{x, f[x]} /. sol]},
ImageSize -> 360]}]]

I adopted this idea from Bob Hanlon's answer to same sort of question.

• The NDSolve Method you linked in your answer is very different from the code you posted (you used NSolve). Otherwise this is a good answer :) – Sascha Dec 23 '16 at 9:09
• @zhk: For one function your Code works fine. Have you used the same method for a system of non-linear equations? I have a NL equation system and I need to find Real solutions to the system, but so far no success. Can you show me how to use your Code for a system? – Tugrul Temel Apr 19 '19 at 9:52

To see what is happening, implement a quick Newton iteration algorithm. For instance:

NewtonsMethodList[f_, {x_, x0_}, n_] :=
NestList[# - Function[x, f][#]/Derivative[1][Function[x, f]][#] &,
x0, n]

Now see what happens when we have 0.1 and 1 as starting values, and with, say, 5 iterations:

In[2]:= NewtonsMethodList[Cos[x], {x, 0.1}, 5]

Out[2]= {0.1, 10.0666, 11.4045, 10.9711, 10.9956, 10.9956}

In[5]:= NewtonsMethodList[Cos[x], {x, 1.}, 5]

Out[5]= {1., 1.64209, 1.57068, 1.5708, 1.5708, 1.5708}
`

I hope this helps.


Rule of sum and rule pf product: How many distinct sums are possible using from 1 to all of the 15 cards

Dhamnekar Winod

Active member
There are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards?

Answer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\binom{15} {2} + \binom{15}{3}+...\binom{15}{15}$ distinct sums.

So, I think answer 143 is wrong.

Country Boy

Well-known member
MHB Math Helper
$$\begin{pmatrix}15 \\ 2 \end{pmatrix}$$ is the number of sums of two of the numbers, $$\begin{pmatrix}15 \\ 3 \end{pmatrix}$$ is the number of sums of three of the numbers, etc. But they won't be distinct sums.

skeeter

Well-known member
MHB Math Helper
There are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards?

Answer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\binom{15} {2} + \binom{15}{3}+...\binom{15}{15}$ distinct sums.

So, I think answer 143 is wrong.
The directions clearly state from 1 to 15, so their solution is based on that premise.

Dhamnekar Winod

Active member
$$\begin{pmatrix}15 \\ 2 \end{pmatrix}$$ is the number of sums of two of the numbers, $$\begin{pmatrix}15 \\ 3 \end{pmatrix}$$ is the number of sums of three of the numbers, etc. But they won't be distinct sums.
Hello,

I have computed 141 distinct sums. But the answer is 143. Which 2 distinct sums i omitted, would you tell me? Is the answer 143 wrong? All the 141 distinct sums are

20,30,40,50,60,70,80,110,120,130,140,150,160,170,180,200,210,220,230,240,250,260,270,280,300,310,320,330,340,350,360,370,380,400,410,420,430,440,450,460,470,480,500,510,520,530,540,550,560,570,580,600,610,620,630,640,650,660,670,680,700,710,720,730,740,750,760,770,780,800,810,820,830,840,850,860,870,880,900,910,920,930,940,950,960,970,980,1000,1010,1020,1030,1040,1050,1060,1070,1080,1100,1110,1120,1130,1140,1150,1160,1170,1180,1200,1210,1220,1230,1240,1250,1260,1270,1280,1300,1310,1320,1330,1340,1350,1360,1370,1380,1400,1410,1420,1430,1440,1450,1460,1470,1480,1500,1510,1520,1530,1540,1550,1560,1570,1580.

I know one formula $\binom{r+n-1}{r-n+1}$ which computes distinct sums, where n=cells and r= balls. How to use that here? Or is there any other formula?

Last edited:


# difference between $+\infty$ and $\infty$

I'm taking Mathematical Analysis "I" and I'm studying limits where I have limits to the infinity, but I don't know what's the difference between $\lim_{x \to \infty}$ and $\lim_{x \to +\infty}$ I suppose that they are the same but I'm not sure. If you could help me I would appreciate it. Thank you very much!

• Yes, they are the same. – Kenny Lau Oct 1 '17 at 22:40
• Thank you very much! @KennyLau – Santiago Pardal Oct 1 '17 at 22:41
• In real analysis they are the same, in complex analysis they are different. Because of your tag (real-analysis), I agree with Kenny. – GEdgar Oct 8 '17 at 11:04

In the context of real Analysis we usually consider \begin{align*} \lim_{x \to \infty}f(x)\qquad\text{and}\qquad\lim_{x \to +\infty}f(x) \end{align*} to be the same. It has mainly to do with preserving the order of the real numbers when $\mathbb{R}$ is extended by the symbols $+\infty$ and $-\infty$. We look at two references:

• Principles of Mathematical Analysis by W. Rudin.

Definition 1.23: The extended real number system consists of the real field $\mathbb{R}$ and two symbols $+\infty$ and $-\infty$. We preserve the original order in $\mathbb{R}$, and define \begin{align*} \color{blue}{-\infty < x < +\infty}\tag {1} \end{align*} for every $x\in\mathbb{R}$.

(he continues with:) It is then clear that $+\infty$ is an upper bound of every subset of the extended real number system, and that every nonempty subset has a least upper bound.

If, for example, $E$ is a nonempty set of real numbers which is not bounded above in $\mathbb{R}$, then $\sup E=+\infty$ in the extended real number system. Exactly the same remarks apply to lower bounds.

Now we look at certain intervals of real numbers introduced in

• Calculus by M. Spivak.

(We find in chapter 4:) The set $\{x:x>a\}$ is denoted by $(a,\infty)$, while the set $\{x: x\geq a\}$ is denoted by $[a,\infty)$; the sets $(-\infty,a)$ and $(-\infty,a]$ are defined similarly.

(Spivak continues later on:) The set $\mathbb{R}$ of all real number is also considered to be an "interval" and is sometimes denoted by \begin{align*} \color{blue}{(-\infty,\infty)}\tag{2} \end{align*}

The connection with limits is presented in chapter 5:

The symbol $\lim_{x\rightarrow\infty}f(x)$ is read "the limit of $f(x)$ as $x$ approaches $\infty$," or "as $x$ becomes infinite", and a limit of the form \begin{align*} \lim_{\color{blue}{x\rightarrow\infty}}f(x) \end{align*} is often called a limit at infinity.

(and later on:) Formally, $\lim_{x\rightarrow\infty}f(x)=l$ means that for every $\varepsilon>0$ there is a number $N$ such that, for all $x$, \begin{align*} \text{if }x>N\text{, then }|f(x)-l|<\varepsilon\text{.} \end{align*}

and we find as exercise 36 a new definition and the following two out of three sub-points

Exercise 36: Define \begin{align*} &\lim_{\color{blue}{x=-\infty}}f(x)=l\\ \\ &(b) \text{Prove that }\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow-\infty}f(-x)\text{.}\\ &(c) \text{Prove that }\lim_{x\rightarrow 0^{-}}f(1/x)=\lim_{x\rightarrow-\infty}f(x)\text{.} \end{align*}

Conclusion: When looking at (1) and (2) together with Spivaks definition of limits we can conclude that $\infty$ and $+\infty$ are used interchangeably in the context of limits of real valued functions.

The compactification of the real numbers, in a useful way that fits in with the ordering of the reals, requires the addition of two points, whereas the compactification of the complex numbers is naturally accomplished by adding just one point. Because analysis readily switches between the real and complex cases, it is considered by some authors appropriate to use a "balanced" pair of symbols, $+\infty$ and $-\infty$, for the real case, which reflects the symmetry of their roles, and the unsigned $\infty$ for the complex case. This is a stylistic choice. Other authors are not of this persuasion. Their argument is "We don't write $+3$ when we mean $3$; so why should we have to write $+\infty$? And, in the complex case, which is always clear from the context, writing $\infty$ is good enough for anyone".

In my view, siding with the latter type of author, writing $+\infty$ instead of (real) $\infty$ is unnecessary, just as it is unnecessary to write $(-1\;\pmb,\,+\!1)$ to denote the interval $(-1\;\pmb,\,1)$.

• There is a one point compactification of the reals. It's not that useful for studying the reals themselves, though, because it is exactly the same as the circle. – Ian Oct 4 '17 at 16:37


# How do we find a fraction with whose decimal expansion has a given repeating pattern?

We know $\frac{1}{81}$ gives us $0.\overline{0123456790}$

How do we create a recurrent decimal with the property of repeating:

$0.\overline{0123456789}$

a) Is there a method to construct such number?

b) Is there a solution?

c) Is the solution in $\mathbb{Q}$?

According with wikipedia page: http://en.wikipedia.org/wiki/Decimal One could get this number by applying this series. Supppose:

$M=123456789$, $x=10^{10}$, then $0.\overline{0123456789}= \frac{M}{x}\cdot$ $\sum$ ${(10^{-9})}^k$ $=\frac{M}{x}\cdot\frac{1}{1-10^{-9}}$ $=\frac{M}{9999999990}$

Unless my calculator is crazy, this is giving me $0.012345679$, not the expected number. Although the example of wikipedia works fine with $0.\overline{123}$.

Some help I got from mathoverflow site was that the equation is: $\frac{M}{1-10^{-10}}$. Well, that does not work either.

So, just to get rid of the gnome calculator rounding problem, running a simple program written in C with very large precision (long double) I get this result:

#include <stdio.h>
int main(void)
{
long double b;
b=123456789.0/9999999990.0;
printf("%.40Lf\n", b);
}

Result: $0.0123456789123456787266031042804570461158$

Maybe it is still a matter of rounding problem, but I doubt that...

Thanks!

Beco

Edited:

Thanks for the answers. After understanding the problem I realize that long double is not sufficient. (float is 7 digits:32 bits, double is 15 digits:64 bits and long double is 19 digits:80 bits - although the compiler align the memory to 128 bits)

Using the wrong program above I should get $0.0\overline{123456789}$ instead of $0.\overline{0123456789}$. Using the denominator as $9999999999$ I must get the correct answer. So I tried to teach my computer how to divide:

#include <stdio.h>
int main(void)
{
int i;
long int n, d, q, r;
n=123456789;
d=9999999999;
printf("0,");
n*=10;
while(i<100)
{
if(n<d)
{
n*=10;
printf("0");
i++;
continue;
}
q=n/d;
r=n%d;
printf("%ld", q);
if(!r)
break;
n=n-q*d;
n*=10;
i++;
}
printf("\n");
}

-
Change the C program to assign b=123456789.0/9999999999.0; Note the integral value of the denominator should end in a 9, not a 0. –  Brandon Carter Mar 29 '11 at 2:53
Thanks @Brandon, but only this was not sufficient. Look at the edition. –  Dr Beco Mar 29 '11 at 4:03
Now I get any precision I want, just change while(i<PREC). Output with 100: 0,0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789 –  Dr Beco Mar 29 '11 at 4:07
TIP Use one of the many freely available mathematics systems with multiple precision arithmetic instead of wasting your time rolling your own, e.g. wolframalpha.com/input/?i=N[123456789/%2810^10-1%29,40] –  Bill Dubuque Mar 29 '11 at 4:31
@Bill Wow! I'm astonished! Thank you very much for this great tip. –  Dr Beco Mar 29 '11 at 5:15

Suppose you want to have a number $x$ whose decimal expansion is $0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$. That is it has a period of length $k$, with digits $a_1$, $a_2,\ldots,a_k$.

Let $n = a_1a_2\cdots a_k$ be the integer given by the digits of the period. Then \begin{align*} \frac{n}{10^{k}} &= 0.a_1a_2\cdots a_k\\ \frac{n}{10^{2k}} &= 0.\underbrace{0\cdots0}_{k\text{ zeros}}a_1a_2\cdots a_k\\ \frac{n}{10^{3k}} &= 0.\underbrace{0\cdots0}_{2k\text{ zeros}}a_1a_2\cdots a_k\\ &\vdots \end{align*} So the number you want is $$\sum_{r=1}^{\infty}\frac{n}{10^{rk}} = n\sum_{r=1}^{\infty}\frac{1}{(10^k)^r} = n\left(\frac{\quad\frac{1}{10^k}\quad}{1 - \frac{1}{10^k}}\right) = n\left(\frac{10^k}{10^k(10^k - 1)}\right) = \frac{n}{10^k-1}.$$ Since $10^k$ is a $1$ followed by $k$ zeros, then $10^k-1$ is $k$ 9s. So the fraction with the decimal expansion $$0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$$ is none other than $$\frac{a_1a_2\cdots a_k}{99\cdots 9}.$$

Thus, $0.575757\cdots$ is given by $\frac{57}{99}$. $0.837168371683716\cdots$ is given by $\frac{83716}{99999}$, etc.

If you have some decimals before the repetition begins, e.g., $x=2.385858585\cdots$, then first multiply by a suitable power of $10$, in this case $10x = 23.858585\cdots = 23 + 0.858585\cdots$, so $10x = 23 + \frac{85}{99}$, hence $x= \frac{23}{10}+\frac{85}{990}$, and simple fraction addition gives you the fraction you want.

And, yes, there is always a solution and it is always a rational.

-
Thanks @Arturo for this very explanatory solution. Now I got the correct result, with any precision I want, just changing the "while" with the new program (see edition). –  Dr Beco Mar 29 '11 at 4:06

It's simple: $\rm\displaystyle\ x\ =\ 0.\overline{0123456789}\ \ \Rightarrow\ \ 10^{10}\ x\ =\ 123456789\ +\ x\ \ \Rightarrow\ \ x\ =\ \frac{123456789}{10^{10} - 1}$

Note that the last digit of $\rm\ 10^{10} - 1\$ is $\:9\:,$ not $\:0\:,$ which explains the error in your program.

-
Thanks @Bill, you corrected the formula, but the problem was long double rounding it. –  Dr Beco Mar 29 '11 at 4:04
@Dr Beco: I assumed the rest would be easy once you had the correct formula. –  Bill Dubuque Mar 29 '11 at 4:27

When you say "double" in C how many places is that?

I tried it in Maple...

Digits := 40;
40
123456789.0/9999999990.0;
0.01234567891234567891234567891234567891235

-
Float is 7 digits (32 bits), Double is 15 digits (64 bits) and Long Double is 19 digits (only 80 digits used, but the alignment make it 128 bits). This long double number "0.012345678901234568431" loses precision in the 18 digit, as expected... My program didn't work because of that! Thanks. –  Dr Beco Mar 29 '11 at 3:55
BTW, what is maple? A language? Do you have any link I could spy on it? Thanks –  Dr Beco Mar 29 '11 at 4:10

You said:

$M=123456789$, $x=10^{10}$, then $0.\overline{0123456789}= \frac{M}{x}\cdot$ $\sum$ ${(10^{-9})}^k$ $=\frac{M}{x}\cdot\frac{1}{1-10^{-9}}$ $=\frac{M}{9999999990}$

but since the block of repeating digits is 10 digits long, the summation term should be $\sum{(10^{-10})}^k$, so that $$0.\overline{0123456789}=\frac{M}{x}\cdot\sum{(10^{-10})}^k=\frac{M}{x}\cdot\frac{1}{1-10^{-10}}=\frac{M}{9999999999}$$ and $$\frac{M}{9999999999}=\frac{123456789}{9999999999}=\frac{13717421}{1111111111}.$$

-
Thanks! Using your last fraction, the long double can give at least 2 repetitions before lose precision: 0.0123456789 0123456789 0470 –  Dr Beco Mar 29 '11 at 4:14


# [SOLVED]A probability question: choosing 3 from 25

#### first21st

##### New member
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
A. 21/46
B. 25/117
C. 1/50
D. 3/25

Could you please solve this problem with proper explanation?

Thanks,

James

#### MarkFL

Staff member
Re: A probability question

Here at MHB, we normally don't provide fully worked solutions, but rather we help people to work the problem on their own. This benefits people much more, which is our goal.

Now, what you want to do here is to find the number of ways to choose 1 girl from 10 AND 2 boys from 15, then divide this by the number of ways to choose 3 children from 25. What do you find?

#### first21st

##### New member
Re: A probability question

Thanks for your reply. It's really a great way of learning. I highly appreciate your approach. Is the solution something like:

(15 C 2) * (10 C 1)/ (25 C 3)

If it is correct, could you please explain why did we divide it with the number of ways of choosing 3 students from 25?

Thanks,

James

#### MarkFL

Staff member
Re: A probability question

Excellent! That is correct!

Now you just need to simplify, either by hand or with a calculator.

As probability is the ratio of the number favorable outcomes to the number of all outcomes, we are in this case dividing the number of ways to choose 1 girl from 10 AND 2 boys from 15 by the number of ways to choose 3 of the children from the total of 25. We are told that 3 children are selected at random, and we know there are 25 children by adding the number of boys to the number of girls. Thus, $$\displaystyle {25 \choose 3}$$ is the total number of outcomes.

#### first21st

##### New member
Re: A probability question

Thank you very much man! But I am still wondering why did we MULTIPLY # the number of ways to choose 1 girl from 10 # AND # 2 boys from 15 #, instead of ADDING these two operands?

#### MarkFL

Staff member
Re: A probability question

We multiply because we are essentially applying the fundamental counting principle. When we require event 1 AND event 2 to happen, we multiply. When we require event 1 OR event 2 to happen, we add. Here we require both 2 boys AND 1 girl.

You see, for each way to obtain 2 boys, we have to account for all of the ways to obtain 1 girl. Or conversely, for each way to obtain 1 girl, we have to account for all of the ways to obtain 2 boys. The product of these two gives us all of the ways to get 2 boys and 1 girl.

#### first21st

##### New member
Re: A probability question

ok. Got it! Thanks a lot.

James


## Monday, February 17, 2020

### Reversing Differences

Fellow blogger Håkan Kjellerstrand posted an interesting question on OR Stack Exchange recently. Starting from a list of integers, it is trivial to compute the list of all pairwise absolute differences, but what about going in the other direction? Given the pairwise (absolute) differences, with duplicates removed, can you recover the source list (or a source list)?

We can view the source "list" as a vector $x\in\mathbb{Z}^n$ for some dimension $n$ (equal to the length of the list). With duplicates removed, we can view the differences as a set $D\subset \mathbb{Z}_+$. So the question has to do with recovering $x$ from $D$. Our first observation kills any chance of recovering the original list with certainty:
If $x$ produces difference set $D$, then for any $t\in\mathbb{R}$ the vector $x+t\cdot (1,\dots,1)'$ produces the same set $D$ of differences.
Translating all components of $x$ by a constant amount has no effect on the differences. So there will be an infinite number of solutions for a given difference set $D$. A reasonable approach (proposed by Håkan in his question) is to look for the shortest possible list, i.e., minimize $n$.

Next, observe that $0\in D$ if and only if two components of $x$ are identical. If $0\notin D$, we can assume that all components of $x$ are distinct. If $0\in D$, we can solve the problem for $D\backslash\{0\}$ and then duplicate any one component of the resulting vector $x$ to get a minimum dimension solution to the original problem.

Combining the assumption that $0\notin D$ and the observation about adding a constant having no effect, we can assume that the minimum element of $x$ is 1. That in turn implies that the maximum element of $x$ is $1+m$ where $m=\max(D)$.

From there, Håkan went on to solve a test problem using constraint programming (CP). Although I'm inclined to suspect that CP will be more efficient in general than an integer programming (IP) model, I went ahead and solved his test problem via an IP model (coded in Java and solved using CPLEX 12.10). CPLEX's solution pool feature found the same four solutions to Håkan's example that he did, in under 100 ms. How well the IP method scales is an open question, but it certainly works for modest size problems.

The IP model uses binary variables $z_1, \dots, z_{m+1}$ to decide which of the integers $1,\dots,m+1$ are included in the solution $x$. It also uses variables $w_{ij}\in [0,1]$ for all $i,j\in \{1,\dots,m+1\}$ such that $i \lt j$. The intent is that $w_{ij}=1$ if both $i$ and $j$ are included in the solution, and $w_{ij} = 0$ otherwise. We could declare the $w_{ij}$ to be binary, but we do not need to; constraints will force them to be $0$ or $1$.

The full IP model is as follows:

$\begin{array}{lrlrc} \min & \sum_{i=1}^{m+1}z_{i} & & & (1)\\ \textrm{s.t.} & w_{i,j} & \le z_{i} & \forall i,j\in\left\{ 1,\dots,m+1\right\} ,i\lt j & (2)\\ & w_{i,j} & \le z_{j} & \forall i,j\in\left\{ 1,\dots,m+1\right\} ,i\lt j & (3)\\ & w_{i,j} & \ge z_{i}+z_{j}-1 & \forall i,j\in\left\{ 1,\dots,m+1\right\} ,i\lt j & (4)\\ & w_{i,j} & =0 & \forall i,j\in\left\{ 1,\dots,m+1\right\} \textrm{ s.t. }(j-i)\notin D & (5)\\ & \sum_{i,j\in\left\{ 1,\dots,m+1\right\} |j-i=d}w_{i,j} & \ge 1 & \forall d\in D & (6)\\ & z_{1} & = 1 & & (7) \end{array}$

The objective (1) minimizes the number of integers used. Constraints (2) through (4) enforce the rule that $w_{ij}=1$ if and only if both $z_i$ and $z_j$ are $1$ (i.e., if and only if both $i$ and $j$ are included in the solution).  Constraint (5) precludes the inclusion of any pair $i < j$ whose difference $j - i$ is not in $D$, while constraint (6) says that for each difference $d \in D$ we must include at least one pair $i < j$ for that produces that difference ($j - i = d$). Finally, since we assumed that our solution starts with minimum value $1$, constraint (7) ensures that $1$ is in the solution. (This constraint is redundant, but appears to help the solver a little, although I can't be sure given the short run times.)

My Java code is available from my repository (bring your own CPLEX).

## Tuesday, February 11, 2020

### Collections of CPLEX Variables

Recently, someone asked for help online regarding an optimization model they were building using the CPLEX Java API. The underlying problem had some sort of network structure with $N$ nodes, and a dynamic aspect (something going on in each of $T$ periods, relating to arc flows I think). Forget about solving the problem: the program was running out of memory and dying while building the model.

A major issue was that they allocated two $N\times N\times T$ arrays of variables, and $N$ and $T$ were big enough that $2N^2T$ was, to use a technical term, ginormous. Fortunately, the network was fairly sparse, and possibly not every time period was relevant for every arc. So by creating only the IloNumVar instances they needed (meaning only for arcs that actual exist in time periods that were actually relevant), they were able to get the model to build.

That's the motivation for today's post. We have a tendency to write mathematical models using vectors or matrices of variables. So, for instance, $x_i \, (i=1,\dots,n)$ might be an inventory level at each of $n$ locations, or $y_{i,j} \, (i=1,\dots,m; j=1,\dots,n)$ might be the inventory of item $i$ at location $j$. It's a natural way of expressing things mathematically. Not coincidentally, I think, CPLEX APIs provide structures for storing vectors or matrices of variables and for passing them into or out of functions. That makes it easy to fall into the trap of thinking that variables must be organized into vectors or matrices.

Last year I did a post ("Using Java Collections with CPLEX") about using what Java calls "collections" to manage CPLEX variables. This is not unique to Java. I know that C++ has similar memory structures, and I think they exist in other languages you might use with CPLEX. The solution to the memory issue I mentioned at the start was to create a Java container class for each combination of an arc that actually exists and time epoch for which it would have a variable, and then associate instances of that class with CPLEX variables. So if we call the new class AT (my shorthand for "arc-time"), I suggested the model owner use a Map<AT, IloNumVar> to associate each arc-time combination with the variable representing it and a Map<IloNumVar, AT> to hold the reverse association. The particular type of map is mostly a matter of taste. (I generally use HashMap.) During model building, they would create only the AT instances they actually need, then create a variable for each and pair them up in the first map. When getting a solution from CPLEX, they would get a value for each variable and then use the second map to figure out for which arc and time that value applied. (As a side note, if you use maps and then need the variables in vector form, you can apply the values() method to the first map (or the getKeySet() method to the second one), and then apply the toArray() method to that collection.)

Now you can certainly get a valid model using just arrays of variables, which was all that was available to me back in the Dark Ages when I used FORTRAN, but I think there are some benefits to using collections. Using arrays requires you to develop an indexing scheme for your variables. The indexing scheme tells you that the flow from node 3 to node 7 at time 4 will be occupy slot 17 in the master variable vector. Here are my reasons for avoiding that.
• Done correctly, the indexing scheme is, in my opinion, a pain in the butt to manage. Finding the index for a particular variable while writing the code is time-consuming and has been known to kill brain cells.
• It is easy to make mistakes while programming (calculate an index incorrectly).
• Indexing invites the error of declaring an array or vector with one entry for each combination of component indices (that $N\times N\times T$ matrix above), without regard to whether you need all those slots. Doing so wastes time and space, and the space, as we saw, may be precious.
• Creating slots that you do not need can lead to execution errors. Suppose that I allocating a vector IloNumVar x = new IloNumVar[20] and use 18 slots, omitting slots 0 and 13. If I solve the model and then call getValues(x), CPLEX will throw an exception, because I am asking for values of two variables (x[0] and x[13]) that do not exist. Even if I create variables for those two slots, the exception will occur, because those two variables will not belong to the model being solved. (There is a way to force CPLEX to include those variables in the model without using them, but it's one more pain in the butt to deal with.) I've lost count of how many times I've seen messages on the CPLEX help forums about exceptions that boiled down to "unextracted variables".
So my advice is to embrace collections when building models where variables do not have an obvious index scheme (with no skips).

## Thursday, January 30, 2020

### Generic Callback Changes in CPLEX 12.10

CPLEX 12.10 is out, and there have been a few changes to the new(ish) generic callbacks. Rather than go into them in detail (and likely screw something up), I'll just point you to the slides for a presentation by Daniel Junglas of IBM at the 2019 INFORMS Annual Meeting.

I've written about half a dozen posts about generic callbacks since IBM introduced them (which you can find by typing "generic callback" in the search widget on the blog). A couple of things have been added recently, and I thought I would mention them. The generic callback approach uses a single callback function that can be called from a variety of contexts, including when CPLEX solves a node relaxation ("RELAXATION" context), when if finds a candidate solution ("CANDIDATE" context) and, now, when it is ready to split a node into children ("BRANCHING" context).

The branching context is one of the new features. It brings back most of the functionality of the branch callback in the legacy callback system. Unfortunately, it does not seem to have the ability to attach user information to the child nodes, which was a feature that was occasionally useful in the legacy system. You can get more or less equivalent functionality by creating a data store (array, map, whatever) in your global memory and storing the node information keyed by the unique index number of each child node. The catch is that you are now responsible for memory management (freeing up space when a node is pruned and the associated information is no longer needed), and for dealing with thread synchronization issues.

Another new feature is that you can now inject a heuristic solution (if you have one) from all three of the contexts I mentioned above. CPLEX gives you a variety of options for how it will handle the injected solution: "NoCheck" (CPLEX will trust you that it is feasible); "CheckFeasible" (CPLEX will check feasibility and ignore the solution if it is not feasible); "Propagate" (Daniel's explanation: CPLEX will "propagate fixed variables and accept if feasible"); and "Solve" (CPLEX will solve a MIP problem with fixed variables and accept the result if feasible). I assume the latter two mean that you provide a partial solution, fixing some variables but not others. (Unfortunately I was unable to make it to Daniel's talk, so I'm speculating here.)

I'm not sure if those are the only new features, but they are the ones that are most relevant to me. I invite you to read through Daniel's slides to get a more complete picture, including both the reasons for switching from legacy callbacks to generic callbacks and some of the technical issues in using them.

## Tuesday, January 7, 2020

### Greedy Methods Can Be Exact

We generally sort optimization algorithms (as opposed to models) into two or three categories, based on how certain we are that solutions will be either optimal or at least "good". An answer by Michael Feldmeier to a question I posted on OR Stack Exchange neatly summarizes the categories:
• exact methods eventually cough up provably optimal solutions;
• approximate methods eventually cough up solutions with some (meaningful) guarantee regarding how far from optimal they might be; and
• heuristics provide no worst-case guarantees (but generally are either easy to implement, fast to execute or both).
I should explain my use of "meaningful" (which is not part of Michael's answer). A common way to estimate the "gap" between a solution and the optimum is to take $|z - \tilde{z}|/|z|$, where $z$ is the objective value of the solution produced by the algorithm and $\tilde{z}$ is some bound (lower bound in a minimization, upper bound in a maximization) of the optimal solution. Now suppose that we are minimizing a function known to be nonnegative. If we set $\tilde{s}=0$, we know that any method, no matter how stupid, will have a gap no worse than 100%. To me, that is not a meaningful guarantee. So I'll leave the definition of "meaningful" to the reader.

What brings all this to mind is a question posted on Mathematics Stack Exchange. The author of the question was trying to solve a nonlinear integer program. He approached it by applying a "greedy algorithm". Greedy algorithms are generally assumed to be heuristics, since it seldom is possible to provide useful guarantees on performance. In his case, though, the greedy algorithm is provably optimal, mainly due to the objective function being concave and separable. I'll state the problem and show a proof of optimality below (changing the original notation a bit). Brace yourself: the proof is a bit long-winded.

You start with $N$ workers to be assigned to $M$ work stations. The output of workstation $m$, as a function of the number of workers $x$ assigned to it, is given by $$f_{m}(x)=a_{m}x+b_{m}-\frac{c_{m}}{x},$$
where $a_{m},b_{m},c_{m}$ are all positive constants. Since $f(0)=-\infty$, we can assume that each work station gets at least one worker (and, consequently, that $N>M$). Since $f_{m}'(x)=a_{m}+c_{m}/x^{2}>0$, each $f_{m}()$ is monotonically increasing. Thus, we can safely assume that all $N$ workers will be assigned somewhere. $f_{m}''(x)=-2c_{m}/x^{3}<0$, so $f_{m}()$ is strictly concave (which we will need later). We also note, for future reference, that the impact of adding one worker to a current staff of $x$ at station $m$ is $$\Delta f_{m}(x)=a_{m}+\frac{c_{m}}{x(x+1)}>0.$$
Similarly, the impact of removing one worker at station $m$ is $$\delta f_{m}(x)=-a_{m}-\frac{c_{m}}{x(x-1)}<0.$$We see that $\delta f_{m}(x)$ is an increasing function of $x$ (i.e., it gets less negative as $x$ gets bigger). We also note that $\Delta f_{m}(x)=-\delta f_{m}(x+1)$.

The IP model is easy to state. Let $x_{m}$ be the number of workers assigned to work station $m$. The model is
$$\max\sum_{m=1}^{M}f_{m}(x_{m})$$
subject to
$$\sum_{m=1}^{M}x_{m}\le N$$
with
$$x\in\mathbb{Z}_{+}^{M}.$$

The greedy algorithm starts with a single worker at each station ($x=(1,\dots,1)$) and, at each step, adds one worker to the workstation where that worker produces the greatest increase in objective value (breaking ties arbitrarily). It stops when all $N$ workers are assigned. To prove that it actually finds an optimal solution, I'll use proof by contradiction.

Let $x^{(0)},x^{(1)},\dots,x^{(N-M)}$ be the sequence of solutions constructed by the greedy algorithm, with $x^{(0)}=(1,\dots,1)$, and let $x^{(k)}$ be the last solution in the sequence for which an optimal solution $x^{*}$ exists such that $x^{(k)}\le x^{*}$. The significance of the inequality is that if $x\le x^{*}$, it is possible to extend the partial solution $x$ to the optimal solution $x^{*}$ by adding unassigned workers to work stations. We know that $k$ is well defined because $x^{(0)}\le x^{*}$ for any optimal $x^{*}$. Since we are assuming that the greedy algorithm does not find an optimum, it must be that $k<N-M$.

Now identify the work station $j$ to which the greedy algorithm added a worker at step $k$, meaning that $x_{j}^{(k+1)}=x_{j}^{(k)}+1$ and $x_{i}^{(k+1)}=x_{i}^{(k)}$ for $i\neq j$. Since, by assumption, $x^{(k)}\le x^{*}$ but $x^{(k+1)}\not\le x^{*}$, it must be that $x_{j}^{(k)}=x_{j}^{*}$.

Next, since $x^{(k)}\le x^{*}$ and $x^{(k)}\neq x^{*}$ (else $x^{(k)}$ would be optimal), there is some work station $h\neq j$ such that $x_{h}^{(k)}<x_{h}^{*}$. Let $\tilde{x}$ be the solution obtained from $x^{(k)}$ by adding a worker to station $h$: $\tilde{x}_{h}=x_{h}^{(k)}+1$ and $\tilde{x}_{i}=x_{i}^{(k)}$ for $i\neq h$. Observe that $\tilde{x}\le x^{*}$. The greedy algorithm chose work station $j$ over work station $h$ at $x^{(k)}$, so it must be that
$$\Delta f_{j}(x_{j}^{(k)})\ge\Delta f_{h}(x_{h}^{(k)}). \quad (1)$$

Finally, let $\hat{x}$ be the result of starting from optimal solution $x^{*}$ and shifting one worker from station $h$ to station $j$. Since
$$x_{j}^{(k+1)}=x_{j}^{(k)}+1=x_{j}^{*}+1=\hat{x}_{j},$$ $$x_{h}^{(k+1)}=x_{h}^{(k)}<x_{h}^{*}\implies x_{h}^{(k+1)}\le\hat{x}_{h}$$
and
$$x_{i}^{(k+1)}=x_{i}^{(k)}\le x_{i}^{*}=\hat{x}_{i}\,\forall i\notin\{h,j\},$$
we have $x^{(k+1)}\le\hat{x}$. Under the assumption that $x^{(k)}$ was the last solution in the greedy sequence that could be extended to an optimal solution, it must be that $\hat{x}$ is not optimal. Thus the net change to the objective function at $x^{*}$ when shifting one worker from station $h$ to station $j$ must be negative, i.e.,
$$\Delta f_{j}(x_{j}^{*})+\delta f_{h}(x_{h}^{*})<0.\quad (2)$$

We showed previously that, under our assumptions, $x_{j}^{(k)}=x_{j}^{*}$, from which it follows that
$$\Delta f_{j}(x_{j}^{*})=\Delta f_{j}(x_{j}^{(k)}). \quad (3)$$
We also showed that $\delta f_{h}()$ is an increasing function. Since $\tilde{x}_{h}\le x_{h}^{*}$,
$$\delta f_{h}(x_{h}^{*})\ge\delta f_{h}(\tilde{x}_{h})=-\Delta f_{h}(x_{h}^{(k)}). \quad (4)$$
Combining (4) with (2), we have
$$\Delta f_{j}(x_{j}^{*})-\Delta f_{h}(x_{h}^{(k)})<0,$$
i.e.,
$$\Delta f_{j}(x_{j}^{*})<\Delta f_{h}(x_{h}^{(k)}). \quad (5)$$

Combining (3) with (5) yields
$$\Delta f_{j}(x_{j}^{(k)})<\Delta f_{h}(x_{h}^{(k)})$$