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1. The problem statement, all variables and given/known data | {
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A group can charter a particular aircraft at a fixed total cost. If 36 people charter the aircraft rather than 40 people, then the cost per person is greater by 12$a) What is the fixed total cost to charter the aircraft? b) What is the cost per person if 40 people charter the aircraft? 2. Relevant equations ThirtySixCost = 36 * (costPerPerson + 12) FixedCost = NumberofPersons * costPerPerson (Note if NumberofPersons>40) Not sure about the correctness of equations 3. The attempt at a solution I dont know how to solve it because data is incomplete. ThirtySixCost=? FortyCost=? Somebody please guide me how to solve it. Zulfi. 2. Jul 10, 2017 ### scottdave Reread the problem statement. Your costPerPerson represents the cost per person, when there are 40 people. Your ThirtySixCost is the total cost to rent the plane (but that is a Fixed cost). I did not see anything about greater than 40. You can get to 3 equations and 3 unknowns to solve it. For example (not real numbers), if it costs$360 to rent something (say a lake house), then for 40 people, that is $9 per person, but for 36 people it is$10 per person. It costs $1 more per person if you only have 36 people. Last edited: Jul 10, 2017 3. Jul 11, 2017 ### zak100 Hi, Please tell me how to find costPerPerson? Zulfi. 4. Jul 11, 2017 ### Mark44 ### Staff: Mentor These variable names aren't useful. The fixed cost is the same whether there are 40 people or 36 people. Let x = the cost per person with 40 passengers. Let f = the fixed cost of chartering the airplane. • Write an expression that represents the cost per person with 36 passengers. (Use the information given in the problem.) • Translate the 2nd sentence in the problem statement into an algebraic equation. 5. Jul 11, 2017 ### zak100 Hi, FTC = Fixed Total Cost CPP = Cost Per Person CPP = FTC + 12 I can form only one equation. Please guide me. Zulfi. 6. Jul 11, 2017 ### scottdave The plane costs an amount (call it FTC), whether there is 1 passenger or 36 passenger or 40 | {
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The plane costs an amount (call it FTC), whether there is 1 passenger or 36 passenger or 40 passengers. The cost per person depends on the fixed cost (which does not change, and the number of passengers. How about this?: FTC is total fixed cost. CPPForty = FTC / 40. CPPthirtysix = FTC / 36. The problem statement gives you another relationship between CPPForty and CPPthirtysix, which you can make another equation. You may want to re-look at the example problem I posed in post #2, to see how you can relate your problem to that one. 7. Jul 11, 2017 ### Mark44 ### Staff: Mentor The last equation makes no sense. You are saying that the cost for one person is$12 more than the fixed cost. | {
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8. Jul 11, 2017
### zak100
Hi,
Mr. scottdave:
Do you mean this:
CPPThirtySix = CPPForty + 12
Zulfi.
9. Jul 11, 2017
### scottdave
Yes, now with 3 equations you should be able to solve for CPPThirtySix, CPPForty and FTC.
10. Jul 12, 2017
### zak100
Hi,
I am able to solve it.
CP36 = FTC/36 ---(1)
CP40 = FTC/40 ---(2)
CP36 = CP40 + 12--- (3)
After solving these eq., i am able to get correct answer:
CP40 = 108
CP36 = 120 (Not required in the question)
FTC = 4320 | {
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# Homework Help: Table probability question
1. Jan 29, 2014
### joshmccraney
1. The problem statement, all variables and given/known data
11 people sit at two round tables, one sits 5 and the other sits 6. how many combinations of seating arrangements are there?
2. Relevant equations
i'm not sure about equations, but my solution attempt may have some info.
3. The attempt at a solution
i know if 11 people were to sit at 1 table, we would have $\frac{11!}{11}=10!$ combinations. thus, would we first have to choose who sits at which table? i.e: $$(11C5)(4!)+(11C6)(6!)$$
thanks for your time (and help!)
2. Jan 29, 2014
### haruspex
You're close, but a couple of errors.
You got 11C54! for choosing the 5 to sit at one table and arranging them. For each such arrangement, how can you arrange the remaining 6?
3. Jan 30, 2014
### joshmccraney
wait, do you mean $(11C5)(4!)+(11C6)(5!)$? also, if we've already selected the 5 to be at on table, do we really need to select 6 for the other?
thanks for the response, and let me know what you think.
4. Jan 30, 2014
### joshmccraney
because i definitely should have done $5!$, not $6!$ (brain fart)
5. Jan 30, 2014
### D H
Staff Emeritus
There are two big problems with your result. One is that when you've chosen the five people who will sit at table #1 you have but one choice as to which set of six people will sit at table #2. The other: Why are you adding?
6. Jan 30, 2014
### joshmccraney
yea, i definitely see this problem but if i dont put both "choices" how do i determine which number (5 or 6) to use?
also, i was adding because i figured after i choose who sits at what table, all i need to do is take the possible combos of one table and add them to the combos of another table.
evidently this is wrong; can you direct me as to what to do next?
7. Jan 30, 2014
### D H
Staff Emeritus
There are a number of right ways to address this part of the problem. You chose a wrong way. | {
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One way to look at this part of the problem is that you choose five people to sit at one table, call it table A. There are 11 choose 5 ways to do this. For any give choice, there are 11-5=6 people left. You now need to choose six of these six people to sit at table B. There are 6 choose 6 ways to do this, and that of course is one. The number of ways to choose people to sit at the two tables is the product of these two numbers: (11 choose 5)*(6 choose 6).
You could of course look at it the other way around: You'll choose six people to sit at the larger table and then choose five of the remaining five to sit at the smaller table. This leads to (11 choose 6)*(5 choose 5) ways to split the people into two groups.
Hmmm. One approach yields (11 choose 5)*(6 choose 6), the other (11 choose 6)*(5 choose 5). Is there a problem here? The answer is no. 11 choose 5 and 11 choose 6 are the same number (462), as are 6 choose 6 and 5 choose 5 (which are of course both 1).
Suppose X and Y are independent. Also suppose there are three ways to accomplish X and five ways to accomplish Y. The number of ways to do accomplish X and Y is fifteen, not eight.
The same applies to the problem at hand. You should be multiplying here, not adding.
8. Jan 30, 2014
### CAF123
You can do it either way (i.e choose 6 for one table and have one set of five for the other, or choose 5 for a table and have one set of six for the other). The result is the same.
Consider the case when there are two people situated at one table and one at the other. How many possible combinations are there?
Alternatively, are the combinations around each individual table dependent on each other? What does this tell you? | {
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# Solving the following equation for x: $(x-5)^{-2} = 9$
1. Jun 24, 2016
### malknar
1. The problem statement, all variables and given/known data
An exercise in Lang's Basic Mathematics which asked to solve for x in the equation $(x - 5)^{-2} = 9$
I tried to solve it and got $x = \frac {16} 3$
which is correct according to the book, but I've found that the book states that x also equals \frac {14}3, and I can't figure out how.
2. Relevant equations
3. The attempt at a solution
$(x-5)^{-2} = 9$
This yields: $x-5 = 9^{-1/2} =\frac 1 {9^{1/2}} = \frac13$
Which yields: $x=\frac 13 +5 = \frac {1+15} 3 =\frac {16}3$
But I can't figure out how can x equals $\frac {14}3$ too.
2. Jun 24, 2016
### Staff: Mentor
It's better to write this as $(x - 5)^2 = \frac 1 9$, which leads to $x - 5 = \pm \sqrt{\frac 1 9}$
Quadratic equations can have two solutions.
3. Jun 24, 2016
### mfig
What is $(-\frac{1}{3})^2$?
4. Jun 24, 2016
### late347
I am thinking about that old math movie with young actor Edward James Olmos. He was playing the role of high school teacher.
Stand and deliver
"Negative times negative is positive!!!" The result would be 1/9
For the original poster's benefit. The following is compiled from my own high school era old math factbook.
second degree equations.
• you can evaluate whether the quadratic has two roots, or one root, or none of the real-numbered roots. (roots belonging to ℝ)
• calculate the quadratic into the form of: $$Ax^2 + Bx+C= 0$$
$$x=\frac{-B±\sqrt{B^2-4AC}}{2A}$$
this is called the discriminant part of the equation $D=\sqrt{B^2-4AC}$
if D>0, then the equation has two non-equal roots
if D=0, then the equation has a twin-root
if D<0, then the equation doesn't have real-numbered roots (roots in the real numbers ℝ)
This last fact is because negative inside the squareroot sign is not defined.
E.g. $\sqrt{(-5)}= ~~not ~~defined~~ in~~ real~~ numbers$ | {
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negative inside the squareroot is defined in the complex numbers, I think, but I have no knowledge of the complex numbers myself. (complex numbers symbol is this one ℂ)
5. Jun 24, 2016
### epenguin
In short if A2 = B, then A = ± √B
6. Jun 24, 2016
### malknar
I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering $-3 = 9^{1/2}$
7. Jun 24, 2016
### Staff: Mentor
The symbol $\sqrt{x}$ denotes the positive square root of x, by established convention. In this case $\sqrt{9} = +3$. However, the equation $x^2 = 9$ has two solutions: 3 and -3.
8. Jun 24, 2016
### Ray Vickson
No: $9^{1/2}$ is $+3$, NOT $-3$. Essentially by definition, the functions $x^{1/2}$ and $\sqrt{x}$ mean the positive root.
There is no really good language for it, but saying "a" square root instead of "the" square root comes close to capturing it: there are two roots to the equation $x^2 = a \:(a > 0)$; these are $x = \sqrt{a}$ and $x = -\sqrt{a}$
9. Jun 24, 2016
### late347
• watch this khan academy video it will be helpful for this problem and also for the purpose of increased understanding.
• review what is the binomial square formula such as:
• $(a+b)^2= a^2+2ab+b^2$
• $(a-b)^2 = a^2-2ab+b^2$
10. Jul 6, 2016
### James R
Alternatively you could do this:
$\frac{1}{(x-5)^2}=9$
$9(x-5)^2 = 1$
$9(x^2 - 10 x + 25) = 1$
$9x^2 - 90 x + 224 = 0$
$(3x - 14)(3x - 16)=0$
$x=\frac{14}{3}$ or $x=\frac{16}{3}$. | {
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# Find the differential equations of the following family of curves.
16 posts / 0 new
Bingo
Find the differential equations of the following family of curves.
Differential Equations
Find the differential equations of the following family of curves.
1. Parabolas with axis parallel to the y – axis with distance vertex to focus fixed as a.
2. Parabolas with axis parallel to the x – axis with distance vertex to focus fixed as a.
3. All ellipses with center at the origin and axes on the coordinate axes.
4. Family of cardioids.
5. Family of 3 – leaf roses.
Bingo
1
Jhun Vert
(1) Upwards and downwards parabolas with latus rectum equal to 4a.
$y - k = \pm 4a(x - h)^2$
$y' = \pm 8a(x - h)$
$y'' = \pm 8a$
Bingo
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
Jhun Vert
I made a mistake in there, it should be (x - h)2 = ±4a(y - k). The solution for (1) should go this way:
$(x - h)^2 = \pm 4a(y - k)$
$2(x - h) = \pm 4ay'$
$2 = \pm 4ay''$
$y'' = \pm \frac{1}{2a}$
Bingo
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
Jhun Vert
Bingo
so for number 2,
x-h=±4a(y-k)2
x'=±8a(y-k)
x''=±8a
am i wrong?
Jhun Vert
It is better to express your answer in terms of y' rather than x'. Although x' will do and simpler.
$(y − k)^2 = \pm 4a(x − h)$
$2(y − k)y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
Note: $y - k = \dfrac{y'}{\pm 2a}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$
$y'' = \dfrac{\mp 8a}{y'}$
$y'' \, y' = \pm 8a$
ChaCha Ingal Cortez
our prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay"=0 are they the same with your answer sir ?
Jhun Vert
They are different. I made a mistake in $y - k$ of the above solution.
$(y − k)^2 = \pm 4a(x − h)$
$2(y - k) \, y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
$y - k = \dfrac{\pm 2a}{y'}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$ | {
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$y' = \dfrac{\pm 2a}{y - k}$
$y - k = \dfrac{\pm 2a}{y'}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{\pm 2a}{y'} \right)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$
$y'' = \dfrac{\mp (y')^3}{2a}$
$2a \, y'' = \mp (y')^3$
$\pm (y')^3 + 2a \, y'' = 0$
Bingo
Thank You so much sir.
Bingo
sir, do you know the standard or general equations for the last 3 problems?
what equations should i use?
Shandy
I think you are from ADZU haha
Alegnaaa (guest)
Good evening, sir. Do you have the solution for the last three problems? Thank you!
Ancha (guest)
Sir do you know how to eliminate arbitrary constant here in this equation?
(y-33)²=4a(x-h)?
• Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics. | {
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# How many four digit integers are there that do not have two consecutive equal even digits?
How many four digit integers are there that do not have two consecutive equal even digits?
My Attempt There are $9*10^3$ four digit numbers. We tie two equal even digits and form the four digit numbers with them. This can be done in $5 \cdot \binom{3}{2} \cdot 10 \cdot 10 = 1500$ ways. There are $10 \cdot 10$ numbers starting with $00$ so we must subtract them from $1500$. So answer is $9000 - 1400 = 7600$. But it is not matching with the given answer $7801$. Can any one help me to find out where I am making mistakes? Thanks in advance.
1. You did not account for the fact that there are only $9$ choices for the leading digit when neither number in the pair of consecutive equal even digits is in the thousands place.
2. You have subtracted numbers in which there are three or more consecutive even digits more than once.
First, we observe that there are $9 \cdot 10 \cdot 10 \cdot 10 = 9000$ four-digit positive integers.
Let $A_1$ be the set of four-digit positive integers in which the thousands place and hundreds place contain equal even digits. Let $A_2$ be the set of four-digit positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \cup A_2 \cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is $$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$ | {
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$|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \cdot 10 \cdot 10 = 400$.
$|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \cdot 5 \cdot 10 = 450$.
$|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \cdot 10 \cdot 5 = 450$.
$|A_1 \cap A_2|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \cap A_2| = 4 \cdot 10 = 40$.
$|A_1 \cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \cap A_3| = 4 \cdot 5 = 20$.
$|A_2 \cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \cap A_3| = 9 \cdot 5 = 45$.
$|A_1 \cap A_2 \cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \cap A_2 \cap A_3| = 4$. | {
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Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is \begin{align*} |A_1 \cup A_2 \cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|\\ & = 400 + 450 + 450 - 40 - 20 - 45 + 4\\ & = 1199 \end{align*} Therefore, the number of four-digit positive even integers that do not contain consecutive equal even digits is $9000 - 1199 = 7801$.
• Thanks for pointing the errors and nice step by step solution. – rugi Jul 31 '17 at 9:05
This where you're making a mistake $-$
What you're trying to do is called the inclusion-exclusion principle. Your error is in the fact that you are overcounting numbers such as $4446$. If I understand your method correctly, you are blacklisting this number twice instead of once.
IMO, an easier way to solve the problem is to count the numbers that are not to be included first by going through all possible digit combinations e.g. $aabb$ etc. | {
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# If $\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5$, then what is $\lim_{x \rightarrow 0}f(x)$?
I know the answer to the above question, but I have a question on some of the reasoning.
The way I know how to solve it is $$\lim_{x \rightarrow 0}f(x) = \lim_{x \rightarrow 0}\left(f(x)\cdot \frac{x^2}{x^2}\right) = \lim_{x \rightarrow 0}\left(\frac{f(x)}{x^2}\cdot x^2\right) = \left(\lim_{x \rightarrow 0}\frac{f(x)}{x^2}\right)\left(\lim_{x \rightarrow 0}x^2\right) = 5\cdot0 = 0.$$
I saw another solution elsewhere that gets the right answer, but I am unsure if the steps are actually correct. \begin{align*} &\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5 \\ \Longrightarrow &\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5 \\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot \lim_{x \rightarrow 0}x^2\\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot 0 = 0. \end{align*}
My issue is with that first step. I know that $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}$$, but only when $$\lim_{x \rightarrow a}g(x) \neq 0$$. Since $$\lim_{x \rightarrow 0}x^2 = 0$$, wouldn't this invalidate the above work? However, it still got the same answer, so my real question is why did it work and when will it work in general?
EDIT: Does anyone have a nice example for when the logic in the second method doesn't work?
• You are correct that the solution you saw elsewhere is invalid. It is invalid for the exact reason you think it is. Using false logic to get a correct answer is still false logic. Applied to another problem, it might fail. – InterstellarProbe May 8 at 16:38
• Do you have an example of another problem in which logic from the second method fails? – Smash May 8 at 16:45
• Simply put, you can't divide by zero. If you want examples, there are many "proofs" of $0=1$ where the trick is to divide by zero and hope no one notices. – Théophile May 8 at 16:47 | {
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If $$\lim\limits_{x\to0}\dfrac{f(x)}{x^2}=5$$, then for every $$n\in\Bbb N$$ there is some $$x\neq 0$$ such that $$5-\frac1n<\frac{f(x)}{x^2}<5+\frac1n$$ and hence $$5x^2-\frac{x^2}n If $$x\to 0$$ we get $$f(x)\to 0$$.
$$\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5$$ doesn't work, the LHS is not defined.
But for the limit of $$\dfrac{f(x)}{x^2}$$ to exist, $$\lim_{x\to0}f(x)$$ must be zero, which is also the limit of $$x^2$$.
You can not do this since the denominator on the right is $$0$$.
$$\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = \frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2}$$
But first aproach is perfect.
• You meant "denominator," not numerator. – Mark Viola May 8 at 16:41 | {
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# Probability of winning dice game
#### TheFallen018
##### Member
Hey, so I've got this problem that I'm trying to figure out. I've worked out something that I think is probably right through simulation, but I'm not really sure how to tackle it from a purely mathematical probability perspective. So, would anyone know how I should approach this? I've tried a few different things, but my two answers tend to conflict a little. Thanks,
#### Opalg
##### MHB Oldtimer
Staff member
The starting point (which I guess you already know) is to make a list of the probabilities $p(x)$ of rolling a total of $x$ with the two dice, where $x$ goes from $2$ to $12$. These are $$\begin{array}{r|ccccccccccc}x& 2&3&4&5&6 &7&8&9&10 &11&12\\ \hline p(x) & \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36} & \frac4{36} & \frac3{36} & \frac2{36} & \frac1{36} \end{array}.$$
The probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\dfrac{244}{495} \approx 0.49292929\ldots$.
#### TheFallen018
##### Member
The starting point (which I guess you already know) is to make a list of the probabilities $p(x)$ of rolling a total of $x$ with the two dice, where $x$ goes from $2$ to $12$. These are $$\begin{array}{r|ccccccccccc}x& 2&3&4&5&6 &7&8&9&10 &11&12\\ \hline p(x) & \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36} & \frac4{36} & \frac3{36} & \frac2{36} & \frac1{36} \end{array}.$$ | {
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The probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\dfrac{244}{495} \approx 0.49292929\ldots$.
Wow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks
#### Opalg | {
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#### Opalg
##### MHB Oldtimer
Staff member
Wow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks
In the expression ${\color {red} p(x)}{\color {green}\dfrac{p(x)}{p(x)+p(7)}}$, the red $\color {red} p(x)$ gives the probability that the first roll of the dice gives the value $x$. The green fraction represents the probability of rolling $x$ again before rolling a $7$. My argument for that is that after rolling the first $x$, you can completely disregard any subsequent rolls until either an $x$ or a $7$ turns up. The only question is, which one of those will appear first. The relative probabilities of $x$ and $7$ are in the proportion $p(x)$ to $p(7)$. So out of a combined probability of $p(x) + p(7)$, the probability of an $x$ is $\dfrac{p(x)}{p(x)+p(7)}$, and the probability of a $7$ is $\dfrac{p(7)}{p(x)+p(7)}$.
I hope that makes sense. | {
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I'm Confused's question on Yahoo! Answers involving a recurrence relation
MarkFL
Staff member
Jonah must take an antibiotic every 12 hours. Each pill is 25 milligrams, and after every 12 hours, 50% of the drug remains in his body. What is the amount of antibiotic in his body over the first two days? What amount will there be in his body in the long run?
Here's what I did:
For my recursive formula, I tried inputting u(n)=u(n-1)*(1-0.50). However, I could not seem to find the long run value.,,, It keeps changing. Am I right on this?
I appreciate your help! Thank you so much!
Source: <Advanced Algebra: An Investigative Approach; Murdock Jerald, 2004>
I posted a link to this topic, so the OP could find my response.
Here is a link to the original question:
Recursive formula help on real world situations? - Yahoo! Answers
Let $A(t)$ represent the amount, in mg, of antibiotic in Jonah's bloodstream at time $t$, measured in hours. With a half-life of 12 hours, we may state:
$\displaystyle A(t)=A_02^{-\frac{t}{12}}$
So, for the first 12 hours, but not including the second dose, we find:
$\displaystyle A_0=25$ hence:
$\displaystyle A(t)=25\cdot2^{-\frac{t}{12}}$
$\displaystyle A(t)=25\cdot2^{-\frac{12}{12}}=\frac{25}{2}$
Now, for the second 12 hours, we have:
$\displaystyle A_0=\frac{25}{2}+25=\frac{75}{2}$ hence:
$\displaystyle A(t)=\frac{75}{2}\cdot2^{-\frac{t}{12}}$
$\displaystyle A(12)=\frac{75}{2}\cdot2^{-\frac{12}{12}}=\frac{75}{4}$
For the third 12 hours, we have:
$\displaystyle A_0=\frac{75}{4}+25=\frac{175}{4}$ hence:
$\displaystyle A(t)=\frac{175}{4}\cdot2^{-\frac{t}{12}}$
$\displaystyle A(12)=\frac{175}{4}\cdot2^{-\frac{12}{12}}=\frac{175}{8}$
For the fourth 12 hours, we have:
$\displaystyle A_0=\frac{175}{8}+25=\frac{375}{8}$ hence:
$\displaystyle A(t)=\frac{375}{8}\cdot2^{-\frac{t}{12}}$
$\displaystyle A(12)=\frac{375}{8}\cdot2^{-\frac{12}{12}}=\frac{375}{16}$
So, now we may try to generalize for the $n$th 12 hour period. | {
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So, now we may try to generalize for the $n$th 12 hour period.
Let $I_n$ represent the initial amount for each 12 hour period. We see we must have the recursion:
$I_{n+1}=\frac{1}{2}I_{n}+25$
This is an inhomogeneous recurrence, so let's employ symbolic differencing to obtain a homogeneous recurrence:
$I_{n+2}=\frac{1}{2}I_{n+1}+25$
Subtracting the former from the latter, we obtain:
$2I_{n+2}=3I_{n+1}-I_{n}$
The characteristic equation is:
$2r^2-3r+1=0$
$(2r-1)(r-1)=0$
Hence, the closed-form will be:
$I_n=k_1\left(\frac{1}{2} \right)^n+k_2$
We may use our data above to determine the parameters $k_i$:
$I_1=k_1\left(\frac{1}{2} \right)^1+k_2=25$
$I_2=k_1\left(\frac{1}{2} \right)^2+k_2=\frac{75}{2}$
or
$\frac{1}{2}\cdot k_1+k_2=25$
$\frac{1}{4}\cdot k_1+k_2=\frac{75}{2}$
Solving this system, we find:
$k_1=-50,k_2=50$ and so we have:
$I_n=-50\left(\frac{1}{2} \right)^n+50=50\left(1+\left(\frac{1}{2} \right)^n \right)$
And so, for the $n$th 12 hour period, we have:
$\displaystyle A_n(t)=I_n2^{-\frac{t}{12}}$
Now, for the long run, which we may take as implying as n grows without bound, we find:
$\displaystyle \lim_{n\to\infty}I_n=50$
and so we find that for the long run the initial amount for a 12 hour period is is 50 mg and the final amount for that period is 25 mg.
Last edited:
tkhunny
Well-known member
MHB Math Helper
That is good mathematics, but not so great reality. If it had said I.V. Administration, the assumption of immediate absorption and distribution would be more realistic. Even subcutaneous or intramuscular injection takes a little while. A pill? No. | {
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# Polynomial Sprint: Useful Lemma
$\color{#D61F06} {\text{Unlocked!}}$
Mathematicians have a huge bag of tricks, which provide them with various ways of approaching a problem. In this note, we introduce an extremely useful lemma, which applies to polynomials with integer coefficients.
Lemma. If $f(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, we have $a-b \mid f(a) - f(b)$
1) Show that for any integers $a, b$ and $n$, we have
$a-b | a^n - b^n.$
2) Using the above, prove the lemma.
3) Come up with a (essentially) two-line solution to question 4 from the book.
Note: In Help Wanted, Krishna Ar mentioned that problem 4 from the book was hard to solve. If you looked at the solution, it was also hard to understand what exactly was happening. This provides us with a simple way of comprehending why there are no solutions.
4) Come up with an alternative solution to E6 (page 250, from the book) using this lemma.
5) * (Reid Barton) Suppose that $f(x)$ is a polynomial with integer coefficients. Let $n$ be an odd positive integer. Let $x_1, x_2, \ldots x_n$ be a sequence of integers such that
$x_2 = f(x_1), x_3 =f(x_2), \ldots x_n = f(x_{n-1}), x_1 = f(x_n) .$
Show that all the $x_i$ are equal.
Where did you use the fact that $n$ is an odd positive integer?
Note by Calvin Lin
5 years, 9 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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I really like Polynomial Sprint. It would be nice if you can do some other themes using similar form.
- 5 years, 8 months ago
Yes, I entirely agree! I find these sets fun to do and I want more sets.
- 5 years, 8 months ago
This has been unlocked. Sorry for being late, I was out yesterday.
Staff - 5 years, 9 months ago
What does the $|$ mean?
- 5 years, 9 months ago
$a|b$ means $a$ divides $b$
- 5 years, 9 months ago
1) $a-b|a^n-b^n$, let $f(x)=x^n$, and plug in $a$ and $b$ into the function and we get $a-b|a^n-b^n$, is this correct?
- 5 years, 9 months ago | {
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- 5 years, 9 months ago
You have to show that it is true of all polynomials, not just specially chosen polynomials.
Staff - 5 years, 9 months ago
The lemma is true for polynomials with integer coefficients, so do you mean that I must show that this is true for non-integers too?
- 5 years, 9 months ago
I might have initially misinterpreted what you are trying to do.
The statement in question 1 is about integers (not polynomials), and you are asked to show that for any pair of integers, we have $a - b \mid a^n - b^n$. For example, $10 - 7 \mid 10^3 - 7^3$, or equivalently that $3 \mid 657$.
Note that you are not allowed to use the lemma, since we want to use question 1 to prove the lemma (in question 2).
Staff - 5 years, 9 months ago
Oh, ok
- 5 years, 9 months ago
5) We see that $x_n-x_{n+1}\mid f(x_n)-f(x_{n+1})=x_{n+1}-x_{n+2}$.
Thus we get the following divisibilitites:
$x_1-x_2\mid x_2-x_3\implies x_2-x_3=k_1(x_1-x_2)$
$x_2-x_3\mid x_3-x_4\implies x_3-x_4=k_2(x_2-x_3)$
$\vdots$
$x_{n-1}-x_n\mid x_n-x_1\implies x_n-x_1=k_{n-1}(x_{n-1}-x_n)$
$x_n-x_1\mid x_1-x_2\implies x_1-x_2=k_n(x_n-x_1)$
This means that $x_1-x_2=k_1k_2\cdots k_n(x_1-x_2)$
since $x_1\ne x_2$, we can divide both sides by $x_1-x_2$ to get $k_1k_2\cdots k_n=1$
Since we are working in the integers, this means that $|k_i|=1$ for all $i=1\to n$.
Thus, $x_k-x_{k+1}=\pm (x_{k-1}-x_k)$
What I got so far.
- 5 years, 9 months ago
If some $k_i=-1$, then we have $x_k=x_{k+1}$, which leads to equality of all $x_i$. If all k=1, then we can sum and get $x_1-x_2=0$, so $x_1=x_2$, which contradicts to your assumption that $x_1$ is different from $x_2$. Then that again leads to $x_1=x_2=...=x_n$
- 5 years, 9 months ago
Is it that simple? When did you use the condition that $n$ is odd?
- 5 years, 9 months ago
You just consider it modulo |x2 - x1| : | {
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- 5 years, 9 months ago
You just consider it modulo |x2 - x1| :
Let d equal |x1 – x2| and thus d = |x1 – x2| = |x2 – x3| = |x3 – x4| = … = |x{n-1} – x{n}| = |x{n} – x1|. If d is nonzero then x2 is congruent to (x1 + d) mod 2d. Similarly x3 is congruent to (x2 + d) mod 2d which is congruent to x1 mod 2d. In general x{k} is congruent to (x1 + d) mod 2d if k is even and congruent to x1 mod 2d if k is odd. Thus we have x{n} is congruent to x1 mod 2d because n is odd. However then |x{n} – x1| = 2dq for an integer q. Also d = |x{n} – x1|, hence d = 2dq and 1 = 2q. This contradicts the fact that q is an integer, hence our previous assumption that d is nonzero must be false. Therefore d = |x1 – x2| = |x2 – x3| = |x3 – x4| = … = |x{n-1} – x{n}| = |x{n} – x1| = 0 and x1 = x2 = x3 = …=x{n} and all the x{i} are equal.
- 5 years, 8 months ago
I do not immediately see how we get $k_k = 0 \Rightarrow x_k = x_{k+1}$ in the first scenario. I think we get $x_{k} = x_{k+2}$ instead, but that isn't immediately useful.
@Daniel Liu You are nearly there. Let me phrase the rest of the question in a different way. You start at some point $X$ on the real number line. You take $n$ odd steps of size exactly $S$. If you end back at $X$, show that $S = 0$.
Staff - 5 years, 9 months ago
We want another polynomial sprint! Soon..
- 5 years, 8 months ago
I guess We can write any polynomial a^n+b^n as (a-b)(a^n-1+a^n-2 b^1. .... Till b^n-1) therefore its always divisible by a-b
- 5 years, 9 months ago
Where can I get the book?
- 5 years, 9 months ago
- 5 years, 9 months ago
- 5 years, 9 months ago | {
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Where can I get the book?
- 5 years, 9 months ago
- 5 years, 9 months ago
- 5 years, 9 months ago
1)a^n-b^n=(a-b)(a^(n-1)+a^(n-2) b+...+b^(n-1)) Which leads us to:a-b/a^n-b^n 2) let f be a polynomial f (x)=an.x^n+an-1.x^(n-1)+...+a0 We can apply question1 since it is true for any n For j {0, n}: a-b / a^j-b^j We can sum Leads us to:a-b/f (a)-f (b) 5) x1-x2 =k1 (xn-x1) x2-x3=k2 (x1-x2) . . . xn-1-xn=kn (xn-2 -xn-1) Leads us to ki=1 or =-1 If one ki=-1 We have xi are equals If all ki=1 (x1-x2)+(x2-x3)+...+(xn-1 -xn)=(xn-x1)+ (x1-x2)+...+(xn-2 - xn-1) Leads us to: x1-xn = xn- xn-1 If just xi#xi-1 We have contradiction: (xi-xi-1)=0=(xi-1 -xi-2)#0 We can make recurrence(recurrance in french) We leads to an a contradiction each time Then k#1 Then all xi are equals
- 5 years, 9 months ago
1. Let $f(t)=t^n-b^n$. Then $f(b)=b^n-b^n=0$, so $t-b\mid t^n-b^n$. Let $t=a$ to get $a-b\mid a^n-b^n$.
2. Consider the $k$th coefficient of $f(a)-f(b)$: $a_k(a^k-b^k)$. Since $a-b\mid a^k-b^k$, $a-b$ is a factor of every term in $f(a)-f(b)\implies a-b\mid f(a)-f(b)$.
For the rest, I don't know which book you're talking about.
- 5 years, 9 months ago
Problem Solving Strategies by Arthur Engel. They might be selling online I don't really know. XD
- 5 years, 9 months ago
Google "arthur engel problem solving strategies", and follow the first link, which should be a pdf. The url is below but you have to surround the italicized text with underscores:
- 5 years, 8 months ago
- 5 years, 8 months ago
Oh oops, never mind the url.
- 5 years, 8 months ago
Book? What book?
- 5 years, 9 months ago
Google "arthur engel problem solving strategies", and follow the first link, which should be a pdf. The url is below but you have to surround the italicized text with underscores: http://f3.tiera.ru/2/MMathematics/MSchSchool-level/Engel%20A.%20Problem-solving%20strategies%20(for%20math%20olympiads)(Springer,%201998)(ISBN%200387982191)(O)(415s)MSch.pdf
- 5 years, 8 months ago | {
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- 5 years, 8 months ago
Oh oops, never mind the url.
- 5 years, 8 months ago
Thanks.. will do that. :)
- 5 years, 8 months ago
You may refer to the original post Let's Get Started to understand the context of the Polynomial Sprint set.
Staff - 5 years, 9 months ago | {
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# Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$ | {
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. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
4. Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$ | {
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Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
6. ## ok..
Originally Posted by masters
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
I got the answer from Soroban but I thank you for your effort.
7. ## yes...
Originally Posted by masters
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
Don't feel bad. No one is perfect except God and He is not taking any math courses lately. | {
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# Different results in DEigensystem compared to NDEigensystem for Laplacian eigenvalue problem (-Δu=λu) on unit square
I want to calculate the solution to the Laplacian eigenvalue problem on the unit square with trivial Dirichlet boundary conditions: $$- \Delta u(x,y) = \lambda u(x,y) \text{ on } {[0,1]}^2$$ with $$u(0,y)=0$$,$$u(1,y)=0$$,$$u(x,0)=0$$,$$u(x,1)=0$$.
However, Mathematica 12 reports different eigenfunctions when using NDEigensystem in contrast to DEigensystem using the following codes:
DEigensystem version:
{vals, funs} =
DEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈ Rectangle[], 2];
Table[ContourPlot[funs[[i]], {x, y} ∈ Rectangle[],
PlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> "Minimal",
Axes -> True], {i, Length[vals]}]
NDEigensystem version:
{vals, funs} =
NDEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈ Rectangle[], 2,
Method -> {"PDEDiscretization" -> {"FiniteElement",
"MeshOptions" -> {"MaxCellMeasure" -> 0.0001}}}];
Table[ContourPlot[funs[[i]], {x, y} ∈ Rectangle[],
PlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> "Minimal",
Axes -> True], {i, Length[vals]}]
For the second eigenfunction, the DEigensystem reports the classical textbook eigenfunction, while the numerical solution with NDEigensystem is fundamentally different, although the mesh discretization is set to a very small value.
Why is that? | {
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Why is that?
• For the lowest state, they're just negatives of each other, so that's fine. The second state is doubly degenerate, and so I'm betting that the second state in the second code is a linear combination of the two degenerate states from the first code. – march Aug 20 '20 at 20:59
• @march is right. Just observe the graphics of first 3 eigenfunctions and you'll know what happens. – xzczd Aug 21 '20 at 3:22
• And, of course, the problem will go away if you work on a rectangular domain such as $[0,1] \times [0,1.1]$. It would be interesting to see how close the last of these numbers can get to 1 before NDEigensystem treats them as degenerate & starts showing different results. – Michael Seifert Aug 24 '20 at 19:29
As already pointed out in the comments by @march and @xzczd, the second lowest state with eigenvalue $$\lambda_{1,2} = \lambda_{2,1} = 5 \pi^2$$ is doubly degenerate.
DEigensystem
NDEigensystem
This means that the corresponding eigenfunctions are not only determined up to a scaling (as for the lowest state). They are rather determined to be some orthogonal basis of the eigenspace
$$E_{5 \pi^2} = \{a \phi_{1,2} + b \phi_{2,1} \mid a,b \in \mathbb{R}, \, -\Delta \phi_{1,2} = 5 \pi^2 \phi_{1,2}, \, -\Delta \phi_{2,1} = 5 \pi^2 \phi_{2,1}, \, \phi_{1,2} \perp \phi_{2,1}\}$$
We have $$\text{dim}(E_{5 \pi^2}) = 2$$. The results from NDEigensystem ($$\phi_{1,2,\text{ND}}, \phi_{2,1,\text{ND}}$$) are therefore also valid solutions because they span the same eigenspace:
$$E_{5 \pi^2} = \text{span}\{\phi_{1,2,\text{NDEigen}}, \phi_{2,1,\text{NDEigen}}\} \\ = \text{span}\{\phi_{1,2,\text{DEigen}}, \phi_{2,1,\text{DEigen}}\} \\ = \text{span}\{\sin(1 \pi x)\sin(2 \pi y), \sin(2 \pi x)\sin(1 \pi y)\}$$ | {
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# Mean Value Theorem and Inequality.
Using the mean value theorem prove the below inequality.
$$\frac{1}{2\sqrt{x}} (x-1)<\sqrt{x}-1<\frac{1}{2}(x-1)$$ for $x > 1$.
I don't understand how these inequalities are related. Am I supposed to work out the first one and then the second and so on? I also would be really grateful if anyone had the time to give some insight in what this problem asks to me.
I really wish someone could give a very simple solution.
Apply the mean value theorem to the function $f(t) = \sqrt{t}$ on the interval $[1,x]$ to deduce
$$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$$
for some $c \in (1,x)$. Use the fact that
$$\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$$
to conclude.
• how did you know $$f(t)=\sqrt{t}$$? can you show me a different approach too? if possible. – Sherlock Homies May 30 '15 at 21:27
• @SherlockHomies note $\sqrt{x} - 1 = \sqrt{x} - \sqrt{1} = f(x) - f(1)$, where $f(t) = \sqrt{t}$. If you want to prove the inequalities without the use of MVT, then rationalize the numerator to get $$\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$$ and use the fact that for $x > 1$, $$\frac{1}{2\sqrt{x}} < \frac{1}{\sqrt{x} + 1} < \frac{1}{2}.$$ – kobe May 30 '15 at 21:30
• wait a sec how would you piece up all together to get to end of the problem. Like in the question above . I mean the last step.Using the mean value theorem again in the end? – Sherlock Homies May 30 '15 at 21:49
• @SherlockHomies multiply the inequalities $\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$ by $x - 1$ and use the fact $\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$ to get the desired inequalities. – kobe May 30 '15 at 21:52 | {
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By the MVT, there is $c \in ]1,x[$ such that $$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x-1), \qquad \qquad \left[f(b)-f(a) = f'(c)(b-a)\right]$$ so you use that: $$c < x \implies \sqrt{c} < \sqrt{x} \implies 2 \sqrt{c} < 2\sqrt{x} \implies \frac{1}{2\sqrt{x}}<\frac{1}{2\sqrt{c}}$$ to get one side, and use that: $$c > 1 \implies \sqrt{c} > 1 \implies 2\sqrt{c} > 2 \implies \frac{1}{2\sqrt{c}}< \frac{1}{2}$$ to get the other. | {
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Non-isomorphic graphs with degree sequence $1,1,1,2,2,3$. What if the degrees of the vertices in the two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4, for example)? Second, the transfer vertex equation is established to synthesize 2-DOF rotation graphs. (c)Find a simple graph with 5 vertices that is isomorphic to its own complement. Note − In short, out of the two isomorphic graphs, one is a tweaked version of the other. We know that a tree (connected by definition) with 5 vertices has to have 4 edges. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. 5.1.8. The graph defined by V = {a,b,c,d,e} and E = {{a,c},{6,d}, {b,e},{c,d), {d,e}} ii. Find all non-isomorphic trees with 5 vertices. For example, these two graphs are not isomorphic, G1: • • • • G2: • • • • since one has four vertices of degree 2 and the other has just two. (b) Draw all non-isomorphic simple graphs with four vertices. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Constructing non-isomorphic signless Laplacian cospectral graphs. Solution: Since there are 10 possible edges, Gmust have 5 edges. All simple cubic Cayley graphs of degree 7 were generated. In particular, ( x − 1 ) 3 x {\displaystyle (x-1)^{3}x} is the chromatic polynomial of both the claw graph and the path graph on 4 vertices. For all the graphs on less than 11 vertices I've used the data available in graph6 format here. Yes. We have also produced numerous examples of non-isomorphic signless Laplacian cospectral graphs. Finally, edge level equation is established to synthesize 2-DOF displacement graphs. Show that two projections of the Petersen graph are isomorphic. I would like to iterate over all connected non isomorphic graphs and test some properties. By continuing you agree to the use of cookies. However, the existing synthesis methods | {
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properties. By continuing you agree to the use of cookies. However, the existing synthesis methods mainly focused on 1-DOF PGTs, while the research on the synthesis of multi-DOF PGTs is very limited. Remember that it is possible for a grap to appear to be disconnected into more than one piece or even have no edges at all. We use cookies to help provide and enhance our service and tailor content and ads. The list does not contain all graphs with 8 vertices. ... consist of a non-empty independent set U of n vertices, and a non-empty independent set W of m vertices and have an edge (v,w) … However, notice that graph C also has four vertices and three edges, and yet as a graph it seems di↵erent from the first two. Two non-isomorphic trees with 7 edges and 6 vertices.iv. This thesis investigates the generation of non-isomorphic simple cubic Cayley graphs. 1.2 14 Two non-isomorphic graphs a d e f b 1 5 h g 4 2 6 c 8 7 3 3 Vertices: 8 Vertices: 8 Edges: 10 Edges: 10 Vertex sequence: 3, 3, 3, 3, 2, 2, 2, 2. There are several such graphs: three are shown below. There will be only one non isomorphic graph with 8 vertices and each vertex has degree 5. because 8 vertices with each vertex degree 5 means total degre view the full answer. For example, the parent graph of Fig. For example, all trees on n vertices have the same chromatic polynomial. List all non-identical simple labelled graphs with 4 vertices and 3 edges. Two non-isomorphic trees with 5 vertices. You Should Not Include Two Graphs That Are Isomorphic. For higher number of vertices, these graphs can be generated by a number of theorems and procedures which we shall discuss in the following sections. Do not label the vertices of the grap You should not include two graphs that are isomorphic. Both 1-DOF and multi-DOF planetary gear trains (PGTs) have extensive application in various kinds of mechanical equipment. I About (a) Draw All Non-isomorphic Simple Graphs With Three Vertices. A Google search shows that a paper by | {
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(a) Draw All Non-isomorphic Simple Graphs With Three Vertices. A Google search shows that a paper by P. O. de Wet gives a simple construction that yields approximately $\sqrt{T_n}$ non-isomorphic graphs of order n. Isomorphic Graphs. One example that will work is C 5: G= ˘=G = Exercise 31. Figure 10: Two isomorphic graphs A and B and a non-isomorphic graph C; each have four vertices and three edges. Now I would like to test the results on at least all connected graphs on 11 vertices. The line graph of the complete graph K n is also known as the triangular graph, the Johnson graph J(n, 2), or the complement of the Kneser graph KG n,2.Triangular graphs are characterized by their spectra, except for n = 8. Question: Exercise 8.3.3: Draw All Non-isomorphic Graphs With 3 Or 4 Vertices. Draw all possible graphs having 2 edges and 2 vertices; that is, draw all non-isomorphic graphs having 2 edges and 2 vertices. Hello! Do Not Label The Vertices Of The Graph. A complete bipartite graph with at least 5 vertices.viii. With 4 vertices (labelled 1,2,3,4), there are 4 2 How many of these are not isomorphic as unlabelled graphs? By continuing you agree to the use of cookies. But still confused between the isomorphic and non-isomorphic $\endgroup$ – YOUSEFY Oct 21 '16 at 17:01 A bipartitie graph where every vertex has degree 3. iv. Copyright © 2021 Elsevier B.V. or its licensors or contributors. Their degree sequences are (2,2,2,2) and (1,2,2,3). In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. A bipartitie graph where every vertex has degree 5.vii. Therefore, a large class of graphs are non-isomorphic and Q-cospectral to their partial transpose, when number of vertices is less then 8. An element a i, j of the adjacency matrix equals 1 if vertices i and j are adjacent; otherwise, it equals 0. These can be used to show two graphs are not isomorphic, but can not show that two graphs are isomorphic. 3(b). | {
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used to show two graphs are not isomorphic, but can not show that two graphs are isomorphic. 3(b). https://doi.org/10.1016/j.disc.2019.111783. And that any graph with 4 edges would have a Total Degree (TD) of 8. iii. So, it follows logically to look for an algorithm or method that finds all these graphs. 5.1.10. The Whitney graph theorem can be extended to hypergraphs. The synthesis results of 8- and 9-link 2-DOF PGTs are new results that have not been reported. A graph with degree sequence (6,2,2,1,1,1,1) v. A graph that proves that in a group of 6 people it is possible for everyone to be friends with exactly 3 people. The sequence of number of non-isomorphic graphs on n vertices for n = 1,4,5,8,9,12,13,16... is as follows: 1,1,2,10,36,720,5600,703760,...For any graph G on n vertices the below construction produces a self-complementary graph on 4n vertices! $\endgroup$ – user940 Sep 15 '17 at 16:56 Isomorphic graphs have the same chromatic polynomial, but non-isomorphic graphs can be chromatically equivalent. The synthesis results of 8- and 9-link 2-DOF PGTs, to the best of our knowledge, are new results that have not been reported in literature. 8 vertices - Graphs are ordered by increasing number of edges in the left column. (a) Draw all non-isomorphic simple graphs with three vertices. If all the edges in a conventional graph of PGT are assumed to be revolute edges, the derived graph is its parent graph. Copyright © 2021 Elsevier B.V. or its licensors or contributors. First, non-fractionated parent graphs corresponding to each link assortment are synthesized. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. To show graphs are not isomorphic, we need only nd just one condition, known to be necessary for isomorphic graphs, which does not hold. 1(b) is shown in Fig. Isomorphic and Non-Isomorphic Graphs - Duration: 10:14. The isomorphism of these two different presentations can be seen fairly easily: pick Looking at the | {
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The isomorphism of these two different presentations can be seen fairly easily: pick Looking at the documentation I've found that there is a graph database in sage. edge, 2 non-isomorphic graphs with 2 edges, 3 non-isomorphic graphs with 3 edges, 2 non-isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. As an example of a non-graph theoretic property, consider "the number of times edges cross when the graph is drawn in the plane.'' By Our constructions are significantly powerful. $\endgroup$ – mahavir Feb 22 '14 at 3:14 $\begingroup$ @mahavir This is not true with 4 vertices and 2 edges. Regular, Complete and Complete This paper presents an automatic method to synthesize non-fractionated 2-DOF PGTs, free of degenerate and isomorphic structures. 10:14. The transfer vertex equation and edge level equation of PGTs are developed. graph. There is a closed-form numerical solution you can use. Is it possible for two different (non-isomorphic) graphs to have the same number of vertices and the same number of edges? Use the options to return a count on the number of isomorphic classes or a representative graph from each class. For example, both graphs are connected, have four vertices and three edges. © 2019 Elsevier B.V. All rights reserved. WUCT121 Graphs 32 1.8. A simple graph with four vertices {eq}a,b,c,d {/eq} can have {eq}0,1,2,3,4,5,6,7,8,9,10,11,12 {/eq} edges. (a)Draw the isomorphism classes of connected graphs on 4 vertices, and give the vertex and edge Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. Draw two such graphs or explain why not. They may also be characterized (again with the exception of K 8) as the strongly regular graphs with parameters srg(n(n − 1)/2, 2(n − 2), n − 2, 4). Answer. of edges are 0,1,2. Two non-isomorphic graphs with degree sequence (3, 3, 3, 3, 2, 2, 2, 2)v. A graph that is not | {
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0,1,2. Two non-isomorphic graphs with degree sequence (3, 3, 3, 3, 2, 2, 2, 2)v. A graph that is not connected and has a cycle.vi. A method based on a set of independent loops is presented to precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs. Since isomorphic graphs are “essentially the same”, we can use this idea to classify graphs. Distance Between Vertices and Connected Components - … 1/25/2005 Tucker, Sec. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. https://www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices 3(a) and its adjacency matrix is shown in Fig. The research is motivated indirectly by the long standing conjecture that all Cayley graphs with at least three vertices are Hamiltonian. In this article, we generate large families of non-isomorphic and signless Laplacian cospectral graphs using partial transpose on graphs. 5. (Start with: how many edges must it have?) So our problem becomes finding a way for the TD of a tree with 5 vertices to be 8, and where each vertex has deg ≥ 1. Solution. Also, I've counted the non-isomorphic for 7 vertices, it gives me 11 with the same technique as you explained and for 6 vertices, it gives me 6 non-isomorphic. by a single edge, the vertices are called adjacent.. A graph is said to be connected if every pair of vertices in the graph is connected. 1 , 1 , 1 , 1 , 4 An unlabelled graph also can be thought of as an isomorphic graph. Sarada Herke 112,209 views. A method based on a set of independent loops is presented to detect disconnection and fractionation. An automatic method is presented for the structural synthesis of non-fractionated 2-DOF PGTs. Find three nonisomorphic graphs with the same degree sequence (1,1,1,2,2,3). More than 70% of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of vertices is ≤8. Their edge connectivity is retained. But as to the | {
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partial transpose when number of vertices is ≤8. Their edge connectivity is retained. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. For an example, look at the graph at the top of the first page. Two graphs G 1 and G 2 are said to be isomorphic if − Their number of components (vertices and edges) are same. $\begingroup$ with 4 vertices all graphs drawn are isomorphic if the no. Two graphs with different degree sequences cannot be isomorphic. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Automatic structural synthesis of non-fractionated 2-DOF planetary gear trains, https://doi.org/10.1016/j.mechmachtheory.2020.104125. Previous question Next question Transcribed Image Text from this Question. The NonIsomorphicGraphs command allows for operations to be performed for one member of each isomorphic class of undirected, unweighted graphs for a fixed number of vertices having a specified number of edges or range of edges. In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v.Otherwise, they are called disconnected.If the two vertices are additionally connected by a path of length 1, i.e. • Isomorphic Graphs ... Graph Theory: 17. We use cookies to help provide and enhance our service and tailor content and ads. Figure 5.1.5. The Whitney graph isomorphism theorem, shown by Hassler Whitney, states that two connected graphs are isomorphic if and only if their line graphs are isomorphic, with a single exception: K 3, the complete graph on three vertices, and the complete bipartite graph K 1,3, which are not isomorphic but both have K 3 as their line graph. Graph from each class © 2021 Elsevier B.V. or its licensors or.... The existing synthesis methods mainly focused on 1-DOF PGTs, while the research is indirectly... To nine links is automatically generated and isomorphic structures tree ( connected by | {
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indirectly... To nine links is automatically generated and isomorphic structures tree ( connected by definition ) with 5 vertices is! The first page research on the number of edges in the left column, look at top... Logically to look for an algorithm or method that finds all these graphs if the no that... Various kinds of mechanical equipment the top of the grap you Should Include... Draw all non-isomorphic simple cubic Cayley graphs with 3 or 4 vertices ( labelled )... Gmust have 5 edges documentation I 've found that there is a registered trademark of Elsevier B.V. its... A count on the number of edges different ( non-isomorphic ) graphs to the... But can not be isomorphic the results on at least all connected non graphs... Do not label the vertices of the Petersen graph are isomorphic first page link assortment synthesized. Of edges in the left column the top of the Petersen graph isomorphic! Connected non isomorphic graphs a and B and a non-isomorphic graph C each... ( Start with: how many edges must it have? each link are... Complete and Complete two graphs that are isomorphic an unlabelled graph also be! Work is C 5: G= ˘=G = Exercise 31 is it possible for two different ( non-isomorphic ) to! Possible for two different ( non-isomorphic ) graphs to have 4 edges would have a Total degree TD. 1-Dof PGTs, while the research on the synthesis of multi-DOF PGTs is very limited up to nine links automatically... Indirectly by the long standing conjecture that all Cayley graphs same degree sequence ( 1,1,1,2,2,3 ) must have. Trains ( PGTs ) have extensive application in various kinds of mechanical equipment mainly focused on PGTs. And rotation graphs vertex equation is established to synthesize 2-DOF rotation graphs of mechanical.... 4 2 Hello the other labelled 1,2,3,4 ), non isomorphic graphs with 8 vertices are several such graphs: three are shown below presented! Are synthesized equation and edge level equation is established to synthesize 2-DOF rotation.. Four | {
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Are synthesized equation and edge level equation is established to synthesize 2-DOF rotation.. Four vertices and three edges of edges of PGTs are new results that have not been reported Next question Image! Their degree sequences can not show that two projections of the Petersen graph are.! Sequences can not show that two projections of the two isomorphic graphs, one is a tweaked of... Isomorphic to its own complement are synthesized if the no 9-link 2-DOF PGTs are developed graph at the documentation 've... The options to return a count on the number of vertices and three edges idea!, both graphs are isomorphic all Cayley graphs of any given order not as much said! 3. iv a ) Draw all non-isomorphic graphs of degree 7 were generated of vertices is ≤8 PGTs very. 8- and 9-link 2-DOF PGTs with up to nine links is automatically generated extensive application in various kinds of equipment. Agree to the construction of all the non-isomorphic graphs having 2 edges and 2 vertices ; that is Draw. Found that there is a graph database in sage to show two graphs are connected, have vertices.: Exercise 8.3.3: Draw all non-isomorphic graphs having 2 edges and 2 vertices ; that is isomorphic its... $\begingroup$ with 4 vertices all graphs drawn are isomorphic to links... Exercise 31 degree 5.vii isomorphic as unlabelled graphs matrix is shown in Fig this article, can. 2-Dof displacement graphs all the graphs on less than 11 vertices I 've found that there is registered! More than 70 % of non-isomorphic signless Laplacian cospectral graphs polynomial, but non-isomorphic graphs with three vertices synthesis. Degree ( TD ) of 8 graphs, one is a closed-form numerical solution you can use idea! Have four vertices and the same ”, we can use loops presented! Results of 8- and 9-link 2-DOF PGTs are developed link assortment are synthesized a registered trademark of Elsevier B.V. non-isomorphic! Four vertices and fractionation graphs to have the same chromatic polynomial, but can not show | {
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Four vertices and fractionation graphs to have the same chromatic polynomial, but can not show two! Of edges in the left column matrix is shown in Fig planetary gear trains ( )! 5 vertices.viii and ( 1,2,2,3 ), non-fractionated parent graphs and rotation graphs and its adjacency matrix shown! ) graphs to have the same number of vertices and the same chromatic polynomial, out of the other trains! Mainly focused on 1-DOF PGTs, free of degenerate and isomorphic structures a... Looking at the top of the other 4 vertices has to have the same,... To precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs each class Petersen graph are.! Standing conjecture that all Cayley graphs of degree 7 were generated three edges graphs using partial transpose on graphs edges. Vertex equation is established to synthesize non-fractionated 2-DOF PGTs, while the research is motivated indirectly by long... − in short, out of the other, it follows logically to look for an algorithm method. Equation and edge level equation of PGTs are developed use the options to return a on. Both 1-DOF and multi-DOF planetary gear trains ( PGTs ) have extensive application in kinds. Edges would have a Total degree ( TD ) of 8 list all non-identical labelled... Mechanical equipment of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of isomorphic or... Constructing non-isomorphic signless Laplacian cospectral graphs using partial transpose on graphs labelled 1,2,3,4 ), there are possible. Does not contain all graphs with three vertices a ) Draw all non-isomorphic simple graphs 4... List all non-identical simple labelled graphs with three vertices are Hamiltonian sequence ( ). On less than 11 vertices Complete bipartite graph with 4 vertices ( 1,2,3,4. Vertices have the same number of isomorphic classes or a representative graph from each class degree 5.vii the isomorphic! List does not contain all graphs with | {
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graph from each class degree 5.vii the isomorphic! List does not contain all graphs with three vertices the left column I would like to test the results at! All possible graphs having 2 edges and 2 vertices ; that is, Draw all non-isomorphic simple with... Graph are isomorphic n vertices have the same number of vertices is ≤8 for different. 'Ve used the data available in graph6 format here non isomorphic graphs with 8 vertices including parent graphs and test some properties B a! Look at the graph at the top of the other have 5 edges Complete bipartite with. Is isomorphic to its own complement edge level equation of PGTs are developed simple graph with at all! Grap you Should not Include two graphs are connected, have four vertices and edges! Found that there is a closed-form numerical solution you can use this to. Article, we generate large families of non-isomorphic simple graphs with 8 vertices indirectly by the standing... Short, out of the two isomorphic graphs, one is a graph database in sage any with... 2 edges and 2 vertices ; that is, Draw all non-isomorphic graphs having 2 edges and vertices... Exercise 8.3.3: Draw all non-isomorphic simple graphs with the same degree (! ˘=G = Exercise 31 ), there are 4 2 Hello than 70 % non-isomorphic! At the graph at the graph at the top of the other available graph6! Its licensors or contributors \begingroup \$ with 4 edges would have a Total degree ( TD ) of 8 tailor! Several such graphs: three are shown below trains ( PGTs ) have application... Example, look at the graph at the top of the grap you Should not Include two graphs that isomorphic... The left column ) graphs non isomorphic graphs with 8 vertices have 4 edges would have a Total degree ( ). C ) Find a simple graph with 4 vertices ( labelled 1,2,3,4 ) there... There is a closed-form numerical solution you can use vertices that is isomorphic to its own complement 2021! Shown in Fig research on the number of edges based on a set independent! Figure 10: two | {
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2021! Shown in Fig research on the number of edges based on a set independent! Figure 10: two isomorphic graphs a and B and a non-isomorphic graph C ; each four! Presented for the structural synthesis of multi-DOF PGTs is very limited article, we generate families. 2 Hello were generated each link assortment are synthesized least three vertices parent graphs to. Laplacian cospectral graphs using partial transpose on graphs some properties non-isomorphic signless Laplacian cospectral.. Degree sequences can not be isomorphic, have four vertices and three.. Must it have? have a Total degree ( TD ) of.! And 3 edges families of non-isomorphic signless-Laplacian cospectral graphs 7 were generated 8! Edges and 2 vertices with at least 5 vertices.viii vertices has to have 4 edges different ( )... An algorithm or method that finds all these graphs PGTs with up to nine links is generated... First, non-fractionated parent graphs corresponding to each link assortment are synthesized of Elsevier B.V. or its or! Any graph with 4 vertices, it follows logically to look for algorithm... Graphs can be thought of as an isomorphic graph B.V. Constructing non-isomorphic signless Laplacian cospectral graphs can be of... 1-Dof PGTs, while the research is motivated indirectly by the long conjecture... Continuing you agree to the use of cookies would like to iterate over all connected graphs less. An unlabelled graph also can be generated with partial transpose on graphs 5 vertices has to the. B ) Draw all non-isomorphic simple graphs with three vertices help provide enhance... Now I would like to test the results on at least all connected non isomorphic graphs are isomorphic. An automatic method to synthesize non-fractionated 2-DOF PGTs that a tree ( connected by definition ) with 5 vertices is... Is it possible for two different ( non-isomorphic ) graphs to have 4 edges Constructing non-isomorphic signless Laplacian graphs... | {
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# Math Help - Determine is a line is tangent to a curve or not?
1. ## Determine is a line is tangent to a curve or not?
Sorry if it sounds silly but I am thinking if there is a way to find out if a line is a tanget of a curve?
Say given a curve of $x^2 + 3xy + y^2 + 4 = 0$, how do I determine if any one given line is a tangent of this curve or not? The line can be y=x+7 or y=-2x-4, etc. And if it is a tangent of the curve, how do I find the points that this line is tangent to the curve?
I am thinking that I differentiate the equation of the curve to have dy/dx. Then I can get the gradient at any one point of the curve to compare with a line. But this isn't enough because same gradient doesn't mean is a tangent to the curve. And how I find the points which the line is tangent to the curve without any given points at all?
Thanks for any help!
2. Do a simultaneous equation to determine first whether the line meet the curve or not.
Once done, you need to find if the gradient at the point of intersection is the same as the gradient of the line.
If they meet and share a common gradient, you have your tangent. If one of those two conditions are not met, the line is not a gradient.
3. ## q.[/math]
Hello, xEnOn!
I have a method but it's quite long.
Maybe someone has a shorter approach?
Is there a way to find out if a line is a tanget to a curve?
Say, given a curve: . $x^2 + 3xy + y^2 + 4 \:=\: 0$
how do I determine if, say, $y \:=\:x+6$ is a tangent of this curve or not?
And if it is a tangent of the curve, how do I find the points of tangency?
I am thinking that I differentiate the equation of the curve to have dy/dx.
Then I can get the gradient at any one point of the curve to compare with the line.
But this isn't enough because same gradient doesn't mean it is tangent to the curve.
And how I find the points which the line is tangent without any given points at all? | {
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We have: . $\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}$
Find the points of intersection, substitute [2] into [1]:
. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0$
. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0$
. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$
. . $(x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4$
Substitute into [2]: . $y \:=\:4,\:2$
The hyperbola and the line intersect at: . $P(-2,4)\,\text{ and }\,Q(-4,2)$
If the line is tangent to the hyperbola, their slopes will be equal at $\,P$ and $Q.$
The slope of the line is: . $\boxed{m \,=\,1}$
Find the slope of the hyperbola.
Differentiate implicitly: . $2x + 3x\dfrac{dy}{dx} + 3y + 2y\dfrac{dy}{dx} \:=\:0$
. . $(3x + 2y)\dfrac{dy}{dx} \:=\:-2(x+y) \quad\Rightarrow\quad \boxed{\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(x+y)}{3x+2y} }$
$\text{At }P(\text{-}2,4)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}2+4)}{3(\text{-}2) + 2(4)} \:=\:\dfrac{\text{-}4}{2} \:=\:-2 \;\ne \;1$
$\text{At }Q(\text{-}4,2)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}4+2)}{3(\text{-}4)+2(2)} \:=\:\dfrac{4}{\text{-}8} \:=\:-\dfrac{1}{2} \;\ne\;1$
The slopes are not equal at the intersection points.
Therefore, the line is not tangent to the hyperbola.
Edit: Too slow again!
. . . .Unknown008 already gave an excellent game plan.
4. Originally Posted by Soroban
I have a method but it's quite long.
Maybe someone has a shorter approach?
We have: . $\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}$
Find the points of intersection, substitute [2] into [1]:
. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0$
. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0$
. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$ | {
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. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$
. . $(x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4$
A somewhat shorter method is to follow Soroban's approach by substituting the line equation into the conic equation, as above, and finding the values of x where the line meets the conic. These turned out to be –2 and –4. Those values are different, so you can tell straight away that the line cannot be a tangent to the conic. If it were, then there would be a repeated root for x (because the point at which the line touches the conic would be a double root of the equation). As it is, the line crosses the conic at two distinct points, so it cannot touch it.
Suppose that we have the conic equation $x^2 + 3xy + y^2 + 4 = 0$, as before, but that the equation of the line is $4x+y+4=0$. Substitute $y = -4x-4$ into the conic equation, as in Soroban's method. You get
$x^2 + 3x(-4x-4) + (-4x-4)^2 + 4 = 0$,
$x^2 - 12x^2 - 12 + 16x^2 + 32x + 16 + 4 = 0$,
$5x^2 + 20x + 20 = 0$,
$x^2+4x+4=0$,
$(x+2)^2 = 0$.
This time, there is a repeated root x = –2, which tells you that the line touches the conic at that value of x. So in that case, the line is tangent to the conic.
5. By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?
I suppose I will still get a result in someway no matter when I do simultaneous?
Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation. | {
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6. Originally Posted by xEnOn
By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?
If you're talking about binomials, (x+a)(x+b): If the simultaneous equation gives 0 or 2 answers, then it is not tangent.
In other cases, it is quite lengthy - you have to find the co-ordinates of where they meet and then check if their gradients are the same for both equations at that point.
You cannot easily determine whether it is tangent or not, just from the intersection points. (Unless there are no intersection points)
Originally Posted by xEnOn
I suppose I will still get a result in someway no matter when I do simultaneous?
No. Well not real results anyway (but who wants to get into imaginary numbers?). Sometimes when you do simultaneous equations, you won't get any answers and this will tell you straight away that the 2 graphs don't meet.
Originally Posted by xEnOn
Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation. | {
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# A certain stock exchange designates each stock with a one-,
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A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A. 2951
B. 8125
C. 15600
D. 16302
E. 18278
OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-stock-exchange-designates-each-stock-with-a-86656.html
[Reveal] Spoiler: OA
Last edited by Bunuel on 20 Jul 2013, 09:43, edited 1 time in total.
Edited the question.
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Remember that Numbers can be repeated.
Number of 1 Digit Codes = 26
Number of 2 Digit Codes = 26 * 26 = 676
Number of 2 Digit Codes = 26 * 26 * 26 = 17576
Total = 26 + 676 + 17576 = 18278
In future do not put OA with the question. As people might solve it back wards to reach the solution.
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You have 3 different sets of Permutations available.
First set is of stocks that have only 1 letter, or 26 possibilities
Second set is of stocks with 2 letters, or 26 * 26 possibilities
Third set is of stocks with 3 letters, or 26 * 26 * 26 possibilities.
This equal 26 +676 + 17576 = 18,278.
japped187 wrote:
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A) 2951
B) 8125
C) 15600
D) 16302
E) 18278
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abhijit_sen wrote:
In future do not put OA with the question. As people might solve it back wards to reach the solution.
I think the moderator/owner of the forum should add a "spoiler" function on the forum. That function is the perfect fit for this type of forum, but this forum does not have it.
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japped187 wrote:
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A) 2951
B) 8125
C) 15600
D) 16302
E) 18278
As discussed above, the solution is indeed 26 + 26*26 + 26*26*26. I noticed that all the digits were different in the options.
So I simply calculated the unit digit of each product. 6 +6 + 6. = 18 = Unit digit of answer = 8. Hence, E.
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20 Jul 2013, 09:45
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A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A. 2951
B. 8125
C. 15600
D. 16302
E. 18278
1 letter codes = 26
2 letter codes =2 6^2
3 letter codes = 26^3
Total=26+26^2+26^3
The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E. | {
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# What is the perimeter of a square with area 9p^2/16 ?
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What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
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What is the perimeter of a square with area 9p^2/16 ?
Area of square, (side)^2 = (3p/4)^2
Therefore side of the square = 3p/4
Perimeter of square = 4*side = 4* (3p/4) = 3p
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Hi All,
We're told that the area of a SQUARE is (9P^2)/16. We're asked for the PERIMETER of the square. This question can be solved by TESTing VALUES.
Since the denominator is 16, we can 'cancel' it out if we make P=4... | {
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Since the denominator is 16, we can 'cancel' it out if we make P=4...
IF... P=4, then the area of the square is 9(16)/16 = 9. This means that each side of the square = 3. Thus, we're looking for an answer that equals 4(3) = 12 when we plug P=4 into the answer choices. There's only one answer that matches...
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30 Nov 2017, 06:36
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
Find $$s$$.
Area of square = $$s^2 = \frac{9\pi^2}{16}$$
$$\sqrt{s^2} = \frac{(\sqrt{9})*(\sqrt{\pi^2})}{\sqrt{16}}$$
$$s= \frac{3\pi}{4}$$
Perimeter = $$4s=(\frac{3\pi}{4})*4 = 3\pi$$
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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30 Nov 2017, 07:04
Area = 9p^2/16
So a^2 = 9p^2/16 ( assumed side of a square as a)
So a = 3p/4
And perimeter of square is sum of all sides
So perimeter = 4a
So ans is 4(3p/4) = 3p
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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30 Nov 2017, 12:15
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Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3 | {
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A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
We know that the area of square = (side)^2
Thus, we can write :
$$(Side)^2 = \frac{9p^2}{16}$$
$$(Side)^2 = (\frac{3p}{4})^2$$
Since the side of square is always positive, we can conclude that the length of each side $$= \frac{3p}{4}$$
And the perimeter of the square $$= 4 * side = 4 * \frac{3p}{4} = 3p$$
The correct answer is Option C.
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
### Show Tags
03 Dec 2017, 18:47
1
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
Since a side of the square is equal to √(9p^2/16) = 3p/4, then the perimeter of the square is 4 x 3p/4 = 12p/4 = 3p.
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Re: What is the perimeter of a square with area 9p^2/16 ? &nbs [#permalink] 03 Dec 2017, 18:47
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What's new
# Miller - Chapter 4 - Distributions - Study Notes, Pg 75, question #2
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi David,
In question (2) referenced as above,
(2) At the start of the year, a stock price is $100. A twelve-step binomial model describes the stock price evolution such that each month the extreme volatility price with either jump from S(t) to S(t)*u with 60.0% probability or down to S(t)*d with 40.0% probability. The up jump (u) = 1.1 and the down jump (d) = 1/1.1; note these (u) and (d) parameters correspond to an annual volatility of about 33% as exp[33%*SQRT(1/12)] ~= 1.10. At the end of the year, which is nearest to the probability that the stock price will be exactly$121?
(a) 0.33%
(b) 3.49%
(c) 12.25%
(d) 22.70%
There are 13 outcomes at the end of the 12-step binomial, with $100 as the outcome that must correspond to six up jumps and six down jumps. Therefore,$121 must be the outcome due to seven up jumps and five down jumps: $100*1.1^7*(1/1.1)^5 =$121
Such that we want the binomial probability given by:
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%
Clarification
Although, intuitively, 13 outcomes at the end of the 12-step binomial seems right (ie expected value of stock price at the end of 12 steps ??), how do you figure out six up jumps and six down jumps from the Binomial model to get \$100 as the outcome? Also how do you get :
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%
I guess I am missing something that I should understand.
Thanks!
Jayanthi
#### ShaktiRathore
##### Well-Known Member
Subscriber
Hi
Consider n up jumps and 12-n down jumps to get 121 in the end
Therefore,100*(1.1)^n*(1/1.1)^12-n=121=>1.1^(2n-12)=1.1^2=>2n-12=2=>2n=14=>n=7 therefore there are 7 up and 12-7=5 down jumps.
Binomial probability is 12C7(.6)^7*(.4)^5=22.69%
Thanks
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi Shakti,
Yes - that is indeed great! Thanks a tonne for making it so simple to understand!
Jayanthi | {
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Yes - that is indeed great! Thanks a tonne for making it so simple to understand!
Jayanthi
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @Jayanthi Sankaran the source question thread contains a visual that should help clarify, too. Please see https://www.bionicturtle.com/forum/...ributions-i-miller-chapter-4.7025/#post-28837
ie..,
#### Dr. Jayanthi Sankaran
##### Well-Known Member
This is neat, David
Thanks!
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi David,
Does the above visual come with its own spreadsheet?
Thanks
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Thanks David - will go through it, at leisure
Warm regards
Jayanthi
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!
#### eva maria | {
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#### eva maria
##### New Member
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!
Many thanks David
Last edited by a moderator: | {
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# Characterizing discontinuous derivatives
Apparently the set of discontinuity of derivatives is weird in its own sense. Following are the examples that I know so far:
$1.$ $$g(x)=\left\{ \begin{array}{ll} x^2 \sin(\frac{1}{x}) & x \in (0,1] \\ 0 & x=0 \end{array}\right.$$ $g'$ is discontinuous at $x=0$.
$2.$ The Volterra function defined on the ternary Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Cantor set ,that is on a nowhere dense set of measure zero.
$3.$ The Volterra function defined on the Fat-Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Fat-Cantor set ,that is on a set of positive measure, but not full measure.
$4.$ I am yet to find a derivative which is discontinuous on a set of full measure.
Questions:
1.What are some examples of functions whose derivative is discontinuous on a dense set of zero measure , say on the rationals?
2.What are some examples of functions whose derivative is discontinuous on a dense set of positive measure , say on the irrationals?
Update: One can find a function which is differentiable everywhere but whose derivative is discontinuous on a dense set of zero measure here. | {
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• The Volterra function is not defined just on the Cantor set, but the whole interval. It equals $0$ on the Cantor set. I know you know this. But perhaps "defined relative to the Cantor set" might be more accurate. – zhw. Feb 24 '18 at 17:32
• @zhw Yeah, but I think its pretty obvious though. :P – yasir Feb 24 '18 at 17:44
• "Some good discussion about this can be found here and here." The first link you gave gives several references to the fact that any $F_{\sigma}$ first category set (equivalently, any countable union of closed nowhere dense sets) is the discontinuity set of some derivative. Those references that give a proof of this result do so in a constructive way, in the sense that given any countable union of closed nowhere dense sets (note that $\mathbb Q$ is a countable union of singleton sets), a differentiable function is constructed whose derivative is discontinuous exactly on that countable union. – Dave L. Renfro Feb 24 '18 at 21:08
There is no everywhere differentiable function $f$ on $[0,1]$ such that $f'$ is discontinuous at each irrational there. That's because $f',$ being the everywhere pointwise limit of continuous functions, is continuous on a dense $G_\delta$ subset of $[0,1].$ This is a result of Baire. Thus $f'$ can't be continuous only on a subset of the rationals, a set of the first category.
But there is a differentiable function whose derivative is discontinuous on a set of full measure.
Proof: For every Cantor set $K\subset [0,1]$ there is a "Volterra function" $f$ relative to $K,$ which for the purpose at hand means a differentiable function $f$ on $[0,1]$ such that i)$|f|\le 1$ on $[0,1],$ ii) $|f'|\le 1$ on $[0,1],$ iii) $f'$ is continuous on $[0,1]\setminus K,$ iv) $f'$ is discontinuous at each point of $K.$
Now we can choose disjoint Cantor sets $K_n \subset [0,1]$ such that $\sum_n m(K_n) = 1.$ For each $n$ we choose a Volterra function $f_n$ as above. Then define
$$F=\sum_{n=1}^{\infty} \frac{f_n}{2^n}.$$ | {
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$$F=\sum_{n=1}^{\infty} \frac{f_n}{2^n}.$$
$F$ is well defined by this series, and is differentiable on $[0,1].$ That's because each summand above is differentiable there, and the sum of derivatives converges uniformly on $[0,1].$ So we have
$$F'(x) = \sum_{n=1}^{\infty} \frac{f_n'(x)}{2^n}\,\, \text { for each } x\in [0,1].$$
Let $x_0\in \cup K_n.$ Then $x_0$ is in some $K_{n_0}.$ We can write
$$F'(x) = \frac{f_{n_0}'(x)}{2^{n_0}} + \sum_{n\ne n_0}\frac{f_n'(x)}{2^n}.$$
Now the sum on the right is continuous at $x_0,$ being the uniform limit of functions continuous at $x_0.$ But $f_{n_0}'/2^{n_0}$ is not continuous at $x_0.$ This shows $F'$ is not continuous at $x_0.$ Since $x_0$ was an aribtrary point in $\cup K_n,$ $F'$ is discontinuous on a set of full measure as desired.
• @ zhw So, if $f'$ is discontinuous on a set $D$, then $D$ can be dense only if its Lebesgue measure is zero? – yasir Feb 25 '18 at 3:04
• No, certainly not. A set of full measure is dense. – zhw. Feb 25 '18 at 7:09
• @ zhw. Is the set on which $F'$ is continuous dense, as per your claim in first paragraph? – yasir Feb 25 '18 at 9:26
• @yasir Yes, it has to be. – zhw. Feb 25 '18 at 17:14 | {
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Small Solver Program
02-14-2019, 05:25 AM
Post: #1
Gamo Senior Member Posts: 518 Joined: Dec 2016
Small Solver Program
Recently I try to make a small but versatile solver program specifically
for HP Voyager Series calculator.
This solver solve for f(x) = 0 and use some type of counter for iteration.
Due to the iteration process this program will run slow on Original Calculator.
With modern 12C, 15C, and all other emulator this will run extremely fast.
This program is intended for HP-11C but this example will use
HP-15C emulator so that this solver program result can be compare
with the 15C build-in SOLVE function.
---------------------------------------------------------
To run program.
Enter 2 guesses closest to the root.
X1 [ENTER] X2 [A] display answer // Use this Solver
X1 [ENTER] X2 f [SOLVE] [E] display answer // Use 15C build-in SOLVE
---------------------------------------------------------
Example for X^X = Y // X to the power of X equal to Y
X on R1 // Register Storage 1 // For this Sover
Y on R5 // Register Storage 5
X on R2 // Register Storage 2 // For 15C SOLVE
Y on R5 // Register Storage 5
--------------------------------------------------------
Program:
Code:
LBL A // This Solver STO 0 Rv STO 1 90 STO I CLx LBL 1 RCL 0 / CHS RCL 1 + STO 1 DSE I GTO B RCL 1 RTN ------------------------------ LBL B // f(x) program for X^X = Y RCL 1 LN RCL 1 x RCL 5 LN - GTO 1 ------------------------------- LBL E // 15C SOLVE STO 2 // f(x) program for X^X = Y LN RCL 2 x RCL 5 LN - RTN
-------------------------------------------
Example: X^X = 1000
STORE 1000 to R5 // [STO] 5
This Solver
4 [ENTER] 6 f [A] display 4.555535704
-------------------------------------------------------
15C SOLVE
4 [ENTER] 6 f [SOLVE] [E] display 4.555535705
-------------------------------------------------------
Remark:
For this solver answer is in R1
For 15C SOLVE answer is in R2 | {
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Gamo
02-14-2019, 07:06 AM
Post: #2
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-14-2019 05:25 AM)Gamo Wrote: Example: X^X = 1000
You can rewrite this equation such that $$x$$ is a fixed point of $$f(x)$$, that is $$x=f(x)$$:
$$x=\frac{3}{LOG(x)}$$
Then you can iterate the following program with a starting value, say $$x=4$$:
Code:
001- 43 13 : LOG 002- 3 : 3 003- 34 : x<>Y 004- 10 : ÷
Example:
4
R/S
4.982892143
R/S
4.301189432
R/S
4.734933901
R/S
4.442378438
R/S
4.632377942
R/S
4.505830646
R/S
4.588735607
R/S
4.533824357
R/S
4.569933524
R/S
4.546075273
R/S
4.561789745
R/S
4.551417837
R/S
4.558254212
R/S
4.553744141
R/S
4.556717747
R/S
4.554756405
R/S
4.556049741
R/S
4.555196752
R/S
4.555759258
R/S
4.555388285
R/S
4.555632930
R/S
4.555471588
R/S
4.555577990
R/S
4.555507820
R/S
4.555554095
R/S
4.555523578
R/S
4.555543703
R/S
4.555530431
R/S
4.555539183
R/S
4.555533412
R/S
4.555537217
R/S
4.555534708
4.555535704
R/S
4.555535706
R/S
4.555535704
R/S
4.555535706
From then on the result flips between these last two values.
Cheers
Thomas
02-14-2019, 07:15 AM (This post was last modified: 02-14-2019 07:16 AM by Thomas Klemm.)
Post: #3
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
If you don't want to press R/S repeatedly:
Code:
001- 43 13 : LOG 002- 3 : 3 003- 34 : x<>Y 004- 10 : ÷ 005- 42 31 : PSE 006- 43 40 : x=0
02-15-2019, 12:07 AM (This post was last modified: 02-17-2019 12:16 PM by Albert Chan.)
Post: #4
Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
(02-14-2019 07:06 AM)Thomas Klemm Wrote: $$x=\frac{3}{LOG(x)}$$
4
R/S
4.982892143
R/S
4.301189432
...
It would be nice if we can temper the oscillation, or slow convergence.
Let x0 = 4, x1, x2 = the first two iterated values.
Rate = (x2-x1)/(x1-x0) ~ (4.30 - 4.98) / (4.98 - 4) = -0.694 | {
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Rate = (x2-x1)/(x1-x0) ~ (4.30 - 4.98) / (4.98 - 4) = -0.694
If the same trend continued, we expect final % = 1/(1-r) ~ 60%
Use weighted fixed-point equation x = 0.4x + 0.6 * 3/log10(x):
4.6
4.555924149
4.555537395
4.555535712
4.555535705 (converged)
Edit: compare with Newton's method, x = (ln(1000) + x) / (ln(x) + 1)
4
4.571001573
4.555546101
4.555535705 (converged)
02-15-2019, 06:26 PM
Post: #5
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-15-2019 12:07 AM)Albert Chan Wrote: It would be nice if we can temper the oscillation, or slow convergence.
We can also use Aitken's delta-squared process to accelerate the speed of convergence:
$$a_{n}=x_{n+2}-\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$
This leads to the following program for the HP-11C:
Code:
001-42,21, 1 : ▸LBL 01 002- 36 : ENTER 003- 32 0 : GSB 00 004- 30 : - 005- 43 36 : LSTx 006- 36 : ENTER 007- 32 0 : GSB 00 008- 30 : - 009- 43 36 : LSTx 010- 34 : x<>y 011- 43 33 : R↑ 012- 34 : x<>y 013- 30 : - 014- 43 36 : LSTx 015- 43 11 : x² 016- 34 : x<>y 017- 10 : ÷ 018- 30 : - 019- 43 32 : RTN 020-42,21, 0 : ▸LBL 00 021- 43 12 : LN 022- 45 0 : RCL 00 023- 34 : x<>y 024- 10 : ÷ 025- 43 32 : RTN
Example:
Solve: $$x^x=1000$$
1000 LN
STO 0
4.6
GSB 1
4.555665779
R/S
4.555535706
R/S
4.555535705
R/S
Error 0
Feel free to add a check for 0 after line 13 to prevent the error and loop back to label 1 instead of returning in line 19.
Cheers
Thomas
02-15-2019, 09:16 PM (This post was last modified: 02-15-2019 09:31 PM by Albert Chan.)
Post: #6
Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
(02-15-2019 06:26 PM)Thomas Klemm Wrote:
(02-15-2019 12:07 AM)Albert Chan Wrote: It would be nice if we can temper the oscillation, or slow convergence. | {
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We can also use Aitken's delta-squared process to accelerate the speed of convergence:
$$a_{n}=x_{n+2}-\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$
Amazingly, my rate formula is same as Aitken extrapolation formula !
Assuming we have 3 values, x0,x1,x2 and tried to guess where it should end up.
My rate analysis: r = (x2-x1)/(x1-x0) = Δx1 / Δx0
We need this for the proof:
(Δx0)²
= ((Δx0 - Δx1) + Δx1)²
= (Δx0 - Δx1)² + 2 * Δx1 (Δx0 - Δx1) + (Δx1)²
If same trend continue, where it will ends up
= x0 + Δx0 * 1/(1-r)
= x0 + (Δx0)² / (Δx0 - Δx1)
= x0 + (Δx0 - Δx1) + 2 * Δx1 + (Δx1)² / (Δx0 - Δx1)
= x0 + (x1-x0) + (x1-x2) + 2*(x2-x1) − (Δx1)² / (Δx1 - Δx0)
= x2 − (Δx1)² / (Δx1 - Δx0)
02-16-2019, 03:58 AM
Post: #7
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-15-2019 09:16 PM)Albert Chan Wrote: Amazingly, my rate formula is same as Aitken extrapolation formula !
That's not really a surprise, is it?
It's essentially the same idea: geometric series
$$1\,+\,r\,+\,r^{2}\,+\,r^{3}\,+\,\cdots \;=\;{\frac {1}{1-r}}$$
02-16-2019, 12:24 PM (This post was last modified: 02-16-2019 12:34 PM by Csaba Tizedes.)
Post: #8
Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-14-2019 05:25 AM)Gamo Wrote: a small but versatile solver program specifically for HP Voyager Series calculator.
Example for X^X = Y // X to the power of X equal to Y
Yes, the classic problem: which number you can power to itself to get the largest number on your calculator: x^x=10^100. Do LOG() of both sides: x×LOG(x)=100, then do the iteration: x_0:=50, then x_i+1:=100/log(x_i). The simple fixed point iteration and converges.
By the way, if you interested, here is my SOLVE(i) for HP-12C, which can solve any equation for ANY variable like on 15C, without rewriting the equation. Looks like indirect addressing without indirect addressing. 28 steps only. | {
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Yes, my English is terrible, but more convenient than read the full material in Hungarian.
At the end of pdf you can find lots of applications which solved with this little solver (like ODE solve with EULER+SOLVE(i) on 12C (!!!), pressure vessel pneumatic conveying pipe sizing program, humidity and dew point solving, outer temperature calculation of a concrete silo wall or comparison of present values of educated and not educated people salaries with the little SOLVE combined with 12C's cash flow calculation capabilities - I guess this is shows clearly the power of this little routine possibilities and why I say everywhere, the SOLVER is MUST to implement ALL current calculators, I hope somebody read and understand what I want to say).
The small secant method in this paper for CASIO fx-50F meantime reduced to 13 steps, I guess the smallest solver for these blind machines. You can find attached.
Csaba
Attached File(s)
02-16-2019, 01:42 PM
Post: #9
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-16-2019 12:24 PM)Csaba Tizedes Wrote: By the way, if you interested, here is my SOLVE(i) for HP-12C
Steps of secant method from 03 to 18:
Code:
03- 45 2 : RCL 2 04- 45 0 : RCL 0 05- 30 : - 06- 43 36 : LSTx 07- 44 2 : STO 2 08- 35 : CLx 09- 45 3 : RCL 3 10- 45 1 : RCL 1 11- 30 : - 12- 43 36 : LSTx 13- 44 3 : STO 3 14- 33 : R↓ 15- 10 : ÷ 16- 45 1 : RCL 1 17- 20 : × 18-44 30 0 : STO+ 0
This can be shortened to:
Code:
03- 45 2 : RCL 2 04- 45 0 : RCL 0 05- 44 2 : STO 2 06- 30 : - 07- 45 3 : RCL 3 08- 45 1 : RCL 1 09- 44 3 : STO 3 10- 30 : - 11- 10 : ÷ 12- 45 1 : RCL 1 13- 20 : × 14-44 30 0 : STO+ 0
Don't hesitate to publish your programs in this forum. You can use Google to translate to English. | {
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Don't hesitate to publish your programs in this forum. You can use Google to translate to English.
Cheers
Thomas
02-16-2019, 03:24 PM
Post: #10
Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-16-2019 01:42 PM)Thomas Klemm Wrote: Steps of secant method from 03 to 18 can be shortened to.
Thanks, this was eons before, I am sure this is only the first iteration and required to fine tuning.
(02-16-2019 01:42 PM)Thomas Klemm Wrote: Don't hesitate to publish your programs in this forum. You can use Google to translate to English.
OK, sometimes I feel myself as Don Quixote... Frankly speaking, who want to discuss about calculators and calculator programming now? 15 years before I wrote something useful, today I am only (an angry) user...
Csaba
02-17-2019, 02:57 AM (This post was last modified: 02-17-2019 03:20 AM by Gamo.)
Post: #11
Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
I just fine tune this solver program for HP-11C
Added the iterations limit to 28 loops count
Added the condition test to end program when root is found.
If iteration is over 28 loops count this indicate that user must use a closer guess.
-------------------------------------------
Procedure:
Use R1 for unknown X in your equation. End your formula with [RTN]
X1 ENTER X2 [A] display Answer
------------------------------------------------------------------------
Program:
Code:
LBL A STO 0 Rv STO 1 28 STO I // Store Loop Count Limit CLx --------------------- LBL 0 GSB B GSB 1 STO 1 GSB B GSB 1 RCL 1 X=Y // End if Root is found GTO 2 DSE // Loop Limit Counter GTO 0 CLx FIX 9 // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 GTO 2 --------------------- LBL 1 // Solver Routine RCL 0 ÷ CHS RCL 1 + RTN -------------------- LBL 2 RCL 1 // Answer RTN -------------------- LBL B . // f(x) equation start here . . // X = Register 1 [R1] . . RTN | {
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Gamo
02-17-2019, 09:06 AM
Post: #12
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-17-2019 02:57 AM)Gamo Wrote: I just fine tune this solver program for HP-11C
Your program doesn't work for this simple example: $$x^2+x=x(x+1)=0$$
I've entered the following program:
Code:
LBL B 1 RCL 1 + LSTx × RTN
Example:
-2
ENTER
-0.5
A
9.999999 99
The reason is that the function value is divided by -0.5, making it larger and larger:
Code:
LBL 1 // Solver Routine RCL 0 ÷ CHS RCL 1 + RTN
Another thing is this in this loop:
Code:
LBL 0 GSB B GSB 1 STO 1 GSB B GSB 1 RCL 1 X=Y // End if Root is found GTO 2 DSE // Loop Limit Counter GTO 0
When you return to label 0, the same value in register 1 will be used as before.
This function evaluation can be avoided if you save the result of the previous evaluation.
You could move this block:
Code:
LBL 2 RCL 1 // Answer RTN
Over here and avoid the jump GTO 2:
Code:
CLx FIX 9 // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 LBL 2 RCL 1 // Answer RTN
Take a look at Csaba's solution for the HP-12C using the secant method.
Cheers
Thomas
02-17-2019, 02:33 PM
Post: #13
Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
Thanks Thomas Klemm
I try single guess and it work.
CHS 2 [A] display -1
If using two negative guesses and result is
Overflow then use Single guess instead.
Formula Routine is
RCL 1
1
+
RCL 1
X
Gamo
02-17-2019, 04:57 PM
Post: #14
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-17-2019 02:33 PM)Gamo Wrote: If using two negative guesses and result is Overflow then use Single guess instead.
Then try this function $$f(x)=4x(x+1)$$:
Code:
LBL B RCL 1 1 + RCL 1 × 4 × RTN
Examples:
-0.5
ENTER
0.5
A
9.999999 99
-0.5
ENTER
A
9.999999 99
0.5
ENTER
A
9.999999 99
Even initial guesses very close to the solutions lead to the same result:
-1.00001
ENTER
-0.99999
A
9.999999 99 | {
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-1.00001
ENTER
-0.99999
A
9.999999 99
-0.00001
ENTER
0.00001
A
9.999999 99
Cheers
Thomas
02-18-2019, 03:49 AM
Post: #15
Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
Your're right that only work for certain situations.
Gamo
02-18-2019, 05:20 AM
Post: #16
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
If you compare your algorithm with Newton's method:
$$x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}$$
you may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.
And with this in mind indeed the solution can now be found:
$$f'(x) = \frac{d}{dx}4 x (x + 1) = 8 x + 4$$
$$f'(-1) = -4$$
$$f'(0) = 4$$
Examples:
-1.5
ENTER
-4
A
-1.00000
0.5
ENTER
4
A
0.00000
Thus your 2nd parameter is an estimate of the slope at the root.
And then you don't really have to call the function twice with the same value:
Code:
LBL A STO 0 Rv STO 1 28 STO I // Store Loop Count Limit LBL 0 GSB B RCL 0 ÷ STO - 1 X=0 // End if Root is found GTO 1 DSE // Loop Limit Counter GTO 0 CLx FIX 9 // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 LBL 1 RCL 1 // Answer RTN LBL B . // f(x) equation start here . . // X = Register 1 [R1] . . RTN
So the question remains how to approximate the derivation at the roots.
If you keep record of the previous function evaluation you can get an estimate of the slope using the secant method.
Cheers
Thomas
02-18-2019, 07:46 PM
Post: #17
Dieter Senior Member Posts: 2,398 Joined: Dec 2013
RE: Small Solver Program
(02-18-2019 05:20 AM)Thomas Klemm Wrote: If you compare your algorithm with Newton's method:
$$x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}$$
you may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof. | {
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Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.
For more details take a look at this post: http://www.hpmuseum.org/forum/thread-123...#pid111680
Gamo, you really should correct and update the instructions. Here it still says:
Quote:Enter 2 guesses closest to the root.
But again: that's NOT true. The first input is NOT a guess for the root. You now have seen a variety of cases that prove it.
Dieter
02-18-2019, 10:22 PM
Post: #18
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-18-2019 07:46 PM)Dieter Wrote: Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.
You could think him stubborn.
But it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Which I still do not understand in detail.
So all in all I'm happy with the outcome.
Cheers
Thomas
02-19-2019, 01:10 AM (This post was last modified: 02-19-2019 01:22 AM by Albert Chan.)
Post: #19
Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Linear regression on slope had another benefit, which my Casio cannot do.
Doing Quadratic Regression on Casio, you don't really know if it is any good.
There is no "correlation coefficient" to lookup. (Why?)
Linear regression on slope can show whether a Quadratic curve is warranted.
I think the code add slope data, like this: | {
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I think the code add slope data, like this:
Data points (0,90) and (30, 90.2), add to linear regression: (30+0)/2, (90.2-90)/(30-0)
Next point (60, 86.4), add to linear regression: (60+30)/2, (86.4-90.2)/(60-30)
...
N points (sorted) thus generate N-1 slopes to regress.
02-19-2019, 08:39 AM
Post: #20
Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-18-2019 10:22 PM)Thomas Klemm Wrote:
(02-18-2019 07:46 PM)Dieter Wrote: Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.
You could think him stubborn.
But it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Which I still do not understand in detail.
So all in all I'm happy with the outcome.
Cheers
Thomas
I feel myself like a star (at least for 15 minutes...)
Yes, this is my "linear regression on slope", as Chan wrote. When I was in my "Summer Practice" at a company I have modelled lots of water network flows between pump houses (sources), consumers and reservoirs, and sometimes the characteristics curves given only in paper form. Thats why I wrote this little routine.
As you can see, if the measured points has significant error, the slope will be very inaccurate then the regression also inaccurate and the coefficients can be meaningless.
But as a programming question this was an interesting "game" for me. | {
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But as a programming question this was an interesting "game" for me.
To the original question from Gamo: I will check carefully the lists here, but I have another idea: what if, if we just simple walking along on the function's curve until the sign is changing, then we turning back, decreasing the stepsize and walking again. If the sign of the function changed, we are turning back again, decreasing the size of the steps and so on... If the stepsize is less than a small value, we found the root.
This "marching method" not so elegant, not so fast, but good to play with it, just for a try - maybe it can be shorter than the other methods. (I have an exactly same method for finding a minimum point of a function for CASIO fx-3650P/Sencor SEC103; perfectly works, like a ball at the bottom of a hole, slowly find the deepest point, when swings around the extremum).
Csaba
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# Representing displacement vectors in cylindrical coordinates and finding the distance in cylindrical coordinates?
In cartesian coordinates, we can derive the vector $\vec v_3$ by vector subtraction $\vec v_2-\vec v_1$. We then get the distance between $P$ och $Q$ by taking the absolute value of $\vec v_3$ which then is: $$\lvert \vec v_3\rvert = \lvert \vec v_2-\vec v_1 \rvert= \lvert (x_2,y_2,z_2)-(x_1,y_1,z_1) \rvert = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
But how can we do this completely in cylindrical coordinates without converting to cartesian coordinates? We only have, and can only use, $\rho_1$ and $\theta_1$ for $\vec v_1$ as well as $\rho_2$ and $\theta_2$ for $\vec v_2$ in cylindrical coordinates $(\rho,\theta,z)$, where $z=0$ in this example. | {
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Your problem for $z=0$ reduces to the polar coordinate problem already answered here. $$|\vec v_3|^2=(\vec v_2-\vec v_1)\cdot(\vec v_2-\vec v_1)=|\vec v_1|^2+|\vec v_2|^2-2\vec v_2\cdot \vec v_1$$ Now using cylindrical coordinates, $|\vec v_i|^2=r_i^2+z_i^2$ and $\vec v_2\cdot \vec v_1=r_1r_2\cos(\theta_2-\theta_1)+z_1z_2$. Then the final answer will be $$|\vec v_3|^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)+(z_2-z_1)^2$$
• Yes, I've seen this solution, but isn't it based in terms of cartesian coordinates/base vectors, $\hat e_x$, $\hat e_y$ and $\hat e_z$ after all? This since, I guess, you must express a distance in constant base vectors? I'm a bit confused about how to interpret the problem I have to admit. How would it look if I want to express the solution completely in cylindrical coordinates with $\vec v_1=\rho_1 \hat e_\rho (\theta_1)$ and base vectors $\hat e_\rho$, $\hat e_\theta$, and $\hat e_z$ etc? Would that be possible? Sep 6, 2018 at 5:46
• I did not use Cartesian coordinates. The $z$ coordinate is there in cylindrical system. The only other reference to the Cartesian system is the angle $\theta$. The issue that you have is that the basis of the cylindrical coordinate system changes with the vector, therefore equations will be more complicated. Sep 6, 2018 at 6:38
• If you choose one of the vectors as the reference, the cylindrical coordinate system is just a Cartesian system rotated with respect to the original. $\hat e_\rho$ is the new $\hat e_x$, and $\hat e_\theta$ is the new $\hat e_y$. Sep 6, 2018 at 7:19 | {
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# How to determine the curve?
In the figure above, segment $PQ$ is determined by two points: $P: (t,0)$ and $Q: (1,t)$, where $t\in [0,1]$ continuously increases and decreases between $0$ and $1$.
Then this gives a close region swept by $PQ$, the upper edge of which is a curve.
How to determine the equation of the curve (maybe implicit form)?
My own method
Suppose the curve is $y=y(x)$, then for any fixed $x_0\in (0,1)$, there is a vertical line $x=x_0$, which intersects a bundle of such $PQ(t)$ segments:
$$y=\frac{t(x-t)}{1-t}$$
Easy to conclude that, the desired $y_0$ on the curve corresponding to $x_0\left(\in(0,1)\right)$ is the maximum of:
$$y_0= \max\limits_{t\in (0,1)}\frac{t(x_0-t)}{1-t}=2-2\sqrt{1-x_0}-x_0$$
How to use the envelope concept? Is there any elementary method since the area is $\frac{1}{6}$?
• Do you know derivatives, tangent lines and integrals? – Tomas Feb 6 '15 at 9:47
• Yes, I know; I am not sure whether calculus is necessary in order to determine the area, but I am pretty sure in order to determine the curve, calculus would be a help – LCFactorization Feb 6 '15 at 9:53
• Look at the wiki entry for Envelope, it teach you how to determine the curve. – achille hui Feb 6 '15 at 10:29
• I use optimization method and obtains $y=2-2\sqrt{1-x}-x$, then the area is $1/6$. Is there any elementary method in order to determine the area? – LCFactorization Feb 6 '15 at 11:12
I know that the result will be a conic section. I'll prove that later on, but start with it. I come from a background of projective geometry, so I'd homogenize your points by appending a $1$, then find coordinates for the line joining them by computing the cross product.
$$\begin{pmatrix}t\\0\\1\end{pmatrix}\times \begin{pmatrix}1\\t\\1\end{pmatrix}= \begin{pmatrix}-t\\1-t\\t^2\end{pmatrix}$$ | {
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This could also be written as $-tx + (1-t)y + t^2=0$ in usual Cartesian coordinates. It is a special case of a line equation $ax+by+c=0$. Now I want to describe a conic section in the dual sense, i.e. not as a set of points but instead a set of tangent lines. That means I need to find a homogeneous quadratic form in $a,b,c$ which is zero for the line given above. For that, consider all quadratic coefficients:
\begin{align*} a^2 &= t^2 & ab &= t^2-t & b^2 &= 1-2t+t^2 \\ ac &= -t^3 & bc &= t^2-t^3 & c^2 &= t^4 \end{align*}
The $c^2$ expression is the only one with degree $4$, and the $b^2$ expression is the only one with constant term. So these two can't be part of the quadratic equation, since they have nothing to cancel against. After removing them, the $ab$ expression is the only one with linear term, so we drop that as well. Now the relation is easy to see:
$$a^2 + ac - bc = 0$$
Written as a matrix:
$$(a,b,c)\cdot\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = 0$$
Now, as I said, that's the matrix for the dual conic. The primary conic is represented by the inverse matrix, or any multiple thereof:
$$(x,y,1)\cdot\begin{pmatrix}1&1&0\\1&1&-2\\0&-2&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix} = 0$$
Since the determinant of the upper left $2\times2$ matrix is zero, this is a parabola. It can also be described by the equation
\begin{align*} x^2 + 2xy + y^2 &= 4y \\ (x+y)^2 &= 4y \end{align*}
Since this is the primal conic to the dual one we computed before, and that dual one was deduced from a relation between the terms of your lines, this ensures that your whole family of lines will be tangent to this curve. If you like, you can compute the point of tangency as
$$\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}-t\\1-t\\t^2\end{pmatrix} =-\begin{pmatrix}2t-t^2\\t^2\\1\end{pmatrix}$$
So the point $(2t-t^2, t^2)$ lies on both the parabola and the line.
HINT
I would say | {
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So the point $(2t-t^2, t^2)$ lies on both the parabola and the line.
HINT
I would say
$F(x,y,t)=0\equiv t(x-t)-y(1-t)=0,\quad t\in\langle 0,1 \rangle$
$F'_t(x,y,t)=0\equiv x-2t+y=0$
Solving this system of equations for the x, y:
$x=t(2-t), y=t^2, \quad t\in\langle 0,1 \rangle$ is a parametric curve of equation to search.
Plot
• Very simple and neat answer. All three are acceptable. I choose the first one that also uses projective geometry. Thank you very much! @MvG – LCFactorization Feb 7 '15 at 2:40
I have a simpler solution.
1. Find the equation of the line with $(t,0,1)\times(1,t,1)=(-t,1-t,t^2)$ or $$(-t)x+(1-t)*y+(t^2)=0$$
2. Find the extema $t$ with $$\frac{{\rm d}}{{\rm d}t}\left(-t\,x+(1-t)\,y+t^2\right)=0$$ $$\left. -x-y+2 t=0 \right\}\; t=\frac{x+y}{2}$$
3. Plug $t$ into line equation for $$-\frac{(x+y)^2}{4}+y=0$$ which is the resulting curve
Hint: The envelope of an implicit curve $f(x,y,t)=0$ is found by solving for $t$ in $\frac{\partial f}{\partial t}=0$ and using it into the original equation.
My other answer describes how I myself think about this. But I realize that there is a lot of background knowledge in there, which might make this less accessible to members of other communities than projective geometry. So here is my attempt at a more generic solution.
Suppose you have found the equation of the line to be $tx + (t-1)y = t^2$. Take a second line from your family, using $u$ instead of $t$ as the parameter, i.e. $ux + (u-1)y = u^2$. These two intersect in a point $(t+u-tu, tu)$. Now if you consider $\lim_{u\to t}$ then strictly speaking the point of intersection becomes undefined, but the above will simply become $(2t-t^2,t^2)$. So that's the one point that your line does not have in common with any other line of the family (extended from $t\in[0,1]$ to $t\in\mathbb R$). So at that point this line is the one defining the curve. Therefore that's your parametric equation of the envelope curve. | {
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# Question Solved1 Answer1. 2. 3. Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
CW7XUD The Asker · Calculus
1.
2.
3. | {
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CW7XUD The Asker · Calculus
1.
2.
3.
Transcribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
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Transcribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
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1)the average velocity,Average Velocity =(" Total Displacement ")/(" Total Time ")Average Velocity =(d(b)-d(a))/(b-a)2)we ... See the full answer | {
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# Spanning forests of bipartite graphs and distinct row/column sums of binary matrices
Let $$F_{m,n}$$ be the set of spanning forests on the complete bipartite graph $$K_{m,n}$$. Let $$S_{m,n} = \{(r(M), c(M)), M \in B_{m,n} \}$$ where $$B_{m,n}$$ is the set of $$m \times n$$ binary matrices and $$r(M), c(M)$$ are the vectors of row sums and column sums of $$M$$, respecitvely. That is, $$S_{m,n}$$ is the set of the distinct row-sum, column-sum pairs of binary matrices. (The term spanning forest here refers to a forest that spans all of the vertices of the given graph; it doesn't have to be a maximal acyclic subgraph.)
Q: Is it true that $$|F_{m,n}| = |S_{m,n}|$$? It is true for $$m,n \leq 4$$. For $$m=n$$ this is OEIS A297077.
There is an obvious mapping from $$F_{m,n} \rightarrow B_{m,n}$$ given by taking the reduced adjacency matrix, so if $$U = \{u_1, \ldots, u_m\}$$, $$V = \{v_1, \ldots, v_m\}$$ are the color categories we set $$M_{ij} = 1$$ if $$v_i \sim u_j$$ in a forest $$F$$, else $$M_{ij} = 0$$. However, this does not help because multiple forests may have the same row and column sums - and not every row-sum, column-sum pair is represented by a forest under this mapping.
The numbers $$|F_{m,n}|$$ are given here:
$$\begin{array}{|c|c|}\hline m\backslash n & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 2 & \\\hline 2 & 4 & 15 \\\hline 3 & 8 & 54 & 328 \\\hline 4 & 16 & 189 & 1856 & 16145 \\\hline 5 & 32 & 648 & 9984 & 129000 & 1475856 \\\hline\end{array}$$
For more see this answer (sum each row.) | {
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For more see this answer (sum each row.)
• The linked MSE thread calculates $|F_{m,n}|$, but how did you calculate $|S_{m,n}|$ to verify they are the same for $m,n\le 4$? Did you just generate all $2^{mn}$ binary matrices? – antkam May 24 '19 at 19:24
• Yes, I did a brute force count. – Jair Taylor May 24 '19 at 22:32
• Just curious: what made you do the brute force count to find out they're equal? Nothing in the "obvious mapping" would suggest (to me) that they are equal, so it would never have occurred to me to even try. – antkam May 25 '19 at 22:54
• @antkam I was interested in counting forests and found that the relevant OEIS entry only mentioned row-sum / column-sums. – Jair Taylor May 26 '19 at 1:01
• In my Answer, when interpreting $F_{m,n}$, I wrote "In your definition, a spanning forest is any subset $T$ of edges of $K_{2,n}$ which is acyclic." So e.g. the empty $T=\emptyset$ qualifies, and in general nodes can be isolated/unconnected. First of all, is my interpretation correct? Second, if I'm correct, then IMHO your use of the word "spanning" is a little confusing. I understand you probably want to make sure unconnected nodes are still included, and simply saying "forest" might also be confusing in a different way... – antkam May 30 '19 at 12:49
I can prove the conjecture, although I can't give a bijection. I've left up a previous answer which does the case $$m=3$$, since it serves as an example of many of the ideas here.
Here is the key observation:
Theorem: There is a unique array $$f(m,n)$$, for $$m$$ and $$n$$ nonnegative integers, satisfying:
1. $$f(m,0) = f(0,n)=1$$
2. If we fix $$m \geq 1$$, then $$f(m,n)$$ is of the form $$(m+1)^n p_m(n)$$ where $$p_m(x)$$ is a polynomial of degree $$\leq m-1$$.
3. If we fix $$n \geq 1$$, then $$f(m,n)$$ is of the form $$(n+1)^m p_n(m)$$ where $$p_n(x)$$ is a polynomial of degree $$\leq n-1$$. | {
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Uniqueness: Let's suppose that we have already found unique values of $$f(m,n)$$ whenever $$\min(m,n) < k$$. In particular, we know the values of $$f(k,x)$$ and $$f(x,k)$$ for $$0 \leq x \leq k-1$$. This is enough values to uniquely determine a degree $$k-1$$ polynomial, so the values we already know determine the values of $$f(k,x)$$ and $$f(x,k)$$ for all $$x$$.
Existence: The proof of uniqueness gives an algorithm to compute $$f$$. The only potential issue is that $$f(k,k)$$ is determined twice, as a polynomial in its first variable and in its second variable. However, the algorithm is symmetric in $$m$$ and $$n$$, so the two interpolating polynomials coincide. $$\square$$
For example, we have $$f(1,0) = 1 = 2^0 \cdot 1$$ so we must have $$p_1(n)=1$$ and $$f(1,n) = 2^n$$. Similarly, $$f(2,0)=1=2^0 \cdot 1$$ and $$f(2,1) = 4=3^1 \cdot (4/3)$$, so we must have $$p_2(n) = 1+n/3$$ and $$f(2,n) = 3^n(1+n/3) = 3^{n-1} (n+3)$$. Using this algorithm, I computed $$f(m,n)$$ for $$0 \leq m,n \leq 10$$. Here is the data, and some Mathematica code, if you'd like to play with it.
f[m_, n_] := f[m, n] =
If[m == 0 || n == 0, 1,
If[m <= n,
(m + 1)^n * InterpolatingPolynomial[Table[{k, f[m, k]/(m + 1)^k}, {k, 0, m - 1}], x] /. x -> n,
(n + 1)^m * InterpolatingPolynomial[Table[{k, f[k, n]/(n + 1)^k}, {k, 0, n - 1}], x] /. x -> m]]
Values of $$f(m,n)$$ for $$0 \leq m,n \leq 10$$. | {
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Values of $$f(m,n)$$ for $$0 \leq m,n \leq 10$$.
1,1,1,1,1,1,1,1,1,1,1
1,2,4,8,16,32,64,128,256,512,1024
1,4,15,54,189,648,2187,7290,24057,78732,255879
1,8,54,328,1856,9984,51712,260096,1277952,6160384,29229056
1,16,189,1856,16145,129000,968125,6925000,47690625,318500000,2073828125
1,32,648,9984,129000,1475856,15450912,151201728,1403288064,12479546880,107152782336
1,64,2187,51712,968125,15450912,219682183,2862173104,34828543449,401200569280,4418300077219
1,128,7290,260096,6925000,151201728,2862173104,48658878080,760774053888,11122973573120,153936323805184
1,256,24057,1277952,47690625,1403288064,34828543449,760774053888,15047189968833,274908280855680,4707029151392121
1,512,78732,6160384,318500000,12479546880,401200569280,11122973573120,274908280855680,6199170628499200,129708290461760000
1,1024,255879,29229056,2073828125,107152782336,4418300077219,153936323805184,4707029151392121,129708290461760000,3283463201858585471
So, we can prove the conjecture by showing that both $$F_{m,n}$$ and $$S_{m,n}$$ obey conditions 1, 2 and 3. For condition 1, we can just define $$S_{m,0}$$, $$S_{0,n}$$, $$F_{m,0}$$, and $$F_{0,n}$$ to be singletons (and I claim this is actually the most natural definition); then we just need to check that our verification of polynomiality goes all the way down to the zero case.
Since conditions 2 and 3 are symmetric, and the definitions of $$F_{m,n}$$ and $$S_{m,n}$$ are symmetric, we just check condition $$2$$.
Verification of condition 2 for forests: Let $$A_{m,b}$$ be the set of isomorphism classes of bicolored forests whose white vertices are labeled $$\{1,2,\ldots,m \}$$ and which have $$b$$ black vertices, each of degree $$\geq 2$$. We claim that $$|F_{m,n}| = \sum_b |A_{m,b}| n(n-1)(n-2) \cdots (n-b+1) (m+1)^{n-b} . \qquad (1)$$ | {
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Proof: Take a forest in $$F_{m,n}$$, with the $$m$$ vertices colored white and the $$n$$ vertices colored black. Delete all black vertices of degree $$0$$ and $$1$$ and forget the numbering of the remaining black vertices. This gives a forest in $$A_{m,b}$$. To undo this process, we first must take the $$b$$ black vertices and decide which of the $$n$$ vertices of $$K_{m,n}$$ they will be; there are $$n(n-1)(n-2) \cdots (n-b+1)$$ ways to do this. (We used that trees in $$A_{m,b}$$ have no automorphisms.) Then we must take the $$n-b$$ remaining black vertices and either join them to one of the $$m$$ white vertices or leave them unconnected; there $$(m+1)^{n-b}$$ ways to do this.
Formula (1) is clearly $$(m+1)^n$$ times a polynomial, it remains to check that the polynomial has degree at most $$m-1$$. We must check that $$A_{m,b}$$ is empty if $$b \geq m$$. Indeed, since every black vertex has degree at least $$2$$, a forest in $$A_{m,b}$$ has at least $$2b$$ edges. But, since it is a forest, it also has at most $$m+b-1$$ vertices. So $$2b \leq m+b-1$$ and $$b \leq m-1$$. $$\square$$
Verification of condition 2 for margins of $$(0,1)$$ matrices: Consider a vector $$C = (C_1, \ldots, C_n) \in \{ 0,1,\ldots, m \}^n$$ of columns sums, and consider how many row sums $$(R_1, \ldots, R_m)$$ are compatible with it. Let $$c_j = \# \{ k : C_k = j \}$$, so $$c_0+c_1+\cdots + c_m=n$$.
By the Gale-Ryser theorem, $$(R_1, \ldots, R_m)$$ is compatible with $$(c_0, \cdots, c_m)$$ if and only if $$\sum R_i = \sum j c_j$$ and, for all index sets $$1 \leq i_1 < i_2 < \cdots < i_k \leq m$$, we have $$R_{i_1} + R_{i_2} + \cdots + R_{i_k} \leq \sum_j \min(j,k) c_j .$$ This is the defining list of inequalities of the permutahedron, so this condition can alternately be stated as saying that $$(R_1, \ldots, R_m)$$ is in the convex hull of the $$m!$$ permutations of $$(c_1+c_2+\cdots+c_m, c_2+\cdots + c_m, \cdots, c_{m-1}+c_m, c_m)$$. | {
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By Theorem 11.3 of Postnikov's Permutohedra, associahedra, and beyond, the number of such $$(R_1, \ldots, R_m)$$ is a polynomial $$g_m(c_0,c_1, \ldots, c_m)$$ in the $$c_j$$, of degree $$m-1$$. So $$|S_{m,n}| = \sum_{c_0+\cdots + c_m=n} \frac{n!}{c_0! c_1! \cdots c_m!} \ g_m(c_0,c_1, \ldots, c_m). \qquad (2)$$ Let $$\partial_j$$ be $$\tfrac{\partial}{\partial x_j}$$. There is some polynomial $$h_m$$ in the $$\partial_j$$, of degree $$m-1$$, such that $$\left. h_m(\partial_0, \ldots, \partial_m) \left( x_0^{c_0} \cdots x_m^{c_m} \right) \right|_{(1,1,\ldots,1)} = g_m(c_1, \ldots, c_m). \qquad (3)$$ Combining (2), (3) and the binomial expansion of $$(x_0+x_1+\cdots+x_m)^n$$, we obtain $$|S_{m,n}| = \left. h_m(\partial_0, \ldots, \partial_m) \left( x_0+x_1+\cdots+x_m \right)^n \right|_{(1,1,\ldots,1)} . \qquad (4)$$ Let $$h_m(t,t,\ldots,t) = \sum_{k=0}^{m-1} h_{m,k} t^k$$. Then $$(4)$$ is $$\sum_{k=0}^{m-1} h_{m,k} n(n-1)(n-2) \cdots (n-k+1) (m+1)^{n-k}$$, which is a polynomial of the desired form. $$\square$$
• Very cool! I think you need another index on your coefficients $c_k$ (say $c_{m,k}$). Then you've shown $c_{m,k} = |A_{m, k}|$ which is interesting as well. I'd be curious to see if this could be generalized more - e.g., to forests with a specified number of components. But I don't know what the appropriate statistic on row-sums/column-sums would be. – Jair Taylor Jun 2 '19 at 21:16
• @JairTaylor Nice observation. Your comment made me realize I had used $c_j$ to mean two things, so I changed the second one to $h_{m,k}$; your observation then is that I have shown $h_{m,k} = |A_{m,k}|$. – David E Speyer Aug 27 '19 at 15:38
This conjectured equivalence (if true) is still amazing to me. However, here is a baby step: The conjecture is indeed true for $$m=2$$. Perhaps someone else will find this useful as a base for generalization.
Claim: $$|F_{2,n}| = |S_{2,n}|$$ for all $$n$$ | {
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Claim: $$|F_{2,n}| = |S_{2,n}|$$ for all $$n$$
First, lets consider $$F_{2,n}$$. In your definition, a spanning forest is any subset $$T$$ of edges of $$K_{2,n}$$ which is acyclic. Let $$u,v$$ be the two nodes on the $$m=2$$ side, and let $$S(u)$$ be the neighbors of $$u$$, i.e. the subset of nodes (on the $$n$$-node side) which $$u$$ is connected to in $$T$$. Similarly for $$S(v)$$.
Next, note that $$T$$ contains a cycle iff $$S = S(u) \cap S(v)$$ contains $$2$$ nodes (or more). So, $$F_{2,n}$$ can be counted by counting all cases where $$S$$ contains $$0$$ or $$1$$ node.
• $$|S|=0$$ case: Each of $$n$$ nodes can be connected to $$u$$, or to $$v$$, or to neither. Total no. $$= 3^n$$.
• $$|S|=1$$ case: There are $$n$$ ways to pick the unique $$w \in S$$. After that, each of the remaining $$n-1$$ nodes can be connected to $$u$$, or to $$v$$, or to neither. Total no. $$= n\, 3^{n-1}$$
• Summing up, $$|F_{2,n}| = 3^n + n \,3^{n-1} = 3^{n-1} (3+n),$$ which matches your table.
Second, lets consider $$S_{2,n}$$. There are $$n$$ column sums, each of which can be $$0, 1, 2$$. Let $$x,y,z$$ be the number of columns whose sum $$=0, 1, 2$$ respectively. The columns whose sum $$=0$$ or $$2$$ dictate their elements. In the $$y$$ columns whose sum $$=1$$, the number of $$1$$s in the top row can be any integer $$\in [0,y]$$, and so the top row sum can be any integer $$\in [z,z+y]$$, and in short there are $$(y+1)$$ possibilities. Obviously, once the top row sum is known that determines the bottom row sum $$= (y+2z) \,-$$ top row sum.
So $$S_{2,n}$$ can be counted by (i) summing over all possible $$y$$, and for each $$y$$: (ii) picking the $$y$$ columns whose sum $$=1$$, and (iii) for the remaining $$n-y=x+z$$ columns assign them $$0$$ or $$2$$ arbitrarily. I.e.:
$$|S_{2,n}| = \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y}$$ | {
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$$|S_{2,n}| = \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y}$$
This can be evaluated many ways but my favorite is to recognize a slightly transformed sum as an explicit formula for the expected value of a Binomial random variable:
\begin{align} |S_{2,n}| &= \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y} \\ &= 3^n \times \sum_{y=0}^n (1+y) {n \choose y} (\frac13)^y (\frac23)^{n-y} \\ &= 3^n \times \mathbb{E}[1 + Bin(n, \frac13)] \\ &= 3^n (1 + {n \over 3}) = 3^{n-1} (3 + n) = |F_{2,n}| & QED \end{align}
As I said, this is a baby step only. The key step in $$F_{2,n}$$ is that $$T$$ has a cycle iff $$|S| \ge 2$$, and the key step in $$S_{2,n}$$ is that the $$y$$ ones can be all in the top row or all in the bottom row or anywhere in between, for $$(1+y)$$ possibilities. These key steps make the counting easy. However, as far as I can see, neither key step has any easy way to generalize to $$m=3$$, let alone arbitrary $$m$$. (E.g. for $$m=3$$, a cycle in $$T$$ can involve $$2$$ or $$3$$ nodes on the $$m=3$$ side, and I don't know any good way to count that.)
• Nice start. I would like to see a bijective proof. I wonder if, in general, if $m$ is fixed, $F_{m,n}$ has a rational generating function. – Jair Taylor May 30 '19 at 19:32
• I would like to see a bijective proof even just for $m=2$, instead of this algebraic mess. Also, I am no good at GFs, but in this case, I couldn't even find nice recursions for $F$ or $S$. Do you have recursions for either of them? – antkam May 30 '19 at 19:54
I have a brute force proof that, for $$m=3$$, both counts are $$4^{n - 2} (3 n^2 + 13 n + 16)$$. It would be possible to extend this method for any fixed $$m$$. I'll write it up for $$m=3$$ in the hope it gives insight. As you'll see, many steps look the same on both sides, but I can't figure out how to give a general proof. | {
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Counting margins of $$3 \times n$$ binary matrices Consider a vector $$C \in \{ 0,1,2,3 \}^n$$ of column sums. Let $$C$$ have $$c_j$$ entries equal to $$j$$ (so $$c_0+c_1+c_2+c_3=n$$). Let's consider how many row sums $$(R_1, R_2, R_3)$$ are compatible with $$C$$. By the Gale-Ryser theorem, these are the vectors with $$R_1+R_2+R_3 = c_1+2c_2+3c_3$$, $$\min(R_1+R_2, R_1+R_3, R_2+R_3) \geq c_2+2c_3$$ and $$\min(R_1, R_2, R_3)\geq c_3$$. Geometrically, this is the lattice points in a hexagon whose vertices are the permutations of $$(c_1+c_2+c_3, c_2+c_3, c_3)$$. I computed that the number of lattice points in this hexagon is $$\binom{c_1}{2} + 2 c_1 c_2 + \binom{c_2}{2} + 2 c_1 + 2 c_2 + 1.$$
The generating function for $$\sum_C x_0^{c_0} x_1^{c_1} x_2^{c_2} x_3^{c_3}$$ is $$(x_0+x_1+x_2+x_3)^n$$. Let $$\partial_j$$ be differentiation with respect to $$x_j$$. Let $$D$$ be the differential operator $$\tfrac{1}{2} \partial_1^2 + 2 \partial_1 \partial_2+\tfrac{1}{2} \partial_2^2 + 2 \partial_1 + 2 \partial_2 + 1.$$ Then $$D$$ applied to the monomial $$x_0^{c_0} x_1^{c_1} x_2^{c_2} x_3^{c_3}$$, and evaluated at $$(1,1,1,1)$$, gives $$\binom{c_1}{2} + 2 c_1 c_2 + \binom{c_2}{2} + 2 c_1 + 2 c_2 + 1$$. So the number of pairs $$(C,R)$$ is $$D (x_0+x_1+x_2+x_3)^n$$ evaluated at $$(1,1,1,1)$$.
This gives $$3 n(n-1) 4^{n-2} + 4 n 4^{n-1} + 4^n = 4^{n - 2} (3 n^2 + 13 n + 16)$$.
Counting forests Given a forest on $$3 \times n$$, define a sequence $$(v_1, v_2, \ldots, v_n)$$ in $$\{ 0,1,2,3 \}^n$$ by saying that $$v_j$$ is $$0$$ if vertex $$j$$ has degree $$0$$, and otherwise $$v_j$$ is the least neighbor of $$j$$. Let $$c_j$$ be the number of $$v_j$$'s equal to $$j$$.
The generating function for the $$(c_0, c_1, c_2, c_3)$$ sequences is $$(x_0+x_1+x_2+x_3)^n$$ (again!). We count how many forests give rise to each $$(c_0, c_1, c_2, c_3)$$, by listing all possiblilities.
One vertex joined to $$(1,2)$$ and another joined to $$(1,3)$$ This can happen in $$c_1 (c_1-1)$$ ways. | {
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One vertex joined to $$(1,2)$$ and another joined to $$(2,3)$$ This can happen in $$c_1 c_2$$ ways.
One vertex joined to $$(1,3)$$ and another joined to $$(2,3)$$ This can happen in $$c_1 c_2$$ ways.
One vertex joined to $$(1,2,3)$$ This can happen in $$c_1$$ ways.
One vertex joined to $$(1,2)$$ This can happen in $$c_1$$ ways.
One vertex joined to $$(1,3)$$ This can happen in $$c_1$$ ways.
One vertex joined to $$(2,3)$$ This can happen in $$c_2$$ ways.
None of the $$n$$-vertices joined to more than one neighbor This can happen in $$1$$ way.
So, this time, we want to sum up $$c_1 (c_1-1) + 2 c_1 c_2 + 3 c_1 + c_2 +1.$$ The corresponding differential operator is $$E=\partial_1^2 + 2 \partial_1 \partial_2 + 3 \partial_1 + \partial_2+1.$$
The differential operators are different, but once again I get $$3 n(n-1) 4^{n-2} + 4 n 4^{n-1} + 4^n$$ at the end of the day.
A thought for the future Why do the differential operators $$\tfrac{1}{2} \partial_1^2 + 2 \partial_1 \partial_2+\tfrac{1}{2} \partial_2^2 + 2 \partial_1 + 2 \partial_2 + 1$$ and $$\partial_1^2 + 2 \partial_1 \partial_2 + 3 \partial_1 + \partial_2+1$$ give the same thing when applied to $$(x_0+x_1+x_2+x_3)^n$$? Because $$(x_0+x_1+x_2+x_3)^n$$ is solely a function of $$z:=x_0+x_1+x_2+x_3$$, so all the $$\partial_j$$'s evaluate to $$\tfrac{d}{dz}$$. Writing $$\partial$$ for $$\tfrac{d}{dz}$$, both operators are $$3 \partial^2 + 4 \partial + 1$$. Any hope to generalize this?
• Nice. I was not aware of Gale-Ryser. If I understand correctly, this could give a method of verifying the conjecture for any fixed $m$ by counting lattice points of appropriate polytopes (to count $S_{m,n}$) and enumerating all possible families of subsets of $[m]$ sharing neighbors (to count $F_{m,n}$), both reducing to finite counting problems. Still, it seems like this problem is crying out for an explicit bijection. – Jair Taylor Jun 1 '19 at 13:43 | {
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2.2k views
A deck of $5$ cards (each carrying a distinct number from $1$ to $5$) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?
1. $\left(\dfrac{1}{5}\right)$
2. $\left(\dfrac{4}{25}\right)$
3. $\left(\dfrac{1}{4}\right)$
4. $\left(\dfrac{2}{5}\right)$
edited | 2.2k views
The number on the first card needs to be One higher than that on the second card, so possibilities are :
$\begin{array}{c} \begin{array}{cc} 1^{\text{st}} \text{ card} & 2^{\text{nd}} \text{ card}\\ \hline \color{red}1 & \color{red}-\\ 2 & 1\\ 3 & 2\\ 4 & 3\\ 5 & 4\\ \color{red}- & \color{red}5 \end{array}\\ \hline \text{Total$:4$possibilities} \end{array}$
Total possible ways of picking up the cards $= 5 \times 4 = 20$
Thus, the required Probability $= \dfrac{\text{favorable ways}}{\text{total possible ways}}= \dfrac{4}{20} = \dfrac 15$
Option A is correct
selected
0
How to solve the question if it would be only HIGHER and not ONE HIGHER?
+4
How to solve the question if it would be only HIGHER and not ONE HIGHER?
Possible combinations :-
First card second card 5 4,3,2,1 4 3,2,1 3 2,1 2 1 1 none
$P(First\ card\ is\ HIGHER\ than\ second\ one\ )=\left [ \left ( \frac{1}{5} \times \frac{4}{4}\right ) + \left ( \frac{1}{5} \times \frac{3}{4}\right ) + \left ( \frac{1}{5} \times \frac{2}{4}\right ) + \left ( \frac{1}{5} \times \frac{1}{4}\right ) \right]\\ =\frac{1}{2}$
+1
Got it
Here we should consider without replacement, since "removed one at a time" means the card has been removed from the deck.
Prob of picking the first card = 1/5
Now there are 4 cards in the deck. Prob of picking the second card = 1/4
Possible favourable combinations = 2-1, 3-2, 4-3, 5-4
Probability of each combination = (1/5)*(1/4) = 1/20 | {
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Probability of each combination = (1/5)*(1/4) = 1/20
Hence answer = 4*1/20 = 1/5
with 5 cards to choose we can only fulfil the condition if we pick 2,3,4,5 in our choice else theres no way to get the same
the probability of choosing first no's is =(2,3,4,5)/(1,2,3,4,5)=4/5
the second time, we have only one option to choose out of four option=1/4
so,the total probability=(4/5)*(1/4)=1/5
1
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Zero rows in echelon matrix
In the MIT lecture notes (pg. 1) for Linear Algebra, it states;
The third row is zero because row 3 was a linear combination of rows 1 and 2; it was eliminated.
Here, $$A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \end{bmatrix}$$ and $$U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$.
Does this mean that if the row of the echelon matrix is all zeros, the corresponding row in A can be represented as a linear combination of the other rows in A?
And if the above statements is true, does this also apply to reduced row echelon matrix?
(1) Yes, if a row of the echelon matrix is all zeros, then that row (but remember to consider possible partial pivoting) is a linear combination of other rows in the original matrix. Generally, the coefficients such that forms this linear combination can be found solving the following linear system for $$a$$ and $$b$$: $$\begin{bmatrix}1 & 2 \\ 2 & 4 \\ 2 & 6 \\ 2 & 8 \end{bmatrix}\cdot\begin{bmatrix}a \\ b \end{bmatrix}=\begin{bmatrix}3 \\ 6 \\ 8 \\ 10\end{bmatrix}$$ | {
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1. ## Question about solving eigenvector problems
I understand how to get eigenvalues, but I have a problem understanding eigenvectors.
I'll demonstrate with an example:
I am given A = $\begin{bmatrix}3 & 1.5 \\1.5 & 3 \end{bmatrix}$
Solving for lambda, I get lambda = 1.5 and 4.5
Now my matrix equations become:
$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$
dividing by 1.5 I get: $x_1-x_2=0$
so I get $x_1=x_2$
Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1
If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
2. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
I understand how to get eigenvalues, but I have a problem understanding eigenvectors.
I'll demonstrate with an example:
I am given A = $\begin{bmatrix}3 & 1.5 \\1.5 & 3 \end{bmatrix}$
Solving for lambda, I get lambda = 1.5 and 4.5
Now my matrix equations become:
$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$
dividing by 1.5 I get: $x_1-x_2=0$
so I get $x_1=x_2$
Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1
If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
There are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.
{1,1} doesn't span R2. It spans the line y = x. Likewise {1,-1} spans the line y = -x
together these two vectors span R2
3. ## Re: Question about solving eigenvector problems
Originally Posted by romsek
{1,1} doesn't span R2. It spans the line y = x.
In that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?
for example, {2,2}, {5,5}, {1000,1000}, etc.
Originally Posted by romsek
There are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.
How did you get this? My book has this answer:
1.5, [1 1], 45°; 4.5, [1 1], 45°
Does that make sense? | {
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Does that make sense?
4. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
so I get $x_1=x_2$
Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1
If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
First, the notation [ℝ ℝ], where the objects inside brackets are sets rather than numbers, is not universally accepted. Even if it were, [A B] would probably mean {[x y] | x ∈ A, y ∈ B}, i.e., A × B. There is no coordination between the first and second coordinate here. It is clear that you mean {[x x] | x ∈ ℝ}, but I don't see an easier way to write this. By the way, even if you delimit matrices with square brackets, I would use parentheses for vectors (i.e., 1 × n matrices): e.g., (x, x), because square brackets are also used to denote closed segments.
Second, you are right, any (x, x) is an eigenvector of A as is (x, -x). Indeed, you solve the equation det(A -λI) = 0 for λ, so if λ is an eigenvalue, the matrix A - λI is singular by definition. Therefore, the system of equations (A - λI)x = 0 has infinitely many solutions.
Edit: If x is an eigenvector with value λ, then so is (ax) for any a ∈ ℝ such that a ≠ 0. Indeed, A(ax) = a(Ax) = a(λx) = λ(ax).
5. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
In that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?
for example, {2,2}, {5,5}, {1000,1000}, etc.
How did you get this? My book has this answer:
1.5, [1 1], 45°; 4.5, [1 1], 45°
Does that make sense?
there are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is "spanned" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.
Your book is missing a minus sign somewhere. The answer I got was
1.5 [1,-1]/sqrt(2)
4.5 [1,1]/sqrt(2) | {
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1.5 [1,-1]/sqrt(2)
4.5 [1,1]/sqrt(2)
6. ## Re: Question about solving eigenvector problems
Originally Posted by romsek
there are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is "spanned" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.
Your book is missing a minus sign somewhere. The answer I got was
1.5 [1,-1]/sqrt(2)
4.5 [1,1]/sqrt(2)
Whoops, that was my bad, there was a minus sign.
where do you get the 1/sqrt(2) from?
is it a scalar that you can multiply throughout the matrix?
7. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
Whoops, that was my bad, there was a minus sign.
where do you get the 1/sqrt(2) from?
is it a scalar that you can multiply throughout the matrix?
it just normalizes the vector to unit length. You tend to want the unit eigenvectors | {
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# Singing Bird Problem
For my algorithms class, the professor has this question as extra credit:
The Phicitlius Bauber bird is believed to never sing the same song twice.
Its songs are always 10 seconds in length and consist of a series of notes that are either high or low pitched and are either 1 or 2 seconds long.
How many different songs can the Bauber bird sing?
This strikes me as a combinatorial problem, but I'm having trouble figuring out the exact formula to use to fit the included variables of 10 seconds length, 2 notes and 2 note periods.
At first, I thought that $\binom{10}{2}*\binom{10}{2}$ can be a solution, since it fits finding all combinations of both and multiplies the total.
My issue is this results in a total of 2025 possible songs, and this just seems a little on the low side.
Any assistance is appreciated.
-
I'm going to work under the assumption that two successive 1-second notes of the same pitch is distinct from a single 2-second note of that pitch.
Let $f(n)$ be the number of distinct $n$-second songs the bird can sing. The bird has four options for starting the song: 1-second low, 1-second high, 2-second low, and 2-second high. This gives the recursion $$f(n) = 2f(n-1) + 2f(n-2),$$ with $f(0) = 1$ and $f(1) = 2$.
The associated characteristic equation is $x^2 - 2x - 2 = 0$, which has roots $x = 1 \pm \sqrt{3}$. Thus, we get the general solution $$f(n) = A(1 + \sqrt{3})^n + B(1 - \sqrt{3})^n.$$
Using the initial conditions, we have the system \begin{align*} 1 &= A + B\\ 2 &= A(1 + \sqrt{3}) + B(1 - \sqrt{3}), \end{align*} which has the unique solution $A = \frac{3 + \sqrt{3}}{6}$ and $B = \frac{3 - \sqrt{3}}{6}$.
Finally, we have $$f(n) = \frac{3 + \sqrt{3}}{6}(1 + \sqrt{3})^n + \frac{3 - \sqrt{3}}{6}(1 - \sqrt{3})^n.$$
For the problem at hand, we want $$f(10) = 18,272.$$ | {
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"lm_label": "1. YES\n2. YES",
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"url": "http://math.stackexchange.com/questions/109032/singing-bird-problem?answertab=oldest"
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