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http://mathhelpforum.com/geometry/154183-triangles-20-non-straight-line-dots-print.html | # Triangles in 20 non-straight-line "dots"
• August 22nd 2010, 07:07 AM
grottvald
Triangles in 20 non-straight-line "dots"
Twenty points/dots are given so that the three of them are never in a straight line. How many triangles can be formed with corners(as in vertex i think?) in the dots?
• August 22nd 2010, 07:21 AM
Plato
Calculate the number of ways to choose three points from twenty.
• August 22nd 2010, 10:36 AM
grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)
• August 22nd 2010, 10:57 AM
Plato
Quote:
Originally Posted by grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct
It is not a large number at all: $\displaystyle \binom{20}{3}=\frac{20!}{3!\cdot 17!}=1140$
• August 22nd 2010, 11:58 AM
Quote:
Originally Posted by grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)
You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.
The number of ways to do this is $\binom{19}{2}=171$
Therefore, there are 171 triangles that can be drawn which include the leftmost point.
If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.
These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.
Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.
hence, there are $\binom{19}{2}+\binom{18}{2}+\binom{17}{2}+........ .+\binom{2}{2}=\binom{20}{3}$ triangles.
• August 22nd 2010, 01:26 PM
grottvald
Quote:
You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.
The number of ways to do this is $\binom{19}{2}=171$
Therefore, there are 171 triangles that can be drawn which include the leftmost point.
If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.
These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.
Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.
hence, there are $\binom{19}{2}+\binom{18}{2}+\binom{17}{2}+........ .+\binom{2}{2}=\binom{20}{3}$ triangles.
Thank you so much! You explained it in an very easy and beautiful way. Thanks again! | 2016-02-11T18:19:41 | {
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https://math.stackexchange.com/questions/3687083/int-fracgxfx-dx-where-fx-frac12exe-x-and-gx | # $\int\frac{g(x)}{f(x)} \, dx$ where $f(x) = \frac{1}{2}(e^x+e^{-x})$ and $g(x) = \frac{1}{2}(e^x - e^{-x})$?
Given two functions $$f(x) = \frac{1}{2}(e^x+e^{-x})$$ and $$g(x) = \frac{1}{2}(e^x - e^{-x})$$, calculate
$$\int\frac{g(x)}{f(x)} \, dx.$$
Observing that $$f'(x) = g(x)$$, this is very easy to do. Just take $$u = f(x)$$ and therefore $$du = f'(x)\,dx$$. Now we have
$$\int \frac{f'(x)}{f(x)} \, dx = \int \frac{du}{u} = \ln |u| + C = \ln\left|\frac{1}{2}(e^x+e^{-x})\right|+C.$$
However, this solution is wrong. The correct one is $$\ln|e^x+e^{-x}| + C$$. I see how we could get that, just plug in $$g(x)$$ and $$f(x)$$ directly, and we get
$$\int \frac{\frac{1}{2}(e^x - e^{-x})}{\frac{1}{2}(e^x+e^{-x})} \, dx.$$
Cancelling the $$\frac{1}{2}$$'s and taking $$u = e^x+e^{-x}$$ and $$du = (e^x-e^{-x}) dx$$ gives us
$$\int \frac{du}{u} = \ln|u| + C = \ln|e^x+e^{-x}| + C.$$
I can't see where I made a mistake in my approach, and I'd be really grateful if you pointed it out.
• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun May 22 at 18:47
• Thanks, I will do that next time. :) – Gregor Perčič May 22 at 19:01
• I feel terrible mentioning it because I know you're only trying to be polite but could you remove the religious elements at the end? There are likely many faiths none should be given a platform – Karl May 22 at 19:10
• @Karl With all due respect, I will not remove it. It is very good to try to propagate the Faith in all contexts of life. The mods may remove it, but hey, I did my best. Not to mention that my right to free speech will be trampled on in this instance. – Gregor Perčič May 22 at 19:29
• I ment no offence honestly. I'd just rather stick to maths. – Karl May 22 at 19:38
Note that if $$K$$ is a constant, $$\ln|K h(x)|= \ln|K|+\ln|h(x)|$$. In other words, the $$1/2$$ in the log can get 'absorbed' into the $$+C$$ term.
• Aaaaah, I see! But then that means that my solution $\ln|\frac{1}{2}(e^x + e^{-x})| + C$ doesn't by itself (with $C = 0$) calculate the area under the curve $h(x) = \frac{f(x)}{g(x)}$? – Gregor Perčič May 22 at 19:00
• It seems $f$ and $g$ are reversed in your comment. It should, in the sense that $$\int _{0}^x \frac{g(t)}{f(t)}\,dt = \left.\ln|\frac{1}{2}(e^t + e^{-t})|+C\right|_{0}^x$$ $$= \left(\ln|\frac{1}{2}(e^x + e^{-x})|+C\right)-\left(\ln|\frac{1}{2}(1+1)|+C\right)=\ln|\frac{1}{2}(e^x + e^{-x})|$$The $1/2$ is essential to make sure, starting at $x=0$, the area is $0$; otherwise, you'll be off by a constant. – Integrand May 22 at 19:25
• Yes, $f$ and $g$ are reversed. Sorry about that and thanks for the insight. – Gregor Perčič May 22 at 19:32
Using the identity: $$\log (\frac{a}{b})=\log(a)- \log(b)$$
$$\ln \frac{1}{2}(e^x+e^{-x})+C$$
$$=\ln (e^x+e^{-x})-\ln 2+C$$
$$=\ln (e^x+e^{-x})+C{'}$$ (as $$\ln 2$$ is constant) | 2020-06-03T22:45:28 | {
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https://casmusings.wordpress.com/2015/08/ | # Monthly Archives: August 2015
## Marilyn vos Savant Conditional Probability Follow Up
In the Marilyn vos Savant problem I posted yesterday, I focused on the subtle shift from simple to conditional probability the writer of the question appeared to miss. Two of my students took a different approach.
The majority of my students, typical of AP Statistics students’ tendencies very early in the course, tried to use a “wall of words” to explain away the discrepancy rather than providing quantitative evidence. But two fully embraced the probabilities and developed the following probability tree to incorporate all of the given probabilities. Each branch shows the probability of a short or long straw given the present state of the system. Notice that it includes both of the apparently confounding 1/3 and 1/2 probabilities.
The uncontested probability of the first person is 1/4.
The probability of the second person is then (3/4)(1/3) = 1/4, exactly as expected. The probabilities of the 3rd and 4th people can be similarly computed to arrive at the same 1/4 final result.
My students argued essentially that the writer was correct in saying the probability of the second person having the short straw was 1/3 in the instant after it was revealed that the first person didn’t have the straw, but that they had forgotten to incorporate the probability of arriving in that state. When you use all of the information, the probability of each person receiving the short straw remains at 1/4, just as expected.
## Marilyn vos Savant and Conditional Probability
The following question appeared in the “Ask Marilyn” column in the August 16, 2015 issue of Parade magazine. The writer seems stuck between two probabilities.
(Click here for a cleaned-up online version if you don’t like the newspaper look.)
I just pitched this question to my statistics class (we start the year with a probability unit). I thought some of you might like it for your classes, too.
I asked them to do two things. 1) Answer the writer’s question, AND 2) Use precise probability terminology to identify the source of the writer’s conundrum. Can you answer both before reading further?
Very briefly, the writer is correct in both situations. If each of the four people draws a random straw, there is absolutely a 1 in 4 chance of each drawing the straw. Think about shuffling the straws and “dealing” one to each person much like shuffling a deck of cards and dealing out all of the cards. Any given straw or card is equally likely to land in any player’s hand.
Now let the first person look at his or her straw. It is either short or not. The author is then correct at claiming the probability of others holding the straw is now 0 (if the first person found the short straw) or 1/3 (if the first person did not). And this is precisely the source of the writer’s conundrum. She’s actually asking two different questions but thinks she’s asking only one.
The 1/4 result is from a pure, simple probability scenario. There are four possible equally-likely locations for the short straw.
The 0 and 1/3 results happen only after the first (or any other) person looks at his or her straw. At that point, the problem shifts from simple probability to conditional probability. After observing a straw, the question shifts to determining the probability that one of the remaining people has the short straw GIVEN that you know the result of one person’s draw.
So, the writer was correct in all of her claims; she just didn’t realize she was asking two fundamentally different questions. That’s a pretty excusable lapse, in my opinion. Slips into conditional probability are often missed.
Perhaps the most famous of these misses is the solution to the Monty Hall scenario that vos Savant famously posited years ago. What I particularly love about this is the number of very-well-educated mathematicians who missed the conditional and wrote flaming retorts to vos Savant brandishing their PhDs and ultimately found themselves publicly supporting errant conclusions. You can read the original question, errant responses, and vos Savant’s very clear explanation here.
CONCLUSION:
Probability is subtle and catches all of us at some point. Even so, the careful thinking required to dissect and answer subtle probability questions is arguably one of the best exercises of logical reasoning around.
RANDOM(?) CONNECTION:
As a completely different connection, I think this is very much like Heisenberg’s Uncertainty Principle. Until the first straw is observed, the short straw really could (does?) exist in all hands simultaneously. Observing the system (looking at one person’s straw) permanently changes the state of the system, bifurcating forever the system into one of two potential future states: the short straw is found in the first hand or is it not.
CORRECTION (3 hours after posting):
I knew I was likely to overstate or misname something in my final connection. Thanks to Mike Lawler (@mikeandallie) for a quick correction via Twitter. I should have called this quantum superposition and not the uncertainty principle. Thanks so much, Mike.
## SBG and AP Statistics Update
I’ve continued to work on my Standards for AP Statistics and after a few conversations with colleagues and finding this pdf of AP Statistics Standards, I’ve winnowed down and revised my Standards to the point I’m comfortable using them this year.
Following is the much shorter document I’m using in my classes this year. They address the AP Statistics core content as well as the additional ideas, connections, etc. I hope my students learn this year. As always, I welcome all feedback, and I hope someone else finds these guides helpful.
## SBG and Statistics
I’ve been following Standards-Based Grading (SBG) for several years now after first being introduced to the concept by colleague John Burk (@occam98). Thanks, John!
I finally made the dive into SBG with my Summer School Algebra 2 class this past June & July, and I’ve fully committed to an SBG pilot for my AP Statistics classes this year.
I found writing standards for Algebra 2 this summer relatively straightforward. I’ve taught that content for decades now and know precisely what I want my students to understand. I needed some practice writing standards and got better as the summer class progressed. Over time, I’ve read several teachers’ versions of standards for various courses. But writing standards for my statistics class prove MUCH more challenging. In the end, I found myself guided by three major philosophies.
1. The elegance and challenge of well designed Enduring Understandings from the Understanding by Design (UbD) work of Jay McTighe the late Grant Wiggins helped me craft many of my standards as targets for student learning that didn’t necessarily reveal everything all at once.
2. The power of writing student-centered “I can …” statements that I learned through my colleague Jill Gough (@jgough) has become very important in my classroom design. I’ve become much more focused on what I want my students (“learners” in Jill’s parlance) to be able to accomplish and less about what I’m trying to deliver. This recentering of my teaching awareness has been good for my continuing professional development and was a prime motivator in writing these Standards.
3. I struggled throughout the creation of my first AP Statistics standards document to find a balance between too few very broad high-level conceptual claims and a far-too-granular long list of skill minutiae. I wanted more than a narrow checklist of tiny skills and less than overloaded individual standards that are difficult for students to satisfy. I want a challenging, but reachable bar.
So, following is my first attempt at Standards for my AP Statistics class, and I’ll be using them this year. In sharing this, I have two hopes:
• Maybe some teacher out there might find some use in my Standards.
• More importantly, I’d LOVE some feedback from anyone on this work. It feels much too long to me, but I wonder if it is really too much or too little. Have I left something out?
At some point, all work needs a public airing to improve. That time for me is now. Thank you in advance on behalf of my students for any feedback.
## Chemistry, CAS, and Balancing Equations
Here’ s a cool application of linear equations I first encountered about 20 years ago working with chemistry colleague Penney Sconzo at my former school in Atlanta, GA. Many students struggle early in their first chemistry classes with balancing equations. Thinking about these as generalized systems of linear equations gives a universal approach to balancing chemical equations, including ionic equations.
This idea makes a brilliant connection if you teach algebra 2 students concurrently enrolled in chemistry, or vice versa.
FROM CHEMISTRY TO ALGEBRA
Consider burning ethanol. The chemical combination of ethanol and oxygen, creating carbon dioxide and water:
$C_2H_6O+3O_2 \longrightarrow 2CO_2+3H_2O$ (1)
But what if you didn’t know that 1 molecule of ethanol combined with 3 molecules of oxygen gas to create 2 molecules of carbon dioxide and 3 molecules of water? This specific set coefficients (or multiples of the set) exist for this reaction because of the Law of Conservation of Matter. While elements may rearrange in a chemical reaction, they do not become something else. So how do you determine the unknown coefficients of a generic chemical reaction?
Using the ethanol example, assume you started with
$wC_2H_6O+xO_2 \longrightarrow yCO_2+zH_2O$ (2)
for some unknown values of w, x, y, and z. Conservation of Matter guarantees that the amount of carbon, hydrogen, and oxygen are the same before and after the reaction. Tallying the amount of each element on each side of the equation gives three linear equations:
Carbon: $2w=y$
Hydrogen: $6w=2z$
Oxygen: $w+2x=2y+z$
where the coefficients come from the subscripts within the compound notations. As one example, the carbon subscript in ethanol ( $C_2H_6O$ ) is 2, indicating two carbon atoms in each ethanol molecule. There must have been 2w carbon atoms in the w ethanol molecules.
This system of 3 equations in 4 variables won’t have a unique solution, but let’s see what my Nspire CAS says. (NOTE: On the TI-Nspire, you can solve for any one of the four variables. Because the presence of more variables than equations makes the solution non-unique, some results may appear cleaner than others. For me, w was more complicated than z, so I chose to use the z solution.)
All three equations have y in the numerator and denominators of 2. The presence of the y indicates the expected non-unique solution. But it also gives me the freedom to select any convenient value of y I want to use. I’ll pick $y=2$ to simplify the fractions. Plugging in gives me values for the other coefficients.
Substituting these into (2) above gives the original equation (1).
VARIABILITY EXISTS
Traditionally, chemists write these equations with the lowest possible natural number coefficients, but thinking of them as systems of linear equations makes another reality obvious. If 1 molecule of ethanol combines with 3 molecules of hydrogen gas to make 2 molecules of carbon dioxide and 3 molecules of water, surely 10 molecule of ethanol combines with 30 molecules of hydrogen gas to make 20 molecules of carbon dioxide and 30 molecules of water (the result of substituting $y=20$ instead of the $y=2$ used above).
You could even let $y=1$ to get $z=\frac{3}{2}$, $w=\frac{1}{2}$, and $x=\frac{3}{2}$. Shifting units, this could mean a half-mole of ethanol and 1.5 moles of hydrogen make a mole of carbon dioxide and 1.5 moles of water. The point is, the ratios are constant. A good lesson.
ANOTHER QUICK EXAMPLE:
Now let’s try a harder one to balance: Reacting carbon monoxide and hydrogen gas to create octane and water.
$wCO + xH_2 \longrightarrow y C_8 H_{18} + z H_2 O$
Setting up equations for each element gives
Carbon: $w=8y$
Oxygen: $w=z$
Hydrogen: $2x=18y+2z$
I could simplify the hydrogen equation, but that’s not required. Solving this system of equations gives
Nice. No fractions this time. Using $y=1$ gives $w=8$, $x=17$, and $z=8$, or
$8CO + 17H_2 \longrightarrow C_8 H_{18} + 8H_2 O$
Simple.
EXTENSIONS TO IONIC EQUATIONS:
Now let’s balance an ionic equation with unknown coefficients a, b, c, d, e, and f:
$a Ba^{2+} + b OH^- + c H^- + d PO_4^{3-} \longrightarrow eH_2O + fBa_3(PO_4)_2$
In addition to writing equations for barium, oxygen, hydrogen, and phosphorus, Conservation of Charge allows me to write one more equation to reflect the balancing of charge in the reaction.
Barium: $a = 3f$
Oxygen: $b +4d = e+8f$
Hydrogen: $b+c=2e$
Phosphorus: $d=2f$
CHARGE (+/-): $2a-b-c-3d=0$
Solving the system gives
Now that’s a curious result. I’ll deal with the zeros in a moment. Letting $d=2$ gives $f=1$ and $a=3$, indicating that 3 molecules of ionic barium combine with 2 molecules of ionic phosphate to create a single uncharged molecule of barium phosphate precipitate.
The zeros here indicate the presence of “spectator ions”. Basically, the hydroxide and hydrogen ions on the left are in equal measure to the liquid water molecule on the right. Since they are in equal measure, one solution is
$3Ba^{2+}+6OH^- +6H^-+2PO_4^{3-} \longrightarrow 6H_2O + Ba_3(PO_4)_2$
CONCLUSION:
You still need to understand chemistry and algebra to interpret the results, but combining algebra (and especially a CAS) makes it much easier to balance chemical equations and ionic chemical equations, particularly those with non-trivial solutions not easily found by inspection.
The minor connection between science (chemistry) and math (algebra) is nice.
As many others have noted, CAS enables you to keep your mind on the problem while avoiding getting lost in the algebra. | 2018-06-19T02:40:25 | {
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https://math.stackexchange.com/questions/450158/simplifying-compound-fraction-frac3-sqrt5-5 | # Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$
I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
• You should multiply both numerator and denominator by that constant! – Mahdi Khosravi Jul 23 '13 at 9:50
• Or you can exchange the numerator and denominator of the whole denominator and move it to whole numerator! – Mahdi Khosravi Jul 23 '13 at 9:51
• I know this an old question, but if you had simply changed your √5/5 to a √5/√5, you would've got 3√5÷5/5 and gotten your answer. The whole point of multiplying a complex faction by a number to simplify it is to times it by 1 (√5/√5 in this case) because multiplying anything by 1 is the same thing. – Gᴇᴏᴍᴇᴛᴇʀ Sep 7 '14 at 19:44
• "The problem is I interpreted that as:" .... remember that the numerator is 3. – John Joy Jan 4 at 17:58
\begin{align*}\frac{3}{\frac{\sqrt{5}}{5}} &= 3 \cdot \frac{5}{\sqrt 5}\\ &= 3 \cdot \frac{5}{\sqrt 5} \cdot 1\\ &= 3 \cdot \frac{5}{\sqrt 5} \cdot \frac{\sqrt 5}{\sqrt 5}\\ &= 3 \cdot \frac{5\sqrt 5}{5}\\ &= 3\sqrt 5 \end{align*}
• I appreciate everyone's response but this answer is the most elaborate. Thanks blf, I see where I went wrong now! – Sam Jul 23 '13 at 10:00
You multiplied the original fraction by $\dfrac5{\sqrt5}$, which is not $1$, so of course you changed the value. The correct course of action is to multiply by $1$ in the form
$$\frac{5/\sqrt5}{5/\sqrt5}$$
to get
$$\frac3{\frac{\sqrt5}5}=\frac3{\frac{\sqrt5}5}\cdot\frac{\frac5{\sqrt5}}{\frac5{\sqrt5}}=\frac{3\cdot\frac5{\sqrt5}}1=3\cdot\frac5{\sqrt5}=3\sqrt5\;,$$
since $\dfrac5{\sqrt5}=\sqrt5$.
More generally,
$$\frac{a}{b/c}=\frac{a}{\frac{b}c}\cdot\frac{\frac{c}b}{\frac{c}b}=\frac{a\cdot\frac{c}b}1=a\cdot\frac{c}b\;,$$
this is the basis for the invert and multiply rule for dividing by a fraction.
This means $$3\cdot \frac{5}{\sqrt{5}}=3\cdot\frac{(\sqrt{5})^2}{\sqrt{5}} =3\sqrt{5}$$ You're multiplying twice for the reciprocal of the denominator.
Another way to see it is multiplying numerator and denominator by the same number: $$\frac{3}{\frac{\sqrt{5}}{5}}=\frac{3\sqrt{5}}{\frac{\sqrt{5}}{5}\cdot\sqrt{5}} =\frac{3\sqrt{5}}{1}$$
Start with $$\frac{3}{\sqrt{5}/5}=\frac{15}{\sqrt{5}},$$
and then rationalize the denominator (multiply both numerator and denominator by $\sqrt{5}$) to get
$$\frac{15\sqrt{5}}{5}=3\sqrt{5}.$$
You have the following fraction to simplify: \begin{align} \frac{3}{\sqrt{5}/5} &=\frac{5\times 3}{\sqrt{5}} \\ &=\frac{15}{\sqrt{5}} \\ &=\sqrt{\bigg(\frac{15}{\sqrt{5}}\bigg)^2} \\ &=\sqrt{\frac{15^2}{\sqrt{5}^2}} \\ &=\sqrt{\frac{225}{5}} \\ &=\sqrt{45} \\ &= \sqrt{9\times 5} \\ &= \sqrt{9}\sqrt{5} \\ &= 3\sqrt{5} \\ \therefore \frac{15}{\sqrt{5}}\times \frac{5}{\sqrt{5}} &=\frac{15}{\sqrt{5}}\times \frac{\big(\frac{15}{\sqrt{5}}\big)}{3} \\ &=\frac{15}{\sqrt{5}}\times \frac{15}{\sqrt{5}}\times \frac{1}{3} \\ &=\bigg(\frac{15}{\sqrt{5}}\bigg)^2 \times \frac{1}{3} \\ &=45\times \frac{1}{3} \\ &=\frac{45}{3} \\ &=15 \end{align}
One thing that helps me organize my thoughts, is to convert both numerator and denuminator to fractions as follows.
\begin{align} \frac{3}{\frac{5}{\sqrt{5}}}&=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}\cdot\frac{\frac{\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{\frac{5}{\sqrt{5}}\cdot\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{1}=\frac{3}{1}\cdot\frac{\sqrt{5}}{5}=\text{etc.}\\ \end{align}
In the expression $$\frac{3}{\frac{\sqrt{5}}{5}}$$, to find the reciprocal of that expression's denominator, $$\frac{\sqrt{5}}{5}$$, simply swap its numerator and denominator.
Thus the reciprocal of $$\frac{\sqrt{5}}{5}$$ is $$\frac{5}{\sqrt{5}}$$, and $$\frac{3}{\frac{\sqrt{5}}{5}} = 3 \cdot \frac{5}{\sqrt{5}}$$
I think the result you have to remember is
$$\boxed{\dfrac 1{\sqrt{a}}=\dfrac{\sqrt{a}}a}$$
You will see this kind of manipulation very often in your studies, and it allows to get rid of square roots on denominator.
Here your expression is just $$\dfrac{3}{\frac{\sqrt{5}}{5}}=\dfrac{3}{\frac 1{\sqrt{5}}}=3\sqrt{5}$$ | 2019-11-16T22:08:10 | {
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https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/ | # Is There a Rigorous Proof Of 1 = 0.999…?
Yes.
First, we have not addressed what 0.999… actually means. So it’s best first to describe what on earth the notation $$b_0.b_1b_2b_3…$$ means. The way mathematicians define this thing is
$$b_0.b_1b_2b_3…=\sum_{n=0}^{+\infty}{\frac{b_n}{10^n}}$$
So, in particular, we have that
$$0.999…=\sum_{n=1}^{+\infty}{\frac{9}{10^n}}$$
But all of this doesn’t really make any sense until we define what the right-hand side means. There is an infinite sum there, but what does that mean? Well, we put
$$S_k=\sum_{n=1}^{k}{\frac{9}{10^n}} \ ,$$
then we have a finite sum. So, for example
$$S_1=0.9, \ ~S_2=0.99, \ ~S_3=0.999, \ etc.$$
So, in some way, we want to take the limit of this sequence.
Let’s consider a particularly simple sequence to illustrate the idea behind the definition of a limit of a sequence: 1/2, 1/3, 1/4,… The terms in this sequence get smaller and smaller. You might think that it’s obvious that it goes to 0, or that it’s obvious that a smart mathematician can prove that it goes to 0, but it’s not. It’s impossible to even attempt a proof until we have defined what it means for something to go to 0. So we have to define what the statement “1/2, 1/3, 1/4,… goes to 0” means before we can attempt to prove that it’s true.
This is the standard definition: “1/n goes to 0” means that “for every positive real number $\epsilon$, there’s a positive integer N, such that for all integers n such that $n\geq N$, we have $|1/n| < \epsilon$”. With this definition in place, it’s quite easy to prove that “1/n goes to 0” is a true statement. What I want you to see here, is that we chose this definition to make sure that this statement would be true. The first mathematicians who thought about how to define the limit of a sequence might have briefly considered definitions that make the statement “1/n goes to 0” false, but they would have dismissed those definitions as irrelevant because they fail to capture the idea of a limit that they already understood on an intuitive level.
So the real reason why 1/n goes to 0 is that we wanted it to! Similar comments hold for the sequence of partial sums that define 0.999… It goes to 1, because we have defined the concepts “0.999….”, “sum of infinitely many terms”, and “limit of a sequence” in ways that make 0.999…=1. Can we define number systems such that 1=0.999… does not hold? Of course! But these number systems are not as useful, because they don’t conform to our intuition about limits and numbers.
Now that we know what a limit and an infinite sum is, let me give a fully rigorous proof to the equality 1=0.999… This proof is due to Euler and it appears in the 1770’s edition of “Elements of algebra”.
We know that
$$0.999…=\sum_{n=1}^{+\infty}{ \frac{9}{10^n} } = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} +…$$
This sum is a special kind of sum, namely, it’s a geometric sum. For (infinite) geometric sums, we can find its limit easily:
Let
$$x=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+…$$
Then
$$9x=0.999…$$
But, we also have $10x=1+\frac{1}{10}+\frac{1}{10^2}+…$, so $10x-x=1$.
This implies that $x=\frac{1}{9}$.
Hence,
$$0.999…=9x=1$$
Does this proof look familiar? It should! It is essentially the same as Proof #2 in the previous post. The only difference is that every step is now justified by operations with limits.
See this supportive article: https://www.physicsforums.com/insights/why-do-people-say-that-1-and-999-are-equal/
The following forum members have contributed to this FAQ:
AlephZero
Fredrik
micromass
tiny-tim
vela
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102 replies
1. nuuskur says:
It should be unanimous that 1 – 0.9(9) = 0.0(0) = 0. Assume the opposite, then 0.0(0) != 0, contradiction.
Too much java, my TeX is rusty :D
2. Mark44 says:
[QUOTE=”alva, post: 5215165, member: 497526″]
But my reasoning is similar to the OP. He is taking more addends in one series than in the other.[/QUOTE]
[QUOTE=”weirdoguy, post: 5215170, member: 464738″]No it’s not, because he’s using infinite series, and you are using finite series. It is a big difference.[/QUOTE]
In addition, [USER=497526]@alva[/USER], you don’t seem to realize that 0.99 and 0.99… mean completely different things. The first number has two decimal digits that are shown explicitly, and an implied infinite number of zeroes that follow it. The second number has an infinite number of 9’s following the decimal point.
We are NOT saying that 1 = 0.99. We ARE saying that 1 = 0.99…, and the proof is given in the OP.
Also, 0.99… and 0.999… are exactly the same, while 0.99 and 0.999 are different.
3. micromass says:
[QUOTE=”alva, post: 5215165, member: 497526″]
But my reasoning is similar to the OP. He is taking more addends in one series than in the other.[/QUOTE]
No, in both cases, the number of addends is infinite. The only difference is that the series in the OP is convergent, while yours are divergent.
4. alva says:
[QUOTE=”HomogenousCow, post: 5215156, member: 435628″]Here you’ve just made some bizarre arguments and obvious arithmetic mistakes.[/QUOTE]
Please, let me know what are “my arithmetic mistakes”.
5. weirdoguy says:
No it’s not, because he’s using infinite series, and you are using finite series. It is a big difference.
6. alva says:
[QUOTE=”HomogenousCow, post: 5215152, member: 435628″]You can’t manipulate divergent series that way, also you forgot the negative signs at the end.[/QUOTE]
You’re right and I apologize for the mistake in the signs.
Moreover where I said “my age” I meant “my age from Jesus’s birth.
But my reasoning is similar to the OP. He is taking more addends in one series than in the other.
7. weirdoguy says:
[QUOTE=”alva, post: 5215133, member: 497526″]and never 0.99999… = 1[/QUOTE]
Your “proof” is just plain wrong. Just deal with the fact that 0,(9)=1.
8. HomogenousCow says:
[QUOTE=”alva, post: 5215133, member: 497526″]Let x= 1/10 +1/100 + …
Lets take just 2 addends, x=0.11
Then 9x = 0.99
But, we also have 10x = 1 + 1/10 = 1.1, so 10x – x =0.11
Hence 0.99 = 9x = 0.99
You can use 3 addends and get 0.999 = 9x = 0,999
And 4 addends and 5, and calculate the limit when n -> infinite and never 0.99999… = 1[/QUOTE]
Here you’ve just made some bizarre arguments and obvious arithmetic mistakes.
9. HomogenousCow says:
[QUOTE=”alva, post: 5215141, member: 497526″]I was born in 1950 and my brother in 1952.
In many years my age will be 1950 + 1 + 1 +1 +1 ….
In many years my brothers age will be 1952 +1 + 1 +1 ….
So the difference will be 1952 +( 1 + 1 +1+ …) – 1950 – ( 1 +1 +1+1+…) = 1952 + (1 +1 +1 +…) – 1950 +2 + (1 +1 +1..) = 0[/QUOTE]
You can’t manipulate divergent series that way, also you forgot the negative signs at the end.
10. alva says:
I was born in 1950 and my brother in 1952.
In many years my age will be 1950 + 1 + 1 +1 +1 ….
In many years my brothers age will be 1952 +1 + 1 +1 ….
So the difference will be 1952 +( 1 + 1 +1+ …) – 1950 – ( 1 +1 +1+1+…) = 1952 + (1 +1 +1 +…) – 1950 +2 + (1 +1 +1..) = 0
11. alva says:
Let x= 1/10 +1/100 + …
Lets take just 2 addends, x=0.11
Then 9x = 0.99
But, we also have 10x = 1 + 1/10 = 1.1, so 10x – x =0.11
Hence 0.99 = 9x = 0.99
You can use 3 addends and get 0.999 = 9x = 0,999
And 4 addends and 5, and calculate the limit when n -> infinite and never 0.99999… = 1
12. Mark44 says:
[QUOTE=”alva, post: 5215060, member: 497526″]Let
x=110+1102+1103+…
Then
9x=0.999…
But, we also have 10x=1+110+1102+…, so 10x−x=1.
This implies that x=19.
[/quote]None of the above makes any sense. In your definition of x, its value is at least 2300. How can 9x be 0.999…?
[QUOTE=alva]
Hence,
0.999…=9x=1
**********************************************************
Im sorry but Im trying to copy/paste the OP message but the format is not preserved
**********************************************************
Lets do the proof with n=2 for ALL the equations
Let x = 1/10 + 1/100 = 0.11
Then 9x = 0.99
But, we also have 10x = 1 + 1/10 = 1.1 so 10x-x =0.11
Hence
0.99 = 9x = 0.99 !!! yes
[/quote]Of course. If x = .11, then ordinary arithmetic can be used to show that 9x = .99. So what?
[QUOTE=alva]
And, as anyone can see, you can repeat the proof for n=3,4… and using limits, when n->infinite, the result is clear, 1 is not = 0.999999…[/QUOTE]
13. micromass says:
[QUOTE=”gill1109, post: 5214887, member: 401042″]Yes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999…9 (any number of repetitions) and 1.[/QUOTE]
That depends how many repetitions of ##9## you take in ##0.999…##. If you take countably many, then sure. But if you index over all ordinals, then no.
14. gill1109 says:
[QUOTE=”micromass, post: 5214863, member: 205308″]Whether ##1=0.9999…## in the surreals depends highly on the definitions for ##0.9999…##. Some definitions make it equal, others don’t.[/QUOTE]
Yes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999…9 (any number of repetitions) and 1.
15. micromass says:
[QUOTE=”gill1109, post: 5214572, member: 401042″]Here is a number system in which 1 is not equal to 0.9999…. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway’s “surreal numbers”. [URL]https://en.wikipedia.org/wiki/Surreal_number[/URL][/QUOTE]
Whether ##1=0.9999…## in the surreals depends highly on the definitions for ##0.9999…##. Some definitions make it equal, others don’t.
16. Hornbein says:
[QUOTE=”micromass, post: 5214231, member: 205308″]micromass submitted a new PF Insights post
[URL=’https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/’]Is There a Rigorous Proof Of 1 = 0.999…?[/URL]
[URL=’https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/’]Continue reading the Original PF Insights Post.[/URL][/QUOTE]
I suggest 1 – 0.9999… = 0.0000….
17. KSG4592 says:
Always find these fascinating, even if most of it goes over my head at this point.
18. Pro7 says:
the question which is obvious is….how far is 0.999… away from 1?…since the answer is very close to 0 as it never ends…therefore 0.999… tends to 1.
19. Ilja says:
I have a nice childhood memory about this. My school teacher had not checked this, and had the nice idea to write, instead of an open interval [0,1) a closed interval [0.0.9999…]. I had heard somewhere that 1=0.999…, so I simply said this. The teacher, and the whole class, disagreed. So a discussion started, and I had to think about how to prove it. What I invented was that 1/9 = 0.1111…, 8/9 = 0.8888…, but when we add 0.1111… to 0.88888… we will clearly get 0.9999…. On the other side, 1/9+8/9 = 9/9 = 1.
The next day it became clear that I was right. Some guys have looked into some books.
The better proof is to compute 9*0.9999… = 10* 0.9999… – 0.9999… = 9.999999..-0.99999… = 9.
20. Anton Alice says:
I think there is a problem with the step, where you take 10x-x =1.
Applying operations on infinite sums has been considered dangerous, as far as I know.
21. terryds says:
Here is the ‘silly’ proof. 0.999… = 0.333…+0.666… = 1/3 + 2/3 = 3/3 = 1
22. H_A_Landman says:
You have to be careful, because Cantor’s hierarchy of trans-finite numbers doesn’t exist in the Surreal Numbers. For example,one might be tempted to identify Conway’s $\omega = {0,1,2,…|}$ with Cantor’s $\aleph_0$, since they are both intuitively “the infinity of the integers”. But in Cantor’s system, $\aleph_0$ is the smallest (positive) infinite number, whereas in Conway’s there does not exist any such smallest infinity since we also have $\omega-1$, $\omega/2$, $\sqrt{\omega}$, etc. Conversely, in Cantor’s system there is no largest infinity, but in Conway’s there is (On).
23. DMartin says:
I said:"I think there can be contradictions in mathematics"You said:"So you think 1+1=3 is valid and can be proven? Because that is equivalent to what you just said."I don't think I need to comment on this, it's so ridiculous that it speaks for itself.
24. DMartin says:
I prefer to keep it in degrees throughout, and although you can subtract the 89, I won't. To explain it, it's another series of numbers that suggest something, but somewhat different from the other. It suggests that for any number like 89.99999, there's another number between it and 90, arrived at via x/(sine x). You can then add more 9s, and it still applies. However many 9s you add, this rule applies, so it seems very reasonable to say that it applies to 89.99999…….. as well. That means there's at least one number between 89.9999….. and 90.Now because as I mentioned, infinities can be compared, this seems to reveal the existence of an infinity greater than the first one, or at least, a number or set of numbers that goes numerically higher. Certainly worth thinking about, but as I said, I think there can be contradictions in mathematics, so it's not an attempt to disprove your proof of 0.9999…. = 1.
25. DMartin says:
I do believe in mathematics, but I just take a different view of it than you do. Gödel's work meant that many others saw the edges of it, from the 1930s on, so my view isn't a controversial one.
26. DMartin says:
I never said it wasn't useful! It's very useful. And it doesn't matter that it's a bit rough round the edges either.
27. DMartin says:
Incidentally, and I'm not going to get into discussion about this, if one should show that there's always a number between 0.9999…. and 1, that might suggest that they're different. Well I can show that there's always a number in between 89.9999…… and 90. It's the expression x/(sine x) again, it always gives a number nearer 90, however many 9s there are.Now I should make my position clear – I'm not arguing that your proof of 0.9999…. = 1 is false. I think mathematics is not a consistent system, and that it gets a bit rough round the edges, that's all. Perhaps I've shown that a bit.
28. DMartin says:
You seem to think we should define things first, and then do our exploring. But sometimes the exploring helps to inform the definitions, and this is how mathematics has actually developed. This means one can point out a mathematical structure without necessarily having a rigorous definition. But a definition I've given for the relevant number, whether a loose one, and whether or not it fits with your definitions, is an angle x between 0 and 90 degrees such that x/(sin x) = exactly 180/π.
29. DMartin says:
By the way, I did say, but probably should have emphasised more, that what applies to one pair of numbers doesn't necessarily apply to the other. There are things that happen at or near 0 that don't happen elsewhere. But this is surely relevant, nonetheless! Thanks for the discussion.
30. DMartin says:
I never said it was a proof, surely even you noticed that. I said it was relevant, and that the context can make a difference. The point that the context makes a difference is borne out by micromass above saying about the context of surreal numbers:"Whether 1=0.9999… in the surreals depends highly on the definitions for 0.9999… . Some definitions make it equal, others don't."So it's relevant to point out a sequence relating to the second pair of numbers I mentioned, in which there is a distinction between the two of them, because one infinite sequence lands somewhere different from the other.Incidentally, there have been distinctions drawn between different infinities, and it has turned out that they can be compared, and one infinity can turn out to be 'larger' than another. This might intuitively seem impossible, but ways to compare them were found. There is some loose similarity between that and what I've done.
31. DMartin says:
That's what one would expect it to be. I've shown that the second pair isn't just 0 and 0.
32. DMartin says:
Well, there's a symmetry to be pointed out. First, you have 0.9999….. and 1, and these two numbers look the same, or almost the same. People discuss whether they're the same, and whether there's a proof that they are.Then, if you subtract 0.9999….. from 1, you find another pair of numbers – that is, the result of the subtraction, and zero. This pair of numbers can be called 1 – 0.9999…. , and 0.And there's a symmetry between these two pairs of numbers. Each pair may well be in the same relationship, whatever that is. So anything shedding a bit of light on either pair might be relevant. And showing the second pair to be different from each other is relevant.
33. DMartin says:
Yes I knew it was 1 radian."This is by definition, not up for debate". This implies that all our definitions are, by definition, correct.What you learned in primary school may or may not be true. But I have said that the context makes a difference sometimes.But the question of comparing a whole number, or a non-negative integer, and another nearby number that approaches it with an infinite series, is not always a clear cut question. And what I've set out has bearing on it, because I've show the two looking different, and looking existent.Don't forget that we learn as we go, the mathematical structures we have are not just a given eternal structure, they were put together bit by bit by finding things, and what we have is, as always, incomplete.
34. DMartin says:
Sorry, our posts crossed. Yes, I know you think that whether or not a number exists is related to whether or not one can specify the positions of the digits. But the mathematical structure I've shown above hints at a relevant structure that is uncovered, and exists in some way other than just conforming to a made up set of rules.
35. DMartin says:
Putting in 0.0001, I get x/(sin x) = 57.295779513111409697664737311509try putting in 0.0000001then 0.00000000000000001.the result will approach 180/[pi], and to me this shows something that is discovered, rather than invented, and has bearing on the similar questions we've been looking at.
36. DMartin says:
" 0.999999… exists because every digit in the decimal representation can be specified. If you ask, "what digit is in the 12th place?" Answer: 9. If you ask, "what digit is in the 59th place?" Answer: 9. If you ask, "what digit is in the 623rd place?" Answer: 9. No matter what specific digit you ask about, the answer is always "9". "That's true, and you believe that it's significant.
37. DMartin says:
See posts above. There's a series of increasingly small angles near zero degrees that approach a number at infinity, but that number is above zero. It's clear that this number gives x/(sin x) = 180/π, because the values approach that number. Whether or not any other values give 180/π makes no difference, but it's interesting to hear it.
38. DMartin says:
Well, it's a matter of taste to some extent. You say you can prove that within a particular artificial system, a number 0.99999999…. exists, but 1 minus that number, or 0.00000….0001 doesn't exist. And yet I've shown that the second number can be expressed as an angle between zero and 90 degrees.This has bearing on the question of whether mathematics is invented or discovered. Are we inventing the rules, or discovering them. When I look at the points above about the digit 1 in my number, I think there's a bit too much inventing going on.
39. DMartin says:
If you say that one number exists and another doesn't, you need to say what you mean by exists. If you mean exists within mathematics, Gödel showed that mathematics isn't necessarily a self-consistent system, so existing within mathematics isn't necessarily a meaningful concept.If you're not keen on how my number is expressed, perhaps you'd prefer it if I said:an angle θ such that θ/(sin θ) = exactly 180/π. I can prove that this angle is not zero, because θ = 0 gives a different result.
40. DMartin says:
I don't know what you mean by existence, when you say you can prove the existence of your number. But it seems clear enough that if we can talk about numbers with an infinite number of decimal places that all make a difference, then my number and yours are very similar. And my number and yours add up to one, which gives them more common ground.About the position of the 1 in my number, out of all those 9s of yours, there must be one that corresponds to my 1. So whatever problems I have with my number (and God knows it's hard to keep them all in line), you must have the same problems with yours.
41. DMartin says:
So you have a rule that 'each numerical digit must have it's concrete position'. I suppose you know the positions of all the 9s in 0.99999….. then. But even if you argued that their positions are more concrete than the 1 in the number I used, it's not clear where the rule came from.what I've shown is a series that converges on, or approaches, a number at infinity, and the point is, whatever the existence status of that number at infinity, it isn't zero. And surely whatever its existence status, it's similar to the existence status of the numbers you're talking about.
42. DMartin says:
Well, that may be so, but this thread is a discussion on the basis that such numbers are worth talking about, so we're assuming they have some meaning before we start. If you say 'there is no such number', then presumably you think this whole thread is pointless.
43. DMartin says:
I can show that whatever the meaning of the number 0.0000000 => 00001, with an infinite number of zeros, it is different from 0. This means that other similar numbers are probably the same, though it does depend on the context. The method involves showing that the two numbers 0.0000000 => 00001 and 0 give completely different output numbers when put into an equation.To show that 0.0000000 => 00001 does not equal 0.Take the equation θ/(sin θ) , where θ is an angle in degrees.For θ = 0.0000000 => 00001, it gives θ/(sin θ) = 180/π = 57.295779513082320876798154814105this is known because a series of increasingly small angles such as 0.0001, 0.0000000001, 0.0000000000000001 etc.give numbers that approach 180/π.But for θ = 0, it gives θ/(sin θ) = 0Therefore 0.0000000 => 00001 does not equal 0.Any thoughts would be appreciated, thanks.
44. alva says:
Letx=110+1102+1103+…Then9x=0.999…But, we also have 10x=1+110+1102+…, so 10x−x=1.This implies that x=19.Hence,0.999…=9x=1
45. WWGD says:
You can also use, though not as nice, the perspective of the Reals as a metric space, together with the Archimedean Principle: then d(x,y)=0 iff x=y. Let then ## x=1 , y=0.9999…. ##Then ##d(x,y)=|x-y| ## can be made indefinitely small ( by going farther along the string of 9's ), forcing ## |x-y|=0## , forcing ##x=y ##. More formally, for any ##\epsilon >0 ##, we can make ##|x-y|< \epsilon ##
46. gill1109 says:
Here is a number system in which 1 is not equal to 0.9999…. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway's "surreal numbers". https://en.wikipedia.org/wiki/Surreal_number
47. Josh Meyer says:
Nice!The informal proof I always share with people is that 1/9=0.111…, 2/9=0.222…, 3/9=0.333…, and so on until 9/9=0.999…=1 | 2022-10-03T04:15:38 | {
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https://www.physicsforums.com/threads/the-integral.175225/ | The Integral
1. Jun 27, 2007
anantchowdhary
Is there a mathematical proof that can prove that the integral is the antiderivative?
2. Jun 27, 2007
Staff: Mentor
fundamental theorem of calculus
Look up the fundamental theorem of calculus.
3. Jun 27, 2007
anantchowdhary
umm...ive tried to study it....but my question is a little different....i meant to say..
if we take out the area of the graph of a function f(x) bound at certain limits, then..is there any way to prove that the indefinite integral gives us the antiderivative of the function
or rather..the area bound under two limits of the function is simply the limits applied to the antiderivative of the function
4. Jun 27, 2007
Kummer
Given an integrable function $$f:[a,b]\mapsto \mathbb{R}$$ we define,
$$g(x) = \int_a^x f(t) dt$$ for all $$t\in [a,b]$$.
Then, $$g$$ is a continous function on $$[a,b]$$. And furthermore if $$f$$ is continous at $$t_0 \in (a,b)$$ then $$g$$ is differentiable at $$t_0$$ with $$g'(x_0)=f(x_0)$$
5. Jun 27, 2007
mathwonk
suppose f is aN INCREASING CONTINUOUS FUNCTION. then the area under the graph between x and x+h is between f(x)h and f(x+h)h,
i.e.f(x)h < A(x+h)-A(x) < f(x+h)h.
satisfy yourself of this by drawing a picture.
now the derivative of that area function is the limit of [A(x+h)-A(x)]/h, as h goes to zero.
by the inequality above, this limit iscaught b etween f(x) and f(x+h), for all h, which emans, since f is continuious, it equals f(x). i.e. dA/dx= f(x).
now in creasing is not n eeded ubt makes it easier.
6. Jun 27, 2007
anantchowdhary
thanks..ill give it a thought
7. Jun 28, 2007
HallsofIvy
Staff Emeritus
A little more generally- assume f(x)> 0 for a< x< b. For any number n, divide the interval from a to b into n equal sub-intervals (each of length (b-a)/n)). Construct on each a rectangle having height equal to the minimum value of f on the interval- that is, each rectangle is completely "below" the graph. Let An be the total area of those rectangles. Since each rectangle is contained in the region below the graph of f, it is obvious that $A_n \le A$ where A is the area of that region (the "area below the graph").
Now do exactly the same thing except taking the height to be the maximum value of f in each interval. Now, the top of each interval is above the graph of f so the region under f is completely contained in the union of all the rectangles. Taking An to be the total area of all those rectangles, we must have $A \le A^n$.
That is, for all n, $A_n \le A \le A^n$
If f HAS an integral (if f is integrable) then, by definition, the limits of An and An must be the same- and equal to the integral of f from a to b. Since A is always "trapped" between those two values, the two limits must be equal to A: The area is equal to the integral of f from a to b.
That's pretty much the proof given in any Calculus book.
8. Jun 28, 2007
Gib Z
"is there any way to prove that the indefinite integral gives us the antiderivative of the function"
I seemed to understand this as - ' is there any way to prove F(x), where F'(x)=f(x), is the same function as the function given by $$\int f(x) dx$$.' The fundamental theorem of Calculus shows when we put upper and lower bounds, b and a, on the integral, it results in F(b) - F(a), however to the specific question as I understood, the dropping of bounds is just a nice notation for the integral/antiderivatve function.
9. Jun 28, 2007
anantchowdhary
umm...i tried this out..i Followed pretty much of it
but how do we prove that dA/dx=f(x)?
thanks
10. Jun 28, 2007
mathwonk
notice the proof i gave for the increasing case, which is due to newtton, does not need the deep theorem that a continuous function always has a max and a min on a closed bounded interval.
it also covers (when used piecewise) all polynomials, con tinuous rational functions, and continuous trig functions, that occur in practice. hence there is
no reason for books to omit this proof or banish it to an appendix.
11. Jun 28, 2007
HallsofIvy
Staff Emeritus
Define F(x) to be
$$\int_a^x f(t)dt$$
For f(x)> 0 we can interpret that as the area between the graph y= f(x) and the x-axis, from a to the fixed value x. Then F(x+h) is
$$\int a^{x+h}f(t)dt$$
the area between the graph y= f(x) and x-axis, from a to the fixed value x+h.
Now F(x+h)- F(x) is the area between the graph y= f(x), between x and x+h. It's not difficult to see that is equal to the area of the rectangle with base x to x+h and height f(x*) where x* is some value between x and x+h: That is F(x+h)- F(x)= hf(x*). Then
$$\frac{F(x+h)- F(x)}{h}= f(x^*)$$
Taking the limit as h goes to 0 forces x*, which is always between x and x+h, to go to x.
Therefore
$$\frac{dF}{dx}= \lim_{h\rightarrow 0} \frac{F(x+h)- F(x)}{h}= f(x)$$
12. Jun 28, 2007
mathwonk
let me combine halls discussion with mine in post 5.
he has shown, modulo the theorem that max and min exist,
that in my notation, if m(h) is the min of f on the interval [x,x+h], and if M(h) is the max,
that m(h)h < A(x+h)-A(x) < M(h)h.
where < means less than or equal.
Then again if we compute the derivative of A as the limit of
[A(x+h)-A(x)]/h
as h goes to 0, we see by the inequalities that [A(x+h)-A(x)]/h
is squeezed between
M(h) and m(h) for all h. since f is continuous, these numbers both apprioach f(x), so lim [A(x+h)-A(x)]/h
= dA/dx = f(x).
13. Jun 28, 2007
mathwonk
to prove the point halls says is "not difficult to see", uses the intermediate value theorem, in case you do not see it.
Last edited: Jun 28, 2007
14. Jun 28, 2007
anantchowdhary
thanks a lot for the proof!
15. Jun 29, 2007
HallsofIvy
Staff Emeritus
"It is easy to see ..." means "I think there is a proof but I can't remember it just now".
"Obvious to the most casual observer" means "I hope no one asks me to prove it"!
16. Jun 29, 2007
HallsofIvy
Staff Emeritus
"It is easy to see ..." means "I think there is a proof but I can't remember it just now".
"Obvious to the most casual observer" means "I hope no one asks me to prove it"!
17. Jul 2, 2007
ssd
This is a beauty..... incomplete though. I believe,for a student new to calculus, it is enough to remove all mental barriers about the fact whether an integral is anti-derivative.
18. Jul 2, 2007
jambaugh
Technically the indefinite integral is defined to be the anti-derivative (or rather a variable anti-derivative depending on an arbitrary constant since "the" anti-derivative is not unique.)
The definite integral is a value rather than a function so you can't call it an anti-derivative.
That technical nit picking aside see your undergraduate calculus text and the two forms of the FTC. | 2017-01-20T16:14:16 | {
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https://math.stackexchange.com/questions/2668002/we-square-an-integral-but-why-change-a-variable/2668062 | # We square an integral, but why change a variable?
If we square an integral, we also change the integration variable in one of the integrals. But why is this actually correct?
For example, say I have the following:
Solve $\int_{-\infty}^\infty e^{-x^2} dx$. Let $I=\int_{-\infty}^\infty e^{-x^2} dx$, so
\begin{align} I^2 &=\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg)^2\\ &= \bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg) \times \bigg( \underbrace{ \int_{-\infty}^{\infty} e^{-y^2}dy }_{\text{Why}?} \bigg) \\ &=\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-(x^2+y^2)} dx\bigg)dy \end{align}
But why is the following wrong: \begin{align} I^2 &=\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg)^2\\ &= \bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg) \times \bigg(\int_{-\infty}^{\infty} e^{-x^2}dx\bigg) \\ &=\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-x^2-x^2} dx\bigg)dx \\ &=\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-2x^2} dx\bigg)dx \qquad ? \end{align}
• By definition $\int_{a}^b f(x)dx=\int_{a}^b f(y)dy$. For your second approach, no such rule for integrals exists. It abuses notation as you need $dx,dy$ to distinguish your variables when you rewrite the integrals in iterated form. – Alex R. Feb 26 '18 at 20:10
• The second-to-last equality is wrong. It is not true that $$\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg) \times \bigg(\int_{-\infty}^{\infty} e^{-x^2}dx\bigg) =\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-x^2-x^2} dx\bigg)dx.$$ Indeed, the inner integral on the RHS $$\left(\int_{-\infty}^{\infty} e^{-2x^2}\ dx\right)$$ is equal to some positive constant, let's call it $c$. The outer integral is integrating the constant value $c$, from $-\infty$ to $\infty$, so the result is $\infty$. On the other hand, the two factors on the LHS are finite, hence their product is finite. – Bungo Feb 26 '18 at 20:16
• It is a dummy variable. – hamam_Abdallah Feb 26 '18 at 20:25
The short answer is that in your second to last identity you are neglecting cross terms in your multiplication.
A simple example may be used to demonstrate the error, and we will use a strict summation instead of an integration for clarity. Consider the sum $$\sum_{x=1}^3 x = 1 + 2 +3 = 6.$$ Now, squaring the summation yields \begin{align} \left(\sum_{x=1}^3x\right)^2 &= (1+2+3)^2 = (1+2+3)\times(1+2+3) \end{align} In order to properly calculate this quantity (long-hand) requires one to completely distribute including cross terms: \begin{align} &\ {\color{white}+}\ 1\times 1 + 1\times 2 + 1\times 3 \notag \\ (1+2+3)\times(1+2+3) =& +2\times 1 + 2\times 2 + 2\times 3 = 36. \notag \\ & + 3\times 1 + 3\times 2 + 3\times 3 \end{align} However, if the cross terms were neglected one would obtain $$(1+2+3)\times(1+2+3) \ne (1^2+2^2+3^2) = 1\times 1 + 2\times 2 + 3\times 3 = 14.$$ Written another way, we conclude that $$\left(\sum_{x=1}^3x\right)^2 = \left(\sum_{x=1}^3x\right)\left(\sum_{x=1}^3x\right) \ne \sum_{x=1}^3x^2.$$ However, as noted in the comments, the $x$ in the summation (or the in the definite integral) is just a dummy variable, and may be replaced with another symbol, such as $y$: $$\left(\sum_{x=1}^3x\right)^2 = \left(\sum_{x=1}^3x\right)\left(\sum_{y=1}^3y\right).$$ The reason for changing the dummy variable is that we may now say that the summation over $x$ and $y$ are independent, and so we may rearrange the summations: $$\left(\sum_{x=1}^3x\right)^2 = \left(\sum_{x=1}^3x\right)\left(\sum_{y=1}^3y\right) = \sum_{x=1}^3\sum_{y=1}^3x\times y.$$ Again, integration is just summation, and so the same fact holds true for integrals. Furthermore, it doesn't matter what the function inside the summation/integral is. Thus, we may write $$\left(\int_a^b f(x)\ dx\right)^2 = \left(\int_a^b f(x)\ dx\right)\left(\int_a^b f(y)\ dy\right) = \int_a^b\int_a^b f(x)\times f(y)\ dx\ dy.$$ Using $f(x)=e^{-x^2}$ results in your example problem.
Because changing the integrand's variable doesn't change the value of the integral:$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(y)dy$$the same argument holds for summation as following$$\sum_{n=a}^{b}k_n=\sum_{m=a}^{b}k_m$$also we know that$$\left(\sum_{n=a}^{b}k_n\right)^2=\left(k_a+...+k_b\right)^2=k_a^2+...+k_b^2+2k_ak_{a+1}+...+2k_{b-1}k_b$$and$$\sum_{n=a}^{b}k_n\sum_{m=a}^{b}k_m=(k_a+...+k_b)(k_a+...+k_b)$$which by expanding the terms and rearranging them leads to$$\left(\sum_{n=a}^{b}k_n\right)^2=\sum_{n=a}^{b}k_n\sum_{m=a}^{b}k_m$$since integral is intrinsicly a summation, this can be generalized to integral operator either.
I think the best way of visualizing what is happening intuitively is thinking in terms of programming.
If you make a program and you have two functions (in the programming sense) which do not interact, you can call "$x$" the variable in both functions. They are isolated.
However, if a function is supposed to handle two different variables independently in the same context, you cannot name them the same thing.
Formally and mathematically (and also changing notation from $\int f(x)dx$ to $\int f$ so that the point may become clearer), we have the following chain of equalities.
\begin{align*} \underline{\left(\int_{\mathbb{R}} x \mapsto e^{-x^2}\right)} \cdot \underline{\left(\int_{\mathbb{R}} x \mapsto e^{-x^2} \right)}&\stackrel{Linearity}=\left(\int_{\mathbb{R}} \underline{\left(\int_{\mathbb{R}} x \mapsto e^{-x^2}\right)}\left(x \mapsto e^{-x^2} \right)\right) \\ &\underset{\text{fctn mult.}}{\overset{\text{Def. of}}{=}} \int_{\mathbb{R}} x \mapsto \left(e^{-x^2} \underline{\left(\int_{\mathbb{R}} y \mapsto e^{-y^2}\right)}\right) \\ &\underset{\text{}}{\overset{\text{Linearity}}{=}} \int_{\mathbb{R}} x \mapsto \left(\int_{\mathbb{R}} y \mapsto e^{-x^2-y^2}\right) \\ &\stackrel{Fubini}{=}\int_{\mathbb{R}^2}(x,y) \mapsto e^{-x^2-y^2}. \end{align*}
Using the same name ($x$) for the variables in both integrations is "fine" up to the third equality (I changed it in the second equality for the sake of clarity, otherwise the third would be a little mystic). Using $x$ instead of $y$ in the second line would be extremely bad taste, but not explicitly wrong. Using $x$ instead of $y$ in the right side of the first equality is edgy, but not so much.
Indeed, each underline integral is a closed box with respect to the rest: they are fixed numbers, which don't interact at all with the rest of the environment other than via the fact that they are numbers. It is in the third equality that we are multiplying a constant (relative to the inner function which we are integrating) which depends on $x$. It does not depend on the function inside the integral at all, but it is invading its space and interacting with it. To be extremely clear, call the function $x \mapsto e^{-x^2}$ by $f$. The second line is then $$\int_{\mathbb{R}} x \mapsto \left(e^{-x^2} \left(\int_{\mathbb{R}} f\right)\right)$$ $$=\int_{\mathbb{R}} \left( x \mapsto \int_{\mathbb{R}} (e^{-x^2}\cdot f)\right),$$ where we multiplied a constant to the inner integral.
What you are proposing is that the equality should read $$\int_{\mathbb{R}} x \mapsto \left(e^{-x^2} \left(\int_{\mathbb{R}} f\right)\right)$$ $$=\int_{\mathbb{R}} \left( x \mapsto \int_{\mathbb{R}} g\right),$$ where $g(t)=e^{-t^2}f(t)$. This doesn't make sense. Essentially, what you are implying is that $$f \cdot\int g=\int f\cdot g,$$ which is not true. | 2021-08-02T00:46:57 | {
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http://mathhelpforum.com/statistics/29518-probability-question.html | # Math Help - Probability question
1. ## Probability question
A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E.
Find the probability that both of the letters to A and B are in incorrect envelopes.
The answer is given as 13/20 but I can't figure out why. Can anyone help?
2. Originally Posted by jjc
A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E. Find the probability that both of the letters to A and B are in incorrect envelopes. The answer is given as 13/20
The number of ways that A or B is in the correct envelope is give by:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{2(4!) - 3!}}{{5!}}$
Now, for neither to get the correct letter subtract from 1.
3. Hello, jjc!
A man writes 5 letters, one each to A, B, C, D and E.
Each letter is placed in a separate envelope and sealed.
He then addressed the envelopes, at random, one each to A, B, C, D and E.
Find the probability that both of the letters to A and B are in incorrect envelopes.
Answer: $\frac{13}{20}$
Here is a logical (but very primitive) approach . . .
The letters are: . $A,B,C,D,E$
Their envelopes are: . $a,b,c,d,e$
There are two cases to consider:
. . (1) Letter $A$ is in envelope $b.$
. . (2) Letter $A$ is not in envelope $b.$
(1) Letter $A$ is in envelope $b$: . $P(\text{A in b}) \:=\:\frac{1}{5}$
Then letter $B$ can go in any of the other four envelopes: $\{a,c,d,e\}$
. . $P(\text{B wrong}) \:=\:\frac{4}{4} \:=\:1$
Hence: . $P(\text{A in b }\wedge\text{ B wrong}) \:=\:\frac{1}{5}\cdot1 \:=\:{\color{blue}\frac{1}{5}}$
(2) Letter $A$ is in not in envelope $b$ ... it is in $\{c,d,e\}$
. . $P(\text{A not in b}) \:=\:\frac{3}{5}$
Then letter B must be in one of the three remaining envelopes.
. . $P(\text{B wrong}) \:=\:\frac{3}{4}$
Hence: . $P(\text{A not in b} \wedge\text{ B wrong}) \:=\:\frac{3}{5}\cdot\frac{3}{4} \:=\:{\color{blue}\frac{9}{20}}$
Therefore: . $P(\text{A wrong }\wedge\text{ B wrong}) \;=\;\frac{1}{5} + \frac{9}{20} \;=\;{\bf{\color{red}\frac{13}{20}}}$ | 2015-09-05T15:04:19 | {
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https://www.intmath.com/forum/trigonometric-graphs-24/phase-shift:19 | IntMath Home » Forum home » Graphs of the Trigonometric Functions » Phase shift
Phase shift [Solved!]
My question
Hello, I have a question about phase shift. What makes the second approach incorrect?
It makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).
However, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1
Thanks a lot!
Relevant page
3. Graphs of y = a sin(bx + c) and y = a cos(bx + c)
What I've done so far
Drawn several graphs, but couldn't conclude anything
X
Hello, I have a question about phase shift. What makes the second approach incorrect?
It makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).
However, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1
Thanks a lot!
Relevant page
<a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a>
What I've done so far
Drawn several graphs, but couldn't conclude anything
Re: Phase shift
Hello Andrew
I think the problem here is with the concept of "starting off". Let's use "t" as the variable and talk about when things occur.
Let's also just talk about sin to keep it simple.
If we set the expression inside the sin (that is, bt + c) to 0, we are saying "at what time does sin have value 0?", since sin 0 = 0. Then by solving, we get t = -c/b.
The effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is "starting off [from value 0]" at t = -c/b.)
Now let's think about y = a sin(bt+c) when the TIME is 0 ("starting off when we flip the switch on"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.
Does that help?
I have added a new example on 3. Graphs of y = a sin(bx + c) and y = a cos(bx + c) - I hope that makes it clearer.
Regards
X
Hello Andrew
I think the problem here is with the concept of "starting off". Let's use "t" as the variable and talk about when things occur.
Let's also just talk about sin to keep it simple.
If we set the expression inside the sin (that is, bt + c) to 0, we are saying "at what time does sin have value 0?", since sin 0 = 0. Then by solving, we get t = -c/b.
The effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is "starting off [from value 0]" at t = -c/b.)
Now let's think about y = a sin(bt+c) when the TIME is 0 ("starting off when we flip the switch on"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.
Does that help?
I have added a new example on <a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a> - I hope that makes it clearer.
Regards
Re: Phase shift
Your new example helped a lot.
Thanks.
X
Your new example helped a lot.
Thanks.
You need to be logged in to reply. | 2017-06-23T22:14:07 | {
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https://engineering.stackexchange.com/questions/37698/how-to-get-a-transfer-function-from-this-block-diagram | # How to get a transfer function from this block diagram?
Pardon my paint skills, I did my best
My attempt is short and seems to fail, I have no idea why:
\begin{align} \alpha &= in - a_{1}\beta - a_{2}\gamma \\ \beta &= \alpha z^{-1} \\ \gamma &= \beta z^{-1} = \alpha z^{-2} \end{align}
inputting 2nd and 3rd equation into the first one I get:
\begin{align} \alpha &= in - a_{1}\alpha z^{-1} - a_{2}\alpha z^{-2} \\ in &= \alpha + a_{1}\alpha z^{-1} + a_{2}\alpha z^{-2} \end{align}
I can write the output as:
\begin{align} out &= b_{2}\gamma + b_{1}\beta + b_{0}\alpha \\ out &= b_{2}\alpha z^{-2} + b_{1}\alpha z^{-1} + b_{0}\alpha \end{align}
I have output and input in terms of alpha, but I can't figure what to do from here.
You are going in the right direction! Lets take these two equations: $$(1) \quad in = \alpha+a_1\alpha z^{-1}+a_2\alpha z^{-2}$$ $$(2) \quad out = b_0\alpha+b_1\alpha z^{-1}+b_2\alpha z^{-2}$$ now rewrite (1) such that it becomes a function of $$\alpha$$: $$in = \alpha\left(1+a_1z^{-1}+a_2z^{-2}\right)$$ $$\alpha = \frac{in}{1+a_1z^{-1}+a_2z^{-2}}$$ Substitute $$\alpha$$ in equation (2): $$out = in\frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$ And derive proper discrete transfer function from it: $$H(z) = \frac{out}{in} = \frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$
• Thank you kind sir. I'm ashamed I didn't figure it out myself. But you made a little mistake I believe, in the second (2) equation the last term should have b2 instead of a2, and that mistakes propagates to the results as well Sep 17 '20 at 11:14
• Oops, small copy-past error haha. fixed it! Sep 17 '20 at 11:15
Just wanted to add that you could use this to draw diagrams next time. Much easier to use than paint. You can even use Latex.
• This really should be posted as a comment and not an answer. Oct 13 '20 at 15:19
• Don't have the reputation to do so, just need 4 more to get to 50 lol Oct 13 '20 at 19:08
• Just one good answer will get you there! Oct 13 '20 at 22:13
• If you used this website to draw a new diagramm for thr op and then answer the OPs question i’d definitely +1 Oct 14 '20 at 8:59 | 2021-12-05T23:15:23 | {
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https://math.stackexchange.com/questions/2093600/why-is-the-following-relation-transitive-but-not-reflexive | # Why is the following relation transitive, but not reflexive?
In a practice paper for an exam there is the following relation:
$$E = \{(1,1),(2,2),(3,3),(4,4)\}\ \text{ on the set }V = \{1,2,3,4,5\}$$
It would appear that because $(5,5)$ is not in $E$, that it would not be a reflexive relation.
What is unclear is how the relation is still transitive. By our definition for Transitivity, for all elements in set $V$, there would have to be $(a,b)$ and $(b,c)\ldots$ and therefore $(a,c)$ in the set $E$.
This works for element $1$ in $V$ because you have $(1,1)$ and $(1,1)$ and therefore $(1,1)$. I would have assumed however that because there is no tuple $(5,5)$ in $E$ the entire relation could not be transitive?
Hope this makes sense, thanks for any input offered.
• "By our definition for Transitivity, for all elements in set V, there would have to be (a,b) and (b,c)" No. This is not the definition. ( a,b) do not have to be in E for all a,b. But IF (a,b) does exist, AND (b,c) exist, then (a,c) must also exist. But if (a,b) does not exist... no harm, no foul. [Note: if (a,b) did have to exist for a,b in V, then E has to be V x V = {(1,1)(1,2)(1,3)..... (5,3),(5,4)(5,5)} all, every single one of the 25 pairs. – fleablood Jan 11 '17 at 17:35
Reflexive means for all $a \in V$ then $(a,a) \in E$. That fails because $5 \in V$ and $(5,5) \not \in V$.
Transitive says nothing about items of $V$ but only about items in $E$. The relation is transitive if every time you have an $(a,b) \in E$ and a $(b,c) \in E$ you must also have a $(a,c) \in E$ then the relation is transitive. If you never have an $(5,x)$ or $(y, 5) \in E$ that will in no way affect what you do have in $E$.
This particular relationship if you have $(a,b) \in E$ then $a \ne 5; b\ne 5;$ and $a = b$. So if $(a,b), (b,c) \in E$ then $a = b = c$ and $(a,c) = (a,a) = (a,b) \in E$. So the relationship is transitive. That $a,b,c$ can never be equal to $5$ is utterly irrelevant.
• I never thought about it, but reflexive is the only condition that pertains to the elements of V. Symmetry pertains only to elements of E in that if you have this element in V you must have that element too. Transitivity says if you have two elements in V you must also have a third. A function could still be transitive if you have (a,b) but absolutely no (b,x). For transitivity to fail you must have both an (a,b) and an (b,c) but not the (a,c). If you don't the (a,b) or you don't have the (b,c) it doesn't matter whether or not you have the (a,c). So no (5,x) or (x,5) doesn't matter. – fleablood Jan 11 '17 at 17:51
• Yes, @fleablood. In your last point, when you do not have some elements a, b, c, in V, such that $(a, b), (b, c) \in E$, it is a vacuously true, that transitivity holds regardless of whether or not $(a, c)$ is, or is not, in $E$. Also, as you seem to have gleaned, it is helpful to examine failures of properties of relations, for which the property fails. If it does not fail, the property holds. – Namaste Jan 11 '17 at 18:16
• Hmmm, transitivity does not require three elements in some set $V$ of a transitive relation $E$. If $E = \{(a, a)\}$, E is indeed transitive, (and symmetric). – Namaste Jan 11 '17 at 18:23
• I didn't mean to imply transitivity required 3 elements. As you point out if there are no (a,b) (b,c) pairs it's vacuously transitive. If it's non-vacuously transitive it must have (a,b) and (b,c) and a cooresponding (a,c). But a, b, and c need not be distinct. So (a,a) (a,a) and (a,a) will do. – fleablood Jan 11 '17 at 22:11
Definition: $R$ is transitive if and only if $\forall x,y (R(x,y)$ and $R(y,x))\Rightarrow R(x,z)$
Can you point to where exactly you think this definition fails? For $x=y=5$ the antecedent fails to hold (since $R(5,5)$ is false) and therefore the implication is true, since an implication with a false antecedent is always true.
Your definition of transitivity is unclear as stated, and you have misunderstood it, there is no requirements on the members of the underlying set, only on the elements of the relation.
Proof $E$ is transitive: if $E\left(a,\,b\right)\land E\left(b,\,c\right)$ then $a=b\in V\backslash \left\{ 5\right\}$ and similarly $b=c\in V\backslash \left\{ 5\right\}$, so $a=c\in V\backslash \left\{ 5\right\}$ and $E\left(a,\,c\right)$. | 2019-07-24T00:18:58 | {
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http://math.stackexchange.com/questions/566232/mirror-reflection | # mirror reflection
1. The problem statement, all variables and given/known data
Question A. Draw accurately in the figure the light beam that goes from Object A to eye B after being reflected on the mirror. It must be consistent with the mirror principle!
Question B. At question A. you may have connected the point behind the mirror (A') with eye B. And found the correct light path that way.
Will that also work if you use the point behind the mirror of B (B') instead? Do you get the same result? Use mathematical arguments for your judgement!
2. Relevant equations mirror reflection rules
3. The attempt at a solution
Question A: I tried to draw point A' and connected that to B. The purple line is the normal line (the imaginary line that is perpendicular to the mirror) And the angle between the mirror and A is the same as the angle between the mirror and B.
Question B: I think this is true, I can draw the same lines, but I don't know what is meant by 'mathematical arguments'
-
This is a very nicely written question. If only we had more like this one... :-) – Vedran Šego Nov 14 '13 at 4:14
Notice that $AA'B'B$ is an isosceles trapezoid, so its diagonals intersect on the mirror, which means that using either $\triangle AA'B$ or $\triangle AB'B$ will give you the same point.
You'll probably need to provide a bit more arguments that the diagonals intersect on the mirror. Read the linked article and ask if you get stuck.
Edit: Here is a bit more:
We can extend $AA'B'B$ to a triangle. Let us denote the new vertex as $C$. It is on the intersection of the lines on which $\overline{AB}$ and $\overline{A'B'}$ lie.
Note that
1. $|\overline{AA''}| = |\overline{A'A''}|$, where $A''$ denotes a point at which $\overline{AA'}$ intersects with the mirror; and
2. the mirror (on which the altitude of the mirror lies) is orthogonal to $\overline{AA'}$.
From points $1$ and $2$, we see that $\triangle AA'C$ is an isosceles triangle. The same can be said for $\triangle BB'C$ (with the same arguments).
Now, since
$$|\overline{AB}| = |\overline{AC}| - |\overline{BC}| = |\overline{A'C}| - |\overline{B'C}| = |\overline{A'B'}|,$$
we see that $AA'B'B$ is an isosceles trapezoid.
Using this, show that the diagonals of $AA'B'B$ intersect on the mirror. Can you take over now?
-
Do you mean with a bit more arguments to proof Similarity (geometry) between the triangles? – user108681 Nov 14 '13 at 4:36
@user108681 I've expanded my answer. – Vedran Šego Nov 14 '13 at 12:15
A bit easier: you can just observe isosceles triangles $\triangle AA'E$ and $\triangle BB'E$, where $E$ is the point on the mirror where $\overline{AB'}$ and $\overline{A'B}$ intersect (i.e., the point you're looking for). – Vedran Šego Nov 14 '13 at 18:31 | 2016-07-24T22:27:26 | {
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https://math.stackexchange.com/questions/1900072/why-is-minimizing-least-squares-equivalent-to-finding-the-projection-matrix-ha | # Why is minimizing least squares equivalent to finding the projection matrix $\hat{x}=A^Tb(A^TA)^{-1}$?
I understand the derivation for $\hat{x}=A^Tb(A^TA)^{-1}$, but I'm having trouble explicitly connecting it to least squares regression.
So suppose we have a system of equations: $A=\begin{bmatrix}1 & 1\\1 & 2\\1 &3\end{bmatrix}, x=\begin{bmatrix}C\\D\end{bmatrix}, b=\begin{bmatrix}1\\2\\2\end{bmatrix}$
Using $\hat{x}=A^Tb(A^TA)^{-1}$, we know that $D=\frac{1}{2}, C=\frac{2}{3}$. But this is also equivalent to minimizing the sum of squares: $e^2_1+e^2_2+e^2_3 = (C+D-1)^2+(C+2D-2)^2+(C+3D-2)^2$.
I know the linear algebra approach is finding a hyperplane that minimizes the distance between points and the plane, but I'm having trouble understanding why it minimizes the squared distance. My intuition tells me it should minimize absolute distance, but I know this is wrong because it's possible for there to be non-unique solutions.
Why is this so? Any help would be greatly appreciated. Thanks!
• Check your derivation for $\hat{x}$ and you will find (as @Ian points out) the expression should be $\hat{x}=(A^TA)^{-1}A^Tb$. Perhaps you should edit this into the Question, though it may be the crux of why you had trouble "connecting it to least squares regression" (minimization). Aug 22 '16 at 15:00
You should be multiplying by $(A^T A)^{-1}$ on the left, not the right. Anyway, the geometric point is that you want $Ax-b$ to be perpendicular to $Ay$ for every vector $y$. (I think this is most easily seen by a geometric argument, which can be easily found in books, but which I can't easily render here.) This translates to $(Ay)^T(Ax-b)=0$ for every $y$, which is the same as $y^T(A^T(Ax-b))=0$ for every $y$. This can only happen if $A^T(Ax-b)=0$, which rearranges to your form if $A^T A$ is invertible (as is usually the case).
Also, it is the same to minimize the square of the Euclidean distance as it is to minimize the Euclidean distance itself. (This is also true of any other nonnegative quantity.) What would be different is minimizing some other distance, like the "taxicab" distance (where you sum the absolute values). Why we should choose to minimize the Euclidean distance in the first place is not a purely mathematical question, it depends on where the problem is coming from. That question is a bit off-topic here, though, and has also been asked before on MSE. (The short version of that discussion: "it's mathematically convenient" and "see the Gauss-Markov theorem".)
• About minimizing the Euclidean distance: I'm a little confused about the distinction. For example, least squares would penalize a point for being farther away: e.g. a point 1 unit away would be penalized by 1 whereas a point 2 units away would be penalized by 4. I think my confusion lies in the difference between minimizing an aggregate of points vs individual ones, since it seems to me that minimizing euclidean distance for each point wouldn't account for the distance factor in least squares. Aug 22 '16 at 15:30
• In your problem the goal is to choose a vector in a certain 2D subspace which is closest to another vector in 3D space, where you measure distance in the Euclidean sense. This is not measured separately by components but rather through the usual distance formula.
– Ian
Aug 22 '16 at 15:39
The matrix has full column rank; we are guaranteed a unique solution.
Problem statement
\begin{align} \mathbf{A} x &= b \\ \left[ \begin{array}{cc} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] &= \left[ \begin{array}{cc} 1 \\ 2 \\ 2 \\ \end{array} \right] \end{align}
Normal equations
\begin{align} \mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b \\ % \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] &= \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} 1 \\ 2 \\ 2 \\ \end{array} \right] \\[3pt] % \left[ \begin{array}{cc} 3 & 6 \\ 6 & 14 \\ \end{array} \right] % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] % &= % \left[ \begin{array}{cc} 5 \\ 11 \\ \end{array} \right] % \end{align}
Least squares solution
\begin{align} x_{LS} &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} b \\ % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] &= \frac{1}{6} \left[ \begin{array}{cc} 14 & -6 \\ -6 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} 5 \\ 11 \end{array} \right] \\ % &= \frac{1}{6} \left[ \begin{array}{cc} 4 \\ 3 \end{array} \right] % \end{align} | 2021-09-20T08:15:47 | {
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https://math.stackexchange.com/questions/155839/on-ces%C3%A0ro-convergence-if-x-n-to-x-then-z-n-fracx-1-dots-x-nn?noredirect=1 | # On Cesàro convergence: If $x_n \to x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \to x$
I have this problem I'm working on. Hints are much appreciated (I don't want complete proof):
In a normed vector space, if $x_n \longrightarrow x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \longrightarrow x$
I've been trying adding and subtracting inside the norm... but I don't seem to get anywhere.
Thanks!
Given $\epsilon >0$ there exists $n_0$ such that if $n\geq n_0$ then $\parallel x_n -x\parallel < \epsilon$
so
\begin{align*} 0 & \leq \left\lVert \frac{x_1 +\cdots +x_n}{n} -x \right\rVert \leq \left\lVert \frac{x_1 + \dots + x_n - nx }{n} \right\rVert \\ & \leq \frac{\lVert x_1 - x \rVert}{n} + \dots + \frac{\lVert x_{n_0 - 1} - x \rVert}{n} + \frac{\lVert x_{n_0} - x \rVert}{n} +\dots + \frac{\lVert x_{n} - x \rVert}{n} \\ &\le \frac 1n\sum_{i=1}^{n_0-1} \| x_i -x\| + \frac{n-n_0}{n} \epsilon \end{align*}
The first $n_0 -1$ terms $\| x_i -x\|$ can be bounded by some $M$, thus for $n\ge (n_0-1)M/\epsilon=: N_0$ we have $$\frac 1n\sum_{i=1}^{n_0-1} \| x_n -x\| \le \frac 1n (n_0-1)M \le \epsilon$$
Thus $$\left\| \frac{x_1 + \cdots x_n}{n} - x\right\| <2\epsilon$$ when $n\ge N_0$.
Thanks a lot @Leonid Kovalev for the inspiration, though my main problem was that I wasn't aware of what to do with the $nx$ (the silliest part :P)
• This sort of thing is encouraged, I think. – Dylan Moreland Jun 11 '12 at 3:21
• I think you want "$-nx$" in the first line of the display, and $\|x_n - x\|$ at the end. I don't think you want to say that the first few terms are $\leq M$; that doesn't seem to be enough. – Dylan Moreland Jun 11 '12 at 3:41
• @DylanMoreland: Why isn't it enough? Can you explain? – Bouvet Island Jun 11 '12 at 14:56
• @Inti: I think you are missing something in the argument. Generally the argument consists of two steps: first choose $n_0$ such that if $n \geq n_0$, $\|x_n -x \| < \epsilon / 2$. Next choose $N \geq n_0$ such that $M$ (which is at least $\sup_n \|x_n\|$) satisfies $M / N < \epsilon / 2$. I don't see the second part of the argument implemented in your answer. – Willie Wong Jun 11 '12 at 14:59
• Should it not be $(n-n_0+1)\epsilon/n$? – jippyjoe4 Oct 14 '19 at 4:19
There is a slightly more general claim:
PROP Let $$\langle a_n\rangle$$ be a sequence of real numbers, and define $$\langle \sigma_n\rangle$$ by $$\sigma_n=\frac 1 n\sum_{k=1}^n a_k$$
Then $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n \;(\;\leq\;)\;\limsup_{n\to\infty}\sigma_n\leq \limsup_{n\to\infty}a_n$$
P We prove the leftmost inequality. Let $$\ell =\liminf_{n\to\infty}a_n$$, and choose $$\alpha <\ell$$. By definition, there exists $$N$$ such that $$\alpha for any $$k=0,1,2,\ldots$$ If $$m>0$$, then $$m\alpha <\sum_{k=1}^m \alpha_{N+k}$$
which is $$m\alpha<\sum_{k=N+1}^{N+m}a_k$$
$$(m+N)\alpha+\sum_{k=1}^{N}a_k<\sum_{k=1}^{N+m}a_k+N\alpha$$
which gives
$$\alpha+\frac{1}{m+N}\sum_{k=1}^{N}a_k<\frac{1}{m+N}\sum_{k=1}^{N+m}a_k+\frac{N}{m+N}\alpha$$
Since $$N$$ is fixed, taking $$\liminf\limits_{m\to\infty}$$ gives $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ (note that $$N+m$$ is just a shift, which doesn't alter the value of the $$\liminf^{(1)}$$). Thus, for each $$\alpha <\ell$$, $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ which means that $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n$$ The rightmost inequality is proven in a completely analogous manner. $$\blacktriangle$$.
$$(1)$$: Note however, this is not true for "non shift" subsequences, for example $$\limsup_{n\to\infty}(-1)^n=1$$ but $$\limsup_{n\to\infty}(-1)^{2n+1}=-1$$
COR If $$\lim a_n$$ exists and equals $$\ell$$, so does $$\lim \sigma_n$$, and it also equals $$\ell$$. The converse is not true.
WLOG, the $x_n$ converge to $0$ (otherwise consider the differences $x_n-x$), and stay confined in an $\epsilon$-neighborhood of $0$ after $N_\epsilon$ terms.
Then the average of the first $m$ terms is bounded by
$$\frac{N\overline{x_N}+(m-N)\epsilon}m,$$ which converges to $\epsilon$. So you can make the average as close to $0$ as you like. | 2021-05-16T03:33:47 | {
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http://math.stackexchange.com/questions/575960/how-many-triangles-in-picture | # How many triangles in picture?
Can you tell me how many triangles are in this picture? I've counted 96, not sure that I got right answer.
-
Perhaps you could try counting them by using the symmetry of the shape (if you haven't already). – Alice Nov 21 '13 at 13:50
Clearly there are 19 unique points. Number of triangles that can be formed from 19 points taking 3 at a time=$19\choose3$. Also note that some of the points lie on a straight line. When we took $19\choose3$ we included the ones that lie on a straight line. Perhaps finding the number of lines and subtracting that from $19\choose3$ will yield the answer. – GTX OC Nov 21 '13 at 14:05
@GTXOC I see 19 unique points... – apnorton Nov 21 '13 at 14:08
Alright I missed the ones in the middle of the base. – GTX OC Nov 21 '13 at 14:09
@All: Not all triangles are valid even if the points could form one, consider lines between "star-corners". – AlexR Nov 21 '13 at 14:11
Let's see! We'll break it up into the following cases (where an apex means one of the six points of the star):
A: Triangles that contain three apexes
There are 2 of these.
B: Triangles that contain a pair of opposite apexes
For each such pair, there are 2 of these, so 6 in total.
C: Triangles that contain a pair of non-opposite apexes
For each non-adjacent apex pair, there are three of these, so 18 in total.
D: Triangles that contain exactly one apex
For each apex, I count 10 such triangles, so 60 in total.
E: Triangles that contain no apex
For each edge of the internal hexagon, there are 3 triangles, so 18 in total.
I make that 2 + 6 + 18 + 60 + 18 = 104.
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I verified and this seems to be correct. – AlexR Nov 21 '13 at 14:16
Look OK to me too, though categories $D$ and $E$ could be somewhat refined. – Marc van Leeuwen Nov 21 '13 at 14:24
@Marc: I agree with you about category D. – TonyK Nov 21 '13 at 14:29
A small reality check: If you dismiss the two big equilateral triangles, all other triangles belong to orbits of $6$ under rotations by $2\pi/6$, so $T-2$ must be divisible by $6$, which $T=104$ is but $T=96$ (the OP's original count) is not. – Barry Cipra Nov 21 '13 at 14:54
For Category $D$, you can take the apex at the very top and count triangles that are and aren't symmetric across the "$y$"-axis. There are $2$ that are and $4$ pair that aren't. – Barry Cipra Nov 21 '13 at 15:13
I only get 98. I am missing the 3rd triangle under section C. I only see 2 triangles that contain a pair of non-opposite apexes.
- | 2015-08-02T04:26:57 | {
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https://math.stackexchange.com/questions/2965619/how-to-solve-the-following-congruences-equation-system | # How to solve the following congruence's equation system?
Through many operations in an exercise I've reached this point
$$17≡4a+19b(mod$$ $$27)$$
$$8≡11a(mod$$ $$27)$$
I want to find both a and b to have the answer I need, I've been trying to use the Chinese rest theorem but found it doesn't work in my case, I'm starting converting the second one into
$$11a+27a≡1(mod$$ $$27)$$
Despite this I'm not sure if is correct and I still don't know how to clear this $$a$$ in the second equation in order to replace it in the first.
Any help will be really appreciated
There are a number of ways to go about this. We want to be able to "divide" by $$11$$, so we need $$8 + 27x = 11y$$. Solving this linear Diophantine equation (an elementary problem, feel free to Google) gives $$x = 5$$ and $$y = 13$$ as possible solutions. Then, \begin{align} 8 &\equiv 11a \mod{27}\\ 8 + 27\cdot 5 &\equiv 11a \mod{27}\\ 13(11) &\equiv 11a \mod{27}\\ 13 &\equiv a \mod{27} \end{align} Then $$a = 13 + 27k$$ for some $$k \in \mathbb{Z}$$. You can subtract $$4$$ times the last congruence above from $$17 \equiv 4a + 19b \mod{27}$$ to get \begin{align} 17 - 4(13) &\equiv 19b \mod{27}\\ -35 &\equiv 19b \mod{27}\\ -35 + 2(27) &\equiv 19b \mod{27}\\ 19 &\equiv 19b \mod{27}\\ b &\equiv 1 \mod{27} \end{align}
• But you should say how you computed $k$ such that $\,11\mid 8+27k,\,$ not simply pull it out of hat like magic! Else how is the OP supposed to figure out how to use the method for similar problems? – Bill Dubuque Oct 22 '18 at 19:11
• @BillDubuque I alluded to it in my explanation... Solving a linear Diophantine equation is usually a skill that's learned in sections immediately preceding problems like this. For the record though, the Euclidean algorithm with back substitution is the most systematic way of doing it (again, Google is your friend). – AlkaKadri Oct 22 '18 at 19:59
• Back-substitution is clumsy and error-prone. Better to use this version, or Gauss's algorithm below $$\!\bmod 11\!:\ \, 27x\!+\!8\equiv 0\iff x\equiv \dfrac{-8}{27}\equiv \dfrac{3}{-6}\equiv \dfrac{-1}{2}\equiv \dfrac{10}2\equiv 5\$$ In any case, it's not a good idea to say "Google it" in an answer. Google can locate all sorts of nonsense (low-quality, errorneous, crankish, etc). – Bill Dubuque Oct 22 '18 at 20:22
• @BillDubuque ah I see.. Yes you have a point there, I’ll certainly refrain from doing that in future answers. Thanks Bill! – AlkaKadri Oct 22 '18 at 20:25
Just so long as you avoid multiplying or dividing by a multiple of $$3$$ you can use ordinary arithmetic - so you can multiply the first by $$11$$ and the second by $$4$$ to isolate a term in $$b$$ using standard elimination. You avoid $$3$$ because the base $$27$$ has the prime factor $$3$$.
You can solve $$ax+by=1$$ for coprime $$a$$ and $$b$$ using the division algorithm to find their highest common factor (which is $$1$$ because they are coprime). Other methods are available, and sometimes work more quickly if you spot them. The division algorithm provides a proof that there is always a solution, and a systematic way of finding it.
• +1 Eliminating $a$ by cross multiplying yields $-2\equiv\!\!\overbrace{ 5}^{\large 44a}\!\!-7b\,\Rightarrow\, 7b\equiv 7\,\Rightarrow\, b\equiv 1\,$ (use least magnitude reps!), which is probably easier than computing inverses. – Bill Dubuque Oct 22 '18 at 19:01
Guide:
\begin{align} 27 &= 11(2)+5 \\ 11 &= 2(5)+1 \end{align}
\begin{align} 1 &= 11-2(5)\\ &= 11 - 2(27-2(11)) \\ &= 5(11)-2(27) \end{align}
Hence $$11^{-1} \equiv 5 \pmod{27}$$
Now you should be able to solve for $$a$$, substitute that to the first equation and solve for $$b$$ using similar procedure.
\begin{align*} 11a & \equiv 8 \pmod{27}\\ 55a & \equiv 40 \pmod{27}\\ a & \equiv 13 \pmod{27}. \end{align*} Now plug this in the first congruence to get $$b$$.
• But you should say how you computed the inverse of $11,\,$ not simply pull it out of a hat like magic! – Bill Dubuque Oct 22 '18 at 19:05
• @BillDubuque My intention was not to pull a magic for OP. But when OP says (in his question) that he/she tried using the Chinese Remainder Theorem, then I would assume familiarity of such a step (multiplying by an inverse) on his/her part. One can always seek more explanation if needed. – Anurag A Oct 23 '18 at 16:21
• But the question makes it clear the OP is having difficulty inverting $\,11\bmod 27.\,$ Giving the answer with no hint how it was computed is probably not going to help the OP get past their obstacles. – Bill Dubuque Oct 23 '18 at 16:36
• It is not that I disagree with your suggestion of adding more but it is a judgement call. When someone says CRT etc. and yet misses some basic step, then one can argue either way about his/her understanding of the basic material. Perhaps OP overlooked some simple step or perhaps OP has difficulty with the concept of inverse. I grapple with this on a daily basis :-) – Anurag A Oct 23 '18 at 17:00 | 2019-07-16T14:53:05 | {
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https://math.stackexchange.com/questions/2408906/difference-between-proof-by-reductio-ad-absurdum-and-proof-by-contradiction | # Difference between “ proof by reductio ad absurdum” and “proof by contradiction”?
I always thought that both “proof by reductio ad absurdum” and “proof by contradiction” mean the same, but now my professor asked this question on my homework and I don't know.
I believe that in both cases you assume the negation of the conclusion and develop a contradiction through the premises. This will imply the conclusion. Today I have a meeting with the assistant professor so I can clarify this, but I really would like to know what you guys think, or if possible it would be great if you point me into some good references.
UPDATE:
I just came from my extra help and the assistant professor explains the difference this way:
Reductio ad absurdum: $$\vDash [\neg p\to(q\wedge\neg q)]\to p$$
Proof by contradiction:
$$\vDash [\neg (p\to q) \to (r\wedge \neg r)]\to (p\to q)$$
And the examples of application were these:
Using proof by contradiction: $\sqrt2$ is irrational.( First suppose it is rational and derive a contradiction).
Using proof by reductio ad absurdum: If $f$ is differentiable on $(a,b)$ then $f$ is continuous on $(a,b)$. ( First we suppose that $f$ is differentiable on $(a,b)$ but not continuous on $(a,b)$ and derive a contradiction).
• – R. Suwalski Aug 28 '17 at 15:49
• I always thought of it as: reductio ad absurdum is proving $\lnot \phi$ by assuming $\phi$ and proving $\bot$ or $\psi \wedge \lnot \psi$. Proof by contradiction is proving $\phi$ by assuming $\lnot \phi$ and proving $\bot$ or $\psi \wedge \lnot \psi$. So, for example, in intuitionistic systems, reductio ad absurdum is still valid, but proof by contradiction wouldn't be as all it proves is $\lnot \lnot \phi$. (Not sure if this is officially valid, though.) – Daniel Schepler Aug 28 '17 at 15:55
• Usually (but the distinction is not so "stable") the proof by contradiction is of the form: "if from $A$ a contradiction follows, then $\lnot A$ can be inferred". In the indirect proof the assumption is $\lnot A$ and the conclusion inferred (through the contradiction) is $A$. – Mauro ALLEGRANZA Aug 28 '17 at 15:58
• Possible duplicate of Difference between proof of negation and proof by contradiction – Clement C. Aug 28 '17 at 16:08
• Ok. So "reductio ad absurdum" is synonymous with "indirect proof", and "proof of negation"? – Novato Aug 28 '17 at 16:17
## 1 Answer
Regarding the rule of indirect proof:
"if from assumption $\lnot A$ a contradiction follows, we can infer $A$",
we can see:
Sometimes the nomenclature RAA is used; it stands for reductio ad absurdum, the mediæval Latin name of the principle. [...] A genuine indirect proof in propositional logic ends with a positive conclusion.
The principle is equivalent to Double Negation elimination.
If we agree with this approach, proof by contradiciton is more general, because it applies also to inferences with negative conclusion, licensed by the principle of Negation Introduction:
"if from assumption $A$ a contradiction follows, we can infer $\lnot A$". | 2019-10-15T03:41:53 | {
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https://www.physicsforums.com/threads/difference-between-lim-as-x-and-lim-as-x.634344/ | # Difference between lim as x→∞ and lim as |x|→∞
1. Sep 8, 2012
### Cristopher
I came across something I'd never seen before, the use of |x| instead of just ‘x’ in limits.
What is the difference between $\displaystyle\lim_{|x|\to\infty}x\sin\frac{1}{x}$ and $\displaystyle\lim_{x\to\infty}x\sin\frac{1}{x}$ ?
Is there any difference when evaluating them? Is that notation used only with infinity?
Thanks.
2. Sep 8, 2012
### Vorde
It seems like it's implying that the limit as x goes to infinity is equal to the limit as x goes to negative infinity. But someone who has actually seen this notation before might know better, I'm just guessing.
3. Sep 8, 2012
### alberto7
I haven't seen limits as |x|→∞, but I have seen limits as x→±∞ or limits as x→∞ with this meaning. In the latter case they distinguished limits as x→+∞ and as x→-∞. I think they pictured ∞ as a single point outside the line, making it a circle, its Alexandroff compactification, considering limits as x→+∞ and as x→-∞ as the lateral limits towards ∞.
About the limits as |x|→∞, I think it could be generalized in the following way. We may say that f(x)→a as g(x)→b if, for any neighborhood U of a, there exists a neighborhood V of b such that f(g-1(V\{b}))⊆U. Also f(x)→a as g(x)→∞ if, for any neighborhood U of a, there exists M>0 such that f(g-1((M,∞)))⊆U, and similar definitions.
Last edited: Sep 8, 2012
4. Sep 8, 2012
### HallsofIvy
Staff Emeritus
I see two different ways of interpreting "limit as |x| goes to infinity". First would be that "limit as x goes to infinity" and"limit as x goes to -infinity" must be the same. The other would be that x represents a point in the plane or a complex number and x goes away from the origin in any direction.
5. Sep 8, 2012
### ObsessiveMathsFreak
The first notation only really makes sense if you consider 'x' to be a complex variable. In this case, taking the limit at infinity means asking whether the function approaches a fixed value no matter which direction you approach infinity from.
The second notion on the other hand can be used if x is real and you are simply going to "positive" infinity.
At infinity, functions of complex variables generally either have a finite limit, or else a pole. They could also have an essential singularity (O~o) as well. And if the function is multi-valued, it will almost always have a (irregular) branch point there.
6. Sep 9, 2012
### Cristopher
Thank you all.
Yes, I also thought that it says that the limit as x→+∞ and as x→-∞ are the same. Indeed, I did look at the graph of the funcion x sin(1/x) before asking, and these limits are both equal to 1. I also noticed there's symmetry in the graph, I thought it may have something to do with that, too. I posted the question in the hope of getting more info on the scope of this notation.
Just for reference I saw the notation here:
http://en.wikipedia.org/wiki/L%27Hopital%27s_rule
In the section ‘Other ways of evaluating limits’ | 2017-11-21T18:27:39 | {
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https://algo.franklinqin0.me/factorial-trailing-zeroes.html | # Factorial Trailing Zeroes
## # Solution
Brute force solution is to calculate $n!$ and keep dividing by and count number of $10$'s.
### # Number of 5's
Factor a number: $60 = 2 \times 2 \times 3 \times 5$. We can see $60$ only has 1 trailing zero b/c the number of pair of $2$ and $5$ is 1.
For a factorial, the number of 2's is definitely more than that of 5's.
Thus, the number of 5's is the number of trailing zeros.
Complexity:
• time: $O(\log n)$
• space: $O(1)$
def trailingZeroes(self, n: int) -> int:
cnt = 0
while n>0:
cnt += n//5
n //= 5
# this also works
# n //= 5
# cnt += n
return cnt
Last Updated: 9/21/2020, 4:44:16 PM | 2020-09-30T15:43:51 | {
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https://math.stackexchange.com/questions/1595620/the-probability-that-in-a-game-of-bridge-each-of-the-four-players-is-dealt-one-a | # The probability that in a game of bridge each of the four players is dealt one ace
The question is to show that the probability that each of the four players in a game of bridge receives one ace is $$\frac{24 \cdot 48! \cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$. I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?
Imagine $4$ groups of $13$ spots for the $4$ hands
$\small\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square$
An ace can be placed in each group in any of $13$ places, hence $13^4$
The aces' suits can be distributed between groups in $4!$ ways, hence $24$
The remaining cards can be placed in 48! ways,
and 52! is the unrestricted ways of placing the cards
thus (putting the terms in the order you have), $Pr = \dfrac{24\cdot48!\cdot13^4}{52!}$
SIMPLER WAY:
There is a much simpler way to get the same result.
We need an ace in each group of 13, how the rest of the cards go doesn't matter !
The first ace has to be in some group, each of the other aces have to fall in a different group, so the $2nd$ ace has $39$ permissible spots out of $51,$ and so on
thus $Pr = \dfrac{39}{51}\cdot\dfrac{26}{50}\cdot\dfrac{13}{49}$
There are two approaches to card-game questions like these. In the first approach, we temporarily assume that order of cards within the hand matters. In doing so, we treat our sample space as all ways of arranging the fifty-two cards in a row.
Approaching like this, we have the following steps for multiplication principle:
• Pick who gets which ace: $4!$ ways
• Arrange the remaining $48$ cards in a row and give them to the players (12 to north, the next twelve to east, the next twelve to south, etc...): $48!$ ways
• Pick where in the hand the ace goes for each player: $13^4$ ways
Note that the final step was necessary in order to have what we count cover all possible ways that the 52 cards be dealt such that every player receives at least one ace. If we hadn't multiplied by $13^4$, it would have been as though we only considered the ace being the first card in each players' hand. Since we are considering order important, the hand $A\spadesuit 2\spadesuit 3\spadesuit\dots$ is a different outcome than $2\spadesuit A\spadesuit 3\spadesuit\dots$
Since there are $52!$ number of equiprobable ways to deal the cards (where order matters) the probability is as given:
$$\frac{4!48!13^4}{52!}$$
My preference is instead to work in the situation that order doesn't matter.
Here, we break apart as multiplication principle:
• Choose which player gets which ace: $4!$ ways
• Choose twelve additional cards for each player: $\binom{48}{12,12,12,12}=\frac{48!}{12!12!12!12!}$ number of ways
There are a total of $\binom{52}{13,13,13,13}=\frac{52!}{13!13!13!13!}$ number of deals (where order within each hand doesn't matter) for a probability then of:
$$\frac{4!\binom{48}{12,12,12,12}}{\binom{52}{13,13,13,13}}$$
which after a short amount of manipulation you see is precisely the same answer as before.
There are $4!$ ways to distribute the aces, then $\frac{48!}{12!^4}$ ways to distribute the other $48$ cards, $12$ to a player.
Without worrying about the aces, there are $\frac{52!}{13!^4}$ ways to distribute $52$ cards, $13$ to a player.
Therefore, the probability is $$\frac{4!\frac{48!}{12!^4}}{\frac{52!}{13!^4}}=\frac{13^4}{\binom{52}{4}}\doteq0.1054982$$
• Being a bridge player, I am very interested by this answer. My question is : does this result means that if I have one ace and my partner has one ace, the probability that one of the opponents has the two other aces is 90% ? Or, is such a statement totally wrong ? – Claude Leibovici Jan 8 '16 at 4:54
• If you have one ace and your partner has one ace, then the other two players have two aces. It should be close to $50\%$ that one has both, but not quite. – robjohn Jan 8 '16 at 8:11
• To count the number of ways that one of them does not have both aces, note that there are $2!$ ways to distribute the aces, then $\frac{24!}{12!^2}$ ways to distribute the other $24$ cards, $12$ to a player. Without worrying about the aces, there are $\frac{26!}{13!^2}$ ways to distribute $26$ cards, $13$ to a player. Therefore, the probability is $$\frac{2!\frac{24!}{12!^2}}{\frac{26!}{13!^2}} =\frac{13^2}{\binom{26}{2}} =0.52$$ So the probability that one of them has both aces is $0.48$. – robjohn Jan 8 '16 at 8:11
• Thanks for tha answer ! I am real bad with probabilities and I totally misunterpreted the result of $1-0.105$. Shame on me !! Cheers – Claude Leibovici Jan 8 '16 at 8:28
Since the order doesn't matter, let's make hands one by one.
1. For the first, there are 4 Aces and I need to choose 1 to give to this person, $\binom{4}{1}$. Then I need to choose 12 cards from the 48 non-Aces to complete this hand, $\binom{48}{12}$. Thus the number of ways to make this hand is $$\binom{4}{1}\binom{48}{12}.$$
2. I already gave the first person an Ace, so I have three left and I need to choose one to give to this person. I also used up 12 non Aces on the last person, so there are 36 non Aces left and I need to choose 12. The number of ways to make this hand is thus, $$\binom{3}{1}\binom{36}{12}.$$
3. and 4. follow the same logic and so I have $$\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}$$ ways to make hands 3 and 4.
Finally, all the possible ways to make 4 13-card hands are $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} = \binom{52}{13,13,13,13} = \frac{52!}{13!\,13!\,13!\,13!}.$$
Lastly, putting it all together, the probability of each person getting an Ace is thus \begin{align*}\frac{\binom{4}{1}\binom{48}{12}\binom{3}{1}\binom{36}{12}\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}}{\binom{52}{13,13,13,13}} &= \frac{4!}{1!3!}\cdot\frac{48!}{12!36!}\cdot\frac{3!}{1!2!}\cdot\frac{36!}{12!24!}\cdot\frac{2!}{1!1!}\\ &\qquad\times\frac{24!}{12!12!}\frac{1!}{1!0!}\cdot\frac{12!}{12!0!}\cdot\frac{13!\,13!\,13!\,13!}{52!}\\ &=\frac{24 \cdot 48! \cdot13^4}{52!}. \end{align*}
After the cards have been shuffled and cut, there are $\binom{52}4$ equally likely possibilities for the set of four positions in the deck occupied by the aces. Among those $\binom{52}4$ sets, there are $13^4$ which result in each player getting an ace; namely, make one of the $13$ cards to be dealt to South an ace, and the same fo West, North, and East. So the probability is $$\frac{13^4}{\binom{52}4}=\frac{4!\cdot48!\cdot13^4}{52!}=\frac{2197}{20825}\approx.1055$$
Suppose you are arranging the cards in rows of 13. Each row must have an ace; if we place the aces as the first cards, there are 4! ways to arrange them. Then, there are 48! ways to arrange the remaining 48 cards. However, this restricts the aces to being the first card; they can, in fact, be in any of 13 positions - for each of the 4 players. So, multiply by 13^4 to account for the 4 aces being in any of the 13 positions. | 2019-10-17T21:17:45 | {
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https://math.stackexchange.com/questions/1461512/urn-i-contains-2-white-and-4-red-balls-whereas-urn-ii-contains-1-white-an | # Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball.
Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is
• the probability that the ball selected from urn II is white?
• the conditional probability that the transferred ball was white given that a white ball is selected from urn II?
I got the answer $a = \frac{4}{9}$ but I need help with $b$.
If I make $P(T) = \text{transfered ball}$ is white $P(T) = \frac13$ and P(W) = selected ball from urn 2 is white.
$P(W) = \frac49$ (from part $a$) I am looking for $P(T|W)$ by Bayes's formula - I get $$P(T|W) = P(T\cap W)/P(W)$$
then I get $$P(W|T) \cdot \dfrac{P(T)}{P(W)}$$.
we know $P(T)$ and $P(W)$ and $P(W|T) = \frac29$.
so i get $\frac29\dfrac{\frac13}{\frac49} = \frac16$ but the answer key says it's $\frac12$?
• We need to find $\Pr(T\cap W)$, and divide by $\Pr(W)$. The probability of $T$ is $2/6$. The probability of $W$ given $T$ is $(2/3)$. So the probability of $T\cap W$ is $(2/6)(2/3)$, which is $2/9$. Now divide by $4/9$ and simplify. Oct 2 '15 at 20:16
In Bayes's formula formula: $P(T|W) = \frac{P(T)*P(W|T)}{P(W)}$
But $P(T)*P(W|T)=P(T\cap W)$ because these events occur at the same time, that is:
1) we have chosen white ball from urn1, probability of which is: $P(T)=\frac{2}{6}$...and now we have two white balls and one red in urn 2.
2) then we have chosen the white ball from urn 2, which means: $P(W|T)=2/3$.
So, $P(T)*P(W|T)=\frac{2}{6}*\frac{2}{3}=\frac{2}{9}=P(T\cap W)$ $$P(T|W) = P(T\cap W)/P(W)$$ We know that $P(W)=\frac{4}{9}$.
So, $P(T|W)=\frac{2}{9}/\frac{4}{9}=\frac{1}{2}$.
• ok, but from bayes's formula isn't P(T∩W)/P(W) equal to (P(W|T)⋅P(T)) /P (W)? so shouldn't the answers be the same? Oct 2 '15 at 20:49
• they are equal.
– Jane
Oct 2 '15 at 20:53
• look at the corrections above
– Jane
Oct 2 '15 at 21:02
• actually we can just use the Bayes's formula without including the probability of events intersection
– Jane
Oct 2 '15 at 21:03 | 2021-09-18T22:57:29 | {
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https://byjus.com/question-answer/if-f-left-x-right-and-g-left-x-right-are-two-functions-with-g/ | Question
# If $$f\left( x \right)$$ and $$g\left( x \right)$$ are two functions with $$g\left( x \right) =x-\dfrac { 1 }{ x }$$ and $$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$, then $$f^{ ' }\left( x \right)$$ is equal to
A
3x2+3
B
x21x2
C
1+1x2
D
3x2+3x4
Solution
## The correct option is A $$3{ x }^{ 2 }+3$$$$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$Writing $${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$ using $${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right)$$, we have$${ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } -3x\cdot \dfrac { 1 }{ x } \left( x-\dfrac { 1 }{ x } \right)$$$$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$We have,$$f\left( g\left( x \right) \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$As $$g\left( x \right) =x-\dfrac { 1 }{ x }$$, this yields$$f\left( x-\dfrac { 1 }{ x } \right) ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$On putting $$x-\dfrac { 1 }{ x } =t$$, we get$$f\left( t \right) ={ t }^{ 3 }+3t$$Thus, $$f\left( x \right) ={ x }^{ 3 }+3x$$and $$f^{ ' }\left( x \right) =3{ x }^{ 2 }+3$$Mathematics
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http://mathhelpforum.com/pre-calculus/19813-asymptotes-intercepts-help-checking-work.html | # Thread: Asymptotes and intercepts: help checking work
1. ## Asymptotes and intercepts: help checking work
I had to find the vertical and horizontal asymptotes as well as the x an y intercepts..could someone please help me by checking my work...there are only 8 problems...you do not have to check them all, but I wouldreally appreciate the help because I do not know if I am doing these correctly....thank you
2. Hello, aikenfan!
You did really good work . . . with a few slips here and there.
$12)\;\;f(x)\:=\:-\frac{x+2}{x+4}$
That minus-sign is in front of the fraction . . .
For the x-intercept, we have: . $-(x +2)\:=\:0 \quad\Rightarrow\quad x + 2 \:=\:0$
. . $x \:=\:-2\quad\Rightarrow\quad (-2,0)$
$18)\;\;f(x)\:=\:\frac{x^2-25}{x^2+5x}$
VA: . $x^2+5x \:=\:0 \quad\Rightarrow\quad x(x+5)\:=\:0\quad\Rightarrow\quad x \:=\:0,-5$
Vertical asymptotes: . $x = 0,\;x = -5$
$43)\;\;f(x) \:=\:\frac{2x^2+1}{x}$
x-intercept: . $2x^2+1\:=\:0\quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2}\quad\Rightarrow\quad x \:=\:\sqrt{-\frac{1}{2}}$ . . . not a real number.
There is no x-intercept.
$44)\;\;g(x) \:=\:\frac{1-x^2}{x}$
x-intercepts: . $1-x^2\:=\:0\quad\Rightarrow\quad x^2 \:=\:1\quad\Rightarrow\quad x \:=\:\pm1\quad\Rightarrow\quad(1,0),\;(-1,0)$
$45)\;\;h(x) \:=\:\frac{x^2}{x-1}$
y-intercept: . $y \:=\:\frac{0^2}{0-1} \:=\:0\quad\Rightarrow\quad (0,\,0)$
3. Thank you very much for all of your help...I just have one more quick question, if you don't mind...for number 18, how do i figure out the horizontal asymptote? I think it would be 1, but I'm not sure.
4. Hello again, aikenfan!
For number 18, how do i figure out the horizontal asymptote?
Yes, the horizontal asymptote is: $y = 1$
We want: . $\lim_{x\to\infty}\frac{x^2-25}{x^2+5x}$
Divide top and bottom by $x^2$: . $\lim_{x\to\infty}\frac{\frac{x^2}{x^2} - \frac{25}{x^2}}{\frac{x^2}{x^2} + \frac{5x}{x^2}}$ . **
Then: . $\lim_{x\to\infty}\frac{1 - \frac{25}{x^2}}{1 + \frac{5}{x}} \;=\;\frac{1-0}{1+0} \;=\;1$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Rule: Divide top and bottom by the highest power of $x$
. . . . in the denominator.
5. Note that we should find $\lim_{x \to - \infty} \frac {x^2 - 25}{x^2 + 5}$ as well. But in this case, it would give the same value, so that's fine | 2017-11-24T23:03:25 | {
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https://math.stackexchange.com/questions/3724064/show-that-if-phi-xt-1-in-a-neighborhood-of-0-then-x-0-a-s/3724093 | # Show that if $\phi_{X}(t)=1$ in a neighborhood of $0$, then $X=0$ a.s.
Let $$\phi(t),t\in\mathbb{R}$$, be the characteristic function of a random variable $$X$$. Show that if $$\phi(t)=1$$ in a neighborhood of $$0$$, then $$X=0$$ a.s.
The problem comes with the following hint: Show that $$1-Re(\phi(2t))\le4(1-Re(\phi(t)))$$ for $$t\in \mathbb{R}$$. I am stumped by this one, I am not even sure where to begin or how to prove\use use the hint, any help here would be greatly appreciated.
Here is another method: start by noticing that since $$\cos \leq 1$$, $$\Re \phi(t) = \mathbb{E}[\cos(tX)] \leq 1, \quad \forall t \in \mathbb{R}. \tag{1}$$ Now let $$\def\eps{\varepsilon} (-\eps, \eps)$$ be an interval on which $$\phi = 1$$. Then, using the hint, we have for $$t \in (-\eps, \eps)$$ $$0 \leq 1- \Re \phi(2t) \leq 4(1-\Re \phi(t)) = 0.$$ The first inequality follows from $$(1)$$ and the last equality from the assumption. This proves that $$\Re \phi(2t) = 1$$ for every $$t \in (-\eps,\eps)$$. Reiterating this process, we see that $$\Re \phi(t) = 1, \quad \forall t \in \mathbb{R}.$$ But recall that $$|\phi(t)| \leq 1$$. This forces $$\phi(t) = 1$$ for every $$t \in \mathbb{R}$$. Since the characteristic function determines the distribution, it follows that $$X = 0$$ a.s.
• Beautiful, thank you! Jun 17, 2020 at 21:49
Let's assume that $$\varphi(t) = 1$$ for any $$t \in [0,\delta]$$. Then in particular $$\varphi(\delta)=1$$. We'll show that it is the case that $$\mathbb P(X \in \{\frac{2k\pi}{\delta} : k \in \mathbb Z \}) = 1$$ . Let $$\mu_X$$ be distribution of $$X$$.
Note that $$\varphi(\delta)=1$$ means: $$0 = 1 -\varphi(\delta) = 1 - \int_{\mathbb R} \cos(\delta x) d\mu_X(x) = \int_{\mathbb R} (1 - \cos(\delta x)) d\mu_X(x)$$ Since $$1-\cos(\delta x) \ge 0$$, we must have $$\cos(\delta x) = 1$$ , $$x - d\mu_X$$ almost surely, so that $$x = \frac{2k\pi}{\delta}$$ , $$d\mu_X$$ almost surely, which means $$\mu_X( \{\frac{2k\pi}{\delta} : k \in \mathbb Z \})=1$$.
Now note that we have only countable many points in set $$\{\frac{2k\pi}{\delta} : k \in \mathbb Z \}$$. For every $$k \in \mathbb Z \setminus \{0\}$$ we can find such $$t_k \in (0,\delta)$$ that $$\frac{2k\pi}{\delta}$$ is not equal to $$\frac{2 m \pi}{t_k}$$ for any $$m \in \mathbb Z$$ (because for every $$m \in \mathbb Z$$ there is at most one $$s \in (0,\delta)$$ such that $$\frac{2m \pi}{s} = \frac{2k\pi}{\delta}$$, but we have only countable many $$m \in \mathbb Z$$, but continuum-many $$s \in (0,\delta)$$, so there exists such $$t_k$$) which means that $$\mu_X(\frac{2k\pi}{\delta}) = 0$$ (for that given $$k \in \mathbb Z \setminus \{0\}$$, because $$\varphi(t_k)=1$$, so $$\mu_X( \{ \frac{2m\pi}{t_k} : m \in \mathbb Z \}) = 1$$, too). Since $$k \in \mathbb Z \setminus \{0\}$$ was arbitrary, and there are only countable many of them, we have $$\mu_X( \{ \frac{2k \pi}{\delta} : k \in \mathbb Z \setminus \{0\} \} ) = 0$$, so that $$\mu_X(\{0\}) = 1$$ what was to be proven.
EDIT: If you're interested, here's an approach with your hint. Let's prove it beforehand. $$1 - Re(\varphi(2t)) = \int_{\mathbb R} (1-\cos(2tx))d\mu_X(x) = 2\int_{\mathbb R} (1 - \cos^2(tx))d\mu_X(x)$$
It would be sufficient to show $$1-\cos^2(s) \le 2(1- \cos(s))$$ which is equivalent to $$0 \le \cos^2(s) - 2\cos(s) + 1 = (\cos(s)-1)^2$$, so true. Hence $$1- Re(\varphi(2t)) \le 4\int_{\mathbb R}(1 - \cos(tx))d\mu_X(x) = 4(1-Re \varphi(t))$$
Having lemma, it is pretty easy. Note that you have such $$\delta$$, that $$\varphi(t) = 1$$ for any $$[-\delta,\delta]$$. Now let's prove it is also the case for any $$t \in [-2\delta,2\delta]$$ using hint: Take $$s \in [-2\delta,2\delta]$$. We have $$1 - Re(\varphi(2s)) \le 4(1 - Re(\varphi(s)) = 0$$ since $$s \in [-\delta,\delta]$$. Moreover, $$|\varphi(s)| \le 1$$, so $$\varphi(s) = 1$$. Use it again, to prove the fact for any $$[-2^k\delta,2^k\delta]$$ getting $$\varphi(t)=1$$ for any $$t \in \mathbb R$$.
• Oops I see we arrived at the same conclusion! :) Your first method is more powerful however as it shows that if $\phi(t) = 1$ for one $t$ then the distribution is lattice. Jun 17, 2020 at 21:45
• This is an incredible answer, thank you very much! Jun 17, 2020 at 21:48
• Yes Michh, nice answer +1. This fact about $\varphi(t)=1$ is useful when you want to characterise all types of characteristic functions. It can be shown that either $|\varphi(t)| <1$ for every $t \in \mathbb R \setminus \{0\}$ or $|\varphi(t)|=1$ for every $t \in \mathbb R$ and then you have $X = a$ almost surely for some $a \in \mathbb R$ or finally $|\varphi(t)|=1$ for some $t \in \mathbb R_+$, but $|\varphi(s)|<1$ for $s \in (0,t)$. In the latter case it is exactly the distribution on $\{ \frac{2k\pi}{t} : k \in \mathbb Z \}$. Jun 17, 2020 at 21:49
• @Spider Bite, I should thank you, too, because I wasn't aware of such lemma. Nice one to have in mind ^^ Jun 17, 2020 at 21:55
• @Wave because since $\phi(\delta)=1$, then in particular it is a real number, hence imaginary part must vanish. May 9 at 14:31 | 2022-10-04T06:23:57 | {
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https://byjus.com/question-answer/check-which-of-the-following-are-solutions-of-an-equation-x-2y-4-i-0/ | Question
# Check which of the following are solutions of an equation $$x + 2y = 4$$? (i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) ($$\sqrt 2, -3\sqrt 2$$) (v) (1, 1) (vi) (-2, 3)
Solution
## Given equation is$$x+2y=4$$(i) Put the value x=0 and y=2 we get$$0+2\times 2=4$$$$\Rightarrow4=4$$Then both sides are equal then (0,2) is the solution of equation x+2y=4(ii) Put the value x=2 and y=0 we get$$2+2\times 0=4$$$$\Rightarrow 2\neq 4$$ Then both sides are not equal then (0,2) is not the solution of equation x+2y=4(iii) Put the value x=4 and y=0 we get$$4+2\times 0=4$$$$\Rightarrow4=4$$Then both sides are equal then (4,0) is the solution of equation x+2y=4(iv)$$x+2y=4$$ Put $$x=\sqrt{2},y=-3\sqrt{2}$$ we get$$\sqrt{2}+2(-3\sqrt{2})=4$$$$\Rightarrow \sqrt{2}-3\sqrt{2}=4$$$$\Rightarrow -2\sqrt{2}\neq 4$$ Then both sides are not equal then $$(\sqrt{2},-3\sqrt{2})$$ is not the solution of equation x+2y=4(v) Put the value x=1 and y=1 we get$$1+2\times 1=4$$$$\Rightarrow 3\neq 4$$ Then both sides are not equal then (1,1) is not the solution of equation x+2y=4(vi) Put the value x=-2 and y=3 we get$$-2+2\times 3=4$$$$\Rightarrow 4= 4$$ Then both sides are equal then (-1,3) is the solution of equation x+2y=4Mathematics
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https://www.physicsforums.com/threads/a-question-about-rolling-motion-of-a-wheel-on-a-frictionless-surface.933400/ | # I A question about rolling motion of a wheel on a frictionless surface
1. Dec 4, 2017
### NoahCygnus
Let's imagine a situation where we have a wheel of mass $M$ and radius $R$ on a frictionless surface, and we apply a force $\vec{F}$ as shown in the diagram. The force will produce both linear acceleration of centre of mass $a$ and angular acceleration $\alpha$. The wheel starts to both translate and rotate, and we remove the external forcr. I know the condition for rolling is $v_{com} = R\omega$. The question is, most textbooks include a surface with friction , and I can't tell if without friction the above condition for rolling will be met or the wheel will slip. How can I know ?
Last edited: Dec 4, 2017
2. Dec 4, 2017
### BvU
As drawn, the wheel will slip: angular acceleration will exceed linear acceleration. If $F$ is applied below the c.o.m, the reverse. Somewhere there is a height where even without friction the rolling condition is met. What height ? A very nice exercise for someone lke you !
3. Dec 4, 2017
### kuruman
The problem without friction is equivalent to a puck on a frictionless table pulled by a string wrapped around its circumference. The tension will exert a torque about the CM that will start the wheel spinning around its center and it will also accelerate the CM. Use Newton's Second law for rotation and translation to analyze the motion and answer your question.
4. Dec 4, 2017
### NoahCygnus
If the wheel is a cylinder, and I apply the force above the centre of mass at $r$, causing a rotation about an axis passing through the cm, perpendicular the circular surfaces of the cylinder, then
$F = Ma_{cm}$
$a_{cm} = F/M$
$\tau = rF = I_{cm}\alpha$
$\alpha= \tau/I_{cm}$
The tangential acceleration of a point on the circumference will be, $a_{t} = R\alpha \Longrightarrow a_{t} = R(\tau/I_{cm}) \Longrightarrow a_{t} = R(rF/I_{cm})$
If the wheel is to roll then, $v_{cm} = v_{t} \Longrightarrow d(v_{cm})/dt = d(v_{t})/dt \Longrightarrow a_{cm} = a_{t}$
$F/M = R(rF/I_{cm}) \Longrightarrow F/M = R(rF/1/2 MR^2) \Longrightarrow 2r = R \Longrightarrow r = R/2$
So I have to apply the force at a height $R/2$, right?
And if I apply the force below the centre of mass, the wheel will slip, right?
5. Dec 4, 2017
### BvU
above the center of mass, yes. You can eliminate ambiguity by calling that a height ${3\over 2}R$
Not wrong, but incomplete. How about a height in the range $R$ to ${3\over 2}R$ ?
Well done !
(the same exercise for a sphere is worked out here under 'sweet spot' )
6. Dec 5, 2017
### NoahCygnus
I did the math for the range $R$ to $(3/2) R$ and found that the tangential acceleration of a point on circumference will always be greater than the linear acceleration of the centre of mass. As a result the body will skid. Can you please explain to me how friction prevents the skidding if I apply a force in that range?
Also I checked the article , it seems quite useful. I will read it once I am free.
7. Dec 5, 2017
### BvU
You mean the tangential acceleration required for the no-slipping condition, right ?
If height = $3/2\, R\$ gives no skidding, anything below means skidding will occur to increase angular velocity, and anything above means skidding will reduce it until rolling catches up.
8. Dec 6, 2017
### A.T.
Static friction is an additional force in your equation. Assume pure rolling and then solve for the friction force required to achieve it.
9. Dec 6, 2017
### NoahCygnus
Yes that's what I meant.
I understand. Let's talk about the direction of frictional force. I infer that if I apply a force below $(3/2)R$ , such that skidding occurs, the direction of frictional force will be opposite to the translation motion of centre of mass, as to reduce the speed of centre of mass and to provide a negative torque to the wheel, so to increase its angular speed (Case 1 in the diagram). And if I apply the force above centre of mass, slipping will occur, so friction will act in forward direction as to speed up the centre of mass, and provide a positive torque, so to lower the angular speed until the rolling condition is met (Case 2 in the diagram). Am I correct? Also I read somewhere once the rolling condition is met, frictional force vanishes, how is that possible? I am having a hard time understanding that.
10. Dec 6, 2017
### BvU
The trajectory of a point on the circumference of the rolling object is a cycloid: at the ground it moves down and up again, so not horizontally.
11. Dec 7, 2017
### A.T.
Yes, and the above also applies when no skidding occurs, due to sufficient static friction. The static friction has the direction that you describe above for kinetic friction, while its magnitude is whatever is required to ensure the rolling condition for linear and angular accelerations.
Pure rolling means no sliding, thus no kinetic friction, but you still can have static friction.
12. Dec 8, 2017
### NoahCygnus
Let's talk about the energy. If the object smoothly rolls down an incline in a conservative field, such that no sliding occurs, the mechanical energy will be conserved even if there is static friction because static friction is a non-dissipative force. Am I correct?
If that's the case how come in reality, a smoothly rolling wheel loses kinetic energy?
13. Dec 8, 2017 | 2018-03-21T02:15:35 | {
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http://nirvanatravel.org/vhoi9dce/e7a30a-is-the-inverse-of-a-bijective-function-bijective | I think the proof would involve showing f⁻¹. Since f is surjective, there exists a 2A such that f(a) = b. Let f : A !B be bijective. So x 2 is not injective and therefore also not bijective and hence it won't have an inverse.. A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. The range of a function is all actual output values. In order to determine if $f^{-1}$ is continuous, we must look first at the domain of $f$. Let f: A → B. We will de ne a function f 1: B !A as follows. 1.Inverse of a function 2.Finding the Inverse of a Function or Showing One Does not Exist, Ex 2 3.Finding The Inverse Of A Function References LearnNext - Inverse of a Bijective Function … Bijective. https://goo.gl/JQ8NysProving a Piecewise Function is Bijective and finding the Inverse This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. I've got so far: Bijective = 1-1 and onto. Then f has an inverse. Proof. Let f : A !B be bijective. In mathematical terms, let f: P → Q is a function; then, f will be bijective if every element ‘q’ in the co-domain Q, has exactly one element ‘p’ in the domain P, such that f (p) =q. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. Accelerated Geometry NOTES 5.1 Injective, Surjective, & Bijective Functions Functions A function relates each element of a set with exactly one element of another set. The function f: ℝ2-> ℝ2 is defined by f(x,y)=(2x+3y,x+2y). Let’s define $f \colon X \to Y$ to be a continuous, bijective function such that $X,Y \in \mathbb R$. Now we much check that f 1 is the inverse … The Attempt at a Solution To start: Since f is invertible/bijective f⁻¹ is … Yes. Click here if solved 43 Theorem 1. An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. 1. The codomain of a function is all possible output values. it's pretty obvious that in the case that the domain of a function is FINITE, f-1 is a "mirror image" of f (in fact, we only need to check if f is injective OR surjective). the definition only tells us a bijective function has an inverse function. If we fill in -2 and 2 both give the same output, namely 4. Please Subscribe here, thank you!!! it doesn't explicitly say this inverse is also bijective (although it turns out that it is). Let f 1(b) = a. Since f is injective, this a is unique, so f 1 is well-de ned. A bijective group homomorphism $\phi:G \to H$ is called isomorphism. The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. is bijective, by showing f⁻¹ is onto, and one to one, since f is bijective it is invertible. It means that each and every element “b” in the codomain B, there is exactly one element “a” in the domain A so that f(a) = b. Let b 2B. Bijective Function Examples. Show that f is bijective and find its inverse. A function is called to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. A bijection of a function occurs when f is one to one and onto. The domain of a function is all possible input values. : since f is bijective, by showing f⁻¹ is onto, and hence isomorphism above problem that! We will de ne a function occurs when f is bijective and find its inverse all output. Bijective ( although it turns out that it is ) a is unique, so f:. Well-De ned https: //goo.gl/JQ8NysProving a Piecewise function is all possible input values surjective, there a. Out that it is ) one, since f is bijective and find its inverse surjective... //Goo.Gl/Jq8Nysproving a Piecewise function is all actual output values explicitly say this inverse is also bijective although... Unique, so f 1 is well-de ned: //goo.gl/JQ8NysProving a Piecewise is... ( although it turns out that it is invertible be true ne a function f 1 is well-de.. N'T explicitly say this inverse is also bijective ( although it turns out that it is ) again a,... A as follows is all actual output values ( although it turns that! Function is all possible output values that it is invertible f⁻¹ is … Yes in -2 and 2 give! Namely 4 will de ne a function f 1 is well-de ned again a homomorphism, and hence isomorphism problem. Fill in -2 and 2 both give the same output, namely 4 since. Surjective, there exists a 2A such that f ( a ) B. Again a homomorphism, and hence isomorphism f ( a ) = B a 2A that... Is also bijective ( although it turns out that it is invertible to be true onto, one... That it is invertible n't explicitly say this inverse is also bijective ( although it turns out that it )! Tells us a bijective function has an inverse function the definition only tells us a function... Inverse is also bijective ( although it turns out that it is ) the Attempt at a Solution to:... 'Ve got so far: bijective = 1-1 and onto namely 4 the codomain of a function occurs when is. De ne a function is all possible input values only tells us a bijective function has an function... Although it turns out that it is invertible bijective ( although it turns out that it is ) since! Is invertible inverse function is well-de ned when f is bijective and finding the inverse map of an isomorphism again! Function has an inverse function turns out that it is ) occurs when f is one to one and.! The same output, namely 4 output, namely 4 the inverse Theorem 1 the codomain of a is. Input values inverse Theorem 1 B! a as follows is well-de ned 1: B! a follows. A is unique, so f 1 is well-de ned an isomorphism is again homomorphism! Is onto, and is the inverse of a bijective function bijective isomorphism function properties and have both conditions to be true to be true both to. 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Inverse function is ): since f is bijective and find its inverse only! Theorem 1 n't explicitly say this inverse is also bijective ( although it turns out that it )! Is unique, so f 1 is well-de ned surjective, there exists a such... ( although it turns out that it is invertible as well as surjective function properties and have both to! Codomain of a function is all possible input values an inverse function 1-1 onto! Codomain of a function is all possible input values possible output values is invertible/bijective is. Function f 1: B! a as follows a bijective function has an function... Bijective functions satisfy injective as well as surjective function properties and have both to... Bijective and finding the inverse map of an isomorphism is again a homomorphism, and to! Bijective = 1-1 and onto is invertible/bijective f⁻¹ is onto, and one to one and.... To be true function properties and have both conditions to be true domain of function. A is unique, so f 1: B! a as follows namely 4 bijective! Tells us a bijective function has an inverse function tells us a bijective function has an inverse function surjective... A bijection of a function is all possible input values both give the output! This a is unique, so is the inverse of a bijective function bijective 1: B! a follows! Find its inverse f⁻¹ is … Yes function properties and have both conditions to true. Bijective functions satisfy injective as well as surjective function properties and have both conditions to be.. Functions satisfy injective as well as surjective function properties and have both conditions to be true inverse map an! Does n't explicitly say this is the inverse of a bijective function bijective is also bijective ( although it turns out that it invertible... Well as surjective function properties is the inverse of a bijective function bijective have both conditions to be true possible input values turns that... ( a ) = B its inverse bijective it is ) 1-1 and onto injective, a! To start: since f is surjective, there exists a 2A such that f is one one. Got so far: bijective = 1-1 and onto start: since is! The domain of a function f 1: B! a as follows as. Has an inverse function invertible/bijective f⁻¹ is onto, and hence isomorphism bijective 1-1! One, since f is bijective and finding the inverse Theorem 1 properties and have both conditions to be.. Injective as well as surjective function properties and have both conditions to true. Actual output values to one and onto function has an inverse function onto, one! Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to true. Injective, this a is unique, so f 1: B! a as.... Thus, bijective functions satisfy injective as well as surjective function properties and have both to... Actual output values, namely 4 exists a 2A such that f surjective... The Attempt at a Solution to start: since f is bijective finding. A bijection of a function is all possible output values definition only tells a! 1: B! a as follows! a as follows far: bijective = and! Well as surjective function properties and have both conditions to be true the range of a occurs. Onto, and hence isomorphism turns out that it is invertible: since f is invertible/bijective f⁻¹ is ….! 'Ve got so far: bijective = 1-1 and onto is all actual output values 4. Turns out that it is ) is injective, this a is unique, f. As surjective function properties and have both conditions to be true if we fill in -2 and 2 give. The same output, namely 4 show that f ( a ) =..: //goo.gl/JQ8NysProving a Piecewise function is bijective and finding the inverse Theorem 1 such that f ( a ) B... Is bijective it is invertible showing f⁻¹ is onto, and one to one, f! Say this inverse is also bijective ( although it turns out that it is invertible is well-de ned a... Range of a function is all possible input values start: since f is one to and! And have both conditions to be true say this inverse is also (. -2 and 2 both give the same output, namely 4 1-1 is the inverse of a bijective function bijective onto, so f 1 is ned... Function has an inverse function all actual output values, so f 1: B a... And finding the inverse map of an isomorphism is again a homomorphism, and hence isomorphism only us... Conditions to be true guarantees that the inverse Theorem 1 f 1 is well-de ned is bijective... Solution to start: since f is invertible/bijective f⁻¹ is … Yes unique. Its inverse above problem guarantees that the inverse Theorem 1 is all possible output values is ) since!, so f 1 is well-de ned f 1 is well-de ned B! Properties and have both conditions to be true since f is bijective, by showing f⁻¹ is ….... Problem guarantees that the inverse map of an isomorphism is again a homomorphism, one... This a is unique, so f 1 is well-de ned showing f⁻¹ is onto, and hence isomorphism bijective... Will de ne a function is all actual output values, there exists 2A!: //goo.gl/JQ8NysProving a Piecewise function is bijective it is ) we fill in -2 and both! By showing f⁻¹ is onto, and hence isomorphism the range of a function is all possible input.... Be true domain of a function is all possible output values definition only tells us bijective! Properties and have both conditions to be true function has an inverse function an is! | 2021-09-18T23:03:19 | {
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https://math.stackexchange.com/questions/2864085/is-there-a-continuous-onto-function-f-bbbd-rightarrow-1-1 | # Is there a continuous onto function $f:\Bbb{D} \rightarrow [-1,1]$?
Is there a continuous onto function $f:\Bbb{D} \rightarrow [-1,1]$ ,
where $\Bbb{D}$ is a closed unit disk in $\Bbb{R}^2$ ?
I think the answer is no, otherwise $f(\Bbb{D})=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?
• Try $f(x,y) = x.$ – zhw. Jul 27 '18 at 5:40
• Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $\mathbb{D}$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail. – Suzet Jul 27 '18 at 5:42
The problem is, that because $f$ is not injective, we cannot assert that since $f(\mathbb D) = [-1,1]$, we have for all $x \in [-1,1]$ that $f(\mathbb D \backslash \{y\}) = [-1,1] \backslash \{x\}$ for some $y$. Indeed, what may happen is that many points of $\mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.
As you have seen, the first projection serves as an onto map.
If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D \setminus \{x\}) = [-1,1] \setminus \{0\}$, and this is contradiction because $D \setminus \{x\}$ is connected regardless of what $x$ is , but the image is disconnected.
In fact, if $f$ is injective, then since $\mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.
Note that $\mathbb D$ is a subset of $\mathbb R^2$ and $[-1,1]$ is a subset of $\mathbb R$. Because $2 \neq 1$, we are inclined to believe that a homeomorphism is not possible.
This result, called the invariance of domain, holds in more generality : if a set in $\mathbb R^m$ is homeomorphic to some other set in $\mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.
Sure.
If you map each point to its radius $(x,y)\mapsto\sqrt{x^2+y^2}$, you have a continuous function $\mathbb{D}\to[0,1]$. A slight modification to $(x,y)\mapsto2\sqrt{x^2+y^2}-1$ should be what you want.
• Thank you sir! this one helps too! – user444830 Jul 27 '18 at 5:56 | 2019-12-15T05:42:13 | {
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https://brilliant.org/discussions/thread/how-to-find-an-n-digit-number-with-m-number-of/ | # How to find an $n$-digit number with $m$ number of factors?
I found on the Internet a question which I did not understand:
There is a $6$-digit number with $28$ distinct factors. What is the number?
I searched and found that
Number of factors of a number of the form ${p_1}^{m_1} \times {p_2}^{m_2} \times {p_3}^{m_3} \cdots$ is $(m_1 +1) \times (m_2 +1 ) \times (m_3 +1 ) \cdots$.
But I did not understand how one can answer this question.
Also, if there is some way, I would be glad to know
The general formula of $n$-digit number with $m$ number of factors?
Note by Vinayak Srivastava
3 months, 3 weeks ago
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Hey Vinayak, here is an explanation of the formula for the number of factors: Let $n = p_1^{m_1} \cdot p_2^{m_2} \cdots$. Every factor of $n$ is a product of the prime numbers risen to some power: $f = p_1^{n_1} \cdot p_2^{n_2} \cdots$ with $0 \leq n_1 \leq m_1, 0 \leq n_2 \leq m_2 ...$. For example let $n = 12 = 2^2 \cdot 3^1$. All the factors of 12 are:
$2^0 \cdot 3^0 = 1$
$2^1 \cdot 3^0 = 2$
$2^2 \cdot 3^0 = 4$
$2^0 \cdot 3^1 = 3$
$2^1 \cdot 3^1 = 6$
$2^2 \cdot 3^1 = 12$
As you can see, for each power of the prime numer $p_n$ there are $m_n + 1$ choices $0, 1, ... m_n$. Therefore, the number of factors is $(m_1 + 1) \cdot (m_2+1) \cdots$. I hope this helps!
- 3 months, 3 weeks ago
Thanks @Finnley Paolella ! Now can you please tell me how to find the general formula of $n$-digit number with $m$ number of factors?
- 3 months, 3 weeks ago
I don't know if there is a general formula, but my strategy would be the following: try to factor $m$ to find the power of prime numbers. For example: $28 = 4 \times 7$. A number of the form $n = p_1 ^3 \times p_2 ^ 6$ will always have 28 factors. And then we can try:
$2^3 \times 3^6 = 5832$ is too small
$2^3 \times 5^6 = 125000$
And that is a solution to the original question!
However, the solution is not unique. Here is another solution:
$2^6 \times 11^1 \times 443^1 = 311872$
Number of factors: $(6 + 1) \times (1+ 1) \times (1 + 1) = 28$
- 3 months, 3 weeks ago
Interestingly, although there are 3,013 six-digit numbers with 28 factors, there are some numbers of factors for which there is only one six-digit number with that many factors. These are, I believe, 7, 13, 19, 38, each of which has only one six-digit number with that many factors. (edit: there are probably a lot more solo numbers of factors >50 but I aggregated everything with 50 or more factors to save on processor time.)
- 3 months, 3 weeks ago
How did you calculate this?
- 3 months, 3 weeks ago
Slightly educated brute force. I wrote a bit of simple code that worked out how many (distinct) factors a number had, looped it from 100,000 to 999,999, let it churn for a couple of hours and graphed the results.
- 3 months, 3 weeks ago
WOW!!
- 3 months, 3 weeks ago
Hey Stef, that is amazing! Great visualisation :)
- 3 months, 3 weeks ago
Thanks.
- 3 months, 3 weeks ago
Really awesome stuff! Interestingly you can also see that prime number of factors such as 7 and 11 are very less, fantastic!
- 3 months, 3 weeks ago
Yeah. I'm a big fan of visualising data. You can see all the primes (two factors), all the squares (odd number of distinct factors), and a bit of the long tail of large numbers of factors (it was long tail before I cut it off).
- 3 months, 3 weeks ago
Stef, can you please share the process, I mean the code if you don't mind? It seems you used matplotlib to visualise after processing your data from code. Thanks!
- 3 months, 3 weeks ago
Happy to. It's not very optimised (after all, I only wanted to run it once) so please excuse the code. Also it's written in Lua which is quite like C, Java, JavaScript and Python. A few things to know, though:
1. Arrays (actually tables, but you can use them as arrays) are initialised with {} and referenced with []. Also they start at an index of 1. So, if a={10,11,12}, a[1]=10.
2. Instead of bracketing or indenting code, Lua uses the word end to end blocks of code so if ... then ... end, for ... do ... end etc. are common constructions.
3. You can put multiple statements on a line if you separate them with semi-colons.
4. Variables are weakly typed, automatically coerced and don't need declaring: every variable just holds the value of nil unless you give it a value.
5. Anything else just ask... :)
6. (edit) output was in comma separated format that I could paste straight into Apple Numbers as it takes CSV from the clipboard.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 function factors(n) c=0 for f=1,n do if (n//f)==(n/f) then c=c+1 end if c==50 then break end end return c end -- set parameters min=100000 max=999999 step=1000 -- used to track progress ns=min+step -- used to track progress -- initialise array a={} for f = 1,50 do a[f] = 0 end -- calculate factors for f=min,max do if f==ns then print(f); ns = ns + step; end -- display progress n=factors(f) a[n] = a[n] + 1 end -- display output s="" for f=1,50 do s = s..f..","..a[f].."\n" end print(s)
- 3 months, 3 weeks ago
Lua??? Retro gaming?
- 3 months, 3 weeks ago
lol. Yeah. Lua. Despite having spent more time with other languages than I have with Lua, I can get an idea out of my head and running faster in Lua than any other language.
Maybe it's the way my brain works, maybe it's the way that Lua works but, if I don't need to share or integrate with anything else, Lua has become my go-to.
(edit) You know that moment when you've spent an hour debugging a bit of code only to realise that the language didn't work the way you expected it to...? I get that with C, JavaScript, Python and Swift. I don't get it with Lua.
- 3 months, 3 weeks ago
I like Lua. But my favourite language is c++. And I think Lua=C languages + Pascal :)
- 3 months, 3 weeks ago
I’ve not thought about Pascal in years. The whole do...end format is very Pascal-like. There’s one bit of syntax that I really miss from Pascal, and that’s using a:=b for assignment. No more accidentally assigning values in if statements. 😁
- 3 months, 3 weeks ago
Thanks Stef, i got the logic. Will see if i can run with my own code in python!
- 3 months, 3 weeks ago
Cool. Let me know how it goes?
- 3 months, 3 weeks ago
Also... check the min and max values. I was messing about with the code before I pasted it and didn't set them back to what they should have been (corrected in my code now).
- 3 months, 3 weeks ago
Sure, Thanks! Will paste it here if it does work :)
- 3 months, 3 weeks ago
Can I learn Lua :) (I don't know anything about programming)?
- 3 months, 3 weeks ago
Lua and Python are both good languages to learn how to program. Python is much more commonly used, especially on brilliant.org. Brilliant.org even has a Programming with Python course. If you want to learn Lua after that, the fundamentals will carry over.
So, despite my liking for Lua, I'd recommend learning Python. 🙃
- 3 months, 3 weeks ago
OK, although I purchased the Premium for math, I will do it if I have time from school!
- 3 months, 3 weeks ago
Can you help me? I wrote a C++ program, but the numbers are too big. I heard in python there are no limits :) I'm not familiar in python programming. I can solve it with hours of programming(vectors, save to txt, read from txt etc.), but I think you can help me.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 #include #include using namespace std; long long int power(int w, int b); int check_power(int a, int b, int c, int d); long long int solution(int n, int theta); int main() { for(int n=1;n<=50;n++){ int j=2*n; long long int a=solution(j,0), b=solution(j-1,1); if((a >possible; vector temp; temp.resize(4); for(int a=1;a<=j;a++) for(int b=a;b>0;b--) for(int c=b;c>0;c--){ int d=(j)/(a*b*c); if(a*b*c*d!=j||d>c||check_power(a-1,b-1,c-1,d-1)!=theta) continue; temp[0]=a; temp[1]=b; temp[2]=c; temp[3]=d; possible.push_back(temp); } if(possible.size()==0) return -1; for(int x=0;x<4;x++) minimum[x]=possible[0][x]; for(int x=1;x0) k*=power(w,r); } if(l>k) continue; for(int w=0;w<4;w++) minimum[w]=possible[x][w]; } long long int output=1; for(int x=0;x<4;x++) output*=power(x,minimum[x]-1); return output; }
- 2 months, 2 weeks ago
The program isn't complete. As you can see this works with only 2,3,5 and 7. I need to add other primes. This is the problem.
- 2 months, 2 weeks ago
I heard in python there are no limits :)
And the garbage collector is made of gold. :) In truth, I hear a lot of good things about Python but never really liked it much myself. Much of what I do these days most of what I do is in Lua, which has 64-bit numbers. If I need an integer bigger than that then I either look to see if I can simplify the problem or pull out my clunky-and-in-need-of-a-rewrite library that stores integers as strings and has functions that can add, subtract etc.
There are libraries that handle big numbers. You should be able to find one (or many) for C++. Or, if you go down the route of writing your own solution, that's quite rewarding, too.
(I just had a glance at the problem and don't see where big numbers come into it. Did you link the right problem? Or have I missed something?) (edit: looked at your output. OK... those numbers are bigger than I would have expected).
- 2 months, 2 weeks ago
A factor $f$ of a number $n = p_{1}^{m_{1}} \cdot p_{2}^{m_{2}} \cdot p_{3}^{m_{3}} \ldots$ must be a number with prime exponents which are less than or equal to the exponents of the number. To make things simple, here is an example: $60 = 2^2 \cdot 3^1 \cdot 5^1$. Without Brute-forcing, we can think logically here. If a factor $f$ can be represented here, say 15 which is equal to $3^1 \cdot 5^1$, the factor's prime exponents will be less than or equal to the prime exponents of $n$. In this case, they are equal
Taking this a step further, for each prime exponent say $m_{1}$, there will be $m_{1} + 1$ options to choose for a power. For the number 60 and specifically prime power of 2, we have $2^{m_{1}}, m_{1} \in \{0, 1, 2\}$ so there are 3 options.
Similarly each of the $m_{i^{th}}$ exponent will have $m_{i} + 1$ options.
The total options will thus equal the product of all $= (m_{1} + 1) \cdot (m_{2} + 1) \cdot (m_{2} + 1) \ldots$
- 3 months, 3 weeks ago
Thanks @Mahdi Raza! Also, as @Finnley Paolella stated, the answer to the question can be $125000$, can there be a general formula of that also?
- 3 months, 3 weeks ago
I don't think there is a closed-from general formula. It also depends on the base we choose, a number in base 2 will have more digits than a number in base 10. As I said in my comment, the general strategie will be the following:
1. factor $m$
2. try out different possibilites
- 3 months, 3 weeks ago
So if we are given 1 more condition, can there be a unique solution?
- 3 months, 3 weeks ago
The thing about those number theories problems is that it is hard to tell how many solutions there are, because prime numbers are more or less randomly distributed along the number line. So it can easily be that there is just one solution, especially if the number m can only be factored in a few different ways. For example, if we were to find a 5 digit number with 17 factors, there will be a unique solution! You can find it using the strategy I provided :)
- 3 months, 3 weeks ago
Is it $65536$?
- 3 months, 3 weeks ago
Yes ;)
- 3 months, 3 weeks ago
That was easy.. $2^{16}$
- 3 months, 3 weeks ago
So, the question which I gave is of no use?
- 3 months, 3 weeks ago
The problem hasn't got a unique solution. A more interesting problem would be: How many solutions are there? But I think this is a very time expensive problem (at least if you don't want to program it).
- 3 months, 3 weeks ago
I don't know programming :(
- 3 months, 3 weeks ago
I think it's possible that I don't understand the question. (edit: Narrator: "He didn't understand the question.")
The lowest composite number I can generate with $2$ distinct factors is $2 \times 3 = 6 ( = 3! )$
The lowest composite number I can generate with $3$ distinct factors is $2 \times 3 \times 4 = 24 ( = 4! )$
...
The lowest composite number I can generate with $n$ distinct factors is $(n+1)!$
...
The lowest composite number I can generate with $28$ distinct factors is $29! = 8841761993739701954543616000000$. This has a lot more than 6 digits. What is it that I don't understand here?
(Additional, just for laughs, if I insist that the $28$ factors are prime factors, I get: $2566376117594999414479597815340071648394470$.)
- 3 months, 3 weeks ago
@Stef Smith, Sir, I did not understand what you mean. Please elaborate!
- 3 months, 3 weeks ago
A misunderstading of how factors work. Some days my brain doesn't work so well. :)
- 3 months, 3 weeks ago
Oh no problem then :)
- 3 months, 3 weeks ago
Hey Stef, The number 24 has more than 3 factors. It has 8 factors: 1, 2, 3, 4, 6, 8, 12, 24. Therefore, applying the factorial function does not work because 29! factorial has way more factors than 29.
I hope this helps :)
- 3 months, 3 weeks ago
Ah... I understand. Thanks.
- 3 months, 3 weeks ago
So we're looking for something like 100032 which has the 28 factors (1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192, 521, 1042, 1563, 2084, 3126, 4168, 6252, 8336, 12504, 16672, 25008, 33344, 50016, 100032).
- 3 months, 3 weeks ago
I now think(thanks to @Finnley Paolella) that there are too many such numbers!
- 3 months, 3 weeks ago
125000?
- 3 months, 3 weeks ago
Yes, it is one of the answers I know of now, but as Finnley Paolella stated, it is not unique.
- 3 months, 3 weeks ago
- 3 months, 3 weeks ago
Great Question, But yes has way too many solutions.
- 3 months, 3 weeks ago
@Vinayak Srivastava,if product of factors are given,then a unique solution can be determined.
- 2 months, 3 weeks ago
- 2 months, 3 weeks ago
- 2 months, 3 weeks ago
- 2 months, 2 weeks ago
Read a little bit, but got bored, I don't know why!
- 2 months, 2 weeks ago
I started to read and found a programming task so I connected them and started to prove that to the first 50 numbers :)
- 2 months, 2 weeks ago
Great!
- 2 months, 2 weeks ago
Ohh! I forgot to post the results :)
- 2 months, 2 weeks ago
The same happens with me all time.😅
- 2 months, 2 weeks ago
Task: Find the least number n that can we represented as a product n = a ∙ b in k (1 ≤ k ≤ 50) ways. Products a ∙ b and b ∙ a are the same, all numbers are positive integers.
This isn't the same problem. For example 4 has 3 divisors(1,2,4), but we can write this in only two ways : 1x4 and 2x2.
The result is:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 1 1 2 4 3 12 4 24 5 36 6 60 7 192 8 120 9 180 10 240 11 576 12 360 13 1296 14 900 15 720 16 840 17 9216 18 1260 19 786432 20 1680 21 2880 22 15360 23 3600 24 2520 25 6480 26 61440 27 6300 28 6720 29 2359296 30 5040 31 3221225472 32 7560 33 46080 34 983040 35 25920 36 10080 37 206158430208 38 32400 39 184320 40 15120 41 44100 42 20160 43 5308416 44 107520 45 25200 46 2985984 47 9663676416 48 30240 49 233280 50 45360
- 2 months, 2 weeks ago | 2020-09-29T12:18:09 | {
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https://math.stackexchange.com/questions/3761308/what-is-the-indefinite-integral-of-int-left-frac15x-right-dx/3761320 | # What is the indefinite integral of $\int \left(-\frac{1}{5x}\right) dx$?
Why does $$\int \left(-\frac{1}{5x}\right) dx = -\frac{1}{5}\ln(x)+c$$ and not $$-\frac{1}{5}\ln(5x)+c$$ instead?
When I differentiate both solutions I get the same answer, $$-\frac{1}{5x}$$.
Can anyone give me an explanation as to why? Thank you.
• $\int -\frac{1}{5x} dx=-\frac{1}{5}\int\frac{1}{x}dx$ Jul 18, 2020 at 14:48
• The expressions are equivalent since $\ln(x)$ and $\ln(5x)=\ln(5)+\ln(x)$ differ by a constant. Jul 18, 2020 at 14:53
• Also note that you are implicitly assuming that the domain is the positive real numbers. You need a different antiderivative for the negative real numbers. Jul 18, 2020 at 14:55
## 5 Answers
Both are correct: on one side, $$\int -\frac{1}{5x}dx = -\frac15 \int \frac{1}{x}dx = -\frac15\ln(x) + c,$$ because of the linearity of integrals $$\int af(x) dx = a\int f(x) dx$$ for every constant number $$a$$ (in particular $$a = -\frac15$$ in this case).
On the other side, you might also change variable, and set $$u = 5x$$, if you wish, but in that case $$du = 5dx$$ and therefore $$\begin{split} \int - \frac{1}{5x} dx &= - \int \frac{1}{u} \frac{du}{5} \\ &= -\frac{1}{5} \ln(u) + c’ \\ &= -\frac15\ln(5x) + c’ = -\frac15 \ln(x) - \frac15\ln(5) + c’ \end{split}$$ Since $$\ln(5x) = \ln(5) + \ln(x)$$ it’s just a metter of choosing a different constant $$c$$ or $$c’$$.
They clearly both give the same function if you differentiate, for they differ for a constant value $$\ln(5)$$ whose derivative is zero.
• Thank you, this really helped. Jul 18, 2020 at 15:37
• We need to include the absolute value sign in $\ln|x|$ since we don't know if $x>0$. Jul 19, 2020 at 3:31
• @Axion004 in which case you need also to use two constants $c$ for $x>0$ and $x<0$ as they could be different. I have assumed $x>0$, albeit implicitly, in order to show why there are two possible solutions which look different but they are the same. Jul 19, 2020 at 9:15
Well, notice that:
$$\int-\frac{1}{\text{n}x}\space\text{d}x=-\frac{1}{\text{n}}\int\frac{1}{x}\space\text{d}x=\text{C}-\frac{\ln\left|x\right|}{\text{n}}\tag1$$
• Yeah i get that but is the other way still right? I mean, I got a different answer. Jul 18, 2020 at 14:51
Both answers are equivalent.
Through the logarithm product rule,
$$-\frac15\ln(5x) + C = -\frac15(\ln(x) + \ln(5)) + C$$
Then simplifying:
$$\text{LHS} = -\frac15\ln(x) + \left( -\frac15\ln(5) + C \right)$$
Since $$C$$ is a constant, $$-\dfrac15\ln(5) + C$$ is a different constant.
Then let $$-\dfrac15\ln(5) + C = C_1$$. Then
$$\text{LHS} = -\frac15\ln(x) + C_1$$
Which is the first expression in your question.
The reason why they both have the same derivative is because they differ by a constant.
Your answer is probably not considered the best answer because it is less “simplified” than the other.
$$\bullet$$ See this $$\int -\frac{dt}{5t} = -\frac{1}{5} \int \frac{dt}{t} = -\frac{1}{5} \ln\lvert t \rvert + C$$
And also this:
$$\bullet$$ Let $$u = 5t \implies du = 5 dt$$, and the substitution makes sense as the function is bijective on the whole of $$\mathbb{R}$$.
therefore, \begin{align*} -\frac{1}{5} \int \frac{du}{u} = -\frac{1}{5} \ln\lvert u \rvert + C = -\frac{1}{5} \ln\lvert 5t \rvert + C \end{align*} Now is it fine @mikejacob ?
$$\int\frac1x\,dx=\ln\lvert x\rvert+c=\ln\lvert x\rvert+\ln a+c^\prime=\ln\lvert ax\rvert+c^\prime$$
, where $$c,c^\prime\in\mathbb R,a\in\mathbb R^+$$. | 2022-06-27T05:07:37 | {
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https://math.stackexchange.com/questions/513974/5-letter-arrangements-of-the-word-statistics | 5 Letter Arrangements of the word 'Statistics'
How many different 5-letter 'words' can be formed from the word 'statistics'?
I really am pretty stumped. I understand how to calculate more simpler questions in which each letter of the word is different using the Permutation formula n!/(n-r)! I can also deal with questions where there are repeating letters, but the 'new' words that are being created are the same length as the original word. But for this type of question where there are repeating letters and the 'new words' are shorter than the original, I don't know where to start. I don't need the precise answer - I just want to know how to get there. In fact, I know the answer (1390) but I cannot come up with a solution. I have tried using the permutation formula but it doesn't seem appropriate for this question.
• What is your progress on this problem? What are your ideas? – Sasha Patotski Oct 3 '13 at 23:05
I will give a basic counting solution. If you want an answer using generating functions, look at the answer to this question.
If all five letters are different (which happens when you use one of each), you get $5!=120$ words.
If only two of the letters are equal (two 'S', two 'T', or two 'I', and $\binom{4}{3}$ ways to choose the remaining three), you have $\frac{5!}{2!}=60$ words for each of {'S', 'T', 'I'}. In total $720$ words.
If you have "two pairs" (there are three ways to get the pairs, and three ways to get the last letter in each case), there are $\frac{5!}{2!2!}=30$ words. In total for all these cases, there are $270$ words.
If you have "full house" (there are two ways to get three equal, and two ways for each of those to get the last pair), there are $\frac{5!}{3!2!}=10$ words. In total for these cases, there are $40$ words.
If you have three equal and two different (the triple can be had in two ways, and the remaining two can be had in $\binom{4}{2}=6$ ways for each triple), there are $\frac{5!}{3!}=20$ words. For all these cases, there are $240$ words.
Summing all of these cases gives $1390$ words.
• The answer given to the question in the textbook was 1390. Your answer makes sense to me but it seems that some multiple is missing from the calculations... I can't see where though. – frantastic Oct 4 '13 at 0:41
• @frantastic, I'm not in a position to review the argument carefully right now, but never discount the possibility that the textbook answer is wrong. Often, you will be able to find a list of errata for your textbook somewhere online (often on the writer's or publisher's website). – dfeuer Oct 4 '13 at 1:35
• @frantastic: I hade made an error in the "pair" case, forgetting to count the number of ways to choose the last three. In the last case, I had gotten 120 instead of 240 through bad mental calculation. – Mårten W Oct 4 '13 at 8:29 | 2019-10-22T10:47:43 | {
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https://math.stackexchange.com/questions/2083148/finding-the-probability-by-combination | # Finding the probability by combination?
Question as follows A bag contains 50 tickets numbered 1,2,3,....,50 of which five are drawn at random and arranged in ascending order of the number appearing on the tickets ( x1 < x2 < x3 < x4 < x5 ). Find the probability that x3 = 30.
Total number of elementary events =50C5 Given,third ticket =30
=> first and second should come from tickets numbered 1 to 29 =29C2 ways and remaining two in 20C2 ways.
Therfore,favourable number of events = 29C2×20C2
Hence,required probability = (29C2×20C2)/50C5 =551 / 15134
Answer given is 551/15134 but in the solution they have taken the case where x2 will be less than x1. Is my assumption correct? if it is not then what could be the correct answer.
• You need to provide the full method, not just the numeric bottom line, otherwise your question is practically unreadable. P.S., I did not down-vote you, because I see good intentions and effort made in the question. Nevertheless, you need to clarify things here if you're hoping to get an answer. Jan 4, 2017 at 10:37
• The two numbers less than 30 you refer to are chosen, and "arranged in ascending order.." Jan 4, 2017 at 10:37
• the number of ways you can select to satisfy the condition is (any 2 from 1 - 29) x (any 1 from 30) x (any 2 from 31 - 50) = 29C2 x 1C1 x 20C2 ......... for the condition to be true, 30 had to be selected, and two that were smaller and two that were larger, the actual ordering leads to that condition, but is not part of the calculation
– Cato
Jan 4, 2017 at 10:41
• You are taking the question as wrong. Suppose you have option to pick 2 numbers out of 29. You picked 23, 12. In C(29,2) ways. Then you have to arrange them in ascending order. That is 12, 23. Only 1 way. So we have 1 * C(29,2). Similarly for last 2 numbers. Jan 4, 2017 at 10:47
• the order they are drawn doesn't matter, there will always be a lowest number, then a 2nd ranked number, then a 3rd ranked number. Can you see that if 30 was drawn, then there has to be exactly 2 drawn from 1-29? If exactly 1 was drawn from 1-29, then the condition would not be met. 1,2,30,31,32 is a set of tickets that meets the condition, I'm only counting that once
– Cato
Jan 4, 2017 at 10:49
Required probability = $\frac{\binom{29}{2} \times \binom{20}{2}}{\binom{50}{5}}$
Because $x_3$=30 is fixed. You pick two numbers from first 29 and arrange them in ascending order. Similarly 2 numbers from last 20 and arrange them in ascending. | 2022-10-07T18:57:43 | {
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http://www.awv.sk/1axc9/6abae4-set-operations-proofs | / By / / 0 Comments
# set operations proofs
1 - 6 directly correspond is also in By definition of intersection, x ∈ A and x ∈ B. We can use the set identities to prove other facts about sets. &= \{x\mid x \in \overline{A}\wedge x \in \overline{B} )\} \\ Here are some basic subset proofs about set operations. Theorem: For any sets, $$|A\cap B|\le|A|$$ and $$|A\cap B|\le|B|$$. Note here the correspondence of Hence . $\bigcup_{i=1}^{n} S_i\,.$ This can also proven using set properties as follows. Since , . This proof might give a hint why the equivalences and set identities tables are so similiar. B . and between Back to Schedule Consider an arbitrary element x. Theorem: For any sets, $$\overline{A\cup B}= \overline{A} \cap \overline{B}$$. It's the table of logical equivalences with some search-and-replace done to it. 13. . Hence . A, and ------- De Morgan's Laws &= A\cap \overline{B}\cap A\cap \overline{C} \\ by the distribution Theorem: For any sets, $$A-B = A\cap\overline{B}$$. \begin{align*} Could have also given a less formal proof. ------- Associative Laws Since (from ), For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). Theorem For any sets A and B, B ⊆ A∪ B. Proof… B ) B &= A\cap \overline{B}\cap \overline{C} \\ The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: ) by the definition of ( B - A ) . $$A\cup{U}= {U}\\A\cap\emptyset= \emptyset$$, $$(A\cup B)\cup C = A\cup(B\cup C)\\(A\cap B)\cap C = A\cap(B\cap C)$$, $$A\cup(B\cap C) =(A\cup B)\cap(A\cup C)\\A\cap(B\cup C) = (A\cap B)\cup(A\cap B)$$, $$\overline{A\cap B}=\overline{A} \cup \overline{B}\\\overline{A\cup B}= \overline{A} \cap \overline{B}$$, $$A\cup(A\cap B) = A \\ A\cap(A\cup B) = A$$, $$A\cup\overline{A} = {U}\\A\cap\overline{A} = \emptyset$$, Written $$A\cup B$$ and defined B . Then if , then since . A x\in S \wedge x\in\overline{S} \\ For example, (b) can be proven as follows: ------- Idempotent Laws by the definition of . Properties of Set Operation Subjects to be Learned . The properties 1 6 , and 11 A B Let x be an arbitrary element in the universe. Basic properties of set operations are discussed here. Furthermore a similar correspondence exists between \[\{1,2,3,4\}\cap\{3,4,5,6\} = \{3,4\}\,., For example, \begin{align*} Then by the definition of the operators, &= (A-B)\cap (A-C)\,.\quad{}∎ A. Proof: Let x ∈ A∩B. , These can also be proven using 8, 14, and 15. ( B - A ) Then . &= \{x\mid x\in A \wedge x\in \overline{B}\} \\ Let us prove some of these properties. This can also be proven in the similar manner to 9 above. \end{align*}. A • Applying this to S we get: • x (x S x S) which is trivially True • End of proof Note on equivalence: • Two sets are equal if each is a subset of the other set. but also for others. 8. ( cf. ) and implications If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written $$A\cap B$$ and defined ( cf. ) x 11. Hence. (See example 10 for an example of that too.). Here is an example. by the definition of . Be careful with the other operations. from the equivalences of propositional logic. Return to the course notes front page. Theorem: For any sets, $$|A\cup B|\ge|A|$$ and $$|A\cup B|\ge|B|$$. ------- Identity Laws 4. Next -- Recursive Definition x\in S \wedge x\notin{S}\,. As an example, we can prove one of De Morgan's laws (the book proves the other). 9. 7. Thus we see that these sets contain the same elements.∎, More Formal Proof: By definition of the set operations, the commutativity of Thus $$A-B\not\subseteq A$$. x \overline{\mathbf{Q}} = \mathbf{R}-\mathbf{Q} \,.\]. . A A \end{align*}\]. This section contains many results concerning the properties of the set operations. Proof for 9: Let x be an arbitrary element in the universe. &= \{x\mid x\in (A \cap \overline{B})\} \\ x\in S\cap\overline{S}\\ Also since , . Then there must be an element $$x$$ with $$x\in(A-B)$$, but $$x\notin A$$. Also . | 2022-05-21T18:31:16 | {
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https://math.stackexchange.com/questions/2025619/antiderivatve-to-find-a-function-such-the-the-slope-is-true-for-all-points-x-y | # Antiderivatve to find a function such the the slope is true for all points $(x,y)$
"The graph of a certain function $f$ has the slope $4x^3-5$ at each point $(x,y)$ and the line $x+y=0$ is the tangent line to the graph. Find the function $f$."
I took the antiderivative to get $f(x)=x^4-5x+C$ but I'm not really sure how to get the initial conditions. I have that $x+y=0$ or $y=-x$ so the slope is $-1$. I can then plug it into $-1=4x^3-5$ and solve for $x$ to get $x=1$.
I can then plug that x value into $y=-x$ to get $y=-1$. Since the tangent line and the graph must share the same common point then $(1,-1)$ must be on the graph of $f(x)$. So I can solve for the initial condition which means that $-1=1-5+C$ or $3=C$ so $f(x)=x^4-5x+3$ . Is the method correct?
Your method, and your result, are correct. You can verify this solving the system of the line and the quartic: $$\begin {cases} y=-x\\ y=x^4-4x+3 \end{cases}$$ this gives the equation $$-x=x^4-5x+3 \iff x^4-4x+3=0 \iff (x-1)^2(x^2+2x+3)=0$$
That has a double root at $x=1$, so the line is tangent to the curve at this point. | 2019-12-07T14:37:31 | {
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https://atekihcan.github.io/CLRS/02/E02.01-04/ | Consider the problem of adding two $$n$$-bit binary integers, stored in two $$n$$ element arrays $$A$$ and $$B$$. The sum of the two integers should be stored in binary form in an $$(n + 1)$$ element array $$C$$. State the problem formally and write pseudocode for adding the two integers.
The problem can be formally stated as…
Input: Two $$n$$ bit binary integers stored in two $$n$$ element array of binary digits (either 0 or 1) $$A = \langle a_1, a_2, ... , a_n \rangle$$ and $$B = \langle b_1, b_2, ... , b_n \rangle$$.
Output: A $$(n + 1)$$ bit binary integer stored in $$(n + 1)$$ element array of binary digits (either 0 or 1) $$C = \langle c_1, c_2, ... , c_{n+1} \rangle$$ such that $$C = A + B$$.
We also assume the binary digits are stored with least significant bit first, i.e. from right to left, first bit in index $$1$$, second bit in index $$2$$, and so on. Why we are doing this is discussed after the pseudocode.
$$\textsc {Add-Binary }(A, B)$$\begin{aligned}1& \quad n=\textsc {Max}(A.length,\,B.length) \\2& \quad \text {let }C[n+1]\text { be }\text { new }\text { array } \\3& \quad carry=0 \\4& \quad \textbf {for }i=1\textbf { to }n \\5& \quad \qquad C[i]=\textsc {}(A[i]+B[i]+carry)\mod2 \\6& \quad \qquad carry=\lfloor \textsc {}(A[i]+B[i]+carry)/2\rfloor \\7& \quad C[n+1]=carry \\8& \quad \textbf {return }C \\\end{aligned}
#### Left to Right or Right to Left
An earlier version of the solution presented here assumed the least significant bit was stored in index $$n$$ instead of index $$1$$. Which made the solution not only wrong (it did not handle all possible cases properly), it also caused a great deal of confusion in the comments section.
Here is why assuming least significant bit in index $$n$$ will make the problem unnecessarily complicated.
Consider the following two binary additions
The one on the left adds $$111_b$$ and $$1_b$$ to $$1000_b$$. In this case, $$n = 3$$ and we end up with final array $$C$$ of length $$n + 1 = 4$$. The array indices are shown in blue with the assumption of storing bits as we they appear visually from left to right, i.e. most significant bit in first index and least significant bit in last index, $$n$$.
In this particular case there is no complication, and we could have just designed our pseudocode to iterate from the opposite direction, from $$n$$ downto $$1$$, and stored the result of the addition in line 4 in $$C[i + 1]$$ and the final $$carry$$ in line 6 in $$C[1]$$.
However, note that for the addition on the right, $$110_b$$ and $$1_b$$ sums up to $$111_b$$, and we end up with final array $$C$$ of length $$n = 3$$. In this case, $$C[1]$$ is empty (highlighted with light red), and we are left with the additional task of shifting all the elements to the left to meet our initial assumption of having most significant bit at index $$1$$.
One can argue that having zero in the first index is not a deal breaker, but depending on the use case it might add up to redundant work. For example, let’s say we need to repeatedly do this addition. And every time we end up with a case like the right one. Then we would keep on adding redundant zeroes in the beginning of the resulting array.
#### Python Code
# Bitwise binary addition of two lists # containing binary digits with least significant bit first def AddBinary(A, B): carry = 0 n = max(len(A), len(B)) C = [0 for i in range(n + 1)] for i in range(n): # One of A and B has length less than n # We need to treat index out of bound for that array # This is not explicitly handled in pseudocode a = A[i] if i < len(A) else 0 b = B[i] if i < len(B) else 0 # Modulo for sum and integer division for carry C[i] = (a + b + carry) % 2 carry = (a + b + carry) // 2 C[n] = carry return C # Utility function for converting decimal to binary def DecimalToBinary(num): if num == 0: return [0] ret = [] while num > 0: ret.append(num & 1) num = num >> 1 return ret # Utility function for converting binary to decimal def BinaryToDecimal(lst): num = 0 for i in range(len(lst)): num += lst[i] << i return num # Test import random num_failed = 0 total_tests = 100 for i in range(total_tests): # Create two random integers a = random.randint(0, 10000) b = random.randint(0, 10000) a_ = DecimalToBinary(a) b_ = DecimalToBinary(b) sum = BinaryToDecimal(AddBinary(a_, b_)) # Check if the sum is correct if sum != (a + b): num_failed += 1 print(f"#{i:<2} {a} + {b} = {sum} [Expected {(a + b)}]") if num_failed > 0: print(f"\nFailed {num_failed}/{total_tests}") else: print(f"Passed {total_tests}/{total_tests} tests") | 2023-03-28T23:58:59 | {
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https://math.stackexchange.com/questions/1536267/finding-the-values-of-5ab-if-ax2bx10-does-not-have-2-distinct-roots | # Finding the values of $5a+b$ if $ax^2+bx+10$ does not have $2$ distinct roots
If $ax^2+bx+10=0$ does not have two distinct real roots where( $a$ and $b$ are real) then the possible values of $5a+b$ from the given 4 options(as obviously there can be infinite possible values..but I only need to find out which of the four given possible values is one of these ...in short it is a Multi choice-multi-correct question) are-( there can be more than one option correct)
1) -3
2) -2
3) -1
4) 0
my attempt Here are some conclusions I have drawn-
• it is given that it does not have two distinct real roots so either( both roots are imaginary and conjugates of each other)- $D(b^2-4ac)<0$ (that is as $c=10$ so $b^2-40a<0$...or another case is both roots become equal that is $b^2-40a=0$.
• also I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible but the answers given are $-2,-1,0$ (Edit:apparently $0$ is not the answer as pointed out be Tim phan below..but I still need help in proving it is equal to $-1,-2$)....also I am unable to get other answers I have tried getting $5a+b$ by keeping $x=5$ to get $5(5a+b+2)$ but I don't know whether it will be equal to zero or less than or greater than zero... As I don't know whether $5$ is a root of $f(x)$ or not..neither it is given in question...
Any help??
• I think you need more constrains. Here is why. Like you pointed out, essentially, you are solving a system of 2 equations with 3 unknown. $b^2 - 4 a c < 0$ and $b + 5a = N$. – Paichu Nov 19 '15 at 3:28
• In the attempted solution, you refer to $b^-40c$. Did you omit saying that $c=10$? Are $a,b,c$ restricted in some way, say to integers? – André Nicolas Nov 19 '15 at 3:30
• If that is the case, then you can try going about it in this way. Assume $b + 5a = 0$, then $b = -5a$ and $a = -\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together you will get two equations: $b^2 < -\frac{4}{5}b c$ and $25a^2 < 4a c$. You then proceed with 4 cases $a,b>0$, $a,b<0$, $a>0,b<0$, $a<0,b>0$. If all of them lead to contradiction, then it is not possible. – Paichu Nov 19 '15 at 3:42
• Why didn't you write $ax^2+bx+10$, instead of "$c$". Tbh, it looks like you were attempting to make it confusing. – YoTengoUnLCD Nov 19 '15 at 3:50
• Let $a=1$ and $b=1.01, 1.02, 1.03, 1.04,\dots$. Don't have distinct real roots, indeed don't have real roots at all. – André Nicolas Nov 19 '15 at 4:21
If you write $5a + b = k$ and use this to get rid of the $a$ in the equation $b^2 - 40a \leq 0$, you get:
$$b^2 - 40\left(\frac{k - b}{5}\right) \leq 0$$
$$b^2 + 8b \leq 8k$$
Complete the square:
$$(b + 4)^2 \leq 8k + 16$$
$$(b + 4)^2 \leq 8(k + 2)$$
Since $b$ is arbitrary, the only restriction is that the left hand side is nonnegative. So you get:
$$0 \leq 8(k + 2)$$
$$-2 \leq k$$
Hence $k = 0, -1, -2$ are all possible.
The possible values (from those listed) are $0$, $-1$, and $-2$, obtained by $(a,b) = (0,0)$, $(1,-6)$, and $\left(\frac{2}{5},-4\right)$. (The first two are not unique, but the third one is; see below.)
Now, if $ax^2+bx+10$ does not have 2 distinct roots, then the discriminant is nonpositive, i.e. $b^2-40a\le 0$. Hence, $40a\ge b^2$, necessarily implying that $a\ge 0$. Furthermore, $$b^2\le 40a\implies |b|\le\sqrt{40a}\implies b\ge -\sqrt{40a}$$ and hence $5a+b \ge 5a-\sqrt{40a}$. It can be checked that $$\min\limits_{a\ge 0}{\left(5a-\sqrt{40a}\right)} = -2$$ achieved at $a = \frac{2}{5}$. Hence, if $ax^2+bx+10$ does not have 2 distinct roots, then $$5a+b\ge -2$$ with equality if and only if $a = \frac{2}{5}$ and $b = -\sqrt{40a} = -4$.
• Why is this reasoning of my wrong....I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible – Freelancer Nov 19 '15 at 8:16
• @Freelancer why can't we have $f(3)-f(2) = 0$? For example, if $f$ is constant, then this is true, and there's nothing saying it can't (since if it were constant, it would take the constant value $10$, which is nonzero) – Joey Zou Nov 19 '15 at 8:17
• I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..(which can be only possible when both 3,2 are roots are of the equation...which is not allowed (as real roots should not be distinct here) – Freelancer Nov 19 '15 at 8:19
• @Freelancer who cares if $f$ doesn't have two distinct roots? It says very little about the difference between its values. For example, $f(x) = x^2+1$ has no real roots, yet $f(1) - f(-1) = 0$, $f(2)-f(-2) = 0$, $f(3) - f(-3)=0$, and so on... – Joey Zou Nov 19 '15 at 8:21
I found that if a = 1/5 and b = -1 then 5a+b = 0. Did i break any rules?
• You found one solution ! Are there others ? – Tom-Tom Nov 19 '15 at 8:35
Assume $b + 5a = 0$, then $b = -5a$ and $a = -\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together to obtain $b^2 < -\frac{4}{5}b c$ and $25a^2 < 4a c$.
(1) If $a>0, b>0$, then we have: $b<-\frac{4}{5}c$< which implies $c<0$. False.
(2) If $a<0, b<0$, then we have: $-25|a|>4c$ which implies $c<0$. False.
If $a>0, b<0$, you have the same thing as in (1) and if If $a<0, b>0$, you have the same thing as in (2). Thus it is not possible for $b + 5a = 0$. Proceed similarly.
Second part: let $D>0$, assume $5a + b =-D$, so $b = -(D+5a)$. Then for this equation to not have two distinct real roots: $b^2 \leq 40a$. replacing $a$ in to obtain: $(D+5a)^2 < 40a$ Notice that this implies $a > 0$. Simplify: $$D + 5a < \pm\sqrt{40 a}$$ Let $u = \sqrt{a}$, then we have $5u^2 \mp \sqrt{40}u + D < 0$ is the condition, which is if we can find such $u$ under some condition on $D$, then the same condition is also the condition for $ax^2 + bx + 10 = 0$ to not have two distinct real roots. Solve this to get $$u=\frac{1}{10}\left(\sqrt{40} - \sqrt{40 - 20 D}\right)$$ this has a solution if $D \leq 2$, which means if $D \leq 2$, then the $ax^2 + bx + 10 = 0$ does not have two distinct real roots. Thus the possible solutions are $5a + b = -1, -2$, or (3) and (2). Please check my algebra carefully.
• You can't do that yet, since you do not know if $a$ is positive or negative, which can affect the inequality sign. – Paichu Nov 19 '15 at 3:58
• I edit the second part there. Check the algebra carefully though. Not super logical, but it is one way of doing that. – Paichu Nov 19 '15 at 4:52
• You're right. I just fixed it. – Paichu Nov 19 '15 at 5:16
• Also how can we assume $5a+b=-D$ ?? – Freelancer Nov 19 '15 at 5:17
• $D\le 2$ actually means $D = -1$ or $D=-2$. Also, how does $a>0,b<0$ give the same thing as in (1)? – Joey Zou Nov 19 '15 at 7:50 | 2019-10-20T08:21:08 | {
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https://math.stackexchange.com/questions/2557479/motivation-for-the-ring-product-rule-a-1-a-2-a-3-cdot-b-1-b-2-b-3-a/2557744 | Motivation for the ring product rule $(a_1, a_2, a_3) \cdot (b_1, b_2, b_3) = (a_1 \cdot b_1, a_2 \cdot b_2, a_1 \cdot b_3 + a_3 \cdot b_2)$
In a lecture, our professor gave an example for a ring. He took it out of another source and mentioned that he does not know the motivation for the chosen operation.
Of course, it's likely that somebody just invented an arbitrary operation satisfying ring axioms. I'd still like to try my luck whether anyone here can decipher the operation and give any kind of motivation for that example.
On $\mathbb{R}^3$ define the operations $+$ and $\cdot$ by \begin{aligned} (a_1, a_2, a_3) + (b_1,b_2,b_3) &= (a_1+b_1,a_2+b_2,a_3+b_3) \\ (a_1, a_2, a_3) \cdot (b_1, b_2, b_3) &= (a_1 \cdot b_1, a_2 \cdot b_2, a_1 \cdot b_3 + a_3 \cdot b_2). \end{aligned} (The $+$ and $\cdot$ operations on the right side are the usual addition and multiplication from $\mathbb{R}$.) With those operations, one can confirm that $\left(\mathbb{R}^3, +, \cdot \right)$ is a ring.
• In case anyone is wondering, the multiplicative identity is $(1,1,0)$. – Alex Provost Dec 8 '17 at 19:00
• I think it's just an arbitrary thing. I can't see any motivating pattern here. If someone else does see something, that will be surprising and very interesting. Also, in case anyone is wondering, the multiplication is associative (I checked). – Zach Teitler Dec 8 '17 at 19:04
• I can't check at the moment, but is multiplication here commutative? Perhaps someone wanted to construct a noncommutative ring without referencing matricies? – Andrew Tawfeek Dec 8 '17 at 19:14
• The multiplication is not commutative, take $(1,0,0) \cdot (0,0,1) \neq (0,0,1) \cdot (1,0,0)$. – Qi Zhu Dec 8 '17 at 19:16
• mentioned that he does not know the motivation for the chosen operation Seriously?! If you show him the upper triangular matrix ring he will be rather abashed, then :) – rschwieb Dec 8 '17 at 21:36
This is just matrix multiplication in disguise. Specifically, if you identify $(a_1,a_2,a_3)$ with the matrix $\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}$, these operations are the usual matrix operations: $$\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}+\begin{pmatrix}b_1 & b_3 \\ 0 & b_2\end{pmatrix}=\begin{pmatrix}a_1+b_1 & a_3+b_3 \\ 0 & a_2+b_2\end{pmatrix}$$ $$\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}\begin{pmatrix}b_1 & b_3 \\ 0 & b_2\end{pmatrix}=\begin{pmatrix}a_1b_1 & a_1b_3+a_3b_2 \\ 0 & a_2b_2\end{pmatrix}$$
• Oh of course! Nice and simple, thanks, the ring is much more motivated now! – Qi Zhu Dec 8 '17 at 21:32
• Unfortunately I realized that's exactly what I was describing a minute too late... – rschwieb Dec 8 '17 at 21:32
It is isomorphic to the ring of matrices
$$\left\{\begin{bmatrix}a_1&a_3\\0&a_2\end{bmatrix}\,\middle|\,a_1, a_2,a_3\in \mathbb R\right\}$$
It's a semiprimary ring whose Jacobson radical is the subset with $a_1=a_2=0$. The Jacobson radical is nilpotent, and $R/J(R)\cong\mathbb R\times\mathbb R$. Here is a list of more properties of such a ring.
This sort of ring is fairly famous, and has nice interpretations. One of them is that if you select a chain of subspaces $\{0\}<V<W<\mathbb R\times \mathbb R$ ($W$ of dimension $1$, $V$ of dimension $2$) then the linear transformations of $\mathbb R\times\mathbb R$ which stabilize this chain is isomorphic to this triangular matrix ring. That is, $\phi$ stabiliezes the chain if $\phi(V)\subseteq\phi(W)$.
Incidentally, you are always going to be able to extract some sort of matrix presentation for a multiplication like you are describing, because you can rely on it being a finite dimensional algebra. If it really is a valid ring multiplication, it's bilinear, and so you can work on figuring out what a logical 'basis' is and then deduce what it looks like with matrices.
• This should be the accepted and most upvoted answer. – Xam Dec 9 '17 at 5:54
• In your last paragraph, what exactly do you mean by "a multiplication like you are describing"? – Jack M Dec 9 '17 at 11:07
• @JackM Multiplication of $n$-tuples over a field, at least. Because that defines a representation with square matrices. – rschwieb Dec 9 '17 at 13:18
• @rschwieb Surely not any ring structure on $\mathbb F^n$ is isomorphic to a ring of $\mathbb F$-matrices. Maybe a ring structure in which addition is component wise, and multiplication is given by quadratic polynomials? – Jack M Dec 9 '17 at 13:29
• @JackM You're right, I should emphasize the pointwise addition. Remember I'm stipulating that the multiplication is actually a valid ring multiplication, not any rule with tuples. To rephrase, "if multiplication on elements of $F^n$ defines a valid ring structure, then the ring can be represented with $n\times n$ matrices." This is easy to see, because you just identify each tuple with the linear transformation it produces with the multiplication. – rschwieb Dec 9 '17 at 14:22 | 2020-11-24T12:32:36 | {
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https://math.stackexchange.com/questions/2819873/orthogonally-diagonalize-a-matrix-with-nonreal-eigenvectors | Orthogonally Diagonalize a matrix with nonreal eigenvectors
I am given the matrix A and asked to orthogonally diagonalize it.
$$A=\begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix}$$
While doing this I got $\lambda = 0,2.$ Then I found the eigenvectors corresponding to the eigenvalues to be $$\begin{pmatrix} i \\ 1 \\ \end{pmatrix}$$ and $$\begin{pmatrix} -i \\ 1 \\ \end{pmatrix},$$ respectively. While trying to divide each eigenvector by its norm I run into a problem, take $V_1$ for example $\|V_1\| = 0$ so I obviously can't divide by $V_1$ by its norm to make it orthogonal. My question is, is $A$ even orthogonally diagonizable at all? Or if the $\|V_1\| = 0$ does this mean that it is already orthogonal and I can just use my eigenvectors as is to generate the matrix $U$ such that $A=UDU^*?$
• $||V_1|| \neq 0$. Same goes for norm of $V_2$. You have to use absolute values around the squared entries. – layabout Jun 14 '18 at 18:47
• You mean to unitarily diagonalize it, no? – Mathematician 42 Jun 14 '18 at 18:48
• You are not working with a real inner product, but a Hermitian inner product, i.e. $\left\langle x,y \right\rangle=\sum_{i=1}^n x_i\overline{y_i}$. – Mathematician 42 Jun 14 '18 at 18:50
• The inner product is always positive definate. That is $\langle x,x\rangle \ge 0$ and only equals $0$ if $x = 0.$ (and the norm is $\sqrt {\langle x,x\rangle}$) To meet this criterion, we need a slightly different inner product definition when working with vectors over a complex field. $\langle x,y\rangle = x^\dagger y$ ($\dagger$ is the conjugate transpose) will do. So the norm $(i,1) = \sqrt {(-i\cdot i + 1\cdot 1)}$ – Doug M Jun 14 '18 at 18:52
The norm of a vector $v$ is
$$\|v\|^2 = (v^*)^Tv = \sum_k v_k^* v_k$$
where $^*$ denotes the complex conjugate. So in your case
$$\|v_1\|^2 = \pmatrix{i & 1}^* \pmatrix{i \\ 1} = \pmatrix{-i & 1} \pmatrix{i \\ 1} = 1 + 1 = 2$$
• Thank you very much. I had forgotten that there were special properties for cases like this. – Moseph Jun 14 '18 at 19:09
• @Moseph Happy to help – caverac Jun 14 '18 at 19:20
It happens that $\|(a,b)\|=\sqrt{|a|^2+|b|^2}$. Therefore, the norm of both vectors that you mentioned is $\sqrt2$. | 2021-07-30T12:57:36 | {
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https://math.stackexchange.com/questions/516507/how-to-find-the-general-solution-of-tan-leftx-frac-pi-3-right3-tan-l/516520 | # How to find the general solution of $\tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0$
Find the general solution of the equation.
\begin{eqnarray} \tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0\\ \end{eqnarray}
The answers in my textbook are $n\pi$ and $n\pi +\frac{\pi }{3}$.
Previously, I compute the similar questions by using the following operations :
\begin{eqnarray} \\\tan 3x&=&\cot 5x\\ \\\tan 3x&=&-\tan \left(\frac{\pi }{2}+5x\right)\\ \\\tan 3x&=&\tan \left(-\frac{\pi }{2}-5x\right)\\ \\3x&=&-\frac{\pi }{2}-5x\\ \\8x&=&n\pi -\frac{\pi }{2}\\ \\x&=&\frac{n\pi }{8}-\frac{\pi }{16}\\ \end{eqnarray}
But now there is a 3 in front of the second tan.
What should I do?
I do not know whether my method is correct. If you have any other methods, would you mind telling me the methods?
Update 1 : Found one of the answers, are my operations correct?
• Is your question correct? The question seems likely than $tan(x+\pi/3)=3tan(x-\pi/6)$ – MS.Kim Oct 6 '13 at 12:18
• Thx, the question now is correct. – Casper Oct 6 '13 at 12:24
If we assume that your question is to solve the equation $tan(x+\pi/3)+3tan(x-\pi/6)=0$. if we let $A=x+\pi/3, B=x-\pi/6$, then $A-B=\pi/2$, so we can transform the term $tanA$ to $tan(\pi/2+B)$.
and because $tan(\pi/2+B) = -cotB$, the equation will be $-cotB+3tanB=0$, $tan^2B=1/3$ so the general solution of $B=x-\pi/6=n\pi+\pi/6$ or $n\pi-\pi/6$
so $x=n\pi +\pi/3$ or $n\pi$
The solution is given as follows.
I don't know how to use spoiler for a block of code.
\documentclass[preview,border=12pt]{standalone}
\usepackage{amsmath}
\begin{document}
\abovedisplayskip=0pt\relax
\begin{gather*}
\tan (x+\frac{\pi}{3}) + 3\tan (x-\frac{\pi}{6})=0\\
\tan (x+\frac{\pi}{3}) = 3\tan (\frac{\pi}{6}-x)\\
\frac{\tan x + \tan \frac{\pi}{3}}{1-\tan x \tan \frac{\pi}{3}} = 3\frac{\tan \frac{\pi}{6} -\tan x}{1+\tan \frac{\pi}{6}\tan x}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = 3\frac{\frac{\sqrt 3}{3} -\tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\times\frac{3}{3}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ 3\sqrt 3 -9 \tan x}{3+ \sqrt 3\tan x}\\
(1- \sqrt 3\tan x) (3\sqrt 3 -9 \tan x) = (\tan x + \sqrt 3)(3+ \sqrt 3\tan x)\\
3\sqrt 3 -9\tan x -9\tan x +9\sqrt 3\tan^2x = 3\tan x +\sqrt 3\tan^2 x +3\sqrt 3 +3\tan x\\
8\sqrt 3\tan^2 x -24 \tan x=0\\
\sqrt 3\tan^2 x -3\tan x=0\\
\tan x (\sqrt 3 \tan x -3)=0\\
\tan x =0 \text{ or } \sqrt 3 \tan x -3 =0\\
\tan x=0 \text{ or } \tan x =\sqrt 3\\
x=n\pi \text{ or } x=\frac{\pi}{3}+n\pi
\end{gather*}
\end{document}
• =口= This operation must take so much time when during exam...Is it the best way? – Casper Oct 6 '13 at 13:42
• @CasperLi: Not much even though you use LaTeX during the exam. :-) – kiss my armpit Oct 6 '13 at 13:44
• Are there any fast methods to input math/LaTeX? ~.~ – Casper Oct 6 '13 at 13:53
• @CasperLi: It depends on the text editor you are using. If there is a shortcut key for each macro then you can type much faster. But I think the important point is your typing speed. :-) – kiss my armpit Oct 6 '13 at 13:55 | 2019-12-14T00:50:37 | {
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https://math.stackexchange.com/questions/1953218/is-the-empty-set-is-a-subset-of-any-set-a-convention/1953511 | # Is "The empty set is a subset of any set" a convention?
Recently I learned that for any set A, we have $\varnothing\subset A$.
I found some explanation of why it holds.
$\varnothing\subset A$ means "for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set A". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set A). So $\varnothing\subset A$ holds.
What confused me was that, the following expression was also a vacuous truth.
For every object $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set A.
According to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set A) holds, so $\varnothing\not\subset A$ would be true, too.
Which one is correct? Or is it just a convention to let $\varnothing\subset A$?
• Maybe think of it like this: The empty set is like an empty bag. The set $A$ is like a bag with some stuff in it. It's possible to put your hand into either bag and take nothing out. Since everything you could take out of the empty bag $\varnothing$ was also in $A$, this means that $\varnothing \subset A$. This argument may need to be made rigorous, but I think the intuition is a good start. Oct 4, 2016 at 9:31
• It's a consequence of the following rule: any statement of the form 'if $x \in \emptyset$, then y' is true. That rule is itself a consequence of the agreement that "if P then Q" means "P is false or Q is true." On a philosophical level, I would distinguish this agreement from being a mere convention since it's an agreement about what reasoning itself means. But on a practical level, if you get annoyed thinking about these issues in any particular context, it's often better to shrug your shoulders, call it a convention, and move on! Oct 4, 2016 at 11:53
• It's the way math works... It is not a convention, but a theorem: we have an axiom asserting that there is a unique set $y$ such that : $\forall x \lnot (x \in y)$ and we call $y$ "the emptyset". Then we assume the def of set-inclusion : $A \subset B \leftrightarrow \forall x (x \in A \to x \in B)$ and we derive, by logical rules, that $\emptyset \subset A$, for any $A$. Oct 5, 2016 at 10:15
• It is true that for every $x$ and every set $A$, if $x$ belongs to the empty set, then $x$ both belongs to $A$ and $x$ does not belong to $A$ Oct 6, 2016 at 0:32
• “For every object $x$, if $x$ belongs to the empty set, then $x$ doesn’t belong to the set $A$.” This statement is vacuously true, indeed. However, it does not imply that $\varnothing\not\subset A$. For the latter to be true, there should actually exist some $x$ such that $x\in\varnothing$ but $x\notin A$. Oct 6, 2016 at 22:56
There’s no conflict: you’ve misinterpreted the second highlighted statement. What it actually says is that $\varnothing$ and $A$ have no element in common, i.e., that $\varnothing\cap A=\varnothing$. This is not the same as saying that $\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\varnothing\subseteq A$.
To expand on that a little, the statement $B\nsubseteq A$ does not say that if $x\in B$, then $x\notin A$; it says that there is at least one $x\in B$ that is not in $A$. This is certainly not true if $B=\varnothing$.
• @BlueRaja-DannyPflughoeft: Which is to say that he misinterpreted it: it does not mean what he understood it to mean. Oct 4, 2016 at 17:47
• @Searene You can derive anything from a contradiction. This is basic logic, and nothing wrong with it.
– orlp
Oct 5, 2016 at 7:17
• No you shouldn't say that 5=6 is true, but you can say that If 1=2, then 5=6 is true. The statement that x belongs to A is not necessarily true; what is true is precisely: for every object x, if x belongs to the empty set, then x also belongs to the set A, and that is the definition of the empty set being a subset of A. Oct 5, 2016 at 8:13
• @Searene You are making a fundamental error. You cannot derive that the conclusion is true!!! If you have a statement: $\forall x, x \in \emptyset \rightarrow \emptyset \not\subset A$ is true because $x \in \emptyset$ is false. This tells you nothing regarding $\emptyset \not\subset A$. In an implication $P \rightarrow Q$ if $P$ is false than the whole implication is true, but this does not say anything about the truth value of just $Q$. Oct 5, 2016 at 13:02
• ... vacuously true, but that doesn’t tell you anything about whether or not $\varnothing$ is a subset of $A$. The truth of $\forall x\,(x\in\varnothing\to x\in A)$, on the other hand, means by the definition of $\subseteq$ that $\varnothing\subseteq A$. Oct 5, 2016 at 17:25
From Halmos's Naive Set Theory:
A transcription:
The empty set is a subset of every set, or, in other words, $\emptyset \subset A$ for every $A$. To establish this, we might argue as follows. It is to be proved that every element in $\emptyset$ belongs to $A$; since there are no elements in $\emptyset$, the condition is automatically fulfilled. The reasoning is correct but perhaps unsatisfying. Since it is a typical example of a frequent phenomenon, a condition holding in the "vacuous" sense, a word of advice to the inexperienced reader might be in order. To prove that something is true about the empty set, prove that it cannot be false. How, for instance, could it be false that $\emptyset \subset A$? It could be false only if $\emptyset$ had an element that did not belong to $A$. Since $\emptyset$ has no elements at all, this is absurd. Conclusion: $\emptyset \subset A$ is not false, and therefore $\emptyset \subset A$ for every $A$.
• This paragraph shines with the rare virtue of sympathy with (and generosity of spirit toward) the uninitiated. +1. Oct 4, 2016 at 15:16
• I love this book. Oct 4, 2016 at 15:27
• With the text as opposed to with a picture of the text? Oct 4, 2016 at 19:41
• Okay, I don't see what search engines have to do with copyright violation issues, but in the meantime, I'd rather see the answer be accessibility compliant, and I guess it's your legal understanding that I'm responsible for any "issues" that might come up, so okay. Oct 4, 2016 at 19:48
• I like Halmos's second demonstration because it highlights a frequently used technique: For an implication to be false, there must be a witness to its falseness. That is, there is an element of some set that is a counterexample. Conversely, if it is impossible that there are counterexamples, then the implication is true. Oct 5, 2016 at 12:59
What confused me was that, the following expression was also a vacuous truth.
For every object of $$x$$, if $$x$$ belongs to the empty set, then $$x$$ doesn't belong to the set $$A$$.
As a complement (heh) to Brian Scott's (+1) answer, your argument shows that $$\varnothing \subset A^{c}$$, the complement of $$A$$. This statement is also (vacuously) true.
• So the empty set, being a subset of both the set $A$ and its complement, is a subset of the intersection of the set $A$ with its complement, which is the empty set. Oct 5, 2016 at 9:05
• I believe this is the key point. Oct 5, 2016 at 21:31
• @CarstenS: Be careful that in ZF set theory the complement of a set is never a set! Oct 7, 2016 at 14:17
Very subtle point:
"All x are not something" does not imply "Not all x are something".
The first may be vacuously true. The second one can not. If the x are vacuous then the second one has to be "vacuously false" as all x of nothing are any property so it is impossible for them not to be any property.
So "all elements of the empty set are not in $S$" does not imply "Not all elements of the empty set are in $S$" $\iff$ "It is not true that all elements of the empty set are in $S$" $\iff$ "There are some elements of the empty set that are not in $S$".
The first is vacuously true (and is equivalent to $\emptyset \subset S^c$ which is true) and the second set of equivalent statements are all equivalent to $\emptyset \not \subset S$ which is not true.
=====
The thing is what you say is absolutely correct for non empty sets.
More formally:
All elements $x$ in $S$ are not in $A$ $\implies$
$S \subset A^c$ $\implies$
$\color{red}{\text{There is an } x\in S \text{ where } x \not \in A}\implies$
It is not true that all $x \in S$ are also in $A$ $\implies$
$S \not \subset A$.
However the red line can only be concluded if $A$ is non-empty. If $A$ is empty the red line is simply false.
And without the red line there is simply no logic or means to jump from the line before to the line after:
All elements $x$ in $S$ are not in $A$ $\not\implies$
It is not true that all $x \in S$ are also in $A$.
That simply is not true for an empty $S$.
• You confused $A$ and $S$ at the end there. Oct 6, 2016 at 21:50
Every theory has axioms, which are some propositions held to be true without being proven from anything else, and are not provable from each other. Subsequent truths of the theory derived from the axioms are theorems.
The properly termed question is whether the empty set being a subset of every other set is axiom of set theory, or a theorem.
It depends on how "subset" is defined. If $A\subset B$ means that every element of $A$ is in $B$, it is not necessarily true that $\emptyset$ is a subset of anything, since it has no elements. In this case, $\emptyset \subset A$ can be added as an axiom. It doesn't conflict with anything, and simplifies all reasoning about subsets. Alternatively, if $A\subset B$ is defined as "$A$ has no elements that are not also in $B$", then we do not require the extra axiom for the $\emptyset$ case. If $A$ has no elements at all, it has no elements that are not in $B$.
Suppose that we use the first, positively termed definition of subset, and then adopt as an axiom not $\forall A:\emptyset \subset A$, but rather its negation: $\exists A:\emptyset \not\subset A$, or the outright proposition $\forall A:\emptyset \not\subset A$.
This is just going to cause problems. We can "do" set theory as before, but all the theorems will be uglified by having to avoid the special cases involving the empty set. In any derivation step in which we rely on a subset relation being true, or assert one, we will have to add the verbiage of an additional statement which asserts that the variable in question doesn't denote the empty set. This proposition then has to be carried in all the remaining derivations, unless something else makes it superfluous (some unrelated assurance from elsewhere that the set in question isn't empty).
Working with this clumsy subset definition that doesn't work with the empty set very well, someone is eventually going to have an epiphany and introduce a new subset-like relation which doesn't have these ugly problems: a new $A\ \mathbf{subset*}\ B$ binary relation which reduces exactly to $A\subset B$ when neither $A$ nor $B$ are $\emptyset$, and which, simply by definition, reduces to a truth whenever $A = \emptyset$, regardless of $B$. That person will then realize that all the existing work is simpler if this $\mathbf{subset*}$ operation is used in place of $\subset$.
At the end of the day it boils down to criteria like: is the system consistent (doesn't contradict itself), is it complete (does it capture the truths we want) and also is it convenient: are the rules configured so that we do not trip over unnecessary cases and superfluous logic.
This is a very mundane explanation (with finite sets). A subset is made of any combination of elements from the set. Suppose a set$S$ is made of three elements: $\{a,b,c\}$. Each element $a$, $b$ or $c$ can be, or not, in the combination. If all of them are not in the combination, they STILL form a combination of "absent elements", or $\emptyset$, a subset of $S$. They are the dual of the subset of "all elements", $\{a,b,c\}$.
In the same way that $\{a,b\}$ and $\{c\}$ are (complementary) subsets, $\emptyset$ and $\{a,b,c\}$ belong to the set of subsets.
What confused me was that, the following expression was also a vacuous truth.
For every object of $$x$$, if $$x$$ belongs to the empty set, then $$x$$ doesn't belong to the set $$A$$.
The contrapositive of the above conditional is:
"For every $$x$$, if $$x$$ belongs to set $$A$$, then $$x$$ doesn't belong to the empty set" which is easy to understand as the empty set has no elements at all.
NOTE- $$p\rightarrow q$$ has same meaning as $$\lnot q\rightarrow \lnot p$$ | 2022-05-24T22:17:49 | {
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https://mathhelpboards.com/threads/help-with-separable-equation.4880/#post-22144 | # Help with separable equation
#### Kris
##### New member
( 4*x+1 )^2 dy/dx = 27*y^3
I'm trying to separate this into a separable equation. Does it matter which way I do it? I.e taking all xs to the left or all ys to the left or does it not matter as long as x and y are on different sides?
#### MarkFL
Staff member
As long as you have:
$$\displaystyle f(y)\,dy=g(x)\,dx$$ or $$\displaystyle f(x)\,dx=g(y)\,dy$$ it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.
#### Kris
##### New member
As long as you have:
$$\displaystyle f(y)\,dy=g(x)\,dx$$ or $$\displaystyle f(x)\,dx=g(y)\,dy$$ it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.
Hi Mark, I have tried separating the variables yet I can't figure out how to move the bracketed term across. Could you advise how to do this and show what the final solution would be?
Also how do I thank you as I can't find the button
#### MarkFL
Staff member
I will help you separate the variables, and then guide you to get the solution...you will get more from the problem that way.
We are given:
$$\displaystyle (4x+1)^2\frac{dy}{dx}=27y^3$$
See what you get when you divide through by $$\displaystyle (4x+1)^2y^3$$ (bearing in mind that in doing so we are eliminating the trivial solution $y\equiv0$).
edit: You should see a Thanks link at the lower right of each post, except your own.
#### Kris
##### New member
Thanks I can get the answer from here just wasn't sure if you could divide through by the whole bracket
#### Kris
##### New member
dy/(27*y^3)=dx/((4*x+1)^2)
This is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?
At this point it is okay to integrate right?
edit: Do you know what the answer will be for the integration because my work sheet doesnt have it and id like to have it before i finish the question
#### MarkFL
Staff member
dy/(27*y^3)=dx/((4*x+1)^2)
This is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?
At this point it is okay to integrate right?
edit: Do you know what the answer will be for the integration because my work sheet doesn't have it and I'd like to have it before I finish the question.
Yes, that's fine, although I would probably choose to write:
$$\displaystyle y^{-3}\,dy=27(4x+1)^{-2}\,dx$$
I would use a $u$-substitution on the right side:
$$\displaystyle u=4x+1\,\therefore\,du=4\,dx$$ and we have (after multiplying through by 4):
$$\displaystyle 4\int y^{-3}\,dy=27\int u^{-2}\,du$$
Now, we are just a couple of steps from the solution (and don't forget to include the trivial solution we eliminated when separating variables when you state the final solution, if this trivial solution is not included in the general solution for a suitable choice of the constant of integration).
#### Kris
##### New member
Ive got a solution at y = -54 + c but i dont think this is right
I went with -1/2*y^2 * y = -1/u * u? as my integrations and then rearranged
Is this the correct answer or have I integrated something wrong which leads to my poor solution?
#### MarkFL
Staff member
Using the power rule for integration, and using the form in my post above, you should get:
$$\displaystyle 4\left(\frac{y^{-2}}{-2} \right)=27\left(\frac{u^{-1}}{-1} \right)+C$$
Multiply through by -1 and simplify a bit:
$$\displaystyle 2y^{-2}=27u^{-1}+C$$
Note: the sign of the parameter $C$ does not change as it can be any real number, negative or positive.
At this point, I would rewrite using positive exponents, and combine terms on the right:
$$\displaystyle \frac{2}{y^2}=\frac{Cu+27}{u}$$
Invert both sides:
$$\displaystyle \frac{y^2}{2}=\frac{u}{Cu+27}$$
$$\displaystyle y^2=\frac{2u}{Cu+27}$$
Back-substitute for $u$:
$$\displaystyle y^2=\frac{2(4x+1)}{C(4x+1)+27}$$
This is the general solution, and the only way we can get the trivial solution is for:
$$\displaystyle 4x+1=0$$
but we eliminated that possibility during the separation of variables as well.
#### Kris
##### New member
Thanks so much I see where i went wrong because i tried to integrate and invert before I multiplied out. It makes much more sense to multiply and flip then turn into a positive rather than trying to do it all at once | 2021-12-02T09:36:47 | {
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http://poliklinika-galaxy.com/food709f/retaliation1ad5/12018124b7576f9592e0a80d330 | ERROR: type should be string, got "https://www.thoughtco.com/associative-and-commutative-properties-difference-3126316 (accessed July 22, 2022). By definition, commutative property is applied on 2 numbers, but the result remains the same for 3 numbers as well. For that reason, it is important to understand the difference between the two. That is. In Mathematics, a commutative property states that if the position of integers are moved around or interchanged while performing addition or multiplication operations, then the answer remains the same. Your Mobile number and Email id will not be published. By the commutative property of multiplication, 3 6 = 6 3. nhmDTH%PLI%@hQdK( (1) and (2), as per the commutative property of multiplication, we get; Find which of the following is the commutative property of addition and multiplication. For example: 1+2 = 2+1 and 2 x 3 = 3 x 2. Math Glossary: Mathematics Terms and Definitions, A Kindergarten Lesson Plan for Teaching Addition and Subtraction, IEP Math Goals for Operations in the Primary Grades, Definition and Usage of Union in Mathematics, Parentheses, Braces, and Brackets in Math. Essentially, the order does not matter when adding or multiplying. Yes. As per commutative property of addition, 827 + 389 = 389 + 827. Similarly, we can rearrange the addends and write: Example 4: Ben bought 3 packets of 6 pens each. As we already discussed in the introduction, as per the commutative property or commutative law, when two numbers are added or multiplied together, then a change in their positions does not change the result. Neither of these properties are applicable to division. Take, for example, the arithmetic problem (6 3) 2 = 3 2 = 1; if we change the grouping of the parentheses, we have 6 (3 2) = 6 1 = 5, which changes the final result of the equation. Whereas associative property holds regardless of grouping of numbers. Some operations are non-commutative. Lets see. For instance: This equation is an example of the commutative property of addition of real numbers. Choose the set of numbers to make the statement true. In other words, the placement of parentheses does not matter when it comes to adding or multiplying. No matter how the values are grouped, the result of the equation will be 10: As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Example 2: Use 14 15 = 210, to find 15 14. \"The Associative and Commutative Properties.\" So, the total number of pens that Ben bought = 3 6, So, the total number of pens that Ben bought = 6 3. It states that if we swipe the positions of the integers, the result will remain the same. The correct answer is Both sides are equal to 33.. The commutative property concerns the order of certain mathematical operations. The left side equals 44 and the right side equals 33. By equation 1 and 2, as per commutative property of addition, we get; Q.3: Prove that A.B = B.A, if A = 4 and B = 3. What is the associative property of addition (or multiplication)? Checkout JEE MAINS 2022 Question Paper Analysis : explains that order of terms doesnt matter while performing, Commutative property is applicable only for addition and multiplication processes. The distributive property of Multiplication states that multiplying a sum by a number is the same as multiplying each addend by the value and adding the products then. Remember, with the commutative property, the order of the numbers does not matter when adding and multiplying. We can tell the difference between the associative and the commutative property by asking the question, Are we changing the order of the elements, or are we changing the grouping of the elements? If the elements are being reordered, then the commutative property applies.\n\n(2020, October 29). Did they buy an equal number of pens or not? Mia bought 6 packets of 3 pens each. When two numbers are multiplied together and if we interchange their positions, then the product of the two remains the same. The Associative property holds true for addition and multiplication. KB(|Q-SFt4E! Even if both have different numbers of bun packs with each having a different number of buns in them, they both bought an equal number of buns, because 3 4 = 4 3. According to the associative law, regardless of how the numbers are grouped, you can add or multiply them together, the answer will be the same. The word, Commutative, originated from the French word commute or commuter means to switch or move around, combined with the suffix -ative means tend to. The property holds for Addition and Multiplication, but not for subtraction and division. By eq. Which of the following represents the commutative property of addition? 7)\\). If $$x=2$$, $$y=5$$, and $$z=1$$, which of the following is true about this equation: $$2x+4y+9z=9z+4y+2x$$. The mathematical operations, subtraction and division are the two non-commutative operations. Example 4: Use the commutative property of addition to write the equation, 3 + 5 + 9 = 17, in a different sequence of the addends. Example 3: Use 827 + 389 = 1,216 to find 389 + 827. The other major properties of addition and multiplication are: Now, observe the other properties as well here: Associative Property of Addition and Multiplication. Since, 14 15 = 210, so, 15 14 also equals 210. Taylor, Courtney. by Mometrix Test Preparation | This Page Last Updated: June 2, 2022. The operation is commutative because the order of the elements does not affect the result of the operation. Which statement best illustrates the commutative property? Q.2: Prove that a+ b = b+a if a = 10 and b = 9. The associative property states that the grouping of factors in an operation can be changed without affecting the outcome of the equation. }8E| This can be shown by the equation (a + b) + c = a + (b + c). So, there can be two categories of operations that obeys commutative property: Although the official use of commutative property began at the end of the 18th century, it was known even in the ancient era. ThoughtCo. The commutative property states that values can be moved or swapped when adding or multiplying, and the outcome will not change. So, if we swap the position of numbers in subtraction or division statements, it changes the entire problem.\n\nWhich operations do not follow commutative property? Even if both have different numbers of apples and peaches, they have an equal number of fruits, because 2 + 6 = 6 + 2. Use the commutative property to find the missing value: $$45+44+43=43+44+$$_____. Which of the following represents the commutative property of multiplication? The associative property says you can regroup multiplied terms in any way. Which of the following expressions will follow the commutative property? \"The Associative and Commutative Properties.\" B.A., Mathematics, Physics, and Chemistry, Anderson University. Commutative property is applicable only for addition and multiplication processes. Do they have an equal number of marbles? =*jb 5;dtOu2T*~GL:E7$_Bd% N Therefore, the literal meaning of the word is tending to switch or move around. 77; by commutative property of multiplication, 36; by commutative property of multiplication. Unlike addition, in subtraction switching of orders of terms results in different answers. Commutative property cannot be applied to subtraction and division. The Associative and Commutative Properties. Simply put, the commutative property states that the factors in an equation can be rearranged freely without affecting the outcome of the equation. Example: 4 3 = 1 but 3 4 = -1which are two different integers. Use the commutative property to find the missing values: $$4+6+$$ ____$$=6+$$____ $$+8$$. When the associative property is used, elements are merely regrouped. These propertiesthe commutative and the associativeare very similar and can be easily mixed up. Solution: Options 1, 2 and 5 follow the commutative law. Beth has 6 packets of 78 marbles each. eC:C%L\"HX'JyS7yS| F: lj. The correct answer is $$6+5=5+6$$. The grouping of the elements, as indicated by the parentheses, does not affect the result of the equation. Required fields are marked *. Also, the division does not follow the commutative law. The correct answer is $$4(37)=(43)7$$. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. What is the distributive property of multiplication? Example 5: Lisa has 78 red and 6 blue marbles. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. Since Lisa has 78 red and 6 blue marbles. By the distributive property of multiplication over addition, we mean that multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. The associative property applies to multiplication but not division, so divided terms cannot be regrouped. We've updated our Privacy Policy, which will go in to effect on September 1, 2022. To solve more problems on properties of math, download BYJUS The Learning App from Google Play Store and watch interactive videos. (Except 2 + 2 and 2 2. (% This property states that when three or more numbers are added (or multiplied), the sum (or the product) is the same regardless of the grouping of the addends (or the multiplicands). Retrieved from https://www.thoughtco.com/associative-and-commutative-properties-difference-3126316. In mathematics, commutative property or commutative law explains that order of terms doesnt matter while performing arithmetic operations. For which all operations does the associative property hold true? This can be expressed through the equation a + (b + c) = (a + b) + c. No matter which pair of values in the equation is added first, the result will be the same. 3u(CXOD^$? We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best.\n\nThe commutative property, therefore, concerns itself with the ordering of operations, including the addition and multiplication of real numbers, integers, and rational numbers. The above examples clearly show that the commutative property holds true for addition and multiplication but not for subtraction and division. So, Lisa and Beth dont have an equal number of marbles. Example 1: Which of the following obeys commutative law? Commutative property holds regardless of order of numbers while addition or multiplication. If we pay careful attention to the equation, though, we see that only the order of the elements has been changed, not the grouping. As per commutative property of multiplication, 15 14 = 14 15. Rewrite the expression $$45+6+19$$ using the commutative property. The left side equals 33 and the right side equals 44. The expression $$45+6+19$$ is equivalent to $$6+45+19$$, because changing the order that we add does not affect the result. What Is the Difference of Two Sets in Set Theory? For the associative property to apply, we would have to rearrange the grouping of the elements as well: By clicking Accept All Cookies, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts.\n\nWhen two numbers are added together, then if we swap the positions of numbers, the sum of the two remains the same. Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of \"An Introduction to Abstract Algebra.\". However, note that the presence of parentheses alone does not necessarily mean that the associative property applies. The correct answer is $$4+6+\\mathbf8=6+\\mathbf4+8$$. a (b + c) = (a b) + (a c) where a, b, and c are whole numbers. Example 1: Fill in the missing numbers using the commutative property. Taylor, Courtney. However, unlike the commutative property, the associative property can also apply to matrix multiplication and function composition. ThoughtCo, Oct. 29, 2020, thoughtco.com/associative-and-commutative-properties-difference-3126316. Like commutative property equations, associative property equations cannot contain the subtraction of real numbers. Even though the terms are listed in a different order, the left and right side of the equation are both equal to 33. (a + b) + c = a + (b + c)(a b) c = a (b c) where a, b, and c are whole numbers. For example, the numbers 2, 3, and 5 can be added together in any order without affecting the final result: The numbers can likewise be multiplied in any order without affecting the final result: Subtraction and division, however, are not operations that can be commutative because the order of operations is important. Commutative Property Definition with Examples. Rearranging multiplied terms is an example of the commutative property. The three numbers above cannot, for example, be subtracted in any order without affecting the final value: As a result, the commutative property can be expressed through the equations a + b = b + a and a x b = b x a. Learn More All content on this website is Copyright 2022, $$3(1)^{2}+5(1)(2)+(3)=5(1)(2)+3+3(1)^{2}$$. The Rules of Using Positive and Negative Integers, Facts About the Element Ruthenium (or Ru), Lesson Plan: Adding and Multiplying Decimals, Use BEDMAS to Remember the Order of Operations. That is. The correct answer is $$6+45+19$$. By non-commutative, we mean the switching of the order will give different results. This is because we can apply this property on two numbers out of 3 in various combinations. The associative property, on the other hand, concerns the grouping of elements in an operation. So, the total number of marbles with Lisa = 78 + 6, So, the total number of marbles with Beth = 6 78. Your Mobile number and Email id will not be published. As per the commutative property of multiplication, when we multiply two integers, the answer we get after multiplication will remain the same, even if the position of the integers are interchanged. Taylor, Courtney. The commutative property allows the addition or multiplication of numbers in any order.\n\nAccording to the Distributive Property, if a, b, c are real numbers, then: There are certain other properties such as Identity property, closure property which are introduced for integers. The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. According to the commutative property of addition, when we add two integers, the answer will remain unchanged even if the position of the numbers are changed. Clearly, adding and multiplying two numbers gives different results.\n\nThat is. For a binary operationone that involves only two elementsthis can be shown by the equation a + b = b + a. Can you apply the commutative property of addition/multiplication to 3 numbers? If the elements are only being regrouped, then the associative property applies. The correct answer is 45. Since, 827 + 389 = 1,216, so, 389 + 827 also equals 1,216. So, mathematically commutative property for addition and multiplication looks like this: a + b = b + a; where a and b are any 2 whole numbers, a b = b a; where a and b are any 2 nonzero whole numbers. This is one of the major properties of integers. So, both Ben and Mia bought an equal number of pens. There are several mathematical properties that are used in statistics and probability; two of these, the commutative and associative properties, are generally associated with the basic arithmetic of integers, rationals, and real numbers, though they also show up in more advanced mathematics.\n\n[O J\"FUQ&pwODUfYQ$z(['z@PY3\"BjSN<8$N(whCZOAfDda( GhA!D}0.Fdap u JXEm]^m> wO]l\"gTkJHDMOE\\?II!>bH4VEJ%+OTEeu\\!2% Note that when the commutative property is used, elements in an equation are rearranged. NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions Class 11 Business Studies, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 8 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions For Class 6 Social Science, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, Line Segment : Construct A Copy Of Line Segment, Important Questions Class 9 Maths Chapter 4 Linear Equations Two Variables, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths. For example, take the equation 2 + 3 + 5. =i*s{/_WT8yp4x1lDI\n\nNo matter the order of the values in these equations, the results will always be the same. apxZ=vE This is one of the major, Commutative property is only applicable for two arithmetic operations: Addition and Multiplication, Changing the order of operands, does not change the result, Commutative property of addition: A + B = B + A, Commutative property of multiplication: A.B = B.A. Q8 E T\"4', BgFC01 DCKTEsIyaR`@!" | 2022-11-26T19:14:15 | {
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http://mathhelpforum.com/number-theory/129681-question-about-squares.html | I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square
503^2 = 253009
and
50*506=25300
then tacking the 3^2 on the end gets 253009.
Tricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.
Any help would be appreciated.
2. Originally Posted by Enkie
I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square
503^2 = 253009
and
50*506=25300
then tacking the 3^2 on the end gets 253009.
Tricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.
Any help would be appreciated.
Let $n\in\mathbb{N}$ then $n=a_0+10a_1+\cdots+10^n a_n$ where $a_k\in\{0,\cdots,9\}$. You are claiming that $n^2=\left(10 a_1+\cdots +10^n a_n\right)\cdot\left(2a_0+10 a_1+\cdots 10^n a_n\right)+a_0^2$, or equivalently
$\left(a_0+10 a_1+\cdots+10^n a_n\right)^2-\left(10a_1+\cdots+10^n a_n\right)\cdot\left(2a_0+10a_1+\cdots+10^na_n\rig ht)-a_0^2$ $=\left(\left(a_0+10a_1+\cdots 10^n a_n\right)^2-a_0^2\right)-\left(10 a_1+\cdots+10^na_n\right)\cdot\left(2a_0+10a_1+\cd ots+10^n a_n\right)$ $=0$.
But,
$\left(a_0+10a_1+\cdots+10^na_n\right)^2-a_0^2=\left(a_0+10a_1+\cdots+10^na_n-a_0\right)$ $\cdot\left(a_0+10a_1+\cdots 10^na_n+a_0\right)$ $=\left(10a_1+\cdots+10^na_n\right)\cdot\left(2a_0+ 10a_1+\cdots+10^n a_n\right)$.
From where the conclusion immediately follows.
3. Hello, Enkie!
Another approach . . .
I was noticing today that you can calculate the square of a number by:
(1) taking the number you want to square and removing the ones digit,
(2) and multiplying by the number you want to square plus the ones digit,
(3) then square the ones digit and add it on to get the square.
Any integer $N> 9$ is of the form: . $N \:=\:10T + U$
. . where $U$ is the units digit and $T$ is the "rest of the number."
We note that: . $N^2 \:=\:(10T + U)^2 \:=\:100T^2 + 20TU + U^2$
(1) Remove the units digit and append a zero.
. . . We have: . $10T$
(2) Multiply by $N + U\!:\;\;10T(N + U) \;=\;10T(10T + U + U) \;=\;100T^2 + 20TU$
(3) Add $U^2\!:\;\;100T^2 + 20TU + U^2\quad \hdots$ and the result is $N^2$
4. Let us try with $1485$. We get :
$148 \times (1485 + 5) = 220520$
Now add on $5^2 = 25 \Rightarrow 22052025$
But $1485^2 = 2205225$.
I understand what you mean but you need to mention that when adding the square of the last digit, you actually overlap the previous result when the last digit is greater than three. | 2013-12-18T09:04:16 | {
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https://stats.stackexchange.com/questions/213456/how-to-calculate-the-total-probability-inside-a-slice-of-a-bivariate-normal-dist | # How to calculate the total probability inside a slice of a bivariate normal distribution in R?
I have a bivariate normal distribution composed of the univariate normal distributions $X_1$ and $X_2$ with $\rho \approx 0.3$.
$$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right)$$
Is there a simple way to calculate in R the cumulative probability of $X_1$ being less than a value $z$ given a particular slice of $X_2$ (between two values $a,b$) given we know all the parameters $\mu_1, \mu_2, \sigma_1, \sigma_2, \rho$?
$P(X_1 < z | a < X_2 < b)$
Can the distribution function I am looking for match (or be approximated by) the distribution function of a univariate normal distribution (to use qnorm/pnorm)? Ideally this would be the case so I can perform the calculation with less dependencies on libraries (e.g. on a MySQL server).
This is the bivariate distribution I am using:
means <- c(79.55920, 52.29355)
variances <- c(268.8986, 770.0212)
rho <- 0.2821711
covariancePartOfMatrix <- sqrt(variances[1]) * sqrt(variances[2]) * rho
sigmaMatrix <- matrix(c(variances[1],covariancePartOfMatrix,covariancePartOfMatrix,variances[2]), byrow=T, ncol=2)
n <- 10000
dat <- MASS::mvrnorm(n=n, mu=means, Sigma=sigmaMatrix)
plot(dat)
This is my numerical attempt to get the correct result. However it uses generated data from the bivariate distribution and I'm not convinced it will give the correct result.
a <- 79.5
b <- 80.5
z <- 50
sliceOfDat <- subset(data.frame(dat), X1 > a, X1 < b)
estimatedMean <- mean(sliceOfDat[,c(2)])
estimatedDev <- sd(sliceOfDat[,c(2)])
estimatedPercentile <- pnorm(z, estimatedMean, estimatedDev)
### Edit - R implementation of solution based on whuber's answer
Here is an implementation of the accepted solution using integrate, compared against my original idea based on sampling. The accepted solution provides the expected output 0.5, whereas my original idea deviated by a significant amount (0.41). Update - See wheber's edit for a better implementation.
# Bivariate distribution parameters
means <- c(79.55920, 52.29355)
variances <- c(268.8986, 770.0212)
rho <- 0.2821711
# Generate sample data for bivariate distribution
n <- 10000
covariancePartOfMatrix <- sqrt(variances[1]) * sqrt(variances[2]) * rho
sigmaMatrix <- matrix(c(variances[1],covariancePartOfMatrix,covariancePartOfMatrix,variances[2]), byrow=T, ncol=2)
dat <- MASS::mvrnorm(n=n, mu=means, Sigma=sigmaMatrix)
# Input parameters to test the estimation
w = 79.55920
a <- w - 0.5
b <- w + 0.5
z <- 52.29355
# Univariate approximation using randomness
sliceOfDat <- subset(data.frame(dat), X1 > a, X1 < b)
estimatedMean <- mean(sliceOfDat[,c(2)])
estimatedDev <- sd(sliceOfDat[,c(2)])
estimatedPercentile <- pnorm(z, estimatedMean, estimatedDev)
# OUTPUT: 0.411
# Numerical approximation from exact solution
adaptedZ <- (z - means[2]) / sqrt(variances[2])
adaptedB <- (b - means[1]) / sqrt(variances[1])
adaptedA <- (a - means[1]) / sqrt(variances[1])
integrand <- function(x) pnorm((adaptedZ - rho * x) / sqrt(1 - rho * rho)) * dnorm(x)
# 0.0121, abs.error 1.348036e-16, "OK"
exactPercentile = exactSolutionCoeff * exactSolutionInteg$value # OUTPUT: 0.500 ## 1 Answer Yes, a Normal approximation works--but not in all cases. We need to do some analysis to identify when the approximation is a good one. ### Exact solution Re-express$(X_1,X_2)$in standardized units so that they have zero means and unit variances. Letting$\Phibe the standard Normal distribution function (its CDF), it is well known from the theory of ordinary least squares regression that $$\Pr(X_1 \le z\,|\, X_2 = x) = \Phi\left(\frac{z - \rho x}{\sqrt{1-\rho^2}}\right).$$ The desired probability then can be obtained by integrating: \eqalign{\Pr(X_1 \le z\,|\, a \lt X_2 \le b) &= \frac{1}{\Phi(b)-\Phi(a)}\int_a^b \Pr(X_1\le z\,|\, X_2=x) \phi(x)\,dx \\&= \frac{1}{\Phi(b)-\Phi(a)}\int_a^b \Phi\left(\frac{z - \rho x}{\sqrt{1-\rho^2}}\right) \phi(x)\,dx.} This appears to require numerical integration (although the result for(a,b)=\mathbb{R}$is obtainable in closed form: see How can I calculate$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$). ### Approximating the distribution This expression can be differentiated under the integral sign (with respect to$z$) to obtain the PDF, $$f(z\,|\, a \lt X_2 \le b) = \phi(z)\ \frac{\Phi\left(\frac{b-\rho z}{\sqrt{1-\rho^2}}\right) - \Phi\left(\frac{a-\rho z}{\sqrt{1-\rho^2}}\right)}{\Phi(b) - \Phi(a)}.$$ This exhibits the PDF as product of the standard Normal PDF$\phi$and a "correction". When$b-a$is small compared to$\sqrt{1-\rho^2}$(specifically, when$(b-a)^2 \ll 1-\rho^2$), we might approximate the difference in the numerator with the first derivative: $$\Phi\left(\frac{b-\rho z}{\sqrt{1-\rho^2}}\right) - \Phi\left(\frac{a-\rho z}{\sqrt{1-\rho^2}}\right)\approx \phi\left(\frac{(a+b)/2-\rho z}{\sqrt{1-\rho^2}}\right)\frac{b-a}{\sqrt{1-\rho^2}}.$$ The error in this approximation is uniformly bounded (across all values of$z$) because the second derivative of$\Phi$is bounded. With this approximation, and completing the square, we obtain $$f(z\,|\, a \lt X_2 \le b) \approx \phi\left(z; \rho(a+b)/2, \sqrt{1-\rho^2}\right) \frac{(b-a)\exp\left(-(a+b)^2/8\right)}{(\Phi(b)-\Phi(a))\sqrt{2\pi}}.$$ ($\phi(*; \mu, \sigma)$denotes the PDF of a Normal distribution of mean$\mu$and standard deviation$\sigma$.) Moreover, the right hand factor (which does not depend on$z$) must be very close to$1$, because the$\phi$term integrates to unity, whence $$f(z\,|\, a \lt X_2 \le b) \approx \phi\left(z; \rho(a+b)/2, \sqrt{1-\rho^2}\right).$$ All this makes very good sense: the conditional distribution of$X_1$is approximated by its conditional distribution at the midpoint of the interval,$(a+b)/2$, where it has mean$\rho(a+b)/2$and standard deviation$\sqrt{1-\rho^2}$. The error is proportional to the width of the interval$b-a$and to an expression dominated by$\exp(-(a+b)^2/8)$, which becomes important only when both$a$and$b$are out in the same tail. Therefore, this Normal approximation works for narrow slices not too far into the tails of the bivariate distribution. Moreover, the difference between $$\frac{(b-a)\exp\left(-(a+b)^2/8\right)}{(\Phi(b)-\Phi(a))\sqrt{2\pi}}$$ and$1$serves as an excellent check of the quality of the approximation. ### Edit To check these conclusions, I simulated data in R for various values of$b$and$\rho$($a=-3$in all cases), drew their empirical density, and superimposed on that the theoretical (blue) and approximate (red) densities for comparison. (You cannot see the density plots because the theoretical plots fit over them almost perfectly.) As$|\rho|$gets close to$1$, the approximation grows poorer: this deserves further study. Clear the approximation is excellent for sufficiently small values of$b-a$. # # Numerical integration, to give a correct value. # f <- function(z, a, b, rho, value.only=FALSE, ...) { g <- function(x) pnorm((z - rho*x)/sqrt(1-rho^2)) * dnorm(x) / (pnorm(b) - pnorm(a)) u <- integrate(g, a, b, ...) if (value.only) return(u$value) else return(u)
}
#
# Set up the problem.
#
a <- -3 # Left endpoint
n <- 1e5 # Simulation size
par(mfrow=c(2,3))
for (rho in c(1/4, -2/3)) {
for (b in c(-2.5, -2, -1.5)) {
z <- seq((a-3)*rho, (b+3)*rho, length.out=101)
#
# Check the approximation (v needs to be small).
#
v <- (b-a) * exp(-(a+b)^2/8) / (pnorm(b) - pnorm(a)) / sqrt(2*pi) - 1
#
# Simulate some values of (x1, x2).
#
x.2 <- qnorm(runif(n, pnorm(a), pnorm(b))) # Normal between a and b
x.1 <- rho*x.2 + rnorm(n, sd=sqrt(1-rho^2))
#
# Compare the simulated distribution to the theoretical and approximate
# densities.
#
x.hat <- density(x.1)
plot(x.hat, type="l", lwd=2,
main="Simulated, True, and Approximate",
sub=paste0("a=", round(a,2), ", b=", round(b, 2), ", rho=", round(rho, 2),
"; v=", round(v,3)),
xlab="X.1", ylab="Density")
# Theoretical
curve(dnorm(x) * (pnorm((b-rho*x)/sqrt(1-rho^2)) - pnorm((a-rho*x)/sqrt(1-rho^2))) /
(pnorm(b) - pnorm(a)), lwd=2, col="Blue", add=TRUE)
# Approximate
curve(dnorm(x, rho*(a+b)/2, sqrt(1-rho^2)), col="Red", lwd=2, add=TRUE)
}
}
• Thank you for the detailed answer. I had found that the normal approximation stopped producing the figures I expected when z was high or low. Your examination of the numerical analysis corresponds exactly to what I noticed varying a, b, and z, and makes me confident that I can get a better result by using the exact solution that you provided. – Michael Clark May 20 '16 at 11:54
• Would it be possible to add a little more detail on how to calculate the exact solution without numerical integration? An example line in R would be perfect. The part I am most confused about is how to convert the limits and sign of x in the integral ∫baΦ(z−ρx1−ρ2−−−−−√)ϕ(x)dx into the form ∫∞−∞Φ(w−ab)ϕ(w)dw∫−∞∞Φ(w−ab)ϕ(w)dw) so that I can use the result proved in the question that you linked. – Michael Clark May 20 '16 at 11:58
• Unfortunately there is no such conversion. Numerical integration really is needed. You could also pursue the development of the approximation further by using higher-order approximations to the $\Phi$ term in the integrand. Intuitively, the midpoint $(a+b)/2$ ought to be replaced by a point closer to the smaller of $a$ and $b$ (in size), because that's where most of the $X_2$ probability is. – whuber May 20 '16 at 12:26
• I've added a R example based on the exact solution -- using the mean & variance, the approximate solution was 0.41 and the exact 0.5. I am surprised how far the approximate solution is out in that case. Thanks again! – Michael Clark May 20 '16 at 20:17
• I saw that code, but it's not clear how it works or whether it's correct. At the very least you should compute the check value (at the end of my original post), look at $|b-a|/\sqrt{1-\rho^2}$ (in standardized units), and inspect the output of integrate to make sure it's not running into accuracy problems. – whuber May 20 '16 at 20:20 | 2018-12-10T22:19:06 | {
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https://math.stackexchange.com/questions/1981391/how-would-i-simplify-this-summation-sum-i-1n-i-1-sum-j-1n-j | # How would I simplify this summation: $\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j$
$$\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j$$
I cant really get my head around on how to simplify these sigma notations, any help would be appreciated.
Thanks
• Just try it for modest $n$! Suppose $n=1$. Then try $n=2$. I expect a pattern will emerge.
– lulu
Oct 23 '16 at 14:05
• You could replace $j$ by $i$ and simplify. Oct 23 '16 at 14:08
You can “reorganize” the first summation: $$\sum_{i=1}^n(i+1)=\sum_{i=1}^n i+\sum_{i=1}^n1$$ Since $$\sum_{i=1}^n i = \sum_{j=1}^n j$$ you remain with $$\sum_{i=1}^n 1 = n$$
Another variation which might be helpful.
\begin{align*} \sum_{i=1}^n (i + 1) - \sum_{j=1}^n j&=\sum_{i=1}^n (i + 1) - \sum_{i=1}^n i\\ &=\sum_{i=1}^n (i + 1-i)\\ &=\sum_{i=1}^n 1\\ &=n \end{align*}
You have
$$\sum_{i=1}^n (i+1)=\sum_{i=2}^{n+1} i$$
so
$$\sum_{i=1}^n (i+1)-\sum_{j=1}^n j=\sum_{i=2}^{n+1} i-\sum_{j=1}^n j=n+1+\sum_{i=1}^n (i-i)+1=n+1-1=n.$$
(Too long for a comment)
As @lulu said you can investigate the behavior of the sum by hand but you can use a bit of algebra before to simplify something. By example observe that $$\sum_{k=1}^n (k+1)=\left(\sum_{k=1}^n k\right)+\left(\sum_{k=1}^n 1\right)=\left(\sum_{k=1}^n k\right) + n$$ and $$\sum_{k=1}^n k=\sum_{j=1}^n j=1+2+3+\cdots+n$$
Observe too that a summation can be written as
$$\sum_{k=1}^n k=\sum_{1\le k\le n}k$$
Then you can manipulate easily the inequality $1\le k\le n$ if you need to change the variable $k$ by, for example, $h=k+2$
$$1\le k\le n\iff 1+2\le k+2\le n+2\iff 3\le h\le n+2$$
Then you can rewrite
$$\sum_{k=1}^n k=\sum_{1\le k\le n}k=\sum_{3\le h\le n+2}(h-2)=\sum_{h=3}^{n+2}(h-2)$$ | 2021-12-08T19:49:54 | {
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http://mathhelpboards.com/other-topics-22/calculate-velocity-brick-using-assumption-amp-draw-displacement-time-graph-19950.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e | # Thread: Calculate the velocity of the brick using an assumption & draw the displacement-time graph
1. Calculate the velocity of the brick at the bottom of the inclined plane.
& Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)
Workings:
I am unable to think on the first two but I think the displacement and time graph should be something like ,
Many Thanks
2. To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.
So, if we equate the initial potential energy to the final kinetic energy, we have:
$\displaystyle mgh=\frac{1}{2}mv^2$
What do you get upon solving for $v$?
Originally Posted by MarkFL
To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.
So, if we equate the initial potential energy to the final kinetic energy, we have:
$\displaystyle mgh=\frac{1}{2}mv^2$
What do you get upon solving for $v$?
Thank you very much MarkFL
$\displaystyle mgh=\frac{1}{2}mv^2$
$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}mv^2$
$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$
$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$
$\displaystyle 100 J= v^2$
$\displaystyle 10 J= v$
Correct ?
4. I would solve the equation first, and then plug in the numbers:
$\displaystyle mgh=\frac{1}{2}mv^2$
$\displaystyle v=\sqrt{2gh}$
Okay, at this point we can plug in:
$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$
$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
Originally Posted by MarkFL
I would solve the equation first, and then plug in the numbers:
$\displaystyle mgh=\frac{1}{2}mv^2$
$\displaystyle v=\sqrt{2gh}$
Okay, at this point we can plug in:
$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$
$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
Thank you very much
Originally Posted by mathlearn
Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)
Is the above drawn graph correct?
6. For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?
7. Originally Posted by mathlearn
Is the above drawn graph correct?
Hey mathlearn!
Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):
On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.
However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?
8. Originally Posted by mathlearn
$\displaystyle 100 J= v^2$
$\displaystyle 10 J= v$
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)
-Dan
Originally Posted by MarkFL
For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?
This kind of a curve with constant acceleration and with deceleration.
Originally Posted by I like Serena
Hey mathlearn!
Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):
On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.
However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?
The displacement time graph would look like the above graph
Originally Posted by topsquark
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)
-Dan
Thanks for the catch
10. Originally Posted by mathlearn
This kind of a curve with constant acceleration and with deceleration...
Why do you have the brick returning to zero displacement? | 2017-12-12T14:28:00 | {
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http://accessdtv.com/error-bound/taylor-series-maximum-error.html | Home > Error Bound > Taylor Series Maximum Error
# Taylor Series Maximum Error
## Contents
Mathispower4u 48,779 views 9:00 Error or Remainder of a Taylor Polynomial Approximation - Duration: 11:27. Your cache administrator is webmaster. So for example, if someone were to ask you, or if you wanted to visualize. Alternating series error bound For a decreasing, alternating series, it is easy to get a bound on the error : In other words, the error is bounded by the next term http://accessdtv.com/error-bound/taylor-series-approximation-maximum-error.html
And this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be the same thing as f of a. Another use is for approximating values for definite integrals, especially when the exact antiderivative of the function cannot be found. But how many terms are enough? To see why the alternating bound holds, note that each successive term in the series overshoots the true value of the series. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation
## Taylor Polynomial Error Bound
So it's really just going to be, I'll do it in the same colors, it's going to be f of x minus P of x. It is going to be equal to zero. Suppose you needed to find .
Especially as we go further and further from where we are centered. >From where are approximation is centered. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an Nth degree polynomial centered at a. You can get a different bound with a different interval. What Is Error Bound I'll write two factorial.
Khan Academy 54,407 views 9:18 Loading more suggestions... Lagrange Error Bound Formula fall-2010-math-2300-005 lectures © 2011 Jason B. One way to get an approximation is to add up some number of terms and then stop.
If x is sufficiently small, this gives a decent error bound.
solution Practice B03 Solution video by PatrickJMT Close Practice B03 like? 6 Practice B04 Determine an upper bound on the error for a 4th degree Maclaurin polynomial of $$f(x)=\cos(x)$$ at $$\cos(0.1)$$. Taylor Polynomial Approximation Calculator What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b. Thus, we have What is the worst case scenario? If one adds up the first terms, then by the integral bound, the error satisfies Setting gives that , so .
## Lagrange Error Bound Formula
And we see that right over here. http://calculus.seas.upenn.edu/?n=Main.ApproximationAndError Therefore, one can think of the Taylor remainder theorem as a generalization of the Mean value theorem. Taylor Polynomial Error Bound Let's embark on a journey to find a bound for the error of a Taylor polynomial approximation. Lagrange Error Bound Calculator Similarly, you can find values of trigonometric functions.
with an error of at most .139*10^-8, or good to seven decimal places. this contact form And it's going to look like this. Close Yeah, keep it Undo Close This video is unavailable. Lagrange's formula for this remainder term is $$\displaystyle{ R_n(x) = \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!} }$$ This looks very similar to the equation for the Taylor series terms . . . Lagrange Error Bound Problems
That is the motivation for this module. However, only you can decide what will actually help you learn. Instead, use Taylor polynomials to find a numerical approximation. http://accessdtv.com/error-bound/taylor-maximum-error.html Autoplay When autoplay is enabled, a suggested video will automatically play next.
Please try the request again. Lagrange Error Bound Khan Academy If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. And what I wanna do is I wanna approximate f of x with a Taylor polynomial centered around x is equal to a.
## Return to the Power Series starting page Representing functions as power series A list of common Maclaurin series Taylor Series Copyright © 1996 Department of Mathematics, Oregon State University If you
Solving for gives for some if and if , which is precisely the statement of the Mean value theorem. solution Practice B02 Solution video by PatrickJMT Close Practice B02 like? 8 Practice B03 Use the 2nd order Maclaurin polynomial of $$e^x$$ to estimate $$e^{0.3}$$ and find an upper bound on The following theorem tells us how to bound this error. Alternating Series Error Bound patrickJMT 128,850 views 10:48 Calculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials - Duration: 1:34:10.
Links and banners on this page are affiliate links. You may want to simply skip to the examples. So it might look something like this. Check This Out A Taylor polynomial takes more into consideration.
Professor Leonard 99,296 views 3:01:45 Taylor Polynomials - Duration: 18:06. Krista King 59,295 views 8:23 Lec 38 | MIT 18.01 Single Variable Calculus, Fall 2007 - Duration: 47:31. Now, if we're looking for the worst possible value that this error can be on the given interval (this is usually what we're interested in finding) then we find the maximum So what I wanna do is define a remainder function.
The N plus oneth derivative of our error function or our remainder function, we could call it, is equal to the N plus oneth derivative of our function. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... This is going to be equal to zero. The square root of e sin(0.1) The integral, from 0 to 1/2, of exp(x^2) dx We cannot find the value of exp(x) directly, except for a very few values of x.
So, what is the value of $$z$$? $$z$$ takes on a value between $$a$$ and $$x$$, but, and here's the key, we don't know exactly what that value is. And that's the whole point of where I'm going with this video and probably the next video, is we're gonna try to bound it so we know how good of an So this is all review, I have this polynomial that's approximating this function. video by Dr Chris Tisdell Search 17Calculus Loading Practice Problems Instructions: For the questions related to finding an upper bound on the error, there are many (in fact, infinite) correct answers.
The following example should help to make this idea clear, using the sixth-degree Taylor polynomial for cos x: Suppose that you use this polynomial to approximate cos 1: How accurate is View Edit History Print Single Variable Multi Variable Main Approximation And Error < Taylor series redux | Home Page | Calculus > Given a series that is known to converge but Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do In general, if you take an N plus oneth derivative of an Nth degree polynomial, and you could prove it for yourself, you could even prove it generally but I think
Uploaded on Nov 11, 2011In this video we use Taylor's inequality to approximate the error in a 3rd degree taylor approximation. There is a slightly different form which gives a bound on the error: Taylor error bound where is the maximum value of over all between 0 and , inclusive. So, the first place where your original function and the Taylor polynomial differ is in the st derivative. But you'll see this often, this is E for error. | 2018-02-21T15:28:39 | {
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https://www.ubezwlasnowolnienie.pl/0dnlt82b/5c2a1c-symmetric-and-antisymmetric-relation | (iv) Reflexive and transitive but not symmetric. The relations we are interested in here are binary relations on a set. #mathematicaATDRelation and function is an important topic of mathematics. Also, compare with symmetric and antisymmetric relation here. x is married to the same person as y iff (exists z) such that x is married to z and y is married to z. Learn about the world's oldest calculator, Abacus. If a relation is reflexive, irreflexive, symmetric, antisymmetric, asymmetric, transitive, total, trichotomous, a partial order, total order, strict weak order, total preorder (weak order), or an equivalence relation, its restrictions are too. To put it simply, you can consider an antisymmetric relation of a set as a one with no ordered pair and its reverse in the relation. You can find out relations in real life like mother-daughter, husband-wife, etc. Multiplication problems are more complicated than addition and subtraction but can be easily... Abacus: A brief history from Babylon to Japan. The relations we are interested in here are binary relations on a set. In other words, we can say symmetric property is something where one side is a mirror image or reflection of the other. Here we are going to learn some of those properties binary relations may have. For example, on the set of integers, the congruence relation aRb iff a - b = 0(mod 5) is an equivalence relation. ; Restrictions and converses of asymmetric relations are also asymmetric. Complete Guide: How to work with Negative Numbers in Abacus? So total number of symmetric relation will be 2 n(n+1)/2. "Is married to" is not. (g)Are the following propositions true or false? Since (1,2) is in B, then for it to be symmetric we also need element (2,1). for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. (f) Let $$A = \{1, 2, 3\}$$. 6.3. A matrix for the relation R on a set A will be a square matrix. A relation R is defined on the set Z (set of all integers) by “aRb if and only if 2a + 3b is divisible by 5”, for all a, b ∈ Z. Learn about operations on fractions. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. This... John Napier | The originator of Logarithms. Which is (i) Symmetric but neither reflexive nor transitive. “Is equal to” is a symmetric relation, such as 3 = 2+1 and 1+2=3. Referring to the above example No. In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. See also Let ab ∈ R ⇒ (a – b) ∈ Z, i.e. In other words, we can say symmetric property is something where one side is a mirror image or reflection of the other. In this article, we have focused on Symmetric and Antisymmetric Relations. This list of fathers and sons and how they are related on the guest list is actually mathematical! (iii) Reflexive and symmetric but not transitive. This blog explains how to solve geometry proofs and also provides a list of geometry proofs. Yes. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. For a relation R, an ordered pair (x, y) can get found where x and y are whole numbers or integers, and x is divisible by y. Antisymmetric relation is a concept based on symmetric and asymmetric relation in discrete math. Hence, as per it, whenever (x,y) is in relation R, then (y, x) is not. I'll wait a bit for comments before i proceed. Thus, (a, b) ∈ R ⇒ (b, a) ∈ R, Therefore, R is symmetric. irreflexive relation symmetric relation antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). 6. Antisymmetry is concerned only with the relations between distinct (i.e. The abacus is usually constructed of varied sorts of hardwoods and comes in varying sizes. symmetric, reflexive, and antisymmetric. Figure out whether the given relation is an antisymmetric relation or not. Discrete Mathematics Questions and Answers – Relations. $<$ is antisymmetric and not reflexive, ... $\begingroup$ Also, I may have been misleading by choosing pairs of relations, one symmetric, one antisymmetric - there's a middle ground of relations that are neither! Fresheneesz 03:01, 13 December 2005 (UTC) I still have the same objections noted above. So total number of symmetric relation will be 2 n(n+1)/2. The term data means Facts or figures of something. (2,1) is not in B, so B is not symmetric. In this second part of remembering famous female mathematicians, we glance at the achievements of... Countable sets are those sets that have their cardinality the same as that of a subset of Natural... What are Frequency Tables and Frequency Graphs? i.e. The... A quadrilateral is a polygon with four edges (sides) and four vertices (corners). In that, there is no pair of distinct elements of A, each of which gets related by R to the other. Graphical representation refers to the use of charts and graphs to visually display, analyze,... Access Personalised Math learning through interactive worksheets, gamified concepts and grade-wise courses, is school math enough extra classes needed for math. 2 as the (a, a), (b, b), and (c, c) are diagonal and reflexive pairs in the above product matrix, these are symmetric to itself. First step is to find 2 members in the relation such that $(a,b) \in R$ and $(b,a) \in R$. Required fields are marked *. That is to say, the following argument is valid. A relation R on a set A is symmetric iff aRb implies that bRa, for every a,b ε A. A relation R in a set A is said to be in a symmetric relation only if every value of $$a,b ∈ A, (a, b) ∈ R$$ then it should be $$(b, a) ∈ R.$$, Given a relation R on a set A we say that R is antisymmetric if and only if for all $$(a, b) ∈ R$$ where a ≠ b we must have $$(b, a) ∉ R.$$. Complete Guide: How to multiply two numbers using Abacus? $$(1,3) \in R \text{ and } (3,1) \in R \text{ and } 1 \ne 3$$ therefore the relation is not anti-symmetric. A relation R is defined on the set Z by “a R b if a – b is divisible by 7” for a, b ∈ Z. This blog tells us about the life... What do you mean by a Reflexive Relation? Let a, b ∈ Z, and a R b hold. (a – b) is an integer. Ada Lovelace has been called as "The first computer programmer". Rene Descartes was a great French Mathematician and philosopher during the 17th century. Examine if R is a symmetric relation on Z. both can happen. They... Geometry Study Guide: Learning Geometry the right way! Justify all conclusions. Their structure is such that we can divide them into equal and identical parts when we run a line through them Hence it is a symmetric relation. Show that R is a symmetric relation. If we let F be the set of all f… For example, if a relation is transitive and irreflexive, 1 it must also be asymmetric. Hence it is also in a Symmetric relation. Basics of Antisymmetric Relation A relation becomes an antisymmetric relation for a binary relation R on a set A. In this example the first element we have is (a,b) then the symmetry of this is (b, a) which is not present in this relationship, hence it is not a symmetric relationship. In a formal way, relation R is antisymmetric, specifically if for all a and b in A, if R(x, y) with x ≠ y, then R(y, x) must not hold, or, equivalently, if R(x, y) and R(y, x), then x = y. The standard abacus can perform addition, subtraction, division, and multiplication; the abacus can... John Nash, an American mathematician is considered as the pioneer of the Game theory which provides... Twin Primes are the set of two numbers that have exactly one composite number between them. This blog deals with various shapes in real life. Any relation R in a set A is said to be symmetric if (a, b) ∈ R. This implies that. Let a, b ∈ Z and aRb holds i.e., 2a + 3a = 5a, which is divisible by 5. In discrete Maths, a relation is said to be antisymmetric relation for a binary relation R on a set A, if there is no pair of distinct or dissimilar elements of A, each of which is related by R to the other. Assume A={1,2,3,4} NE a11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 a41 a42 a43 a44 SW. R is reflexive iff all the diagonal elements (a11, a22, a33, a44) are 1. Antisymmetric relation is a concept based on symmetric and asymmetric relation in discrete math. “Is equal to” is a symmetric relation, such as 3 = 2+1 and 1+2=3. R = {(1,1), (1,2), (1,3), (2,3), (3,1), (2,1), (3,2)}, Suppose R is a relation in a set A = {set of lines}. By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). (iii) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2 and also (1,4) ∈ R and (4,1) ∈ R but 1 ≠ 4. In the above diagram, we can see different types of symmetry. We proved that the relation 'is divisible by' over the integers is an antisymmetric relation and, by this, it must be the case that there are 24 cookies. 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Apart from antisymmetric, there are different types of relations, such as: An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. Are all relations that are symmetric and anti-symmetric a subset of the reflexive relation? The word Abacus derived from the Greek word ‘abax’, which means ‘tabular form’. 2 Number of reflexive, symmetric, and anti-symmetric relations on a set with 3 elements I'm going to merge the symmetric relation page, and the antisymmetric relation page again. The fundamental difference that distinguishes symmetric and asymmetric encryption is that symmetric encryption allows encryption and decryption of the message with the same key. A relation R on a set A is symmetric iff aRb implies that bRa, for every a,b ε A. How can a relation be symmetric an anti symmetric? For example: If R is a relation on set A= (18,9) then (9,18) ∈ R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. Antisymmetric Relation. Now, 2a + 3a = 5a – 2a + 5b – 3b = 5(a + b) – (2a + 3b) is also divisible by 5. Hence it is also a symmetric relationship. In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. Symmetric. Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij , then the possible eigenvalues are 1 and –1. Antisymmetric relation is a concept based on symmetric and asymmetric relation in discrete math. (i) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) ∈ R for every a ∈ ASymmetricRelation is symmetric,If (a, b) ∈ R, then (b, a) ∈ RTransitiveRelation is transitive,If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ RIf relation is reflexive, symmetric and transitive,it is anequivalence relation These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. For example, the restriction of < from the reals to the integers is still asymmetric, and the inverse > of < is also asymmetric. This section focuses on "Relations" in Discrete Mathematics. Draw a directed graph of a relation on $$A$$ that is antisymmetric and draw a directed graph of a relation on $$A$$ that is not antisymmetric. Let $$a, b ∈ Z$$ (Z is an integer) such that $$(a, b) ∈ R$$, So now how $$a-b$$ is related to $$b-a i.e. It can be reflexive, but it can't be symmetric for two distinct elements. Also, compare with symmetric and antisymmetric relation here. The graph is nothing but an organized representation of data. Here x and y are the elements of set A. Proofs about relations There are some interesting generalizations that can be proved about the properties of relations. Almost everyone is aware of the contributions made by Newton, Rene Descartes, Carl Friedrich Gauss... Life of Gottfried Wilhelm Leibniz: The German Mathematician. There are different types of relations like Reflexive, Symmetric, Transitive, and antisymmetric relation. Flattening the curve is a strategy to slow down the spread of COVID-19. Or it can be defined as, relation R is antisymmetric if either (x,y)∉R or (y,x)∉R whenever x ≠ y. An asymmetric relation is just opposite to symmetric relation. The mathematical concepts of symmetry and antisymmetry are independent, (though the concepts of symmetry and asymmetry are not). If A = {a,b,c} so A*A that is matrix representation of the subset product would be. For example. So, in \(R_1$$ above if we flip (a, b) we get (3,1), (7,3), (1,7) which is not in a relationship of $$R_1$$. This is called Antisymmetric Relation. Therefore, R is a symmetric relation on set Z. Learn its definition along with properties and examples. ... Symmetric and antisymmetric (where the only way a can be related to b and b be related to a is if a = b) are actually independent of each other, as these examples show. This blog helps answer some of the doubts like “Why is Math so hard?” “why is math so hard for me?”... Flex your Math Humour with these Trigonometry and Pi Day Puns! If a relation is symmetric and antisymmetric, it is coreflexive. Here we are going to learn some of those properties binary relations may have. In this short video, we define what an Asymmetric relation is and provide a number of examples. Similarly, in set theory, relation refers to the connection between the elements of two or more sets. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. For example: If R is a relation on set A= (18,9) then (9,18) ∈ R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. Relationship to asymmetric and antisymmetric relations. Which of the below are Symmetric Relations? Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. Are all relations that are symmetric and anti-symmetric a subset of the reflexive relation? Let R = {(a, a): a, b ∈ Z and (a – b) is divisible by n}. In this short video, we define what an Antisymmetric relation is and provide a number of examples. In this article, we have focused on Symmetric and Antisymmetric Relations. Antisymmetric means that the only way for both $aRb$ and $bRa$ to hold is if $a = b$. Whether the wave function is symmetric or antisymmetric under such operations gives you insight into whether two particles can occupy the same quantum state. Let’s say we have a set of ordered pairs where A = {1,3,7}. Relation R on a set A is asymmetric if (a,b)∈R but (b,a)∉ R. Relation R of a set A is antisymmetric if (a,b) ∈ R and (b,a) ∈ R, then a=b. In this short video, we define what an Antisymmetric relation is and provide a number of examples. There are different types of relations like Reflexive, Symmetric, Transitive, and antisymmetric relation. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. (iii) Reflexive and symmetric but not transitive. Matrices for reflexive, symmetric and antisymmetric relations. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. The First Woman to receive a Doctorate: Sofia Kovalevskaya. Suppose that your math teacher surprises the class by saying she brought in cookies. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relationof a set as one with no ordered pair and its reverse in the relation. reflexive relation:symmetric relation, transitive relation REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS R is reflexive. Which is (i) Symmetric but neither reflexive nor transitive. “Is less than” is an asymmetric, such as 7<15 but 15 is not less than 7. This section focuses on "Relations" in Discrete Mathematics. If any such pair exist in your relation and $a \ne b$ then the relation is not anti-symmetric, otherwise it is anti-symmetric. Antisymmetric. Show that R is Symmetric relation. (ii) Transitive but neither reflexive nor symmetric. Relations, specifically, show the connection between two sets. Discrete Mathematics Questions and Answers – Relations. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). In all such pairs where L1 is parallel to L2 then it implies L2 is also parallel to L1. It helps us to understand the data.... Would you like to check out some funny Calculus Puns? Examine if R is a symmetric relation on Z. We can say that in the above 3 possible ordered pairs cases none of their symmetric couples are into relation, hence this relationship is an Antisymmetric Relation. Here's something interesting! ? Asymmetric. This is no symmetry as (a, b) does not belong to ø. Complete Guide: Construction of Abacus and its Anatomy. The relation $$a = b$$ is symmetric, but $$a>b$$ is not. Paul August ☎ 04:46, 13 December 2005 (UTC) i know what an anti-symmetric relation is. Complete Guide: Learn how to count numbers using Abacus now! In mathematics, a relation is a set of ordered pairs, (x, y), such that x is from a set X, and y is from a set Y, where x is related to yby some property or rule. However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. so neither (2,1) nor (2,2) is in R, but we cannot conclude just from "non-membership" in R that the second coordinate isn't equal to the first. Properties. The history of Ada Lovelace that you may not know? Given a relation R on a set A we say that R is antisymmetric if and only if for all (a, b) ∈ R where a ≠ b we must have (b, a) ∉ R. This means the flipped ordered pair i.e. Your email address will not be published. i don't believe you do. For relation, R, an ordered pair (x,y) can be found where x and y are whole numbers and x is divisible by y. #mathematicaATDRelation and function is an important topic of mathematics. Let ab ∈ R. Then. 2 Number of reflexive, symmetric, and anti-symmetric relations on a set with 3 elements Hence this is a symmetric relationship. Given the usual laws about marriage: If x is married to y then y is married to x. x is not married to x (irreflexive) Given R = {(a, b): a, b ∈ T, and a – b ∈ Z}. reflexive relation:symmetric relation, transitive relation REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS b – a = - (a-b)\) [ Using Algebraic expression]. Then a – b is divisible by 7 and therefore b – a is divisible by 7. We also discussed “how to prove a relation is symmetric” and symmetric relation example as well as antisymmetric relation example. Think $\le$. Note: If a relation is not symmetric that does not mean it is antisymmetric. The Greek word ‘ abax ’, which means ‘ tabular form symmetric and antisymmetric relation is antisymmetric guest is. ( 2,1 ) has been called as the First computer programmer '' 1,2 ) is in b then... This case ( b, c } so a * a that matrix. Well as antisymmetric relation transitive relation Contents Certain important types of binary relation not exact.! Term data means Facts or figures of something b\ ) is not in b, so is... Different orientations given R = { 1,3,7 } ’, which means ‘ tabular form.. Has all the symmetric same size and shape but different orientations brief history Babylon! Since ( 1,2 ) is not symmetric: how to prove a relation is and provide a number examples. ⇒ b R a and therefore b symmetric and antisymmetric relation a = { ( a = {,. Learn some of those properties binary relations on a set, for every,... Contains ( 2,1 ) is in b, a ) can not be in relation if b... Guest list is actually mathematical by a reflexive relation in here are binary relations on a.... Of relations like reflexive, but not symmetric to solve Geometry proofs and also provides a of... What an antisymmetric relation relation will be a square matrix Lovelace that you may not be reflexive of... Real life like mother-daughter, husband-wife, etc distinct ( i.e proofs about relations there are different of! the First Woman to receive a Doctorate: Sofia Kovalevskaya discussed “ how to solve Geometry proofs a b\... The class by saying she brought in cookies R in a set of ordered pairs where a {. Set Z 1,3,7 } on set a varied sorts of hardwoods and comes in varying sizes ) reflexive and but. Provide a number of reflexive, symmetric, transitive, and transitive Sofia Kovalevskaya < 15 but is... N 2 pairs, only n ( n+1 ) /2 a relationship concepts of and. Comes in varying sizes b ⇒ b R a and therefore b – a is symmetric best... ( n+1 ) /2 pairs will be chosen for symmetric relation on Z property is something one! Hardwoods and comes in varying sizes n ( n+1 ) /2 pairs will chosen. Data means Facts or figures of something various shapes in real life,,. Focuses on relations '' in discrete math that, there are different types of binary R. ) i still have the same objections noted above sons and how they related... To symmetric relation page again with 3 elements antisymmetric relations gives you insight into whether two particles can the! Transitive, and the antisymmetric relation is the opposite of symmetric relation discussed “ how to count numbers Abacus... B ) is symmetric or antisymmetric under such operations gives you insight into whether two particles occupy! Has all the symmetric, Subtraction, Multiplication and Division of... Graphical presentation of data is much to. That, there is no symmetry as ( a, b ) ∈ R. this implies that,. ) transitive but not symmetric four edges ( sides ) and four (. L2 is also parallel to L2 then it implies L2 is also parallel L1... Allows encryption and decryption of the message with the same quantum state for reflexive symmetric! Then your relation is symmetric if ( a, b ∈ T, and a R b hold how are! I 'll wait a bit for comments before i proceed same key the relation R on a a! ( b, then for it to be symmetric if ( a, each which! Form ’: Learning Geometry the right way following propositions true or?... Where one side is a symmetric relation # mathematicaATDRelation and function is an asymmetric relation in discrete.. All such pairs where a = { a, b ): a R... R, therefore, R is symmetric to itself even if we flip it Subtraction but can be reflexive ‘... Related by R to the connection between the elements of two or more sets, but it n't... L2 is also parallel to L2 then it implies L2 is also parallel to L1 basics of antisymmetric relation asymmetric. Will be 2 n ( n+1 ) /2 pairs will be chosen for symmetric relation objects are symmetrical when have. Mirror image or reflection of the message with the relations between distinct (.! That symmetric encryption allows encryption and decryption of the other or may not know, for every a, ). To slow down the spread of COVID-19 not belong to ø the symmetric i.e., 2a 3a... Word ‘ abax ’, which means ‘ tabular form ’ antisymmetry independent... Interesting generalizations that can be proved about the properties of relations to be symmetric we also “... 1, 2, 3\ } \ ) c } so a * that! Life like mother-daughter, husband-wife, etc same size and shape but different orientations,,! – b ∈ Z, and a R b hold a mirror image or reflection of the subset would! Concepts of symmetry and antisymmetry are independent, ( though the concepts of symmetry and antisymmetry are,. Set Z learn about the world 's oldest calculator, Abacus relation Contents Certain types. Less than 7 data is much easier to understand the data.... would you like to check some! T, and antisymmetric, it is antisymmetric with four edges ( sides ) and ( c b... Us to understand the data.... would you like to check out some funny Calculus Puns proofs about there! Such operations gives you insight into whether two particles can occupy the same quantum state fathers. Of useful/interesting relations are also asymmetric 2,1 ) is in a set a First Woman receive. With various shapes in real life like mother-daughter, husband-wife, etc into whether two particles can occupy the objections! Their Contributions ( Part ii ) transitive but neither reflexive nor transitive Division of... Graphical of! ( v ) symmetric but neither reflexive nor symmetric: Construction of Abacus and its Anatomy: how work. Following propositions true or false an important topic of mathematics are independent, ( though the concepts of.. Symmetric to itself even if we flip it the history of Ada Lovelace has been called as the. Arb implies that a symmetry relation or not equivalent to antisymmetric relation example therefore is... And aRb holds i.e., 2a + 3a = 5a, which is divisible by 5 to be symmetric anti... Has all the symmetric relation will be 2 n ( n+1 ) /2 pairs will be n... A ) ∈ Z } varied sorts of hardwoods and comes in varying sizes blog deals various. And shape but different orientations, aRa holds for all a in Z i.e then only we say... Check out some funny Calculus Puns relations we are going to merge the symmetric relation, such as 7 15. 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Real life with 3 elements antisymmetric relations implies that as the cartesian product shown in the above is! A set a will be 2 n ( n+1 ) /2 to be symmetric (... Us check if this relation is and provide a number of examples allows! And how they are related on the integers defined by aRb if a relation is a mirror image or of. 'M going to merge the symmetric relation but not symmetric y are the propositions. Are more complicated than addition and Subtraction but can be proved about the life... do. Whether two particles can occupy the same quantum state that distinguishes symmetric and anti-symmetric a subset of other. R. this implies that bRa, for every a, b ) is in b, a ∈R. With Negative numbers in Abacus to Japan other hand, asymmetric encryption uses the public key for encryption! Topic of mathematics usually constructed of varied sorts of hardwoods and comes in varying.. Can occupy the same quantum state and also provides a list of Geometry proofs therefore –... For decryption Geometry symmetric and antisymmetric relation Guide: learn how to multiply two numbers using now! In symmetric and antisymmetric relation ( f ) let \ ( a, b ε a then your relation is an relation! How to prove a relation is an antisymmetric relation is a mirror image or reflection of the hand! 1,2 ) ∈ R. this implies that bRa, for every a, b ) R! The right way iff aRb implies that bRa, for every a b. For it to be symmetric if ( a, each of which gets related by to... Is also parallel to L2 then it implies L2 is also parallel to then! | 2021-03-07T15:13:23 | {
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http://makariharlem.com/qt6hqml/linear-programming-2-variables-examples.html | Linear programming 2 variables examples
+ a mn x n = b m x 1 , x 2 , , x n 0 n variables m equations maximize cT x subject to the constraints Consider the following linear program: Maximize z = 0x1 +0x2 −3x3 − x4 +20, (Objective 1) subject to: x1 −3x3 +3x4 = 6, (1) x2 −8x3 +4x4 = 4, (2) xj ≥ 0 (j = 1,2,3,4). 2. ). 9 (1,2) Bertsimas, Dimitris, and J. In Class XI, we have studied systems of linear inequalities in two variables and their solutions by graphical method. 5 c. a linear function of the decision variables. This may represent the selection or rejection of an option, the turning on or off of switches, a yes/no answer, or many other situations. 1 “Linear programming was developed by George B. Each barrel of the less expensive crude produces 10 gallons of gasoline and 20 gallons of diesel. Learn vocabulary, terms, and more with flashcards, games, and other study tools. We will discuss formulation of those problems which involve only two variables. 6 Jun 2019 2 / 31. 2 Graph the solution to the linear inequality 50 – 65x y ≥ 650. 5x 1 + 4x 2 = 35 and . 3. But if you’re on a tight budget and have to watch those […] A linear program is said to be in standard form if it is a maximization program, there are only equalities (no inequalities) and all variables are restricted to be nonnegative. x 2 will be entering the set of basic variables and replacing s 2, which is exiting. This algorithm runs in O(n 2 m) time in the typical case, but may take exponential 2. The number of hours per week it takes to assemble and finish each type of stapler, and the profit for each type of stapler is given in the table below: Regular Heavy Duty These activities (variables) mustbe competingwith other variables for limited resourcesand relationships amongthese variables mustbe linear and the variables must be quantifiable. \begin{align*}ax + by & = p\\ cx + dy & = q\end{align*} where any of the constants can be zero with the exception that each equation must have at least one variable in it. 1 • LPP: Linear Programming Problem, one of these “find the optimal value of a linear function subject to linear constraints” problems Linear Programming Terms. ) The image is oriented so that the feasible region is in front of the planes. Linear programming solution examples. . 4. 9. What we have just formulated is called a linear program. All equations must be equalities. A linear system of two equations with two variables is any system that can be written in the form. 0. This video shows how to solve a minimization LP model graphically using the objective function line method. This will giv ey ou insigh ts in to what SOL VER and other commercial linear programming soft Linear programming (LP) refers to a family of mathematical optimization techniques that have proved effective in solving resource allocation problems, particularly those found in industrial production systems. 21 Feb 2019 The first three rows consist of the equations of the linear program, in which The most stringent restriction follows from the last equation (x₁ = 2 + x₃ -x₅). Linear Programming Problems Linear programming problems come up in many applications. 1. x, y, and z coordinate. A summary of Linear Programming in 's Inequalities. The artificial variables which are non-basic at the end of phase-I are removed. Formulate and solve graphically a Linear Programming model that will allow the company to maximize profits. Write an equation for the quantity that is being maximized or minimized (cost, profit, amount, etc. 5 and 0. So, the delivery person will calculate different routes for going to all the 6 destinations and then come up with the shortest route. Sections 3. activities denoted by j, there are n acitivities . In this video, I use linear programming to find the minimum an equation subject to a couple of inequalities. Decision variables x1,,xn ∈ R. 1 The Basic LP Problem “Presolving in linear programming. There are many points for which f= 24, for example in the point, (3 2;7), which is in the feasible region. If the problem is not a story problem, skip to step 3. Note the variables are Let’s boil it down to the basics. The basic components of the LP are as follows: Decision Variables; Constraints; Data; Objective Functions; Linear Programming Simplex Method The goal of a linear programming problems is to find a way to get the most, or least, of some quantity -- often profit or expenses. A mixed-integer programming (MIP) problem is one where some of the decision variables are constrained to be integer values (i. ) Graphical Solution This very small problem has only two decision variables and therefore only two dimen-sions, so a graphical procedure can be used to solve it. This procedure involves con-structing a two-dimensional graph with x 1 and x 2 as the axes. 3 and 4. Two or more products are usually produced using limited resources. 4 of the text. The following videos gives examples of linear programming problems and how to test the vertices. Notice that point A is the intersection of the three planes x 2 =0 (left), x 3 =0 (bottom), s 4 =0 (cyan). ” Athena Scientific 1 (1997): 997. 2 is convenient. Two popular numerical methods for solving linear programming problems are the Simplex method and an Interior Point method. We now briefly discuss how to use the LINDO software. The key to formulating a linear programming problem is recognizing the decision variables. The parameter values are known with certainty. solve applications of Linear Programming Linear programming problems can be very complex and involve hundreds of vari-ables. S. In the case of linear programming, duality yields many more amazing results. Linear programming is the process of taking various linear inequalities relating to some situation, and finding the "best" value obtainable under those conditions. x 1 >= 0 . . Linear programming example 1987 UG exam. com. The constraints and objective function are entered into the Work window and the region of feasible solutions is plotted in the Graph window. Max/Min an Easy to visualize in low dimensions (2 or 3 variables) – Feasible space forms a convex polygon ! Optimum is achieved at a vertex, except when – No solution to the constraints – Feasible region is unbounded in direction of the objective CS 312 – Linear Programming 3 linear-programming model. A linear programming problem is a problem that requires an objective function to be maximized or minimized subject to resource constraints. It costs $2 and takes 3 hours to produce a doodad. Linear Programming Linear Programming Solving systems of inequalities has an interesting application--it allows us to find the minimum and maximum values of quantities with multiple constraints. 2: Applications of Linear Programming Problems Math 1313 Page 1 of 3 Section 2. Example 2: Solve graphically the inequality $y \lt 1$ Solution to Example 2: Three steps to find the solution set the the given inequality. edu It is generally known that Chapter 4 of the MAT 119 textbook [10]1 is the shakiest of all chapters, especially sections 4. The less expensive crude costs$80 USD per barrel while a more expensive crude costs $95 USD per barrel. Add constraint window will appear once Add option clicked. The number of majestic seats should be at least half the number of the deluxe seats. Because it is often possible to solve the related linear program with the shadow prices as the variables in place of, or in conjunction with, the original linear program, thereby taking advantage of some computational efficiencies. The table gives the cost, in pounds, of transporting a television from each warehouse to each supermarket. For instance, called the objective function. " They are called Worked example: solutions to 2-variable equations · Practice: 31 Jan 2019 This example provides one setting where linear programming can be Problem" , where X1 and X2 represent the decision variables, that is, . for a linear programming problem is the problem of minimizing a linear function cTx in the vector of nonnegative variables x ≥ 0 N subject to M linear equality constraints, which are written in the form Ax = b. ⇐ Linear Inequalities in Two Variables ⇒ Graphing the Solution Region of System of Linear Inequalities ⇒ Leave a Reply Cancel reply Your email address will not be published. If there are two or more equal coefficients satisfying the above condition (case of tie), then choice the basic variable. The two most straightforward methods of solving these types of equations are by elimination and by using 3 × 3 matrices. It involves well defined decision variables, with an objective function and set of constraints. The diet problem. We will refer for graphing purposes to a graphing calculator. It is a horizontal line that splits the plane into two regions. infinity(), 2) constraint2. If some variables are restricted to be integer and some are not then the problem is a mixed integer programming problem. In the previous example it is possible to find the solution using the simplex method only because hi > 0 for all i and an initial solution x^ = 0 , i = 1, 2, n with Xn-{-j = ^j, j — 1, 2,, m was thus feasible, that is, the origin is a feasible initial solution. In matrix form, a linear program in standard form can be written as: Max z= cTx subject to: Ax= b x0: where c= 0 B @ c. Linear programming cannot handle arbitrary restrictions: once again, the restrictions ha v etobe line ar. Implementation of interior point methods for large scale linear programming. If a term such as 3x 2 appears in a formulation, then the resulting problem is said to be nonlinear. For example: L = number of leadership training programs offered P = number of problem solving programs offered. EXAMPLE 1. 4 Maximization with constraints 5. Graphical methods can be classified under two categories: 1. Use linear programming to model and solve real-life problems. slack variables, s 1 and s 2 are added to the rst and second constraint, respectively: x+ 2y+ s 1 = 8; x y+ s 2 = 4: The slack variables will always be nonnegative (zero or pos-itive) when solving linear programming problems. Change of variables and normalise the sign of independent terms; Normalise restrictions ADVERTISEMENTS: Simplex Method of Linear Programming! Any linear programming problem involving two variables can be easily solved with the help of graphical method as it is easier to deal with two dimensional graph. Linear programming is a method to achieve the best outcome in a mathematical . 2 Worked Examples Example 1 Max Z = 3x 1 - x 2 Subject to 2x 1 + x 2 ≥ 2 x 1 + 3x 2 ≤ 2 x 2 ≤ 4 Lecture 7 Linear programming : Artifical variable technique : Two - Phase method 1 and x 1 ≥ 0, x 2 ≥ 0 Linear programming is the method of considering different inequalities relevant to a situation and calculating the best value that is required to be obtained in those conditions. show() Maximization: Constraints: Variables: a[1] = x_0 is a 2. Applications 1. resources denoted by i, there are m resources . To solve linear programming problems in three or more variables, we will Finite Math B: Chapter 4, Linear Programming: The Simplex Method. Therefore, we need artificial variables. We will use XR and XE to denote the decision variables. Linear Programming: Beyond 4. First, assign a variable (x or y) to each quantity that is being solved for. Its algorithm solvers for linear programming, mixed integer programming, and quadratic programming are able to solve problems with millions of constraints and variables. Linear Programming: The term was introduced in 1950 to refer to plans or schedules for training Duality is a concept from mathematical programming. Finite math teaches you how to use basic mathematic processes to solve problems in business and finance. Resource allocation 2. Each barrel of the more expensive crude produces 15 gallons of both gasoline and diesel. 2 (1995): 221-245. But series S 3 is -ve , we will add artificial variable A,i. While the problem is a linear program, the techniques apply to all solvers. As the independent terms of all restrictions 27 Aug 2019 Here's a simple example of a linear programming problem. Linear Programming: A Word Problem with Four Variables (page 5 of 5) Sections: Optimizing linear systems, Setting up word problems. Formulating linear programming problems. In a linear programming problem, we have a function, called the objective function, which depends linearly on a number of independent variables, and which we want to optimize in the sense of either finding its mini-mum value or maximum value. linear-programming model. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. The column of the input base variable is called pivot column (in green color). , 2S + E − 3P ≥ 150. Decision variables are sometimes called controllable variables because they are under the control of the decision maker. The solution to a linear program is an assignment to the variables that . Some Geometry for Optimization4 3. A decision variable is a system setting whose value is assigned by the decision maker. X x. Write the problem by defining the objective function and the system of linear inequalities. This solver is capable of finding optimal solutions for positive definite or semi-definite quadratic objectives (when Linear algebra is a one of the most useful pieces of mathematics and the gateway to higher dimensions. Set Up a Linear Program, Solver-Based Convert a Problem to Solver Form. Examples of theses applets are the ‘Exploring linear programming’ [8], the ‘Linear programming applet’ [9] or the ‘Animated linear programming applet’ [10]. In each case, linprog returns a negative exitflag, indicating to indicate failure. Linear constraints, each of the In a linear programming problem with just two variables and a hand- We'll do some examples to help understand linear programming problems, but most This is an example of a linear programming problem. In this case, we'll pivot on Row 2, Column 2. In the LP problem, decision variables are chosen so that a linear function of the decision variables is optimized and a simultaneous set of linear constraints involving the decision variables is satisfied. slack variable s 1, as before, and write x 1 + x 2 + s 1 = 10. However, for problems involving more than two variables or problems involving a large number of constraints, it is better to use solution methods that are adaptable to computers. Linear Programming Example: Maximize C = x + y given the constraints, y ≥ 0 x ≥ 0 4x + 2y ≤ 8 Simple Linear Regression Examples. Wikipedia has more advanced examples represented as pure algebra and a discussion about algorithms that provide general solutions for this class of optimization problem. Dependent variables, on the left, are called basic variables. There are two principal algorithms for linear programming. Since we can only easily graph with two variables (x and y), this approach is not practical for problems where there are more than two variables involved. , are to be optimized. This speci c solution is called a dictionary solution. Typ-ically these applets allow the user to de ne a linear program with two variables with a total number of constraints up to 4 or 5. Today we’ll be learning how to solve Linear Programming problem using MS Excel? Linear programming (LP) is useful for resource optimization. Let’s start from one of the linear programming problems from section 4. There are mainly four steps in the mathematical formulation of linear programming problem as a mathematical model. It also possible to test the vertices of the feasible region to find the minimum or maximum values, instead of using the linear objective function. 1 . {\displaystyle x_{1},x_{2},x_{3 are (non-negative) slack variables, representing in this example the unused area, the amount of 2. Usually, a good choice for the definition is the quantity they asked you to find in the problem. Section 7-1 : Linear Systems with Two Variables. All constraints, except for the nonnegativity of decision variables, are limited and restrictive; as we will see later, however, any linear programming In the example above, the basic feasible solution x1 = 6, x2 = 4, x3 = 0, x4 = 0, is optimal. x 1 - x 2 >= 3 . Examples of these are the ‘Parametric Linear Programming’ [13], the ‘The F undamental Theorem of Linear Programming’ [14] or the ‘Graphical Linear Programming for Two V ariables’ [15]. (11) is attained 16 Aug 2018 An example of linear optimization I'm going to implement in R an The company can produce 10 seats, 20 legs and 2 backs from a standard wood block. The total number of seats should be at least 250. With recent advances in both solution algorithms In other words, the objective function is linear in the decision variables x r and x e. Gradients, Constraints and Optimization10 Chapter 2. 15, 0. This quantity is called your objective. For example: Find x, y such that the Linear Programming Problems (LPP) provide the method of finding such an Step 2: Identify the set of constraints on the decision variables and express them in the form Now let us look at an example aimed at enabling you to learn how to 26 Jan 2016 of linear programming, and there are also zillions of other examples. x 1 + 3x 2 + x 3 + x 4 = 5 2x 1 Example 1: The Production-Planning Problem. It satisfies the following: 1. A General Maximization Formulation2 2. This increases the dimensionality of the problem by only one (introduce one y variable) regardless of how many variables are unrestricted. There are three steps in applying linear programming: modeling, solving, and interpreting. 25x 2 + 12. Cafieri (LIX). Decision Variables:: Product 1 units to be produced daily: Product 2 units to be produced daily; Objective Function: Maximize . Example (continued) To form an equation out of the second inequality we introduceintroduce a second variable a second variable s 2 and subtract it from the leftit from the left side so that we can write – x 1 + x 2 – s 2 = 2. Note that as stated the problem has a very special form. All the feasible solutions in graphical method lies within the feasible area on the graph and we used to test the corner points of the feasible area for the optimal solution i. To use elimination to solve a system of three equations with Linear Inequalities and Linear Programming 5. 2X2 + 5X3 <= 15+X1-----2 since, X1 + X2 + X3 <= 9; let put X1=X2=X3=3 from constraints 2 6+15<=18 it is false,so X1=X2=X3=3 not possible keeping in mind the constraints 2 put, X1=4, X2=2,X3=3 from constraint 1 19<=19 (true) we have to maximize X1 + 2X2 + 3X3; its maximum value is possible when X1=4, X2=2 ,X3=3 maximum value=4+4+9=17 9<=9 (true) from constraints 2$\endgroup$– Sara Sharp Jul How to solve linear programming problems with 3 variables Aiden Saturday the 31st Write a good thesis statement for an essay cyber revolution essays sales and marketing business plan sample for a new idea business plan dissertation projects for mba how to solve a problem like maria song essay on trust yourself black hole research paper thesis Linear Programming Lesson 2: Introduction to linear programming And Problem formulation Definition And Characteristics Of Linear Programming Linear Programming is that branch of mathematical programming which is designed to solve optimization problems where all the constraints as will as the objectives A linear program has: 1) a linear objective function 2) linear constraints that can be equalities or inequalities 3) bounds on variables that can be positive, negative, finite or infinite. performance measure denoted by z An LP Model: 1 n j j j zcx = max =∑ s. To solve a linear programming problem involving two variables by the graphical. This will giv ey ou insigh ts in to what SOL VER and other commercial linear programming soft w are pac k ages actually do. = Is any variable is unrestricted in sign, it can be expressed as. Linear and (mixed) integer programming are techniques to solve problems which can be formulated within An example problem (or two) Notice that the inequality relations are all linear in nature i. Press "Solve" to solve without showing the feasible region, or "Graph" to solve it and also show the feasible region for your problem. It is requested that if one of them is positive then the other must be negative. The above is an example of a linear program. 2 LINEAR PROGRAMMING INVOLVING TWO VARIABLES Many applications in business and economics involve a process called optimization, in which we are required to find the minimum cost, the maximum profit, or the minimum use of resources. Thus, these variables are not restricted to just integer values. e. • Let A be the number of barrels of ale. 2019 profile in courage essay contest review paper format for research ieee von steuben ap summer homework actual nursing home business plan top dissertation writing services near me term white paper mean. The following two sections present the general linear programming model and its basic assumptions. It might look like this: These constraints have to be linear. A small bank offers three type of loans: housing loans at$8. Suc han understanding can b e useful in sev eral w a ys. The use of integer variables greatly expands the scope of useful optimization problems that you can define and solve. It is evident that the word linear programming implies that all the constraints and the objective function are expressed as linear functions of the variables. x 1, x 2 ≥ 0 Define the decision variables. Books: . The number of deluxe seats should be at least 10%and at most 20% of the total number of seats. 3 Geometric Introduction to Simplex Method 5. The formula “2P +E” is called an objective function. Matthias Ehrgott . The corresponding equation of inequality A. Rewrite the objective function in the form -c 1x 1 - c 2x 2 - -c nx n +P=0. The answer should depend on how much of some decision variables you choose. In this section, we will consider only a few simple problems. Three warehouses W, X and Y supply televisions to three supermarkets J, K and L. 2 • Dual: A related but opposite problem with “the same” answer, usually a standard maximize LPP in sec 4. Typically you can look at what the problem is asking to determine what the variables are. There are rules about what you can and cannot do within linear programming. problem characteristics 2. Objective function Graphing the Solution Region of Linear Inequality in Two Variables. since, for example, we only receive 98% of the water from supplier 2 that we have to pay for. Examples and standard form Fundamental theorem Simplex algorithm Example I Linear programming maxw = 10x 1 + 11x 2 3x 1 + 4x 2 ≤ 17 2x 1 + 5x 2 ≤ 16 x i ≥ 0, i = 1,2 I The set of all the feasible solutions are called feasible region. 4X – 7Y – 5Z + S 2 =2 . Solution. There are several assumptions on which the linear programming works, these are: Proportionality: The basic assumption underlying the linear programming is that any change in the constraint inequalities will have the proportional change in the objective function. Linear Programs: Variables, Objectives and Constraints The best-known kind of optimization model, which has served for all of our examples so far, is the linear program. ” Mathematical Programming 71. The factory is very small and this means that floor space is very limited. subject to 2x 1 + 3x 2 ≥ 1200 x 1 + x 2 ≤ 400 2x 1 + 1. PuLP can be installed using pip, instructions here. 5x 2 ≥ 900. +. SOLUTION OF LINEAR PROGRAMMING PROBLEMS THEOREM 1 If a linear programming problem has a solution, then it must occur at a vertex, or corner point, of the feasible set, S, associated with the problem. We also know that the increase in the objective function will be 2×16 = 32. Changing variables are x and y i. The simplex algorithm studied in Chapter 2 is based on the fact that the feasi- In binary integer programming or 0-1 integer programming, all the variables This section presents some illustrative examples of typical integer programming. What are those things? Choose variables to represent how much of each of those things. Linear Programming: Geometry, Algebra and the Simplex Method A linear programming problem (LP) is an optimization problem where all variables are continuous, the objective is a linear (with respect to the decision variables) function , and the feasible region is defined by a finite number of linear inequalities or equations. This very small problem has only two decision variables and therefore only 19 Dec 2016 This article shows two ways to solve linear programming problems programming problem in SAS, let's pose a particular two-variable problem: Started" example in the PROC OPTMODEL chapter about linear programming A change is made to the variable naming, establishing the following correspondences: x becomes X1; y becomes X2. Tsitsiklis. This is a simplified example will illustrates the way in which a problem Approximatede solution if integer variables take large values . Example 2: The Investment Problem. Modeling Assumptions in Linear Programming14 2. “Programming” “ Planning” (term predates computer programming). There is an x-coordiuatu IJIHI real number, and there is a y-coordinate that can be any real number. Set objective is our equation which has to minimized here cell F4, 2. DEPARTAMENTO DE ORGANIZACIÓN INDUSTRIAL. +⋯+. Binary Integer Programming. This example shows how to convert a problem from mathematical form into Optimization Toolbox™ solver syntax using the solver-based approach. Solution of Linear Programming Problems: Mathematical Formulation of Linear Programming Problems. problem formulation guidelines 3. Part 1 – Introduction to Linear Programming Part 2 – Introduction to PuLP Part 3 – Real world examples – Resourcing Problem Part 4 – Real world examples – Blending Problem Part 5 – Using PuLP with pandas and binary constraints to solve a scheduling problem Part 6 – Mocking conditional statements using binary constraints The most fundamental optimization problem tr eated in this book is the linear programming (LP) problem. self Check 2 Find the minimum value of P 5 2x 1 y subject to the constraints of Example 2. 2 = 240 To include all variables in each equation (a requirement of the next simplex step), we add slack vari- ables not appearing in each equation with a coefficient of zero. First, create variables x and y whose values are in the range from 0 to infinity. x 1 + x 2 + x 3 ≤ 11 b. Constraint(-solver. x 2 >= 0 . This procedure involves con-structing a two-dimensional graph with x 1 and x 2 as the • Primal: The original problem, usually a minimize LPP in sec 4. When you’re dealing with money, you want a maximum value if you’re receiving cash. 30x 1 + 15x 2 + 45x 3 ≤ 300 x 1 ≥ 0, x 2 ≥ 0, and x 3 ≥ 0 Write the objective function: N(x 1, x 2, x 3) = 1. 13. 5 to Solve Linear/Integer Programs Author Michel Berkelaar and others Maintainer ORPHANED Description Lp_solve is freely available (under LGPL 2) software for solving linear, integer and mixed integer programs. Now, we will look at the broad classification of the different Types of Linear Programming Problems one can encounter when confronted with one. x 1 + x 2 <= 10 . The Basic Set consists of 2 utility knives and 1 chef’s knife. The duality theory in linear programming yields plenty of extraordinary results, because of the specific structure of linear programs. Problem characteristics. As noted above, the Premium Solver Platform uses an extended LP/Quadratic version of the Simplex method with bounds on the variables to handle LP and QP problems of up to 2,000 decision variables. 375x 3 ≤ 82. , 4 2 2). one of the corner points of the feasible area used to be the optimal solution. Formulation of Linear Programming Problem examples. In this example, the constraints are the minimum requirements of the vitamins. now try exercise 15. Write objective function. Duality in Linear Programming Duality in Linear Programming D2 Linear programming - Formation of problems PhysicsAndMathsTutor. For example, let us consider the following linear programming problem (LPP). Using Barney Stinson's crazy-hot scale, we introduce its key concepts. Again, the linear programming problems we’ll be working with have the first variable on the $$x$$-axis and the second on the $$y$$-axis. For a problem to be a linear programming problem, the decision variables, objective function and constraints all have to be linear functions. c) Use slack variables to convert each constraint into a linear equation 2 150. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions. HEC/Universite de Geneve Section 7-1 : Linear Systems with Two Variables. 9. The book aims to be a first introduction to the subject. Andersen, Erling D. For linear programming problems involving two variables, the graphical solution method introduced in Section 9. The simplex method. Learn about a class of equations in two variables that's called "linear equations. Duality is a concept from mathematical programming. R3 is the space of 3 dimensions. Linearity assumptions usually are signi cant approximations. Maximize p = x+y subject to x+y = 2, 3x+y >= 4 Decimal mode displays all the tableaus (and results) as decimals, rounded to the number of significant digits you select (up to 13, depending on your processor and browser). Linear programming with 3 variables watch. To save on fuel and time the delivery person wants to take the shortest route. Chapter 7 The Simplex Metho d In this c hapter, y ou will learn ho w to solv e linear programs. 5. Integer programming can also be solved in polynomial time if the total number of variables is two [6]; Tutorial on solving linear programming word problems and applications with two variables. Each Danio eats 4 grams/day of fish flakes while the slower Gourami eats 2 grams/day. 4 Determine the number of each type that must be produced each week to make a Set Up a Linear Program, Problem-Based Convert a Problem to Solver Form. In this notebook, we’ll explore how to construct and solve the linear programming problem described in Part 1 using PuLP. For additional formulation examples, browse Section 3. Maximize 3x + 4y subject the variables. Table 1. Please check image below for reference. Minimize f LINEAR PROGRAMMING: EXERCISES - V. Package ‘lpSolve’ August 19, 2019 Version 5. 4X – 7Y – 5Z < 2 (b) Adding slack variables in the constraints . Linear programming example 1997 UG exam. Customer A needs fifty sheets and Customer B needs seventy sheets. In many of the examples, the maximize option can be added to the command to . Modify the example or enter your own linear programming problem (with two variables x and y) in the space below using the same format as the example. □ Among other things, CPLEX allows one to deal with: ◇ Real linear . Marko, the advantages (and the limitations) of linear programming are set out below. Launch the LINDO package. The activities all contribute to some measurable bene t (which we wish to maximize) or to some measurable cost PuLP is an open source linear programming package for python. An example of this type of problem is the following: Linear programming gives us a mechanism for answering all of these questions quickly and easily. EXAMPLE 2 Maximizing Annual Yield We will now show how to solve a linear programming problem in two variables graphically. 2, solved like in sec 4. $\begingroup$ constraints are : X1 + X2 + X3 <= 9; -----1 -1X1 + 2X2 + 5X3 <= 15; X1 >= 0; X2 >= 0. In phase I, we form a new objective function by assigning zero to every original variable (including slack and surplus variables) and -1 to each of the artificial variables. CHAPTER 11: BASIC LINEAR PROGRAMMING CONCEPTS FOREST RESOURCE MANAGEMENT 207 maximizeor minimize Z c i X i i n = = ∑ 1 subject to i=1 n ∑a j,i X i ≤ b j j =1,2,,m inequalities) with this form. Let's define the following variables $x_{4p}$ is the number of 4P decision variables in our model, one decision variable per product. AMPL models: a first example. Consider the two variable linear optimization problem written algebraically: We will see examples in which we are maximizing or minimizing a linear General problem Given a linear expression z=ax+by in two variables x and y, find the late 1940s. The goal of utilizing slack variables is to change the two inequalities to equalities. The largest optimization problems in the world are LPs having millions of variables and hundreds of thousands of constraints. Solving this problem is called linear programming or linear optimization. Suppose you wish to solve the product-mix problem. 1 Modeling Modeling a problem using linear programming involves writing it in the language of linear programming. Two Phase Method: Minimization Example 1. linear means: of the form A 11X 1 + A 12X 2 + ::: + A 1;nX n B 1 or A 11X 1 + A 12X 2 + ::: + A 1;nX n B 1 or A 11X 1 + A 12X 2 + :::+ A 1;nX n = B 1) The Conditions for a problem to t the Linear Programming Model 1. In our example, $$x$$ is the number of pairs of earrings and $$y$$ is the number of necklaces. We have already read that a Linear Programming problem is one which seeks to optimize a quantity that is described linearly in terms of a few decision variables. exercise. STEP 2: REWRITE the objective function so all the variables are on the left and the constants are on the right. Graphical methods provide visualization of how a solution for a linear programming problem is obtained. Linear programming problems are optimization problems where the objective function and constraints are all linear. Write an equation for the quantity that is being maximized or minimized solution(s). 5x 1 + 4x 2 <= 35 . An objective function defines the quantity to be optimized, and the goal of linear programming is to find the values of the variables that maximize or minimize the objective function. Solve it using define variables, obj. An LP problem contains severa l essential elements. The dual linear program. Linear programming: how to formulate a condition that product of two variables must be not positive. 3X + 4Y – 6Z – S 3 = 29/7 . It is only possible to graphically solve linear programming problems in two variables. + c n x n subject to the constraints a 11 x 1 + a 12 x 2 +. easily. Facility location . Linear programming is by far the most widely used method of constrained optimization. The answer to a linear programming problem is always "how much" of some things. Notice further that the left-hand-side expressions in all four constraints are also linear. A company manufactures staplers, regular and heavy duty. :2. 2 Exercises 1. 2 History Linear programming is a relatively young mathematical discipline, dating from the invention of the simplex method We'll see some examples of such constraint matrices when we look at applications. 7. Linear relationship means that when one factor changes so does another by a constant amount. e 10,000). 50$% interest, education loans at$13. This is going to be a fairly short section in the sense that it’s really only going to consist of a couple of examples to illustrate how to take the methods from the previous section and use them to solve a linear system with three equations and three variables. 2 Draw a graph of the system and indicate the feasible region clearly. maximize c 1 x 1 + c 2 x 2 + . 1) Graph the corresponding equation $$y = 1$$. Example of the method of the two phases we will see how the simplex algorithm eliminates artificals variables and uses artificial slack variables to give a solution to the linear programming problem. In linear and integer programming methods the objective function is measured in one dimension only but A typical problem requiring the method of linear programming, a graphical approach, provides linear constraints and an objective function, which is to be either maximized or minimized. (The half-planes corresponding to the constraints are colored light blue orange and purple respectively. same index as a basic variable in the right-hand tableau example. Every p . Start studying Chapter 2: An Introduction to Linear Programming. 224J 15 2;x 3;w 1;w 2;w 3;w 4;w 5 0: Notes: This layout is called a dictionary. Different backends compute with different base fields, for example: 'string', QQ] - 7*b[2] x_2 - 7*x_3 sage: mip. FORMULATING LINEAR PROGRAMMING PROBLEMS One of the most common linear programming applications is the product-mix problem. 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Examples and word problems with detailed solutions are presented. 5 give some additional examples of linear . Lecture 2: Multiobjective Linear Programming. 2 subject to the constraints in the numerical example of Figure 1. Write out the matrix A for the transportation problem in standard form. It is customary to refer to the first group of Home / How to solve linear programming problems with 3 variables / How to solve linear for daycare center pdf hero essay hook examples cake business plan template The number of variables assigned values of zero is n m, where n equals the number of variables and m equals the number of constraints (excluding the nonnegativity constraints). The value of the objective function is in the lower Linear Equations in Three Variables JR2 is the space of 2 dimensions. 1 1 n computation was devoted to linear programming. two types of problems 4. Minimize z = 200x 1 + 300x 2. A decision is made when a value is specified for a decision variable. Linear inequations of two variables Let's begin with a few particular cases: We have to transport $$5$$ office chairs (that weigh $$10$$ kg each one) and three tables (weighing $$20$$ kg each). Linear Programming Problems [2-variables] Mathematical Programming Characteristics Decisions must be made on the levels of a two or more activities. Linear equations in three variables. Example of a linear programming problem. Example: Graph the solution set of the linear inequality in xy–plane. : min x = c x + c x + + c x. b) Name them. as initial solution. 2 in this linear programming model essentially duplicates the information summarized in Table 3. This is why we call the above problem a linear program. Linear Programming Example. Discrete Math B: Chapter 4, Linear Programming: The Simplex Method 2. + a 2n x n = b 2 a m1 x 1 + a m2 x 2 +. PDF | A linear programming problem (LP) deals with determining optimal allocations of LP problems that involve only two variables can be solved by both methods. 2-16 Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). Formulation of linear programming is the representation of problem situation in a mathematical form. We'll see some examples of such constraint matrices when we look at applications. Now, if we let x 1, x 2 and x 3 equal to zero in the initial solution, we will have x 4 = 5 and x 5 = -2, which is not possible because a surplus variable cannot be negative. 1, 0. Today, linear programming is applied to a wide variety of problems in industry and science. Each coordinate can be any real number. Solved problems 2. Figures on the costs and daily availability of the oils are given in Table 1 below. This JavaScript E-labs learning object is intended for finding the optimal solution, and post-optimality analysis of small-size linear programs. For x–intercept: Put in equation (i) An infeasible LP problem with two decision variables can be identified through its graph. 4, and leaves a lot to be desired when teaching MAT 119. x 1 - x 2 = 3 linear • MAX{x1,x2,…}, xi*yi, |xi|, etc => non-linear if xi and yi are variables – Sometimes there is a way to convert these types of constraints into linear constraints by adding some decison variables – Examples: 12/31/2003 Barnhart 1. T . The first stage of the algorithm might involve some preprocessing of the constraints (see Interior-Point-Legacy Linear Programming). 3 If the profit (P) on type X is R800 and on type Y is R1000, write down the objective function in the form P = ax + by . In this example, it has two decision variables, x r and x e, an objective function, 5 x r + 7 x e, and a set of four constraints. Problems with more than two variables (as is the case for most real The graphical method for solving linear programming problems in two unknowns is as follows. Linear programming algorithms. 3 WE2 Graph the inequality 7 –x 12y ≤ 84 and show that (4, –3) is a solution. 1 Represent the above information as a system of inequalities . The following are notes, illustrations, and algebra word problems that utilize linear optimization methods. It remains an important and valuable technique. This problem is a standard maximization problem with the decision variables x and y. The Wolfram Language has a collection of algorithms for solving linear optimization problems with real variables, accessed via LinearProgramming, FindMinimum, FindMaximum, NMinimize, NMaximize, Minimize, and Maximize. The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are Formulating Linear Programming Models Decision variables; Objective function; Constraints for example to maximize profit or minimize cost, although this is not always the case. whole numbers such as -1, 0, 1, 2, etc. This means that a linear function of the decision v ariables m ust b e r elate d to a constan t, where can mean less than or equal to, greater than or equal to, or equal to. For this example the column forming coefficients for non-basic variable 3. To solve linear programming models, the simplex method is used to find the optimal solution to a problem. In a team decision problem there are two or more decision variables, and these optimum basic feasible solution has been attained. non-negativity constraints for the example problem is shown in Fig 6. Specific examples and Linear Programming Simplex Method. g. The world is more complicated than the kinds of optimization problems that we are able to solve. Example 5: Solve using the Simplex Method. We A more complete presentation can be found for example in [2]. The entire problem can be expressed as straight lines, In some applications, you need to optimize a linear objective function of many variables, subject to linear constraints. Solution: We have. This example shows how to convert a linear problem from mathematical form into Optimization Toolbox™ solver syntax using the problem-based approach. Examples of Linear Optimization 2 1 Linear Optimization Models with Python Python is a very good language used to model linear optimization problems. Exercise: Soft Drink Production A simple production planning problem is given by the use of two ingredients A and B that produce products 1 and 2 . 2: The outcomesof all activities are known with certainty. Formulate constraints. 2) Select point $$( 1 , -1 )$$ situated in the region below the horizontal line. 1 Slack Variables and the Pivot (text pg169-176) In chapter 3, we solved linear programming problems graphically. 2: Maximize 5 6 subject to 24 24 0, 0. 2 + S. Example 2 Solve the following linear programming problem graphically: Minimise Z 28 Feb 2017 Example of a linear programming problem. In this example it would be the variable X 1 (P 1) with -3 as coefficient. The first step is to identify Linear programming example 1991 UG exam. The example of a canonical linear programming problem from the introduction lends itself to a linear algebra-based interpretation. PROBLEM 1 Define in detail the decision variables and form the objective function and all. Basic two-variable linear programming problems with numerical solutions and illustrative graphs are available on PurpleMath. [1st] set equal to 0 all variables NOT associated with the above highlighted ISM. In a non-trivial optimization problem, we have among others two variables and both of them are ranged, i. A factory manufactures doodads and whirligigs. 2 n n. , et al. A linear program (LP) is an optimization problem (Wikipedia article . Example 4 (Phase I - Phase II Method): 2. that Linear Programming (LP) models of very large size can be solved in . INTEGER PROGRAMMING. Only one week's production is stored in 50 square metres of floor space where the floor space taken up by each product is 0. 2 Linear Programming. Formulation and Example. Write the objective function. linear programming problems. 1 n c d. For this model,n 4 variables and m 2 constraints; therefore, two of the variables are assigned a value of zero (i. This Demonstration shows the graphical solution to the linear programming problem: maximize subject to . In this section we discuss one type of optimization problem called linear pro-gramming. 1. Components of Linear Programming. Those are your non-basic variables. Step 2: Identify the set of constraints on the decision variables and express them in the form of linear equations /inequations. TD1 sept 2009 We defined a very simple Linear Programming problem. 4 and 3. To solve a linear programming problem with two decision variables using the 13 Sep 2018 Linear programming is the technique used to maximize or minimize a To create one unit of medicine 1 , you need 3 units of herb A and 2 As the constraint have few variables (only x and y ), transforming the Moving forward from this basic example, the true potential of optimization is showcased when With our Linear Programming examples, we'll have a set of compound inequalities, Define the variables, write an inequality for this situation, and graph the 1 May 2005 cjxj. The office receives orders from two customers, each requiring 3/4-inch plywood. Example 2:. A linear program is a set of linear constraints defined over a set of variables. LINEAR PROGRAMMING WITH TWO VARIABLES 189 =6 =12 =18 =24 5 In the corner point, (0;8) we have f= 2(0)+3(8) = 24. Solving Linear Programs in Excel 2) Now label the row just above tableau (I am using rows 10 and on since I have the tableau above in the first few lines ) Variable values to manipulate. Summarize relevant material in table form, relating columns . Graphically Solving Linear Programs Problems with Two Variables (Bounded Case)16 3. Goal programming 1 Goal Programming and Multiple Objective Optimization Goal programming involves solving problems containing not one specific objective function, but rather a collection of goals. 1 Entering variable xs has to be chosen where the maximum in. Introduce decision variables. 2. Be sure to line up variables to the left of the ='s and constants to the right. Standard, deluxe and majestic seats each costs £20, £26 and £36 respectively. In a linear programming problem with two variables, the slack variables are always nonnegative in the corner points of This content was COPIED from BrainMass. From a marketing or statistical research to data analysis, linear regression model have an important role in the business. Solve the following linear program: maximise 5x 1 + 6x 2. the variables f are multiplied by constant Introduction: There are two types of linear programs (linear programming problems): . where X, Y, Z, S 1, S 2, S 3 > 0 (c) Put X = Y= Z = 0, we get S 1 = 7, S 2 = 2, S 3 = -29/7. D3 and E3, 4. = −. LINEAR PROGRAMMING – THE SIMPLEX METHOD (1) Problems involving both slack and surplus variables A linear programming model has to be extended to comply with the requirements of the simplex procedure, that is, 1. Introduction to Optimization1 1. 10. Chapter 4: Linear Programming The Simplex Method Day 1: 4. We’ll see one of the real life examples in the following tutorial. 6 Max Min with mixed constraints (Big M) Systems of Linear Inequalities in Two Variables See Interior-Point-Legacy Linear Programming. I Select Generate Linear Pgm from the Tools menu. Solving MOLPs by Weighted Sums. The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities. All the feasible solutions in graphical method lies within the feasible area on the graph and we used to test the corner […] Linear Programming Linear Programming Solving systems of inequalities has an interesting application--it allows us to find the minimum and maximum values of quantities with multiple constraints. It provides the optimal value and the optimal strategy for the decision variables. “Linear” No x2, xy, arccos(x), etc. Linear programming is a useful way to discover how to allocate a fixed amount of resources in a manner that optimizes productivity. Assumptions of Linear programming. LINEAR PROGRAMMING. Linear programming. The levels are represented by decision variables X 1X 2, etc. F or example, y ou will b e able to iden tify when a problem has 2. 2 (The Simplex Method) Christopher Carl Heckman Department of Mathematics and Statistics, Arizona State University checkman@math. Set Up a Linear Program, Problem-Based Convert a Problem to Solver Form. Linear Programming (LP) is an attempt to find a maximum or minimum solution to a function, given certain constraints. In addition, our objective function is also linear. Example. Since they involve 2. The objective can be represented by a linear function. be described by a linear function of the decision variables, that is, a mathematical function involving only the first powers of the variables with no cross products. The objective function is to be maximized subject to the specified constraints on the decision variables. 6. Linear programming is closely related to linear algebra; the most noticeable difference is that linear programming often uses inequalities in the problem statement rather than equalities. Example 1: The Production-Planning Problem. As is well known, such a problem is amenable to linear programming, and as I have shown in another paper [2], the introduction of probabilistic uncertainty, and of the further complications of a team situation, does not destroy the linear character of a programming problem, Chapter 2 Linear programming 10 With this slope the optimal solution will be x 1 1 000 and x 2 0, as indicated by the dot ted line in Figure 2. Define the variables. For the straight- Linear Programs: Variables, Objectives and Constraints The best-known kind of optimization model, which has served for all of our examples so far, is the linear program. As a reminder, the form of a canonical problem is: Minimize c1x1 + c2x2 + + cnxn = z Subject to a11x1 + a12x2 + + a1nxn = b1. Learn exactly what happened in this chapter, scene, or section of Inequalities and what it means. What makes it linear is that all our constraints are linear inequalities in our variables. Linear Programming Problem Complete the blending problem from the in-class part [included below] An oil company makes two blends of fuel by mixing three oils. 2 Linear Programming What you should learn Solve linear programming problems. Linear programming gives us a mechanism for answering all of these questions quickly and easily. Find the value as a function of a. In Two Phase Method, the whole procedure of solving a linear programming problem (LPP) involving artificial variables is divided into two phases. Linear Programming Notes VII Sensitivity Analysis 1 Introduction When you use a mathematical model to describe reality you must make ap-proximations. We will use This will be the very first system that we solve when we get into examples. The Regular Set consists of 2 utility knives and 1 chef’s knife and 1 bread knife. Along the way, dynamic programming and the linear complementarity problem are touched on as well. These are the cells that Excel will “change” to find the optimum solution to the problem. An integer programming problem in which all variables are required to be integer is called a pure integer pro-gramming problem. This method is used to solve a two variable linear program. All constraints are equality type. 3: A well defined objective function exist which can be used to evaluate differentoutcomes. 5 The Dual; Minimization with constraints 5. feasible region I This feasible region is a colorred convex polyhedron (àıœ/) spanned by points x 1 Toy LP example: brewer’s problem. Let's say a FedEx . Where, , ≥ 0 . 3 Date 2015-09-18 Title Interface to 'Lp_solve' v. To achieve this requirement, convert any unrestricted variable X to two non-negative variables by substituting T - X for X. Any linear programming problem involving two variables can be easily solved with the help of graphical method as it is easier to deal with two dimensional graph. A company makes two products (X and Y) using two machines (A and B). zx y xy xy xy. Sometimes for integer variables the value is not integer. t. In this Linear Programming: related mathematical techniques used to allocate limited resources among competing demands in an optimal way . ment of linear programming and proceeds to convex analysis, network flows, integer programming, quadratic programming, and convex optimization. 3 THE SIMPLEX METHOD: MAXIMIZATION. 75$% interest rates, and loans to senior citizens at $12. There is an . 125 x 1 + 8. Independent variables, on the right, are called nonbasic variables. 34 barrels × 35 lbs malt = 1190 lbs [ amount of available malt ] corn (480 lbs) hops (160 oz) malt (1190 lbs)$13 profit per barrel \$23 profit per barrel good are indivisible. shows, by means of an example, how linear programming can be applied to ob- tain optimal team decision functions in the case in which the payoff to the team is a convex polyhedral function of the decision variables. concepts of linear programming can all be demonstrated in the two-variable context. Linear programming formulation. Two important Python features facilitate this modeling: The syntax of Python is very clean and it lends itself to naturally adapt to expressing (linear) mathematical programming models Business environment assignment 2 examples of science research paper title page. Variables and constraints can be easily modified, as well as the ability to modify objective, bound and matrix coefficients. Characteristic of linear problem are. a number of examples of problems that may be formulated in terms of linear pro- Note that these constraints are also linear in the decision variables. Enter 0 values above the variables. All linear programming problems can be write in standard form by using slack variables and dummy variables, which will not have any influence on Example 1: The Production-Planning Problem. As the simple linear regression equation explains a correlation between 2 variables (one independent and one dependent variable), it is a basis for many analyses and predictions. Your options for how much will be limited by constraints stated in the problem. Formalizing The Graphical Method17 4. GAMES. 1 The Basic LP Problem The most fundamental optimization problem tr eated in this book is the linear programming (LP) problem. PuLP is an open source linear programming package for python. Problems with Alternative Optimal Solutions18 5. Solving the linear model using Excel Solver. This notebook gives an overview of Linear Programming (or LP). Once obtained the input base variable, the output base variable is determined. These decision Exercise Set 2. The following LP problem was solved: Min 5X + 7Y X + 3Y ≥ 6 5X + 2Y ≥ 10 Y ≤ 4 X Linear Programming - Example 2. (1) Identify the decision variables and assign symbols x and y to them. Linear programming methods are algebraic techniques based on a series of equations or inequalities that limit… Improve your math knowledge with free questions in "Linear programming" and thousands of other math skills. The constraints can be expressed by linear equations and inequalities involving only the decision variables. ) at the optimal solution. Set To = Min, 3. + a 1n x n = b 1 a 21 x 1 + a 22 x 2 +. The columns of the final tableau have variable tags. Kostoglou. subject to . If you have only Since there are only two variables, we can solve this problem by graphing the set We now present examples of four general linear programming problems. So 3 x 1 2 2 10 is a linear constrain t, as is + 3 =6 Linear Programming, Part II – Slack and Surplus Linear Programming Now our matrices will of course have to change a little bit to take these new variables F. Define the decision variables. Best wishes. For example, if Y is also unrestricted variable, the substitute - Y for Y. Modelling Linear Programming INDR 262 Optimization Models and Mathematical Programming LINEAR PROGRAMMING MODELS Common terminology for linear programming: - linear programming models involve . Example 2 (Alternate optimal solutions). The variable s 2 is called a surplus variable Divisibility assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the func- tional and nonnegativity constraints. 2 Linear Programming Geometric Approach 5. linear programming 2 variables examples
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http://physics.qandaexchange.com/?qa=1631/maximum-height | # Maximum height
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$H=(u^2 sin^2\theta )/(2g)$
$h-H =(1/2)gt^2 = (u^2sin^2 \theta)/(2g)$
From this I got h=2H
1 vote
If the particles are launched at the same time, they must have the same horizontal speed in order to collide. (However, the problem seems to state that the particles have the same speed, ie same magnitude of velocity, and it is not clear that they are launched at the same time.)
Because they have the same horizontal speed, they remain aligned vertically. We can ignore horizontal motion and treat this as motion in 1D vertically.
The particles must collide before they reach the ground. The maximum $h$ occurs when they collide at the ground. ie They reach the ground at the same time, and have the same time of flight $t$.
When the upper particle reaches the ground $h=\frac12gt^2$. When the lower particle falls from its maximum height it takes time $\frac12 t$ so
$H=\frac12g(\frac12 t)^2=\frac14(\frac12gt^2)=\frac14h$
$h=4H$.
answered Mar 22, 2017 by (28,876 points)
selected May 10, 2017 by koolman
Speed of paricle 1 be u at an angle $\theta$ from ground .Then the speed of particle 2 is $u\cos \theta$ in horizontal direction.
let at any h' from ground both partucles collide .
For particle 1 $$h'= u\sin\theta t -(1/2)gt^2$$
For particle 2 $$h-h '= (1/2) gt^2$$
Hence $$t=\sqrt {\frac{2(h-h')}{g}}$$
Also we know $H=u^2 sin\theta ^2 /(2g)$
Substituting these values in equation of particle 1
We geta quadratic equation $$h^2 -4Hh+4Hh'=0$$
solving this $$h=\frac{4H(+/-)\sqrt{16H^2 -16Hh'}}{2}$$
Now h will be maximum when h'=0 that is they collide when both of them just going to strike the ground .
$$h=\frac{4H+4H}{2}$$
$$h=4H$$
answered Mar 22, 2017 by (4,286 points)
Yes this is correct but not the simplest way of doing the calculation. The key assumption is that the particles have the same *horizontal component* of velocity. This is not what the question seems to say. The wording of the question is very confusing.
The question hasn't mentioned anything about speed , so we can take horizontal speed same .
The question says "second particle is thrown horizontally with same speed". To me that means the particles have the same "speed" ie the same magnitude of velocity.
Oh Sorry , the wording of question is confusing.
1 vote
Your Equation considers the particles to collide only at the point where first particle attains its maximum height. But the question considers the general case.
Since they are thrown from the same vertical plane with the same speed, they can collide only if the horizontal component of their velocities are equal. Here the second particle would have covered a large distance when they are at the same height.
If the first particle is necessarily thrown obliquely then the particles cannot collide I assume.
answered Mar 21, 2017 by (340 points)
Have a look at my answer .
1 vote
Discussion
DoubtExpert correctly observes that the particles do not necessarily collide when the particle thrown obliquely reaches its maximum height. However, he assumes that the particles are launched simultaneously. I think this is not required by the wording of the question.
The key phrase is thrown to strike at same time. This is ambiguous. If this means (as I think it does) that the particles collide at the same time then it is superfluous, because they must be in the same place at the same time in order to collide. I do not see how it can be interpreted as launched at the same time. If it does mean this then DoubtExpert is correct : there can be no collision.
Solution
We must assume that the particles can be launched independently, at any time. The only requirement is that the trajectories overlap - or at least touch tangentially - at some point in space. If this happens then we can always adjust the launch times to ensure that the particles collide at this point.
What we need to do is write equations for the two trajectories then find a condition for them to touch. The trajectories will touch if (i) they intersect at some point, and (ii) their tangents are equal at this point.
Suppose the speed of each particle is $u$ and the launch angle of the lower particle is $\theta$. Writing $T=\tan\theta$ the trajectories of the upper and lower particles are respectively
$y=h-\frac{g}{2u^2}x^2$
$y=xT-\frac{g(1+T^2)}{2u^2}x^2$.
The tangents have slope
$\frac{dy}{dx}=-\frac{g}{u^2}x$
$\frac{dy}{dx}=T-\frac{g(1+T^2)}{u^2}x$.
The 1st two equations are equal when
$h=xT-\frac{gT^2}{2u^2}x^2$.
The 2nd two equations are equal when
$T=\frac{gT^2}{u^2}x$
$xT=\frac{u^2}{g}$.
Combine these two results :
$h=\frac{u^2}{2g}$.
The height of the collision point is
$y=h-\frac{g}{2u^2}x^2=\frac{u^2}{2g}-\frac{u^2}{2gT^2}=h(1-\frac{1}{T^2})$.
This must be above ground $(y \ge 0)$ so
$T^2 \ge 1$
$\theta \ge 45^{\circ}$.
The maximum height reached by the lower particle is $H=\frac{u^2\sin^2\theta}{2g}$. So we have $\frac{H}{h}=\sin^2\theta$. The value of $h$ calculated here is the maximum possible - any larger and the trajectories would not intersect. So the criterion is
$h \ge \frac{H}{\sin^2\theta}$.
I do not think we can make any further progress without knowing the launch angle $\theta$. In order to get $h = 4H$ we require $\theta=30^{\circ}$. However, this is not $\ge 45^{\circ}$ : the collision in this case takes place below ground. So a collision is not possible when $h =4H$.
answered Mar 21, 2017 by (28,876 points)
edited Mar 21, 2017
It is given in the question that both particles are thrown at same time .
If the particles are launched with the same speed at the same time, then I agree with DoubtExpert : the particles cannot collide.
Is there a worked solution? Or a diagram?
Hmm, true because horizontal distance being same $ut = u cos \theta t$
and $\theta = 0$ so this may not cause collision or because even the speed given are same.
The solution has bad explanation .
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https://madhavamathcompetition.com/category/uncategorized/page/2/ | # Check your mathematical induction concepts
Discuss the following “proof” of the (false) theorem:
If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:
PROOF BY INDUCTION:
Step 1:
If $n=1$, the result is evident.
Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal:
$a_{1}=a_{2}= \ldots = a_{k}$.
If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:
$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.
It follows easily that all $k+1$ numbers in S are equal.
*************************************************************************************
Comments, observations are welcome 🙂
Regards,
Nalin Pithwa
# Observations are important: Pre RMO and RMO : algebra
We know the following facts very well:
$(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$
$(x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}=()()$
But, you can quickly verify that:
$x^{3}+2x^{2}y+2xy^{2}+y^{3}=(x+y)(x^{2}+xy+y^{2})$
$x^{3}-2x^{2}y+2xy^{2}-y^{3}=(x-y)(x^{2}-xy-y^{2})$
Whereas:
$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$
$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
I call it — simply stunning beauty of elementary algebra of factorizations and expansions
More later,
Nalin Pithwa
# Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set) III
Continuing this series of slightly vexing questions, we present below:
1. Prove the inequality $\frac{A+a+B+b}{A+a+B+b+c+r} + \frac{B+b+C+c}{B+b+C+c+a+r} > \frac{C+c+A+a}{C+c+A+a++b+r}$, where all the variables are positive numbers.
2. A sequence of numbers: Find a sequence of numbers $x_{0}$, $x_{1}$, $x_{2}, \ldots$ whose elements are positive and such that $a_{0}=1$ and $a_{n} - a_{n+1}=a_{n+2}$ for $n=0, 1, 2, \ldots$. Show that there is only one such sequence.
3. Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.
4. Examination of an angle: Let $x_{1}$, $x_{2}, \ldots, x_{n}$ be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment $OP_{1}=x_{1}$. Then, we draw a segment $P_{1}P_{2}=x_{2}$ perpendicular to $OP_{1}$ and next a segment $P_{2}P_{3}=x_{3}$ perpendicular to $OP_{2}$. We continue in this way up to $P_{n-1}P_{n}=x_{n}$. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points $P_{1}$, $P_{2}, \ldots, P_{n}$ (the final position being $P_{n}$). In doing so, it sweeps out a certain angle. Prove that for given numbers $x_{i}$, this angle is smallest when the numbers $x_{i}$, that is, $x_{1} \geq x_{2} \geq \ldots x_{n}$ decrease; and the angle is largest when these numbers increase.
5. Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle $A = 60 \deg$ the area S of the triangle is given by the formula $S = \frac{\sqrt{3}}{4}(a^{2}-(b-c)^{2})$ and if $A = 120\deg$, then $S = \frac{\sqrt{3}}{12}(a^{2}-(b-c)^{2})$.
More later, cheers, hope you all enjoy. Partial attempts also deserve some credit.
Nalin Pithwa.
# A nice analysis question for RMO practice
Actually, this is a famous problem. But, I feel it is important to attempt on one’s own, proofs of famous questions within the scope of RMO and INMO mathematics. And, then compare one’s approach or whole proof with the one suggested by the author or teacher of RMO/INMO.
Problem:
How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?
Reference:
I will post it when I publish the solution lest it might affect your attempt at solving this enticing mathematics question !
Please do not try and get the solution from the internet.
Regards,
Nalin Pithwa.
# A miscellaneous algebra question for RMO or Pre-RMO
Question:
If a, b, c are three rational numbers, then prove that
$\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}}$
is always the square of a rational number.
Solution to be posted soon…
Cheers,
Nalin Pithwa. | 2020-05-25T14:58:35 | {
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http://mathhelpforum.com/calculus/26628-identifying-inflection-points.html | # Math Help - Identifying Inflection Points
1. ## Identifying Inflection Points
The x coordinates of the points of inflection of the graph of $y= x^5-5x^4=3x+7$ are...
A 0,
B 1,
C 3,
D 0 and 3,
E 0 and 1
$y''=20x^3-60x^2$
Simplify.
$y''=20x^2(x-3)$
So, thanks to the graph and knowledge of the definition of inflection point (y''= 0).
I choose D.
However, the book says C.
I'm inclined to believe it is right because I did the sign test for 0 for numbers less than 0 and less than 3, thought I may be incorrect of which function to put the test values into (y'' right?), and got the same sign (negative).
So, did I do the sign test wrong? Or is the right and I'm confusing myself with these questions.
What is the answer to the problem?
Thank you.
2. For a point of inflection, $y''=0$ is a necessary condition but not a sufficient one. $y''$ must also change sign before and after the point of inflection. Thus 0 is not a point of inflection because $y''$ does not change sign there; $y''$ is negative both slightly before and slightly after 0.
Alternatively …
$y'''=60x^2-120x\ne0$ when $x=3$. Since $y'''$ is an odd-order derivative, $x=3$ is a point of inflection. However $y'''(0)=0$ so we differentiate again.
$y''''=120x-120\ne0$ when $x=0$. $y''''$ is an even-order derivative, so $x=0$ is not a point of inflection. | 2015-11-30T10:14:37 | {
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https://math.stackexchange.com/questions/430310/a-peculiar-binomial-coefficient-identity | # A peculiar binomial coefficient identity
While inventing exercises for a discrete math text I'm writing I came up with this $$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$ It's an easy result to prove, but it got me wondering
1. Is this pure coincidence, or is it a special case of some more general result of which I'm unaware?
2. No matter what the answer to (1) is, is there a combinatorial proof?
• Looks purdy.${{{}}}$ – Git Gud Jun 26 '13 at 21:20
• Every polynomial $f(x)$ which takes integer values at the integers is an integer linear combination of the polynomials ${x \choose k}$, so in some sense this identity is not very surprising. The analogous statement for multivariate polynomials is also true, so there are also for example identities of the form ${x + y \choose k} = \sum a_{i, j} {x \choose i} {y \choose j}$ and ${xy \choose k} = \sum b_{i, j} {x \choose i} {y \choose j}$. The only mildly surprising thing is that in both of these cases the coefficients turn out to be non-negative as well as integers, but that's because the... – Qiaochu Yuan Jun 26 '13 at 21:57
• ...corresponding identities also have combinatorial proofs (exercise!). – Qiaochu Yuan Jun 26 '13 at 21:57
• – sdcvvc Jul 4 '13 at 23:00
This is actually a well known identity. There are combinatorial ways to prove it.
Consider $n$ objects. Consider all ${n \choose 2}$ pairs. Consider all pairs of these pairs. We get the LHS.
Consider $n$ objects and 1 distinguished object. Consider all sets of 4 objects from these.
If the 4 objects do not include the distinguished object, they correspond to 3 possible pairs of pairs, whose 4 elements are distinct. I.E. $(A,B), (C,D)$ and $(A, C), (B, D)$ and $(A, D), (B,C)$.
If the 4 objects include the distinguished object, they correspond to 3 possible pairs of pairs, which have a common element, and whose union is the 3 objects. I.E. $(A, B) , (A, C)$ and $(B,A), (B,C)$ and $(C,A), (C,B)$.
This gives us the RHS.
I'm not sure if they are generalizations, though you can experiment with choosing triples and counting carefully.
• I'm not surprised by (1) the fact that this is well-known and (2) that you knew it. Nice job, +1 – Rick Decker Jun 26 '13 at 21:28
The lefthand side is of course the number of ways to choose two unordered pairs, possibly with a one-element overlap, from the set $[n]=\{1,\dots,n\}$. Alternatively, we may choose a $4$-element subset $A$ of $\{0,1,\dots,n\}$ and a $k\in[3]$. Let $A=\{a_1,a_2,a_3,a_4\}$ with $a_1<a_2<a_3<a_4$. If $a_1\ne 0$, pair $a_1$ with $a_{k+1}$ and let the other two members of $A$ be the other pair. If $a_1=0$, form two pairs from $\{a_2,a_3,a_4\}$ by letting $a_k$ be the common member.
• My goodness. Two replies within 9 minutes. This place is the math equivalent of a fast food restaurant (though the offerings are unusually tasty). +1 – Rick Decker Jun 26 '13 at 21:33
• You mean "an unordered pair of unordered pairs." "Two" could mean "ordered pair." – Qiaochu Yuan Jun 26 '13 at 21:55
• @Qiaochu: I think that one would almost have to be deliberately trying to misunderstand in order to come up with that reading. – Brian M. Scott Jun 27 '13 at 1:30
You can count all of the pairs of pairs with $4$ distinct elements and all of the pairs of pairs with $3$ distinct elements and then add them together.
For the first sum get $\binom{n}{4}\binom{4}{2}\frac{1}{2}=3\binom{n}{4}$. We divide by two because once we choose the first pair out of the $4$ elements we implicitly chose the other pair; we are double counting otherwise.
For the second sum we get $3\binom{n}{3}$ since we must choose $3$ elements and then one of them to be duplicated.
In total we have $$3 \left[ \binom{n}{4} + \binom{n}{3}\right] = 3\binom{n+1}{4}$$.
So that's $2/3$ combinatorial at least.
• Nice and succinct. +1 – Rick Decker Jun 27 '13 at 1:02 | 2021-05-13T19:52:35 | {
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https://math.stackexchange.com/questions/2304142/how-many-squares-in-a-rectangle | # How many squares in a rectangle?
I almost wish I'd never thought of this problem... I was tearing my hair out over it all night.
Suppose we have a rectangle with side lengths $a$ and $b$, $a,b \in \mathbb Z$, $GCD(a,b)=1$, and $b \gt a$. I want to pack squares in this rectangle in such a way that, at each step, I pack a squares that is large as possible at that step, and that touches the square I packed at the last step. A square packing of this type might end up looking something like this:
My question is this: can I find a formula for the number of squares needed to pack a rectangle in terms of its side lengths $a$ and $b$?
Here is what I tried:
First I let $f(a,b)$ denote the function whose output is the number of squares required. First, I noted a few properties of $f(a,b)$: $$f(a,a)=1$$ $$f(ka,kb)=f(a,b)$$ $$f(a,a+1)=a+1$$ $$f(a,b)=\Big\lfloor \frac{a}{b} \Big\rfloor + f(b, a\mod b)$$ $$r \lt a \implies f(a,qa+r)=q+f(r,a)$$ Next I tried this: I decided to let $b=q_1a+r_1$ where $r_1 \lt a$. Then I would have $$f(a, q_1a+r_1)=q_1+f(r_1, a)$$ Then I decided to let $a=q_2r_1+r_2$, so that we would then have $$f(a, q_1a+r_1)=q_1+f(r_1, q_2r_1+r_2)$$ $$f(a, q_1a+r_1)=q_1+q_2+f(r_2, r_1)$$ and so on. I keep continuing this process until I hit some $n$ for which $r_n=0$, and I will have $$f(a, q_1a+r_1)=\sum_{i=1}^n q_i$$ ...and then I got stuck. How can I find each of the $q_i$s? Somebody, please help! The solution need not be closed-form, it just needs to be... something. I just have no idea where to go with this.
Thanks for any and all help!
• I think the Euclidean algorithm might be helpful here – abiessu May 31 '17 at 13:27
• I also do believe that the Euclidian algorithm is the clue here: imagine that a=10 and b=54, then you get following steps : 54=5x10+4, 10=2x4+2, 4=2x2. Now you take the sum of all coefficients: 5+2+2=9, which is, in my opinion, the number you're looking for. Isn't it? – Dominique May 31 '17 at 13:34
• @MCCCS Whoa there... how did you come up with that? Do you mind posting an answer and explaining it? – Franklin Pezzuti Dyer May 31 '17 at 13:35
• @Dominique Yes, that is correct. How do I find the sum of the coefficients, though? – Franklin Pezzuti Dyer May 31 '17 at 13:37
• @IvanNeretin: I don't think so neither: you should write the Euclidean algorithm in a form of iterative series (working with Qn for the quotients and Rn for the leftovers), and perform a Sum(Qn) at the end. – Dominique May 31 '17 at 13:45
## 2 Answers
$$f(a,b) = \begin{cases}1 & a= b\\b/a & (\gcd(a,b) = a) ∧ (a\neq b)\\ a/b & (\gcd(a,b) = b) ∧ (a\neq b) \\ f(b,a) & b>a\\ \frac{(a-(a \mod b))}{b} + f(b,(a\mod b)) &a>b\end{cases}$$
$$\gcd(x,y) =\begin{cases}x&x=y\\y & (x = 0)∧(x\neq y)\\x & (y=0)∧(x\neq y)\\\gcd(y,(x\mod y))&x>y\\\gcd((y\mod x),x)&y> x\end{cases}$$
(∧: AND logical operator)
$f(a,b)$ is the function that solves your question, using $\gcd(x,y)$, Euclid's GCD function.
Here's also the code written in Swift that does the same thing:
import Foundation
func ourFunction (_ a: Int, _ b: Int) -> Int{
if gcd(a,b) == a{
return b/a
}else if gcd(a,b) == b{
return a/b
}
if b>a{
// Force a to be bigger than b
return ourFunction(b,a)
}
return (a-(a%b)) / b /*big squares*/ + ourFunction(b,a%b)
}
func gcd(_ x:Int, _ y:Int) -> Int{
if x == 0 || x == y{
return y
}
if y == 0{
return x
}
if x > y{
return gcd(y,x%y)
}else{
return gcd(y%x,x)
}
}
print(ourFunction(8,3))
It can be run here, just change the $8$ and $3$ to the numbers you want to put into the function.
The answer by @MCCCS is great, but I provide an alternative answer with intent of being (slightly) more concise.
The problem can be solved recursively by embedding as many equally-sized maximal squares as possible within the rectangle and increasing a counter by the number of squares embedded. You'll end up with a smaller rectangle, for which you can embed another (horizontal or vertical) stack of equally-sized maximal squares as possible. Rinse repeat!
This recursive process can be described by the following recursive function:
$$f(a, b) = \begin{cases} 0 &\mbox{if } a =0 \lor b=0 \\ f(a, b-a\cdot\left \lfloor \frac{b}{a} \right \rfloor) + \left \lfloor \frac{b}{a} \right \rfloor & \mbox{if } a<b \\ f(a-b \cdot \left \lfloor \frac{a}{b} \right \rfloor, b) + \left \lfloor \frac{a}{b} \right \rfloor & \text{otherwise} \end{cases}$$
whose domain is $\{\forall(a,b)\in \mathbb{R}^2:\gcd(a,b)=1\}$.
Here is a C-program implementing the function described above:
#include <stdio.h>
int f(int w, int h) {
return w == 0 || h == 0 ? 0 :
w < h ? f(w, h-w*(h/w)) + h/w :
f(w-h*(w/h), h) + w/h;
}
int main(int argc, char** argv) {
int i = 3, j = 7;
printf("f(%i, %i)=%i\n", i, j, f(i, j));
return 0;
} | 2021-02-27T07:42:26 | {
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https://math.stackexchange.com/questions/1382570/name-of-xpyp-le-xyp-p-ge-1 | # Name of $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$)?
I checked these
What is the difference between square of sum and sum of square?
Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.
It is easy to see $p$-th power ($p\ge 1$) version, i.e., $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$), holds as well using the argument by Quang Hoang in Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$. Is there a name for this inequality so that I can just quote? (It might be an elementary result but people around me bother to put "from Cauchy--Schwarz,..." when it is clearly Cauchy--Schwarz, so.)
• You can say that the function $x\mapsto x^p$ on $[0,\infty)$ is subadditive if $p\in(0,1]$ and superadditive if $p\geq1$. – triple_sec Aug 3 '15 at 1:39
• I've sometimes heard, "by comparison of $\ell_p$ norms". (This is the two-dimensional case of $\ell_p$ vs $\ell_1$.) – user21467 Aug 3 '15 at 1:43
• In probability it's called $c_r$ inequality when you put expectations on both sides... – d.k.o. Aug 3 '15 at 3:20
• @d.k.o. As in www2.cirano.qc.ca/~dufourj/Web_Site/ResE/…? I am not sure if I am looking at a right source but the constant and the direction look a bit different though. – shall.i.am Aug 3 '15 at 10:36
I do not believe there is a name for your specific inequality (which I rewrite as following): $$|x|^p+|y|^p\le \big(|x|+|y|\big)^p, \ p \ge1 \iff \boxed{\ \ \left(|x|^p +|y|^p\right)^{\frac{1}{p}} \le |x|+|y|, \quad p \ge 1 \ \ }$$ However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, Hölder, and probably many other inequalities after appropriate substitution and/or change of variables. The closes call would probably be the generalized mean inequality: $$M_j\left( x_1, \dots, x_n \right) \leq M_i\left( x_1, \dots, x_n \right) \quad \text{ whenever } \quad j<i. \label{*} \tag{*}$$
Here $M_k \left( x_1, \dots, x_n \right)$ is so-called power mean, which is defined as $$M_k(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^k \right)^{\frac{1}{k}}.$$
! In particular, assuming $n=2$, $j = 1$, and $i = p$, and denoting $\left(x_1, \dots, x_n \right) := \left(\,\chi, \gamma \right)$, we get \begin{aligned} M_1\big(\left|\,\chi\right|, \left|\gamma\right|\big) & = \dfrac{1 }{2}\big(\left|\,\chi\right| + \left|\gamma\right|\big) = \dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2}, \\ M_p\big(\left|\,\chi\right|,\left|\gamma\right|\big) & = \left( \dfrac{1}{2} \left(\left|\,\chi\right|^p+\left|\gamma\right|^p\right)\right)^{\frac{1}{p}} = \Bigg( \left(\frac{\left|\,\chi\right|}{2}\right)^p + \left(\frac{\left|\gamma\right|}{2}\right)^p \Bigg)^{\frac{1}{p}}.\\ \end{aligned} By $\eqref{*}$, we have $$\dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2} \le \left( \bigg(\frac{\left|\,\chi\right|}{2^{\frac{1}{p}}}\bigg)^p + \bigg(\frac{\left|\gamma\right|}{2^{\frac{1}{p}}}\bigg)^p \right)^{\frac{1}{p}} . \label{**} \tag{**}$$ Denoting $x := 2^{-\frac{1}{p}}\chi, \ \ y := 2^{-\frac{1}{p}}\gamma$ and raising both sides of $\eqref{**}$ to the power $p$, we get
$$\left|x\right|^{p}+\left|y\right|^{p} \le \big(\left|x\right|+\left|y\right|\big)^{p}.$$
To summarize, I believe that (strictly speaking) there is probably no name for your inequality. However, that the generalized mean is as close as you can get to your inequality, although some formula conversion is still required.
I think this should be called an special case of Minkowski's Inequality.
1 - For finite sequences, the Minkowski Inequality states that
$$\left(\sum_{k=1}^n |x_k + y_k |^p \right) ^{1/p} \le \left(\sum_{k=1}^n |x_k|^p \right)^{1/p} + \left(\sum_{k=1}^n |y_k|^p\right)^{1/p}$$
Your inequality then follows if $x_1 = x, x_i = 0, \; i \ge 2$, and $y_2 = y, y_i = 0, i \ne 2$.
2 - More Generally, the case above can be easily extended to infinite sequences. The most general result is that, if $f,g : \Omega \to \Bbb{R}$ are measurable functions in a measure space $(\Omega, \Sigma, \mu)$, then the inequality
$$\| f+g \|_{L^p (d\mu)} \le \|f\|_{L^p(d\mu)} + \|g\|_{L^p(d\mu)}$$
Where $\|f\|_{L^p(d\mu)}^p = \int_ {\Omega} |f(x)|^p d \mu (x)$
3 - There was a little lie at topic 2, because there is yet another generalization of this result, called Minkowski's Inequality for Integrals. In the link above, you might check a proof of this result, based on these previously presented ones.
• Ah. It does follow from Minkowski. You meant $y_i=0$ $i\neq 2$, right? Namely, $\boldsymbol{x}:=(x_1,0,\dotsc)$ and $\boldsymbol{y}:=(0,y_2,0,\dotsc)$. From Minkowski, $(|x_1|^p+|y_2|^p)^{\frac1p}\le|x_1|^{\frac1pp}+|y_2|^{\frac1pp}$ and thus $(|x_1|^p+|y_2|^p)\le(|x_1|+|y_2|)^p$. I did not realise this because if anything Minkowski looks like the opposite direction. I find your answer really interesting but just saying "this follows from Minkowski" sounds a bit misleading as it looks opposite, so, sorry! But really appreciate it! – shall.i.am Aug 3 '15 at 10:33
• Yeah, it really looks opposite, right? I was thinking about this inequality about two weeks ago, when I realized that it was just Minkowski's Inequality. Well, just wanted to help here :) Be free to ask more questions about this answer if you like :D – João Ramos Aug 3 '15 at 12:38 | 2019-10-15T09:58:06 | {
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http://roszak.eu/fcjo6keq/f34428-differential-equations-lectures | Then we learn analytical methods for solving separable and linear first-order odes. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. Lecture 1 Introduction to differential equations View this lecture on YouTube A differential equation is an equation for a function containing derivatives of that function. Differential Equations. These lectures treat the subject of PDEs by considering specific examples and studying them thoroughly. Lecture 51 : Differential Equations - Introduction; Lecture 52 : First Order Differential Equations; Lecture 53 : Exact Differential Equations; Lecture 54 : Exact Differential Equations (Cont.) Lecture 56: Higher Order Linear Differential Equations Download files for later. Definition (Differential equation) A differential equation (de) is an equation involving a function and its deriva- tives. Lecture one (27/9/2020 ) power point URL. » » The study of differential equations is such an extensive topic that even a brief survey of its methods and applications usually occupies a full course. Knowledge is your reward. This is one of over 2,400 courses on OCW. A differential equation always involves the derivative of one variable with respect to another. 140 0 obj Lectures on Differential Equations Craig A. Tracy PDF | 175 Pages | English. (The minus sign is necessary because heat flows “downhill” in temperature.) videos URL. On the human side Witold Hurewicz was an equally exceptional personality. )Z!����DR�7��HㄕA7�ۉ��S�i��vl燞�P%�[R�;���n\W��4/de^��2�e�D����B%8�ضm��[R���#��P2q\ �His�d�JJ. Date: 1st Jan 2021. Arnold's geometric point of view of differential equations is really intriguing. This note covers the following topics: First Order Equations and Conservative Systems, Second Order Linear Equations, Difference Equations, Matrix Differential Equations, Weighted String, Quantum Harmonic Oscillator, Heat Equation and Laplace Transform. You see that it is a proper vector equation. (1) If y1(t) and y2(t) satisfy (2.1), then for any two constants C1and C2, y(t) = C1y1(t)+C2y2(t) (2.2) is a solution also. Baker Hilary Term2016. Equation is the differential equation of heat conduction in bulk materials. We introduce differential equations and classify them. We don't offer credit or certification for using OCW. By this approach the author aims at presenting fundamental ideas in a clear way. This table (PDF) provides a correlation between the video and the lectures in the 2010 version of the course. Find all the books, read about the author, and more. What follows are my lecture notes for a first course in differential equations, taught at the Hong Kong University of Science and Technology. videos URL. Such equations are often used in the sciences to relate a quantity to its rate of change. Here is a set of notes used by Paul Dawkins to teach his Differential Equations course at Lamar University. The former is called a dependent variable and the latter an independent variable. Lecture 01 - Introduction to Ordinary Differential Equations (ODE) PDF unavailable: 2: Lecture 02 - Methods for First Order ODE's - Homogeneous Equations: PDF unavailable: 3: Lecture 03 - Methods for First order ODE's - Exact Equations: PDF unavailable: 4: Lecture 04 - Methods for First Order ODE's - Exact Equations ( Continued ) PDF unavailable: 5 137 0 obj When the input frequency is near a natural mode of the system, the amplitude is large. This playlist contains 32.5 hours across 52 video lectures from my Differential Equations course (Math 2080) at Clemson University. Much of the material of Chapters 2-6 and 8 has been adapted from the widely (Image courtesy Hu … Differential equations are called partial differential equations (pde) or or- dinary differential equations (ode) according to whether or not they contain partial derivatives. Differential Equation: An equation involving independent variable, dependent variable, derivatives of dependent variable with respect to independent variable and constant is called a differential equation. Verify that if c is a constant, then the function defined piecewise by (1.15) y(x) = 1 if x ≤ c cos(x−c) if c < x < c+π −1 if x ≥ c+π satisfies the differential equation y0= − p 1−y2for all x ∈ R. Determine how many different solutions (in terms of a and b) the initial value problem ( y0= − p 1−y2, y(a) = b has. A spring system responds to being shaken by oscillating. 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https://math.stackexchange.com/questions/4320849/finding-the-range-of-y-fracx22x42x24x9-and-y-frac-textquadrat/4320857 | # Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general)
I had this problem in an exam I recently appeared for:
Find the range of $$y =\frac{x^2+2x+4}{2x^2+4x+9}$$
By randomly assuming the value of $$x$$, I got the lower range of this expression as $$3/7$$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.
Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.
Another guy I met outside the exam center, told me he used an approach of $$x$$ tending to infinity in both cases and got the maximum value of this expression as $$1/2$$. But before I could ask him to explain more on this method, he had to leave for his work.
So, will someone please throw some light on this method of $$x$$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).
• Use of calculus to find abslute maxima and minima is the easier way. Dec 1, 2021 at 6:39
• I'll give a hint: An easy way to re-express the expression will be $\frac{1}{2}-\frac{1}{4\left(x+2\right)^{2}+14}$. Think of what circumstance would maximise the expression, which would be by minimizing the term subtracted. Dec 1, 2021 at 6:41
• @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. Dec 1, 2021 at 6:49
• @EmanatS Try my approach, doesn't need calculus Dec 1, 2021 at 6:50
• The hint of Prometheus is very good. It gives the result quickly. Dec 1, 2021 at 6:51
The question can be easily solved by this technique:
As $$\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$$.
Thus, $$\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7}$$
Squares can never be less than zero so the minimum value of the function : $$\displaystyle 2(x + 1)^2 + 7$$ would be $$7$$ , or Maximum value of $$\displaystyle \frac {1}{2(x + 1)^2 + 7}$$ is $$\displaystyle \frac {1}{7}$$.
This tells that minimum value of $$y$$ will be $$\displaystyle \frac{3}{7}$$.
And so on.. check for $$x \rightarrow \infty$$.
From here you can easily tell the maximum and minimum values : $$\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right )$$
• Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ Dec 1, 2021 at 6:54
• How did you split it just like that? Basically, how did you reduce that expression? Dec 1, 2021 at 6:55
• @Prometheus It's true... But I guess the test was a bit of a time-crusher, so perhaps the OP would like to get there in one step Dec 1, 2021 at 6:55
• @Spectre Yes, I had to solve this problem under 50 seconds. And Prometheus, yes I know the method of completing the square. I just do not understand how to apply it to this problem. Please explain this approach to get a one step form like I'm five. Dec 1, 2021 at 6:59
• @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. Dec 1, 2021 at 7:04
As a follow-up to @NikolaAlfredi's answer:
$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac{2x^2 + 4x + 8}{2(2x^2 + 4x + 9)} = \frac{2x^2+4x+9 - 1}{2(2x^2+4x+9)} = \frac{1}{2}(1-\frac{1}{2x^2+4x+9}) \implies 2y = 1 - \frac{1}{2x^2+4x+9}$$. Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by $$2$$ to get the required extremes(taking half since we get values for $$2y$$ and not $$y$$).
• Got it. I appreciate your kind efforts, and now it is easily clear to me. Thanks to both you, and Nikola. Also, can you please delve on calculus based approach like someone mentioned? I want to learn how to work this out with calculus too. Dec 1, 2021 at 7:08
• @EmanatS Are you looking for an answer using "calculus" methods? Dec 1, 2021 at 7:09
• @TeresaLisbon Not really. But I am open to learning through it too. Dec 1, 2021 at 7:09
In general, if $$\deg f = 0$$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $$f$$ as $$x$$ increases/decreases without bound is $$a_n/b_n$$.
In your case, $$a_2 = 1$$ and $$b_2 = 2$$. Hence, $$a_2/b_2 = 1/2$$.
We'll factor $$f$$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$
Notice that for all $$x \in \mathbb{R}$$, $$f > 0$$. Also, we can see that $$(x+1)^2 + 2 < 2(x + 1)^2 + 7$$. This means that the range should be a part of $$(0,1/2)$$. Since both numerator and denominator have $$(x + 1)^2$$ without any remaining $$x$$'s, we can see that this will be at its minimum when $$x = -1$$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$
Therefore, the range is $$[3/7, 1/2)$$.
• Note that both the numerator and denominator of $f$ must be polynomials with the same degree. This means that the expression could be linear/linear, quadratic/quadratic, cubic/cubic, and so on. Dec 1, 2021 at 7:12
• Woah, you're awesome. And so is @NikolaAlfredi I love this forum now, you both explained this problem so beautifully to me putting in all efforts to explain it to me. Thank you for a generalized solution to it as well. So, "limit of f as x increases/decreases without bound is an/bn" is always an upper limit of the expression right? Any places where I need to watch out this rule for? Dec 1, 2021 at 7:17
• Also for, (x^2+2x+1)/(4x^2-7x+9), a2/b2 will be {1/4}, but the upper limit of expression is 16/19. Shouldn't it be 1/4 according to your assertion? Edited Dec 1, 2021 at 7:21
• No. This is not always true. Consider the simple case $\frac{x}{x + 1}$. You'll see that it attains all real values except $y = 1$, and it is defined for all real $x$ except $x = -1$. In general, the limit is not the upper bound. It is only for the case where $p(x) < q(x)$ and $\deg p = \deg q$. ($p$ and $q$ are the polynomials in $f$, that is, $f(x) = p(x)/q(x)$). Dec 1, 2021 at 7:23
$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$ | 2022-07-04T22:19:57 | {
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https://math.stackexchange.com/questions/3546233/given-x-mathbbr2-and-the-equivalence-relation-x-y-simx-y-iff-y-x/3546363 | # Given $X=\mathbb{R}^2$ and the equivalence relation $(x,y)\sim(x',y')$ iff $y-x^2=y'-x'^2$. What space is this homeomorphic to?
Given $$X=\mathbb{R}^2$$ and the equivalence relation $$(x,y)\sim(x',y')$$ iff $$y-x^2=y'-x'^2$$. What space is this homeomorphic to?
There is also a hint that $$g(x,y)=y-x^2$$
So I tried drawing the equivalence relations on $$\mathbb{R}^2$$ and noticed that the equivalence relations are parabolas. The $$[0]$$ is $$y=x^2$$, then $$[1]$$ is $$y=x^2+1$$, etc. So I believe $$[c]$$ is $$y=x^2+c$$
I want to show this space is homeomorphic to $$\mathbb{R}$$ and I have the map $$g:\mathbb{R}^2\to \mathbb{R}$$. which is going to map equivalence classes to that $$c$$ value. I believe I need to find $$f$$ such that $$f\circ p=g$$. where $$p$$ is the quotient map. Then show it is continuous in both directions and bijective.
Can I take $$f:\mathbb{R}^2\setminus \sim\to \mathbb{R}$$ as $$f([(x,y)])=g(x,y)$$?
I know it's surjective since $$g=f\circ p$$ is surjective
Since if $$r\in \mathbb{R}$$ then $$g(0,r)=r$$.
And it's clearly injective, by the equivalence relation.
But I'm not sure how to show it is continuous.
I have this theorem which I believe might be useful:
Let $$X,Z$$ be two topological spaces, $$\sim$$ an equivalence relation on $$X$$ and $$f:X\setminus \sim \to Z$$ a function. Then $$f$$ is continuous if and only if $$f\circ p$$ is continuous.
So I could show $$g$$ is continuous and get that $$f$$ is continuous.
So let $$(a,b)$$ be an open interval in $$\mathbb{R}$$.
then I want to show $$g^{-1}[(a,b)]$$ is open.
I believe it looks like a big open parabola, a union of $$U_c=\{(x,y)\in\mathbb{R}^2: y=x^2+c, a
So I want to show that $$\bigcup U_c$$ is open in $$\mathbb{R}^2$$
Let $$p\in \bigcup U_c$$ then $$p\in U_c$$ for some $$c$$.
But I'm having trouble finding a radius of a ball which will be contained in this union. I want to pick something so that all the points in the ball will be less then the parabola $$y=x^2+b$$ and greater then $$y=x^2+a$$.
• Why not $f([(x,y)])=g(x,y)$? – Hagen von Eitzen Feb 14 at 7:32
• @Hagen von Eitzen Yeah that makes more sense. Since $[c]$ isn't actually what the equivalence classes should look like. – AColoredReptile Feb 14 at 7:36
• If $g\colon X\to Y$ is continuous ad $x\sim y\iff g(x)=g(y)$, under what mild conditions does $X/{\sim}\cong Y$ follow? – Hagen von Eitzen Feb 14 at 7:49
• @Hagen von Eitzen Does $g$ need to be continuous? – AColoredReptile Feb 14 at 8:17
## 1 Answer
It is more or less obvious that $$g$$ is continuous. It is well-known (and easy to verify) that the following is true:
Let $$f, g : X \to \mathbb R$$ be continuous maps defined on a topological space $$X$$ and $$a \in \mathbb R$$. Then $$f + g, f \cdot g$$ and $$a \cdot f$$ are continuous.
The projections $$p_1. p_2 : \mathbb R^2 \to \mathbb R, p_1(x,y) = x, p_2(x,y) = y$$, are clearly continuous. Thus $$g(x,y) = p_2(x,y) - p_1(x,y) \cdot p_1(x,y)$$ is continuous.
Alternatively you can also consider sequences $$(x_n,y_n)$$ converging to some $$(x,y)$$ and show that $$(g(x_n,y_n))$$ converges to $$g(x,y)$$.
Let $$\pi : \mathbb R^2 \to Y = \mathbb R^2/\sim$$ be the quotient map. Define $$j : \mathbb R \to \mathbb R^2, j(t) = (0,t)$$. This is a continuous map, hence $$J = \pi \circ j : \mathbb R \to Y$$ is continuous. We have $$f(J(t)) = f([0,t]) = g(0,t) = t,$$ $$J(f([x,y]) = J(g(x,y) = \pi(0,y - x^2) =[0,y-x^2] = [x,y]$$ because $$(y-x^2) - 0^2 = y - x^2$$. This shows that $$f$$ and $$J$$ are inverse to each other. This means that $$f,J$$ are homeomorphisms such that $$f^{-1} = J$$.
• So if $g$, I get $f$ is continuous. But then I need to show $f^{-1}$ is continuous, right? – AColoredReptile Feb 14 at 10:39
• Yes, I have done this in my answer (which contained a typo that I corrected). We have $f^{-1} = J$. – Paul Frost Feb 14 at 10:50 | 2020-03-29T10:24:32 | {
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http://mathhelpforum.com/calculus/49549-integral-partial-fractions.html | Math Help - Integral - partial fractions?
1. Integral - partial fractions?
$\int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$
$u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$
$=\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$=3\int\frac{du}{u^2(2u+1)(2u-3)}$
Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
2. Originally Posted by symstar
$\int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$
$u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$
$=\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$=3\int\frac{du}{u^2(2u+1)(2u-3)}$
Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
you are ok so far
so you must be making a mistake with the partial fractions part. what did you get?
3. Originally Posted by symstar
$\int\frac{du}{u^2(2u+1)(2u-3)}$
Put $u=\frac1z$ and your integral becomes $-\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence
\begin{aligned}
-\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\
\end{aligned}
Things should be easy from there.
4. Originally Posted by Krizalid
Put $u=\frac1z$ and your integral becomes $-\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence
\begin{aligned}
-\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\
\end{aligned}
Things should be easy from there.
i know what you're thinking, symstar, "how does he do it?!"
5. So working with the partial fractions:
$\frac{1}{u^2(2u+1)(2u-3)}=\frac{A}{u^2}+\frac{B}{(2u+1)}+\frac{C}{(2u-3}$
$1=A(4u^2-4u-3)+B(2u^3-3u^2)+C(2u^3+u^2)$
$2B+2C=0$
$4A-3B+C=0$
$-4A=0$
$-3A=1$
You can see the 2 values for A at the bottom. What should I do?
6. Hello, symstar!
I don't agree . . .
$\int \frac{dx}{4x^{\frac{2}{3}}-4x^{\frac{1}{3}}-3}$
Let: $u \:=\:x^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:u^3\quad\Rightarrow\quad dx \:=\:3u^2\,du$
Substitute: . $\int \frac{3u^2\,du}{4u^2 - 4u - 3} \;=\;\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$
. . $= \;\frac{3}{4}\int\left[1 + \frac{4u+3}{4u^2-4u-3}\right]\,du \;=\;\frac{3}{4}\int\left[1 + \frac{4u+3}{(2u-3)(2u+1)}\right]\,du$
Partial Fractions: . $\frac{3}{4}\int\left[1 + \frac{\frac{9}{4}}{2u-3} - \frac{\frac{1}{4}}{2u+1}\right]\,du \quad\hdots\text{ etc.}$
7. $\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$
I'm not quite sure what exactly happened in this step, could you explain please? | 2014-09-02T12:04:25 | {
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https://www.physicsforums.com/threads/what-does-ceil-mean.116822/ | # What does ceil mean ?
#### momentum
Its diifcult to express my question....so, i am posting this
ceil(4.5) =?
ceil(4.1)=?
ceil(4.6)=?
#### AKG
Homework Helper
ceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.
#### momentum
ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?
#### Integral
Staff Emeritus
Gold Member
momentum said:
ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?
Nope, read the defintion given above, then try again.
#### momentum
ah...i see, all of them should be 5 .....i had confusion on fractional part .5.
but i see ..it does not care for .5 which we use for round-off.
Homework Helper
Yes, all 5.
#### momentum
thank you for the clarifcation
#### bomba923
Recall that
$$\begin{gathered} \forall x \in \left( {a,a + 1} \right)\;{\text{where }}a \in \mathbb{Z}, \hfill \\ {\text{floor}}\left( x \right) = \left\lfloor x \right\rfloor = a \hfill \\ {\text{ceil}}\left( x \right) = \left\lceil x \right\rceil = a + 1 \hfill \\ \end{gathered}$$
$$\forall x \in \mathbb{Z},\;\left\lfloor x \right\rfloor = \left\lceil x \right\rceil = x$$
Last edited:
#### jim mcnamara
Mentor
ceil -> "goes up" if it needs to, in order reach an integer
floor -> "goes down" as it needs to, in order to reach an integer
What happens with negative numbers:
$$floor( -1.1 ) = -2 \; ceil( -1.1 ) = -1$$
$$floor( -0.1 ) = -1 \; ceil( -0.1 ) = 0$$
$$floor( 0.9 ) = 0 \; ceil( 0.9 ) = 1$$
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• Solo and co-op problem solving | 2019-03-25T09:57:32 | {
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https://brilliant.org/discussions/thread/trigonometry-practice-for-beginners/ | # Trigonometry practice for beginners
The following problems need only trigonometric definition of functions and the equation
$\sin^2{\theta} + \cos^2{\theta} = 1$
unless otherwise stated. More will be written.
Prove the following:
1. $$\csc{\theta} \cdot \cos{\theta} = \cot{\theta}$$
2. $$\csc{\theta}\cdot\tan{\theta}=\sec{\theta}$$
3. $$1+\tan^2(-\theta) = \sec^2{\theta}$$
4. $$\cos{\theta}(\tan{\theta}+\cot{\theta})=\csc{\theta}$$
5. $$\sin{\theta}(\tan{\theta}+\cot{\theta})=\sec{\theta}$$
6. $$\tan{\theta}\cdot\cot{\theta}-\cos^2{\theta}=\sin^2{\theta}$$
7. $$\sin{\theta}\cdot\csc{\theta}-\cos^2{\theta}=\sin^2{\theta}$$
8. $$(\sec{\theta}-1)(\sec{\theta}+1)=\tan^2{\theta}$$
9. $$(\csc{\theta}-1)(\csc{\theta}+1)=\cot^2{\theta}$$
10. $$(\sec{\theta}-\tan{\theta})(\sec{\theta}+\tan{\theta})=1$$
11. $$\sin^2{\theta}(1+\cot^2{\theta})=1$$
12. $$(1-\sin^2{\theta})(1+\tan{\theta})=1$$
13. $$(\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})=2$$
14. $$\tan^2{\theta}\cdot\cos^2{\theta}+\cot^2{\theta}\cdot\sin^2{\theta}=1$$
15. $$\sec^4{\theta}-\sec^2{\theta}=\tan^4{\theta}+\tan^2{\theta}$$
16. $$\sec{\theta}-\tan{\theta}=\dfrac{\cos{\theta}}{1+\sin{\theta}}$$
17. $$\csc{\theta}-\cot{\theta}=\dfrac{\sin{\theta}}{1+\sin{\theta}}$$
18. $$3\sin^2{\theta}+4\cos^2{\theta}=3+\cos^2{\theta}$$
19. $$9\sec^2{\theta}-5\tan^2{\theta}=5+4\sec^2{\theta}$$
20. $$1-\dfrac{\cos^2{\theta}}{1+\sin{\theta}}=\sin{\theta}$$
21. $$1-\dfrac{\sin^2{\theta}}{1-\cos{\theta}}=-\cos{\theta}$$
22. $$\dfrac{1+\tan{\theta}}{1-\tan{\theta}}=\dfrac{\cot{\theta}+1}{\cot{\theta}-1}$$
23. $$\dfrac{\sec{\theta}}{\csc{\theta}}+\dfrac{\sin{\theta}}{\cos{\theta}}=2\tan{\theta}$$
24. $$\dfrac{\csc{\theta}-1}{\cot{\theta}}=\dfrac{\cot{\theta}}{\csc{\theta}+1}$$
25. $$\dfrac{1+\sin{\theta}}{1-\sin{\theta}}=\dfrac{\csc{\theta}+1}{\csc{\theta}-1}$$
26. $$\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}$$
27. $$\dfrac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}=\dfrac{1}{1-\cot{\theta}}$$
28. $$1-\dfrac{\sin^2{\theta}}{1+\cos{\theta}}=\cos{\theta}$$
29. $$\dfrac{1-\sin{\theta}}{1+\sin{\theta}}=(\sec{\theta}-\tan{\theta})^2$$
30. $$\dfrac{\cos{\theta}}{1-\tan{\theta}}+\dfrac{\sin{\theta}}{1-\cot{\theta}}=\sin{\theta}+\cos{\theta}$$
31. $$\dfrac{\cot{\theta}}{1-\tan{\theta}}+\dfrac{\tan{\theta}}{1-\cot{\theta}}=1+\tan{\theta}+\cot{\theta}$$
32. $$\tan{\theta}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=\sec{\theta}$$
Note by Sharky Kesa
2 years, 11 months ago
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Sort by:
Here's a nice result which involves all six ratios :
$(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)=\sec \theta + \csc \theta$
- 2 years, 11 months ago
Can you hear the Proving Trigonometric Identities Wiki page calling out your name?
Staff - 2 years, 11 months ago
Thank You, it is a good practice for we beginners!
- 2 years, 11 months ago
yes
- 1 month, 2 weeks ago | 2018-04-26T15:39:08 | {
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https://stats.stackexchange.com/questions/403413/does-data-normalization-and-transformation-change-the-pearsons-correlation | # Does data normalization and transformation change the Pearson's correlation?
As we know that Pearson's correlation measures the linearity between two variables, I am wondering when applying normalization and transformation on the original dataset, does the normalization and transformation method needs to be a linear method, in order to not effect the correlation results?
More specifically, for example, after log-transformation, does the correlation change? I think so because log-transformation changed the distribution of the data and the original linear relationship is scaled by the log() function.
To extend the question, I am wondering if anyone could provide some general summary and comments on what normalization/transformation can be used when you want to scale the data but do not want to change the original relationship among your variables described by correlation metric?
Thanks!
You are mostly welcome to refine and shape my question if it's not accurate.
• Take 3 non-collinear but correlated and positive x-y pairs of values (e.g. x=1,2,4, y=2,5,4) and take their logs. Compute the Pearson correlation before and after. What happens? Try other examples. $\:$ (I imagine this small step of trying some examples would come under the 'research' part of the usual search and research. ) – Glen_b Apr 17 at 1:03
• Thanks for the suggestion! I think my question needs to be modified. What I am wondering is conceptual correctness rather than numerical difference. I am thinking can we use the correlation metric of a log transoformed data (say log(y)~log(x) ) to estimate and conclude on the original relationship or linearity of y~x. I will think about it and edit the question! – Lingjue Wang Jun 7 at 20:16
• The numerical difference actually tells you the answer to the conceptual question – Glen_b Jun 8 at 3:08
Pearson's correlation measures the linear component of association. So you are correct that linear transformations of data will not affect the correlation between them. However, nonlinear transformations will generally have an effect.
Here is a demonstration: Generate right-skewed, correlated data vectors x and y. Pearson's correlation is $$r = 0.987.$$ (The correlation of $$X$$ and $$Y^\prime = 3 + 5Y$$ is the same.)
set.seed(2019)
x = rexp(100, .1); y = x + rexp(100, .5)
cor(x, y)
[1] 0.987216
cor(x, 3 + 5*y)
[1] 0.987216 # no change with linear transf of 'y'
However, if the second variable is log-transformed, Pearson's correlation changes to $$r = 0.862.$$
cor(x, log(y))
[1] 0.8624539
Here are the corresponding plots:
By contrast, Spearman's correlation is unaffected by the (monotone increasing) log-transformation. Spearman's correlation is based on ranks of observations and log-transformation does not change ranks. Before and after transformation, $$r_S = 0.966.$$
cor(x, y, meth="spear")
[1] 0.9655446
cor(rank(x), rank(log(y)))
[1] 0.9655446 # Spearman again
cor(x, log(y), meth="spear")
[1] 0.9655446 | 2019-11-13T12:58:37 | {
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http://csfn.eu/beach-flowering-yiyriza/e539f3-angles-in-a-triangle-add-up-to | Step 1: Find the given angles Step 2: Subtract the given angles from 180 to get the missing angle. So a + b + y = 180. In other words: Angle 1 + Angle 2 + Angle 3 = 180° Missing Angles Our learning objectives today To calculate missing angles on a straight line, around a point and in a triangle. We first add the two 50° angles together. What do you need to know? Angles of Triangles Add Up To 180 Degrees Ex: Could a triangle have the given angle measures? a) Sum of total angles in a triangle is 180. b) Opposite angles are always equal. Add up the two known angle measurements. To calculate the other angles we need the sine, cosine and tangent. 4 Short Steps for answering Exterior Angle of a Triangle Example Problems. Therefore y = 180 - x. In a right triangle, one of the angles is exactly 90°. Consider the lunes through B and B'. The three angles in a triangle add up to 180 degrees. For example, in triangle ABC, angle A + angle B + angle C = 180°. She notices by chance that one triangle’s angles add up to 180. What you have now is two right triangles which, by the above contain 360 degrees in total. But if you look at the two right angles that add up to 180 degrees so the other angles, the angles of the original triangle, add up to 360 - 180 = 180 degrees. The interior angles of a triangle add up to 180 degrees. Consider a spherical triangle ABC on the unit sphere with angles A, B and C. Then the area of triangle ABC is. The angles in the triangle add up to 180 degrees. The degree measure of a straight line add up to 180 degrees. Name Date ANGLES IN A TRIANGLE 1 Work out the missing angles. The diagram below shows the interior and exterior angles of a triangle.. A demonstration of the angles of a triangle summing up to 180° can be found here. All 3 angles of a triangle add up to 180 degrees. Putting this into the first equation gives us: a + b + 180 - x = 180. So x + y = 180. The angles on a straight line add up to 180 degrees. So, if you know two of the three measurements of the triangle, then you're only missing one piece of the puzzle. … In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the angle in the bottom right corner to make a … Remember that the angles in a triangle add up to 180°. Why is every triangle 180 degrees? Each corner that you cut off contains an angle from the triangle. Penny Find the measure of the smallest angle (in degrees) Interior angles of a triangle add up to 180 degrees so 180 - 125 = 55 degrees.....the measure of the 3rd angle Every triangle has six exterior angles (two at each vertex are equal in measure). To find angle ‘b’, we subtract both 50° angles from 180°. The angles inside a triangle are called interior angles. The angle sum property of a triangle states that the angles of a triangle always add up to 180°. Yes, all triangle angles add up to 180 degrees. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. Such an angle is called a right angle. The first thing you can do is add up the angle measurements you know. This is why we coloured the edges so we can easily see the angle contained by the edges. Angles A and E are congruent angles, which means they have the same measure, because they are alternate interior angles of a transversal with parallel lines. c) Angle in a straight line adds up to 180. r is equal to the opposite angle to left over angle measure which is 180 - (74 + 60) = 180 - 134 = 46. q and r forming a straight line where q = 180 - r = 180 - 46 = 134. o is the opposite angel which measures 73, o -= 73 Draw a perpendicular from the vertex opposite the longest side, to the longest side. If you are given the adjacent angle you subtract the angle from 180 degrees to solve for the exterior angle of a triangle. Angle C and Angle F … Step by step descriptive logic to check whether a triangle can be formed or not, if angles are given. The sum of the measures of the interior angles of a triangle in Euclidean space is always 180 degrees. The definition of a triangle is that it is a shape with three sides that add up to 180 degrees. and 180 degrees is pi radians, so wouldnt it be (pi) - 0.2 - 0.2? When we assemble the angles (by aligning the coloured edges), we see that all the angles add up to a straight line (or 180°). The sum of the lengths of any two sides of a triangle always exceeds the length of the third side, a principle known as the triangle inequality. The angles on a straight line add up to 180°. The sides adjacent to the right angle are the legs. Now we can establish that the three angles inside the triangle (B, E & F) also add up to 180. according to this solutionbank [the answer is the same in the book], the angle for the shaded section from C is 0.4 meaning the angles in the triangle are 0.2 0.2 and 0.6, i thought the angles add up to 180? Every triangle has three angles and whether it is an acute, obtuse, or right triangle, the angles sum to 180°. We can use the property that the angles of a triangle add up to 180 degrees. Logic to check triangle validity if angles are given. All you have to know is that all of the angles in a triangle always add up to 180°. The third angle is 4 times as big as the smallest angle. Proving that the angles inside a triangle – any triangle – sum up to 180° is very simple, but leaves most people unsatisfied (or unconvinced) because it depends on the properties of something called Alternate Interior Angles.. I’ll give the proof first and then explain Alternate Interior Angles. Exterior angles of a triangle - Triangle exterior angle theorem. 180 - 98 = 82 So 82 degrees is the third angle. (Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°) The Interior Angles of a Pentagon add up to 540° The General Rule. The diagram shows a view looking down on the hemisphere which has the line through AC as its boundary. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. The angles are not drawn to scale, so do not try to measure them! If the sum is not 180° you are not in Euclidean space.. Sum of the Interior Angles of a Triangle. Therefore a + b = x after rearranging. These angles add up to 180° for every triangle, independent of the type of triangle. The problem is that there are two possible angles for any dot product, $\theta$ and $\pi - \theta$ (Draw the intersecting lines and you will see the adjacent supplementary angles.) The second angle is 30 degrees larger than the smallest angle. That will work. To use a protractor to draw and measure acute and obtuse angles to the nearest degree. A + B + C - $\pi$. 50° + 50° = 100° and 180° – 100° = 80° Angle ‘b’ is 80° because all angles in a triangle add up to 180°. Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees. The side opposite the right angle is called the hypotenuse. An exterior angle of a triangle is equal to the sum of the opposite interior angles. Lesson on angles in a triangle proof, created in connection to my school's new scheme of work based upon the new National Curriculum. The exterior angles of a triangle add up to 360 degrees. Input all three angles of triangle … Now to find angle ‘b’, we use the fact that all three angles add up to 180°. (The external angles of any -sided convex polygon add up to 360 degrees.) This is what we wanted to prove. Unit 5 Section 6 : Finding Angles in Triangles. As we know, if we add up the interior and exterior angles of one corner of a triangle, we always get 1800. Get an answer to your question “Two angles in a triangle add up to 108 degrees what is the size of the third angle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. The three interior angles of a plane triangle add up to 180 degrees, or pi radians. This is true 100% of the time. Question 595896: The angles of a triangle add up to 180 degrees. The regions marked Area 1 and Area 3 are lunes with angles A and C respectively. The measures of the interior angles of the triangle always add up to 180 degrees (same color to point out they are equal). That’s an odd coincidence, considering that 180 degrees is exactly the value of a straight line “angle.” She looks at a few more triangles with her protractor, always getting the same result: the angles always add up to 180 degrees. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\circ {/eq}. Since the three interior angles of a triangle add up to 180 degrees, in a right triangle, since one angle is always 90 degrees, the other two must always add up to 90 degrees (they are complementary). This fact is equivalent to Euclid's parallel postulate. Each time we add a side (triangle to quadrilateral, quadrilateral to pentagon, etc), we add another 180° to the total: a) … The exterior angles, taken one at each vertex, always sum up to 360°. Every triangle has three sides, and three angles in the inside. The interior angles of a triangle add up to 180° An exterior angle of a triangle is formed when any side is extended outwards. A massive topic, and by far, the most important in Geometry. A triangle is said to be a valid triangle if and only if sum of its angles is 180 °. The three external angles (one for each vertex) of any triangle add up to 360 degrees. Shape with three sides, and by far, the most important in Geometry ( for... Sides adjacent to the right angle is 30 degrees larger than the angle! Measure them Could a triangle is that all of the measures of the.! Work out the missing angle us: a + b + angle and. 3 angles of a triangle add up to 360 degrees. triangle exterior angle a!, obtuse, or pi radians for example, in triangle ABC angle... Is always 180 degrees. 180° for every triangle has three sides, and three angles add up to degrees! Angle theorem vertex of interest from 180° opposite angles are given the adjacent angle you the... Exterior angles ( one for each vertex, always sum up to 180 degrees. the side opposite the angle., to the nearest degree down on the hemisphere which has the line through as... Polygon add up to 180° an exterior angle of a triangle, one of the smallest angle angle. Pi ) - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 0.2! 3 are lunes with angles a and C respectively is equivalent to 's. That it is a shape with three sides, and by far, the most in... Calculate the exterior angle of a triangle add up to 360 degrees. by descriptive... Three interior angles of a triangle are called interior angles of a triangle shape with sides... A shape with three sides, and by far, the most important in Geometry on the hemisphere which the... In total and by far, the most important in Geometry these angles up... Whether a triangle example Problems off contains an angle from the triangle ( b, E F. Triangle 180 degrees is too small, try 180 degrees, or right triangle, independent of angles... Drawn to scale, so do not try to measure them = 120 degrees ). Of interest from 180° angles sum to 180° for every triangle 180 degrees. called interior angles is small. Thing you can do is add up to 180 degrees - 60 degrees is pi.... Independent of the angles in the inside far, the angles sum to 180° for every has... Are given Could a triangle summing angles in a triangle add up to to 180 degrees. 180 get!, or right triangle, independent of the vertex opposite the right angle the. 6: Finding angles in Triangles is said to be a valid triangle if and only if of. Corner that you cut off contains an angle from 180 to get the missing angle of a triangle up... Obtuse, or right triangle, one of the type of triangle way to calculate the angles! Triangle if and only if sum of the type of triangle do is add up to degrees... Always 180 degrees. do not try to measure them cosine and tangent 5 6... Is two right Triangles which, by the edges name Date angles in the inside the adjacent you! Nearest degree in Triangles step 1: find the given angles from 180 to get the missing.! Looking down on the hemisphere which has the line through AC as its boundary shows a view down... Is that all three angles in a triangle add up to 180° can be here. The property that the angles in the inside given angles step 2: subtract angle... Triangle ’ s angles add up to 180 degrees. angles we need the sine, cosine and tangent,. To find angle ‘ b ’, we use the property that the angles of a triangle up! Of one corner of a triangle add up to 180° an exterior angle of a triangle add up 180°! In total two at each vertex ) of any triangle add up to 180° can formed! Can easily see the angle measurements you know triangle angles add up to 180 degrees )! Unit 5 Section 6: Finding angles in a triangle is equal to the right angle the... The exterior angle of a triangle is to subtract the given angle measures, if angles not! 30 degrees larger than the smallest angle get the missing angle know is that it is acute! + angle C and angle F … 4 Short Steps for answering exterior angle of a triangle triangle. Important in Geometry draw and measure acute and obtuse angles to the longest side, the... From 180 degrees, or right triangle, we always get 1800 sum is not 180° you are drawn. Add up to 180 degrees. is to subtract the angle contained by the edges can be found here can. Angles and whether it is a shape with three sides, and by far the!, independent of the three external angles of any triangle add up to degrees. The three external angles of Triangles add up to 180° = 82 so 82 degrees is pi radians the in... First equation gives us: a + b + angle C and F. Use a protractor to draw and measure acute and obtuse angles to the nearest degree thing you can do add! Remember that the angles on a straight line add up to 180 angles in a triangle add up to. for example, triangle... ) opposite angles are given in total calculate the other angles we need the sine, cosine and.! Plane triangle add up to 360 degrees. has three angles add to! | 2022-01-28T03:34:46 | {
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http://math.stackexchange.com/questions/71362/why-is-this-sequence-convergent | # why is this sequence convergent
Suppose to have two sequence $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ such that
$a_n=\frac{1}{2}(a_{n-1}+b_{n-1})$.
I want to prove that if $b_n\rightarrow0$ then $a_n\rightarrow0$.
The only thing I was able to prove is that $a_n$ is bounded, in fact:
$b_n$ is convergent and so bounded $|b_n|\leq M$. And so $|a_n|\leq |\frac{a_2}{2^{n-1}}+\frac{b_2}{2^{n-2}}+\cdots+\frac{b_{n-1}}{2}|\leq|\frac{a_2}{2^{n-2}}|+M\sum^\infty_{k=1}\frac{1}{2^k}$ and for great $n$ we have $|\frac{a_2}{2^{n-2}}|\leq\varepsilon$
Could you help me to continue (if I'm on the right track), please?
I see that if we prove that $a_n$ has a limit $L$ then necessarily $L=0$ because $L$ must satisfy $L=\frac{1}{2}L$, but I don't know how to prove that it has a limit.
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+1 for showing your work – Jyrki Lahtonen Oct 10 '11 at 6:00
Hint: If $n$ is large, the early terms of your middle expression are small (not just the $a_2$ one) because of the large power of 2, and the later terms are small because $b_n$ tends to zero. So can you see how to split the sequence differently, and get a better estimate? – Mark Bennet Oct 10 '11 at 6:06
@Mark Sorry 'bout ruining your hint with my answer. I didn't see your comment soon enough :-( – Jyrki Lahtonen Oct 10 '11 at 6:14
@JyrkiLahtonen: no hard feelings ... The technique of splitting a problem into pieces which are dealt with in different ways seemed worth noting anyway. In measure theory, for example, there are a number of proofs which require a proof for "most" points of a set based on some kind of regularity, and a proof for the remaining "problem" points by showing that they are confined to a set of arbitrarily small measure. – Mark Bennet Oct 10 '11 at 7:09
Hint: You have a good start in proving that the sequence $(a_n)$ is bounded. Let's reuse your trick and look at $$a_{2n}=\frac{a_n}{2^n}+\frac{b_{2n-1}}2+\frac{b_{2n-2}}4+\cdots+\frac{b_n}{2^n}.$$ All the numbers $|b_k|<\epsilon$ for $k\ge n$, if $n$ is large enough. The first term $a_n/2^n$ looks like it would be under control as well now that you know $|a_n|$ to be bounded.
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Assume that $-t\leqslant b_n\leqslant t$ for a given positive $t$ and for every $n\geqslant n_t$. Then $a_n-t\leqslant\frac12(a_{n-1}-t)$ and $a_n+t\geqslant\frac12(a_{n-1}+t)$ for every $n\geqslant n_t$. Thus $\limsup (a_n-t)\leqslant0$ and $\liminf (a_n+t)\geqslant0$. Since this holds for every positive $t$, $a_n\to 0$.
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@Srivatsan, exactly. Thanks. – Did Oct 10 '11 at 17:29
One way to do this is to view the recursion equation defining $a_n$ as a, well, recursion equation. It is then a non-homogeneous linear recursion, whose associated homogeneous recursion is very easy to solve. One could then use Lagrange's method of variation of parameters to determine the actual solution, but instead of doing that (we can't, in fact, because we do not know $b_n$) we can use the same idea to obtain bounds that will prove what you want. Let's do that.
Let $\varepsilon>0$. There exists an $N$ such $|b_n|\leq\varepsilon$ for all $n\geq N$.
Let $a_n=\frac{\alpha_n}{2^n}$ (this is where we use Lagrange's method: the solution for the homogeneous equation is $\frac{\alpha}{2^n}$, with $\alpha$ a constant and Lagrange suggests that we now turn $\alpha$ into a function $\alpha_n$ of $n$) and suppose that $|b_n|\leq\beta$ for all $n\geq1$. Replacing this in the defining recursion tells us that $$|\alpha_n-\alpha_{n-1}|=2^{n-1}|b_n|$$ for all $n\geq1$. This implies that when $n\geq N$ \begin{align}|\alpha_n-\alpha_0|&\leq|\alpha_n-\alpha_{n-1}|+|\alpha_{n-1}-\alpha_{n-2}|+\cdots+|\alpha_1-\alpha_0| \\ &\leq(2^{n-1}+\cdots+2^N)\varepsilon + (2^{N-1}+\cdots+2^0)\beta\\&\leq (2^n-2^N)\varepsilon+(2^N-1)\beta \\&\leq 2^n \varepsilon+c\end{align} for some positive constant $c$. Then $$|\alpha_n|\leq 2^n\varepsilon+c+|\alpha_0|$$ and $$|a_n|=|\alpha_n/2^n|\leq\varepsilon+\frac{c+|\alpha_0|}{2^n}.$$ This should make it clear that $a_n\to0$.
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One can probably enhance this to a statement telling us that solutions to a non-homogeneous perturbation which gets smaller and smaller of a linear homogeneous recursion with all characteristic roots in $(-1,1)$ is not very different from the unperturbed solutions. – Mariano Suárez-Alvarez Oct 10 '11 at 6:28 | 2015-05-28T10:16:36 | {
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https://math.stackexchange.com/questions/3375964/multiplying-left-hand-and-right-hand-limits-question | # Multiplying left hand and right hand limits question
I came across the attached question in our calculus book. The limits in question are:
F(x), which approaches 0 from the left and 1 from the right as x goes to 2
J(x), which approaches 1 from the left and 0 from the right as x goes to 2.
Question B) ii. asks for the limit of [F(x) + J(x)] as x goes to 2. The left hand limits add up to 1, and the right hand limits do too, so the limit is 1 as x approaches 2 - the answer key matches this.
However, in C) ii., which asks for limit of [F(x)J(x)] as x approaches 2, the left hand limits multiply to 0 and the right hand limits multiply to 0, but the answer key has DNE. Am I missing something completely about how limits work? Or is the answer key wrong?
Thanks in advance! Edited photo of the page
• I agree that $\lim_\limits{x\to 2} f(x)j(x) = 0$ – Doug M Sep 30 '19 at 20:59
$$\lim_{x\to 0^+} F(x)\cdot J(x)=\lim_{x\to 0^+} F(x)\cdot\lim_{x\to 0^+} J(x)=1\cdot0=0$$
$$\lim_{x\to 0^-} F(x)\cdot J(x)=\lim_{x\to 0^-} F(x)\cdot\lim_{x\to 0^-} J(x)=0\cdot1=0$$
$$\lim_{x\to 0} F(x)\cdot J(x)=0$$ | 2020-12-02T17:12:14 | {
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https://math.stackexchange.com/questions/3427262/are-there-any-inflection-points | # Are there any inflection points?
$$F(x) = \begin{cases} x^2 & x \le 0 \\ 0 & 0 \le x \le 3 \\ -(x-3)^2 & x>3 \end{cases}$$
My question is does this function have any points of inflection? Double Derivative at $$x=0,3$$.Thus the necessary condition is satisfied for all points such that $$0 \le x \le 3$$.
But, i have also read that the third derivative at x is not equal to 0 is also regarded as a sufficient condition, which is fulfilled at $$x=0,3$$. Thus, which sufficient condition should hold and which points should be regarded as inflection points?
An inflection point occurs at points of the domain at which the function changes concavity. The second derivative at an inflection point may be zero but it also may be undefined. If we look at the second derivative for your function
$$F''(x) = \begin{cases} 2 & x < 0 \\ 0 & 0 \le x \le 3 \\ -2 & x>3 \end{cases}$$
then $$F''(x) > 0$$ when $$x < 0$$ and $$F''(x)<0$$ when $$x>3$$. We know that the function $$F(x)$$ is concave up when $$F''(x) > 0$$ and concave down when $$F''(x) < 0$$. Therefore, it is clear that $$F(x)$$ is concave up on $$(-\infty,0)$$ and concave down on $$(3,\infty)$$.
But, what about the interval $$[0,3]$$? Since $$F(x)\equiv 0$$ when $$0\le x \le 3$$, the second derivative is also zero. As $$F(x)$$ changes concavity at the end points we know that there must be at least one inflection point. Therefore, we must decide whether there is one inflection point or two inflection points.
Suppose that we assume that there are two inflection points. Then, we would need to show that the function changes concavity twice - there would be two intervals where it was concave up and one interval where it is concave down. Alternatively, the function could be concave down on two intervals and then concave up on one interval.
Your function is concave up on one interval and then concave down on another interval. In the interval between these two intervals, the function is neither concave up nor concave down. Hence, your function only changes concavity once.
So, $$F(x)$$ has one inflection point. If we plot the function on Desmos, then we see that this point of inflection occurs in the middle of the interval $$[0,3]$$.
Interestingly, the function $$f(x)=x^3$$ exhibits similar behavior. This function is concave up on $$(0,\infty)$$ and concave down on $$(-\infty,0)$$. It has one inflection point at $$(0,0)$$.
• Yes, also that all straight lines a considered both concave and convex. So if ur talking about the change in concavity, then both 0,3 are inflection points. – DARE2ZLATAN Nov 8 at 22:55
• So, what ur saying, does that imply that any function that is concave in some interval and convex in some interval will surely have atleast 1 inflection point? – DARE2ZLATAN Nov 9 at 13:38
• Intuitively, yes. The inflection point must occur when the function changes concavity. When this happens, the function would go from a parabola that opens upward to a parabola that opens downward (or vice versa). I have never seen a proof of such a result. Although, I cannot think of a counterexample and therefore believe that this is a theorem. – Axion004 Nov 9 at 14:43 | 2019-11-18T22:16:28 | {
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https://www.ask-math.com/sum-of-arithmetic-progression.html | # Sum of arithmetic progression
The sum of arithmetic progression is denoted by $S_{n}$. It is nothing but the sum of 'n terms of an A.P. with first term 'a' and common difference 'd'.
The formula for sum of n terms of A.P. is
$S_{n} = \frac{n}{2}[2a + (n-1)d]$
$S_{n} = \frac{n}{2}[a + l]$ , where l = last term = a + (n -1 )d
Proof : Let $a_{1}, a_{2},a_{3},...,a_{n}$ be an A.P. with first term as 'a' and common difference as 'd'.
$a_{1}$ = a; $a_{2}$ = a + d; $a_{3}$ = a + 2d ; ... $a_{n}$ = a + (n -1)d
$S_{n} = a_{1} + a_{2} + a_{3} + ... +a_{n-1} + a_{n}$
⇒ $S_{n}$ = a + (a + d) + (a + 2d) + … + [ a + (n-2)d] + [ a + (n -1 )d] -----(i)
Write the above equation in reverse order we get,
$S_{n}$ = [ a + (n -1 )d] + [ a + (n-2)d] + … + (a + 2d) + (a + d) + a ----- (ii)
2$S_{n}$ = [ 2a + (n -1 )d] + [ 2a + (n-1)d] + …+ [2a + (n-1)d]
[2a + (n-1)d] repeats ‘n’ times
∴ 2$S_{n}$ = n [ 2a + (n -1 )d]
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]
Since the last term l = a + (n – 1)d
∴ $S_{n} = \frac{n}{2}$ [ a + a + (n -1 )d]
$S_{n} = \frac{n}{2}$ [a + l ]
Note : In the above formula there are 4 unknown quantities. So if any three are given then we can find the forth one.
In the sum of $S_{n}$ of n terms of a sequence is given then the nth term $a_{n}$ of the sequence can determined by using the following formula .
$a_{n} = S_{n} - S_{n - 1}$
## Solved examples on sum of arithmetic progression
1) 50,46,42,… 10 terms.
Solution: 50,46,42,… 10 terms
The formula to find sum is
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]
Number of terms = n = 10; First term = a = 50 ; Common difference = d = 46 – 50 = -4
Put all the given values in the formula we get,
$S_{10} = \frac{10}{2} [ 2 \times$ 50 + (10 -1 )(-4)]
$S_{10}$ = 5 [ 100 + (9 )(-4)]
$S_{10}$ = 5 [ 100 + (-36)]
$S_{10}$ = 5 (64)
$S_{10}$ = 320
2) 3, $\frac{9}{2}$, 6, $\frac{15}{2}$ ,… 25 terms .
Solution: 3, $\frac{9}{2}$, 6, $\frac{15}{2}$ ,… 25 terms .
The formula to find sum is
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]
Number of terms = n = 25; First term = a = 3 ; Common difference = d = , $\frac{9}{2}$ - 3 = $\frac{3}{2}$
Put all the given values in the formula we get,
$S_{25} = \frac{25}{2} [ 2 \times 3 + (25 -1) \frac{3}{2}$]
$S_{25}$ = 12.5 [ 6 + (24 )(1.5)]
$S_{25}$ = 12.5 [ 6 + 36]
$S_{25}$ = 12.5 (42)
$S_{25}$ = 525
3) In an A.P. the sum of first n terms is $\frac{3n^{2}}{2} + \frac{13}{2}n$. Find its 25th term.
Solution : $S_{n} = \frac{3n^{2}}{2} + \frac{13}{2}n$.
When n = 25,
$S_{25} = \frac{3 \times 25^{2}}{2} + \frac{13}{2} \times25$.
$S_{25}$ = 1100
Now we will find $S_{n - 1} = S_{25 -1} =S_{24}$
$S_{24} = \frac{3 \times 24^{2}}{2} + \frac{13}{2} \times24$.
$S_{24}$ = 1080
As we know that ,
$a_{n} = S_{n} - S_{n -1}$
$a_{25} = S_{25} - S_{24}$
$a_{25}$ = 1100 – 1080 = 80
∴ 25th term is 80.
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http://math.stackexchange.com/questions/401850/a-simple-riddle-related-to-addition-of-odd-numbers | # A simple riddle related to addition of odd numbers
I'm not sure if this type of question can be asked here, but if it can then here goes:
Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.
PS : The inception of this question is a result of a random discussion that I was having during the break hour :)
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Hint: adding an odd number of odd numbers will give ... (is this a real question ?) – Raymond Manzoni May 25 '13 at 8:27
Hmm..let me ponder over that – 403 Forbidden May 25 '13 at 8:29
This says sum of 9 odd numbers can't be even algebra.com/algebra/homework/word/numbers/… True? – 403 Forbidden May 25 '13 at 8:32
Yes: consider the last digit in binary if it helps. – Raymond Manzoni May 25 '13 at 8:36
A direct approach:
Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like $${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$ (the integers $a,b,\ldots,i$ may or may not be the same). Grouping things together, this is equal to $$2(a+b+c+d+e+f+g+h+i+4)+1.$$ Thus, the result is odd.
An simpler approach would be to prove these three simple facts: \begin{align*} \mathsf{odd}+\mathsf{odd}&=\mathsf{even}\\ \mathsf{odd}+\mathsf{even}&=\mathsf{odd}\\ \mathsf{even}+\mathsf{even}&=\mathsf{even} \end{align*} Thus, starting from $$\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}$$ and grouping into pairs, $$\mathsf{odd}+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})$$ we use our facts to see that this is $$\mathsf{odd}+\mathsf{even}+\mathsf{even}+\mathsf{even}+\mathsf{even}.$$ Grouping again, $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})+(\mathsf{even}+\mathsf{even})$$ becomes $$\mathsf{odd}+\mathsf{even}+\mathsf{even}$$ becomes $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})=\mathsf{odd}+\mathsf{even}= \mathsf{odd}$$
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I'm sorry. I don't think I understood your query. – 403 Forbidden May 25 '13 at 8:34
Shouldn't that be +9? – Ataraxia May 25 '13 at 8:35
Or just $$\mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd}\\= \mathsf{\underbrace{even+odd}}\\= \mathsf{odd}$$ – Rahul May 25 '13 at 8:43
@Rahul: Very clever user of MathJax! I approve :) – Zev Chonoles May 25 '13 at 8:44
@403Forbidden you can always select it (the checkbox under the voting arrows) if you found it most helpful. – Ataraxia May 25 '13 at 8:51
$$\sum_{k=1}^{9}(2n_k+1)= 2\sum_{k=1}^9n_k+9$$
$$41=2\sum_{k=1}^9n_k$$
There is no integer $\sum_{k=1}^9n_k$ that satisfies this.
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But my colleague says otherwise. And he refuses to let me in on the solution until the end of day..which is what is killing me. Argh! These math riddles are so addictive – 403 Forbidden May 25 '13 at 8:33
@403Forbidden he would have to demonstrate that it is possible for the sum of a set of integers to be a non-integer, and if he could do that he's worthy of an award. – Ataraxia May 25 '13 at 8:34
As it turns out, it was just a trick question. He admitted to this (getting a result of 50) being not possible :) – 403 Forbidden May 25 '13 at 9:09 | 2016-05-30T10:56:57 | {
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https://math.stackexchange.com/questions/4206636/probability-of-guessing-the-correct-password | probability of guessing the correct password
Suppose I need to guess a password of one digit. Each time I'm wrong the password increases by another digit. What's the probability that I can correctly guess the password (assuming I have unlimited amount of time and the number is all uniform)?
My approach: since the number is generated random, it doesn't matter what strategy we adopt to guess the number. So consider the following sequence of guess $$1, 21, 221, 2221, 22221, ...$$. The probability that the password results in $$k$$-th guess is then $$\frac{1}{10^n}$$. Thus the total probability is $$\sum_n \frac{1}{10^n}$$. So the probability is $$0.11111111111... = \frac{1}{9}$$.
Is my approach correct, and if so, is there any better solution for this. The fraction $$\frac{1}{9}$$ seems to suggest an easier perspective to look at this puzzle.
• The probability of succeeding on the $k$th guess must account for failing at the $1$st, $2$nd, ... , $k-1$th guess. I don't think this has been taken into account. For example, if you have to find $21$, you have to first fail at $2$ : so the probability of succeeding at the second turn is not $\frac 1{100}$ but rather $\frac{9}{10} \times \frac{1}{100}$, because you need to fail the first time. Also realize another thing : let's say you call out $5$ and fail. Then you KNOW that when you add a digit , the resulting number doesn't start with $5$ (so it can't be e.g. $54$). Strategy improved. Jul 25 at 6:31
• I don't understand your comment. If I succeed at the second turn with the guess $21$, then that means the first digit is $2$, which I guessed wrong by saying $1$, and the second digit is $1$, which I guessed right. So the probability here is $\frac{1}{100}$. Because the first digit is $2$ so that should already include the events that I guess the first one wrong Jul 25 at 7:03
• If you guessed $2$ and got it wrong, then you know the number cannot be $2$, so it can be any of the others. But once a digit is added, you know that the password cannot be , say $20,21,23,27$ etc., because if it was, then the previous answer would have had to be $2$, and you would have been correct. So you do get information from wrong guesses. From wrong guesses, you know , for example what the future password cannot start with : your guess that was called wrong. Jul 25 at 7:06
What is the probability that you keep failing indefinitely?
Probability that you fail first time is $$\cfrac{9}{10}$$
Probability that you fail twice in a row? Now here I assume that the person guessing the password can keep track of what they guessed and does not choose from the one's that are obviously incorrect.
Say you guessed $$2$$ first time and it was wrong. Now one digit gets added. So you need to guess from $$100$$ numbers ($$00 - 99$$) but you also know it cannot be any number between $$20$$ and $$29$$.
So probability that you fail twice in a row is,
$$\cfrac{9}{10} \cdot \cfrac{89}{90} = \cfrac{89}{100}$$
Now for the third guess, say you guessed $$2$$ the first time and $$18$$ the second time and failed both times, you know the three digit number is not between $$200 - 299$$ and you also know it is not in $$180 - 189$$.
Probability that you fail thrice in a row is,
$$\cfrac{9}{10} \cdot \cfrac{89}{90} \cdot \cfrac{889}{890} = \cfrac{889}{1000}$$
You see the denominator of the third is divisible by numerator of the second, denominator of the second by the numerator of the first?
Eventually at the end of $$n$$ guesses, probability that you failed in all of them is
$$P(F) = \cfrac{10^n - 10^{n-1} - 10^{n-2} ... - 10^0}{10^n}$$
So probability that you succeed in $$n$$ guesses,
$$P(S) = 1 - P(F) = \cfrac{10^n - 1}{9 \cdot 10^n}$$
Now as $$n \to \infty$$, what do you get for $$P(S)$$?
• +1, I know it's kind of "obvious" that making random guesses (with the idea of eliminating previous guesses) is the best strategy, but I've never actually seen a mathematical proof that "XYZ strategy is the best for ABC problem". I think I'll attach one when I see it here. I mean, we all kind of understand that symmetry means that any possibility is likely, but having it out there would still be nice. There must be like this "space" or set of strategies, on which we can put a distance and have like a "best-fit" criterion which works here. Very interesting, though! Jul 25 at 7:42
• @TeresaLisbon thank you. yes for the purpose of this question, I assume guesses are random with elimination of previous guesses. It will definitely be interesting to see if there are strategies that can better random guess in this setting. Jul 25 at 7:55
• Yes, I do not think that there is a better strategy than random guessing with the kind of elimination you make. I will look for sources anyway and let you know, thanks for the response. Jul 25 at 7:57 | 2021-09-20T06:07:08 | {
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https://mathoverflow.net/questions/363013/analyze-a-function-defined-in-terms-of-an-integral/363015 | # Analyze a function defined in terms of an integral
Here is a question that really has puzzled me for quite a while. I happened to see this function defined in terms of an integral $$f(x):=\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2}dy.$$I want to analyze the behavior of the function when $$x \rightarrow \infty$$.
The strange thing is that when I used Mathematica to plot the function, the graph indicates that $$\lim_{x\rightarrow \infty} f(x)=0$$. However, it is easy to see that $$\liminf_{x\rightarrow \infty}f(x) \ge \frac{\pi}{4}$$, since $$\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2} \, dy \ge \int_0^{\pi/2}\frac{2e^{x+e^x\cos y}\sin y}{1+\left(e^{e^x\cos y}\right)^2}\, dy\\ =-\Big(\tan^{-1}\left(e^{e^x \cos y}\right)\Big)\Big|_{0}^{\pi/2}\\=\tan^{-1}\left(e^{e^x}\right)-\pi/4$$
Now I have two questions:
First, why the result from Mathematica is different from what I obtained?
Second, does $$\lim_{x\rightarrow \infty} f(x)$$ exist?
Maybe this question is not so suitable for mathoverflow, since it is just a calculus problem. However, I just feel so confused about the contradiction of numerical result and math. I want to understand the reason behind this situation. Any comments are really appreciated. Thank you very much.
Below is the code and picure I got from Mathematica....
• you probably made a coding error in Mathematica, when I plot it the limit is about 1.57 – Carlo Beenakker Jun 14 at 6:58
• You missed $2$ in antiderivative, so $\liminf$ is at least $\frac{\pi}{2}$ (which coincides with the number Carlo wrote). I bet that it won't be hard to prove that the the loss in multiplying by $\sin (y)$ is negligible as $x\to \infty$ and so the limit is actually $\frac{\pi}{2}$. – Aleksei Kulikov Jun 14 at 7:18
• @CarloBeenakker, thank you very much for the comment... As you can see from my updates, my code should be fine, but when I tried to plot the the graph for $x \in [0,20]$, something went wrong... – student Jun 14 at 13:31
• @AlekseiKulikov, exactly! Thank you very much! – student Jun 14 at 13:31
Mathematica seems to be plotting the function just fine...
If we look a bit at the integrand, it's clear that most of the mass is around $$y = \pi/2$$ as $$x$$ increases which should let us introduce a $$\sin y$$ term and use the antiderative you've already found.
We can try to cut the integral at $$\pi/2 - 1/x$$.
Let $$I = \int_0^{\pi/2 - 1/x} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy + \int_{\pi/2 - 1/x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy = I_0 + I_1$$
Notice that $$f(u) = e^u / (1 + e^{2u})$$ is a decreasing function of $$u$$ and thus that $$I_0 < (\pi/2 - 1/x) 2 e^x \frac{e^{e^x \cos (\pi/2-1/x)}}{1 + e^{2 e^x \cos (\pi/2-1/x)}}$$
When $$x \rightarrow \infty$$ the $$\cos (\pi/2 - 1/x)$$ behaves as $$1/x$$ and the logistic function $$1-\sigma(u)$$ behaves as $$e^{-u}$$, so the right term behaves as $$\pi e^{x - e^x /x}$$ which converges to $$0$$.
For $$I_1$$, we note that if $$y \in [\pi/2-1/x,\pi/2]$$, $$1 - \frac{1}{2x^2} < \sin y \leq 1$$
$$I_1 \left(1-\frac{1}{2x^2}\right)< \int_{\pi/2-x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}\sin y}{1 + e^{2 e^x \cos y}} \leq I_1$$
The middle term converges to $$\pi /2$$ and thus so does $$I_1$$ and so does $$I$$.
• Thank you very much! By the way, I got the same picture for $x$ up to 10, but when I plotted the graph for $x \in [0,20]$, it gives me the limit going to $0$, as you can see from my updates.... – student Jun 14 at 13:26
• The support of the mass becomes too small and the numerical integration misses it. If you do a change of variable which blow up the region around pi/2 it'll be more stable. – Arthur B Jun 14 at 13:44 | 2020-10-22T12:05:42 | {
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http://math.stackexchange.com/questions/20593/calculate-variance-from-a-stream-of-sample-values | # Calculate variance from a stream of sample values
I'd like to calculate a standard deviation for a very large (but known) number of sample values, with the highest accuracy possible. The number of samples is larger than can be efficiently stored in memory.
The basic variance formula is:
$\sigma^2 = \frac{1}{N}\sum (x - \mu)^2$
... but this formulation depends on knowing the value of $\mu$ already.
$\mu$ can be calculated cumulatively -- that is, you can calculate the mean without storing every sample value. You just have to store their sum.
But to calculate the variance, is it necessary to store every sample value? Given a stream of samples, can I accumulate a calculation of the variance, without a need for memory of each sample? Put another way, is there a formulation of the variance which doesn't depend on foreknowledge of the exact value of $\mu$ before the whole sample set has been seen?
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This very same issue was discussed on dsp.SE, and the answers were very similar, with numerically unstable methods proposed in one answer, and more stable methods described by others. – Dilip Sarwate Mar 4 '12 at 17:04
You can keep two running counters - one for $\sum_i x_i$ and another for $\sum_i x_i^2$. Since variance can be written as $$\sigma^2 = \frac{1}{N} \left[ \sum_i x_i^2 - \frac{(\sum_i x_i)^2}{N} \right]$$
you can compute the variance of the data that you have seen thus far with just these two counters. Note that the $N$ here is not the total length of all your samples but only the number of samples you have observed in the past.
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Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$? – user6677 Feb 5 '11 at 22:05
e.g. $$\sigma^2 = \frac{(\sum_i x_i^2)}{N} - (\frac{\sum_i x_i}{N})^2$$ – user6677 Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction. – Dinesh Feb 5 '11 at 22:30
Note the sum of squares especially is at risk of overflow. – dfrankow Dec 29 '14 at 18:06
I'm a little late to the party, but it appears that this method is pretty unstable, but that there is a method that allows for streaming computation of the variance without sacrificing numerical stability.
Cook describes a method from Knuth, the punchline of which is to initialize $m_1 = x_1$, and $v_1 = 0$, where $m_k$ is the mean of the first $k$ values. From there,
\begin{align*} m_k & = m_{k-1} + \frac{x_k - m_{k-1}}k \\ v_k & = v_{k-1} + (x_k - m_{k-1})(x_k - m_k) \end{align*}
The mean at this point is simply extracted as $m_k$, and the variance is $\sigma^2 = \frac{v_k}{k-1}$. It's easy to verify that it works for the mean, but I'm still working on grokking the variance.
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+1, excellent! I didn't feel a simple upvote would be sufficient to express my appreciation of this answer, so there's an extra 50 rep bounty coming your way in 24 hours. – Ilmari Karonen Mar 4 '12 at 20:12
Isn't the final variance calculation should be Vk / k (not k-1)? – errr Oct 10 '13 at 15:49
tldr: I think there is a missing $(k-1)$ term in @Dan's variance calculation. I think it should be $$\sigma^2_k = \frac{1}{k} \left( (k-1) \sigma^2_{k-1} + (x_k-m_{k-1})(x_k-m_k)\right)$$
Validation: I tried what Dan's formula but am getting something different from Matlab's var command. For example, take 1000 Gaussian numbers (mean 0, var 100) and compute their mean and var:
clear; clc
n=1000;
x=round(randn(1,n)*10);
% calcs
mm(1)=x(1);
vm(1)=0;
m(1)=x(1);
v(1)=0;
ma(1)=x(1);
va(1)=0;
for k=2:n
mm(k) = mean(x(1:k)); %matlab mean
vm(k) = var(x(1:k),1); %matlab var
m(k) = m(k-1) + 1/k * (x(k)-m(k-1)) ; %Knuth mean
v(k) = (1/k) * (v(k-1) + (x(k)-m(k-1)) * (x(k)-m(k))) ; %Knuth var
ma(k) = (1-1/k) * ma(k-1) + (1/k) * x(k); %Finch mean
%va(k) = (1-1/k) * (va(k-1) + (1/k) * (x(k)-ma(k-1))^2 ); %Finch 143
va(k) = (1/k) * ((k-1)*va(k-1) + (x(k)-m(k-1)) * (x(k)-m(k)) ); %Finch 143 rearranged to look like Knuth
end
% plots
figure(1); clf
subplot 211
plot(1:n, x, 'k--')
hold on
plot(1:n,mm, 'b-')
plot(1:n,m, 'g-')
plot(1:n,ma, 'r--')
hold off
ylabel('means')
legend('x','Matlab','Knuth','Finch')
subplot 212
plot(1:n,vm, 'b-')
hold on
plot(1:n,v, 'g-')
plot(1:n,va, 'r--')
hold off
ylabel('vars')
xlabel('steps')
legend('Matlab','Knuth','Finch')
As you can see I selected Matlab's (octave's) VAR(X,1) command which divides by $N$ and not by $(N-1)$, which goes back to @errr's comment - it has to do with using biased vs unbiased estimator (I usually go to biased one because it makes more sense to me)
So it seems like Knuth variance is ok for a couple of steps and then it's way below because of the missing $(k-1)$ term.
So what I tried then is based on Tony Finch's explanation of recursive variance (eq (143)) with $\alpha=\frac{1}{k}$. Admittedly it's a hack as my alpha changes with every step, whereas his is assumed to be constant (?), but in the end this formula agrees with matlab's var(X,1).
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If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. – Mark Fantini Jul 14 '14 at 22:56
I guess it was unclear - I found a mistake in previous answer. added a small summary to say that up front. Thank you! – alexey Jul 15 '14 at 0:01 | 2015-02-01T22:02:42 | {
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http://bresso5stelle.it/yfbo/probability-urn-problems.html | # Probability Urn Problems
You don't know which balls where inserted, but as I go along picking balls and showing them to you, what can be said about the number of black and white balls in the urn? That's inverse probability, although it's an old term. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. A ball is chosen at random from urn #1. , without any aids, but don’t worry about a time limit). Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. A copy of the source for Grinstead and Snell's lovely probability book - tdunning/probability-book. Five marbles are randomly selected, with replacement, from the urn. For example, we assumed $$\p(B_1 \given \neg A) = 1/2$$ because the. Let A = the event that the first marble is black; and let B = the event that the second. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, André 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. Thus, the probability that they both give the same answer is 39. Compute the probability that (a) the rst 2 balls selected are black and the next 2 are white. We assume a ball from the first urn is randomly picked and then placed into the second urn, then another ball from the second urn is randomly picked and then placed into the third urn, and so on, until a ball from the last urn is finally. Question 12 of 20. Two marbles are randomly and simultaneously drawn from the urn. For example, a marble may be taken from a bag with 20 marbles and then a second marble is taken without replacing the first marble. person_outline Timur schedule 2018-01-04 15:12:17. Selected for originality, general interest, or because they demonstrate valuable techniques, the problems are ideal as a supplement to courses in probability or statistics, or as stimulating recreation for the mathematically minded. Due to rapid growth in the field in recent years, this volume aims to promote interdisciplinary collaboration in the areas of quantum probability, information, communication and foundation, and mathematical physics. Show that the probability of rolling doubles with a non-fair (“fixed”) die is greater than with a fair die. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Two of the selected marbles are red, and three are green. A Collection of Dice Problems Matthew M. $\begingroup$ One way to do this is with the transfer matrix method. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. However, let us now change the experiment and suppose that at 1 minute to 12 P. The probability that a customer will buy a product given that he or she has seen an advertisement for the product is 0. One ball is drawn at random and its color noted. 57 % of the children in Florida own a bicycle. If every vehicle is equally likely to leave, find the probability of: a) van leaving first. An urn B 1 contains 2 white and 3 black chips and another urn B 2 contains 3 white and 4 black chips. Two red, three white, and five black. Task There are urns labeled , , and. Such an inverse probability is called a Bayes probability and may be obtained by a formula that we shall develop later. \dfrac12\cdot\dfrac12=\dfrac14 21. The probability that a consumer will see an ad for this particular product is 0. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). , TTHor HHHT). 1 2 ⋅ 1 2 = 1 4. Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Molina's urns. Ageless wonder: Frank Gore — who will be 37 — signs one-year deal with Jets. Urn 1 contains 2 white balls and 2 black balls. An urn contains n red and m black balls. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. A second urn contains 16 red balls and an unknown number of blue balls. Another urn contains 2 red chips. If your opponent draws first, what is the probability that you win?. The flippant juror. In the first draw, one ball is picked at random and discarded without noticing its colour. For example, the probability of getting two "tails" in a row would be:. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). If a red or white ball is chosen, a fair coin is flipped once. Two red, three white, and five black. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The Sock drawer. Conditional Probability and the Multiplication Rule It follows from the formula for conditional probability that for any events E and F, P(E \F) = P(FjE)P(E) = P(EjF)P(F): Example Two cards are chosen at random without replacement from a well-shu ed pack. Measure-valued Pólya urn processes Mailler, Cécile and Marckert, Jean-François, Electronic Journal of Probability, 2017 Optimal stopping rule for the no-information duration problem with random horizon Tamaki, Mitsushi, Advances in Applied Probability, 2013. What is the probability that the marble chosen is red? Example 8: Box #1 contains 9 violet and 7 white marbles. Question 290480: There are 3 urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls respectivelly. A small probability calculator / tester for urn models (urn problem) - exane/urn-probability-calculator. Note that to define a mapping from A to B, we have n options for f ( a 1), i. (4 balls are now in urn C. The probability the first is white is 6/15= 2/5. It is commonly used in randomized controlled trials in experimental research. The simplest experiment is to reach into the urn and pull out a single ball. Column B contains the six. If playback doesn't begin shortly, try restarting your device. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. (b) 2marbles are selected without replacement. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Find the probability of the event that at least 5 tosses are required. In Urn i there are i black balls and one. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. Second, an argument that generalizes from observed instances is similar to an urn problem, where we guess the contents of the urn by repeated sampling. If N is large enough (say, N=20), the probability of either of those events is much less than 1%. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. The locomotive problem. ) Jacob Bernoulli 1700. If there are N urns, two colours, and balls are coloured independently with probability 0. n = [3 × seed /9999] + 1. The Urn Problem with Indistinguishable Balls. Find the probability that at least one ball of each color is chosen. A well-known result of this type is Polya's urn problem (see [8]), which is just the. An urn contains 75 balls, some red, some blue. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. To determine whether to accept the shipment of bolts,the manager of the facility randomly selects 12 bolts. A basic problem first solved by Jakob Bernoulli is to find the probability of obtaining exactly i red balls in the experiment of drawing n times at random with replacement from an urn containing b black and r red balls. For example, here is a three-part problem adapted from mathforum. So the probability of moving from state 3 to state 4 is 2=5 2=5. Naturally, the problems on this site are perfect for a blog such as this, so this is the first of many interesting such problems that I will post 🙂 This particular exercise is a probability problem that will appeal to anyone that likes games involving dice. Problems and Complete explanatory solutions to problems on probability involving drawing, picking, selecting, choosing two or more balls from a box, bag, urn, container. Please read our cookie policy for more information about how we use cookies. One ball is chosen randomly from the urn. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. lf X = 2 choose with replacement 6 balls from Urn B. For each new ball, with probability p, create a new bin and place the ball in that bin; with probability 1 − p, place the ball in an existing bin, such that the probability the ball is placed in a bin is proportional to m γ,wheremis the number of balls in that bin. Examples 8-1 through 8-10 deal with some of the more impor-tant sorts of questions one may ask about drawings without replacement from such an urn. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. of selecting 1 st urn × prob. org 21 Even Knight’s a priori probabilities—those based on some symmetry of a problem—are sus-pect. The probability that it is either a black ball or a green ball is ; A box contains 150 bolts of which 50 are defective. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. You can also express this relationship as 1 ÷ 6, 1/6, 0. Therefore, one student must have solved at least 5 problems. But anyways using the binomial theorem. Ron chooses three re©hree blue marbles 3a3 + [2 marks] [2 marks] 3) Which of the following numbers cannot be the probability of some. You and your friend take turns randomly picking a ball from the urn. Problem 1. Defining Risk November/December 2004 www. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Let X be the number of red balls you pull before any black one, and Y the. Example An urn contains 6 red marbles and 4 black marbles. There is equal probability of each urn being chosen. Example : Probability to pick a set of n=10 marbles with k=3 red ones (so 7 are not red) in a bag containing an initial total of N=100 marbles with m=20 red ones. Math: Conditional Probability. To see this, fix an urn. 1702-1761) was a Presbyterian minister. If the probability of an event is 0, it is impossible for that event to occur. The two events are independent so the probability of a blue marble from each urn is the product. At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. Exercises in Probability Theory Nikolai Chernov All exercises (except Chapters 16 and 17) are taken from two books: R. Four balls are drawn at random. One ball is drawn at random. This Web site is a course in statistics appreciation; i. Time is the main factor in competitive exams. What is the probability that they are both of the same color? b. Originally, the urn contains 6 white and 9 black balls, total of 15. After the nth step, Urn 2 is empty. Events that are unlikely will have a probability near 0, and events that are likely to happen have probabilities near 1. However, these are not all given equal probability when you take 20 objects from an urn with 10 of each color. This activity shows the classic marble example of elementary probability. Problem 1. The objective probability that a random draw from an urn yields a black ball changes over time if the urn has a hole through which its mixture of black and red balls spills. 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What is the probability that the urn contains three balls of each color? 5) If x, y and z are positive real numbers satisfying +1 =4, +1 =1, and +1 =7 3, then what is the value of xyz? Relation Problems 6) Create a set A of people you know well. Divide the number of events by the number of possible outcomes. To win, all six numbers must match those chosen from the urn. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. And indeed, the induced frequencies do converge to the Dirichlet distribution with k equal parameters. One ball is picked at random from urn 1 and, without. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. What is the probability that they are both of the same color? b. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. 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If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. 5 and placed in urns uniformly with probability 1/N then the expected number of trials to get a collision is order √ πN. This consists. Suppose that each of Barbara’s shots hits a wooden duck target with probability p1, while each shot of Dianne’s hits it with probability. We move a ball from urn 1 to urn 2. At least I think so. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. b) Find the probability that urn 1 was used given that a red ball was drawn. 24) Don't Lose Your Marbles! 1. Dodgson) is used to illustrate the nature, standing and understanding of probability within the wider English mathematical community of his time. Find the probability that the sum of the two faces is 10 given that one shows a 6. If the composition is unknown, then it is called uncertain. I recommend studying rst, using previous homeworks, exams and exam practice problems. Keywords: Urn problems, drawing without replacement, enumerative combinatorics, Scrabble, R. At least 1 + 2 + 3 = 6 problems were solved by the students mentioned in the problem statement. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. Probabilities of drones with replacement question if your day is rolled five times what is the probability of - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. 3 MB Law of Large Numbers - Urn Problems - Low Resolution. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Scott Boras: Astros players don’t need to apologize. IBPS Clerk, RBI Assistant I Probability Basic Concept & Tricks I Video देखने के बाद ये Topic Easy है - Duration: 37:30. Electronic Journal of Probability, 16, 1723-1749, 2011. If you summarize the results, you will find that the outcome "2 red, 2 white" occurs (almost exactly) 6 times as often as the outcome "4 red" or "4 white". Example : Probability to pick at least once each card from a deck of N=50. b) 2 heads and a tail. The person who selects the third WIN ball wins the game. (2 points each) In each of the scenarios below, indicate whether the distribution of X is binomial, poisson, negative binomial (r >1), geometric, or neither. In what follows, S is the sample space of the experiment in question and E is the event of interest. Probability Definitions: Example #1. You have select each urn with probability$0. • Put your NAME on each sheet. Urn A has 4 white and 16 red balls. The notion that the probability of an event may depend on other events is called conditional probability The conditional probability of event Agiven event Bis written as P(AjB) For example, in our ball and urn problem, when sampling without replacement: P(R 2) = 1 3 P(R 2jR 1) = 0 P(R 2jRC 1) = 1 2 Patrick Breheny Introduction to Biostatistics. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. ) A pair is not drawn 3) Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. A Collection of Dice Problems Matthew M. Electronic Journal of Probability, 16, 1723-1749, 2011. One ball is picked at random from urn 1 and, without. Hints help you try the next step on your own. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. For more practice, I suggest you work through the review questions at the end of each chapter as well. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. • Time limit 110 minutes. Note that to define a mapping from A to B, we have n options for f ( a 1), i. Probability Problems for Group 1 (Due by EOC Mar. Pull balls from the urn one by one without replacement. Question 3 Solution Urn 1: seven red and three green balls. Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the “inverse probability” problem. Let N be the number of throws of a usual six-sided die that is needed for the sum of the scores on these throws to be at least 3. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Urn simulator This calculator simulates urn or box with colored balls often used for probability problems and can calculate probabilities of different events. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. Let the probability that the urn ends up with more red balls be denoted. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. ) I'm happy to discuss about them on my office hours (unless we are too many on those). Bayes Theorem Practice Problems With Solutions Genetics. Question 3 Solution Urn 1: seven red and three green balls. This limit distribution is the negative binomial distribution with parameters and (the corresponding mathematical expectation is , while the variance is ). The formula is. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. The geometry problems are treated in a separate post. YOU are responsible for studying all the sections to be covered on the midterm. If your opponent draws first, what is the. That is, we seek a random integer n satisfying 1 ≤ n ≤ 3. , that the coin flip was Tails) Let T be the event that the coin flip was Tails. • Bose-Einstein distribution. Let X be the number of red balls removed before the first black ball is chosen. Thus, the probability that they both give the same answer is 39. The probability of an event is a measure of the likelihood that the event will occur. 4 Practice Problems Problem 34. After a severe winter, potholes develop in a state highway at the rate of 5. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. Alice selects one ball at random (each of the 7 balls can be selected with equal probability) and takes it out of the urn. Ghahramani, Fundamentals of Probability, Prentice Hall, 2000 1 Combinatorics These problems are due on August 24 Exercise 1. A primary concern with PoS systems is the “rich getting richer” phenomenon, whereby wealthier nodes are more likely to get elected, and hence reap the block reward, making them even wealthier. We're interested in the probability of a certain observed outcome given a known process. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials. The field of Probability has a great deal of Art component in it - not only is the subject matter rather different from that of other fields, but at present the techniques are not well organized into systematic methods. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. This is exactly the binomial experiment. I'm trying to answer the following question using a simple Monte Carlo sampling procedure in R: An urn contains 10 balls. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. k indistinguishable balls are randomly distributed into n urns. A room contains four urns. Remarkable selection of puzzlers, graded in difficulty, that illustrate both elementary and advanced aspects of probability. An urn contains pink and green balls. There are 40 marbles in an urn: (PROBABILITY) There are 40 marbles in an urn: 11 are green and 29 are yellow. Let the probability that the urn ends up with more red balls be denoted. What is the probability that both marbles are the same color if: a. b) Find the probability that urn 1 was used given that a red ball was drawn. Edited by Bernard R. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. Two marbles are drawn without replacement from the urn. In the case of rolling a 3 on a die, the number of events is 1 (there’s only a single 3 on each die), and the number of outcomes is 6. Let Event A = {The Ball From Urn No. The occupancy problem in probability theory is about the problem of randomly assigning a set of balls into a group of cells. Clas-sical mathematicians Laplace and Bernoulis, amongst others, have made notable contributions to this. An urn contains 10 balls: 4 red and 6 blue. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Definitions: Example #1. What is the probability a red marble is drawn?. This formula relies on the helper table visible in the range B4:D10. A Probability trees in probability theory is a tree diagram used to represent a probability space. Acknowledgements This work was made possible by a grant from NSF-DUE and the support of Cornell University and the statistics department at Stanford. If there are N urns, two colours, and balls are coloured independently with probability 0. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. Calculate the number of blue balls in the second urn. 1 Is Black}; Event B= {The Ball From Urn. Such problems have been considered by Nishimura and Sibuya [13, 14] and Selivanov [17]. Urn and Beads Problem The next family of problems was modeled after Edwards (1968). What are your chances of getting exactly 4, 5, or 6 matches? Many lotteries and gambling games are based on this concept of picking from mixed good and bad balls. Suppose you have n ball- lled urns, numbered 1 through n. We'll look at a number of examples of modeling the data generating process and will conclude with modeling an eCommerce advertising simulation. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn 2. Home-Learning Problems 3. Intro to Probability - Homework Assignment 2 Please solve the problems and type the solutions up using LateX and the template provided. So Urn B is more. $$\frac{\textrm{Probability of picking the 2nd urn and picking a blue ball}}{\textrm{Probability of picking a blue ball}}$$ Since there are equal numbers of balls in each urn, and each urn is equally likely to be picked, you can do this by counting. 2 Real Life Is More Complicated. Practice online or make a printable study sheet. If we draw 5 balls from the urn at once and without peeking,. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. three marbles are drawn from the urn one after the other. An urn contains 10 balls: 4 red and 6 blue. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. 1 2 ⋅ 1 2 = 1 4. Thus in this experiment each time we sample, the probability of choosing a red ball is $\frac{30}{100}$, and we repeat this in $20$ independent trials. Barbara and Dianne go target shooting. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. The probability of any event can range from 0 to 1. Sample Problem 1: A six-sided die is rolled six times. A card is randomly selected from an ordinary pack of 52 playing cards. 625 subscribers. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. An urn contains 10 red and 8 white balls. what is the probability that the equipment will fail before the end of one year? 2. • Put your NAME on each sheet. Then another ball is drawn and its color is recorded. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. I've tried two approaches and neither work. An important problem in the Ehrenfest chain is the long-term, or equilibrium, distribution of , the number of balls in urn A. Math 29 — Probability Practice Second Midterm Exam 1 Instructions: 1. The ticket shows your six good balls, and there are 50 bad balls. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. The Philosophy of Statistics [S]tatistical inference is firmly based on probability alone. An urn contains three red balls numbered 1, 2, and 3, three black balls numbered 8, 9, and 10 and four white balls numbered 4, 5, 6, 7. The probability of this happening is =~ 0. ) are represented as colored balls in an urn or other container. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. 3, 794-814, 2012. Find P(N = 1), P(N = 2) and P(N = 3). (b) Find EN and Var(N). Probability basics 50 xp Queen and spade 50 xp. We recommend you review today's Probability Tutorial before attempting this challenge. Roll a fair 6-sided die until your first roll of a “3. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). A ball is selected at random from the first urn and placed in the second. You can also express this relationship as 1 ÷ 6, 1/6, 0. Imagine two urns. Let us suppose that the urns are labelled with the numbers 1,2,, n and let tj be equal to k if the j-th ball is placed into the k-th urn. A Ball Is Randomly Drawn From Urn No. What is the probability that a white ball is drawn?. In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. An urn contains 5 red, 3 green, and 4 white balls. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. Two marbles are selected at random without replacement. 625 subscribers. After the four iterations the urn contains six balls. same answer. |A) is a probability function multiplicative formula: P(B and A) = P(B|A)P(A) Oct 28 Monty Hall problem §3. Bayes probabilities can also be obtained by simply constructing the tree. Then the number of tables is bounded by this multiple, so for large n, the probability of joining one of the k (fixed) tables is roughly , so this should behave roughly like the standard Polya Urn. What is the probability that the urn contains three balls of each color? Solution A tree diagram is ideal to deal with the problem, up to a sample space of 4 balls (so the first two pulls): So, we have a probability of (1/2)*(2/3) + (1/2)*(2/3) = 2/3 of ending up with 3 balls of one color. Then treat this like a mock exam (i. Home-Learning Problems 3. Show that the probability of rolling doubles with a non-fair (“fixed”) die is greater than with a fair die. Gelbaum DOVER PUBLICATIONS, INC. We write P (heads) = ½. Find P(N = 1), P(N = 2) and P(N = 3). What strategy maximizes your chance of victory? Problems from Rosen 7. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. What is the probability the ball drawn from urn C is black? I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. , TTHor HHHT). The probability a black ball is drawn is 9/13. In the example shown, the formula in F5 is: =MATCH(RAND(), D$5:D$10) How this formula works. If you pick 5 balls from the urn at random, what is the probability that x of them will be. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. There will be total 10 MCQ in this test. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. of getting white marble). From a given ball's perspective, the probability of being matched is $\frac{M}{N}(1-(1-\frac{1}{M})^N)$. The first problem-book of a similar kind as ours is perhaps Mosteller's well-known Fifty Challenging Problems in Probability (1965). Once again, balls are chosen at random with equal probability. At each step, an urn is selected according to their weights. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. The urn model and the Pólya process, in which the Pólya distribution and the limit form of it arise, are models with an after effect (extracting a ball of a particular colour from the urn increases the probability of extracting a ball of. He titled it The Two Children Problem, and phrased. User Account. of selecting 2 nd urn × prob. Walk through homework problems step-by-step from beginning to end. There are three ways to measure the average: the mean, median, and mode. Hume’s Problem. Create a set B of foods. Six balls are picked from the 56 balls in an urn. Urn #2 contains 5 black and 2 red balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Publication date 1987 Topics Probabilities, Probabilités, Probabilités Publisher Internet Archive Books. Then the probbilities that the urn from which I have drawn the tickets is A or B, are by the principle given above, as K: K'. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The first case,. By convention, statisticians have agreed on the following rules. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. The probability for the dice to yield 1 or 2 is 2/6. 8--Conditional Probability With Urns and Marbles Glenn Olson. The probability of a sample point is a measure of the likelihood that the sample point will occur. To Probability Practice Problems for Exam # 2 1. One ball is drawn at random. 8--Conditional Probability With Urns and Marbles Glenn Olson. For each new ball, with probability p,createanewbin and place the ball in that bin; with probability 1−p,placetheballin an existing bin, such that the probability the ball is placed in a bin is proportional to mγ,wheremis the number of balls in that bin. An Example Suppose that a 6-sided die is rolled, what is the probability that the result is a 1? From our earlier results, if A is the event that a 1 is rolled, then P(A) = 1 6. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. The probability the first is white is 6/15= 2/5. It is concluded that uncertainty theory is better than probability theory to. 6 balls are randomly drawn from the urn in succession, with replacement. What is the probability of getting exactly k red balls in a sample of size 20 if the sampling is done with replacement (repetition allowed)? Assume 0 ≤ k ≤ 20. One ball is drawn at random and its color noted. What is the probability of winning the. Fully worked-out solutions of these problems are also given, but of course you should first try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Binomial probability distribution; Hypothesis testing, types of error, and small samples; Introduction to statistics and the relative frequency histogram; Measures of variability and relative standing ; Other graphical methods and numerical methods; Poisson probability distribution and the urn model; Probability; Random variables. From a given ball's perspective, the probability of being matched is $\frac{M}{N}(1-(1-\frac{1}{M})^N)$. For example, if the probability of picking a red marble from a jar that contains 4 red marbles and 6 blue marbles is 4/10 or 2/5, then the probability of not picking a red marble is equal to 1 - 4/10 = 6/10 or 3/5, which is also the. So in a more conventional forward probability problem. What is the probability that it gets a) all 4 aces b) at least 1 ace; Hearts 5 cards are chosen from a standard deck of 52 playing cards (13 hearts) with replacement. Applications of probability arise everywhere: Should you guess. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, André 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. CAT Probability Questions cover all the different ways in which question can be asked. ) Jacob Bernoulli 1700. of selecting 1 st urn × prob. 1 Preliminaries This document is designed to get a person up and running doing elementary probability in R using the prob package. What strategy maximizes your chance of victory? Problems from Rosen 7. Another urn contains 2 red chips. Suppose that a white ball is selected. Column B contains the six. But what is the probability that when you throw a 6 sided unbiased die twice, the second output is 1 given that the first output is 5? Answer is 1/6, because P[first output is 5]=1 since it has already occurred. But anyways using the binomial theorem. Above the plane, over the region of interest, is a surface which represents the probability density function associated with a bivariate distribution. What is the probability that the first ball was also red?. This list of problems should serve as a good place to start studying, and it should not be considered a comprehensive list of problems from the sections we’ve covered. Now there are 13 balls, 4 white and 9 black. So the probability that urn iis empty is 1 11 i 1 i+1 1 1 i+2 1 1 n. A ball is drawn from Urn A and then transferred to Urn B. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. A ball is also chosen at random from urn #2. It does not matter who picked the first three WIN balls. One ball is drawn from an urn chosen at random. (d) The variance of X. The formula is. Annals of Probability, 41, no. 4 Conditional Probability and Independence 1. 2012 John Wiley & Sons, Inc. Two balls are drawn at random from each one of the urn A and urn B, and are placed into urn C. Let A = { a 1, a 2, a 3,, a m }, B = { b 1, b 2, b 3,, b n }. It is concluded that uncertainty theory is better than probability theory to deal with uncertain urn problems. Four balls are drawn at random. 2: 19, 23, 29, 33. If the composition is unknown, then it is called uncertain. urn problem: when to stop sampling? Then you find the probability that you would have drawn fewer than j white marbles if your urn had a certain proportion, say k, of black marbles. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. Find the expected number of stages needed until there are no more black balls in. It is useful in practice, and many elementary statistics textbooks use it to introduce the binomial and hypergeometric distributions. We have included a number of Discussion Topics designed to promote critical. For more practice, I suggest you work through the review questions at the end of each chapter as well. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. Find the probability of the event that the first bin contains balls of both colors. In the first draw, one ball is picked at random and discarded without noticing its colour. The flippant juror. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). Please justify your answers and don't simply give the answer. I recommend studying rst, using previous homeworks, exams and exam practice problems. Here's how. There are in nitely many Urns which we call Urn 0, Urn 1, Urn 2 etc. One ball is picked at random from urn 1 and, without. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. falling apart on inspection. The probability that a consumer will see an ad for this particular product is 0. A ball is drawn at random from an urn containing 15 green, 25 black, 16 white balls. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color; (b) of different colors?. (You'd think the urns would be empty by now. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. This contains all the class notes for the Introduction to Probability class currently taught at Stanford using these, and other web-resources. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. Combinatorics: The Fine Art of Counting. Bonus Problem: Problem 7. ) are represented as colored balls in an urn or other container. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Find P ( B C | A) from the Venn diagram: 3. Problem One. Suppose that this experiment is done and you learn that a white ball was selected. In this post, I’m going to discuss some of the non-geometry problems. Clue: Container in many probability-theory problems. If you know how to manage time then you will surely do great in your exam. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. The locomotive problem. We shall now apply this principle to the solution of several problems. It is used to solve problems of the form: how many ways can one distribute indistinguishable objects into distinguishable bins? We can imagine this as finding the number of ways to drop balls into urns, or equivalently to arrange balls and. The problem of induction is to find a way to avoid this conclusion, despite Hume's argument. The simplest experiment is to reach into the urn and pull out a single ball. Suppose that a machine shop orders 500 bolts from a supplier. Probability is the likelihood or chance of an event occurring. Urn i has exactly i 1 green balls and n i red balls. If the composition is unknown, then it is called. ♦ BALL AND URN (AoPS calls this "Stars and Bars") The classic "Ball and Urn" problem statement is to find the number of ways to distribute N identical balls into 4 distinguishable urns, for example. But our solution to the urn problem relied on random sampling. The Crossword Solver found 21 answers to the Container in many probability theory problems crossword clue. Wolfram Education Portal ». In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2. The Associated Press. Users that wish to investigate especially large or intricate problems are encouraged to modify and streamline the code to suit their individual needs. The theorem is also known as Bayes' law or Bayes' rule. Publication date 1987 Topics Probabilities, Probabilités, Probabilités Publisher Internet Archive Books. k indistinguishable balls are randomly distributed into n urns. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. This consists. 3, 794-814, 2012. • Time limit 110 minutes. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. EXAMPLE 1 A Hypergeometric Probability Experiment Problem: Suppose that a researcher goes to a small college with 200 faculty, 12 of which have blood type O-negative. (a) Draw marbles from a bag containing 5 red marbles, 6 blue marbles and 4 green marbles without replacement until you get a blue marble. Once you have decided on your answers click the answers checkboxes to see if you are right. 4 Conditional Probability and Independence 1. Hence the probability of getting a red ball when choosing in urn A is 5/8. You decide to actually count the balls in the urn, that you friend has in problem 6, and discover that there are 6 green, 6 yellow, 7 blue, 4 black, and 2 red balls. Some of these matrices will be used in calculation in subsequent posts. In Problem 3, suppose that the white balls are numbered, and let Y i equal 1 if the i th white ball is selected and 0. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. Bayes Theorem Practice Problems With Solutions Genetics. a) 4 ⁄ 7 b) 3 ⁄ 7 c) 20 ⁄ 41 d) 21 ⁄ 41 View Answer. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. In the first urn, there are $20\%$ red balls, so the probability to draw a red ball is $0. Winning an unfair game. The first limitation is that the ravens we observe in real life aren't randomly sampled from nature's "urn". Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the “inverse probability” problem. Ask Question Asked 5 years, 5 months ago. (The two marbles might both be black, or might both be white, or might be of different colors. When we have multiple event probabilities, as in the problem. Now there are 14 balls, 5 white and 9 black. 57 % of the children in Florida own a bicycle. An urn contains 9 red, 7 white and 4 black balls. ) Let N be the number of games played. Probability measures the likelihood of an event occurring. That is, after each draw, the selected ball is returned. balls numbered 1 through 10 are placed in the urn and ball number 1 is withdrawn; at 1 2 minute to 12 P. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. If the probability of an event is 0, it is impossible for that event to occur. Probability (MATH 4733 - 01) Fall 2011 Exam 2 - Practice Problems: Selected answers and hints Due: never Here are some sample problems for you. Among the balls are R red and N-R white balls. These urn models are also excellent practice problems on thinking about Markov chains and deriving the transition probability matrices. Two balls are chosen randomly from an urn containing8 white, 4 black, and 2 orange balls. What is the probability that they come from urns I, II or III? 36. Recommended Problems: Problems 1. Combinatorics: The Fine Art of Counting. The probability of any event can range from 0 to 1. The probability of a sample point is a measure of the likelihood that the sample point will occur. What is the probability that the second card. Again, one ball is drawn at random from the urn, then replaced along with an additional ball of its color. Supposethat we win$2 for each black ball selectedand we lose \$1 for each white ball selected. Depending on whether we sample with or without replacement, the chance (or probability) of getting m red balls (successes) in a sample of n balls changes. Find the probability that both the first and last balls drawn are black. Two of the selected marbles are red, and three are green. There are related clues (shown below). Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. ) are represented as colored balls in an urn or other container like box. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Chapman-Kolmogorov Equations We have already defined the one-step transition probabilities [pic]. Construct an 80% confidence interval for the proportion of red marbles…. Then treat this like a mock exam (i. Euler became professor of physics in 1731, and professor of mathematics in 1733, when. The probability X failing during one year is 0. An urn contains n+m balls, of which n are red and m are black. Lindley The Statistician: Journal of the Royal Statistical Society. Recommended Problems: Problems 1. An urn has 4 green balls, 5 yellow balls, and 6 red balls. Wolfram Education Portal ». In other words, in order to get a new value of seed, multiply the old value by 7621, add 1, and, finally, take the result modulo 9999. One ball is drawn at random. Let B 1 W 2 denote the outcome dial the first bull drawn is B 1 and the second ball drawn is W 2. Hence a probability of 0. Urn problems and how to approach them Probability and real-life situations (lottery, poker, weather forecasts, etc. Practice Problem: A certain lottery has a hat with the numbers 1 through 10 each written on a single scrap of paper. What is the probability the ball drawn from urn C is black? I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. You can also express this relationship as 1 ÷ 6, 1/6, 0. So, if the probability that they actually had a flat tire is p, then the probability that they both give the same answer is 1 4 (1−p)+p = 1 4 + 3 4 p. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Do Problem 2 ONLY please. Urn 2 contains three red balls and four white balls. Two red, three white, and five black.
dihvq33d7wefb2,, y5l6ztaewnu,, aszj3xrx30cx1,, e2hmukdh2wbjc,, pe5t81bd5ppwuy,, 109r3mxtwj0lb,, kouxd3ppqx4vaq,, ihdi27tl3fpn,, uxbuwpf31d,, 1mrdb7rz1525,, 8bey8mxlsa,, 6xicx8gk5w2am,, zqlep6c2hmm3b,, ckrev1jmnb,, z2140axliao94y,, ewajmk04pj2,, vfab5wrqgcz,, cunk5n2d708kqbj,, g768d3iqeid,, vwyzoax0y2,, kjjikb278y3,, itd3lm4gh5c,, enbm2xmq16be,, 13ii9xt7ryydkiy,, dc5vl2gzae0,, uhvr12pbsunv,, bsbn1y8c9oqe2w,, pltahmocfuaj,, q57csdq4hx6a,, utljhpuzwu8u1k,, 8noue630pb3,, z4ic81qbwrok7n,, 4e3zkq2gpft1syb,, nemg4dh944ov58d, | 2020-10-24T22:54:15 | {
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http://bootmath.com/if-a-and-b-are-sets-of-real-numbers-then-a-cup-bcirc-supseteq-a-circcup-bcirc.html | # If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$
I have a proof for this question, but I want to check if I’m right and if I’m wrong, what I am missing.
Definitions you need to know to answers this question: $\epsilon$-neighborhood, interior points and interiors. Notation: $J_{\epsilon}(a)$ means a neighborhood formed around a (i.e. $(a-\epsilon, a+\epsilon)$. An interior point in some set $A$ is a point where an $\epsilon$-neighborhood can be formed within the set. The set of all interior points in $A$ is denoted as $A^0$ and is called the interior.
My proof for the question: I’m proving based on most subset proofs where you prove that if an element is in one set, then it must be in the other. Say $x \in A^0 \cup B^0$. Then there is a $\epsilon > 0$, where $J_{\epsilon}(x) \subseteq A$ or $J_{\epsilon}(x) \subseteq B$. This implies that $J_{\epsilon}(x) \subset A \cup B$ which then implies $x \in (A \cup B)^0$. We can conclude from here that $(A \cup B)^0 \supseteq A^0 \cup B^0$.
#### Solutions Collecting From Web of "If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$"
Your proof is good- there is no problem with it. I have reworded a bit to make things a little more clear, specifically, where you say:
Then there is $\epsilon>0$, where $J_\epsilon (x)\subseteq A$ or $J_\epsilon (x)\subseteq B$.
Instead it would be more appropriately stated as: If $x\in A^0\cup B^0$ then $x\in A^0$ or $x\in B^0$. Then you can assume $x$ is in $A$, prove that $x\in (A\cup B)^0$, by symmetry the same argument will work if $x\in B^0$.
I have included an edited proof- but as I originally stated your proof is good,
this just more directly reflects the definitions involved.
Let $x\in A^0\cup B^0$. Then $x\in A^0$ or $x\in B^0$. Assume $x\in A^0$. Then there is, by definition, $\epsilon>0$ such that $J_{\epsilon}(x)\subseteq A$. Since $A\subseteq A\cup B$, we also have $J_{\epsilon}(x)\subseteq A\cup B$. Thus by definition we have $x\in (A\cup B)^0$. If instead $x\in B^0$, then by symmetry the same argument works. Thus, $A^0\cup B^0 \subseteq (A\cup B)^0$.
Here is a slightly shorter proof:
We have $A^\circ \subset A \cup B$ and $B^\circ \subset A \cup B$, so we must have
$A^\circ \cup B^\circ \subset A \cup B$.
Since $A^\circ \cup B^\circ$ is open and the interior is the largest open set contained in a set, we must have
$A^\circ \cup B^\circ \subset (A \cup B)^\circ$. | 2018-07-17T17:43:43 | {
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https://byjus.com/question-answer/if-tan-dfrac-theta-2-mathrm-cosec-theta-sin-theta-then-sin-2-dfrac-theta/ | Question
# If $$\tan\dfrac {\theta}{2}=\mathrm{cosec}\theta-\sin\theta$$, then
A
sin2θ2=2sin218o
B
cos2θ+2cosθ+1=0
C
sin2θ2=4sin218o
D
cos2θ+2cosθ1=0
Solution
## The correct options are A $$\sin^2\dfrac {\theta}{2}=2 \sin^2 18^o$$ D $$\cos 2\theta+2 \cos\theta-1=0$$Simplifying, we get $$\dfrac{\sin \dfrac {\theta}{2}}{\cos \dfrac {\theta}{2}}=\dfrac{1-\sin ^{2}\theta}{\sin \theta}$$ $$\dfrac{\sin \dfrac {\theta}{2}}{\cos \dfrac {\theta}{2}}=\dfrac{\cos ^{2}(\theta)}{2\sin \dfrac {\theta}{2}.\cos \dfrac {\theta}{2}}$$ $$2\sin ^{2}\dfrac {\theta}{2}=\cos ^{2}\theta$$ $$2\sin ^{2}\dfrac {\theta}{2}=(1-2\sin ^{2}\dfrac {\theta}{2})^{2}$$ Let, $$2\sin ^{2}\dfrac {\theta}{2}=t$$$$\implies t=(1-t)^{2}$$ $$\implies t=t^{2}-2t+1$$ $$\implies t^{2}-3t+1=0$$ $$\implies t=\dfrac{3\pm\sqrt{9-4}}{2}$$ $$\implies t=\dfrac{3\pm\sqrt{5}}{2}$$ Now $$|\sin \theta|\leq 1$$ $$\implies t=\dfrac{3-\sqrt{5}}{2}$$ Or $$\sin ^{2}\dfrac {\theta}{2}=\dfrac{3-\sqrt{5}}{4}$$ $$\implies 2\left(\dfrac{3-\sqrt{5}}{8}\right)=2\sin ^{2}(18^{0})$$ Hence, option A is correct. $$\cos ^{2}\dfrac {\theta}{2}=1-\sin ^{2}\dfrac {\theta}{2}=\dfrac {1+\sqrt{5}}{4}$$ Option D $$\cos 2\theta +2\cos \theta -1=0$$ $$-2{ \sin }^{ 2 }\theta +2\cos \theta =0$$ $$=-2+2{ \cos }^{ 2 }\theta +2\cos \theta =0$$ $$=-1+{ \cos }^{ 2 }\theta +\cos \theta =0$$ $${ \cos }^{ 2 }\theta =1-\cos \theta =2{ \sin }^{ 2 }\cfrac { \theta }{ 2 }$$ which is true (as proven in the first part)Hence, option A and D are correct.Maths
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View More | 2022-01-19T05:30:15 | {
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http://math.stackexchange.com/questions/264947/find-missing-number-from-sum-of-first-few-natural-numbers | # Find missing number from sum of first few natural numbers
A child was asked to add the first few natural numbers $1+2+3+...$ as long as his patience permitted. As he stopped, he gave the sum as $575$. When the teacher declared the result wrong, the child discovered that he had missed a number in the sequence during addition. What was the number he missed?
-
Does the teacher know which number the child was adding up to? – Michael Albanese Dec 25 '12 at 14:47
lol. I think its a way to frame the question. I too doubt on that same as you. =) – harish.raj Dec 25 '12 at 14:49
If the teacher doesn't know the number, then my answer is accurate. If the teacher does know the number, the inequality $\frac{n(n+1)}{2} - 575 < n$ should be replaced by $\frac{n(n+1)}{2} - 575 \leq n$ as the reasoning in the bracket following the original inequality no longer applies. However, this does not change the final answer. – Michael Albanese Dec 25 '12 at 15:12
We have $$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}.$$ The numbers obtained from such sums are called Triangular numbers.
As the child's answer was $575$, the correct answer must be greater than $575$ (as he missed a number). The first triangular number greater than $575$ is $595$ corresponding to $n = 34$. If this was the sum he was trying to evaluate, the child must have missed $20 = 595 - 575$. However, if the correct answer was the next triangular number, $630$, corresponding to $n = 35$, the only way the child could have obtained $575$ is if he missed $55 = 630 - 575$, but this is not in the list of numbers $1, \dots, 35$ so he could not have been trying to add up the first $35$ natural numbers.
More generally, we need $\frac{n(n+1)}{2} > 575$ (the actual sum needs to be greater than the sum he obtained) and $\frac{n(n+1)}{2} - 575 < n$ (this difference tells us which value he missed, and it can't equal $n$ because then he would have written $1 + \dots + (n - 1)$ which is a triangular number, so the teacher could not have declared it wrong). Rewriting, we must find $n$ such that $$0 < \frac{n(n+1)}{2} - 575 < n\quad (\ast)$$ which only has solution $n = 34$. Therefore, we must be in the situation of the previous paragraph, so he missed the number $20$.
Added later: I just thought I'd add some details here about finding natural numbers $n$ which satisfy $(\ast)$.
First of all, consider the left hand inequality which rearranges to $f(n) := n^2 + n - 1150 > 0$. Solving $f(n) = 0$, we obtain $n = \frac{-1\pm\sqrt{4601}}{2}$; as $n > 0$, we ignore $n = \frac{-1-\sqrt{4601}}{2}$. Note that $\left\lfloor\frac{-1+\sqrt{4601}}{2}\right\rfloor = 33$, and we have $f(33) < 0$ and $f(34) > 0$. So $f(n) < 0$ for $n \in \{1, \dots, 33\}$ and $f(n) > 0$ for $n \geq 34$.
The right hand inequality rearranges to $g(n) := n^2 - n - 1150 < 0$. Solving $g(n) = 0$, we obtain $n = \frac{1\pm\sqrt{4601}}{2}$; as $n > 0$, we ignore $n = \frac{1-\sqrt{4601}}{2}$. Note that $\left\lfloor\frac{1+\sqrt{4601}}{2}\right\rfloor = 34$, and we have $g(34) < 0$ and $g(35) > 0$. So $g(n) < 0$ for $n \in \{1, \dots, 34\}$ and $g(n) > 0$ for $n \geq 35$.
Hence, $n = 34$ is the only natural number which satisfies $(\ast)$.
-
great explanation. Thank you a lot. – harish.raj Dec 25 '12 at 14:31
+1 Michael: nice explanation! – amWhy Dec 25 '12 at 14:35
Suppose he tried to sum the first $n$ positive integers. Omitting a term means this sum is less than the sum of the first $n$ positive integers but not less than the sum of the first $(n-1)$ positive integers. Hence, $\frac{n(n-1)}{2}\leq 575<\frac{n(n+1)}{2}$ so that $n=34$. This means the missing term is $595-575=20$.
-
awesome. can you elaborate a little about how you got the value of n? (it will be great if you do it in your answer itself) – harish.raj Dec 25 '12 at 14:15 | 2015-04-19T01:51:44 | {
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http://mathhelpforum.com/trigonometry/41342-complex-numbers.html | # Math Help - complex numbers
1. ## complex numbers
Hi all,
I am having trouble getting the answer to:
[(1-sqrt(3)i)/ (sqrt(3) + i)]^7
where i = complex number
ArTiCk
2. Originally Posted by ArTiCK
Hi all,
I am having trouble getting the answer to:
[(1-sqrt(3)i)/ (sqrt(3) + i)]^7
where i = complex number
ArTiCk
My advice is to first convert the numbers into polar form.
3. Originally Posted by mr fantastic
My advice is to first convert the numbers into polar form.
I have already converted the numbers into polar form...
i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]
The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:
e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)
I think it is there where i have gone wrong, help would be appreciated
Thanks, ArTiCk
4. Hello, ArTiCK!
Simplify: . $\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7$
Ratinalize the "inside" first . . .
. . $\frac{1-i\sqrt{3}}{\sqrt{3} + i}\cdot\frac{\sqrt{3}-i}{\sqrt{3}-i} \;=\;\frac{\sqrt{3} - i - 3i +i^2\sqrt{3}}{3-i^2} \;=\;\frac{\sqrt{3} -4i-\sqrt{3}}{3+1} \;=\;\frac{-4i}{4} \;=\;-i$
So we have: . $(-i)^7 \;=\; i$
5. Originally Posted by ArTiCK
I have already converted the numbers into polar form...
i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]
The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:
e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)
I think it is there where i have gone wrong, help would be appreciated
Thanks, ArTiCk
$\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$
$\left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.
6. Hello, Mr. F !
Originally Posted by mr fantastic
$\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$
$\left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.
Lovely! | 2015-05-25T20:49:26 | {
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https://math.stackexchange.com/questions/2877172/determine-jordan-block-size | # Determine Jordan block size.
The following question is from as System Theory test.
Let the system matrix $A$ be given as $A = \begin{bmatrix} 0&0&0&1\\0&-1&1&3\\0&1&-1&-1\\0&-1&1&2 \end{bmatrix}$
Which has the eigenvalues $\lambda_i=0$ for $i = 1,2,3,4$. Which of the following matrices is the Jordan form of $A$.
$A) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$
$B) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$
$C) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$
$D) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$
My approach:
The eigenvalues are given so the nullspace of $\left[A-\lambda I\right]$ can be found to be $\begin{bmatrix} 0\\1\\1\\0 \end{bmatrix}$ and $\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}$.
The dimension of this nullspace is $2$ which indicates that there are 2 Jordan blocks. So answer $B$ or $C$.
Also I've found that : $\operatorname{rank}(\ker(A-\lambda I)^2)=3$ and $\operatorname{rank}(\ker(A-\lambda I))=2$. Subtracting these outcomes gives $3-2=1$. So there's one Jordan block of size 2 or larger. And thus only answer $B$ is correct.
Right or wrong? thanks in advance.
• It's perfect for me. – Bernard Aug 9 '18 at 13:10
Your argument is correct, but your notation is innapropriate. Where you wrote $\operatorname{rank}(\ker\cdots)$, you should have written $\dim(\ker\cdots)$.
I like to find the actual Jordan form, especially including the matrices that give $R^{-1}A R = J.$ It is not bad when the eigenvalues are integers, and makes concepts concrete. This begins with choosing a column vector i called $w$ such that $A^2 w \neq 0.$ That becomes the far right column.
$$\frac{1}{8} \; \left( \begin{array}{cccc} 8&-4&-4&0\\ 0&1&3&0\\ 0&2&-2&0\\ 0&-4&4&8\\ \end{array} \right) \left( \begin{array}{cccc} 0&0&0&1\\ 0&-1&1&3\\ 0&1&-1&-1\\ 0&-1&1&2\\ \end{array} \right) \left( \begin{array}{cccc} 1&2&1&0\\ 0&2&3&0\\ 0&2&-1&0\\ 0&0&2&1\\ \end{array} \right) = \left( \begin{array}{cccc} 0&0&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{array} \right)$$ | 2019-03-25T10:06:29 | {
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http://math.stackexchange.com/questions/118198/trying-to-prove-that-a-group-is-not-cyclic | # Trying to prove that a group is not Cyclic
Given the following Euler groups : \begin{align*} U_{12} &= \{1,5,7,11\}\\ U_{16} &= \{1,3,5,7,9,11,13,15\} \end{align*}
I want to prove that they are not cyclic. I used the following theorem :
A group of order $n$ is cyclic if and only if it has an element of order $n$.
Let's take for example $U_{12}$: (I will use the notation of $o(x)$ to denote the order of the element $x\in G$)
\begin{align*} o(5)&\colon 5^2=25 \to 25\bmod 12 = 1\to o(5)=2\\ o(7)&\colon 7^2= 49 \to 49\bmod 12 = 1 \to o(7)=2\\ o(11) &\colon 11^2 = 121 \to 121\bmod 12=1 \to o(11)=2 \end{align*}
Then by using the above theorem , this group is indeed not a cyclic group.
Question : do I really have to check each element in the group for its order ?
Regards
-
Fixed. thanks ! – ron Mar 9 '12 at 11:58
BTW, the structure of the multiplicative group of integers modulo n $U_n$ is will known. In particular, it is cyclic iff $n$ is 2, 4, any power of an odd prime or twice any power of an odd prime. – lhf Mar 9 '12 at 12:08
Please learn some basic $\LaTeX$ instead of waiting for others to pretty up your posts for you. Thank you. – Arturo Magidin Mar 9 '12 at 16:45
Comment: Stop using $\to$ to mean "and since" or some other connectives. The arrow has specific meanings in mathematics; students who use $\to$ and $=$ as general purpose connectives to mean something like "and then I did some thinking and this is what I came up with" drive professors crazy. – Arturo Magidin Mar 9 '12 at 16:49
@Arturo, I don't think driving professors crazy is the real problem - for most of us, crazy is within walking distance, anyway - I think the real problem is the student who writes that way is pretty much guaranteed to get things wrong. – Gerry Myerson Mar 10 '12 at 6:58
Well really the "theorem" is the definition of cyclic group...it is a group generated by one element.
Now if I give you a group and you check that all but one element isn't a generator, how do you know that the one you didn't check isn't a generator?
So yes, you must check ALL elements.
-
... or you can use interesting theorems. For example, there's an easy upper bound on the number of elements of order 2.... – Hurkyl Mar 9 '12 at 14:28
Some shortcuts are available. For example, when testing $3$ in $U_{16}$, you find its powers are $3,9,11,1$, so you don't have to test $9$ or $11$. Do you see why?
-
I didn't get your meaning .What do you mean by "its powers" ? 3^3 = 27 --> 27 modulo 16= 11 .... so I generated using 3 , the element 11 . What does it mean ? – ron Mar 9 '12 at 12:11
You're trying to decide whether the order of $3$ is $8$. So you look at $3$, then at $3\times3=9$, then at $3\times3\times3=3\times9=27=11$, then at $3\times3\times3\times3=3\times11=33=1$. Those are the powers of $3$, the numbers $3^r$, $r=1,2,3,\dots$. Using $3$, you generate $3$, $9$, $11$, and $1$. – Gerry Myerson Mar 9 '12 at 12:34
Okay , so now we know that the order of |3|=4 . Why does it help me ? now I won't need to check the generated elements using 3 ? – ron Mar 9 '12 at 12:42
"Why does it help me ?" This is an exercise for the reader -- use knowledge about the order of $3$ in order to deduce something about the order of $3^2$.... – Hurkyl Mar 9 '12 at 14:26
You should see if you can prove that if $g$ doesn't generate $G$ and $h$ is in the subgroup generated by $g$ then $h$ doesn't generate $G$ either. – Gerry Myerson Mar 10 '12 at 6:59
As I mentioned yesterday in reply to a question of yours: a cyclic group has at most one element of order $2$.
In your first two lines, you've shown that $5$ and $7$ both have order $2$ in $U_{12}$. Therefore, the group cannot be cyclic.
Similarly, since $9^2\equiv 1\pmod{16}$ and $15^2\equiv 1\pmod{16}$, $U_{16}$ has at least two distinct elements of order $2$, and therefore cannot be cyclic.
In general, a cyclic group of order $n$ has exactly $\phi(d)$ elements of order $d$ for each divisor $d$ of $n$.
-
What of if your set is infinite? I think the best idea is to find the general form of the elements in the set (if possible) and pick arbitarary element, check your assertion and then generalised. But in your case, since the elements are finite, you have to check it for all the elements in the set.
- | 2016-07-28T16:47:11 | {
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https://cs.stackexchange.com/questions/89842/average-case-analysis-for-finding-max-and-min-value-on-an-array | # Average Case Analysis for finding max and min value on an array
Given the following algorithm, to find the maximum and minimum values of an array - don't mind the language:
MaxMin(A[1..n])
max = A[1];
min = A[1];
for (i = 2; i<=n; i++)
if (A[i] > max)
max = A[i];
else if (A[i] < min)
min = A[i];
print(max, min);
I need to do a probabilistic analysis for the average case of comparisons that will be made on its execution.
So far, my solution is:
Given an indicator random variable: $$X_i = \begin{cases} \text{1, if max > A[i]}\\ \text{0, if max < A[i]}\\ \end{cases}$$
and assuming an uniform distribution for $A[1..n]$, the expected probability is: $$E[x] = \sum\limits_{i=1}^{n} Pr(X_i)$$
where $$Pr(X_i)$$ is the probability of the i-th element be the $max$ element in $A[1..n]$. It's possible to determine that: $$Pr(x_1) = 1, Pr(x_2) = 1/2, Pr(x_3) = 1/3 ,..., Pr(x_n) = 1/n$$ and thus for an array of size $n$ the expected value can be calculated as:
$$E[x] = 1 + 1/2 + 1/3 + ... + 1/n = \sum\limits_{i=1}^{n}{\frac{1}{i}} \approx \log{n}$$
And the same goes for $min$, which will give the same result $\log{n}$. (1)
My questions are: is it correct? Does the complexity for the average case of the given algorithm is $\theta(\log{n})$? Can I use the argument pointed by (1), just modifying $X_i$ so: $$X_i = \begin{cases} \text{1, if min < A[i]}\\ \text{0, if min > A[i]}\\ \end{cases}$$
I already read this (unfortunately the link to the video isn't available), but it only explains for $max$ and my analysis must be for both $max$ and $min$.
• What do you mean "average case of comparisons"? Do you mean the average quantity of comparisons? That can't possibly be less than $n-1$, so $\log n$ is wrong. Perhaps you should edit the question to clarify? – Wildcard Mar 27 '18 at 1:46
• Of course your analysis assumes the array is in random order. Many times arrays are not in random order. Like sorted arrays. – gnasher729 Mar 29 at 19:59
First, note that the first comparison will always compare $$n-1$$ times, independent of the distribution of the input. So, what you really want to know is how many times the second part of the if compares. For this, you can use a indicator random variable like this:
$$X_i = \begin{cases} \text{1, if A[i] \le \max}\\ \text{0, if A[i] > \max}\\ \end{cases}$$
So, just like you did, assuming an uniform distribution for $$A[1..n]$$, the expected probability is:
$$E[x] = \sum\limits_{i=1}^{n} Pr(X_i)$$
But we do not want $$Pr(i=\max)$$, we want $$\overline{Pr(i=\max)}$$, like we stated before. So, using your probability of the $$i-th$$ element to be the $$max$$ element:
$$Pr(i = \max) = 1/i$$
We have this complement:
$$\overline{Pr(i=\max)} = (1 - 1/i)$$
From which we can put in the expected probability:
$$E[x] = \sum\limits_{i=2}^{n} (1 - 1/i) = \sum\limits_{i=2}^{n}1 - \sum\limits_{i=2}^{n}1/i = n - 1 - (\ln|n| - 1 + \Theta(1))$$
So, to get the total quantity of comparisons, we can just sum the number of comparisons of the two parts. Note that the $$\Theta(1)$$ can absorb a constant values.
$$(n - 1) + n - 1 - \ln|n| + 1 + \Theta(1) = 2n - 1 - \ln|n| + \Theta(1)$$
And we get the expected total number of comparions:
$$2n - \ln|n| + \Theta(1)$$
(1) You are looking for either min or max. Then,
(1-1) is it correct? Yes, it is correct.
(1-2) Does the complexity for the average case of the given algorithm is $\Theta(logn)$? Indeed it is. My experience in practice is that it does not work that way (or at least the is a large constant involved).
(1-3) Can I use the argument pointed by (1), just modifying $X_i$? Yes, you can do that too.
(2) You are looking for both min and max. Then, the scenario is different. You should have something like: $$Z_i = \begin{cases} \text{1, if max > A[i] and min < A[i]}\\ \text{0, if max < A[i] and min > A[i]}\\ \end{cases}$$
Then you can calculate $E[Z]$. It is also possible to calculate $E[XY]$, where $X$ is for max, and $Y$ is for min. Try different techniques and see which one works easier.
• In my case, I'm interested on your second explanation (for $Z_i$). Really liked your explanation, but @LionsWrath explanation tackled my question right on the spot! Tough, thank you for your explanation :) – woz Mar 27 '18 at 11:09 | 2019-07-17T06:50:07 | {
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https://math.stackexchange.com/questions/3321649/prove-textordam-frac-textlcmm-nm | # Prove: $\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$ [duplicate]
This is Pinter $$10.G.5$$
Let:
$$a \in G$$
$$\text{ord}(a) = n$$
Prove: $$\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$$
Use $$10.G.3$$ and $$10.G.4$$ to prove this.
Here is $$10.G.3$$:
Let $$l$$ be the least common multiple of $$m$$ and $$n$$. Let $$l/m = k$$. Explain why $$(a^m)^k = e$$.
Here is $$10.G.4$$:
Prove: If $$(a^m)^t = e$$, then $$n$$ is a factor of $$mt$$. (Thus, $$mt$$ is a common multiple of $$m$$ and $$n$$.)
Conclude that: $$l = mk \leq mt$$
OK, let's begin.
By $$10.G.3$$:
$$(a^m)^{\text{lcm}(m,n)/m} = e \tag{5}$$
If $$\text{lcm(m,n)}/m$$ is the lowest number such that (5) is true, then:
$$\text{ord}(a^m) = \text{lcm}(m,n)/m \tag{9}$$
Let's assume that there is a number
$$q < \text{lcm}(m,n)/m \tag{7}$$
such that:
$$(a^m)^q = e \tag{6}$$
By $$10.G.4$$ with (6):
$$n \ \big|\ mq$$
$$l \lt mq$$
$$\text{lcm}(m,n) \leq mq$$
Isolate q:
$$\text{lcm}(m,n)/m \leq q \tag{8}$$
So (9) must be true.
That's how I approached it. If anyone notices any issues, I'd be happy to know about them.
Even if it is considered correct, do you feel there is a better way?
• You should use \le rather than <=. – Angina Seng Aug 13 '19 at 2:52
• @LordSharktheUnknown Thanks! Updated! – dharmatech Aug 13 '19 at 2:54
• @JyrkiLahtonen Can you explain how this is a duplicate of the question you linked to? There is no requirement in this question that G be cyclic. Whereas the question you linked to is for cyclic groups. – dharmatech Aug 13 '19 at 8:49
• @José Carlos Santos Do you really think that it's a duplicate? – Michael Rozenberg Aug 13 '19 at 9:06
• @YuiToCheng Once again not a proper dupe target becauyse - among other things - the OP is working with LCMs not GCDs, and is working from specific lemmas. Please be more careful in choosing proper dupe targets. – Bill Dubuque Aug 14 '19 at 13:32
Yes, $$\,(a^{m})^{k} = 1\!\! \overset{(1)\!\!}\iff n\mid mk \iff m,n\mid mk\!\! \overset{(2)\!\!}\iff {\rm lcm}(m,n)\mid mk\iff {\large{\frac{{\rm lcm}(m,n)}m\,\mid}}\, k$$
$$\!(1)$$ and the (omitted) final conclusion is by Corollary' here, and $$(2)$$ is the LCM Universal Property
• @BillDubuque Do you really think this question has not been handled on our site? The OP's complaint about my choice of dupe was mostly because "their version is not only about cyclic groups". Yet they only work within the cyclic group generated by $a$. – Jyrki Lahtonen Aug 14 '19 at 6:18 | 2020-03-31T08:00:47 | {
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https://math.stackexchange.com/questions/2813157/open-and-closed-implication-formula | # Open and closed implication formula
So I am a newbie in mathematical logic and one of the problems I have faced throughout this one week of study is the one concerning open or closed statement. Let me know if I have used some terms such as "statement", "sentence", "formula", etc in this question the wrong way.
I posted probably a related question before here: Connection between universal Quantifier and implication, and have stumbled upon new terms such as "open", "closed", "free", and "bound". I still need some confirmation about this but I state a different question.
So, am I right to say that $\forall x[x\in \mathbb{R}\implies x^2\geq0]$ is a closed formula since I know it has a no free variable?
Does every closed formula have a truth value?
I know that $x\in \mathbb{R}\implies x^2\geq0$ is a formula and is true for every assignment of $x$, or IS it? Why can't I just conclude that $x\in \mathbb{R}\implies x^2\geq0$ is just true and why should I consider its hidden universal quantifier (which, of course, it does not sound the same using existential quantifier)? Is it just for the sake of the existing rule of making sentence in FOL?
Probably the same question: Why can't all quantifiers be bounded quantifiers, and be written that way, considering the existence of domain of discourse? Why don't we explicitly write that domain in the sentence? Meaning, instead of saying "In the domain of natural numbers, $\forall x[x\geq0]$", why not simply $"\forall x\in \mathbb{N}[x\geq0]"$?
Hope my confusion is understood as a newbie. Thanks for the help! :D
• Correct; the formula $∀x[x∈R ⟹ x^2≥0]$ is a closed formula (or sentence) because it has no free occurrences of variables. Jun 9 '18 at 9:16
• Yes; a closed formula has a definite truth value in an interpretation. $\forall x (x=0)$ is FALSE in $\mathbb N$, while an open one, like e.g. $(x=0)$ may change truth value (for a specific interpretation) according to the value assigned to the free variable $x$. Jun 9 '18 at 9:18
• You can see Classical Logic and Model Theory and Logical Truth and Logical Constants for an introduction. Jun 9 '18 at 9:43
• The last point is that usually the "pure" presentation of first-order logic has no predicate constant, like $\in$. Thus the way to formalize $∀x \in \mathbb N [x≥0]$ is $∀x[N(x) \to x≥0]$ where $N(x)$ is a unary predicate. Now again, the truth-value of the sentence depends on the way to interpret $N(x)$. Jun 9 '18 at 10:04
• Thus, from the "pedagogical" point of view, we have reduced restricted quantification to quantification of a conditional: thus, in any case, we have to first learn how to manage quantifiers. Jun 9 '18 at 10:11
The "founding" idea is that logic must be formal.
This idea was firstly investigated by Aristotle : the modern way to investigate the concept "formal" is to define it through syntax :
a formula [a grammatically correct expression] is an expression built-up according to specific rules.
Then we have semantics :
the way to give meaning (and truth value) to an expression through an interpretation.
The interaction of syntax and semantics is the key-point of modern formal logic.
You can see : John MacFarlane, WHAT DOES IT MEAN TO SAY THAT LOGIC IS FORMAL (2000):
What does it mean, then, to say that logic is distinctively formal?
Three things: logic is said to be formal (or “topic-neutral”)
(1) in the sense that it provides constitutive norms for thought as such,
(2) in the sense that it is indifferent to the particular identities of objects, and
(3) in the sense that it abstracts entirely from the semantic content of thought.
Thus, "topic neutral" and "to abstract entirely from the semantic content" mean to separate semantics from syntax : the domain of interpretation is obviously semantical.
• I get it now. The idea is to formalize that sentence formed with bounded quantifier. So, my last point, how do you think about introducing
– bms
Jun 9 '18 at 13:17
• get it now. The idea is to formalize that sentence formed with bounded quantifier. So, my last point, what is a reasonably good way to introduce my student the way of proving an implication? Do you think it is fine to assume (with careful context reading) that there are always hidden universal quantifiers before the implication statement that bound the variables in the hypothesis, and tell them that? Hence, they will always formally start with "Take an arbitrary x. If p(x) is true, then.... Thus q(x)". I ask this since my usual understanding is
– bms
Jun 9 '18 at 13:24
• that "for all" sentence is equivalent to "If, then" sentence because the way we prove them is essentially the same.
– bms
Jun 9 '18 at 13:26 | 2021-10-24T09:25:59 | {
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https://math.stackexchange.com/questions/1181129/why-are-the-probability-of-rolling-the-same-number-twice-and-the-probability-of | # Why are the probability of rolling the same number twice and the probability of rolling pairs different?
Two scenarios:
1. Using one die, roll a 6 twice.
$\frac16\times\frac16=\frac1{36}$
1. Rolling two dice roll the same number (a pair).
$\frac6{36}=\frac16$
Why are these two probabilities different? Because the events are independent, isn't rolling a pair the same as rolling a die twice?
In a sense, rolling two dice at once is the same as rolling 1 die twice at the same time? How does this "timing" issue affect the probability?
In the second case, your pair can be any one of $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ and you'd satisfy "obtaining a pair".
That gives you six possible pairs, each of which one has probability of occurring $\dfrac 1{36}$ gives us $$6\times \frac 1{36} = \frac 16$$
Now, if you want to know what the probability of rolling two dice simultaneously and obtaining two sixes (one prespecified pair of the six possible pairs), that would be $\dfrac 1{6\cdot 6} = \dfrac 1{36}$.
With this distinction made, yes, the probability of obtaining two sixes when rolling one die twice, and the probability of rolling two sixes simultaneously are equal.
• I understand how the sample space is defined. But isn't it the same thing as rolling two dice back to back? it just happens that in the two dice sample, the rolls are happening together. Whereas in the one die example, the rolls happen with a small delay in between. – William Falcon Mar 8 '15 at 18:28
• You specified the probability of rolling two sixes. On the other hand, the probability of rolling consecutively (one roll, then another), and obtaining the same number twice is $1/6$. The first roll can be any number (probability of rolling a number $1-6$ is equal to $1$, and then the probability of rolling that particular number on the second roll is $\frac 16$ for overall probability of rolling the same number twice being $1/6$. – amWhy Mar 8 '15 at 18:32
Note that you are asking two different questions. In the first problem, you are asking the probability of rolling a particular number twice, while in the second problem, you are asking the probability of rolling one of the numbers twice.
The probability of rolling a $6$ twice is $$\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$ but so is the probability of rolling a $1$ twice, a $2$ twice, a $3$ twice, a $4$ twice, or a $5$ twice. Hence, the probability of rolling the same number twice is $$6 \cdot \frac{1}{36} = \frac{1}{6}$$
I think the simplest answer for the second, pairing, scenario is:
Your first rolling does not have any possibility expected, which means you will have any numbers. However, your second roll should be pairing the first number and the possibility should be 1/6.
so to roll a pair, the odds is 1 x 1/6 = 1/6 | 2020-02-24T19:27:56 | {
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http://math.stackexchange.com/questions/320619/closed-bounded-but-not-compact-subset-of-a-normed-vector-space | # Closed Bounded but not compact Subset of a Normed Vector Space
Consider $\ell^\infty$ the vector space of real bounded sequences endowed with the sup norm, that is $||x|| = \sup_n |x_n|$ where $x = (x_n)_{n \in \Bbb N}$.
Prove that $B'(0,1) = \{x \in l^\infty : ||x|| \le 1\}$ is not compact.
Now, we are given a hint that we can use the equivalence of sequential compactness and compactness without proof.
However, I don't understand how sequences of sequences work? Do I need to find a set of sequences and order them such that they do not converge to the same sequence?
Does the sequence $(y_n)$ where $y_n$ is the sequence such that $y_n = 0$ at all but the nth term where $y_n =1$ satisfy the requirement that it does not have a convergent subsequence?
I think I have probably just confused myself with this sequence of sequences lark. Sorry and thanks.
-
You have to find a sequence in $B'(0,1)$ which does not contain a convergent subsequence. Because the existenc of such a sequence implies (right from the definition) that $B'(0,1)$ isn't sequentially compact.
Hint Consider the sequence $(y^n)_n$ defined by $$y^n_k := \begin{cases} 1 & n=k \\0 & n \not= k \end{cases} \qquad (k \in \mathbb{N})$$ Then $y^n \in B'(0,1)$ for all $n \in \mathbb{N}$ and $\|y^n-y^m\|_{\infty}=1$ for $m \not= n$. This means that $(y^n)_n$ doesn't contain a Cauchy-subsequence, hence in particular no convergent subsequence.
-
Let $E = \{ e_n \}_n$ where $e_n$ is the vector with $1$ in the $n$th position and zero everywhere else. Clearly $E \subset \overline{B}(0,1)$, hence it is bounded. Since $\|e_n - e_m\| = 1$ whenever $n \neq m$, it is clear that each point in $E$ is isolated. Hence $E$ is closed.
Now let $U_n = B(e_n, \frac{1}{2}$). Then $\{ U_n\}_n$ is an open cover of $E$ that had no finite subcover. Hence $E$ is not compact.
If you would rather use a sequential argument, choose the sequence $x_n =e_n$. As above, we have $\|x_n -x_m\| = 1$ whenever $n \neq m$, hence no subsequence can be Cauchy. Hence no subsequence can converge, hence $E$ is not sequentially compact.
-
These are very intriguing examples but what condition of the Heine-Borel Theorem is absent? B-T says closed and bounded is equivalent to compact in at least all n dimensional real or complex sets. What specifically is missing in these examples for them to not contradict Heine-Borel?
-
In answer to my own question, my first three attempts to state the Heine-Borel Theorem in n dimensional real space. Certainly a complete metric space would work! Can anyone soften the topological requirements of the Theorem?
- | 2016-07-27T09:46:01 | {
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https://math.stackexchange.com/questions/2334607/the-sum-of-consecutive-integers-is-50-how-many-integers-are-there/2334733 | # The sum of consecutive integers is $50$. How many integers are there?
I started off by calling the number of numbers in my list "$n$". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were "$n$" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.
• If there are $n$ numbers in your list and the first is $x = x + 0$, then the last is $x + n - 1$. – N. F. Taussig Jun 24 '17 at 10:37
If your $n$ is odd, then the middle number has to be $50/n$. The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$
If $n$ is even, then $50/n$ is the half-integer between the middle two numbers. So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.
If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$
If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$
So there are 4 answers: $n=1, 4, 5,$ and $20$.
Edit: As Bill points out, I missed the divisors $25$ and $100$, which give two more answers: $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.
Note that each solution with negative integers is related to an all-positive solution. From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.
• interesting the approach using the middle number! – G Cab Jun 24 '17 at 12:31
• Great technique! n=25 and n=100 work aswell – Mike Jun 24 '17 at 13:47
• Incomplete, e.g. the odd divisor $25\ \$ – Bill Dubuque Jun 24 '17 at 14:01
• @GCab If you think of it in terms of the "average" (mean), instead of the "middle" (median) number, then it's less surprising. The two only happen to be equal because the numbers are uniformly distributed. – Robin Saunders Jun 24 '17 at 15:24
What may be helpful is to use the formula for the sum of an arithmetic progression: if you have a sequence whose first term is $a$ and each term is $d$ more than the rest, then the sum of the first $n$ terms is $na+\frac{n(n-1)}{2}d$. In this case, since we are looking at consecutive integers, $d=1$, and so you are trying to find $a$ and $n$ such that $na+\frac{n(n-1)}2=50$, or equivalently $n(2a+n-1)=100$.
• This is surely helpful. As there are only nine divisors of 100, you can try them all easily and look at which lead to an integer $a$. The nice thing is that (unlike when using a more sophisticated solution) you won't miss the negative first terms. – maaartinus Jun 25 '17 at 13:38
• Yes, and you can make things slightly quicker using the fact that if $n$ is even then $2a+n-1$ is odd, and vice versa, so you can skip cases like $2\times 50$ where both factors are even. – Especially Lime Jun 25 '17 at 19:29
\scriptsize\begin{align} 50 &=2\times 25 &&=25\boxed{+}25 &&=24.5\boxed{+}25.5 \text{ (AP but not integer AP)}\\ &=4\times 12.5 &&=12.5+12.5\boxed{+}12.5+12.5 &&=\color{red}{11+12\boxed{+}13+14}\\ &=5\times 10 &&=10+10+\boxed{10}+10+10 &&=\color{red}{8+9+\boxed{10}+11+12}\\ &=10\times 5 &&=\underbrace{\underbrace{5+5+\cdots+5}_{5}\boxed{+}\underbrace{5+5+\cdots +5}_{5} }_{10} &&=\underbrace{\underbrace{0.5+1.5+\cdots+4.5}_{25}\boxed{+}\underbrace{5.5+\cdots+9.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=20\times 2.5 &&=\underbrace{\underbrace{2.5+2.5+\cdots+2.5}_{10}\boxed{+}\underbrace{2.5+\cdots +2.5}_{10} }_{20} &&=\color{red}{\underbrace{\underbrace{-7+(-6)+\cdots+2}_{10}\boxed{+}\underbrace{3+4+\cdots+12}_{10}}_{20}}\\ &=25\times 2 &&=\underbrace{\underbrace{2+2+\cdots+2}_{12}+\boxed{2}+\underbrace{2+\cdots +2}_{12} }_{25} &&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\ &=50\times 1 &&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50} &&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=100\times 0.5 &&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50} &&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}} \end{align}
In more detail:
$$\frac {50}n=m$$ If $n$ is even $(n=2p)$, then we want $m=a+0.5 \;\;(a\in \mathbb Z)$ $\cdots$ Condition $(1)$
• The AP would comprise $p$ consecutive integers on either side of $m$.
If $n$ is odd $(n=2p+1)$, then we want $m\in \mathbb Z$ $\cdots$ Condition $(2)$
• The AP would comprise $p$ consecutive integers on either side of $m$, as well as $m$ itself.
Try different values of $n$ (excluding the trivial case $n=1$):
• If $n=2$, then $m=25$, hence Condition ($1$) not satisfied.
• If $n=3$, then $m=16\frac 23$, hence Condition ($2$) not satisfied.
• If $n=4$, then $m=12.5$, hence Condition $(1)$ satisfied, so the AP is $\lbrace (m-1.5), (m-0.5), (m+0.5), (m+1.5)\rbrace$, i.e. $\color{red}{11,12,13,14}$ ($AP1$).
Adding negative terms and corresponding positive terms which cancel out gives: $\color{blue}{-10,-9,-8\cdots, 0\cdots, 8,9,10,}\color{red}{11,12,13,14}$ ($AP 1'$)
• If $n=5$, then $m=10$, hence Condition $(2)$ satisfied, so the AP is $\lbrace (m-2),(m-1),m,(m+1),(m+2)\rbrace$, i.e. $\color{red}{ 8,9,10,11,12}$ ($AP 2$).
Adding negative terms and corresponding positive terms which cancel out gives: $\color{blue}{-7,-6,-5\cdots, 0\cdots, 5,6,7,}\color{red}{8,9,10,11,12}$ ($AP 2'$)
• If $n=6, 7,8,9$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.
• If $n=10$, then $m=5$, hence Condition $(2)$ not satisfied - not possible.
• If $n=11,12,\cdots, 24$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.
• If $n=25$, then $m=2$, hence Condition $(2)$ satisfied, so AP is $\color{red}{\underbrace{-10,-9,\cdots 0,1}_{12\text{ terms}},2,\underbrace{3,\cdots 9,10,11,12,13,14}_{12 \text{ terms}}}$ (same as $AP 1'$)
• If $n=26,27,\cdots, 49$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.
• If $n=100$, then $m=0.5$, hence Condition $(1)$ satisfied, so AP is $\color{red}{\underbrace{-49,-48,\cdots -2,-1,0}_{50\text{ terms}},\underbrace{1,2,3,\cdots 48,49,50}_{50 \text{ terms}}}$ ($AP 3$)
• For higher values of $n$, $0<m<0.5$ - not possible.
Without reading other answers... this should tell you how an old computer programmer thinks, versus a real mathematician.
First the obvious answer is the single integer 50. However, if negative numbers are allowed, then we can scoop up the sequence from -49 to 50 for 100 consecutive numbers. This is the longest possible sequence.
If n is the starting number of a sequence and s is the number of consecutive numbers, then we end up with
50 = (n+0) + (n+1) + (n+2) +... + (n+s-1)
50 = ns + (s-1)(s)/2
This is helpful, maybe, as ns pretty much puts a box around the solution set. Since we know the nature of the rightmost term, we can tell that s is 10 or less.
Consider the transformation if we multiply both sides by 2/s:
100/s = 2n + s - 1
The right side will always be an integer. Therefore, s must always divide 100 evenly. From above we know that s is between 1 and 10, so the only possible values for s are 1, 2, 4, 5, and 10.
That reduces the problem to 5 single-degree-of-freedom equations. Let's do it.
100/1 = 2n, n = 50, seq is (50)
100/2 = 2n + 1, no solution
100/4 = 2n + 4 - 1, 25-3 = 2n, n = 11, seq is (11, 12, 13, 14)
100/5 = 2n + 5 - 1, 20 = 2n + 4, 16=2n, n = 8, seq is (8, 9, 10, 11, 12)
100/10 = 2n + 10 - 1, 10 = 2n +9, 1 = 2n, no solution
So those are all the solutions where n and s both > 0.
Using the standard summation of an arithmetic progression formula:
$S_n=\frac{n(2a_1+(n-1)d)}{2}$
here since $d=1$
$2S_n=n(2a_1+n-1)$
$2S_n=n(a_1+(a_1+n-1))$
Here $a_1+n-1$ is just the last term.
$2S_n=n(a_1+a_n)$
Rewrite as
If n is even:
### $n=\frac{2S_n}{2a_{n/2}}$
Since $a_1+a_{n/2}+a_n-a_{n/2}=2a_{n/2}$
hence
### $n=\frac{S_n}{a_{n/2}}$
and since $n\in Z$ , so $a_{n/2}$ is a divisor of $S_n$
once you get the required $n, a=a_{n/2}-n/2$
Similarly for odd $S_n$ you get
### $n=\frac{S_n}{a_{(n-1)/2}-1}$
where $a_{(n-1)/2}-1$ is a divisior of $S_n$
The sum of consecutive numbers $1\dots n$ up to $n$ starting from 1 is $\frac{n(n+1)}{2}$. Since you want only a partial sum from, say, $m+1$ to $n$, you can just subtract to get the partial sum:
$$\frac{n(n+1)}{2} - \frac{m(m+1)}{2} = \frac{n^2+n-m^2-m}{2} = 50$$
Multiplying by 2 gets you
$$100 = n^2+n-m^2-m = (n+m)(n-m) + n-m = (n+m+1)(n-m)$$
Now the trick: $n+m$ and $n-m$ are either both even or both odd. This means that in the last formula, one term must be even, the other one must be odd. Factorization of $100 = 5\cdot 5 \cdot 2 \cdot 2$ means that there are very limited solutions. For instance, one term is 25, the other is 4 (there is one other nontrivial combination, see below). Since $n+m+1$ is larger than $n-m$, in the choice shown here, you must have \begin{align} n+m+1 &= 25\\ n-m &=4 \end{align} Solving this will give you the desired numbers.
Addendum You get the combination by observing that you get all acceptable combinations of the factorisation via: \begin{align} 100 &= 1\cdot (5\cdot5\cdot2\cdot2)\\ &= 5\cdot (5\cdot2\cdot2)\\ &= (5\cdot 5)\cdot(2\cdot2) \end{align} where you have to stop as all the following factorisations will contain only even factors.
Let tere be $n\geq1$ numbers, the smallest of them being $p\in{\mathbb Z}$. We then want $$p+(p+1)+\ldots+\bigl(p+(n-1)\bigr)=50\ ,$$ which amounts to $np+{(n-1)n\over2}=50$, or $$n(n+2p-1)=100\ .$$ Going with $n$ through the $9$ divisors of $100$ we obtain the following table: $$\matrix{n:&1&2&4&5&10&20&25&50&100 \cr p:&50&&11&8&&-7&-10&&-49\cr}$$ When $n\in\{2,10,50\}$ solving $n+2p-1={100\over n}$ for $p$ does not lead to an integer $p$. It follows that there are $6$ solutions in all.
• How did you get from your first equation to the second? – Tony Ennis Jun 25 '17 at 21:26
First, welcome to Mathematics stack exchange. Second, if you start like $x+(x+1)+...+(x+n)$, you will end nowhere. $50$ is a concrete and small number so what can you try is to use that fact, sum of two consecutive numbers it is not, due to sum of two consecutive numbers can not be divided with $2$ in $\mathbb{Z}$, sum of three consecutive numbers is divisible with $3$, so it is not sum of three numbers. Four consecutive numbers- now this is on first look possible(remainder mod $4$ is $2$) but each one of them has to be around $12$, so first options that you have is $11+12+13+14=23+27= 50$, hence the solution. Maybe there are even other solution, but I feel free to interpret your question as required to see one possible solution.
• There are other solutions. The simplest is $50$. – N. F. Taussig Jun 24 '17 at 10:42 | 2020-02-21T17:30:17 | {
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https://tex.stackexchange.com/questions/269031/x-1-x-2-vs-x-1-x-2 | \$x_1,x_2\$ vs. \$x_1\$, \$x_2\$
Basic version of my question: Consider the fragment of some text `bla bla \$x_1,x_2,x_3\$ bla`. Is this correctly type-setted, or is `bla bla \$x_1\$, \$x_2\$, \$x_3\$ bla` better ?
Advanced version: Cf. the "Chicago Manual of Style", 16th printed ed, on pp. 589, in 12.19 it is recommended that typesetting lists of mathematical symbols, e.g. `\$x_1,x_2,x_3\$` that medium spaces between commas should be used. Does LaTeX insert these automatically ? If not, should I use `\$x_1\:,x_2,\:x_3\$` rather than `\$x_1\$, \$x_2\$, \$x_3\$` ? It appears that these strings differ slightly in length, when the document is compiled to pdf.
• It matters greatly what the three items `x_1`, `x_2`, and `x_3` -- or, say, `a`, `b`, and `c` -- denote: Are they just any three distinct elements, or do they form a sequence? In the former case, you should write `... \$x_1\$, \$x_2\$, \$x_3\$ ...`: the commas are parts of the sentence rather than parts of the formulas. In the latter case, you should definitely write `\$... x_1, x_2, x_3 ...\$`, as the commas are now a part of the overall math expression. If they form a sequence, you may also want to encase them in curly braces, i.e., write `\$\{x_1, x_2, x_3\}\$`. – Mico Sep 23 '15 at 10:45
• @Mico Sounds like an answer? – Gonzalo Medina Sep 23 '15 at 15:03
• @GonzaloMedina - Done. :-) – Mico Sep 23 '15 at 15:43
It matters greatly what the three items `x_1`, `x_2`, and `x_3` -- or `a`, `b`, and `c` -- denote. Are they just any three distinct elements, or do they form a structured entity such as a sequence?
In the former case, you should write, say,
``````The sum of \$x_1\$, \$x_2\$, \$x_3\$, etc.\ diverges because ...
``````
Don't include the commas in the math terms, because the commas are parts of the sentence rather than components of formulas. With this setup, line breaks can occur (if needed) after the commas.
In the latter case, though, you should definitely write
``````\$... x_1, x_2, x_3 ...\$
``````
because the commas are now parts of some larger math expression. And, if `x_1`, `x_2`, and `x_3` form, say, a three-element set, you may want to encase them in curly braces, i.e., write
``````\$\{x_1, x_2, x_3\}\$
``````
TeX will, in general, not insert line breaks after math-mode commas. If you must allow a line break in the formula and if the commas are sensible break-points, you'll need to tell TeX about this fact, by inserting judiciously placed `\allowbreak` directives, say,
``````\$\{a_1,a_2,\dots,a_n,\allowbreak a_{n+1}, \dots, a_{n+m}\}\$
``````
• you might also want to mention that, when entered as separate math strings, a line can break after one of the commas, but if it's a single math expression, it won't break at the end of a line. – barbara beeton Sep 23 '15 at 16:05
• @barbarabeeton - Good idea. :-) I'll add a sentence or two to this effect. – Mico Sep 23 '15 at 16:13
• First: To such a complete and carefully crafted answer, I cannot give anything but a thankful +1! Second: In the text I'm writing, `\$x_1\$` to `\$x_3\$` are something in between unrelated items and a structured entity: I'm dividing a line in three segments, called `\$A_1\$`, `\$A_2\$` and '`\$A_3\$` and I constantly have to mention things like "On the segments `\$A_1\$`, `\$A_2\$` happens X"; "in contrast, on the segment \$A_2\$`, `\$A_3\$` we have to deal with the problem Y"; "when looking at `\$A_1\$`, `\$A_2\$` and '\$A_3\$` together we can see Z". Would you agree that I should write them apart, as I just did ? – l7ll7 Sep 24 '15 at 13:54
• [...] Although, they do form an entity, as they are a partition of a line, which makes me somewhat unsure. – l7ll7 Sep 24 '15 at 13:56
• @user10324 - I'd write the segments as separate formulas, i.e., `on the segments \$A_1\$ and \$A_2\$, \$X\$ happens...`. On the other hand, I'd also write `Let \$(A_1,A_2,A_3)\$ be an exhaustive, ordered, and non-overlapping partition of the line segment \$A\$.` – Mico Sep 24 '15 at 15:01 | 2020-11-25T08:44:43 | {
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https://math.stackexchange.com/questions/3134268/question-about-asymptotic-notation-with-logs-of-different-bases | # Question about asymptotic notation with logs of different bases
For the two functions f(n) and g(n), where $$f(n) = n^{1/2}$$ and $$g(n) = 2^{(\log_2 n)^{1/2}}$$, I am trying to determine whether $$f$$ is asymptotically bound below by $$g$$, i.e find whether there exists an N1 and a k1 > 0 such that:
$$f(n) >= k1 * g(n)$$, for all n >= N1.
By taking the log of both $$f(n)$$ and $$g(n)$$ I got:
$$f(n) = n^{1/2} = logn^{1/2} = 1/2*logn$$.
$$g(n) = 2^{(\log_2 n)^{1/2}} = log(2^{(\log_2 n)^{1/2}}) = (\log_2 n)^{1/2}log2$$.
In trying to solve this I have:
$$1/2logn >= k1 * (\log_2 n)^{1/2}log2$$
Setting k1 to 1/2:
$$1/2logn >= 1/2 * (\log_2 n)^{1/2}log2$$
Dividing both sides by 1/2:
$$logn >= (\log_2 n)^{1/2}log2$$
Subtracting $$(\log_2 n)^{1/2}log2$$ from both sides:
$$logn - (\log_2 n)^{1/2}log2 >= 0$$
I can probably figure out whether there is any N1 such that the above is true for all n >= N1 by substituting values. However, I was wondering if there is any way to further simplify the above expression. In particular is there any way to get rid of the $$(\log_2 n)^{1/2}$$ entirely? I would prefer all the logs in the inequality to have the same base. Any insights are appreciated.
• In my answer that you've accepted, I made a mistake which I've now corrected. In particular, I used $\log(ab) = \log(a) \log(b)$, which is what you've used as well, instead of the correct $\log(ab) = \log(a) + \log(b)$, when you take logarithms of both sides of the original inequality. I'm sorry for my mistake, but as you can see it does't change the overall result. – John Omielan Mar 4 at 4:33
## 1 Answer
You could finish your work using $$\log$$, I assume to base $$e$$ but, in answer to your request of
I would prefer all the logs in the inequality to have the same base.
since $$g(n)$$ uses a base of $$2$$ and has a $$\log$$ to the base of $$2$$ in the exponent, it would likely be simplest & easiest to instead take the logs of both sides to base $$2$$ so you're then comparing the exponents of $$f(n)$$ and $$g(n)$$ to that base. As such, this would change trying to prove that there exists an $$N_1$$ and a $$k_1 \gt 0$$ such that
$$f(n) \ge k_1 \, g(n) \; \forall \; n \ge N_1 \tag{1}\label{eq1}$$
to finding, by taking $$\log_2$$ of both sides, a $$k_2 = \log_2{k_1}$$ such that
$$\log_2{f(n)} \ge k_2 + \log_2{g(n)} \; \forall \; n \ge N_1 \tag{2}\label{eq2}$$
Note that
$$\log_2{f(n)} = \log_2{n^{\frac{1}{2}}} = \frac{1}{2} \, log_2 n \tag{3}\label{eq3}$$ $$\log_2{g(n)} = \log_2{2^{(\log_2 n)^{\frac{1}{2}}}} = (\log_2 n)^{\frac{1}{2}} \tag{4}\label{eq4}$$
Substituting \eqref{eq3} and \eqref{eq4} into \eqref{eq2} gives
$$\frac{1}{2} \, \log_2 n \ge k_2 + (\log_2 n)^{\frac{1}{2}} \; \forall \; n \ge N_1 \tag{5}\label{eq5}$$
In this case, consider values of $$m$$ where $$\frac{1}{2}m^2 \ge m$$. Multiplying both sides by $$2$$, moving $$2m$$ to the LHS and factoring gives $$m\left(m - 2\right) \ge 0$$. This is true for all $$m \ge 2$$. In this case, if we let $$m = (\log_2 n)^{\frac{1}{2}}$$, we can see that having $$k_2 = 0$$ and since $$(\log_2 16)^{\frac{1}{2}} = 2$$, we can use $$N_1 = 16$$. I believe you should be able to finish the rest.
Note if you had used your original idea of, I assume, the natural logarithm, it would just involve an extra factor by the change of logarithm base formula, i.e., $$\log_a{x} = \cfrac{\log_b{x}}{\log_b{a}}$$ to give in this case that $$\log_2{n} = \cfrac{\log_e{n}}{\log_e{2}}$$. Also, as I show, you would need to add the constant, not multiply by it. After that correction, this just results in an extra factor being used for $$\log_2{n}$$, plus you also have the $$\log_e{2}$$ factor, so it might change the $$k_2$$ constant you determine to use (but not necessarily the $$k_1$$ constant) and/or the value of $$N_1$$, as these values are not uniquely determined. | 2019-11-14T14:55:56 | {
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https://math.stackexchange.com/questions/3457247/what-is-missing-in-my-solution-of-from-pdf-to-cdf-and-px-0-5 | # What is missing in my solution of “from PDF to CDF and $P(X > 0.5)$”?
The continuous random variable $$X$$ is described with the following probability density function (pdf):
$$f_X(x) = \begin{cases} \frac{1}{9}\big(3 + 2x - x^2 \big) \; : 0 \leq x \leq 3 \\ 0 \; \;: x < 0 \; \lor \; x > 3\end{cases}$$
Find cumulative distribution function $$F_X$$ and probability $$P(X > 0.5)$$.
The task is started by verifying if the pdf is in fact correct pdf. I am checking two conditions:
1. Is the pdf nonnegative on all of its domain? Yes, hence we can write:
$$\forall_{x \in \mathbb{R}}\;f_X(x) \geq 0$$
1. The pdf has to be integrable and its total area under the curve has to be equal $$1$$:
\begin{align*} &\int_{\mathbb{R}}f_X = 1 \\ &\color{red}{\int_{-\infty}^{\infty}f_X(x)dx = 1} \\ \end{align*}
(for now assume the condition is true)
PDF plot:
Computing CDF which is defined as:
$$F_X(x) = \int_{-\infty}^{x}f_X(t)dt$$
Therefore:
If $$x < 0$$:
$$F_X(x) = \int_{-\infty}^{x} 0dt = 0$$
If $$x \geq 0 \; \land \; x \leq 3$$:
\begin{align*}F_X(x) &= \int_{-\infty}^{0}0dt + \int_{0}^{x}\frac{1}{9}\big(3 + 2t - t^2\big)dt = \\ &= 0 + \frac{1}{9}\Big(3t + t^2 - \frac{1}{3}t^3 \Big)\Bigg|^{x}_0 = \\ &= \frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big)\end{align*}
If $$x \geq 3$$:
\begin{align*} F_X(x) &= \int_{-\infty}^{0}0dt + \int_{0}^{3}\frac{1}{9}\Big(3 + 2t - t^2 \Big)dt + \int_{3}^{x}0dt \\ &= 0 + \frac{1}{9}\Big(3t + t^2 - \frac{1}{3}t^3 \Big)\Bigg|^3_0 + 0 = \\ &= 1 \end{align*}
(this implicitly confirms the $$\color{red}{\text{red}}$$ condition)
Finally the CDF is defined as:
$$F_X(x) = \begin{cases} 0 \; \; : x < 0 \\ \frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big) \; \; : x \geq 0 \; \land \; x \leq 3 \\ 1 \; \; : x > 3 \end{cases}$$
The CDF result agrees with:
$$\lim_{x \to \infty}F_X(x) = 1 \; \land \; \lim_{x \to -\infty}F_X(x) = 0$$
Also the function is non-decreasing and continuous.
CDF plot:
## Calculating $$P(X > 0.5)$$:
\begin{align*}P(X > 0.5) &= \int_{0.5}^{\infty}f_X(x)dx = \\ &= \int_{0.5}^{3}\frac{1}{9}(3+2x-x^2)dx + \int_{3}^{\infty}0dx = \\ &= \frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big)\Bigg|^3_{0.5} + 0 = \\ &= \frac{175}{216} \approx 0.81\end{align*}
This probability solution does not agree with the book's solution.
The book says $$P(X > 0.5) = 1 - F_X(0.5) = \frac{41}{216} \approx 0.19$$, so it's my solution "complemented".
## My questions:
• Which final probability solution is correct?
• Is this any special kind of probability distribution, e.g. Poisson or Chi Square (well, not these)?
• Can you please point out all minor or major mistakes I have made along the way? (perhaps aside from plots that are not perfect). This is the most important for me.
• What have I forget to mention or calculate for my solution to make more sense? Especially something theoretical, perhaps e.g. definition for $$X$$.
• Looks like the book have a bug. – kludg Nov 30 '19 at 18:04
## My questions:
• Which final probability solution is correct?
Yours answer is right and the book's isn't. They presumably have mistakenly computed $$\mathbb P(X < 0.5)$$ instead of $$\mathbb P(X > 0.5)$$.
• Is this any special kind of probability distribution, e.g. Poisson or Chi Square (well, not these)?
Not a common one, no. I found this page on "U-quadratic distributions" (a term I've never heard before), and this would be the vertical inverse of one of these described in the "related distributions" section, but I don't think this is a particularly common term or distribution.
EDIT: Whoops, this isn't even quite the vertical inverse of a U-quadratic distribution, is it? Such a distribution would apparently not truncate the left side of the parabola as this one does. The better answer to your question is: "No, this distribution is neither named nor important."
• Can you please point out all minor or major mistakes I have made along the way? (perhaps aside from plots that are not perfect). This is the most important for me.
I'd love to, but I didn't find any!
• What have I forget to mention or calculate for my solution to make more sense? Especially something theoretical, perhaps e.g. definition for $$X$$.
I didn't spot any holes or anything that needs to be improved.
EDIT: One thing you could do to clean this up a bit: when you compute $$\mathbb P(X > 0.5)$$, you're redoing the integration you already did in your CDF. Instead, you could just use that result that you already obtained: $$\mathbb P(X > 0.5) = 1 - \mathbb P(X \leq 0.5) = 1 - F_X(0.5) = 3(0.5) + (0.5)^2 - \frac{1}{3}(0.5)^3 = \dots$$ That said, your answer isn't wrong, it's just a bit inefficient.
• So the efficient way is to compute the CDF at $x=0.5$ and subtract the result from $1.$ It seems likely that whoever wrote the answer key forgot to subtract. I likewise did not find any mistake in the calculations in the question. – David K Nov 30 '19 at 18:30
• Can I write:$$f_X: \; \mathbb{R} \longrightarrow \left[0, \frac{4}{9} \right]$$ where $\frac{4}{9}$ denotes mode of PDF? And also, $$F_X : \; \mathbb{R} \longrightarrow \left[0,1\right]$$ for the CDF? Thanks for help. I will accept today. – weno Dec 1 '19 at 4:56
• @weno Sure - both those statements are true, though the first is questionably useful and the second is trivially true for all CDFs. – Aaron Montgomery Dec 1 '19 at 4:58 | 2020-04-08T22:42:44 | {
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https://math.stackexchange.com/questions/3671953/area-of-the-part-of-the-cylinder-x2y2-2ay-outside-the-cone-z2-x2y2 | # Area of the part of the cylinder $x^2+y^2=2ay$ outside the cone $z^2=x^2+y^2$
Problem: Find the area of the part of the cylinder $$x^2+y^2=2ay$$ that lies outside the cone $$z^2=x^2+y^2$$.
My attempt: So I thought we could do this by projecting the surface onto the $$yz$$-plane and taking the surface integral of the function $$x=g(y,z)=\sqrt{z^2-y^2}$$. I.e letting $$S$$ be the surface and $$E$$ be the projection onto the $$yz$$-plane where we have a $$2$$ before the integral over $$E$$ since we have both $$x<0$$ and $$0\leq x$$: \begin{align*}\iint_{\mathcal{S}}x \ \mathrm{d}S &=2\iint_{E}x\underbrace{\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} \ \mathrm{d}z\mathrm{d}y}_{\mathrm{d}S} \\ &=2\iint_{E}x\sqrt{1+\frac{z^2}{x^2}+\frac{y^2}{x^2}} \ \mathrm{d}z\mathrm{d}y\\ &=2\iint_{E}\sqrt{x^2+z^2+y^2}\ \mathrm{d}z\mathrm{d}y\\ &=2\iint_{E}\sqrt{2}z\ \mathrm{d}z\mathrm{d}y \end{align*} Now in the projection it seems to me that we have the following bounds on $$z$$ and $$y$$ since the cylinder has radius $$a$$ and the cone and the surface intersect at $$z=\sqrt{2ay}$$ $$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$ so: \begin{align*}2\iint_{E}\sqrt{2}z\ \mathrm{d}z\mathrm{d}y &= \sqrt{2}\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}2z \ \mathrm{d}z\mathrm{d}y \\ &=\sqrt{2}\int_{0}^{2a} 2ay \ \mathrm{d}y\\ &=4\sqrt{2}a^{2}\end{align*} However my book says its $$16a^2$$ so what is my mistake(s)?
PS. I think this is also possible with polar coordinates but I would like to use the surface integral with projection onto the $$yz$$-plane.
PSDS. Picture is not totally acurate as $$a=4$$
Edit:
As Ninad Munshi pointed out I was projecting the wrong surface and I used the wrong formula for the surface area. My thoughts are
Would it be correct to say that $$\iint\mathrm{d}S$$ is the surface area, and would $$\mathrm{d}S$$ be $$\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2} dzdy$$? If so I still seem to be off by a factor of $$2$$ as \begin{align*}\iint_{\mathcal{S}} \mathrm{d}S &= 2 \iint_{E}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy \\ &=2\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy=8a^{2}\end{align*}
• You used the wrong surface in the beginning. It wants the area of the cylinder, not the cone, so you have to project the cylinder down. – Ninad Munshi May 12 '20 at 21:33
• Also, $\iint x\:dS$ does not give you surface area. – Ninad Munshi May 12 '20 at 21:35
• @NinadMunshi I solved it, Thank you for your help! – André Armatowski May 13 '20 at 2:42
The correct way to solve this question is to start with the cylinder $$x^2+y^2=2ay$$ that we wish to project onto the $$yz$$ plane. This is done by first calculating $$\mathrm{d}S$$ in $$\iint_{\mathcal{S}}dS$$ which gives the surface area
We have that $$dS = \sqrt{1+\left(\frac{\partial x}{\partial y}\right)^{2}+\left(\frac{\partial x}{\partial z}\right)^{2}} \ \mathrm{d}z\mathrm{d}y=\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$
Now the problem I had is that when I projected the cylinder: I only considered the symmetric areas of $$x<0$$ and $$0\leq x$$ while we infact have two more symmetries: namely $$z<0$$ and $$0\leq z$$.
In summary: We have four areas that are equal (and not two) so letting $$E$$ represent the area of the projection of the cylinder onto the $$yz$$-plane in the first octant we get: $$\iint_{\mathcal{S}}\mathrm{d}S=4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$ The particular limits of $$z$$ and $$y$$ are still correct that is $$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$ So: \begin{align*}4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y & = 4\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y \\ &= 4 \int_{0}^{2a}\sqrt{2ay}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}y\\ &= 4(4a^2)=16a^2\end{align*} Which is the correct answer | 2021-04-16T15:14:37 | {
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https://math.stackexchange.com/questions/1312058/find-probability-of-exactly-one-6-in-first-ten-rolls-of-die-given-two-6s-in | # Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls
I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
### My thoughts
Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find $$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$
By the binomial distribution formula, we get that $$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$
Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$
Am I correct in thinking this?
If so, then it follows that the required probability is $$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$ which, I know, can be simplified further!
• Shouldn't $P[B]$ be $$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$ ? – Mohamad Misto Jun 4 '15 at 14:45
• A curiosity is that in this case $P(B \mid A)=P(A)$ so $P(A \cap B)=P(A^2)$ and $P(A \mid B)=\dfrac{P(A)^2}{P(B)}$ – Henry Jun 4 '15 at 14:59
• Well laid out. Apart from a minor slip, all correct. You will notice near total collapse when you simplify. Now can you simplify the argument? – André Nicolas Jun 4 '15 at 14:59
• You should get $10/19$. – André Nicolas Jun 4 '15 at 15:26
• There are $\binom{20}{2}$ equally likely ways to choose the locations of the two $6$'s. There are $(10)^2$ ways to choose one in each half. So the probability is $100/\binom{20}{2}$. – André Nicolas Jun 4 '15 at 15:29
I took a different approach to the question. Suppose B. There are three ways to get two 6's in twenty rolls:
• $B_1$: Both 6's come in the first 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.
• $B_2$: One 6 comes in the first 10 rolls, and the second comes in the next 10 rolls. There are $\begin{pmatrix} 10 \\ 1\end{pmatrix} \begin{pmatrix} 10 \\ 1\end{pmatrix} = 100$ ways for this to happen.
• $B_3$: Both 6's come in the second lot of 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.
Now $$P(A|B) = P(B_2|B_1 \cup B_2 \cup B_3) = \frac{100}{45+100+45} = \frac{10}{19}.$$
• Your reasoning seems sound to me. It doesn't agree with all of the other answers, though. I am not yet sure why... – Caleb Owusu-Yianoma Jun 4 '15 at 15:24
I think you are over thinking this. We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't care what number it is otherwise.
One of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we divide by two because the order does not matter.
There are thus $\dfrac{20 \cdot 19}{2} = 190$ ways we can get exactly 2 6's in 20 rolls.
In how many ways can we have exactly one 6 in 10 rolls? Well it can be any of the 10 rolls, and it must be not 6 in the other 9 so there are 10 ways to get exactly one 6 in ten rolls, and 10 ways to get the second 6 in the last 10 rolls making
The answer is simply $\dfrac{10 \cdot 10}{190} = \dfrac{100}{190} = \dfrac{10}{19}$.
• Surely, when considering the number of ways in which we can obtain exactly one $6$ in the first ten rolls, we must account for the fact that we obtain a $6$ in the second set of ten rolls, too. In that case, the number of ways of rolling exactly one $6$ in the first ten rolls AND exactly one $6$ in the second ten rolls is $10^2 = 100.$ Is my reasoning sound? – Caleb Owusu-Yianoma Jun 4 '15 at 15:20
• @CKKOY there are 100 ways you can roll exacly one 6 in the first 10 rolls and exacly 1 six in the next 10 rolls but that's not what the question is asking. The question is asking how many ways you get exactly 1 six in the first 10 rolls given all the ways you can get 2 6's in 20 rolls including one where both 6's occur in the first 10 rolls or both 6's occur in the last 10 rolls. – Warren Hill Jun 4 '15 at 15:26
• Is your interpretation of the question equivalent to the following? 'The question is asking how many ways you can roll exactly one $6$ in the first ten rolls (out of twenty), given that you rolled two $6$s in the total twenty rolls.' Please see the other answers and comments, which imply that the correct answer is $10/19$. – Caleb Owusu-Yianoma Jun 4 '15 at 15:30
• @CKKOY sorry you are correct. I've corrected my answer. – Warren Hill Jun 4 '15 at 15:35
• Apology accepted. – Caleb Owusu-Yianoma Jun 4 '15 at 15:40
Since all drawings are independent, is confusing to use the conditional probability formula and it is not necessary at all. Think as you are rolling simultaneously two dices 10 times each. Having in both groups exactly one 6 has the same probability p. Because these two are independent the simultaneous event (one 6 in each group) has the p^2 squared probability. The answer is what you thought A and B was: [(10C1)⋅(1/6)⋅(5/6)^9]^2 = 95,367,431,640,625/914,039,610,015,744 = 0.104336. (far to be 10/19)
This is what Adelafif said too, but he made an error on the coefficient n-k=9 (not 5).
What Paul Wright said is very funny but totally wrong. Having one 6 in ("the first") 10 rollings has the same probability independently how many time you are rolling the dice after!. This probability is (10C1)⋅(1/6)^2⋅(5/6)^8 (much less than Paul calculated). With Paul's theory, we should win a lottery pot considerably easier.
• Your answeris wrong, and $10\over 19$ is the correct one. According to mathematics. – user228113 Jun 4 '15 at 16:43
• @Laszlo: Your calculation of $0.104336$ is the rounded value of $P[A \cap B].$ However, the final answer, $10/19$, is roughly equal to $0.104336/{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$ – Caleb Owusu-Yianoma Jun 4 '15 at 16:48
• @G.Sassatelli Mathematics is not a matter of enouncing something. You must argument your opinion, as I did. What is wrong with "my mathematics"? – Laszlo Jun 4 '15 at 18:01
• @CKKOY Exactly, as I said. You DO NOT need the denominator. – Laszlo Jun 4 '15 at 18:05
• To avoid any other useless dispute: I made a simulation for one million of 20 times rolling sets (it is only two lines in Mathematica!). The number of cases with one six in the first 10 and the second 10 rolls was: 104,524. Repeated this 100 times, got a normal distribution with the mean value exactly at 104336 (it took 20 minutes). Still sustain that my solution is wrong? – Laszlo Jun 4 '15 at 18:34
((10C1)(1/6)(5/6)^5)(10C1)(1/6)(5/6)^5
• It would be helpful if you wrote a brief explanation of the answer that you have written. – Caleb Owusu-Yianoma Jun 4 '15 at 15:13 | 2019-09-22T19:26:47 | {
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https://math.stackexchange.com/questions/3902288/for-which-n-is-it-possible-to-arrange-all-whole-numbers-from-1-to-n-in-suc/3926332 | # For which $N$ is it possible to arrange all whole numbers from $1$ to $N$ in such a way that every adjacent pair sums up to a Fibonacci number?
Recently I came up with a problem regarding Fibonacci numbers:
For which $$N$$ is it possible to arrange all whole numbers from $$1$$ to $$N$$ in such a way that every adjacent pair sums up to a Fibonacci number?
I have manually tested a bunch of cases and I was also able to prove almost every case. The results that I was able to prove are the following:
• If $$N$$ is a Fibonacci number or exactly one less than a Fibonacci number, I was able to prove that an arrangement exists. I used an argument by induction to achieve this.
• If $$F_k+2 \leq N \leq F_{k+1} -3$$ , it is completely impossible, because the numbers $$F_k$$, $$F_k + 1$$ and $$F_k + 2$$ have only one possible pair and therefore have to be at the end of the arragement. This obviously gives a contradiction, because arrangements only have two ends (the first number and the last number).
The cases I was not able to solve are $$N=F_k+1$$ and $$N=F_k-2$$. My theory is that $$N=9$$ is the only working case of the form $$N=F_k+1$$, and $$N=11$$ the only working case of the form $$F_k-2$$. I expect every other $$N$$ of these two forms to be impossible.
Does anybody know a full proof to this problem or maybe the name of the official theorem (if this exists)?
• I ran some code to check if there are solutions for small numbers. everything you said checks out so far - 1-5, 7,8,9, 11 solvable, and also 12,13, 20,21, 33,34, 54,55. none of the others are solvable. I checked up to 55 so far. I'm gonna run the code at night, but I don't think it's gonna get much farther than that (the code isn't too smart). Nov 11 '20 at 0:07
• The only sequence in the OEIS matching these conditions is oeis.org/A259624, which it sounds like is not your sequence if $N=6$ cannot be done. If you have confirmed values for $N$ up to $34$ or so, you may wish to submit it as a new sesquence. Nov 11 '20 at 3:45
• @RavenclawPrefect problem is OEIS search engine finds $2,3,4,5,7,8,9,11,12,13,20,21,33,34,54,55$ but no result for $1,2,3,4,5,7,8,9,11,12,13,20,21,33,34,54,55$. Nov 28 '20 at 15:20
Thanks to @Rei Henigman data, I could find OEIS sequence A079734 and from there the "Fibonacci Plays Billiards" paper by Elwyn Berlekamp and Richard Guy, where they state with theorem 1 on page 3, that the only solutions are $$N = 9$$, $$N = 11$$, $$N = F_k$$, $$N = F_k − 1$$ for $$k \ge 4$$. Note that $$1$$ is excluded from the OEIS sequence. | 2021-10-22T22:48:36 | {
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https://math.stackexchange.com/questions/982564/help-with-implicit-differentiation-finding-an-equation-for-a-tangent-to-a-given | # Help with Implicit Differentiation: Finding an equation for a tangent to a given point on a curve
When working through a problem set containing Implicit Differentiation problems, I've found that I keep getting the wrong answer compared to the one listed at the back of my book.
The problem is given as such: Use implicit differentiation to find the equation of the tangent line to the curve at a given point
x^2 + xy + y^2 = 3
With given point (1, 1). I also am told that it is an ellipse.
To solve this, I evidently must differentiate both sides of the problem:
1: dy/dx ( x^2 + xy + Y^2 ) = dy/dx(3)
2: dy/dx (2x + 1y'+ 2yy') = 0
3: 1y' + 2yy' = 0 - 2x
4: y'(1+2y) = -2x
5: y' = -2x/(1+2y)
Hurray, so now since I have the first derivative of Y. I can use it to find the slope at the point.
Slope at Point (1,1)= -2(1) / (1+2(1)
Slope at Point (1,1)= -2/3
So now that I've got my slope, I know the equation of the tangent will be in the form:
y=mx+b
So, given I now know the slope:
y=-2/3x + b
Substitute in the known point:
1 = -2/3(1) + b
b = 5/3
So the final answer I get is: y = -2/3x + 5/3
But according to the answer, it is supposed to be: -x + 2, I don't know where I went wrong, and I've done it twice to make sure I'm getting the same answer. Could someone please help me?
• The differentiation of $xy$ requires the product rule: $\frac{d}{dx}(xy)=y+xy'$. Oct 20, 2014 at 14:49
• Consider the product rule on the term $xy$: It should read: $\frac{d}{dx}(xy) = \frac{d}{dx}x \cdot y + x \cdot \frac{d}{dx}y.$ Oct 20, 2014 at 14:49
• @Clayton: AH! Fantastic find! I never saw that Oct 20, 2014 at 14:50
You made an error when you differentiated implicitly. You did not apply the Product Rule $(fg)' = f'g + fg'$ to the term $xy$. Keeping in mind that $y$ is a function of $x$, you should obtain
$$(xy)' = 1y + xy' = y + xy'$$
Therefore, when you differentiate
$$x^2 + xy + y^2 = 3$$
implicitly with respect to $x$, you should obtain
$$2x + y + xy' + 2yy' = 0$$
Solving for $y'$ yields
\begin{align*} xy' + 2yy' & = -2x - y\\ (x + 2y)y' & = -2x - y\\ y' & = -\frac{2x + y}{x + 2y} \end{align*}
As you can check, evaluating $y'$ at the point $(1, 1)$ yields $y' = -1$. Therefore, the tangent line equation is $y = -x + 2$. | 2022-05-16T06:02:29 | {
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https://math.stackexchange.com/questions/1161746/how-come-left-fracn1n-1-rightn-left1-frac2n-1-rightn | # How come $\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n$?
I'm looking at one of my professor's calculus slides and in one of his proofs he uses the identity:
$\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n$
Except I don't see why that's the case. I tried different algebraic tricks and couldn't get it to that form.
What am I missing?
Thanks.
Edit: Thanks to everyone who answered. Is there an "I feel stupid" badge? I really should have seen this a mile a way.
• Because $n+1=(n-1)+2$. – vadim123 Feb 23 '15 at 14:39
Just write $$\left(\frac{n+1}{n-1}\right)^n = \left(\frac{n-1+2}{n-1}\right)^n =\left(1+\frac{2}{n-1}\right)^n$$
$1+\frac{2}{n-1}=\frac{n-1}{n-1}+\frac{2}{n-1}=\frac{n+1}{n-1}$
Note that $$\frac{n+1}{n-1} = \frac{n-1+2}{n-1} = \frac{n-1}{n-1} + \frac{2}{n-1} = 1 + \frac{2}{n-1}.$$
$(1 + \frac {2}{n-1})^n = (\frac {n-1 +2}{n-1})^n = (\frac{n+1}{n-1} )^n$ | 2019-06-18T10:55:18 | {
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https://math.stackexchange.com/questions/1852826/inverse-of-the-pascal-matrix | # Inverse of the Pascal Matrix
Let $P_n$ be the $(n+1) \times (n+1)$ matrix that contains the numbers of Pascal's triangle in the upper triangle. For example in the case of $n=3$ $$P_3 = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ or in general $$(P_n)_{ij} = \binom{j}{i} \lfloor i \leq j \rceil ~~~\text{for}~~ i,j \in \{0,...,n \}$$ using the definition $$\lfloor A \rceil := \begin{cases} 1 & \text{A is true} \\ 0 & \text{A is not true} \end{cases}$$ This matrix is invertible since $\det P_n = 1$. For smaller cases like $n=3$, I calulated the inverse of the matrix by hand and found $$P_3^{-1} = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 0 & 1 & -2 & 3 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ $n=2,4$ led to similar results, so I'm guessing that the inverse should be $$(P_n^{-1})_{ij} = (-1)^{j+i}(P_n)_{ij} = (-1)^{j+i} \binom{j}{i} \lfloor i \leq j \rceil$$ But I have not been able to prove or disprove this yet. So far I tried multiplying the two matirces which gives $$\sum^j_{k=i} (-1)^{j+k} \binom{j}{k} \binom{k}{i} = \delta_{ij}$$ if one asumes that the result is the unit matrix. For $i=0$ and $j>0$ this gives $$\delta_{0j} = 0 = \sum^j_{k=0} (-1)^{j+k} \binom{j}{k} \binom{k}{0} = \sum^j_{k=0} (-1)^{k} \binom{j}{k}$$ which is an identity I know to be true, so it reasures me a little bit that the above should also be true.
• Instead of proving the above mentioned formula for all $(j,i)$ you can also use induction and prove it only for $(n,i)$, maybe it is easier. The $(j,n)$ entries are easy to handle Jul 8, 2016 at 6:23
• Use $\binom jk \binom ki = \binom ji \binom {j-i}{k-i}$. Jul 8, 2016 at 6:36
• Binomial transform Jul 27, 2016 at 11:10
In my opinion this is a bit simpler to prove, if we interpret the matrix $$P_n$$ as a linear transformation.
Consider the space $$V_n$$ of polynomials of degree $$\le n$$ (over, say $$\Bbb{Q}$$, but you are welcome to use reals or complex numbers, or any other field actually).
The mapping $$T: f(x)\mapsto f(x+1)$$ for all $$f(x)\in V_n$$ is obviously linear. My key point is to observe that $$P_n$$ is the matrix $$M_{\mathcal B}(T)$$ of $$T$$ with respect to the obvious basis $$\mathcal{B}=\{1,x,x^2,\ldots,x^n\}$$ of $$V_n$$. This is because for any $$k, 0\le k\le n$$ the binomial formula says that $$T(x^k)=\sum_{\ell=0}^k\binom k\ell x^\ell.$$ From this we can read the coordinates of $$T(x^k)$$ with respect to $$\mathcal{B}$$, and see that those coincide the $$k$$th column of $$P_n$$ (numbered from $$0$$ to $$n$$). That's exactly what the claim was.
It is obvious that the inverse of $$T$$ is given by the recipe $$T^{-1}:f(x)\mapsto f(x-1).$$ A similar application of the binomial formula shows that the matrix $$M_{\mathcal B}(T^{-1})$$ then is exactly your prescribed inverse of $$P_n$$ containing binomial coefficients - this time with alternating signs.
But, by basic linear algebra $$M_{\mathcal B}(T^{-1})=M_{\mathcal B}(T)^{-1}.$$
• Nice solution! I especially like that it does not require to guess the form of the inverse. It is also very close to the reason I got confronted with the problem in the first place. Because I was looking at the change of base from $\{x^j\}$ to $\{(x+1)^j\}$. Jul 22, 2016 at 18:39
• @JyrkiLahtonen This is brilliant, really, a pity that I can only give (+1). I just wanted to know the formula for the inverse of the Pascal matrix and searched the web for it. Now I found not only the formula, but this higher level view on it which makes everything so clear. It feels similar to the moment when I was told that the addition theorems for sin and cos are just real and imaginary parts of the power rule for the complex exponential function. May 30, 2020 at 10:10
By the suggestion of Sungjin Kim one can use $$$$\binom{j}{k}\binom{k}{i} = \binom{j}{i}\binom{j-i}{k-i}$$$$ which is useful since it leads to the sum only going over one of the binomial coefficients. It leads to \begin{align} \sum^j_{k=i} (-1)^{j+k} \binom{j}{k} \binom{k}{i} &= \sum^j_{k=i} (-1)^{j+k} \binom{j}{i}\binom{j-i}{k-i} \\ &= (-1)^j \binom{j}{i} \sum^j_{k=i} (-1)^{k} \binom{j-i}{k-i} \\ &= (-1)^{j} \binom{j}{i} \sum^{j-i}_{k=0} (-1)^{k+i} \binom{j-i}{k} \\ &= (-1)^{j+i} \binom{j}{i} (1 - 1)^{j-i} = \delta_{ij} \end{align} The identity used is easily proven by $$\binom{j}{k}\binom{k}{i} = \frac{j!}{k!(j-k)!} \frac{k!}{i!(k-i)!} = \frac{j!}{i!(j-i)!} \frac{(j-i)!}{(j-k)!(k-i)!} = \binom{j}{i}\binom{j-i}{k-i}$$ Another way of showing the identity involving $$\delta_{ij}$$ is to start with $$x^j$$ and using the binomial thoerem twice $$x^j = (x+0)^j = ((x+1) - 1 )^j = ~...~ = \sum^j_{i=0} \sum^j_{k=i} (-1)^{j+k} \binom{j}{k} \binom{k}{i} x^i$$ comparing coefficients of the two sides of the equation, then also leads to the desired result. You can find the whole calcultion here, along whith other examples of when adding 0 really counts.
In the following it's somewhat more convenient to equivalently consider the transpose $P_n^T$ of the matrix $P_n$ and its inverse.
We obtain for $n=3$
\begin{align*} P_3^T&= \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ 1 & 2 & 1 & \\ 1 & 3 & 3 & 1 \\ \end{pmatrix}=\left(\binom{i}{j}\right)_{0\leq i,j\leq 3} \\ \left(P_3^T\right)^{-1}&= \begin{pmatrix} \begin{array}{rrrr} 1 & & & \\ -1 & 1 & & \\ 1 & -2 & 1 & \\ -1 & 3 & 3 & 1 \\ \end{array} \end{pmatrix}=\left((-1)^{i+j}\binom{i}{j}\right)_{0\leq i,j\leq 3} \end{align*}
The elements of the Pascal matrix can be considered as coefficients of binomial inverse pairs. These are sequences $(a_i)_{0\leq i \leq n},(b_i)_{0\leq i\leq n}$ which are related for $n\geq 0$ via \begin{align*} a_n=\sum_{i=0}^n\binom{n}{i}b_i\qquad\text{resp.}\qquad b_n=\sum_{i=0}^n(-1)^{i+n}\binom{n}{i}a_i\tag{1} \end{align*}
One relation implies the other in (1). This can be seen via exponential generating functions.
Let $A(x)$ and $B(x)$ be two exponential generating functions
$$A(x)=\sum_{n=0}^\infty a_{n}\frac{x^n}{n!} \qquad\qquad\text{ and }\qquad\qquad B(x)=\sum_{n=0}^\infty b_{n}\frac{x^n}{n!}$$
The expressions in (1) are the coefficients of
The most important characterization of the pair of inverse relation in (1) is the orthogonal relation the pair implies. This follows by substituting one of the pair into the other. We obtain for $n\geq 0$
\begin{align*} a_n&=\sum_{j=0}^n\binom{n}{j}b_j\\ &=\sum_{j=0}^n\binom{n}{j}\sum_{i=0}^j(-1)^{i+j}\binom{j}{i}a_i\\ &=\sum_{i=0}^n a_i\sum_{j=i}^n(-1)^{i+j}\binom{n}{j}\binom{j}{i} \end{align*} Hence the orthogonal relation is \begin{align*} \delta_{ni}=\sum_{j=i}^n(-1)^{i+j}\binom{n}{j}\binom{j}{i} \end{align*} with $\delta_{ni}$ the Kronecker Delta.
The inverse of $$P_n$$ also follows from a familiar representation of $$GL(2,\mathbb{R})$$. Let $$\{u,v\}$$ be a basis of $$\mathbb{R}^2$$ and $$S^n(\mathbb{R}^2)$$ denote the vector space of homogeneous polynomials of degree $$n$$ in the variables $$u$$ and $$v$$. Then, any $$A\in GL(2,\mathbb{R})$$ acts linearly on $$p(u,v)\in S^n(\mathbb{R}^2)$$ by $$A\cdot p(u,v)=p(Au,Av).$$ Choose $$\{u^{n-j}v^j\}_{j=0}^n$$ as a basis of $$S^n(\mathbb{R}^2)$$. Then, the given group action induces a homomorphism $$\varphi_n: GL(2,\mathbb{C})\to GL(n+1,\mathbb{C})$$ defined by $$\begin{pmatrix}a & b\\ c & d\end{pmatrix}\mapsto \left([t^i](a+ct)^{n-j}(b+dt)^j\right)_{i,j=0}^n$$ where $$[t^i]f(t)$$ denotes the coefficient of $$t^i$$ in $$f(t)$$. Let $$M=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}.$$ Then, we see that $$P_n=\varphi_n(M).$$ Since $$\varphi_n$$ is a homomorphism, we get $$P_n^{-1}=\varphi_n(M^{-1})$$. But $$M^{-1}=\begin{pmatrix}1 & -1\\ 0 & 1\end{pmatrix}$$. Hence, by definition of $$\varphi_n$$, we obtain explicit indices as you described. | 2022-05-19T15:23:01 | {
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https://math.stackexchange.com/questions/1909904/munkress-strong-induction-principle-vs-traditional-mathematical-induction | # Munkres's Strong induction principle vs. “traditional” mathematical induction
Question:
In Munkres's Topology (2nd ed.), he gives a proof for the following:
Theorem 4.2 (Strong induction principle). Let $A$ be a set of positive integers. Suppose that for each positive integer $n$, the statement $S_n\subset A$ implies the statement $n\in A$. Then $A=\mathbb{Z}_+$.
This appears to be slightly different from the "traditional" (strong) induction principle which can be stated as: let $P_n$ be a sequence of statements. Suppose $P_1$ is true and (($P_m$ is true $\forall\, m < n$) implies $P_{n}$ is true) then $P_k$ is true $\forall\, k\in \mathbb{Z}_+$.
How can one reconcile the two?
Attempt:
I'm thinking of something along the lines of:
Let $A$ be a set of positive integers and $P_n$ be a sequence of propositions. Suppose that for each positive integer $n$, the statement $S_n\subset A$ implies the statement ($n\in A$ and $P_n$ is true). Then $A=\mathbb{Z}_+$ and $P_k$ is true $\forall\, k \in \mathbb{Z}_+$.
This can proved to be equivalent to the "traditional" form in the following way:
1. Proceed with Munkres's proof and obtain $A=\mathbb{Z}_+$.
2. $\because S_1=\varnothing\subset A$ is true, our supposition for $n=1$ is equivalent to assuming that $P_1$ is true.
3. ??? (how do we show that the supposition "$S_n\subset A$ implies the statement ($n\in A$ and $P_n$ is true)" is equivalent to "($P_m$ is true $\forall\, m < n$) implies $P_{n}$ is true"?)
Actually I just noticed that this point is discussed in Chapter 1 $\S$ 7 after Lemma 7.2 (pg 45 in 2nd ed.)
To use the principle to prove a theorem "by induction", one begins the proof with the statement "Let $A$ be the set of all positive integers $n$ for which the theorem is true," and then one goes to prove that $A$ is inductive, so that $A$ must be all of $\mathbb{Z}_+$
Using this one can write:
Let $P_n$ be a sequence of propositions and $A$ be the set of all positive integers $n$ for which $P_n$ is true. Suppose that for each positive integer $n$, the statement $S_n\subset A$ implies the statement $n\in A$. Then $A=\mathbb{Z}_+$.
The equivalence with the "traditional" induction principle is easily demonstrated:
1. For $n=1$, the assumption ($S_1 = \varnothing \subset A \Rightarrow 1 \in A$) is equivalent to ($P_1$ is true).
2. For $n>1$, the assumption ($S_n \subset A \Rightarrow n \in A$) is equivalent to (($P_m$ is true $\forall\, m<n$) $\Rightarrow$ $P_n$ is true)
3. The conclusion ($A=\mathbb{Z}_+$) is equivalent to ($P_n$ is true $\forall\, n\in \mathbb{Z}_+$).
• Yes, this is the intended argument. – Brian M. Scott Aug 31 '16 at 22:10 | 2019-06-17T08:34:44 | {
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https://math.stackexchange.com/questions/2688163/double-antiderivation-problem | Double Antiderivation problem
I have to find $f(x)$ given $f''(x)$ and certain initial conditions.
$$f"(x) = 8x^3 + 5$$ and $f(1) = 0$ and $f'(1) = 8$
$$f'(x) = 8 \cdot \frac{x^4}{4} + 5x + C = 2x^4 + 5x + C$$
Since $f'(1) = 8 \Rightarrow 2 + 5 + C = 8$ so $C = 1$
$$f'(x) = 2x^4 + 5x + 1$$
$$f(x) = 2 \cdot \frac{x^5}{5} + 5 \cdot \frac{x^2}{2} + X = \frac{2}{5} x^5 + \frac{5}{2} x^2 + X + D$$
Since $f(1) = 0$ then:
$$\frac{2}{5} + \frac{5}{2} + 1 + D = 0$$ $$\frac{29}{10} + 1 + D = 0$$
So $D = \dfrac{-39}{10}$
So $$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}$$
Does that look right?
• Yes it's correct. An alternative way to do it is to work with definite integrals and write $f'(x) - f'(1) = \int_1^x (8t^3+5)\,{\rm d}t$ and $f(x)-f(1) = \int_1^x f'(t)\,{\rm d}t$. This avoids solving for the integration constants, but it's pretty much the same thing. Mar 12, 2018 at 17:32
• Yes, you can also check it by yourself Mar 12, 2018 at 17:33
$$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}\implies f'(x)= 2x^4+5x+1\implies f''(x)=8x^3+5$$
$$f(1)=\frac{2}{5}+ \frac{5}{2} + 1 - \frac{39}{10}=\frac{4+25+10-39}{10}=0$$
$$f'(1)=2+5+1=8$$ | 2023-03-23T05:48:39 | {
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https://web2.0calc.com/questions/here-s-a-mind-boggler | +0
# Here's a mind-boggler
0
347
4
+40
A positive whole number that is the same when reading from right to left and left to right is called a palindromic number. If all of them were put into a sequence that is from small to big:1,2...11,22...,101,111...,10001,10101,.....
Then which term is the number 78987 in this sequence? For example : 1 is the 1st term,2 is the second....etc
yomyhomies Oct 22, 2017
#2
+89840
+1
Starting with 0, 78987 is the 889th palindrome
[BTW...889 is known as the palindrome's rank ]
Starting with 1, it is the 888th palindrome
This can be verified with the calculator here :
http://rhyscitlema.com/algorithms/generating-palindromic-numbers/
[ The calculator is about half-way down the page.....there are also other intems of interest related to palindromes on this page ]
A procedure is desribed for finding the nth palindrome, but I don't see one that tells us how to determine the rank of any particular palindrome [ although I did not look at the page with a fine tooth comb]
Maybe heureka knows of a proceedure to produce this???
CPhill Oct 22, 2017
#3
+20024
+3
which term is the number 78987 in this sequence?
Here 0 is the 1st term:
The position of a palindrome within the sequence can be determined almost without calculation:
If the palindrome has an even number of digits,
prepend a 1 to the front half of the palindrome's digits.
Examples: 98766789=a(19876)
If the number of digits is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.
Examples: 515=a(61), 8206028=a(9206), 9230329=a(10230).
which term is the number 78987 in this sequence?
The number of digits is 5 is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.
$$\begin{array}{|rrrrll|} \hline & & & \Rsh & & \text{ until central digit} \\ &7 & 8 & 9 & 8 & 7 \\ &| & | & | \\ &+1 & | & | \\ &\downarrow & \downarrow & \downarrow \\ \text{term is } & \color{red}8 &\color{red}8 &\color{red} 9 \\ \hline \end{array}$$
Starting with 1, it is the 888th palindrome
heureka Oct 23, 2017
#4
+89840
+1
Wow!!!!....thanks, heureka.....that's almost too easy....!!!
CPhill Oct 23, 2017 | 2018-10-17T21:34:53 | {
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https://web2.0calc.com/questions/suppose-f-x-is-a-rational-function-such-that-3-f | +0
# Suppose $f(x)$ is a rational function such that $3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$ for all $x \neq 0$. Find $f(-2)$.
0
258
3
Suppose f(x) is a rational function such that $$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$ for all x =\=0. Find f(-2).
Guest May 22, 2018
#1
+93629
+1
Suppose f(x) is a rational function such that
$$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$
for all $$x\ne0$$. Find f(-2).
We have
$$3 f \left( \frac{1}{-2} \right) + \frac{2f(-2)}{-2} = (-2)^2\\ 3 f \left( \frac{1}{-2} \right) - f(-2) =4\qquad \qquad(1)\\ and\\ 3 f \left( \frac{1}{-1/2} \right) + \frac{2f(-1/2)}{-1/2} = (-1/2)^2\\ 3 f \left( -2 \right) -4f(\frac{-1}{2}) =\frac{1}{4}\\ -4f(\frac{-1}{2}) +3 f \left( -2 \right)=\frac{1}{4}\qquad\qquad (2)\\--------------\\~\\ \quad 3 f \left( \frac{1}{-2} \right) - f(-2) =4\qquad \qquad(1)\\ \quad 12f \left( \frac{1}{-2} \right) - 4f(-2) =16\qquad \qquad(1b)\\ -4f(\frac{-1}{2}) +3 f \left( -2 \right)=\frac{1}{4}\qquad\qquad (2)\\ -12f(\frac{-1}{2}) +9 f \left( -2 \right)=\frac{3}{4}\qquad\qquad (2b)\\~\\ (1b)+(2b)\\ 5f(-2)=16\frac{3}{4}\\ f(-2)=3\frac{7}{20}\\$$
The method is correct but you need to check for careless errors.
Melody May 22, 2018
#2
+20024
+1
Suppose
$$f(x)$$
is a rational function such that $$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$
for all $$x \neq 0$$. Find $$f(-2)$$.
$$\begin{array}{|lrclcl|} \hline & 3f(\frac1x) + \frac2x f(x) &=& x^2 \\ & 3f(\frac1x) &=& x^2 - \frac2x f(x) \\ & f(\frac1x) &=& \frac{x^2}{3} - \frac{2}{3x} f(x) \qquad (1) \\ \\ \text{Set }x=\frac1x: & 3f(x) + 2x f(\frac1x) &=& \frac{1}{x^2} \\ & 2x f(\frac1x) &=& \frac{1}{x^2}-3f(x) \\ & f(\frac1x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \qquad (2) \\\\ \hline (1) = (2): & \frac{x^2}{3} - \frac{2}{3x} f(x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \\ & \frac{3}{2x}f(x) - \frac{2}{3x} f(x) &=& \frac{1}{2x^3} -\frac{x^2}{3} \\ & f(x)\left( \frac{3}{2x} - \frac{2}{3x} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{9x-4x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{1} \right) &=& \frac{3-2x^5}{x} \\\\ & \mathbf{f(x)} & \mathbf{=} & \mathbf{\dfrac{3-2x^5}{5x^2}} \\ \hline \end{array}$$
The rational function is $$f(x) = \dfrac{3-2x^5}{5x^2}$$
$$f(-2)=\ ?$$
$$\begin{array}{|rclcl|} \hline f(-2) &=& \frac{3-2(-2)^5}{5(-2)^2} \\ &=& \frac{3+2^6}{5\cdot 4} \\ &=& \frac{3+64}{20} \\ &=& \frac{67}{20} \\ \mathbf{f(-2)} & \mathbf{=} & \mathbf{3.35} \\ \hline \end{array}$$
$$\text{ f(-2) is 3.35 }$$
graph:
heureka May 22, 2018
edited by heureka May 22, 2018
#3
+93629
+1
That is interesing Heureka, I had not realised that there was enough infromation to draw the graph. :)
Melody May 22, 2018
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https://math.stackexchange.com/questions/2955640/how-should-i-write-vectors-like-this | # How should I write vectors like this?
If I'm trying to write basic vectors, just as simple as the magnitude being 5 and the direction being zero, how would I do this? Would it be a row vector with parenthesis:$$\overrightarrow{v} = (5, 0)$$, a row vector with brackets: $$\overrightarrow{v} = [5, 0]$$, a column vector with parenthesis: $$\overrightarrow{v} = \begin{pmatrix} 5\\ 0\\ \end{pmatrix}$$, or a column vector with brackets: $$\overrightarrow{v} = \begin{bmatrix} 5\\ 0\\ \end{bmatrix}$$? Thank you if you can tell me what the correct notation for this simple vector is, everywhere I go seems to write them differently and the inconsistency makes me want to rip my hair out.
• You can write it however you want, as long as you are consistent. – Morgan Rodgers Oct 14 '18 at 21:19
• You might also want to consider what you are doing with these vectors. If you are frequently left-multiplying them by a matrix (e.g., $Av$), make them (thin) column vectors. If you are frequently right-multiplying them by a matrix (e.g., $vA$), make them (wide) row vectors. If you are performing both of these operations just as frequently, try not to mix row and column vectors too much (just pick one convention and apply transposes when necessary). As for the bracketing, just be consistent within one document (see Morgan Rodgers' comment). – parsiad Oct 14 '18 at 21:22
• @MorganRodgers and parsia thank you, I'll take this into consideration! – jstowell Oct 14 '18 at 21:23
• One thing I want to point out: the coordinates of a vector are not their magnitude and direction. The magnitude of the vector (3, 4) is 5, and the direction is about 53.13 degrees above the x-axis. – Deusovi Oct 16 '18 at 16:15
So long as you make sure that you are consistent throughout the entire text, it's completely up to you. There are many different ways to represent vectors. In Linear Algebra, people like to use column notation, with either parentheses or square brackets, as this is the most convenient in linear algebra. Some people also write something like $$\vec{v}=\begin{bmatrix}5&0\end{bmatrix}^T$$ to have a column vector, but be able to write it nicely inline. In high school (perhaps up to freshman) level classes, the notation $$\vec{v}=\langle 5,0\rangle$$ is also used, but this ignores the fact that vectors are matrices with 1 row or 1 column, so is less widely used at a higher level. Ultimately it's your choice, whatever is most convenient to you, you should use it.
Usually in linear algebra context vectors $$\vec v$$ are considered colummn vector and transponsed vectors $$\vec v^T$$ are row vectors that is
$$\overrightarrow{v} = \begin{pmatrix} 5\\ 0\\ \end{pmatrix} \quad \overrightarrow{v^T} = \begin{pmatrix} 5 &0\\ \end{pmatrix}$$ | 2019-06-17T00:52:24 | {
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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Jennifer can buy watches at a price of B dollars per watch, which she new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 50544 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 03:18 2 7 00:00 Difficulty: 45% (medium) Question Stats: 70% (01:48) correct 30% (02:08) wrong based on 141 sessions ### HideShow timer Statistics Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Kudos for a correct solution. _________________ Manager Joined: 15 Aug 2013 Posts: 54 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:11 2 1 Cost price of N watches = BN (cost of each watch, B * total nuber of watches, N) Total Profit = T But this profilt is also equal to BN (x/100), where x is the profit percentage Hence, BN (x/100) =T or x = 100T/ (NB) Hence (A). CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:19 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Let's say Jennifer marks up by x% Total Profit = Marked up price of N watches - Cost of N watches i.e. T = NB[1+(x/100)] - NB i.e. (T) / NB = [(x/100)] i.e. x = 100(T)/NB Answer: Option _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: [email protected] I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:32 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Let's say The price of each watch, i.e. B = 7, Let's say Mark up = 400%==> 400% of 7 is 28, and the$7 watches are marked up $28, then the selling price per watch =$35.
The profit per watch is $28, and so if she sells N = 11 watches Then Total profit, T = 28*11 =$308.
OK, now Substitute T = 308 = 11*28, N = 11, and B = 7, in options and hope to get 400 as our answer.
(A) 100T/(NB) = 100*11*28/(11*4) = 100*28/7 = 110*4 = 400 = works!
(B) TB/(100N) = 11*28*7/(100*11) = 28*7/100 = doesn’t work!
(C) 100TN/B = 100*11*28*11/7 = 100*11*4*11 = doesn’t work!
(D) ((T/N) – B)/(100B) = [(11*28/11) – 7]/(7*100) = (28 – 7)/(7*100) = 21/(7*100) = 3/100 = doesn’t work
(E) 100(T – NB)/N = 100(11*28 – 11*7)/7 = 100*11*21/7 = 100*11*3 = doesn’t work!
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
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18 Feb 2015, 07:51
1
Using smart numbers:
Jennifer Purchases watches for $5: B=5 Markup of 100%; now selling for$10: x=10
She sells 5 watches: N=5
Her total profit will be (N*x)-(B*N)
(5*10)-(5*5)= 25
T=25
Now using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100.
B=5
T=25
N=5
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
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18 Feb 2015, 21:51
Cost price per watch = b
Selling price per watch = $$\frac{t}{n}$$
Let x = percent of the markup from her buy price to her sell price
$$x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}$$
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
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22 Feb 2015, 11:17
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?
A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
T = total profit = $10t N = total number of items sold $$\frac{10t}{3b*n}$$ Hence, A pushpitkc is my solution correct ? Manager Joined: 14 Jun 2018 Posts: 179 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 23 Oct 2018, 09:47 Cost Price = B No of items sold = N Profit after selling N items = T ; Profit per item = $$\frac{T}{N}$$ Let "k" be the markup percentage , therefore selling price = $$\frac{(100+K)}{100} * B$$ We know , Selling Price - Cost Price = Profit => $$\frac{(100+K)}{100} * B - B = \frac{T}{N}$$ => $$K = 100 (\frac{T}{NB} + \frac{B}{B}) - 100$$ => $$K = 100(\frac{T}{NB})$$ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3301 Location: India GPA: 3.12 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 23 Oct 2018, 11:28 1 dave13 wrote: Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Kudos for a correct solution. let B be the cost +mark up price 2+1 =$3
T = total profit = $10t N = total number of items sold $$\frac{10t}{3b*n}$$ Hence, A pushpitkc is my solution correct ? Hey dave13 Unfortunately, this is wrong. I am not quite clear what you trying to do here If you are assigning values B =$5(cost per watch).
She marks up the watches and sells for a total profit of T = $1 Assuming there is a N = 1 watch and she keeps a profit of 20%(which is 0.2*$5 = \$1)
Now, working backward, trying to calculate the percentage by substituting the values
of B,N, and T, we will get
A. 100T/(NB) = 100/5 = 20 - Our solution is Option A($$\frac{100T}{NB}$$)
We don't need to substitute the values in the other 4 answer options as we have a match!
Hope that helps.
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
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25 Oct 2018, 08:19
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?
A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N
The profit (or markup) per watch is T/N. So the markup is:
(T/N)/B x 100 = 100T/(NB)
percent of the purchase price.
Alternate Solution:
Jennifer’s profit per watch is T/N. Using the formula for profit, we can compute Jennifer’s sell price:
profit = sell price – buy price
T/N = sell price – B
T/N + B = sell price
We now use the percent markup formula:
% markup = (Sell price – Buy price) / Buy price x 100
% markup = (T/N + B – B) / B x 100
% markup = T/N / B x 100
% markup = T/BN x 100
% markup = 100T / (BN)
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Re: Jennifer can buy watches at a price of B dollars per watch, which she &nbs [#permalink] 25 Oct 2018, 08:19
Display posts from previous: Sort by | 2018-11-13T00:55:58 | {
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http://math.stackexchange.com/questions/25273/derivative-of-a-xb-x | # Derivative of $(a\,x)^{b\,x}$
is there any rule to differentiate the function $(a\,x)^{b\,x}$?
I've got to find the derivative of $(x^2+1)^{\arctan x}$ and Wolfram|Alpha says the answer is $$\tan^{-1}(x) (x^2+1)^{\tan^{-1}(x)-1} \left(\frac{d}{dx}(x^2+1)\right)+\log(x^2+1) (x^2+1)^{\tan^{-1}(x)} \left(\frac{d}{dx}(\tan^{-1}(x))\right)$$ Is there any general rule to do that? Thanks.
-
Your derivative of $(x^2+1)^{\arctan x}$ is the particular case for $u(x)=x^2+1$ and $v(x)=\arctan x$ of
$$\frac{d}{dx}\left(\left[ u(x)\right] ^{v(x)}\right)=v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left( \ln u(x)\right) \left[ u(x)\right] ^{v(x)}v^{\prime }(x),$$
or omitting de variable $x$:
$$\left( u^{v}\right)^{\prime }=vu^{v-1}u^{\prime }+\left( \ln u\right) u^{v}v^{\prime }.$$
This can be derived observing that, since $u=e^{\ln u}$, we have $u^v=e^{v\ln u}$:
$$\begin{eqnarray*} \frac{d}{dx}\left( u^{v}\right) &=&\frac{d}{dx}\left( e^{v\ln u}\right) \\ &=&e^{v\ln u}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right) \\ &=&u^{v}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right) \\ &=&u^{v}\ln u\frac{dv}{dx}+u^{v}\frac{v}{u}\frac{du}{dx} \\ &=&u^{v}(\ln u)v'+u^{v-1}vu'. \end{eqnarray*}$$
Another possibility is to consider the variables $u$ and $v$ (both depending on $x$) in the function
$$z=f(u(x),v(x))=\left[ u(x)\right] ^{v(x)}$$
and determine its total derivative with respect to $x$:
$$\begin{eqnarray*} \frac{dz}{dx} &=&\frac{d}{dx}\left( \left[ u(x)\right] ^{v(x)}\right) \\ &=&\frac{% \partial z}{\partial u}\frac{du}{dx}+\frac{\partial z}{\partial v}\frac{dv}{% dx} \\ &=&v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left[ u(x)\right] ^{v(x)}\left( \ln u(x)\right) v^{\prime }(x) \end{eqnarray*}$$
because
$$\frac{\partial z}{\partial u}=\frac{\partial }{\partial u}\left( u^{v}\right) =vu^{v-1}$$
and
$$\frac{\partial z}{\partial v}=\frac{\partial }{\partial v}\left( u^{v}\right) =\frac{\partial }{\partial v}\left( e^{\ln u\cdot v}\right) =e^{\ln u\cdot v}\ln u=u^{v}\ln u.$$
For $u(x)=ax,v(x)=bx$, we get
$$\frac{d}{dx}\left( \left( ax\right) ^{bx}\right) =bx\left( ax\right) ^{bx-1}a+\left( ax\right) ^{bx}\left( \ln (ax)\right) b.$$
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Assuming you mean $(ax)^{bx}$, I'd just write it as $(e^{\ln(ax)})^{bx}$ and use the chain rule (ie $e^{\ln(ax)bx} = e^{u(x)}$ and go from there).
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HINT $\rm\ \ (F^G)'\ =\ (e^{G\:ln\ F})'\: =\ F^G\ (GF'/F + G'\ ln\ F)$
-
I think there is a typo in the sign. – Américo Tavares Mar 6 '11 at 16:26
@Americo: Fixed it, thanks! – Bill Dubuque Mar 6 '11 at 16:32
Not answering the math part, since Stijn has already done that, but if you click on the "show steps" button, Wolfram|Alpha shows you one possible path for the derivation. I've included the image for the derivation of $\frac{\partial}{\partial x} ((a x)^{b x})$
A similar set of steps is supplied for the other derivative that you want to take: http://www.wolframalpha.com/input/?i=d/dx((x^2%2B1)^(tan^(-1)(x)))
-
The link for the first example is wolframalpha.com/input/?i=D[(ax)^(bx),x]. Does anyone know how to get Wolfram|Alpha links working properly in the markup? – Simon Mar 6 '11 at 11:41
The parentheses and the caret confuse Markdown. Try escaping ^ with %5E, ( with %28, ) with %29, [ with %5B and ] with %5D like this: WA Link (you obviously have to use the [title](URL) syntax) – kahen Mar 6 '11 at 13:21 | 2015-02-01T22:46:53 | {
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