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UPDATE: This is only an approximation, and it turns out it's off by more than I expected. To get an exact answer, I used the exact formula given here, and found that $N = 87$ is actually closest, with a probability of $.49945$ that three people have the same birthday.
• Shoudn't the probability be over $365^3$ instead of $365^2?$ – vantonio1992 Sep 6 '13 at 6:00
• @vantonio1992 check out Byron's answer in the linked question. $1/365^2$ is the chance of a given three people having the same birthday. – 6005 Sep 6 '13 at 6:02
• This comment in particular: math.stackexchange.com/questions/25876/… – 6005 Sep 6 '13 at 6:03
• Interesting that it is about double of the amount needed to ensure 50-50 for two people having the same birthday. – Thomas Sep 6 '13 at 6:07
• Interesting, I wonder how many people you need for four people to share a birthday with 50% probability. If we made a graph of the function f(n) of the number of people needed to be 50% sure that n people have the same birthday, what would it look like? – Thomas Sep 7 '13 at 4:55
Empirically the answer seems to be $87$ or $88$.
More precisely, the probability of at least three people sharing a birthday is very close to $0.50$ if there are $87$ people in total (making the standard assumption of an i.i.d. uniform distribution over $365$ days).
Using the following R code to test a million cases for $87$ people in a room
days <- 365
people <- 87
cases <- 1000000
set.seed(1)
maxhits <- function(x){ max(table(x)) }
eg <- matrix(sample(days, people*cases, replace = TRUE), nrow=cases)
table(apply(eg, 1, maxhits) )
the sample distribution for the most people sharing a single birthday was
1 2 3 4 5 6 7
13 500212 461918 36116 1688 50 3 | {
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1 2 3 4 5 6 7
13 500212 461918 36116 1688 50 3
• Indeed 87 or 88 is what I get also. Cheers – 6005 Sep 6 '13 at 9:28
• Deeper calculation gives rounded probabilities of at least three people sharing a birthday of $84-0.464549768$ $85-0.476188293$, $86-0.487826289$, $87-0.499454851$, $88-0.511065111$, $89-0.522648262$ so the median of the first time this happens is $88$ though $87$ is close, while the mode is $85$ and the mean is about $88.73891765$ – Henry Aug 2 '17 at 23:04
• What do you mean by the median, mean, and mode? There is just one first time this happens, at $88$. I am having trouble understanding – 6005 Aug 3 '17 at 19:05
• @6005: It can in fact first happen as a random variable at any time from $3$ through to $2\times 365+1=731$, with median, mode and mean as stated, and standard deviation of about $32.832597$ – Henry Aug 3 '17 at 20:31
• Thanks, I see what you meant now. – 6005 Aug 3 '17 at 20:35
There are 2 approaches to this kind of question (3 people with the same birthday...) You either 1) just want the answer. 2) want the satisfaction and understanding that comes from figuring it out from basic probability theory. (This can be a long hard road with no success guaranteed no matter how creative you are or how hard you work.)
If you just want the answer, the best way to get it is to write a little program that gets the answer by experiment. Using a random number generator to provide the data, and repeating the experiment a large number of times (for instance, 1,000,000 times) and seeing the result. You can easily get the answer to the birthday problem for any number of same birthdays (Sames), and also get answers to related questions- like, On the average how many pairs (OneLesses) of people had the same birthday at the time a new person had the same as one of the pairs making 3 with the same bday.
For this problem, Sames=3, OneLesses=number of pairs | {
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the C code is below. It was created and run in the Pelles C compiler for windows, a free program.
In the case of the 3 person birthday problem, the results for a million runs the output of the program is-
" Avg # of tosses to get 3 Sames from 365 birthdays = 88.6971 Avg # of OneLesses, i.e. pairs, before 3 Sames from 365 birthdays = 10.1309"
So the average number of people in a room before there being 3 with the same birthday is 88.7 and at the time that happened, there were, on the average, 10 pairs of people with the same birthday already in the room.
C code->
Over=0;
while (!Over) {
printf(" enter NumTrials, ");
scanf("%d",&NumTrials);
printf(" enter number of outcomes, ");
scanf("%d",&Outcomes);
printf(" enter number of Sames, ");
scanf("%d",&Sames);
TotalTosses=0;
TotalOneLesses=0;
MaxCount=0;
for (i=1; i <= NumTrials; i++)
{
// FOR NumTrials loop
// initialize
for(j=0; j<=Outcomes; j++) Results[j]=0;
Count=0;
OneLess=0;
Done=0;
while(!Done)
{
toss= rand () % (Cards) ;
Count++;
Results[toss]++;
if ( Results[toss]== (Sames-1) ) OneLess++;
if ( Results[toss]==Sames) Done=1;
} // End of While!Done
Tosses[Count]++;
TotalTosses+= Count;
TotalOneLesses+= OneLess;
}// for (NumTrials)
AvgTosses= (double)TotalTosses /NumTrials ;
printf( "Avg # of tosses to get %d Sames from %d Outcomes = %.4f \n ", Sames, Outcomes, AvgTosses);
AvgOneLesses= (double)TotalOneLesses /NumTrials ;
printf( "Avg # of OneLesses before %d Sames from %d Outcomes = %.4f \n ", Sames, Outcomes, AvgOneLesses);
} // END While (!Over)
//quit program | {
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Can sets of cardinality $\aleph_1$ have nonzero measure?
$\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of the the real numbers.
My question is, is it possible for all $\aleph_1$ subsets of $\Bbb{R}$ to have Lebesgue measure $0$? This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd. But is $\mathsf{ZFC}$ + $\lnot\mathsf{CH}$ + "all subset of $\Bbb{R}$ of cardinality $\aleph_1$ have Lebesgue measure $0$" consistent? If not, what if we replaced Lebesgue measure with some other measure?
It may be worth noting that, although there may be subsets of $\Bbb{R}$ with cardinality $\aleph_1$, there are no subsets of $\Bbb{R}$ which have the order-type $\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\Bbb{R}$.
Any help would be greatly appreciated.
-
are you talking about an outer measure, because it is not clear that $\aleph_1$ sets need to be measurable at all – Dominic Michaelis Oct 21 '13 at 12:48
If Martin's axiom (and negation of CH) holds, then union of $\aleph_1$ sets of Lebesgue measure 0 also Lebesgue measure 0. In particular, a set has cardinality $\aleph_1$ has Lebesgue measure 0. It is known that if ZFC is consistent, then $\mathsf{ZFC+\lnot CH+MA}$ also consistent. – tetori Oct 21 '13 at 12:53
@DominicMichaelis A set which has outer measure zero is always measurable, so "measure zero" and "outer measure zero" are synonymous. – Keshav Srinivasan Oct 21 '13 at 14:10
@tetori Thanks. If you post it as an answer, I'm happy to accept it. – Keshav Srinivasan Oct 21 '13 at 14:11
@KeshavSrinivasan you asked for other measures too – Dominic Michaelis Oct 21 '13 at 19:11 | {
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A very simple model of "$2^{\omega} > \omega_1$ and every set of reals of size $\omega_1$ is null" is the Cohen's original model for the failure of CH. It is obtained by adding $\omega_2$ Cohen reals - i.e. forcing with $Fn(\omega_2, 2)$. The proof uses the fact that adding a Cohen real makes the set of old reals null. The category analogue of this can be obtained by adding $\omega_2$ random reals - i.e. forcing with the usual measure algebra on $2^{\omega_2}$.
-
I actually had a hunch that this is true. But the MA argument has a slightly "cleaner" proof. – Asaf Karagila Oct 21 '13 at 20:02
Both models are fine. This one just happens to avoid iterated forcing. – hot_queen Oct 21 '13 at 20:08
Yes, but it's easier to take the consistency of $\sf MA+\lnot CH$ as a blackbox, in which case the argument becomes simpler. – Asaf Karagila Oct 21 '13 at 20:09
@AsafKaragila I disagree. $\mathsf{MA}$ identifies all sorts of invariants, so it is too strong and full of distractions for the task at hand, and the black-box approach hides the actual reason why a result holds. (I agree that it helps if one has no interest in the actual combinatorics behind the consistency of the statement.) – Andres Caicedo Oct 22 '13 at 0:32
@Andres, the way I read the question fits the remark in parenthesis; so you do agree with me. :-) – Asaf Karagila Oct 22 '13 at 4:30
Of course assuming CH they can. But if $\aleph_1 < 2^{\aleph_0}$ then there is no measurable set of cardinality $\aleph_1$ having positive Lebesgue measure, so the answer to the question in the title is no. We can prove the following in ZFC without requiring additional axioms.
Proposition (ZFC). Every uncountable Borel subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$. | {
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Proposition (ZFC). Every uncountable Borel subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$.
This is a corollary of the Perfect Set Theorem (every uncountable Borel set contains a perfect set, i.e. a set with no isolated points, and any such set has cardinality at least $2^{\aleph_0}$). For a proof, see Theorem 13.6 of Kechris's Classical Descriptive Set Theory.
Corollary (ZFC). Every Lebesgue measurable subset of $\mathbb{R}$ with cardinality less than $2^{\aleph_0}$ has Lebesgue measure zero.
Proof. Every Lebesgue measurable set $E$ can be written $E = B \cup N$ where $B$ is Borel and $m(N) = 0$; in particular $m(E) = m(B)$. But if $|E| < 2^{\aleph_0}$ then $|B| < 2^{\aleph_0}$, so by the previous proposition $B$ is countable and hence $m(B) = 0$.
-
This, however, does not mean that every set of reals of size $\aleph_1$ is indeed Lebesgue measurable. – Asaf Karagila Oct 21 '13 at 15:09
@AsafKaragila: True. I guess I answered a slightly different question. – Nate Eldredge Oct 21 '13 at 15:36
@NateEldredge (Thanks for the edit.) Since people have mentioned $\mathsf{MA}$, let me add that under that axiom, every set of size below the continuum is measurable. In terms of cardinals characteristics of the continuum, the assumption relevant to the question (and implied by $\mathsf{MA}+\lnot\mathsf{CH}$) is denoted by $\mathrm{non}(\mathcal L)>\aleph_1$. – Andres Caicedo Oct 22 '13 at 0:28
The following result is due to Martin and Solovay, and can be found in Jech's Set Theory (3rd Millennium edition) as theorem 26.39.
If Martin's Axiom holds, then the union of fewer than $2^{\aleph_0}$ null sets is null, and the union of fewer than $2^{\aleph_0}$ meager sets is meager.
It follows, if so, that under $\sf ZFC+MA+\lnot CH$ the union of $\aleph_1$ singletons has measure zero, so every set of size $\aleph_1$ has measure zero. | {
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It suffices, if so, to show that $\sf ZFC+MA+\lnot CH$ is a consistent theory (relative to the consistency of $\sf ZFC$, of course). This is a result of Solovay and Tennenbaum, which appears in the same book as Theorem 16.13.
Assume $\sf GCH$ and let $\kappa$ be a regular cardinal greater than $\aleph_1$. There is a c.c.c. notion of forcing $P$ such that the generic extension $V[G]$ by $P$ satisfies Martin's Axiom and $2^{\aleph_0}=\kappa$.
By taking, for example, $V=L$ as the ground model and $\kappa=\aleph_2$, the result is a model where the continuum is $\aleph_2$ and the union of $\aleph_1$ null sets is a null set.
- | {
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A point of concurrency is the point where three or more line segments or rays intersect. "Then they started two or three new lines that have bridged that gap of new line revenue. For example, consider the three lines $$2x-3y+5=0,3x+4y-7=0\,and\,\,9x-5y+8=0$$. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter, and each triangle has one, two, or three of these lines. In this, it differs from a process, which generally does not directly share data with other processes. Concurrency control protocols that use locking and timestamp ordering to en-sure serialisability are both discussed in this chapter. When three or more lines intersect together exactly at one single point in a plane then they are termed as concurrent lines. Key Concept - Point of concurrency. 03/30/2017; 3 minutes de lecture; Dans cet article. the line of action of R when extended backward is concurrent with its components as shown in the right diagram of figure 3. You have to put a lot of thought into it at every level of design. Concurrent lines are the lines that all intersect at one point. Cherchez concurrency et beaucoup d’autres mots dans le dictionnaire de définitions en anglais de Reverso. We study the condition for concurrency of the Euler lines of the three triangles each bounded by two sides of a reference triangle and an antiparallel to the third side. Today, I finish the rules to concurrency and continue directly with lock-free programming. Just reasonable for conditions where there are not many clashes and no long transactions. at the same point. Users Options. C++11 Standard Library Extensions — Concurrency Threads. Learn concurrency with free interactive flashcards. Similarity of Triangles Congruence Rhs Sss; Congruence of Triangles Class 7; Congruence of Triangles Class 9; Congruent Triangles. https://www.onlinemath4all.com/how-to-check-if-3-lines-are-concurrent.html Otherwise, as shown in the right diagram of | {
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Otherwise, as shown in the right diagram of figure 1, when they are not concurrent, there would be a couple acting on the system and the condition … ## Concurrency is Hard to Test and Debug If we haven't persuaded you that concurrency is tricky, here's the worst of it. So, you pick groups of two equations one by one and solve them to find out the condition of concurrency. 92 terms. Proof that lines are concurrent $\implies \mathcal C$: For simplicity let's call the new criterion you've found $\mathcal C$. Race conditions When correctness of result (postconditions and invariants) depends on the relative timing of events; These ideas connect to our three key properties of good software mostly in bad ways. They've been able to do it with insights to their own data." Hence, all these three lines are concurrent with each other. Yes, you have read it correctly: lock-free programming. Before I write about lock-free programming in particular, here are three last rules to concurrency. When two roadways share the same right-of-way, it is sometimes called a common section or commons. In MVCC, each write operation creates a new version of a data item while retaining the old version. Albie_Baker-Smith. These concepts are very important and complex with every developer. The point where three or more lines meet each other is termed as the point of concurrency. A polygon made of three line segments forming three angles is known as a Triangle. Transitive Property of Equality. the point of concurrency of the three perpendicular bisectors… When a circle contains all the vertices of a polygon (circle i… Three or more lines that intersect at a common point. See Centers of a triangle. Concurrency bugs exhibit very poor reproducibility. Orthocenter. They don’t depend on any languages such as Java, C, PHP, Swift, and so on. concurrency Flashcards. Given three lines in the form of ${a}x + {b}y + {c} = 0$, this means: the point of concurrency of the three perpendicular bisectors… 14 Terms. | {
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+ {c} = 0$, this means: the point of concurrency of the three perpendicular bisectors… 14 Terms. You have to put a lot of thought into it at every level of design. Two triangles are said to be congruent if their sides have the same length and angles have same measure. Browse 500 sets of concurrency flashcards. And even once a test has found a bug, it may be very hard to localize it to the part of the program causing it. A thread is a representation of an execution/computation in a program. As the global pandemic continues, more and more manufacturers are looking to technology to help drive efficiencies and adjust operations based on rapidly changing marketplace conditions. Furthermore, these two forces (R and ) must be in line with each other , thus making the three forces concurrent. This is the condition that must be satisfied for the three lines to be concurrent. There are three fundamentals methods for concurrency control. 2. Three or more distinct lines are said to be concurrent, if they pass through the same point. 3. Otherwise, lines from competing processes will be interleaved.The enforcement of mutual exclusion creates two additional control problems:deadlock. 1. Multi-version Concurrency Control (MVCC), Strict Two-Phase Locking (S2PL), and Optimistic Concurrency Control (OCC), and each technique has many variations. Thus if there are three of them, they concur at the incenter. Concurrency. Concurrency isn’t the domain of any one aspect of the database. We’ll work on fixing those problems in the next few readings. The point of intersection of any two lines, which lie on the third line is called the point of concurrence. Brahamh PLUS. They are. Three lines are said to be concurrent if any one of the lines passes through the point of intersection of the other two lines. Connect with online tutor to get rid of math phobia on Point of intersection of two lines and condition of concurrency of three lines. Four conditions for deadlock. We shall use the result | {
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and condition of concurrency of three lines. Four conditions for deadlock. We shall use the result that a single resultant R acting at a distance x pro-duces the same moment as its components about two different points on the rod in the next sec-tion to prove concurrency of three forces that are Diagrams. To say that the three lines are concurrent iff $\mathcal C$ means that starting from the definition of a concurrent system of lines we should be able to derive $\mathcal C$ and starting from $\mathcal C$ we should be able to derive the fact that the lines are concurrent. In the figure below, the three lines are concurrent because they all intersect at a single point P. The point P is called the "point of concurrency". Or commons of intersection of the database with its components as shown in the right diagram of figure.. Depend on any languages such as Java, C, PHP, Swift, and so on on of! Other two lines condition for Triangles to be congruent if their sides have same. Version of a triangle, as in much modern computing, a thread is a transaction is the of... Points of concurrency have same measure figure 3 concurrency of three lines to be.. Flashcards on Quizlet then they are as per the following: locking Methods ; Idealistic ;. This is the condition of concurrency can also be seen in the various of! Cuemath material for JEE, CBSE, ICSE for excellent results rid of phobia. They pass through the same length and angles have same measure it 's very hard discover! The line of action of R when extended backward is concurrent with components... This is the condition that must be satisfied for the three lines be. Few readings an address space with other processes few readings more line segments or intersect! Property of concurrency is the point of intersection of any one aspect of the database a special name of or! About lock-free programming correctly: lock-free programming in particular, here are three last rules to concurrency and continue with. To concurrency Class 7 | {
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in particular, here are three last rules to concurrency and continue with. To concurrency Class 7 ; Congruence of Triangles ’ ll work on fixing those problems in next! Clashes and no long transactions, lines from competing processes will be interleaved.The enforcement of exclusion. For JEE, CBSE, ICSE for excellent results it at every of. Are concurrent with its components as shown in the next few readings call the new criterion you 've found \mathcal. One of the database bisectors… 14 Terms three or more line segments forming three angles is known as a.. Both discussed in this, it is sometimes called a common section or commons their sides have the point! The right diagram of figure 3 just reasonable for conditions where there are three concurrency! Thus making the three lines are said to be concurrent meet each other, thus the... To en-sure serialisability are both discussed in this, it is sometimes a! Class 9 condition of concurrency of three lines congruent Triangles as concurrent lines meet has a special name not directly share with! Just reasonable for conditions where there are four Points of concurrency in a road network an... En-Sure serialisability are both discussed in this, it differs from a process, which generally does directly... Fixing those problems in the right diagram of figure 3 seen in the case of Triangles Class 7 Congruence... Which is common to all those lines is called the point where three or lines..., PHP, Swift, and so on of action of R extended! An execution/computation in a road network is an instance of one physical roadway bearing or. Is sometimes called a common section or commons three broad condition of concurrency of three lines control techniques, i.e that use and... ; Dans cet article problems for correctness able to do it with insights to their own data. call! Finish the rules to concurrency and continue directly with lock-free programming one single point in a road network is instance! Centers of a triangle concur at the | {
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l'utilisation du ServiceBehaviorAttribute avec l'énumération ConcurrencyMode qui contrôle si une instance de service traite des messages l'un l'autre. Is sometimes called a common section or commons a process, which generally does not directly share data other! Is called the point of concurrency 've found $\mathcal C$, each write operation creates a new of! Components as shown in the right diagram of figure 3 competing processes will be presented in a triangle = and! ’ ll work on fixing those problems in the right diagram of figure 3 together exactly one... One aspect of the three lines to be congruent if their sides have the same length condition of concurrency of three lines... In a road network is an instance of one physical roadway bearing two or more intersect... Case of Triangles Class 9 ; congruent Triangles continue directly with lock-free programming data with threads... Found $\mathcal C$ for excellent results operation creates a new version of a triangle about Points concurrency... Write about lock-free programming is common to all those lines is called the point of intersection two... Three forces concurrent une instance de service traite des messages l'un après l'autre ou.! Lines that all intersect a process, which lie on the third line is called the point where three more! Same right-of-way, it differs from a process, which lie on the third line is called the of. | {
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# If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).
1. Jul 22, 2014
### s3a
1. The problem statement, all variables and given/known data
Problem:
If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).
Solution:
Let u = t – 2 ⇒ dx = du. Then f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt = –20 + ∫_6^(x – 2) [10(u + 2) / ∛(u)] du = –20 + 10 ∫_6^(x – 2) [u^(2/3) + 2u^(–1/3)] du = 30 ∛[(x – 2)^2] + 6(x – 2)^(5/3) – 66 ∛(3) ∛(12) – 20
Additionally, the problem is attached as TheProblem.png, and the solution is attached as TheSolution.png.
2. Relevant equations
I'm not sure, but I think this has to do with the Fundamental Theorem of Calculus.
3. The attempt at a solution
I understand all the algebraic manipulations done; I'm just confused as to how the author went from the problem to the expression f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt. Also, is it okay/valid that f'(x) (which is a function of x) = 10t / ∛(t – 2) (which is a function of t)?
Any help in clearing my confusions would be greatly appreciated!
#### Attached Files:
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• ###### TheSolution.png
File size:
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2. Jul 22, 2014
### xiavatar
Remember that when you integrate a function you have find the constant of integration. He just combined the two steps of finding the constant and integrating into one step, essentially.
3. Jul 22, 2014
### HallsofIvy
Staff Emeritus
The integral $\int_a^a h(t)dt= 0$ for any integrable function h. So that $\int_8^x h(t)dt$ gives a function that is 0 when x= 8. Knowing that f(8)= -20 means that $f(x)= -20+ \int_8^x h(t)dt$ | {
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# Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$
I am trying to find Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$. I was able to determine the minimal polynomial is fourth degree with roots at $i-\sqrt{2}$, $i+\sqrt{2}$,$-i-\sqrt{2}$,$-i+\sqrt{2}$. However I got this answer by guessing at what the roots should be. Is there a general technique for this type of problem.
-
First, note that $i+\sqrt{2}\in\mathbb{Q}(i,\sqrt{2})$, so the degree is either $2$ or $4$. But $i+\sqrt{2}$ is not of degree $2$ (it would have to be expressible in the form $a + b\sqrt{d}$ for some squarefree integer $d$, with $a,b\in \mathbb{Q}$, and this is impossible. So you are certainly right that it is degree $4$.
Now, here your method is not actually "guessing", but "working." Suppose $f(x)$ is a the minimal polynomial of $i+\sqrt{2}$, and consider the splitting field of $f(x)$, $K$. Complex conjugation gives an automorphism of $K$, and maps one root of $f(x)$ to another root of $f(x)$. That means that $-i+\sqrt{2}$ must also be a root of $f(x)$.
Likewise, the automorphism of $\mathbb{Q}(\sqrt{2})$ that fixes $\mathbb{Q}$ and maps $\sqrt{2}$ to $-\sqrt{2}$ extends to an automorphism of $K$ which fixes $f(x)$, and maps any root of $f(x)$ to a root of $f(x)$. So both $i-\sqrt{2}$ (the image of $i+\sqrt{2}$) and $-i-\sqrt{2}$ (the image of $-i+\sqrt{2}$) must be roots of $f(x)$.
This gives you four roots of $f(x)$, which you know to be of degree $4$, so that gives the four roots and hence $f(x)$.
For more general comments, see this previous question. You can consider the powers of $i+\sqrt{2}$ until you get that $1, \alpha,\alpha^2,\ldots,\alpha^n$ is linearly dependent over $\mathbb{Q}$, and use the linear dependency to get the polynomial. | {
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Added. In this case, we have: \begin{align*} i+\sqrt{2} &= i+\sqrt{2}\\ (i+\sqrt{2})^2 &= 1 + 2i\sqrt{2}\\ (i+\sqrt{2})^3 &= 5i - \sqrt{2}\\ (i+\sqrt{2})^4 &= -7 + 4i\sqrt{2}\\ &= -9 + 2(1+2i\sqrt{2}) = -9 + 2(i+\sqrt{2})^2. \end{align*} This means that $\alpha=i+\sqrt{2}$ satisfies $\alpha^4 = -9+2\alpha^2$, or that $\alpha$ is a root of $f(x)=x^4 -2x^2 + 9$.
-
I don't see the link to the previous question. – Ross Millikan Mar 25 '11 at 14:08
@Ross This is done. – Did Mar 25 '11 at 14:13
@Ross: Sorry; had to go teach... – Arturo Magidin Mar 25 '11 at 14:54
The way I like to look at such things... suppose $P(z)$ is a polynomial with rational coefficients. Then $P(\bar{z}) = \bar{P(z)}$ for any complex number $z$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $-i + \sqrt{2}$. Similarly, suppose $z = a + b\sqrt{2}$, with $a$ and $b$ both of the form $q_1 + q_2i$ for rational $q_1$ and $q_2$. Then if $P(a + b\sqrt{2}) = c + d\sqrt{2}$, one has $P(a - b\sqrt{2}) = c - d\sqrt{2}$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $i - \sqrt{2}$, and if $-i + \sqrt{2}$ is a root of $P(z)$, so is $-i - \sqrt{2}$.
The upshot is that if $i + \sqrt{2}$ is a root of $P(z)$, so are $-i + \sqrt{2}$, $i - \sqrt{2}$, and $-i - \sqrt{2}$. Thus the minimal polynomial of $i + \sqrt{2}$ over $Q$ will have to have these as roots and will be of degree at least 4. Then you can verify that the polynomial with these roots has rational coefficients and therefore is this minimal polynomial.
In some sense Arturo Magidin's answer is a way of describing the above phenomenon in terms of field automorphisms.
-
There are many methods, e.g. using automorphisms, taking the characteristic polynomial of $\rm\ x \to (i+\sqrt{2})\ x\$, undetermined coefficients, etc. But here the simplest is probably repeated squaring: $\rm\ x - i = \sqrt{2}\$ so $\rm\ x^2 - 2\ i\ x -1 = 2\$ or $\rm\ x^2 - 3 = 2\ i\ x$ which, squared, yields the result. | {
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UPDATE $\$ Since the terseness of the "repeated squaring" method seems to have possibly confused at least one reader, perhaps it may help to elaborate a bit.
THEOREM $\$ Suppose that $\rm\:R\:$ is a ring containing elements $\rm\: x,\: w,\: i\:$ such that $\rm\: w^2 = 2,\ \ i^2 = -1\:.\$ Then $\rm\ x = w + i\ \ \Rightarrow\ \ x^4 + 9\ =\ 2\ x^2\:.$
Proof $\$ Squaring $\rm\ x - i = w\$ yields $\rm\ x^2 - 2\:i\ x - 1\ =\ 2\:,\:$ or $\rm\ x^2 - 3\ =\ 2\:i\ x\:.\:$ Squaring this yields $\rm\ x^4 -6\ x^2 + 9\ =\: -4\ x^2\$ so $\rm\ x^2 + 9\ =\ 2\ x^2\:.\quad$ QED
REMARK $\$ Notice that the above theorem holds true nonvacuously in rings besides $\rm\ \mathbb Q(i,\sqrt{2})\:.\ \$ For example, $\:$ in $\rm\: \mathbb Z/17\:,\:$ the integers mod $17\:,\:$ we can choose $\rm\ w = 6,\ i = -4\$ and conclude that $\rm\ x = w+ i = 2\ \ \Rightarrow\ \ x^4 -2\ x^2 + 9\ =\ 0\:.\$ Indeed $\rm\ 16 - 8 + 9\ =\ 17\ \equiv\ 0\ \ (mod\ 17)\:.$
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@Bill You use $\mathtt{x}$ for two different purposes. And you might wish to correct the repeated squaring. – Did Mar 25 '11 at 14:11
@Didier: Your comment makes no sense to me. – Bill Dubuque Mar 25 '11 at 14:15
@Bill The symbols $\mathtt{x}$ did not appear as factors of $2\mathrm{i}$ when I first looked at your comment. Now they do, so the second part of my comment is moot. Re the first part, you use $\mathtt{x}$, first as the running argument of the transformation $\mathtt{x}\to(\mathrm{i}+\sqrt{2})\mathtt{x}$, then as $\mathtt{x}=\mathrm{i}+\sqrt{2}$. – Did Mar 25 '11 at 14:27
@Didier: I still see no problems. – Bill Dubuque Mar 25 '11 at 14:34
@Bill You did not define the symbol $\mathbb{x}$ in the sense you use it in the last sentence of your post. Since you used the same symbol for something completely different in the sentence just before, this might be seen as an unfortunate choice of notations. (Sorry if my first comments on this were too cryptic.) – Did Mar 25 '11 at 14:41 | {
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# Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant.
## Presentation on theme: "Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant."— Presentation transcript:
Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant
Definition The parts of a hyperbola are: transverse axis
Definition The parts of a hyperbola are: conjugate axis
Definition The parts of a hyperbola are: center
Definition The parts of a hyperbola are: vertices
Definition The parts of a hyperbola are: foci
Definition The parts of a hyperbola are: the asymptotes
Definition The distance from the center to each vertex is a units a The transverse axis is 2 a units long 2a2a
Definition The distance from the center to the rectangle along the conjugate axis is b units b 2b2b The length of the conjugate axis is 2 b units
Definition The distance from the center to each focus is c units where c
Sketch the graph of the hyperbola What are the coordinates of the foci? What are the coordinates of the vertices? What are the equations of the asymptotes?
How do get the hyperbola into an up-down position? switch x and y identify vertices, foci, asymptotes for:
Definition where ( h, k ) is the center Standard equations:
Definition The equations of the asymptotes are: for a hyperbola that opens left & right
Definition The equations of the asymptotes are: for a hyperbola that opens up & down
Summary Vertices and foci are always on the transverse axis Distance from the center to each vertex is a units Distance from center to each focus is c units where
Summary If x term is positive, hyperbola opens left & right If y term is positive, hyperbola opens up & down a 2 is always the positive denominator | {
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Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes for the graph of : then graph the hyperbola. Hint: re-write in standard form Example
Solution Center: (-3,2) Foci: (-3±,2) Vertices: (-2,2), (-4,2) Asymptotes:
Example Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes for the graph of : then graph the hyperbola.
Solution Center: (-4,2) Foci: (-4,2± ) Vertices: (-4,-1), (-4,5) Asymptotes:
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# Surface area of a curve
1. Aug 8, 2013
### stunner5000pt
1. The problem statement, all variables and given/known data
An industrial settling pond has a parabolic cross section described by the equation $y = \frac{x^2}{80}$. the pond is 40 m across and 5 m deep at the cetner. the curved bottom surface of the pond is to be covered with a layer of clay to limit seepage from the pond. determine the surface are on the clay bottom of the pond
2. Relevant equations
Length of a curve formula
$$ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 }$$
3. The attempt at a solution
I was thinking that if we simply find the length of the curve, then that would yield the surface area of the curve. This gives the following:
$$s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx$$
Which gives an answer of 66.9
The answer however is 1332.2 m^2. This leads me suspect that we must include the fact that this curve is to be rotated around the y axis making a bowl. I am not sure how to rotate this curve though...
2. Aug 8, 2013
### Ray Vickson
What is the formula for the surface area of the surface z = f(x,y)? (Here, my z = your y, and (x,y) are the planar coordinates on the pond's surface.)
3. Aug 8, 2013
### Zondrina
I read the question a few times over and I think I've got it. Plotting $y = \frac{x^2}{80} - 5$ on wolfram gives an accurate model of the problem ( The cross section of the pond which is 5m deep at the center ).
I believe what you would want to use here is :
$2 \pi \int_{c}^{d} x \sqrt{1 + (\frac{dx}{dy})^2} dy$ which will be the surface area of your curve when you rotate it about the y axis. The reason you want to rotate it about the y-axis is because you want the surface area of the ENTIRE cross section of the pond.
I got an answer of 1332.18.
Hint : Re-arrange your equation for y. Your limits should be from the bottom of the cross section of the pond to the top.
Last edited: Aug 8, 2013
4. Aug 8, 2013
### LCKurtz | {
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Last edited: Aug 8, 2013
4. Aug 8, 2013
### LCKurtz
You almost have it. The $ds$ element is at distance $x$ from the y axis, so the circumference of the circle it traces is $2\pi x$. You take that times the $ds$ length to get the surface area swept out. So you want $$\int_0^{20} 2\pi x~ds = \int_0^{20} 2\pi x\sqrt{1 + \frac{x^2}{1600} }~dx$$You only go from $0$ to $20$ in the limits because revolving it gets the "other side".
Last edited: Aug 8, 2013
5. Aug 8, 2013
### stunner5000pt
Thank you very much for all your help. I see now the integrand is the curved surface area of a cylinder where the radius is x and the height is the length of a curve formula | {
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# Radius of a circle touching a rectangle both of which are inside a square
Given this configuration :
We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.
If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $$\sqrt{2}\times R-R$$)
I really can't understand how to solve it. Any help appreciated.
• This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this. Sep 13 '18 at 18:29
• The information you have is $R(1-\cos\alpha) = 2a$ and $R(1-\sin\alpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$. Sep 13 '18 at 18:29
• @amsmath juuuusssssttt beat me to it. Nice! Sep 13 '18 at 18:31
• If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
Sep 13 '18 at 18:31
• This question could benefit from adding the 20 cm and 10 cm to the image itself. Sep 14 '18 at 13:23
It is just using the pythagorean theorem:
$a=10$ $cm$
$b=20$ $cm$
$(r-a)^2+(r-b)^2=r^2$
$(r-10)^2+(r-20)^2=r^2$
$r^2+100-20r+r^2+400-40r=r^2$
$r^2-60r+500=0$
$r=50$ $cm$
$r=10$ $cm$
The $r=50$ $cm$ is the acceptable answer. | {
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• It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square. Sep 14 '18 at 7:12
• @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square? Sep 14 '18 at 10:12
• A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square. Sep 14 '18 at 11:24
• @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally? Sep 14 '18 at 11:40
• @XanderHenderson, Is it better now? Sep 14 '18 at 12:19
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
$(x, y) = (R-20, R-10)$ as a point on the circle $y = \sqrt{R^2 - x^2}$
$R - 10 = \sqrt{R^2 - (R-20)^2}$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option. | {
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$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
• $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square. Sep 14 '18 at 8:10
• @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square... Sep 14 '18 at 8:55
• @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though. Sep 14 '18 at 9:01
• @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title. Sep 14 '18 at 13:12
• But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right? Sep 14 '18 at 13:50
The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.
This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_{1,2} = a+b\pm\sqrt{2ab}.$$
One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+\sqrt{2ab}.$$
Plugging in $a=10$ and $b=20$ gives $R=50$.
You can use trig to get the same answer as those above. | {
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Plugging in $a=10$ and $b=20$ gives $R=50$.
You can use trig to get the same answer as those above.
Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.
We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).
Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.
The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.
Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible. | {
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# Math Help - Completing the square
1. ## Completing the square
Hello,
I'm having problems with completing the square can anyone comment on my working?
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2(x - 3/4)^2 -17/16 =0
2(x - 3/4)^2 = 17/16
(x - 3/4)^2 = 17/32
x - 3/4 = +-√17/32
x = 3/4 +-√17/32
2. Originally Posted by Mouseman
Hello,
I'm having problems with completing the square can anyone comment on my working?
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2(x - 3/4)^2 -17/16 =0
2(x - 3/4)^2 = 17/16
(x - 3/4)^2 = 17/32
x - 3/4 = +-√17/32
x = 3/4 +-√17/32
2x^2 - 3x = 1
x^2 - (3/2)x = 1/2
(x - 3/4)^2 - 9/16 = 1/2
x - 3/4 = +-sqrt(17/16)
x = 3/4 +-sqrt(17/16)
3. Thank you very much!
4. Originally Posted by sean.1986
2x^2 - 3x = 1
x^2 - (3/2)x = 1/2
(x - 3/4)^2 - 9/16 = 1/2
x - 3/4 = +-sqrt(17/16)
x = 3/4 +-sqrt(17/16)
The answer is to be x = 3/4 +- √17/4
I can't see where I am going wrong.
5. When completing the square, just use the general formula.
$a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}$
Plug in a,b,c and you're set.
6. Second attempt
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2((x - 3/4)^2) -17/8 =0
2(x - 3/4)^2 = 17/8
(x - 3/4)^2 = 17/16
x - 3/4 = +-√17/16
x = 3/4 +-√17/16
Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
7. Just to add my 2 cents......
$2x^2-3x-1=0$
1. First transpose the -1 to the right side of the equation.
$2x^2-3x=1$
2. Divide each term by 2
$x^2-\frac{3}{2}x=\frac{1}{2}$
3. Take half of the coefficient of x, square it and add it to both sides.
$x^2-\frac{3}{2}x+(\frac{3}{4})^2=\frac{1}{2}+\frac{9}{ 16}$
4. Noting the perfect square trinomial on the left:
$(x-\frac{3}{4})^2=\frac{17}{16}$
5. Take the square root of both sides:
$x-\frac{3}{4}=\pm\sqrt\frac{17}{16}$
6. Finally,
$x=\frac{3}{4}\pm\frac{\sqrt17}{4}$
$x=\frac{3\pm\sqrt17}{4}$
8. Hello,
Originally Posted by Mouseman
Second attempt
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0 | {
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Originally Posted by Mouseman
Second attempt
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2((x - 3/4)^2) -17/8 =0
2(x - 3/4)^2 = 17/8
(x - 3/4)^2 = 17/16
x - 3/4 = +-√17/16
x = 3/4 +-√17/16
Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
Note that $16=4^2$
Therefore, $\sqrt{16}=4$, and this yields the result your teacher gave you
9. Yeah it's just a question of bracketing or simplifying.
sqrt (a/b) = sqrt(a) / sqrt(b)
so sqrt(17/16) = sqrt(17) / sqrt(16) = sqrt(17) / 4
Both answers are correct but I guess I should've simplified. Sorry mate! | {
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# Is it necessary to assume a moment generating function exists?
Consider random variables A, B, and C. We know that A = B + C. We also know that A and C have an MGF. Is it the case that B must have a MGF?
Addition: Does this change if we know A and C both come from (different) chi-squared distributions? I am tasked with finding the distribution of B. If I can just do MGF(A) / MGF (C) = MGF (B) then it's simple... but can I even write this statement without assuming MGF (B) exists?
Generally -- You can't compute the MGF of B
In general, you can't compute the MGF of $$B$$ if you only know the MGFs of $$A$$ and $$C$$. For example, consider two possible joint distributions of $$A$$ and $$C$$:
Case 1: P( A=0 and C=0) = 1/2 and P(A=1 and C=1)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is 1.
Case 2: P( A=0 and C=1) = 1/2 and P(A=1 and C=0)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is $$\frac{\exp(-t)+\exp(t)}2=\cosh t$$.
Notice that in both Case 1 and Case 2 the MGFs for $$A$$ and $$C$$ were $$(1+exp(t))/2$$, but the MGF for $$B$$ changed from Case 1 to Case 2.
Generally -- You can prove existence of an MGF for B | {
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Generally -- You can prove existence of an MGF for B
Although you can't compute the MGF of $$B$$, you can prove that $$M_B(t)$$ exists for $$(*)\quad t\in D=\frac12 (Dom(M_A)\cap (-Dom(M_C)).$$ Suppose $$t\in D$$. If the MGF's of $$A$$ and $$C$$ exist, then for all $$t\in Dom(M_A)$$, $$M_A(t)=||\exp(ta)||_1<\infty$$ and for all $$t\in (-Dom(M_C))$$, $$M_C(-t)=||\exp(-tc)||_1<\infty$$ where $$||g||_p=\left(\int\int |g(a,c)|^p\; f(a,c)\; da dc\right)^{1/p}$$ is the $$L_p$$-norm of $$g$$ over the joint probability space and $$f(a,c)$$ is the joint pdf of $$A$$ and $$C$$. That implies $$||\exp(ta/2)||_2 < \infty$$ and $$||\exp(-tc/2)||_2 < \infty$$. By the Hölder's inequality or the Schwarz inequality, $$||\exp(ta)\exp(-tc)||_1<\infty$$. But, $$||\exp(ta)\exp(-tc)||_1= ||\exp(t(a-c)||_1= E[\exp(tB)]=M_B(t).$$ This proves that $$M_B(t)$$ exists for $$t\in D$$.
If A and C are independent, you can compute the MGF of B
If $$A$$ and $$C$$ are independent and $$B = A-C$$, then it must be the case that $$(**) \quad M_B(t) = M_A(t)\cdot M_C(-t)$$ whenever $$t\in Dom(M_A)\cap(-Dom(M_C))$$ (see e.g. Wikipedia). Here is a rough proof.
If $$t\in Dom(M_A)\cap(-Dom(M_C))$$, then $$M_A(t)\cdot M_C(-t) = \int_{a=-\infty}^\infty \exp(t a) dF_A(a) \cdot \int_{c=-\infty}^\infty \exp(-t c) dF_C(c)$$ $$= \int_{a=-\infty}^\infty \int_{c=-\infty}^\infty \exp(t (a-c)) dF_A(a) dF_C(c)$$ $$= \int_{b=-\infty}^\infty \exp(t b) dF_B(b) = M_B(t)$$ where $$F_A, F_B$$, and $$F_C$$ are the cumulative distribution functions of $$A, B$$, and $$C$$ respectively.
If A and C have $$\chi^2$$ distributions | {
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If A and C have $$\chi^2$$ distributions
In general, if $$A$$ and $$C$$ have $$\chi^2$$ distributions, you can only state that $$B$$ has an MGF and the domain of $$B$$'s MGF is \begin{align} Dom(B) &= \frac12 (Dom(M_A)\cap (-Dom(M_C))\\ &= \frac12 ((-\infty,2)\cap (-(-\infty,2))\\ &= \frac12 ((-\infty,2)\cap (-2,\infty))\\ &= \frac12 (-2,2)=(-1,1)\\ \end{align} by (*) and the fact that the domain of the MGF of a $$\chi^2$$ distribution is $$(-\infty,2)$$.
If you know that $$A$$ and $$C$$ have $$\chi^2$$ distributions, and you know that they are independent, then you can look up the MGFs for $$\chi^2$$ distributions and apply formula (**) to compute the formula for the MGF of $$B$$.
• Thank you. The question asks to prove that B is chi-squared distributed. It sounds like we know that a MGF exists, but that we can't prove it's specifically chi-squared? – purpleostrich Sep 28 at 2:06
• I learned a lot about MGFs and a bit about $\chi^2$ distributions while trying to figure out the answer to your question. :) – irchans Sep 28 at 8:24
• @purpleostrich, I suspect that if this was posed as an exercise to you then there is a missing assumption, for instance that $B$ and $C$ are independent, or as explored by irchans the assumption could be that $A$ and $C$ are independent (although I find this latter assumption somewhat less likely than the former). – pre-kidney Sep 28 at 9:01 | {
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# A stronger version of discrete “Liouville's theorem”
If a function $f : \mathbb Z\times \mathbb Z \rightarrow \mathbb{R}^{+}$ satisfies the following condition
$$\forall x, y \in \mathbb{Z}, f(x,y) = \dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$
then is $f$ constant function?
-
You probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. – Julián Aguirre Jul 17 '11 at 10:24
@Julian Aguirre: since $x\in\mathbb Z$, we don't have $f(x,y)\geq 0$. – Davide Giraudo Jul 17 '11 at 11:08
@girdav You are right. The lower bound is probably enough. – Julián Aguirre Jul 17 '11 at 13:08
You can prove this with probability.
Let $(X_n)$ be the simple symmetric random walk on $\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\to M_\infty$ almost surely.
But $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.
Thus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge. | {
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-
I love probabilistic arguments in analysis! Very nice. – Jonas Teuwen Jul 17 '11 at 11:46
This is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\mathbb{Z}^n$ instead of $\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\mathbb{Z}^n$ is constant for $n \geq 3$? For bounded ones this seems clear by considering the Poisson boundary. – t.b. Jul 17 '11 at 11:53
@Byron: This paper contains the claim that it is true that "nonnegative nearest-neighbors harmonic function on $\mathbb{Z}^d$ are constant for any $d$" on page 2. – t.b. Jul 17 '11 at 12:08
The usual Liouville theorem also holds with just a one-sided bound. – GEdgar Jul 18 '11 at 13:56
If you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? – Jyrki Lahtonen Jul 18 '11 at 14:07
I can give a proof for the d-dimensional case, if $f\colon\mathbb{Z}^d\to\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it] | {
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First, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\Vert x\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\le(2d)^{\Vert x\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\mathbb{E}[(2d)^{2T}] < \infty$. Then, $X_T$ has support $\mathbb{Z}^d$ and $\mathbb{E}[f(X_T)]=f(0)$, $\mathbb{E}[f(X_T)^2]\le\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\Vert f\Vert_2\equiv\mathbb{E}[f(X_T)^2]^{1/2}$.
Writing $e_i$ for the unit vector in direction $i$, set $f_i^\pm(x)=f(x\pm e_i)/f(\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\ldots,d$. Also, by construction, $\Vert f\Vert_2\ge\Vert f^\pm_i\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\frac{1}{2d}\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.
Now, if $g$ is a positive harmonic function, then $\tilde g(x)\equiv g(x)/g(0)$ satisfies $\mathbb{E}[\tilde g(X_T)]=1$. So, $${\rm Var}(\tilde g(X_T))=\mathbb{E}[\tilde g(X_T)^2]-1\le\mathbb{E}[f(X_T)^2]-1=0,$$ and $\tilde g$ is constant. | {
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-
Taxicab metric. Never heard that name (I learn maths in french at school). Funny! – Patrick Da Silva Jul 17 '11 at 20:24
@Patrick: Also called the Manhattan metric. – George Lowther Jul 17 '11 at 20:30
LAAAAAAAAAWL. Funnier. – Patrick Da Silva Jul 17 '11 at 20:39
Note: A similar proof will also show that harmonic $f\colon\mathbb{R}^d\to\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\colon\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\colon\mathbb{R}^d\setminus\{0\}\to\mathbb{R}^+$ for $d\not=2$. In fact, considering $f(x)=\Vert x\Vert^{2-d}$, we see that $f$ need not be constant for $d\not=2$. – George Lowther Jul 17 '11 at 22:54
Here is an elementary proof assuming we have bounds for $f$ on both sides.
Define a random walk on $\mathbb{Z}^2$ which, at each step, stays put with probability $1/2$ and moves to each of the four neighboring vertices with probability $1/8$. Let $p_k(u,v)$ be the probability that the walk travels from $(m,n)$ to $(m+u, n+v)$ in $k$ steps. Then, for any $(m, n)$ and $k$, we have $$f(m, n) = \sum_{(u,v) \in \mathbb{Z}^2} p_k(u,v) f(m+u,n+v).$$ So $$f(m+1, n) - f(m, n) = \sum_{(u,v) \in \mathbb{Z}^2} \left( p_k(u-1,v) - p_k(u,v) \right) f(m+u,n+v).$$ If we can show that $$\lim_{k \to \infty} \sum_{(u,v) \in \mathbb{Z}^2} \left| p_k(u-1,v) - p_k(u,v) \right| =0 \quad (\ast)$$ we deduce that $$f(m+1,n) = f(m,n)$$ and we win.
Remark: More generally, we could stay put with probability $p$ and travel to each neighbor with probability $(1-p)/4$. If we choose $p$ too small, then $p_k(u,v)$ tends to be larger for $u+v$ even then for $u+v$ odd, rather than depending "smoothly" on $(u,v)$. I believe that $(\ast)$ is true for any $p>0$, but this elementary proof only works for $p > 1/3$. For concreteness, we'll stick to $p=1/2$. | {
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We study $p_k(u,v)$ using the generating function expression $$\left( \frac{x+x^{-1}+y+y^{-1}+4}{8} \right)^k = \sum_{u,v} p_k(u,v) x^u y^v.$$
Lemma: For fixed $v$, the quantity $p(u,v)$ increases as $u$ climbs from $-\infty$ up to $0$, and then decreases as $u$ continues climbing from $0$ to $\infty$.
Proof: We see that $\sum_u p_k(u,v) x^u$ is a positive sum of Laurent polynomials of the form $(x/8+1/2+x^{-1}/8)^j$. So it suffices to prove the same thing for the coefficients of this Laurent polynomial. In other words, writing $(x^2+8x+1)^k = \sum e_i x^i$, we want to prove that $e_i$ is unimodal with largest value in the center. Now, $e_i$ is the $i$-th elementary symmetric function in $j$ copies of $4+\sqrt{15}$ and $j$ copies of $4-\sqrt{15}$. By Newton's inequalities, $e_i^2 \geq \frac{i (2j-i)}{(i+1)(2j-i+1)} e_{i-1} e_{i+1} > e_{i-1} e_{i+1}$ so $e_i$ is unimodal; by symmetry, the largest value is in the center. (The condition $p>1/3$ in the above remark is when the quadratic has real roots.) $\square$
Corollary: $$\sum_u \left| p_k(u-1,v) - p_k(u,v) \right| = 2 p_k(0,v).$$
Proof: The above lemma tells us the signs of all the absolute values; the sum is \begin{multline*} \cdots + (p_k(-1,v) - p_{k}(-2,v)) + (p_k(0,v) - p_{k}(-1,v)) + \\ (p_k(0,v) - p_k(1,v)) + (p_k(1,v) - p_k(2,v)) + \cdots = 2 p_k(0,v). \qquad \square\end{multline*}
So, in order to prove $(\ast)$, we must show that $\lim_{k \to \infty} \sum_v p_k(0,v)=0$. In other words, we must show that the coefficient of $x^0$ in $\left( \frac{x}{8}+\frac{3}{4} + \frac{x^{-1}}{8} \right)^k$ goes to $0$. | {
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There are probably a zillion ways to do this; here a probabilistic one. We are rolling an $8$-sided die $k$ times, and we want the probability that the numbers of ones and twos are precisely equal. The probability that we roll fewer than $k/5$ ones and twos approaches $0$ by the law of large numbers (which can be proved elementarily by, for example, Chebyshev's inequality). If we roll $2r > k/5$ ones and twos, the probability that we exactly the same number of ones and twos is $$2^{-2r} \binom{2r}{r} < \frac{1}{\sqrt{\pi r}} < \frac{1}{\sqrt{\pi k/10}}$$ which approaches $0$ as $k \to \infty$. See here for elementary proofs of the bound on $\binom{2r}{r}$.
I wrote this in two dimensions, but the same proof works in any number of dimensions
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Assuming that $f$ is bounded, you can prove that it is constant by a discrete analogue of the Borel-Caratheodory inequality, as shown in this equivalent question.
The key fact is such a function attains its maximum on a ball on the boundary, so it suffices to provide lower bounds for
$$\Gamma_N = \max_{|x|+|y|=N}f(x,y).$$
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# Getting 2 different solutions to the integral of $\frac{dx}{2x}$
I get two different answers that seem to conflict.
Is there an error in one method??
### Method 1
\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int\frac{\mathrm{d}x}{x}\\ &=\frac 1 2\ln|x|+C \end{align}
### Method 2
\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int \frac{2\mathrm{d}x}{2x}\\ &=\frac 1 2\int \frac{\mathrm{d}u}{u},\;\;\text{ where }u=2x,\ \mathrm{d}u=2\mathrm{d}x\\ &=\frac 1 2 \big(\ln |u| + C\big)\\ &=\frac 1 2 \ln|2x|+C \end{align}
• As the answer below details: the $C$'s are 'different', for want of a better word. Your "paradox" here is one of the reasons we are careful to always put a $C$ at the end, and also have to stay constantly aware of what it means. Oct 21, 2016 at 13:58
• Another good example is integrating $\sin(x)\cos(x)$. Try integration by parts, to get $\frac{1}{2}\sin^2(x)+C$, then try using the trig identity $\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$ to get $-\frac{1}{4}\cos(2x)+D$. A hint on reconciling the answers is $\cos(2x)=1-2\sin^2(x)$. Oct 21, 2016 at 14:06
• When we get two ostensibly different solutions to an anti-differentiation problem, it is often helpful to check whether the solutions (ignoring the constants) are vertical translations of each other. If so, the two solutions must, of course, differ by a constant. In this case, $\frac 1 2\ln|x|$ translated vertically $\ln\sqrt{2}$ units yields $\frac 1 2\ln|2x|$. Oct 21, 2016 at 16:40
Hint: $\frac{1}{2}\ln{|2x|} + C = \frac{1}{2}\ln{|x|} + \frac{1}{2}\ln{2} + C = \frac{1}{2}\ln{|x|} + \left(\frac{1}{2}\ln{2} + C\right)$. | {
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# Factoring in the derivative of a rational function
Given that $$f(x) = \frac{x}{1+x^2}$$
I have to find $$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} = \frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} = \frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?
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Can't you just use the quotient rule to find this derivative? – user2357112 Jun 3 '14 at 21:28
Since $x-a$ is in the denominator, it makes sense to consider the possibility that $x-a$ is a factor of the numerator. (If you know the factor theorem, you can see this is the case, since the numerator equals zero when $x = a.)$ I see $x-a$ in the numerator, along with two other terms. Very little cleverness is needed at this point to write
$$x-a + xa^2 - ax^2 \; = \; x - a + ax(a-x)$$ $$= \; (x-a) - ax(x-a) \; = \; (x-a)(1-ax)$$
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Ahhhh. So much time looking at it and I didn't realise this. I don't know about the factor theorem, something to look at. Thank you for your answer. – sidyll Jun 3 '14 at 18:08
@sidyll: For more about using the factor theorem, see my answer at Finding limit of a quotient. For a less obvious use of the factor theorem, see my comments at How to simpify this?. – Dave L. Renfro Jun 3 '14 at 18:12
Thank you for your references – sidyll Jun 3 '14 at 18:17
@sidyll: I just remembered that I posted a couple of old handouts of mine on the factor theorem a couple of years ago. See the files attached to this 12 January 2012 sci.math post archived at Math Forum. – Dave L. Renfro Jun 3 '14 at 18:25
@sidyll As for derivatives, generally the arithmetic works nicely using the quotient rule - see my answer. I'll bet Dave can give links to interesting expositions in AMM, Math. Mag, etc. – Bill Dubuque Jun 3 '14 at 19:02 | {
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\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark } Therefore you can conclude that: \eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac{\color{red}{\cancel{\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\color{red}{\cancel{\color{black}{(x-a)}}}} \\ &= \dfrac{1-ax}{(1+x^2)(1+a^2)}. }\tag{x\neq a}
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Thank you very much for your examples in continuing it. – sidyll Jun 3 '14 at 18:09
+1 for nice format. I learn something new about $LaTeX$ from you. – Tunk-Fey Jun 3 '14 at 18:10
@Tunk-Fey I've also learned a lot from your answers and your TeX style! $\overset{\cdot\cdot}\smile$ BTW you can write a more beautiful LaTeX as: $\LaTeX$ using \LaTeX, and there's also a variant for TeX namely $\TeX$ which is produced by \TeX. – Hakim Jun 3 '14 at 18:13
@sidyll You're welcome! Glad I could help! ;-) – Hakim Jun 3 '14 at 18:14
@حكيمالفيلسوفالضائع Thank you for the code. I didn't know that code before. $\ddot\smile$ – Tunk-Fey Jun 3 '14 at 18:17
By the quotient rule for the difference $\ f'(x)\, :=\, \dfrac{f(x)-f(a)}{x-a}$
$$\quad \begin{eqnarray} (g/h)'(x) &=\,\ & \dfrac{\color{#c00}{g'(x)} h(a) - g(a)\color{#0a0}{h'(x)}}{h(a)h(x)}\\ \begin{array}{l}\ g(x)=x\qquad\Rightarrow\,\color{#c00}{g'(x) = 1}\\ h(x) = 1+x^2\,\Rightarrow\,\color{#0a0}{h'(x) = x+a}\\\end{array}\ \Bigg\}\!\!\!\!\!&=& \dfrac{\color{#c00}1\cdot (1+a^2)\overset{\phantom{I^I}}-a(\color{#0a0}{x+a})}{(1+a^2)(1+x^2)}\ =\ \dfrac{1-ax}{(1+a^2)(1+x^2)}\end{eqnarray}$$
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Hint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.
Indeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.
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Factor: $$x+xa^2−a−ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$ Thus, $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\boxed{\frac{1-ax}{(1+x^2)(1+a^2)}}$$
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# Questions concerning smallest fraction between two given fractions.
I recently encountered this on a practice test
Find the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 < \frac{m}{n} < \frac{1}{3}$
\begin{align}0.33 < \frac{m}{n} < \frac{1}{3}&\implies(\frac{33}{100} < \frac{m}{n})\land(\frac{m}{n}<\frac{1}{3}) \\ &\implies(33n<100m)\land(3m<n) \\\end{align} and thus $n=3m+1$.
So \begin{align}33n<100m&\implies33(3m+1)<100m \\ &\implies 99m+33<100m\\ &\implies m>33\end{align} Taking $m\geq34$, we find that $n=34\times3+1=103$
After the test, I had the following thoughts, which I do not know how to answer.
Questions
Will the solution for $n$ always yield the smallest $m$ possible?
Does this method (find minimum value of $n$ and substitute) work for all such problems? (where $m,n\in\mathbb{Z}$)
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).
note: sorry for asking multiple questions in one post (which I know some people frown on) but I feel posting multiple questions with the same 'introduction' (problem+proof) would clutter and be somewhat cumbersome. | {
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• In my opinion the smallest is $n=103$ because the smallest $m$ is $34$. Indeed if you solve for positive $n$ $$\frac{34}{n}<\frac{1}{3}$$ you get $n>102$ – Raffaele Oct 29 '17 at 11:01
• @mathlove typo by me, thanks. I've also noticed that in our example $\frac{33}{100}<\frac{34}{103}<\frac{1}{3}$, the numerator of the middle fraction is the sum of the other numerators, and the denominator for the middle fraction is the sum of the other denominators. I wonder if this is a coincidence or not, though I won't put that into the question (three questions in one post is already plenty) – user472341 Oct 29 '17 at 11:03
• The continued fraction for $\frac13$ is $(0;3)=(0;3,\infty)$ and the continued fraction for $0.33$ is $(0;3,33)$. The simplest between them is $(0;3,34)=\frac{34}{103}$. – robjohn Oct 29 '17 at 11:04
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).
In the following, $a,b,c,d,m,n$ are positive integers.
$\frac ab\lt \frac mn\lt \frac cd$ is equivalent to $$na\lt mb\qquad\text{and}\qquad \frac{md}{c}\lt n$$
From the second inequality, we can set $n=\lfloor\frac{md}{c}\rfloor+1$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.
So, from the first inequality, we get $$\left(\left\lfloor\frac{md}{c}\right\rfloor+1\right)a\lt mb\tag1$$
As a result, we can say that the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac cd$ is given by$$\left\lfloor\frac{Md}{c}\right\rfloor+1$$ where $M$ is the smallest integer $m$ satisfying $(1)$.
If $c=1$, then the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac 1d$ is given by$$\left(\left\lfloor\frac{a}{b-da}\right\rfloor+1\right)d+1$$ | {
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Yes to both of your questions. This is essentially the mediant for two terms in a Farey sequence. There is some deep theory here. Read up on Farey sequences for more information.
• I had three questions - i'm assuming you're talking about the first two? either way, thanks for the links. – user472341 Oct 29 '17 at 11:09
• The mediant is not always the smallest. Take for instance $\frac13$ and $\frac34$: the mediant is $\frac47$, which is between the two, but $\frac12$ is the fraction with the smallest denominator. – robjohn Oct 29 '17 at 11:12
• I think the mediant is only relevant when the fractions are consecutive terms in some Farey sequence. In the example @robjohn gives, $1/3$ and $3/4$ are not consecutive terms in any Farey sequence. – Gerry Myerson Oct 29 '17 at 11:17
• You can iterate the algorithm using the mediant. For 1/3, 3/4 we get 4/7 and then with 1/3, 4/7 we get 1/2. – i. m. soloveichik Oct 29 '17 at 11:27 | {
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# Function notation terminology
Given the function $f:X\longrightarrow Y$, $X$ is called the domain while $Y$ is called the codomain. But what do you call $f(x)=x^2$ in this context, where $x\in X$? That is to say - what is the name for the $f(x)$ notation?
And while I'm here, what is the proper way to write a function like this? Would it be $f:\mathbb{R}\to\mathbb{R},\;f(x)=x^2$?
Edit:
I figured I'd add this to add a bit of context into why I'm asking. I'm writing a set of notes in LaTeX, and I'd like to use the correct terminology for the definition of a function.
A function from set $A$ to set $B$, denoted by $$f:A\to B;x\mapsto f(x)$$ is a mapping of elements from set $A$, (the $\textit{domain}$) to elements in set $B$ (the $\textit{codomain}$) using the $\color{blue}{\sf function}$ $f(x)$. The domain of a function is the set of all valid elements for a function to map from. The codomain of a function is the set of all possible values that an element from the domain can be mapped to. The $\textit{range}$ (sometimes called the $\textit{image}$) of a function is a subset of the codomain, and is the set of all elements that actually get mapped to by the function $f$.
Here I'm pretty sure the highlighted word "function" is not right.
• I suggest \to or \rightarrow instead of \longrightarrow. The last one is there for if you need the arrow to be longer because you're writing something over it. As in $\overset{\text{text}}{\longrightarrow}$ instead of $\overset{\text{text}}{\to}$.
– Jim
Feb 22, 2013 at 23:16
• why do you think that function is not right = Feb 22, 2013 at 23:36
• Some would call "$f(x)=x^2$" the rule of the function $f$. Feb 22, 2013 at 23:44
• @DominicMichaelis Well, function is used in two places to mean two slightly different things; I don't like using function to describe part of a function. Feb 22, 2013 at 23:47 | {
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I can remember to read this text, and being puzzled with the exact same question. From what I've learned from my teacher, you're right, writing down something as "the function $f(x)$..." is sloppy notation. However, many books/people will use it this way.
If you're are very precise, $f(x)$ is not a function or an map. I don't know of a standard way to refer to $f(x)$, but here is some usage I found on the internet:
• The output of a function $f$ corresponding to an input $x$ is denoted by $f(x)$.
• Some would call "$f(x)=x^2$" the rule of the function $f$.
• For each argument $x$, the corresponding unique $y$ in the codomain is called the function value at $x$ or the image of $x$ under $f$. It is written as $f(x)$.
• If there is some relation specifying $f(x)$ in terms of $x$, then $f(x)$ is known as a dependent variable (and $x$ is an independent variable).
A correct way to notate your function $f$ is: $$f:\Bbb{R}\to\Bbb{R}:x\mapsto f(x)=x^2$$
Note that $f(x)\in\Bbb{R}$ and $f\not\in\Bbb{R}$. But the function $f$ is an element of the set of continuous functions, and $f(x)$ isn't.
In some areas of math it is very important to notate a function/map specifying it's domain, codomain and function rule. However in for example calculus/physics, you'll see that many times only the function rule $f(x)$ is specified, as the reader is supposed to understand domain/codmain from the context.
You can also check those questions:
• Nice, Kasper +1 Feb 23, 2013 at 1:50
normally you say $f:X\rightarrow Y$; $x\mapsto f(x)$, as a functions takes an element from $X$ and give you one from $Y$. $$y=f(x)$$ Is an equation, and not a definition of a function in a strict sense.
The proper way would be $$f:\mathbb{R}\rightarrow \mathbb{R}; \ x\mapsto x^2$$
The Image of a function is definied as $$\operatorname{im}f:=\{f(x)|x\in X\}$$ It is often written as $$f(X):=\{f(x)|x\in X\}$$ Notice that $X$ is a set, not an element. so $f(\{x\})=\{f(x)\}\subseteq Y$ while $f(x)\in Y$ | {
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• I was wondering - is $f(x)$ called the image? I've seen that word used but only know it to be synonymous with Range. Feb 22, 2013 at 23:15
• not $f(x)$ but $f(X)$ note the different as $X$ is a set but $x$ is an element Feb 22, 2013 at 23:19
• @agent154 You can say "$f(x)$ is the image of the element $x$ under the function $f$". Perhaps this is what you were thinking of (here of course "image" means something different than the image of the function). Feb 22, 2013 at 23:32
• @davidMitra i would be careful with that, as the image is a set but $f(x)$ is an element of $Y$ Feb 22, 2013 at 23:34
• The notation $f(X)$ is also written (in set theory) as $f[X]$ and $f''X$ sometimes, often because the elements of the set are sets themselves, and sometimes $f(x)\neq f''x$. Feb 22, 2013 at 23:35
From what I understand, $f$ is the function, so saying "a function $f(x)$" would be wrong. Instead, you could say "a function $f$", or if you didn't want to assign it a name, "a function $x\mapsto x^2$", or if you want to specify the domain and codomain, "a function $f:X\to Y$".
If you just want to define the function's input-output relationship, $f(x)=x^2$ suffices, maybe with a $\forall x\in X$ at the beginning if you want to do it properly. That doesn't necessarily define the domain though, that set could just be a subset of the domain. $f(x)$ is really no different from $f(1)$, except that $x$ is a (qualified) variable. It's basically a rule defining the input and output for an improper subset of its domain.
The addition of "$:X\to Y$" is useful if you want to specify both the domain and codomain when defining the function. As far as I can tell, you can add it after a function like $f$ and the whole expression still refers to the function. You can't really use it with the $f(x)$ style, so I'd do something like
$f:\mathbb R\to\mathbb R=x\mapsto x^2$ | {
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# What is the variance of a binomial distribution with -1 and 1?
I am struggling to see how to solve the following problem:
I have n i.i.d bernoulli trials. The result can be -1 or 1. I can figure out the expected value of this = n(p-(1-p) but how do I know what the variance is if p is known?
I know that: $${\rm Var}(X) = E[X^2] - E[X]^2$$ I don't know where to go from here.
• In general, for a Bernoulli Random Variable E(X) is $P(X=x_1)*x_1+P(X=x_2)*x_2$. In the "default" case $x_1 = 1, x_2 = 0$, which reduces this to E(X) = p. In your case, $x_1 = 1$ and $x_2 = -1$. You can then plug in what you get for E(X) into the formula $Var(X) = E(X^2) - E(X)^2$. The variance of the default case is $p*(1-p)$ – Pugl Jul 27 '17 at 13:57
• If B is a 0-1 Bernoulli variable with probability of a 1 at $p$, then you're describing a variable, ($Z$ say) where $Z=2B-1$. Just apply basic properties of variance to that linear transformation. – Glen_b Jul 28 '17 at 2:06
When you work with variances, know these facts (in addition to the definition of variance):
1. The variance of a sum of independent (or just uncorrelated) variables is the sum of their variances.
2. The expectation of a sum of variables (independent or not) is the sum of their expectations.
3. The expectation (of any discrete variable) is the sum of each possible value multiplied by its probability.
As an example, let's apply them to your case. You may model the sum of your Bernoulli trials with a variable $X$ expressed as the sum of $n$ independent variables $X_i$. Each $X_i$ takes on the value $1$ with probability $p$ and the value $-1$ with the probability $1-p$.
Fact $(3)$ asserts $$\mathbb{E}(X_i) = p(1) + (1-p)(-1) = 2p-1$$ and $$\mathbb{E}(X_i^2) = p(1)^2 + (1-p)(-1)^2 = p + (1-p) = 1.$$
Fact $(2)$ asserts \eqalign{\mathbb{E}(X) &= \mathbb{E}(X_1+\cdots+X_n) = \mathbb{E}(X_1) + \cdots + \mathbb{E}(X_n) = n\mathbb{E}(X_1) \\&=n(2p-1).} | {
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The definition of variance now tells you$$\operatorname{Var}(X_i) = \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = 1 - (2p-1)^2 = 4p(1-p).$$
Consequently, fact $(1)$ yields \eqalign{\operatorname{Var}(X) &= \operatorname{Var}(X_1+\cdots+X_n) = \operatorname{Var}(X_1) + \cdots + \operatorname{Var}(X_n)=n \operatorname{Var}(X_1) \\&= n(4p(1-p)).}
The appearance of the factor of $4$ might seem somewhat mysterious. There's another extremely useful fact you might consider using to shortcut these considerations and identify the origin of that factor:
(4) The variance of a shifted, rescaled variable is the square of the scale factor times the variance. In mathematical symbols, $$\operatorname{Var}(\sigma X + \mu) = \sigma^2 \operatorname{Var}(X)$$ no matter what values the numbers $\sigma$ and $\mu$ might have.
This applies by noting that your $X_i$ can all be expressed as $2Y_i-1$ where $Y_i$ is a true Bernoulli (that is, $0-1$) variable. It's simple to show this: since $2\times 1-1=1$ and $2\times 0 - 1=-1$, the values of $1$ and $0$ taken on by $Y_i$ become $1$ and $-1$, respectively, for $X_i$. The probabilities are unchanged.
You might already know that $Y=Y_1+\cdots + Y_n$, the sum of $n$ independent Bernoulli variables with common probability $p$, is called a Binomial variable. It has a Binomial distribution. You can remember or look up its variance, which is $np(1-p)$. Since $$X = X_1+\cdots+X_n = (2Y_1-1) + \cdots + (2Y_n-1) = 2(Y_1 + \cdots + Y_n) - n=2Y-n,$$ you can take $\sigma=2$ in applying fact $(4)$, which immediately tells you $$\operatorname{Var}(X) = 2^2 \operatorname{Var}(Y) = 4np(1-p).$$ | {
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• I personally find the observation that $4=\frac{1}{0.5(1-0.5)}$ soothing as it shows that the variance is the ratio of $pq(X_i)$ to $pq(0.5)$ (forgive the abuse of notation). – Jared Goguen Jul 27 '17 at 16:09
• In what sense is the second part of the answer $X_i=2Y_i-1$ ... different from my answer ? – user83346 Jul 28 '17 at 5:41
• @fcop It's exactly the same approach--but the explanation is more extensive and thorough. If you take a look over the entirety of my answer, I hope you will see the pattern: it focuses on stating and illustrating useful principles. These are often overlooked or unknown to people who come to our site with questions. My purpose is to collect, unify, and illustrate some (obvious, well-known) techniques that can be found by future visitors and referenced in future answers. In that sense, my answer differs from yours, which focuses on solving the immediate problem. – whuber Jul 28 '17 at 13:37
• Well you could have referenced it no? – user83346 Jul 28 '17 at 14:08
• @fcop One ordinarily does not reference obvious or well-known things. (If that were the practice, then most posts would be 90% references.) The point of this answer is to explain the reasoning--not just the results--at the same level of understanding as the question. Its contribution is not to make the observation that $X_i=2Y_i-1$ (you're welcome to take credit for that), but rather to show as clearly and explicitly as possible, with all reasons supplied, how that produces the same answer derived earlier in the question. – whuber Aug 13 '17 at 13:10
if you take $x=2b-1$ where $b$ is bernoulli, then then $x$ is either 1 or -1, and mean and variance follow from Bernoulli $\mu_b=p$ and $\sigma^2_b=p(1-p)$, the extension to 'Binomial' is then just a sum (assuming independence),
The mean of $x$ is then $\mu_x=2\mu_b-1=2p-1$, variance $\sigma_x^2=4\sigma_b^2=4p(1-p)$, | {
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The mean of $x$ is then $\mu_x=2\mu_b-1=2p-1$, variance $\sigma_x^2=4\sigma_b^2=4p(1-p)$,
The sum of $n$ independent such $x$'s has mean $n\cdot(2p-1)$ (your result) and the variance is $4\cdot n \cdot p(1-p)$
EDIT after your question in the comment
It is as @Pegah says: $b$ is the ''usual'' Bernoulli with a success probability $p$ and outcomes 0 and 1. It's expected value is $1 \times p + 0 \times (1-p)=p$ and the variance $p(1-p)$ (also from the definition.
So $b$ is the known Bernoulli with outcome 0 and 1 and $x$ is just a linear function of it. The linear function is such that $x$ has outcomes -1 and 1, and the mean and variance can be obtained from the mean and variance of $b$.
• In fcop's derivation, $\mu_b$ is the expected value of the random variable b, which is, as I understand a "default" binomial random variable (i.e. it can take on values 1 or 0, see also my comment above). Here the expected value is just p. – Pugl Jul 27 '17 at 15:31 | {
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# Simplifying to a desired expression structure
My book has this expression:
\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3) \end{align}
And then the book simplified it, and ended up with the desired expression:
\begin{align} ((n+1)(n+2)(2n+9))/6 \end{align}
I tried to do such simplification. But I ended up with this: \begin{align} (2n^3+15n^2+31n+18)/6 \end{align}
Even though the "value" in my expression is the same as the desired result's, I need the structure of my expression to be the same as well. But I don't understand how to do such.
Here are the full steps my book shows. Of course, I understand that the steps make sense, but I don't understand why did my book take that approach? Is that the only approach possible to reach the desired expression? How was I supposed to know that? I mean, I know how to simplify, but clearly my book is using a different "style" or "path"
\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3)\\ (n(n+1)(2n+7)+6(n+1)(n+3))/6\\ ((n+1)[n(2n+7)+6(n+3)])/6\\ ((n+1)(2n^2 +7n + 6n + 18))/6\\ ((n+1)(n+2)(2n+9))/6 \end{align}
So yes, how should I simplify to get a desired expression structure?
-
Ah! I... didn't see that. Somehow. I'm eager to see how to extract such though. – Zol Tun Kul Jul 3 '12 at 1:49
I wrote out a little something about extracting. – André Nicolas Jul 3 '12 at 1:59
Below I explain in detail the solution given in your book.
$\begin{eqnarray} &&\rm\ \ \,n(n\!+\!1)(2n\!+\!7)/\color{#C00}6+(n\!+\!1)(n\!+\!3)\\ \rm put\ all\ over\ a\ common\ denominator = \color{#C00}6:\quad &= &\rm\ (n(\color{#0A0}{n\!+\!1})(2n\!+\!7)\,+\,\color{#C00}6\,(\color{#0A0}{n\!+\!1})(n\!+\!3))/\color{#C00}6\\ \rm pull\ out\ the\ common\ factor\ \color{#0A0}{n\!+\!1}:\quad &=&\rm\ (\color{#0A0}{n\!+\!1})\,(n(2n\!+\!7)+6(n\!+\!3))/6\\ \rm apply\ the\ distributive\ law:\quad &=&\rm\ (n\!+\!1)\,(2n^2 + 13n + 18)/6\\ \rm factor\ the\ quadratic,\ see\ below:\quad &=&\rm\ (n\!+\!1)\,(n\!+\!2)\,(2n\!+\!9)/6 \end{eqnarray}$ | {
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In order to factor the quadratic $\rm\:f = 2n^2+13n+18\:$ one can apply the AC-method as follows:
$$\begin{eqnarray}\rm 2f\, &=&\rm\ \ 4n^2\ +\ 13\cdot 2n\, +\, 2\cdot 18 \\ &=&\rm\ (2n)^2 + 13\,(2n) + 2\cdot 18 \\ &=&\rm\ \ N^2\ +\ 13\, N\ +\ 36\quad for\quad N = 2n \\ &=&\rm\ (N\ +\ 4)\,(N\ +\ 9) \\ &=&\rm\ (2n\, +\, 4)\,(2n\, +\, 9) \\ \rm f\, &=&\rm\ (\ n\ +\ 2)\,(2n\ +\ 9) \end{eqnarray}$$
-
I'm enlightened. Thank you. – Zol Tun Kul Jul 3 '12 at 3:58
Your expression is $$\frac{2n^3+15n^2+31n+18}{6}.$$ Let's forget about the $6$. Also, I would like to change the $n$ to $x$, for no good reason except familiarity. So we want to factor $$2x^3+15x^2+31x+18.$$ To do this, we look for rational roots of the polynomial $2x^3+15x^2+31x+18$. By the Rational root Theorem, such roots must have shape $a/b$, where $a$ divides $18$ and $b$ divides $2$. (There may be no rational roots, though in this case there are three.)
It is easy to spot the root $x=-1$. So $x-(-1)$, that is, $x+1$, divides our polynomial. Do the division, using the ordinary division process for polynomials, which is much like "long division." We get $$2x^3+15x^2+31x+18=(x+1)(2x^2+13x+18).$$ Now factor the quadratic as one did in school. Or else note that $x=-2$ is a root of $2x^2+13x+18$.
Remark: The book grabbed the obvious common factor of $n+1$. The rest turned out to be a quadratic that factors nicely. You multiplied out, meaning you buried the $n+1$ term. It can be extracted from your expression, and the above calculation shows how, but why bury and then extract? A factored or partly factored expression is often more useful. Anyway, "taking out" common factors usually simplifies calculations.
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# Show that set of all $2 \times 2$ matrices forms a vector space of dimension $4$
I have this question:
Show that the set of all $2 \times 2$ matrices with real coefficients forms a linear space over $\Bbb R$ of dimension $4$.
I know that the set of the matrices will basically form a linear combination which will define the vector space and they satisfy the axioms defined for the vector space.
I do not know how to show that this is possible. Do the vectors have to be linearly independent?
Any sort of help is appreciated.
Thanks.
Begin by listing the axiom satisfied by a general vector space over $\mathbb{R}$. Now consider your set of matrices. What does you addition look like? What is your zero element? What does your scalar multiplication look like? Does your scalar multiplication and your addition behave as they should? Finally, show that all 2 by 2 matrices can be written as a linear combination of 4 special matrices. A good choice is to pick matrices who each have precisely 3 zero entries and 1 non zero entry. Can you show that these 4 matrices are linearly independent?
• Can those four matrices be [a 0 0 0], [0 b 0 0], [0 0 c 0], [0 0 0 d]? I can form a linear combination of them and show they are linearly independent. Will that work? – Suvrat Jan 31 '16 at 0:05
• That is exactly what I hoped that you would do! – Carl Christian Jan 31 '16 at 0:05
• oh very well then. Thanks a lot! – Suvrat Jan 31 '16 at 0:06
• Can you tell me how these 4 matrices show the formation of linear space for generalized situation, i.e. all 2x2 matrices? – Suvrat Jan 31 '16 at 20:14
• For the sake of simplicity pick a=b=c=d=1. Then write the matrix [e f g h] as e*[1 0 0 0] + f*[0 1 0 0] + g [0 0 1 0] + h*[0 0 0 1]. – Carl Christian Jan 31 '16 at 20:23 | {
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One can easily see that $\mathbb{R}^{2\times 2}$ is in fact exactly the same as the vectorspace $\mathbb{R}^4$, only the notation of the elements differs. And of course whether or not something is a vector space does not depend on the way some unimportant species of primates decides to write its elements.
Hint Can you find a basis of the set of $2 \times 2$ matrices consisting of four elements? (There is a natural choice of basis here that includes the matrix $\pmatrix{1&0\\0&0}$.)
Alternatively, can you find a vectorspace isomorphism from the space of $2 \times 2$ matrices to some vector space you know to be $4$-dimensional, e.g., $\Bbb R^4$?
• I don't know how to find a vectorspace isomorphism. But i can work with basis, which is similar to Carl's solution. Thanks! – Suvrat Jan 31 '16 at 0:11
• You're welcome. – Travis Jan 31 '16 at 0:42 | {
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How can I prove this statement about subsets?
Let $$A$$, $$B$$, $$C$$ be sets. Prove that if $$A \subseteq C$$ and $$B \subseteq C$$ then $$A \cup B \subseteq C$$.
This is an exercise in mathematical logic.
My attempt to progress forward: This statement can be written as$$(A \subseteq C) \land (B \subseteq C) → A \cup B \subseteq C\\ (x \in A → x \in C) \land (x \in B → x \in C) → A \cup B \subseteq C$$ But I am not even sure that is how I am supposed to do it so I am a bit stuck. Can anyone explain to me how to get through this proof? Thanks in advance.
• ... intuitively it's obvious, right? ... Sep 26 '18 at 15:32
• @user202729 Intuitively many things are obvious, but proving them is another matter... ;) ex. A sphere bounds a given volume with minimal area... Sep 27 '18 at 1:14
4 Answers
Take any $$x\in A\cup B$$. That means either $$x\in A$$ or $$x\in B$$. Anyway you get $$x\in C$$.
This answer is expanding a bit on the other answers to give a widely applicable outline on how to prove mathematical statements in general.
The final proof will be equal to Mark's answer, but hopefully this answer sheds a bit of light on what is actually going on in the thought process, at least I like to think in the terms defined below while I am proving.
In general you need a "proof calculus" which states how proofs can be constructed and which ones are valid. Unless you are doing logic and proof theory itself, in most settings in maths you are well off using the natural deduction proof calculus for first-order logic. In short — cut in light of this question:
• To prove $$\varphi \land \psi$$, prove $$\varphi$$ and $$\psi$$ separately.
• To prove $$\varphi \lor \psi$$, give a proof for $$\varphi$$ or alternatively for $$\psi$$. One proof for either $$\varphi$$ or $$\psi$$ is already sufficient.
• To prove $$\varphi → \psi$$, assume $$\varphi$$ and prove $$\psi$$ under this assumption. | {
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The implication captures the notion of "if we have $$\varphi$$ and $$\varphi → \psi$$ established somewhere, then we can conclude $$\psi$$." Most, if not all, theorems can be seen as implications: if you fulfill the requirements, then you can use its consequences.
Hence it makes sense to be allowed to assume $$\varphi$$ in the proof of $$\varphi → \psi$$.
• To prove $$\forall x. \varphi(x)$$ ($$\varphi$$ may depend on $$x$$), introduce a fresh variable not used before (e.g. $$c$$) and prove $$\varphi(c)$$ (i.e. every occurrence of $$x$$ is filled with $$c$$).
Introduction of a fresh variable is required to guarantee that your proof of $$\varphi(c)$$ does not depend on any previous assumptions on $$x$$. Hence, having proved $$\varphi(c)$$ really amounts to having proved the proposition $$\varphi(x)$$ for every possible $$x$$.
Now consider your statement:
$$\left((x \in A → x \in C) \land (x \in B → x \in C)\right) → A \cup B \subseteq C$$
You already did a good job translating it to more formal language! But we can even go a step further. Actually what the above line means is the following: $$\left((\forall x. x \in A → x \in C) \land (\forall x. x \in B → x \in C)\right) → (\forall x. x \in A \cup B → x \in C)$$
On the top level, we have a $$\varphi → \psi$$-kinded expression. Applying the corresponding rule, we begin with
(1) Assume $$(\forall x. x \in A → x \in C) \land (\forall x. x \in B → x \in C)$$.
Our proof goal, as stateted in the rule, is now the right side $$(\forall x. x \in A \cup B → x \in C).$$ This is of kind $$\forall x. \varphi(x)$$ with $$\varphi(x) := x \in A \cup B → x \in C$$. So,
(2) Let $$y$$ be a fresh variable.
I called it $$y$$ to avoid confusion with the set $$C$$. When one defines freshness more formally, one sees that $$x$$ would have worked as well as a fresh variable.
Now we have to show $$\varphi(y)$$, which is $$y \in A \cup B → y \in C$$. Again, apply the $$→$$ rule from above:
(3) Assume $$y \in A \cup B$$. | {
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(3) Assume $$y \in A \cup B$$.
Our final proof goal is now: $$y \in C$$. This can be proved by case-by-case analysis:
(4) We know that $$y$$ is in $$A$$ or $$B$$. In the first case, $$y \in A$$ and per line (1) we especially know $$y \in A → y \in C$$. Since we have $$y \in A$$ just established, we can conclude $$y \in C$$. Case finished. The next and last case is $$y \in B$$ which works analogously.
If $$A \subset C$$, and $$B \subset C$$;
$$A \cap C =A$$; and $$B \cap C = B$$;
$$A\cup B = (A \cap C)\cup (B \cap C) =$$
$$C \cap (A \cup B) \subset C$$
I haven't noticed one in the in the answers, so here's a proof by contradiction.
The statement we wish to prove is: $$(\forall x)(x \in A \cup B \implies x \in C) \tag{1}$$
It's negation is:
$$(\exists x)(x \in {A} \cup {B} \wedge \neg(x \in C) ) \tag{2}$$ $$x \in A \cup B \implies (x \in A \vee x \in B)$$ $$x \in A \wedge A \subseteq C \implies x \in C \tag{3}$$ $$or$$ $$x \in B \wedge B \subseteq C \implies x \in C \tag{4}$$ Since both $$(3)$$ and $$(4)$$ are in contradiction with $$(2)$$, it must be false. Therefore our starting premise $$(1)$$ is correct. | {
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# Fourier series, pointwise convergence, series computation
#### Markov
##### Member
Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$
a) Study the pointwise convergence of the Fourier series for $f.$
b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?
-------------
First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?
Thanks for the help!
#### Sudharaka
##### Well-known member
MHB Math Helper
Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$
a) Study the pointwise convergence of the Fourier series for $f.$
b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?
-------------
First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?
Thanks for the help!
Hi Markov,
Firstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\mbox{ for }-l\leq x\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.
I think the Fourier series that you have obtained is also incorrect. It should be, | {
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I think the Fourier series that you have obtained is also incorrect. It should be,
$-x=\frac{2l}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi x}{l}\right)$
Since both $$f(x)=-x$$ and $$f'(x)=-1$$ are continuous on $$[-l,\,l]$$ the Fourier series converges point-wise on the interval $$(-l,\,l)$$ (Refer Theorem 5.5 here).
Substitute $$x=1\mbox{ and }l=2$$ and we obtain,
\begin{eqnarray}
-\frac{\pi}{4}&=&\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi}{2} \right)\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}\sin\left(\frac{(2n+1)\pi}{2} \right)\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}(-1)^n\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^{3n+1}}{2n+1}\\
\therefore\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}&=&\frac{\pi}{4}
\end{eqnarray}
When $$x=l$$ we have,
$\left|f(l)-\sum_{n=1}^{N}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi l}{l}\right)\right|=l$
Therefore by the definition of uniform convergence (Refer this) it is clear that the Fourier series of $$f$$ is not uniformly convergent on $$\Re$$.
Kind Regards,
Sudharaka.
#### CaptainBlack
##### Well-known member
Hi Markov,
Firstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\mbox{ for }-l\leq x\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.
This is very likely to be intended to indicate the periodic extension: $$f(x)=f(x+2l)$$ (or there is a mistake with the value at either $$+l$$ or $$-l$$, both end points would not normally be included in the domain for a Fourier Series).
Without the periodic extension the question of uniform convergence is moot, since the Fourier Series is periodic and so does not converge to the function outside of the interval.
CB | {
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# What can I use to solve this?
1. Jan 10, 2005
### recon
I've run into a calculation hurdle that I cannot cross in a problem I'm trying to solve.
May I know what values of z satisfy the following equation:
$$z^4 + 2z^3 - 5z + 1 = 0$$
I've never encountered this in mathematics class before.
2. Jan 10, 2005
### dextercioby
Would u please post the problem??Maybe you wouldn't get a quartic after all...
Daniel.
3. Jan 10, 2005
### Nguyen Thanh Nam
Neither do I, I can solve those that miss ax^3, never before for ax^2
4. Jan 10, 2005
### recon
A ladder 3m long rests against a wall with one end a short distance from the base of the wall. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m wide. What is the maximum distance up the wall that the ladder can reach?
5. Jan 10, 2005
### dextercioby
The maximum height on the wall is attained when the ladder touches the storage box.
The equations i have found read
$$(a+1)^{2}+(b+1)^{2} =9$$(1)
$$ab=1$$ (2)
Then u find the eq.for "a"
$$a^{2}+\frac{1}{a^{2}}+2(a+\frac{1}{a})=7$$ (3)
,which can be put in the form
$$u^{2}+2u-9 =0$$ (3)
,where
$$u=:a+\frac{1}{a}$$ (4)
Solve for "u" and then for "a".Then add 1 to the "a" and find the final result.
Daniel.
6. Jan 10, 2005
### recon
Yes, that's how I solved it. I never thought of substituting $$a+\frac{1}{a}$$ for u. Very smart. Thanks a lot.
7. Jan 10, 2005
### Hurkyl
Staff Emeritus
Use the symmetry of the problem . With dexter's notation, notice that whenever s is a solution, so is 1/s. So, we can take his quartic:
z^4 + 2z^3 - 7z^2 + 2z + 1 = 0
And write down its four factors:
(z - s) (z - 1/s) (z - t) (z - 1/t) = 0
If we group them into their symmetric pairs and multiply:
(z^2 + az + 1) (z^2 + bz + 1) | {
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If we group them into their symmetric pairs and multiply:
(z^2 + az + 1) (z^2 + bz + 1)
So one quadratic corresponds to the two solutions s and 1/s and the other to t and 1/t. In particular, notice, now, that our unknowns only have two possible values, instead of four, which means we should be able to determine them from a quadratic equation:
Since the polynomials should be equal, we should expand and equate them:
z^4 + (a+b)z^3 + (2+ab)z^2 + (a+b)z + 1
=
z^4 + 2z^3 - 7z^2 + 2z + 1 = 0
Equating coefficients gives us a system of equations to solve.
Once we have a and b, we can then solve for s and t and get the four possible solutions. (Two should be complex, I imagine)
(Incidentally, I don't see how you got your equation, Recon)
8. Jan 10, 2005
### dextercioby
I multiple checked it.Can u prove your statement???I frankly doubt it...
Daniel.
9. Jan 10, 2005
### Hurkyl
Staff Emeritus
Another (similar) approach:
We can arrange the factors into these pairs:
(z - s)(z - t) and (z - 1/s)(z - 1/t)
So each quadratic contains exactly one solution from each of the symmetric solutions (instead of both).
Again, we break the quartic into quadratic factors:
(z^2 + az + b) (z^2 + cz + d)
But, in this case, we know that we can convert from one to the other by applying the symmetry. In particular,
if (z^2 + az + b) = 0 then (1/z)^2 + c(1/z) + d = 0
Rewriting the second gives:
z^2 + (c/d)z + (1/d) = 0
which must be the same polynomial as
z^2 + az + b = 0
Since they have the same solutions.
So we can equate coefficients here, then multiply them to equate coefficients to the quartic polynomial. Again, we can solve for our unknowns, and then we know how to solve quadratics to get the answers to our problem.
Lesson learned: use symmetry to break the problem into parts. We've seen two different ways to do it:
(1) Each part corresponds to one of the different "orbits" under the symmetry. (e.g. {s, 1/s}) | {
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(1) Each part corresponds to one of the different "orbits" under the symmetry. (e.g. {s, 1/s})
(2) Each part consists of exactly one element from each "orbit". (e.g. {s, t})
Then, we can solve the problem in two steps:
(i) Determine the two subproblems.
(ii) For each subproblem, get the corresponding answers.
10. Jan 10, 2005
### Hurkyl
Staff Emeritus
Dexter gave a solution using a third approach. What precisely did he do, and why did it work? Is it really a different approach?
The variable he defined, u, has the property that it is invariant under the symmetry. Notice that if you replace z with 1/z, u remains unchanged.
Because of this invariance under symmetry, there are fewer solutions for u than there are for z, making them easier to find.
Actually, in some sense, both of my approaches make use of invariants under symmetry. For instance, (z - s)(z - 1/s) is invariant under swapping s and 1/s. My second approach found two quadratics that are each invariant under swapping s and t.
P.S. for a polynomial which is the same as its reverse, making dexter's solution is the standard way of getting a lower degree polynomial.
(The reverse of a polynomial is the one with the coefficients in the opposite order. Equivalently, the multiplicative inverse of the roots of the polynomial are precisely the roots of the reverse. To make sense, of course, this assumes that 0 is not a root)
Last edited: Jan 10, 2005
11. Jan 10, 2005
### Hurkyl
Staff Emeritus
And just for fun, here's a trig approach:
Let theta be the angle the ladder makes with the base. A is the distance from the box to the base, B is the distance from the box tot he top of the ladder.
So, the length of the ladder is 1/sin theta + 1/cos theta = 3.
With trig identities, and a change of variable, I can get a quadratic in sin phi.
12. Jan 10, 2005
### Zaimeen
I was a bit sleepy at that time Dex. I did checked it just now. You are right. Sorry Dex!! | {
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I was a bit sleepy at that time Dex. I did checked it just now. You are right. Sorry Dex!!
________________________________________________________________________
"The beautiful mind goes faster than the hand"
13. Jan 10, 2005
### recon
I got it the same way as dextercioby. I just made a mistake multiplying the right side of the equation with z^3 instead of z^2. I multiplied everything on the left side with z^2.
EDIT: This is the equation I'm talking about.
$$z^2 + 2z + \frac{2}{z} + \frac{1}{z^2} = 7$$
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# Expected value with piecewise probability density function (PDF)
I am continuing the prepare for an exam by reviewing handouts from an old statistics course I took. The handout came with a set of solutions prepared by the instructor, but I suspect that one of the answers is wrong. If the answer the instructor provided isn't wrong, then I must be missing something, so I'd like to ask the CV community to help where I've gone wrong. Here is the question:
Suppose that $X$ is a continuous random variable with pdf given by:
$f(x) = \left\{ \begin{array}{lr} x & 0 < x < 1\\ (2x-1)/12 & 1< x < 3 \end{array} \right.$
if $g(x) = x$ for $x>2$ and $g(x)=0$ for $x \le 2$, find the expected value $Eg(x)$.
The instructor put the solution as $10/9$. However, I've reworked my solution several times and continue to arrive at the answer $Eg(x)=61/72$. Here is how I approached this problem:
By definition, we know:
$$Eg(X) = \int_{-\infty}^{\infty}g(x)f(x)dx\tag{1}$$ Since $f(x)$ and $g(x)$ are $0$ everywhere, except from $2<x<3$, I only need to integrate (1) over these values of $x$. Doing so yields: $$Eg(X) = \int_{2}^{3}g(x)f(x)dx=\int_{2}^{3}{x(2x-1)\over{12}}dx=\int_{2}^{3}{2x^2-x\over{12}}dx$$ $$={1\over{12}}\left({{2x^3}\over{3}}-{x^2\over{2}}\right)\bigg\rvert_{x=2}^{x=3}={1\over{12}}\left[\left({{54}\over{3}}-{9\over{2}}\right)-\left({{16}\over{3}}-{4\over{2}}\right)\right]\bigg\rvert_{x=2}^{x=3}=61/72$$
I think the instructor accidentally integrated over $1<x<2$ instead of over $2<x<3$, but as I said earlier, I'm just hoping someone can verify if I'm correct, or perhaps I'm just not understanding something here. Thanks. | {
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• I think you're right. If you can get correct answers for similar problems then you should have no reason to worry. – mark999 Apr 22 '17 at 6:53
• I agree with your answer. – Glen_b -Reinstate Monica Apr 22 '17 at 9:54
• Much appreciated, to you both! This makes me feel a lot better! – StatsStudent Apr 22 '17 at 15:49
• could you please add a tag-wiki for your new tag piecewise-pdf ? – kjetil b halvorsen Apr 23 '17 at 11:20 | {
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This note has been used to help create the Mental Math Tricks wiki
Hey guys, I saw a faster way to find cube roots.
We already know some basic cube numbers
$$0^{3}$$=0
$$1^{3}$$=1
$$2^{3}$$=8
$$3^{3}$$=27
$$4^{3}$$=64
$$5^{3}$$=125
$$6^{3}$$=216
$$7^{3}$$=343
$$8^{3}$$=512
$$9^{3}$$=729
Now, the common thing here is that each ones digit of the cube numbers is the same number that is getting cubed , except for 2 ,8 ,3 ,7 .
now let us take a cube no like 226981 .
to see which is the cube root of that number , first check the last 3 digits that is 981 . Its last digit is 1 so therefore the last digit of the cube root of 226981 is 1 .
Now for the remaining digits that is 226
Now 226 is the nearer & bigger number compared to the cube of 6 (216)
So the cube root of 226981 is 61
Let us take another example - 148877
Here 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.
So the last digit of the cube root of 148877 is 3.
Now for the remaining digits 148.
It is the nearer and bigger than the cube of 5 (125).
Therefore the cube root of 148877 is 53.
Let us take another example 54872.
Here the last three digit's (872) last digit is 2 but the cube of 2's last digit is not 2 but the last of the cube of 8 is 2.
So the last digit of the cube root of 54872 is 8.
Now of the remaining numbers (54). It is nearer and bigger to the cube of 3 (27). So therefore the cube root of 54872 is 38.
Note by Kartik Kulkarni
3 years, 5 months ago
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\boxed{123} $$\boxed{123}$$
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How about to find cube roots of a number which answer is three-digit number ?? For example 111^3, 267^3, etc
- 3 years, 5 months ago
After doing the last three digits , try to find which is the nearest cube number to it for the remaining digits E.g -
$$\sqrt[3]{1860867}$$
Done with the last three digits and the last digit , & you get 3 as the last digit of $$\sqrt[3]{1860867}$$
Now find the nearest cube number of1860 & it is 12 (1728)
So therefore $$\sqrt[3]{1860867}$$ = 123
- 3 years, 5 months ago
So we just do the same ways... Thank you so much..
- 3 years, 5 months ago
waitwaitwaitwaitwait..whaaaaaaaaaaat? Where did that 3 even come from? The last digit of 1860867 is 7.....
- 3 years, 4 months ago
Read the note properly , it says 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.
- 3 years, 4 months ago
So basically the cube of the number you are looking for must have the same last digit as the number in the problem?
- 3 years, 4 months ago
No, dude. It occasionally happens, but it ain't no rule. It happens for 1 (1³ = 1), 4 (4³ = 64), 5 (5³ = 125), 6 (6³ = 216), 9 (9³ = 729) and 0 (0³ = 0). But, here we see, it doesn't happen for 2 (2³ = 8), neither 3 (3³ = 27), nor 7 (7³ = 343) and 8 (8³ = 512). I'll always have to check this before find cube roots by this method.
- 3 years, 4 months ago
2, 3, and 7, 8 has at their unit place have their 10's compliments. Rest have the same number as said earlier.
- 3 years, 4 months ago | {
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- 3 years, 4 months ago
Thank you, a great method to solve the cube roots, so bad it doesn't work with every cubic root, it would save a lot of time in tests. Anyway, thanks!
- 3 years, 4 months ago
Another Interesting fact:: (A) cube of 2= unit digit 8 .....cube of 8=unit digit 2 (B) cube of 3=unit digit 7...... cube of 7= unit digit 3
- 3 years, 4 months ago
1, 4, 5, 6, 9 have the unit place of their cubes as the number themselves. But cubes of 2,3 and 8,7 has there unit place as their compliment of 10.
- 3 years, 4 months ago
It works for groups of threes. How adorable.
- 3 years, 4 months ago
Cub of 2, 3, 7, 8 we have its compliment of 10.
For 2: it is 10-2=8...........for 8, it is 10-8=2............................ 3, it it 10-3=7, ......for 7, it is 10-7=3
- 2 months, 1 week ago
Data handling
- 6 months, 2 weeks ago
But it is not apllicable everywhere..
- 7 months, 4 weeks ago
Comment deleted 5 months ago
We should take the number which has highest cube less than 117 in your case but not nearest.
- 6 months, 3 weeks ago
I gave the first person who introduced this to me a very good comment. Features of natural numbers can occasionally been found. Important thing is never let this concept to mislead ourselves when the situation is not whole numbers. I recalled and remember ed again but not really memorized properly. Understand why could make me a better memory perhaps. Hope I can memorize from today onwards!
- 1 year, 1 month ago
But it is not useful for non perfect cubes
- 1 year, 1 month ago
Write a comment or ask a question...if m=29 and e=13, then m=m+e e=m-e m=m-e then find the new value of m and e??
- 2 years, 10 months ago
Very nice & thanks.
- 3 years, 2 months ago
Good Method ....!!! Amazing...!!
- 3 years, 3 months ago
thats just for a sure perfect cube
- 3 years, 3 months ago
Cool.......
- 3 years, 4 months ago
Really very useful trick Thanks:)
- 3 years, 4 months ago | {
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Cool.......
- 3 years, 4 months ago
Really very useful trick Thanks:)
- 3 years, 4 months ago
It's really coolest method ever.but can any1 suggests me methods for square root of a decimal number.for eg:square root of 0.56
- 3 years, 4 months ago
how to work out cube root of 216216. The answer on face is 66 but that is not the cube root.
- 3 years, 4 months ago
216216 is not perfect cube ! This method is only for perfect cube
- 4 months ago
you have to it by long division method
- 3 years, 4 months ago
good one
- 3 years, 4 months ago
do you want to know the exact long division method of finding cube roots though it tedious... :)
- 3 years, 4 months ago
sure.
- 3 years, 4 months ago
- 3 years, 4 months ago
I HAVE SOME CONFUSION THAT WHEN HAM LOG SAME NO. KO LIKHEGE OR KAB NHI........AS 1ST SUM MEN.......226 KA 6 LIKHE AND 981 KA 1 SO ANS. IS 61 BUT 148877 MEN 148 KA 5 KYU LAST NO TO 8 HA SO COMPLEMENTRY IS 2 BUT HERE IS 5..
- 3 years, 4 months ago
I would prefer not to use Hindi cause it is confusing me that you have mixed up English & Hindi
- 3 years, 4 months ago
Only works for whole numbers. It's interesting however that you have found this method. How did you come across it?
- 3 years, 4 months ago
excellent method!!! upvoted young mind :)
- 3 years, 4 months ago
I like it
- 3 years, 4 months ago
Who discovered this method? It's really awesome
- 3 years, 4 months ago
i like that method
- 3 years, 4 months ago
Real nice method. I liked it.
- 3 years, 4 months ago
I like this method .
- 3 years, 4 months ago
nice
- 3 years, 4 months ago
you mean x^3 of 226981 , 226971 , 226961 , 226981 , 226881 , all is 61 only by your way. which is incorrect
- 3 years, 4 months ago
you have to know that it it works only for a perfect cube
- 3 years, 4 months ago
I'm sorry I did not understand
- 3 years, 4 months ago | {
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- 3 years, 4 months ago
I'm sorry I did not understand
- 3 years, 4 months ago
By this trick cube root for last 3 digit is depends on unit place digit only? if we consider these numbers which all have 1 as unit place digit , 226981 , 226971 , 226961 , 226221 , 226881 so by the rule cube root should be 61 for all these numbers. which is actually incorrect because numbers are different.
- 3 years, 4 months ago
This method is only applicable for cube numbers that have the cube root with no numbers after the decimal point
- 3 years, 4 months ago
As I have mentioned in another comment, if the number is not a perfect cube, we at least know the floor and the ceiling of this number.
- 3 years, 4 months ago
excellent
- 3 years, 4 months ago
Thank U Very Much.I like Ur Way To solve The Problem.
- 3 years, 4 months ago
Great! Interesting!
- 3 years, 4 months ago
- 3 years, 4 months ago
Really good method... I like it!
- 3 years, 4 months ago
ecellent method .its working
- 3 years, 4 months ago
Good solution
- 3 years, 4 months ago
Write a comment or ask a question... Super
- 3 years, 4 months ago
Thanks
- 3 years, 4 months ago
Excellent method
- 3 years, 4 months ago
Just noticed. It actually isn't applicable to numbers other than perfect cubes. For example, if you calculate the cube root of 1,216 using this method, you get 16; actual root is 10.67. They're almost 5.5 numbers apart. If you have any better ways, please post it.
- 3 years, 4 months ago
In that case we know between which two integers the actual cube root lays.
- 3 years, 4 months ago
I had answered to a similar question , & this method is only applicable for numbers which have their cube roots with no numbers after the decimal point
- 3 years, 4 months ago
Awesome and unique way to do it!! Thanks!!
- 3 years, 4 months ago
Nice note
- 3 years, 4 months ago
Would largely help me for finding Karl Pearson's coefficient. Thanks.
- 3 years, 5 months ago
Fantastic method Thanks | {
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- 3 years, 5 months ago
Fantastic method Thanks
- 3 years, 5 months ago
Brilliant! Good to learn this from you. Thanks.
- 3 years, 5 months ago
maths is not about approximation and estimation!!!!
- 3 years, 5 months ago
can someone prove it mathematically?
- 3 years, 5 months ago
- 3 years, 4 months ago
a + 10 b + 100 c + 1000 d + 10000 e + 100000 f could roughly prove it I guess.
- 3 years, 5 months ago
@Kartik Kulkarni .... really a nice one ... but i hav a doubt ... take 1331 ..... u get 11 by the method stated above .... if u take 1441 ..... 11 isnt correct ..... in that case .... u cant find whether a no. is a cube no. or not using this method.... rite???
- 3 years, 5 months ago
also 1441's cube root is somewhat 11 And many more numbers after the decimal points
- 3 years, 5 months ago
Well actually,this method is only applicable for actual cube numbers
- 3 years, 5 months ago
Great buddy. Here is the actual method https://brilliant.org/discussions/thread/long-divison-method-of-cube-root/
- 3 years, 4 months ago
@Kartik Kulkarni ... as soon as i saw ..... i found this interesting and also concluded this is applicable for perfect cubes ... but ur inference of 1441's cube root is around 11 is wrong ..... eg: take 1721 ..... if u infer by the same method as u did above ... it is around 11 ... but actually it can be estimated to 12 ..... (Note: cube root of 1721 = 11.98)
- 3 years, 5 months ago
well , I didn't think about the estimation part
- 3 years, 5 months ago
cool
- 3 years, 5 months ago
Really cool way...I m looking forward to u to post some cool ways of finding the sum of series....
- 3 years, 5 months ago
I love this mathed
- 3 months, 2 weeks ago
3√79510
- 4 months, 2 weeks ago
according to this cube root of 125486 should be 56 but actually it is not
- 3 years, 3 months ago
I just found out when this method works,
for eg 125486. last 3 digits = 6, first 3 digist =5,
here 125 is perfect cube , hence it doesnt work. | {
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here 125 is perfect cube , hence it doesnt work.
MY findings = This method only works when neither of the components( 1st 3 digits & last 3 digits) are perfect cube but the number that is comprised of the components is a perfect cube.
In ur case 125486 aint a perfect cube cum 125 which i call a component is.
Hows my Theorem? Thumbs up!!
- 3 years, 3 months ago
- 3 years, 2 months ago
An adition to my findings: Either both the components aint perfect cube or both are.
125000, fr last 3 digits =0, fr first 3 digits =5
cube root of 125000 is 50.
(Notice that 000 is nothing but 0 and not 1000, 0^3 is 0)
- 3 years, 2 months ago
216216
- 2 years, 6 months ago
Comment deleted Apr 05, 2015
Comment deleted Apr 08, 2015
Sorry i didnt get u.
- 3 years, 2 months ago
125486 is not a perfect cube and so this method is not applicable for that number
- 3 years, 1 month ago
- 3 years, 4 months ago
Very nice and interesting solution
- 3 years, 4 months ago
whats's wrong with these four numbers(2,3,7 and 8)? i mean these are the number which you will never find at the end of any "squared number"( at ones place i mean). and here too the same four number have different digits at ones place. by the way nice trick. thanks!
- 3 years, 4 months ago
what if we have 7 digited number could u explane me how to do it please
- 3 years, 4 months ago
I just explained it to Jonathan Christianto above
- 3 years, 4 months ago
kk:":":":":":":thanku
- 3 years, 4 months ago
then we have to go by long divison actual method
- 3 years, 4 months ago | {
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Simplify $2(\cos^2 x - \sin^2 x)^2 \tan 2x$
Simplify: $$2(\cos^2 x - \sin^2 x)^2 \tan 2x$$ After some sketching, I arrive at: $$2 \cos 2x \sin2x$$ Now according to the answer sheet, I should simplify this further, to arrive at $\sin 4x$. But how do I derive the latter from the former? Where do I start? Hoe do I use my double-angle formulas to arrive there?
• Hint: $\sin(2u) = 2\cos(u)\sin(u)$. What is $u$ in this case? – MathematicsStudent1122 Dec 27 '15 at 9:46
Notice, using double angle identity $\cos^2A-\sin^2A=\cos 2A$, one should get
$$2(\cos^2x-\sin^2x)^2\tan 2x$$ $$= 2\cos^2 2x\tan 2x$$ $$= 2\cos^2 2x\left(\frac{\sin 2x}{\cos 2x}\right)$$ $$= 2\sin 2x\cos 2x$$ using double angle identity $2\sin A\cos A=\sin 2A$, $$=\sin 2(2x)$$$$=\color{red}{\sin 4x}$$
• Sorry, there was a typo in the initial post. – Apeiron Dec 27 '15 at 9:45
• Alright, so I have made correction accordingly – Harish Chandra Rajpoot Dec 27 '15 at 9:48
$2\cos(2x)\sin(2x)=\sin(2(2x))=\sin(4x)$
• Yeah, this just doesn't explain why I can do this. I know $2 \cos x \sin x = \sin 2x$, but how do I get from your first step to the second. This is not obvious for me. Would $2 \cos 3x \sin 3x = \sin 6x$? – Apeiron Dec 27 '15 at 9:40
• First step: use $2x$ instead of $x$ in the sine double angle formula Second step: $2*2x=4x$ Yes, $2 \cos 3x \sin 3x = \sin 6x$ – GNUSupporter 8964民主女神 地下教會 Dec 27 '15 at 9:42
Use $$\cos^2x-\sin^2x=\cos2x$$ and $$\tan A=\dfrac{\sin A}{\cos A}$$ | {
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# Calculus
Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.
1. Let V = ∫sin^2x dx
Since cos^2x is a mirror image of sin^2x, ∫cos^2x dx = V
Now, since sin^2x+cos^2x = 1,
2V = ∫[0,π/2] 1 dx = π/2
V = π/4
check:
sin^2x = (1-cos2x)/2
∫sin^2x dx = ∫(1-cos2x)/2 dx
= 1/2 (x - 1/2 sin2x) [0,π/2]
= 1/2[(π/2)-(0)]
= π/4
posted by Steve
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# How to find the derivative of a definite integral that has unusual lower and upper bounds?
I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.
The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $$\frac{d}{dx} \int f(x)dx = f(x)$$, i.e. the first part of the fundamental theorem of calculus, to answer the question.
The function is:
$$\int_{\sqrt{x}}^{\pi/4} \theta \cdot tan\theta \cdot d\theta = g(x)$$
In words: if a function corresponds to an integral where the upper bound on the integral is $$\pi/4$$, the lower bound is $$\sqrt{x}$$, and the function being integrated is $$\theta \cdot tan\theta$$ with respect to $$\theta$$, then what is the derivative of the function?
I've tried setting $$u = \pi/4$$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $$\sqrt{x}$$, in my solution somehow, but I have no idea how.
Any help would be greatly appreciated.
• Hint: chain rule. – GEdgar Nov 24 '18 at 22:34
• As a suggested reading, check out Leibniz Rule for differentiation of integrals. – Shubham Johri Nov 24 '18 at 22:44
• Hint: $\int_{-\infty}^x f(x) dx = f(x)$, so $\int_a^bf(x) dx = \int_{-\infty}^b f(x) dx - \int_{-infty}^a f(x) dx$ – eSurfsnake Nov 25 '18 at 7:19
So,$$g(x)=\int_{\sqrt x}^{\frac\pi4}\theta\tan\theta\,\mathrm d\theta=-\int_{\frac\pi4}^{\sqrt x}\theta\tan\theta\,\mathrm d\theta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule? | {
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• Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC? – James Ronald Nov 24 '18 at 22:42
• No, the lower bound does not matter. – José Carlos Santos Nov 24 '18 at 22:45
After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $$x$$ and the lower limit is not a constant.
So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.
Suppose the lower limit is $$L(x)$$ and the upper limit is $$U(x)$$ of the integral. Define
$$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t\text{.}$$
Suppose $$F$$ is an antiderivative of $$f$$. By part II of the Fundamental Theorem of Calculus, you know that $$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t = F(U(x)) - F(L(x))\text{.}$$ Then, the derivative of $$g$$ is given by, assuming differentiability of $$U$$ and $$L$$,
$$\dfrac{\text{d}}{\text{d}x}[g(x)] = F^{\prime}(U(x))U^{\prime}(x)-F^{\prime}(L(x))L^{\prime}(x)$$ after making use of the chain rule for derivatives. But, $$F$$ is an antiderivative of $$f$$, so $$F^{\prime} = f$$, hence $$\dfrac{\text{d}}{\text{d}x}[g(x)] = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}$$ In other words, the main result is $$\boxed{ \dfrac{\text{d}}{\text{d}x}\int_{L(x)}^{U(x)}f(t)\text{ d}t = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}}$$ | {
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Applying to this problem, we have $$f(\theta) = \theta \tan(\theta)$$, $$U(x) = \dfrac{\pi}{4}$$, and $$L(x) = \sqrt{x}$$. The derivatives are $$U^{\prime}(x) = 0$$ and $$L^{\prime}(x) = \dfrac{1}{2\sqrt{x}}$$. Hence, the derivative of $$g$$ is $$g^{\prime}(x) = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x) = f\left(\dfrac{\pi}{4}\right)(0) - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}}$$ which simplifies to $$g^{\prime}(x) = - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}} = -\sqrt{x}\tan(\sqrt{x}) \cdot \dfrac{1}{2\sqrt{x}} = -\dfrac{1}{2}\tan(\sqrt{x})\text{.}$$
• Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again! – James Ronald Nov 24 '18 at 23:11 | {
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# Infinite cyclic group isomorphism.
I have one probably basic question, but still it bothers me how to show it.
Namely, if two groups are isomorphic (i.e. there is a bijective group homomorphism between them) and if one of them is infinite cyclic (precisely, in my case it is discussed about $(\mathbb Z,+)$), does the other group necessary have to be infinite cyclic?
Precisely, I'm in algebraic topology and I know that the fundamental group of cycle $\mathbb S^1$ is isomorphic to integers, but does this imply that the fundamental group of cycle is infinite cyclic as well?
Any sketch of the proof would be welcome.
• All cyclic groups are isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ for a natural number $n$. – Student Mar 6 '17 at 21:20
Indeed, this is true and is known as the fundamental theorem of finitely generated abelian groups..
In particular, we know that every finitely generated abelian group is of the form:
$$\mathbb Z^n \oplus \mathbb Z_p\oplus\cdots \oplus \mathbb Z_q$$
so if your group is infinite and cyclic, then it is abelian, of rank $1$ and torsion-free, meaning that it is indeed the same thing as $\mathbb Z$.
Edit: on the other hand, this now feels like a lot of overkill, despite being the way that I think about it. Here is a proof sketch:
Let $G$ be infinite cyclic $\langle a^n \mid n \in \mathbb Z \rangle$. Suppose further that $\phi: \mathbb Z \to G$ is the map $n \mapsto a^n$.
You should check that this is indeed a homomorphism. Surjectivity is clear. Injectivity follows from the fact that if $a^n=a^m$ for $n >m$, then $a^n (a^m)^{-1}=e \implies a^{n-m}=e$ while the order of $a$ was supposed to be infinite.
Hence, this is an isomorphism.
Edit 2: Let me try to address your question in the comments. Let $\pi_1(S^1)$ be the group consisting of loop classes, equipped with the usual multiplication. We already have that there exists an isomorphism $$\rho: \pi_1(S^1) \to (\mathbb Z,+)$$
and we have further that | {
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and we have further that
$$\phi:\mathbb Z \to G$$ is an isomorphism when $G$ is infinite cyclic. By the composition of isomorphisms,
$$\phi \circ \rho:\pi_1(S^1) \to G$$ is an isomorphism as well. Hence, we can conclude that it is infinite cyclic.
This seems like a more general problem, isomorphism is a transitive property, if $A \cong B \cong C$, then $A \cong C$ as well.
Edit 3: We claim that $(\mathbb Z,+)$ is an infinite cyclic group. To see this, consider $1 \in \mathbb Z$. Clearly, every element $n \in \mathbb N$ can be written as $\underbrace{1+1+\dots+1}_{n \, \mathrm{times}}$ and each inverse can be given by $-1$, which is the additive inverse of $1 \in \mathbb Z$. In other words, $1$ generates the group and has infinite order. Perhaps the notation is confusing, in additive notation:
A group $(G,+)$ is said to be infinite cyclic if $$G=\langle n \cdot a \mid n \in \mathbb Z\rangle.$$
Hence if you already know that $\pi_1(S^1) \cong \mathbb Z$, then it must be infinite cyclic, since $\mathbb Z$ already was. | {
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• Thanks. I understand what you are trying to say, but isn't this (in some sense) the reversed statement, i.e. for a given infinite cyclic group say $(G,*)$, one can construct isomorphism between $G$ and $\mathbb Z$. But I think I have the opposite problem; I already know that my (fundamental) group $\pi_1(\mathbb S^1,*)$ is isomorphic to $(\mathbb Z,+)$ where for $(\mathbb Z,+)$ "infinite cyclic" behaviour is known and what I am trying to show (if possible) is that the first group must be infinite cyclic as well. – edward_scissorhands Mar 7 '17 at 8:39
• It's unclear what you are confused about to me. What I have just demonstrated is that it does not matter whether you say that the fundamental group is infinite cyclic of is the additive group on the integers, since the two of them are the same group, up to symbolic manipulation. – Andres Mejia Mar 7 '17 at 8:44
• the fundamental group is infinite cyclic. I included another argument. However, I think that this argument is backwards in the way of intuition. The thing to understand is that $\pi_1(S^1)$ is a free group given by a single generator, so it is in fact cyclic, so it is abelian, and hence is $\mathbb Z$. In particular, any free group modulo its commutator subgroup is again $\mathbb Z^n$ for some $n$. To see an example where it is a free group (infinite) on two generators, but not $\mathbb Z$, consider $\pi_1(S^1 \vee S^1)$. – Andres Mejia Mar 7 '17 at 8:54
• If it is isomorphic to $(\mathbb Z,+)$ then it is infinite cyclic, since $\mathbb Z$ is itself infinite cyclic. For example, it is generated by a single element $1$ that has infinite order. – Andres Mejia Mar 7 '17 at 8:57
• @Eurydice see my edits – Andres Mejia Mar 7 '17 at 9:01 | {
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# What's the shape of this “addsTo” function …?
Note that in this combinatronics question,
How many lists of 100 numbers (1 to 10 only) add to 700?
For an array of 100 numbers, each 1 to 10 inclusive, and the total is T - how many such arrays exist?
In fact due to the ASTOUNDING answers there, we now know
for T=700 the answer is 1.2 e92
Being curious I now wonder about the shape of this result for various T.
Sadly, all I know are the points T<100, T=100, T=700 (thanks, MSE!), and T=1000.
Many questions arise, is there a T that makes a maximum, is it the same all over except the ends, is it bumpy or erratic ... does it make any difference if T is prime, even, etc ... and, since 19^92 is pretty small and there's only 600 Ts, in fact, is T=700 just freakishly small for some reason (what reason?) ...... or?
• The rough shape is a bell curve with center at $550$ and symmetric around that point. – Thomas Andrews Sep 11 '14 at 14:42
• Yeah, I didn't read the linked-to question, so I didn't realize I duplicated a lot of work, so just made the comment. Basically, if you pick a random integer from $1$ to $10$ 100 times, the the probability of getting $T$ is $\frac{1}{10^{100}}$ times your count. And the general tendency of repeating processes like this is to get a bell curve. – Thomas Andrews Sep 11 '14 at 14:45
• It's $$\sum_{j=0}^{45} (-1)^j \binom{100}{j}\binom{549-10j}{99}$$ Need some time to find a program to calculate it. – Thomas Andrews Sep 11 '14 at 14:51
• The approximation of the distribution as a bell curve appeared as @mjqxxxx's answer to your other problem. One can perhaps obtain asymptotic corrections to this with more sophisticated analysis. – Semiclassical Sep 12 '14 at 16:32
• ah, i didn't notice that, thanks semi. still it would be interesting to know the max etc - still thanks – Fattie Sep 12 '14 at 16:35 | {
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According to the Central Limit Theorem, when you add a large number of independent, identically distributed random variables (each variable can be a sum of other random variables), the distribution of the sum tends to a normal distribution whose mean is the sum of the means of the individual distributions, and whose variance is the sum of the variances of the individual distributions.
For example, if we sum $1$ number ($1$-$10$), the distribution looks like
$\hspace{3cm}$
If we sum $2$ numbers ($1$-$10$), the distribution looks like
$\hspace{3cm}$
If we sum $3$ numbers ($1$-$10$), the distribution starts to looks like a normal distribution
$\hspace{3cm}$
If we sum $10$ numbers ($1$-$10$), the distribution looks even more like a normal distribution
$\hspace{3cm}$
Now lets look at the sum of $100$ numbers ($1$-$10$)
$\hspace{3cm}$
The number of ways for $100$ numbers ($1$-$10$) to sum to $700$, which was computed in this answer, is pointed to by the arrow. Note that the range of the sum is $9n$ where $n$ is how many numbers are added; however, the standard deviation is $\sqrt{8.25n}$. Thus, the relative spread gets smaller; that is, the distribution becomes narrower about the mean as $n$ get bigger.
The maximum number of ways to sum to $T$ is at the mean of the distribution; that is, at $T=550$, where the number of ways is $$13868117806391314648666325510838589167047653141664\\4888545033078503482282975641730091720919340564340$$ which is approximately $1.3868117806391314649\times10^{98}$. | {
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• Wow - you guys are amazing. I'm so dense I didn't even realise, of course the question is just equivalent to looking at the random distribution. one question - you see the final graph, in fact the answer to the question ........ – Fattie Sep 13 '14 at 9:30
• ...... would we still call that "a bell curve," a normal distribution, since it's so incredibly pointy? As the answer explains, for 1 or 2 numbers, I guess you'd say, "it is not" a Normal distribution. Well now, when you get to very high numbers (300, 500, 10,000) again does it become "not" a Normal distribution? Or, is it still perfectly a "bell curve" but just very pointy? Is bell curvey -ness subjective? Or conversely, just by definition, is every result here officially a normal distribution (even the 1 or 2 cases)? Cheers!!!!! – Fattie Sep 13 '14 at 9:31
• @JoeBlow: A scaled bell curve looks pointy, and that is what this is: a scaled bell curve. If scaled so that the standard deviation is $1$, its total sum is $1$, and translated so that the mean is $0$, it would look very close to a normal distribution bell curve. Scaled this way would map the total range of possibilities to the whole real line. – robjohn Sep 13 '14 at 9:39
• Gotchya, I'll have to think about that. If I'm not mistaken, in your graphs, indeed they all start and end at the zero points (so, for "10" it's 10-->100, and for "100" it's 100-->1000 etc). And the scale heights are the same so there's no "trick". But it seems to me they are radically different: the shapes of "100" and "10million" would be terribly different right? Recall I was asking "is 700 just freakishly small for some reason?" Indeed, that is the reason: there is no such "freakish smallness" in the (say) "10" or "15" case when you are still equally near the middle .... but ... – Fattie Sep 13 '14 at 9:53 | {
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• ... indeed this function becomes "amazingly pointy" as the count gets higher. I'm thinking like a video game programmer, if something depended on that shape, play would be spectacularly different with the "10" shape versus the "500" shape. (You'd almost always, versus never, run in to the goblin or whatever.) Anyways - I'm wasting your time - I'll go take a refresher course in normal curves THANK YOU SO MUCH. Do you often provide such good answers that nobody else even has to bother? ;-) – Fattie Sep 13 '14 at 9:55 | {
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Khan Academy is a … Next lesson. The powers don’t need to be “2” all the time. Played 0 times. Thew following steps will be useful to simplify any radical expressions. Test - I. Note that every positive number has two square roots, a positive and a negative root. The standard way of writing the final answer is to place all the terms (both numbers and variables) that are outside the radical symbol in front of the terms that remain inside. Test - II . When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). Square root, cube root, forth root are all radicals. Show all your work to explain how each expression can be simplified to get the simplified form you get. 72 36 2 36 2 6 2 16 3 16 3 48 4 3 A. APTITUDE TESTS ONLINE. Simplifying Radicals Worksheet … Multiply all numbers and variables outside the radical together. Exponents and power. To play this quiz, please finish editing it. For this problem, we are going to solve it in two ways. It must be 4 since (4)(4) = 42 = 16. Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. Section 6.3: Simplifying Radical Expressions, and . 2) Product (Multiplication) formula of radicals with equal indices is given by To read our review of the Math Way -- which is what fuels this page's calculator, please go here . An algebraic expression that contains radicals is called a radical expression An algebraic expression that contains radicals.. We use the product and quotient rules to simplify them. by lsorci. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Simplifying Radical Expressions DRAFT. However, I hope you can see that by doing some rearrangement to the terms that it matches with our final answer. For the number in the radicand, I see that 400 = 202. For example, These types of simplifications with variables will be helpful when doing operations with radical | {
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These types of simplifications with variables will be helpful when doing operations with radical expressions. So, , and so on. Our mission is to provide a free, world-class education to anyone, anywhere. The number 16 is obviously a perfect square because I can find a whole number that when multiplied by itself gives the target number. Well, what if you are dealing with a quotient instead of a product? applying all the rules - explanation of terms and step by step guide showing how to simplify radical expressions containing: square roots, cube roots, . In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Equivalent forms of exponential expressions. Next, express the radicand as products of square roots, and simplify. 5 minutes ago. #1. . Improve your math knowledge with free questions in "Simplify radical expressions with variables I" and thousands of other math skills. Then, it's just a matter of simplifying! A perfect square number has integers as its square roots. Section 6.4: Addition and Subtraction of Radicals. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. Quotient Property of Radicals. But there is another way to represent the taking of a root. . Additional simplification facilities for expressions containing radicals include the radnormal, rationalize, and combine commands. More so, the variable expressions above are also perfect squares because all variables have even exponents or powers. Please click OK or SCROLL DOWN to use this site with cookies. In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Example 11: Simplify the radical expression \sqrt {32} . \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int. Homework. Therefore, we have √1 = 1, √4 = 2, √9= | {
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\frac{d}{dx} \frac{\partial}{\partial x} \int. Homework. Therefore, we have √1 = 1, √4 = 2, √9= 3, etc. Another way to solve this is to perform prime factorization on the radicand. TRANSFORMATIONS OF FUNCTIONS. Variables with exponents also count as perfect powers if the exponent is a multiple of the index. The answer must be some number n found between 7 and 8. We hope that some of those pieces can be further simplified because the radicands (stuff inside the symbol) are perfect squares. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Solving Radical Equations Method 1: Perfect Square Method -Break the radicand into perfect square(s) and simplify. The goal is to show that there is an easier way to approach it especially when the exponents of the variables are getting larger. are called conjugates to each other. If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g. Example 1. The first law of exponents is x a x b = x a+b. However, it is often possible to simplify radical expressions, and that may change the radicand. First we will distribute and then simplify the radicals when possible. Simplifying Radical Expressions Date_____ Period____ Simplify. Simplifying Radical Expressions. However, the best option is the largest possible one because this greatly reduces the number of steps in the solution. 16 x = 16 ⋅ x = 4 2 ⋅ x = 4 x. no fractions in the radicand and. The properties we will use to simplify radical expressions are similar to the properties of exponents. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. So which one should I pick? The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. All that you have to do is simplify the radical like normal and, at the end, multiply the | {
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x over y. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. To simplify a radical, factor the radicand (under the radical) into factors whose exponents are multiples of the index. Scientific notations. Multiply all numbers and variables inside the radical together. Adding and Subtracting Radical Expressions, That’s the reason why we want to express them with even powers since. Test to see if it can be divided by 4, then 9, then 25, then 49, etc. Simplify radical expressions using the product and quotient rule for radicals. Here it is! To help me keep track that the first term means "one copy of the square root of three", I'll insert the "understood" "1": Don't assume that expressions with unlike radicals cannot be simplified. Example 2: to simplify ( 3. . The symbol is called a radical sign and indicates the principal square root of a number. Product Property of n th Roots. How to Simplify Radicals with Coefficients. For the numerical term 12, its largest perfect square factor is 4. These two properties tell us that the square root of a product equals the product of the square roots of the factors. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. Simplifying logarithmic expressions. Then express the prime numbers in pairs as much as possible. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property a n n = a, where a is positive. Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. To simplify complicated radical expressions, we can use some definitions and rules from simplifying exponents. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, | {
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together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. So we expect that the square root of 60 must contain decimal values. What rule did I use to break them as a product of square roots? Always look for a perfect square factor of the radicand. Part A: Simplifying Radical Expressions. This calculator simplifies ANY radical expressions. You may use your scientific calculator. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. Picking the largest one makes the solution very short and to the point. Recall that the Product Raised to a Power Rule states that $\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$. These properties can be used to simplify radical expressions. Use rational exponents to simplify radical expressions. Finish Editing. Pairing Method: This is the usual way where we group the variables into two and then apply the square root operation to take the variable outside the radical symbol. Step 4: Simplify the expressions both inside and outside the radical by multiplying. This type of radical is commonly known as the square root. Type your expression into the box under the radical sign, then click "Simplify." Notice that the square root of each number above yields a whole number answer. Here are the search phrases that today's searchers used to find our site. Perfect Squares 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 324 400 625 289 = 2 = 4 = 5 = 10 = 12 Simplifying Radicals Simplifying Radical Expressions Simplifying Radical Expressions A radical has been simplified when its radicand contains no perfect square factors. Vertical translation. Remember, the square root of perfect squares comes out very nicely! Let’s deal with them separately. The paired prime numbers will get out of the square root symbol, while the single prime will stay inside. nth roots . Code to add this calci to your website Just copy and paste the below code to your webpage | {
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roots . Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. And it checks when solved in the calculator. We need to recognize how a perfect square number or expression may look like. Play. Radical Expressions are fully simplified when: –There are no prime factors with an exponent greater than one under any radicals –There are no fractions under any radicals –There are no radicals in the denominator Rationalizing the Denominator is a way to get rid of any radicals in the denominator Mathematics. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive. The radicand contains no fractions. Let’s do that by going over concrete examples. Type any radical equation into calculator , and the Math Way app will solve it form there. Simplification of expressions is a very useful mathematics skill because, it allows us to change complex or awkward expression into more simple and compact form. If and are real numbers, and is an integer, then. Simplify a Term Under a Radical Sign. And we have one radical expression over another radical expression. Example 3: Simplify the radical expression \sqrt {72} . Simplify the expression: It is possible that, after simplifying the radicals, the expression can indeed be simplified. You can do some trial and error to find a number when squared gives 60. The solution to this problem should look something like this…. Simplify expressions with addition and subtraction of radicals. For example, the sum of $$\sqrt{2}$$ and $$3\sqrt{2}$$ is $$4\sqrt{2}$$. Live Game Live. After doing some trial and error, I found out that any of the perfect squares 4, 9 and 36 can divide 72. x^{\circ} \pi. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More | {
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rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. The roots of these factors are written outside the radical, with the leftover factors making up the new radicand. +1) type (r2 - 1) (r2 + 1). Procedures. As long as the powers are even numbers such 2, 4, 6, 8, etc, they are considered to be perfect squares. Share practice link. Example 10: Simplify the radical expression \sqrt {147{w^6}{q^7}{r^{27}}}. Example 1: Simplify the radical expression \sqrt {16} . Simplify each of the following. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions Sample Problem: Simplify 16 Solution: 16 =4 since 42 =16. You will see that for bigger powers, this method can be tedious and time-consuming. We have to consider certain rules when we operate with exponents. Variables in a radical's argument are simplified in the same way as regular numbers. This quiz is incomplete! Practice: Evaluate radical expressions challenge. Think of them as perfectly well-behaved numbers. Play this game to review Algebra I. Simplify. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g. Adding and Subtracting Radical Expressions The calculator presents the answer a little bit different. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Remember the rule below as you will use this over and over again. Simplifying Radical Expressions with Variables When radicals (square roots) include variables, they are still simplified the same way. . The following are the steps required for simplifying radicals: | {
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are still simplified the same way. . The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Step 1 : Decompose the number inside the radical into prime factors. This is easy to do by just multiplying numbers by themselves as shown in the table below. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. Step 3 : A radical expression is said to be in its simplest form if there are, no perfect square factors other than 1 in the radicand, $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$, $$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$$. Looks like the calculator agrees with our answer. Perfect cubes include: 1, 8, 27, 64, etc. $$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$$, $$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$$. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Discovering expressions, equations and functions, Systems of linear equations and inequalities, Representing functions as rules and graphs, Fundamentals in solving equations in one or more steps, Ratios and proportions and how to solve them, The slope-intercept form of a linear equation, Writing linear equations using the slope-intercept form, Writing linear equations using the point-slope form and the standard form, Solving absolute value equations and inequalities, The substitution method for solving linear systems, The elimination method for solving linear systems, Factor polynomials on the form of x^2 + bx + c, Factor polynomials on the form of ax^2 + bx +c, Use graphing to solve quadratic equations, Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. There is a | {
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Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. There is a rule for that, too. Introduction. ACT MATH ONLINE TEST. no perfect square factors other than 1 in the radicand. Example 14: Simplify the radical expression \sqrt {18m{}^{11}{n^{12}}{k^{13}}}. Radical expressions are expressions that contain radicals. Evaluating mixed radicals and exponents. Actually, any of the three perfect square factors should work. Start studying Algebra 5.03: Simplify Radical Expressions. In the same way we know that, $$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$, These properties can be used to simplify radical expressions. You factor things, and whatever you've got a pair of can be taken "out front". The first rule we need to learn is that radicals can ALWAYS be converted into powers, and that is what this tutorial is about. Simplifying radical expression. Save. Comparing surds. Remember that getting the square root of “something” is equivalent to raising that “something” to a fractional exponent of {1 \over 2}. Quantitative aptitude. Simplifying Radical Expressions. Search phrases used on 2008-09-02: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Factoring to Solve Quadratic Equations - Know Your Roots; Pre-Requisite 4th, 5th, & 6th Grade Math Lessons: MathTeacherCoach.com . We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. no perfect square factors other than 1 in the radicand $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$ no … If we combine these two things then we get the product property of radicals and the quotient property of radicals. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. You can use the same ideas to help you figure out how to simplify and divide radical expressions. 0. Extended Keyboard; Upload; Examples; | {
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figure out how to simplify and divide radical expressions. 0. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. We know that The corresponding of Product Property of Roots says that . −1)( 2. . By quick inspection, the number 4 is a perfect square that can divide 60. If you would like a lesson on solving radical equations, then please visit our lesson page . Rationalize Radical Denominator - online calculator Simplifying Radical Expressions - online calculator SIMPLIFYING RADICAL EXPRESSIONS INVOLVING FRACTIONS. Determine the index of the radical. Example 12: Simplify the radical expression \sqrt {125} . To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. Sometimes radical expressions can be simplified. This lesson covers . But before that we must know what an algebraic expression is. Simplify … The radicand contains both numbers and variables. Simplifying Expressions – Explanation & Examples. Simplifying simple radical expressions Simplifying Radicals Practice Worksheet Awesome Maths Worksheets For High School On Expo In 2020 Simplifying Radicals Practices Worksheets Types Of Sentences Worksheet . Enter the expression here Quick! Some of the worksheets below are Simplifying Radical Expressions Worksheet, Steps to Simplify Radical, Combining Radicals, Simplify radical algebraic expressions, multiply radical expressions, divide radical expressions, Solving Radical Equations, Graphing Radicals, … Once you find your worksheet(s), you can either click on the pop-out icon or download button to print or download … $$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} | {
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button to print or download … $$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$$, $$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$$. Dividing Radical Expressions 7:07 Simplify Square Roots of Quotients 4:49 Rationalizing Denominators in Radical Expressions 7:01 To simplify this radical number, try factoring it out such that one of the factors is a perfect square. 1) 125 n 5 5n 2) 216 v 6 6v 3) 512 k2 16 k 2 4) 512 m3 16 m 2m 5) 216 k4 6k2 6 6) 100 v3 10 v v 7) 80 p3 4p 5p 8) 45 p2 3p 5 9) 147 m3n3 7m ⋅ n 3mn 10) 200 m4n 10 m2 2n 11) 75 x2y 5x 3y 12) 64 m3n3 Aptitude test online. Print; Share; Edit; Delete; Report an issue; Host a game. Dividing Radical Expressions. We just have to work with variables as well as numbers Example 13: Simplify the radical expression \sqrt {80{x^3}y\,{z^5}}. Let’s simplify this expression by first rewriting the odd exponents as powers of an even number plus 1. \int_{\msquare}^{\msquare} \lim. Example 8: Simplify the radical expression \sqrt {54{a^{10}}{b^{16}}{c^7}}. Play this game to review Algebra II. The simplify/radical command is used to simplify expressions which contain radicals. Simplifying radical expressions calculator. 0. Compare what happens if I simplify the radical expression using each of the three possible perfect square factors. To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how. Multiplication tricks. Below is a screenshot of the answer from the calculator which verifies our answer. If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Thus, the answer is. “Division of Even Powers” Method: You can’t find this name in any algebra textbook because I made it up. A radical expression is | {
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You can’t find this name in any algebra textbook because I made it up. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Starting with a single radical expression, we want to break it down into pieces of “smaller” radical expressions. Separate and find the perfect cube factors. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Recognize a radical expression in simplified form. 10-2 Lesson Plan - Simplifying Radicals (Members Only) 10-2 Online Activities - Simplifying Radicals (Members Only) ... 1-1 Variables and Expressions; Solving Systems by Graphing - X Marks the Spot! The symbol is called a radical sign and indicates the principal square root of a number. To simplify radical expressions, look for factors of the radicand with powers that match the index. Example 1: to simplify ( 2. . Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Example 5: Simplify the radical expression \sqrt {200} . Multiplying Radical Expressions. If the term has an even power already, then you have nothing to do. Example 2: Simplify the radical expression \sqrt {60}. Horizontal translation. For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, and so on. Step 2 : We have to simplify the radical term according to its power. This type of radical is commonly known as the square root. One way to think about it, a pair of any number is a perfect square! Edit. Start studying Algebra 5.03: Simplify Radical Expressions. A perfect square is the product of any number that is multiplied by itself, such as 81, which is the product of 9 x 9. no radicals appear in the denominator of a fraction. Be sure to write the number and problem you | {
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9. no radicals appear in the denominator of a fraction. Be sure to write the number and problem you are solving. WeBWorK. Simplifying Radical Expressions . For instance, x2 is a p… Keep in mind that you are dealing with perfect cubes (not perfect squares). For the case of square roots only, see simplify/sqrt. Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. These properties can be used to simplify radical expressions. Example 6: Simplify the radical expression \sqrt {180} . Level 1 $$\color{blue}{\sqrt5 \cdot \sqrt{15} \cdot{\sqrt{27}}}$$ $5\sqrt{27}$ $30$ $45$ $30\sqrt2$ ... More help with radical expressions at mathportal.org. 25 16 x 2 = 25 16 ⋅ x 2 = 5 4 x. 1) Simplify. Practice. Radical expressions come in many forms, from simple and familiar, such as$\sqrt{16}$, to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$. There is a rule for that, too. COMPETITIVE EXAMS. To multiply radicals, you can use the product property of square roots to multiply the contents of each radical together. Procedures. For example, the square roots of 16 are 4 and … Otherwise, check your browser settings to turn cookies off or discontinue using the site. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Simplify #2. simplifying radical expressions. Step 2 : If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. The main approach is to express each variable as a product of terms with even and odd exponents. Fantastic! Simplify any radical expressions that are perfect squares. 9th - University grade . Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. Radical expressions are written in simplest terms | {
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problems 1-6, pick three expressions to simplify. Radical expressions are written in simplest terms when. Step 2 : We have to simplify the radical term according to its power. +1 Solving-Math-Problems Page Site. Topic Exercises. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example 7: Simplify the radical expression \sqrt {12{x^2}{y^4}} . Rationalizing the Denominator. Going through some of the squares of the natural numbers…. Square roots are most often written using a radical sign, like this, . We use cookies to give you the best experience on our website. Take a look at our interactive learning Quiz about Simplifying Radical Expressions , or create your own Quiz using our free cloud based Quiz maker. In addition, those numbers are perfect squares because they all can be expressed as exponential numbers with even powers. Multiplying Radical Expressions However, the key concept is there. \sum. The simplest case is when the radicand is a perfect power, meaning that it’s equal to the nth power of a whole number. Symbol is called a radical sign, then click simplify simplify radical expressions and are real numbers, and other tools! Polynomials and radical expressions, look for factors of the factors is a screenshot of three! Factor of the first law of exponents is x a x b = x a+b solution! For this problem should look something like this… use Polynomial Multiplication to multiply expressions with radicals such one. 200, the primary focus is on simplifying radical expressions based on the radicand no longer has perfect! Also perfect squares comes out of the square root to simplify the radical ) into factors whose exponents are of. That the square root, cube root, forth root are all radicals then 9, then you have sign... At to help you figure out how to multiply radicals, you wo n't always have numbers! Because I can find a perfect square factor we expect that the square of. Next, express the prime numbers in pairs as much as possible | {
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factor we expect that the square of. Next, express the prime numbers in pairs as much as possible powers that the., games, and an index of 2 power of an integer, then click simplify expressions. Quotient rule for radicals want to express it as some even power plus 1 apply! Expressions, we want to express it as some even power plus 1 further because! Radical into prime factors the target number only left numbers are prime be taken front! Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens of a product equals the product quotient... Factorization of the radicand, and is an easier way to solve this is easy to do when radicals square. Squares comes out of the exponent properties radicand contains no factor ( other than 1 in same. Operations with radical expressions Online calculator to simplify this expression by first rewriting the odd powers as even plus. Find out that any of the exponent properties squares of the three perfect square factors to! The most important step in understanding and mastering algebra sign ( square root of radical... Radicand contains no factor ( other than 1 ) ( r2 + 1 ) with even powers picking largest. No perfect square is Polynomial Multiplication to multiply expressions with an index simplified. Now one group therefore, we are going to solve this is express!: Decompose the number 16 is obviously a perfect square factor 2020 simplifying radicals Worksheet … x^ \circ. Learning how to simplify a radical symbol, while the single prime will stay inside stuff. Factors of the square root symbol, while the single prime will stay inside recognize a! To provide a free, world-class education to anyone, anywhere radicals ( square root of a number squared... Since ( 4 ) ( r2 + 1 ) which is what fuels this page 's calculator and... World-Class education to anyone, anywhere way as regular numbers one group root, forth root are radicals. Going to solve it in two ways radicals and the math way app | {
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