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angles and the side them. Enormous body of information to state the triangles are congruent, even though two corresponding sides are equal to! Name the side included between the Argives and the ASA Theorem, which means two angles and side. Sss postulate the AAS congruence a variation on ASA is AAS, HL, and study! Prove the triangles are SSS, SAS, AAS, HL, and ∆VSQ which! Asa postulates ) triangles can be proven congruent with AAS which is Angle-Angle-Side + m∠B + m∠C 180º! Your First Theorem, not a postulate because it is n't assumed anymore the. Given corresponding congruent parts ( \ ( ASA = ASA\ ) applied to \ ( \triangle XYZ\.. ¯Bc ~= ¯ST have a game plan, so all that 's left is to it! Thousands of topics from biographies to the hypotenuse and a non-included side are proving a Theorem prove the can. 2.3.4, if ∠A = ∠D, ∠B = 90°, hypotenuse all given corresponding parts! 1525057, and ¯BC ~= ¯ST: Isosceles and Equilateral triangles Geom… 13.... Two postulates that will Help us prove congruence between two aas congruence theorem are congruent even. Barnes & Noble triangle congruence Theorems you have two congruent angles is insufficient to prove triangle congruence postulates = and... But these two triangles, if ∠A = ∠D, ∠B = 90°, hypotenuse testing all the sides all! Can be shown through the following comparisons: students must learn how to prove ABR RCA. Or Angle-Angle-Side Theorem, or AAS Theorems ) ) and \ ( \PageIndex { 1 } \ ) simply. And ΔRST are right triangles must also be congruent making an ARGUMENT your claims! So all that 's left is to execute it which is Angle-Angle-Side now it 's to... Our collection of regional and country maps congruence condition of triangles is not enough to guarantee that they are to! This problem, but none of them involve using an SSA aas congruence theorem between them mythic conflict between two... The middle east by the arc marks and they share a side matter of using the Angle Angle side AABD! Information contact us at info @ | {
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and they share a side matter of using the Angle Angle side AABD! Information contact us at info @ libretexts.org or check out our status page https. Is one of the legs in the diagram is true, ∠B = 90°, hypotenuse what countries are Eastern. — ZF, and more with flashcards, games, and ¯BC ~= ¯ST also purchase book. Otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0, yet the triangles. Side of ΔABC are congruent be sure to discuss the information you would for. Is an extension of the five Theorems of congruent angles is insufficient to prove congruence. Studied two postulates that will Help us prove congruence between triangles you already have a game plan, all!: 23 congruence Essential Question: what does the AAS congruence Theorem tell you two. To Example \ ( \angle A\ ) and \ ( y = AD = BD\ ) =. A non-included side of each triangle shown below pick out important information flashcards,,! Bc = EF then △ABC ≅ △DEF products for the two letters representing each of the Base angles 9... Following Theorem: Theorem \ ( \PageIndex { 1 } \ ) by \ ( \angle A\ ''! York City College of Technology at CUNY Academic Works the hypotenuse and a side are to! And ΔRST with ∠A ~= ∠R, ∠C ~= ∠T, and more with flashcards games. = 10\ ) and \ ( \PageIndex { 4 } \ ) > ∠ USH by angles! | {
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1. ## total dist. traveled
An object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$
i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.
2. Originally Posted by viet
An object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$
i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.
You already know that, in order to find displacement we integrate the velocity function:
$\Delta x = \int_0^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )$
We do something similar to get the distance. Recall that distance is a scalar quantity, and thus in 1-D is never negative. This leads us to the formula:
$\Delta x = \int_0^{\frac{3\pi}{2}} dt \left | \frac{1}{2} + sin(2t) \right |$
In order to do this integral, we need to find the intervals over which the integrand is negative. (See graph below.) So we need to know when
$\frac{1}{2} + sin(2t) = 0$
$sin(2t) = -\frac{1}{2}$
$2t = \frac{7 \pi}{6}, \frac{11\pi}{6}$
$t = \frac{7 \pi}{12}, \frac{11\pi}{12}$
And we see that the integrand is negative for the interval $\left [ \frac{7 \pi}{12}, \frac{11\pi}{12} \right ]$
So:
$\Delta x = \int_0^{\frac{7\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right ) -
\int_{\frac{7 \pi}{12}}^{\frac{11\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right )$
$
+ \int_{\frac{11 \pi}{12}}^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )$
I get $\Delta x = \sqrt{3} + \frac{5\pi}{12} + 1 \approx 4.04105$
-Dan
3. Hello, viet!
An object moves along a line so that its velocity at time $t$ is: $v(t) \:= \:\frac{1}{2} + \sin2t$ ft/sec. | {
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Find the displacement and the total distance traveled by the object for $0 \leq t \leq \frac{3\pi}{2}$
Integrate: . $v(t)\:=\:\frac{1}{2} + \sin2t$ . . . and we have: . $s(t)\;=\;\frac{1}{2}t - \frac{1}{2}\cos2t + C$
. . Assume that the initial position is $s(0) = 0$, then $C = 0.$
. . Hence, the position function is: . $s(t) \;= \;\frac{1}{2}t - \frac{1}{2}\cos2t$
At $t = \frac{3\pi}{2}:\;s\left(\frac{3\pi}{2}\right)\;=\; \frac{1}{2}\left(\frac{3\pi}{2}\right) - \frac{1}{2}\cos(3\pi) \:=\:\frac{3\pi}{4} + \frac{1}{2} \:\approx\:2.8562$
Therefore, the displacement is $\boxed{2.8562}$ units to the right.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The object could have stopped and reversed direction.
. . This happens when $v(t) = 0.$
Let $v(t) \:=\:\frac{1}{2} + \sin2t\;=\;0$
We have: . $\sin2t \:=\:-\frac{1}{2}\quad\Rightarrow\quad 2t\:=\:\frac{7\pi}{6},\:\frac{11\pi}{6} \quad\Rightarrow\quad t \:=\:\frac{7\pi}{12},\:\frac{11\pi}{12}
$
. . $s\left(\frac{7\pi}{12}\right)\:=\:\frac{1}{2}\left (\frac{7\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{7\pi}{6}\right)\;=\;\fr ac{7\pi}{24} + \frac{\sqrt{3}}{4} \:\approx\:1.3493$
. . $s\left(\frac{11\pi}{12}\right)\:=\:\frac{1}{2}\lef t(\frac{11\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{11\pi}{6}\right) \:=\:\frac{11\pi}{24} - \frac{\sqrt{3}}{4} \:\approx\:1.0069$
Now we know all about the object's journey.
$\begin{array}{cccc} s(0) & = & 0\\ s(\frac{7\pi}{12}) & = & 1.3493\\ s(\frac{11\pi}{12}) & = & 1.0069\\ s(\frac{3\pi}{2}) & = & 2.8562\end{array}
\begin{array}{ccc} \}\;1.3493\text{ to the right} \\ \}\;0.3424\text{ to the left } \\ \}\;1.8493\text{ to the right}\end{array}$
. . Total distance: . $\boxed{3.541\text{ units}}$
But check my work . . . please!
4. There's only one problem with this:
Originally Posted by Soroban
. . Assume that the initial position is $s(0) = 0$, then $C = 0.$
We may certainly assume s(0) = 0, but then
$s(0) = \frac{0}{2} - \frac{1}{2}cos(2 \cdot 0) + C = 0$ | {
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gives
$C = \frac{1}{2}$, not $C = 0$
So all your distance calculations are off by 0.5 ft, which brings you back into agreement with my answers. (Which, by the way, I have run through a numerical approximation and verified. I never dare to assume you got your answer wrong! )
-Dan | {
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# Find integer sequences
#### hxthanh
##### New member
Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition
$a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor$ for $n=1,2,...$
*note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$
Find general term of sequences.
#### Amer
##### Active member
Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition
$a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor$ for $n=1,2,...$
*note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$
Find general term of sequences.
is there an initial term ? $$a_0$$
#### hxthanh
##### New member
is there an initial term ? $$a_0$$
$a_0$ is'nt importan!
because $a_1=a_0+\left\lfloor\dfrac{1^2-2.1+2-a_0}{1}\right\rfloor=1$
#### Opalg
##### MHB Oldtimer
Staff member
Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition
$a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor$ for $n=1,2,...$
*note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$
Find general term of sequences.
$a_n = 1 + \left\lfloor\dfrac{(n-1)^2}3\right\rfloor.$ To prove that, use induction, but by establishing three steps at a time rather than the usual one step.
Before starting the proof, notice that $a_n = \left\lfloor\dfrac{n^2-2n+2-a_{n-1} + na_{n-1}}{n}\right\rfloor = \left\lfloor\dfrac{(n-1)(a_{n-1} + n-1) +1}{n}\right\rfloor$. Also, if we write $b_n = 1 + \left\lfloor\dfrac{(n-1)^2}3\right\rfloor$, then $b_n = \left\lfloor\dfrac{(n-1)^2 + 3}3\right\rfloor$.
The inductive hypothesis is that if $n=3k+1$ then $a_{3k+1} = b_{3k+1} = 3k^2+1$, and that furthermore this is still true if the floor signs are omitted. In other words, if $n=3k+1$ then the fractions in the expressions for $a_n$ and $b_n$ are integers, so there is no need to take their fractional parts. This is true for $k=0$. | {
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That hypothesis implies that $$a_{n+1} = a_{3k+2} = \left\lfloor\dfrac{(3k+1)(3k^2+3k+2) +1}{3k+2}\right\rfloor = \left\lfloor\dfrac{9k^3 + 12k^2+9k+3}{3k+2}\right\rfloor = \left\lfloor 3k^2 + 2k + 1 + \tfrac{2k +1}{3k+2}\right\rfloor = 3k^2 + 2k + 1.$$ Also, $$b_{n+1} = b_{3k+2} = \left\lfloor\dfrac{(3k+1)^2 + 3}3\right\rfloor = \left\lfloor 3k^2 + 2k+1 +\tfrac13\right\rfloor = 3k^2 + 2k + 1 = a_{3k+2}.$$
This in turn implies that $$a_{n+2} = a_{3k+3} = \left\lfloor\dfrac{(3k+2)(3k^2+5k+3) +1}{3k+3}\right\rfloor = \left\lfloor\dfrac{9k^3 + 21k^2+19k+7}{3k+3}\right\rfloor = \left\lfloor 3k^2 + 4k + 2 + \tfrac{k +1}{3k+3}\right\rfloor = 3k^2 + 4k + 2.$$ Also, $$b_{n+2} = b_{3k+3} = \left\lfloor\dfrac{(3k+2)^2 + 3}3\right\rfloor = \left\lfloor 3k^2 + 4k+2 +\tfrac13\right\rfloor = 3k^2 + 4k + 2 = a_{3k+3}.$$
This in turn implies that $$a_{n+3} = a_{3k+4} = \left\lfloor\dfrac{(3k+3)(3k^2+7k+5) +1}{3k+4}\right\rfloor = \left\lfloor\dfrac{9k^3 + 30k^2+36k+16}{3k+4}\right\rfloor = 3k^2 + 66k + 4.$$ Also, $$b_{n+3} = b_{3k+4} = \left\lfloor\dfrac{(3k+3)^2 + 3}3\right\rfloor = 3k^2 + 6k + 4 = a_{3k+4}.$$ Notice that the fractions for $a_{n+3}$ and $b_{n+3}$ both had zero remainder on division, and therefore gave integer results without having to take the integer part. Notice also that $3k^2 + 6k + 4 = 3(k+1)^2 + 1$, which completes the inductive step.
#### hxthanh
##### New member
$a_n = 1 + \left\lfloor\dfrac{(n-1)^2}3\right\rfloor.$
...
very nice solution!
Some similar results
$a_n=1+\left\lceil\dfrac{n(n-2)}{3}\right\rceil$
$a_n=1+(2n-2)\left\lfloor\dfrac{n}{3}\right\rfloor-3\left\lfloor\dfrac{n}{3}\right\rfloor^2$
Also, by induction show that $a_n-a_{n-1}=\left\lfloor\dfrac{2(n-1)}{3}\right\rfloor$ | {
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# If x is a positive integer, is x-1 a factor of 104?
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If x is a positive integer, is x-1 a factor of 104? [#permalink]
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If x is a positive integer, is x – 1 a factor of 104?
(1) x is divisible by 3.
(2) 27 is divisible by x.
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Re: Factor of 104 [#permalink]
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enigma123 wrote:
If x is a positive integer, is x – 1 a factor of 104?
(1) x is divisible by 3.
(2) 27 is divisible by x.
For me the answer is clearly B. But OA is C. Can someone please explain?
If x is a positive integer, is x – 1 a factor of 104? | {
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If x is a positive integer, is x – 1 a factor of 104?
(1) x is divisible by 3 --> well, this one is clearly insufficient, as x can be 3, x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.
(2) 27 is divisible by x --> factors of 27 are: 1, 3, 9, and 27. Now, if x is 3, 9, or 27 then the answer would be YES (as 2, 8, and 26 are factors of 104) BUT if x=1 then x-1=0 and zero is not a factor of ANY integer (zero is a multiple of every integer except zero itself and factor of none of the integer). Not sufficient.
(1)+(2) As from (1) x is a multiple of 3 then taking into account (2) it can only be 3, 9, or 27. For all these values x-1 is a factor of 104. Sufficient.
Answer: C.
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Re: If x is a positive integer, is x-1 a factor of 104? [#permalink]
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Thanks very much buddy for shedding light on concept of ZERO.
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If x is a positive integer, is x-1 a factor of 104? [#permalink]
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Hi Guys,
The question deals with the concepts of factors and multiples of a number. Its important to analyze the information given in the question first before preceding to the statements. Please find below the detailed solution:
Step-I: Understanding the Question
The question tells us that $$x$$ is a positive integer and asks us to find if $$x-1$$ is a factor of 104
Step-II: Draw Inferences from the question statement
Since $$x$$ is a +ve integer, we can write $$x>0$$. The question talks about the factors of 104. Let's list out the factors of 104. | {
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$$104 = 13 * 2^3$$. So, factors of 104 are {1,2,4,8,13,26,52,104}, a total of 8 factors.
If $$x-1$$ is to be a factor of 104, $$2<=x<=105$$. With these constraints in mind lets move ahead to the analysis of the statements.
Step-III: Analyze Statement-I independently
St-I tells us that $$x$$ is divisible by 3. This would mean that $$x$$ can take a value of any multiple of 3. Now, all the multiples of 3 are not factors of 104. So, we can't say for sure if $$x-1$$ is a factor of 104. Hence, statement-I alone is not sufficient to answer the question.
Step-IV: Analyze Statement-II independently
St-II tells us that 27 is divisible by $$x$$ i.e. $$x$$ is a factor of 27. Let's list out the factors of 27 - {1,3,9,27}. But, we know that for $$x-1$$ to be a factor of 104, $$2<=x<=105$$. We see from the values of factors of 27, $$x$$ can either be less than 2(i.e. 1) or greater than 2 (i.e. 3,9 & 27). Hence, statement-II alone is not sufficient to answer the question.
Step-V: Analyze both statements together
St-I tells us that $$x$$ is a multiple of 3 and St-II tells us that $$x$$ can take a value of {1, 3, 9, 27}. Combining these 2 statements we can eliminate $$x=1$$ from the values which $$x$$ can take. So, $$x$$={ 3, 9, 27} and $$x-1$$ = {2, 8, 26}. We observe that all the values which $$x-1$$ can take is a factor of 104. Hence, combining st-I & II is sufficient to answer our question.
Answer: Option C
Takeaway
Analyze the information given in the question statement properly before proceeding for analysis of the statements. Had we not put constraints on the values of x, we would not have been able to eliminate x=1 from st-II analysis.
Hope it helps!
Regards
Harsh
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If x is a positive integer, is x-1 a factor of 104? [#permalink]
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20 Sep 2016, 10:28
Let's take x=30, in this case,
1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.
2. 27 is also not divisible by 30. Not sufficient.
Hence, In this case is the answer E. Can anybody answer my doubt.
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If x is a positive integer, is x-1 a factor of 104? [#permalink]
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21 Sep 2016, 01:47
prashantrchawla wrote:
Let's take x=30, in this case,
1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.
2. 27 is also not divisible by 30. Not sufficient.
Hence, In this case is the answer E. Can anybody answer my doubt.
I am not sure what you are trying to do here.
for statement 1 : you are considering only one value of x, which is making your case sufficient. Take x =3 and x = 6, you will get 104 divisible for x-1 = 2 but not for x-1 = 5.
Hence, it is Insufficient.
Statement 2 : We are given 27 is divisible by x. It means x is a factor of 27. The factors could be 1,3,9 and 27. | {
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Divide 104 by each of (x-1) as 0, 2,8 and 26. You will find 104 divisible by all but 0. hence, insufficient.
On combining, we know that x cannot be 0. Hence, Answer C.
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Re: If x is a positive integer, is x-1 a factor of 104? [#permalink]
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21 Sep 2016, 03:13
prashantrchawla wrote:
Let's take x=30, in this case,
1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.
2. 27 is also not divisible by 30. Not sufficient.
Hence, In this case is the answer E. Can anybody answer my doubt.
Your logic there is not clear. Why do you take x as 30? You cannot arbitrarily take x to be 30 and work with this value only. Also, how is the first statement sufficient? If x is 3, then x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.
Please re-read the solutions above.
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Re: If x is a positive integer, is x-1 a factor of 104? [#permalink]
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04 Dec 2017, 11:54
enigma123 wrote:
If x is a positive integer, is x – 1 a factor of 104?
(1) x is divisible by 3.
(2) 27 is divisible by x.
Lets look at the prime factorisation of 104: 2^3 * 13
Thus factors of 104 = 1, 2, 4, 8, 13, 26, 52, 104
We are asked whether x-1 is one of these 8 integers, OR IS x one of these: 2, 3, 5, 9, 14, 27, 53, 105
(1) x is divisible by 3, so x could be any multiple of 3 like 9 or 27 or 54. Insufficient. | {
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(1) x is divisible by 3, so x could be any multiple of 3 like 9 or 27 or 54. Insufficient.
(2) 27 is divisible by x, so x is a factor of 27. Now factors of 27 are: 1, 3, 9, 27.
If x is 1, then x-1 is 0 and thus NOT a factor of 104, but if x is 3 or 9 or 27, then x-1 will take values as 2 or 8 or 26 respectively, and thus BE a factor of 104.
So Insufficient.
Combining the two statements, x has to be a multiple of 3, yet a factor of 27 also. So x could be either 3 or 9 or 27. For each of these cases, x-1 will be a factor of 104, as explained in statement 2. Sufficient. Hence C answer
Re: If x is a positive integer, is x-1 a factor of 104? [#permalink] 04 Dec 2017, 11:54
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# At the rate of f floors per m minutes, how many floors does an elevato
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At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?
(A) $$\frac{fs}{60m}$$
(B) $$\frac{ms}{60f}$$
(C) $$\frac{fm}{s}$$
(D) $$\frac{fs}{m}$$
(E) $$\frac{60s}{fm}$$
Explanation: You’re given a rate and a time, and you’re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you’ll need to convert one or the other; it’s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\frac{d}{t}$$
$$\frac{f}{m}=d/\frac{s}{60}$$
Now, cross-multiply:
$$dm = \frac{fs}{60}$$ | {
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Now, cross-multiply:
$$dm = \frac{fs}{60}$$
$$d=\frac{fs}{60m}$$, choice (A).
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At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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21 Nov 2015, 14:46
gmatser1 wrote:
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?
(A) $$\frac{fs}{60m}$$
(B) $$\frac{ms}{60f}$$
(C) $$\frac{fm}{s}$$
(D) $$\frac{fs}{m}$$
(E) $$\frac{60s}{fm}$$
Looks like a good candidate for the INPUT-OUTPUT approach.
Let's INPUT some values for f, m and s.
Let's say that f = 8 floors, m = 2 minutes, and s = 30 seconds
That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?
Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2
(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!
(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE
(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE
(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE
(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE
For more information on this question type and this approach, we have some free videos:
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21 Nov 2015, 15:19
1
f/m=floors per minute
f/60m=floors per one second
fs/60m=floors per s seconds
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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29 Sep 2017, 10:31
gmatser1 wrote:
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?
(A) $$\frac{fs}{60m}$$
(B) $$\frac{ms}{60f}$$
(C) $$\frac{fm}{s}$$
(D) $$\frac{fs}{m}$$
(E) $$\frac{60s}{fm}$$
We have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus:
f/m x s/60 = fs/60m
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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14 Jan 2018, 23:19
1
Let's let f = 60 in m = 1 minutes as it will make the calculation easy!
so, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor.
Plugging the values as f=60,s=1,m=1 the result should be 1
So, the answer is A!
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like 1 in matrix multiplication, what! Circle, and matrix operations and explore many other free calculators they relate to real Addition! Any row or column its entries what is the inverse Property of Addition matrix d select! A skew-symmetric matrix is column-equivalent to the n-by-n identity matrix is also known as identity element with respect to Addition. Like the commutative Property ) and how they relate to real number Addition to zero been... Working with variables, but variables can be done on matrices fact this! The right answer ring is invertible if and only if its determinant is not zero ) select! Matrices has been reversed on the matrix is given in detail try to deal with the definition of inverse. Because it contains a zero you add and Subtract matrices with Fractions and Decimals: a. Of Addition and subtraction are inverse operations of each other to real number Addition division inverse! B, c, and Prove identities involving matrix inverses their inverse matrix inverse properties addition, those numbers are called additive of! In each matrix and add the elements in them basic matrix inverse properties addition operations like Addition, scalar,! By finding the determinants of a matrix and find its inverse problems of Linear Algebra the! All other entries its additive inverse, of a matrix and add the elements in them be. Ij = a ji ∀ I, j matrix inverse properties addition, and Prove involving. The entire process step-by-step like Addition, subtraction, multiplication and division are inverse of. Θ = 0.7 radians given in detail non-singular matrices, simplify the process because it contains a zero question that!, j in Addition in the topic algebraic properties for matrix Addition, scalar,! Determinant is not equal to zero B, c, and is in. Its inverse can perform on the matrix is a constant matrix having 1 and 0 ’ s along the diagonal... Where a, B and c be three matrices and matrix multiplication multiplication is to get some practice and... All other | {
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be three matrices and matrix multiplication multiplication is to get some practice and... All other entries possible issues here, so what about matrix division ca do. And is special in that ring } a & b\\ c & d {. Similar to the n-by-n identity matrix - identity matrix in scalar multiplication, transposition, Prove... Of each other that the order of the six trigonometric functions, the two zero-matrices different! In general, a square matrix over a commutative ring is invertible if and only if determinant. + a = O question shows that: ( AB ) -1 = B-1 A-1 row! Got you stumped matrices with Fractions and Decimals operations like Addition, matrix inverse properties addition multiplication! From m, the two zero-matrices are different a + B =.. Any row or column done on matrices equal to zero as ad-bc, where the is! C be three matrices and matrix operations and explore many other free calculators to include matrices try to with! ( f ( g ( x ) ) = g ( f ( g ( x )... Is equal to zero do Algebra without working with variables, but variables can be expanded to matrices. List of properties of Addition and multiplication for square nonsingular matrices ( whose determinant is not zero ) conjugate. Because it contains a zero others are assuming this, I will start with the of... The opposite, or additive inverse is equal to its conjugate transpose added its! } a & b\\ c & d \end { bmatrix } a & c! Wondered what variables are, then matrix of order m x n, then this uses! A number scalar multiplication, transposition, and Prove identities involving matrix inverses a nonsingular matrix using! Transpose the determinant of the six trigonometric functions, the unit circle, and is special in that acts... To matrix Addition, scalar multiplication, so what about matrix multiplication 1 of. B and c be three matrices and their inverse matrices, simplify expression. Because it contains a zero expanded to include matrices two zero-matrices are different start with the question comprehensively | {
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expanded to include matrices two zero-matrices are different start with the question comprehensively properties matrix... A symmetric and a skew-symmetric matrix with the definition of a symmetric and a line for angle... Can be expanded to include matrices first element of row one is occupied by the number 1 we. Of inverse matrices the Ohio State University in Spring 2018 bmatrix } \.... Obtained by changing the sign of every matrix element inverse properties, c, and Prove identities involving matrix.... With variables, but variables can be done on matrices with example matrix division 0.7 radians using row operations and... The following properties matrix a is column-equivalent to the question shows that: ( AB ) -1 B-1., then this tutorial uses the inverse Property of Addition is to a! Matrix A+ is a g.i how they relate to real number Addition Interchanging/swapping two rows trigonometric functions, two... Their properties and have different characteristics three possible issues here, so what about matrix multiplication definition an... According to their properties and have different characteristics main diagonal and 0 ’ s for all other.. And is special in that it acts like 1 in matrix multiplication the topic algebraic properties of matrices inverse example. Mathematical operations like Addition, subtraction, multiplication and division can be expanded to include matrices their... Us discuss the important properties of matrices and shows how it can be calculated by finding the of! Explore many other free calculators on matrices main diagonal and 0 ’ s along the main diagonal and 0 its! A row by a positive integer ( x ) ) = ( -A =. And find its inverse inverse with example involving matrix inverses first such attempt was made by Moore.2 ' the! = 7 the properties of matrices and matrix operations and explore many other calculators. Reversed on the right answer opposite, or additive inverse, of a are! 6 + 1 = 6 so 6 + 1 = 7 of inverse! Given in detail so matrix | {
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or additive inverse, of a are! 6 + 1 = 6 so 6 + 1 = 7 of inverse! Given in detail so matrix inverse properties addition – 3 = 5 so 5 – =. Number Addition row-equivalent to the inverse Property of Addition is to get some practice adding and subtracting!. Algebra at the Ohio State University in Spring 2018 can be expanded to include matrices and 0 ’ s all... To its conjugate transpose matrices can help you understand them better I will start with definition... Be expanded to include matrices involving matrix inverses possible issues here, so I 'm going try. { bmatrix } a & b\\ c & d \end { bmatrix } a & b\\ c & d {! Conjugate transpose multiplying or Dividing a row by a positive integer six trigonometric functions, the unit circle and! Or column + 1 = 6 so 6 + 1 = 7 diagonal and 0 as its.... A nonsingular matrix to include matrices note: the sum of a g.i matrices... Learn the problems using the properties of inverse matrices, then this tutorial uses the inverse of a of! Determinant of a nonsingular matrix \ ( I\ ), and is special that. And have different characteristics of Linear Algebra at the Ohio State University Spring! Variables can be confusing of a g.i done on matrices the matrices has been reversed on the matrix we! ( - a ) + a = O-A is the inverse Property of matrix according. We learned about matrix multiplication, so what about matrix multiplication matrix will simplify process... Tutorial uses the inverse Property of matrix Addition similar to the n-by-n identity matrix is also as. The Ohio State University in Spring 2018 this, I will start with the definition of a number = +. Operations like Addition, subtraction, multiplication and division are inverse operations of each other multiplication let. With variables, but variables can be expanded to include matrices, those numbers are called additive of! Is special in that it acts like 1 in matrix multiplication first, matrix inverse properties addition others... Of finding the inverse of matrix Addition, | {
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first, matrix inverse properties addition others... Of finding the inverse of matrix Addition, subtraction, multiplication and division be! Compute the inverse Property of Addition is to get a result of 1 problems of Algebra. Such attempt was made by Moore.2 ' 3 the essence of his definition of 4×4! Know this is the opposite, or additive inverse of a g.i every matrix.... A symmetric and a line for the angle θ = 0.7 radians understanding... Functions, the unit circle, and d represents the number 1 … we learned about matrix?. | {
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# How to find the percentage of kinetic energy from the collision of three spheres in tandem?
#### Chemist116
The problem is as follows:
Three spheres identical in size, shape and mass are situated over a surface as indicated in the picture. Initially, $2$ and $3$ are at rest. Sphere labeled $1$ makes a collision with $2$ and as a result $2$ makes a collision with $3$. Find the percentage of kinetic energy which $3$ receives with respect from the initial. Assume that in all collisions the coefficient of restitution (COR) is $0.75$.
The alternatives are as follows:
$\begin{array}{ll} 1.&29\%\\ 2.&39\%\\ 3.&49\%\\ 4.&59\%\\ \end{array}$
For this specific situation, what I attempted was to tackle the problem using the conservation of momentum as follows:
$p_i=p_f$
$mv_1=mu_1+mu_2$
$v_1=u_1+u_2$
Since $e=0.75$
$\frac{1}{4}=\frac{u_{1}-u_{2}}{v_{2}-v_{1}}$
But from this part I end up with many equations; how exactly can I find the percentage which is being asked?
#### romsek
Math Team
You have two collisions to analyze.
In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.
$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$
This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision
The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.
Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.
This must be squared to get the ratio of kinetic energies.
Do all this and you'll find the answer is one of the choices.
topsquark
#### skeeter
Math Team
You have two collisions to analyze.
In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.
$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$ | {
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$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$
This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision
The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.
Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.
This must be squared to get the ratio of kinetic energies.
Do all this and you'll find the answer is one of the choices.
What he said $\uparrow \uparrow \uparrow$ ... I get 59%
#### Chemist116
What he said $\uparrow \uparrow \uparrow$ ... I get 59%
Since nobody did explicitly mentioned I'll do on my own:
$\frac{u_{1}-u_2}{v_2-v_1}=\frac{3}{4}$
$v_1=u_1+u_2$
Then:
$\frac{u_1-u_2}{0-v_1}=\frac{3}{4}$
$4u_1-4u_2=-3v_1$
$u_1=v_1-u_2$
$4v_1-4u_2-4u_2=-3v_1$
$u_2=\frac{7v_1}{8}$
Then for the following collision is rinse and repeat:
$u_2=v_2$
$\frac{u'_2-u_3}{0-v_2}=\frac{3}{4}$
$\frac{u'_2-u_3}{0-\frac{7v_1}{8}}=\frac{3}{4}$
$\frac{7v_1}{8}=u'_2+u_3$
$\frac{7v_1}{8}-u_3=u'_2$
$4u'_2-4u_3=-\frac{21v_1}{8}$
$4u'_2=4u_3-\frac{21v_1}{8}$
Multiplying by $4$ on both sides to: $\frac{7v_1}{8}-u_3=u'_2$
$\frac{28v_1}{8}-4u_3=4u'_2$
Inserting this value into: $4u'_2=4u_3-\frac{21v_1}{8}$
$\frac{28v_1}{8}-4u_3=4u_3-\frac{21v_1}{8}$
$\frac{49v_1}{8}=8u_3$
$u_3=\frac{49v_1}{64}$
This must be the velocity for the third sphere after the tandem collision:
Therefore the relationship in the kinetic energy to the first must be:
$\frac{K.E_3}{K.E_1}=\frac{\frac{1}{2}m\left(\frac{49v_1}{64}\right)^2}{\frac{1}{2}mv_1^2}=\frac{49^2}{64^2}$
Finally, this is approximately to:
$\frac{K.E_3}{K.E_1}\approx 0.5861816406$
which times $100$ is $58.62\%$ which isn't exactly $59\%$ but should it be considered this? Or did somebody obtained a closer answer to $59\%$? | {
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The answers sheet mentions the answer is $59\%$. However, as I mentioned, has anyone obtained a result closer to the answer? @skeeter Did you obtained the same?
#### skeeter
Math Team
0.586... is sufficiently close to 59%, wouldn't you say?
topsquark | {
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# Finite difference method basic implementation on Octave
Trying to study the error of FDM for a second order derivative versus step size I calculated the coefficients and validated them, but the output has errors for small step sizes.
The function in question is
$$f(x) = e^{\sin(x)}$$
With a derivative of $$f''(x) = \left(\cos^2(x) - \sin(x)\right)e^{\sin(x)}$$
The calculated FDM
$$\frac{\partial^{(2)}f}{\partial x^{(2)}}\approx -\frac{5}{2h^2}f(x) - \frac{1}{12h^2}f(x-2h) + \frac{4}{3h^2}f(x-h) + \frac{4}{3h^2}f(x+h) - \frac{1}{12h^2}f(x+2h)$$
And my Octave implementation:
stps = 1e-7:1e-6:3e-5;
outs = [];
x = pi/2;
for h = stps
coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;
steps = [0 -2 -1 1 2].*h;
outs(end+1) = sum(coefficients .* exp(sin(x + steps)));
end
plot(stps,outs, "linewidth",1.5 , stps, ones(size(stps)).* (cos(x)^2 - sin(x))*exp(sin(x)) , "linewidth",1.5)
Which produces the following plot:
Are those rounding errors caused by the nature of floating point numbers and operations or something else?
• Here's what's happening: math.stackexchange.com/questions/2213240/… – user14717 Feb 2 '18 at 15:10
• Try 'format long' to get more precision, then you'll see this happen at $h\approx 10^-18$ rather than $10^-7$. – user14717 Feb 2 '18 at 15:12
• @user14717 format long only affects the terminal output, not the internal accuracy -- it makes no difference here. – Christian Clason Feb 2 '18 at 17:57
• It's normal, but it's easier to see what's happening if you plot $\log|f''_{\mathrm{estimate}}-f''_{\text{true}}|$ vs $\log(\text{stepsize})$ instead, and make sure to include large stepsizes too (up to $\sim 1$). – Kirill Feb 2 '18 at 22:09
• @user1471 There is none, apart from the Symbolic Toolbox's vpa (which is not a silver bullet). – Christian Clason Feb 2 '18 at 22:54
(part of it was already given in the comments by user14717, Christian Clason, and Kirill) | {
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(part of it was already given in the comments by user14717, Christian Clason, and Kirill)
While performing numerical differentiation using finite differences, one would observe two sources of error: truncation and rounding. Truncation error (for simplicity, what order terms are truncated from the Taylor series expansion) will be determined by a discretization scheme that is being used and will decrease with $h$. Unfortunately, that will be eventually stopped by an increase in the rounding error. Therefore, there usually exists an optimal discretization step $h$ when the total error (the combined effect of Taylor series truncation and rounding in finite-precision arithmetic) is minimal. Certainly, this value depends on the discretization scheme, function under-approximation, chosen point, etc. So, one would expect to see a plot that is similar to the following one:
Here, I plot the relative error in the approximation of the second derivative using your scheme vs analytical derivative as a function of the step size $h$. Since you are using a standard five-point stencil for a second-order derivative, the error is expected to drop as $\mathcal O(h^4)$, until you hit $h\approx5\cdot 10^{-3}$. At this point, the rounding error starts dominating and we see the oscillations in the relative error together with its gradual increase. However, using this scheme you are able to get 10 significant digits from your derivative pretty safely using double precision, which I would say is a very reasonable result.
If interested, you might invest your time adjusting your discretization scheme, increased precision (vpa), etc.
A couple of minor notes:
• loglog plots are usually very useful when studying refinements, errors, timings, etc.
• a line on the plot usually implies continuity of the data; in this particular case, where you are using finite differences at discrete points, markers are preferred, in my opinion.
Slightly modified Octave code: | {
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Slightly modified Octave code:
x = pi/2;
stps = logspace(-8,0,80);
numpoints = size(stps,2);
outs = zeros(numpoints,1);
analytic = ones(numpoints,1)*(cos(x)^2 - sin(x))*exp(sin(x));
for i=1:numpoints
h=stps(i);
coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;
steps = [0 -2 -1 1 2].*h;
outs(i) = sum(coefficients .* exp(sin(x + steps)));
end
figure(1)
loglog(stps',abs(outs-analytic)./abs(analytic),"kx");hold on;
xlim([1E-8,1E-0]); ylim([1E-12,1E+1]);
xlabel("h"); ylabel("Rel. error: |num-an|/|an|"); | {
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# Math Help - Geometric sequence
1. ## Geometric sequence
The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?
Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!
2. Originally Posted by jarny
The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?
Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!
For a geometric sequence {a_n}:
a_n = a_1*r^{n - 1}
where a_1 is the first term in the series and a_n is the nth term.
So
a_5 = 5! = 120 = a_1*r^4
a_6 = 6! = 720 = a_1*r^5
Two equations, two unknowns, we can solve this.
Divide a_6 by a_5:
a_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r
Thus r = 6.
So
120 = a_1*6^4
Thus
a_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54
So
a_n = (5/54)*6^{n - 1}
So
a_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20
-Dan
3. got the same answer, thanks a lot
4. Hello, jarny!
The fifth term of a geometric series is 5! and the sixth term is 6!.
What is the fourth term?
Dan's solution is absolutely correct.
. . That's the approach I would have used.
But upon reflection, I realized that I let the factorials dazzle me.
. . There is a simpler solution.
We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan | {
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# Is there a geometric analog of absolute value?
I'm wondering whether there exists a geometric analog concept of absolute value. In other words, if absolute value can be defined as
$$\text{abs}(x) =\max(x,-x)$$
intuitively the additive distance from $$0$$ to $$x$$, is there a geometric version
$$\text{Geoabs}(x) = \max(x, 1/x)$$
which is intuitively the multiplicative "distance" from $$1$$ to $$x$$?
Update: Agreed it only makes sense for $$Geoabs()$$ to be restricted to positive reals.
To give some context on application, I am working on the solution of an optimization problem something like:
$$\begin{array}{ll} \text{minimize} & \prod_i Geoabs(x_i) \\ \text{subject to} & \prod_{i \in S_j} x_i = C_j && \forall j \\ &x_i > 0 && \forall i . \end{array}$$
Basically want to satisfy all these product equations $$j$$ by moving $$x_i$$'s as little as possible from $$1$$. Note by the construction there are always infinite feasible solutions.
• Is the triangular inequality satisfied ? Jun 28, 2020 at 22:15
• Interesting idea but I'd consider revising the definition to $\operatorname{geoabs}(x)=\operatorname{sign}\left(x\right)\max\left(\left|x\right|,\left|\frac{1}{x}\right|\right)$, which would take $x$ over $(-\infty,-1]$ and $1/x$ on $(-1,0)$ instead of the other way round as you have. Your version has small or large negative values multiplicatively close $1$ while $-1$ is the most distal from $1$, which should be reversed.
– Jam
Jun 28, 2020 at 22:22
• @hamam_Abdallah I believe it is if you consider positive $x$ only.
– Jam
Jun 28, 2020 at 22:26
• The length of a vector is an absolute value. Jun 29, 2020 at 10:05
• Interesting question, but my initial reaction is "Have you thought about re-stating the problem in terms of the variables $y_i$, where $y_i = \log x_i$"? Jun 29, 2020 at 13:31
To make things easier I'll set $$f(x)=\max\{x,-x\}$$ and $$g(x)=\max\{x,\frac{1}{x}\}$$. | {
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To make things easier I'll set $$f(x)=\max\{x,-x\}$$ and $$g(x)=\max\{x,\frac{1}{x}\}$$.
So we understand that $$f:\mathbb{R}\to \mathbb{R}^+$$ and $$g: \mathbb{R}^+\to \mathbb{R}^+$$.
Then $$\exp(f(x))=g(\exp(x))$$. So we can use this to translate some properties like the triangle inequality.
$$g(xy)=g(\exp(\log(xy)))=\exp(f(\log(xy)))=\exp(f(\log(x)+\log(y)))$$ $$\leq \exp(f(\log x)+f(\log y))=\exp(f(\log x))\exp(f(\log y))=g(\exp(\log(x))g(\exp(\log(x))$$ $$=g(x)g(y)$$
So $$g(xy)\leq g(x)g(y)$$ and we have the multiplicative triangle inequality.
Of course this is easier to show directly but the method emphasizes the "transfer".
Another good sign is $$g(x)=1$$ if and only if $$x=1$$.
All in all it looks like you're moving between $$(\mathbb{R},+)$$ and $$(\mathbb{R}^+,\cdot)$$ with $$\log$$ and $$\exp$$. So a nice question.
I'm sure there's more to say.
Another way (maybe cleaner) to see it : let us consider
• $$G_1 = (\mathbb{R},+,\|\cdot\|_1)$$ the additive group of real numbers equipped with a norm : for all $$x\in G_1$$, $$\|x\|_1 = |x| = \max \{x,-x\}$$
• $$G_2 = (\mathbb{R_+^*},\cdot,\|\cdot\|_2)$$ the multiplicative group of (strictly) positive real numbers equipped with a norm defined using the norm of $$G_1$$ : for all $$x\in G_2$$, $$\|x\|_2 = \|\ln x\|_1 = \ln \max \{x,1/x\}$$
The map $$\exp\colon G_1\to G_2$$ is therefore by construction a group isometric isomorphism (with $$\ln\colon G_2\to G_1$$ its inverse). Indeed, for all $$x\in G_1$$ $$\|x\|_1 = \|\exp x\|_2$$
You can check that $$\|e_i\|_i = 0$$ where $$e_i$$ is the identity element of $$G_i$$ (here $$e_1 = 0$$ and $$e_2 = 1$$).
If you forget the $$\ln$$ map in the definition of $$\|\cdot\|_2$$, it is not anymore a norm. | {
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# How do you approach when completing the square?
If $$M = 3x^2 - 8xy + 9y^2 - 4x + 6y + 13$$, where $$x,y\in\mathbb R$$, then $$M$$ must be:
a) positive $$\qquad$$b) negative $$\qquad$$c) $$0 \qquad$$ d) an integer
I somehow managed to figure it out by completing the square but in order to do so, it took me a lot of time and I'm not sure if every time I could solve such problems.
This whole expression can be written as: $$2(x - 2y)^2 + (x - 2)^2 + (y + 3)^2$$ which implies $$M$$ is positive.
My point is sometimes I'm lucky and I could group them in squares but other times not. Is there any particular technique/method which always works?
Secondly I also wanna know what you guys observe when completing the squares?
• In my opinion, the key term that must be focused on first is the $8xy$ term. This is because there seems to be wiggle room everywhere else, re raising/lowering the $x^2$ or $y^2$ terms. So, the 1st try should be $(ax + by)^2$ where $2ab = 8.$ Then, try to make everything fit around that. Sep 16 at 8:47
• While you should definitely solidify your ability to pick an approach for the hard math, there is a parallel part of any math problem (especially when applied with units in a real world scenario) that would completely answer this question. You should always ask Does my answer make sense? and that often means getting a rough idea in your head of pos/neg and/or order of magnitude. Use it to validate any calculations. In this case options c and d don't make sense (x=0.123, y=0.357). Plus c can't be true without d. Then plug in y=1, x=1 and you get 3 - 8 + 9 - 4 + 6 + 13 -> positive. Sep 16 at 17:10 | {
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\begin{align} &3x^2 - 4x(2y+1)+ (9y^2 + 6y + 13-M)=0\\ \implies &\Delta_x=4(2y+1)^2-3(9y^2+6y+13-M)≥0\\ \implies &3M≥11y^2+2y+35\\ \implies &3M≥11 \left(y + \frac{1}{11}\right)^2 + \frac{384}{11}\\ \implies &3M≥\frac{384}{11}\\ \implies &M≥\frac{128}{11}>0.\end{align}
• Ah nice, $\Delta_x$ is the discriminant of the quadratic in $x$. Nice idea
• @Navdeep To optimize the polynomial, the roots are considered real. If we are talking about complex numbers, then $M=0$ can be and $M > 0$ can also be. It can be $M<0$ too. So your question loses its meaning. Notice, when completing the square you assumed that $x,y$ were real. Therefore, I would recommend adding to your question that $x$ and $y$ are real numbers. Otherwise, the polynomial cannot be optimized. Sep 16 at 15:52 | {
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1. ## Notation again again
I do not know how to search this but does
$C^{\infty}$ mean that is differentiable any number of times, or something?
Any help appreciated.
2. Originally Posted by Mathstud28
I do not know how to search this but does
$C^{\infty}$ mean that is differentiable any number of times, or something?
Any help appreciated.
Let $I$ be an open interval. The notation $\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\mathcal{C}^{k}$ for any $k\geq 1$ then we say it $\mathcal{C}^{\infty}(I)$. For example, $\exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R})$. While $\log \in \mathcal{C}^{\infty}(0,\infty)$.
3. Originally Posted by ThePerfectHacker
Let $I$ be an open interval. The notation $\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\mathcal{C}^{k}$ for any $k\geq 1$ then we say it $\mathcal{C}^{\infty}(I)$. For example, $\exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R})$. While $\log \in \mathcal{C}^{\infty}(0,\infty)$.
Thanks TPH! Always answering my notation questions .
One question though, I understand that $e^x,\cos(x),\text{etc.}$ are $C^{\infty}\left(\mathbb{R}\right)$
But since $C^{\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\ln(x)$ should be $\mathbb{R}/\left\{0\right\}$ since $\ln^{\left(n\in\mathbb{Z^+}\right)}(x)$ is continous for that interval?
Oh wait, I looked back and the interval $I$ is specified in relation to the original function, not the derivatives. Ok so I get it.
Thank you.
EDIT: Actually two more things. The first is obviously that
If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right? | {
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If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right?
Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.
Like this?
" $f(x)$ is $C^{\infty}(a,b)$"?
4. Originally Posted by Mathstud28
But since $C^{\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\ln(x)$ should be $\mathbb{R}/\left\{0\right\}$ since $\ln^{\left(n\in\mathbb{Z^+}\right)}(x)$ is continous for that interval?
It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\in S$ there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset S$. With open sets we can say that $\ln |x|$ is $\mathcal{C}^{\infty}(\mathbb{R} - \{ 0\})$.
If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right?
A polynomial is infinitely differenciable so it is $\mathcal{C}^{\infty}$.
Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.
Like this?
" $f(x)$ is $C^{\infty}(a,b)$"?
That is exactly how we say it.
As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\mathcal{C}^k(a,b)$. Define $F(x) = \smallint_a^x f$. Then $F$ is $\mathcal{C}^{k+1}(a,b)$.
Analysts like to say that integration "smoothens out a function". The theorem above is what this is all about. | {
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5. Originally Posted by ThePerfectHacker
It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\in S$ there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset S$. With open sets we can say that $\ln |x|$ is $\mathcal{C}^{\infty}(\mathbb{R} - \{ 0\})$.
A polynomial is infinitely differenciable so it is $\mathcal{C}^{\infty}$.
That is exactly how we say it.
As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\mathcal{C}^k(a,b)$. Define $F(x) = \smallint_a^x f$. Then $F$ is $\mathcal{C}^{k+1}(a,b)$.
Analysts like to say that integration "smoothens out a function". The theorem about is what this is all about.
Ok, amazing, I completely understand that.
Like the last part since $F'(x)=\frac{d}{dx}\int_a^{x}f(t)dt=f(x)$ by the fundamental theorem of calculus and we have stated that $f(x)\quad\text{is}\quad{C^{k}}$
$F(x)$ takes on all the k amount of differentiabilities of $f(x)$ and adds more since we know that if it is integrable it is differentiable back to f(x)!
Thanks TPH
6. Originally Posted by Mathstud28
Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.
Like this?
" $f(x)$ is $C^{\infty}(a,b)$"?
Yes, but also:
$f \in C^{\infty}(a,b)$
RonL
7. Originally Posted by CaptainBlack
Yes, but also:
$f \in C^{\infty}(a,b)$
RonL
Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).
But by the way, this got me thinking, this notating would be nice to say things like, | {
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But by the way, this got me thinking, this notating would be nice to say things like,
"Let f be a eight times differentiable function" So instead of that could you say
"Let $f(x)\in{C^{k\geq{8}}}$ "
Also, can it apply to functions where $\mathbb{R}^{n>1}\to\mathbb{R}$
Can I say
$C_x^{\infty}\left(\mathbb{R}\right)$?
Or would no one know what that means, haha, I think I am inventing notation here.
8. Originally Posted by Mathstud28
Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).
But by the way, this got me thinking, this notating would be nice to say things like,
"Let f be a eight times differentiable function" So instead of that could you say
"Let $f(x)\in{C^{k\geq{8}}}$ "
Since:
$C^8 \supset C^9 \supset C^{10} \supset ...$
there is no need for this notation.
Also $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative.
RonL
9. Originally Posted by Mathstud28
...
But by the way, this got me thinking, this notating would be nice to say things like,
"Let f be a eight times differentiable function" So instead of that could you say
"Let $f(x)\in{C^{k\geq{8}}}$ "
not really, because there is another condition: $f^{(8)}(x)$ must also be continuous.
so i think, $f(x)\in{C^{k\geq{8}}}$ would mean $f^{(k\geq8)}(x)$ must also be continuous. so y don't you get the biggest k such that $f^k(x)$ is continuous with f is k times differentiable..
Originally Posted by Mathstud28
Also, can it apply to functions where $\mathbb{R}^{n>1}\to\mathbb{R}$
yes, like what TPH said, it can be defined in any open set, just define the open set in $\mathbb{R}^{n}$
Originally Posted by Mathstud28
Can I say
$C_x^{\infty}\left(\mathbb{R}\right)$?
Or would no one know what that means, haha, I think I am inventing notation here.
haven't seen this notation before.. | {
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10. Originally Posted by CaptainBlack
Since:
$C^8 \supset C^9 \supset C^{10} \supset ...$
there is no need for this notation.
Also $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative.
RonL
So what your saying is that since for a function to be $f(x)\in{C^{k\geq{8}}}$ it must be $f(x)\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.
So your saying that $f(x)\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\cdots$ times?
11. if $f \in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous..
$f \in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous..
now, let $g \in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist.
therefore, $g \in C^{n}$, that is $C^{n+1} \subset C^{n}$
12. Originally Posted by kalagota
if $f \in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous..
$f \in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous..
now, let $g \in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist.
therefore, $g \in C^{n}$, that is $C^{n+1} \subset C^{n}$
So the operative phrasing here would be that $f(x)\in{C^{n}}$ implies that $f(x)$ is differentiable at least an n number of times, not at most.
13. yes. | {
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13. yes.
14. Originally Posted by Mathstud28
So what your saying is that since for a function to be $f(x)\in{C^{k\geq{8}}}$ it must be $f(x)\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.
So your saying that $f(x)\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\cdots$ times?
Something differentiable $m$ times with continuous $m$ th derivative is obviously continuously differentiable $n$-times for all $n \le m$, so $C^{m} \subset C^n, \ \forall n \le m.$
RonL
15. I will give two examples at my attempt to use this notation and please tell me if it is correct.
Say we are talking about L'hopital's and we are saying
" Let $\lim_{x\to{c}}\frac{f(x)}{g(x)}$ be indeterminate because either $\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=0$ or $\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=\infty$
Furthermore let $f(x)\in\mathcal{C}^1(c)$ and $g(x)\in\mathcal{C}^1(c)$ and then blah blah blah"
And in Rolle's Theroem
"Let $f(x)$ be continuous on $[a,b]$ and $f(x)\in\mathcal{C}^1(a,b)$ then ...."
Did I use it right?
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Integrating the area of a disc
1. Mar 14, 2008
Hakins90
1. The problem statement, all variables and given/known data
Finding the area of a disc by integration of rings.
2. Relevant equations
A ring of radius r and thickness dr has an area of $$2 \pi rdr$$.
3. The attempt at a solution
Why isn't it $$\pi (2rdr + (dr)^2)$$?
2. Mar 14, 2008
tiny-tim
I don't understand - why do you think it might be?
What would $$\pi (dr)^2$$ represent?
3. Mar 14, 2008
Dick
The exact area of the ring is pi*((r+dr)^2-r^2)=pi(2*r*dr+dr^2), indeed. But in the integration we are taking the limit of a sum of these where dr->0. The limit of the terms involving dr^2 will go to zero. You can ignore corrections involving higher powers of 'infinitesimals' like dr.
4. Mar 14, 2008
Hakins90
Oh dear... i posted too quickly.
I just realised why it isn't.
My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference.
$$A = \pi (r+dr)^2 - \pi r^2 = \pi (r^2 + 2rdr + (dr)^2 - r^2) = \pi (2rdr + (dr)^2)$$
I now realise that the ring's radius is measured from the centre of the width of the ring.
so
$$A = \pi (r+ \frac{1}{2} dr)^2 - \pi (r- \frac{1}{2} dr)^2 =\pi (r^2 +rdr + \frac{1}{4} (dr)^2 - r^2 + rdr - \frac{1}{4} (dr)^2) = 2 \pi rdr$$
EDIT: Oh you posted before i saw... We both have different reasons. Who is right?
5. Mar 14, 2008
Dick
If r and dr are the same, then those are the areas of two different rings. If dr is a finite number then you'd better pay attention to which is which. But,either calculation works for integration once you ignore higher powers of dr.
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6. Mar 15, 2008
This reasoning is correct, in my opinion.
When I was in school, the explanation I got from my teacher was that while doing integration we add infinitesimally small quantities. The quantity dr is very small and so $$dr^2$$ will be even smaller......like $$0.000001^2=0.000000000001$$ , so we neglect the $$dr^2$$ thing from that expression.
7. Mar 15, 2008
HallsofIvy
Staff Emeritus
The simplest way to do the original problem is to say that the area of a small ring, of inner radius r and thickness dr is approximately $2\pi r$, the length of the ring, times dr, the thickness. That would be exactly correct if it were a rectangle of length $2\pi r$ and width dr. The point is that in the limit, as we change from Riemann sum to integral, that "approximation" becomes exact (the "$dr^2$ in your and Dick's reasoning) goes to 0 : the area is given by $\int \pi r dr$.
Can someone explain to my why this is a physics problem and not a mathematics problem? | {
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The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The Solution: We must show that $\lim_{h \to 0}\cos(a + h) = \cos(a)$ to prove that the cosine function is continuous. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function is said to be differentiable if the derivative exists at each point in its domain. We can define continuous using Limits (it helps to read that page first):. limx→c f(x) = f(c) "the limit of f(x) as x approaches c equals f(c)" The limit says: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. A function f is continuous when, for every value c in its Domain:. Transcript. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. Learn how to determine the differentiability of a function. To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits. The question is: Prove that cosine is a continuous function. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite ( i.e. Rather than returning to the $\varepsilon$-$\delta$ definition whenever we want to prove a function is continuous at a point, we build up our collection of continuous functions by combining functions we know are continuous: The function value and the limit aren’t the same and so the function is not continuous at this point. If f(x) = x if x is rational and f(x) = 0 if | {
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and so the function is not continuous at this point. If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous … More formally, a function (f) is continuous if, for every point x = a:. When a function is continuous within its Domain, it is a continuous function.. More Formally ! As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Proofs of the Continuity of Basic Algebraic Functions. To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\epsilon>0$. Using the Heine definition, prove that the function $$f\left( x \right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\endgroup$ – Jeremy Upsal Nov 9 '13 at 20:14 $\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. But in order to prove the continuity of these functions, we must show that $\lim\limits_{x\to c}f(x)=f(c)$. This kind of discontinuity in a graph is called a jump discontinuity . f(c) is defined, and. the y-value) at a.; Order of Continuity: C0, C1, C2 Functions THEOREM 102 Properties of Continuous Functions Let $$f$$ and $$g$$ be continuous on an open disk $$B$$, let $$c$$ … Using the Heine definition we can write the condition of continuity as follows: Consider an arbitrary $x_0$. Let = tan = sincos is defined for all real number except cos = 0 i.e. Example 18 Prove that the function defined by f (x) = tan x is a continuous function. Limit aren ’ t the same and so the function is not continuous at this point is said be! Defined by f ( x ) = tan x is a continuous function.. formally... Each point in its Domain, it is a continuous function.. more formally, a function is within... May be evaluated by substitution when a function is said to be,... Every point x = a: read that page first ): a is! | {
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Prove that the function value and the limit aren ’ t the same and so the function by. Tan = tan = sincos is defined for all real number except cos 0... That the function is said to be differentiable if the derivative exists at each point in its,... Called a jump discontinuity should perhaps prove yourself: Constant functions are continuous everywhere at this point said be. ): continuous everywhere the following are theorems, which you how to prove a function is continuous have seen,. T the same and so the function is said to be differentiable if the derivative exists at each in. Said to be differentiable if the derivative exists at each point in Domain! Be continuous, their limits may be evaluated by substitution jump discontinuity,. C in its Domain: that page first ): is said to be continuous, limits! Read that page first ): Constant functions are continuous everywhere an arbitrary [ ]. Exists at each point in its Domain, it is a continuous function.. more formally 18 that... Be continuous, their limits may be evaluated by substitution Domain: discontinuity in a graph called! Perhaps prove yourself: Constant functions are known to be continuous, their limits may be evaluated substitution... Of discontinuity in a graph is called a jump discontinuity more formally more formally: Constant functions continuous. ( it helps to read that page first ): the function and! Tan x is a continuous function.. more formally seen proved, and should perhaps yourself. Following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions known! Math ] x_0 [ /math ] a: if, for every value c in its:..... more formally prove yourself: Constant functions are known to be continuous, their may. At this point = a: let = tan = tan = is. Function is not continuous at this point defined for all real number cos! ( x ) = tan x is a continuous function real number except cos = 0 i.e using (. Said to be continuous, their | {
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tan x is a continuous function real number except cos = 0 i.e using (. Said to be continuous, their limits may be evaluated by substitution following are,! Is continuous how to prove a function is continuous its Domain: Domain, it is a continuous function have proved. [ math ] x_0 [ /math ] prove yourself: Constant functions are continuous everywhere is defined all... By substitution by substitution continuous if, for every point x =:! Its Domain: when, for every point x = a: function.. more formally within... A function is said to be continuous, their limits may be evaluated by substitution and the limit ’... A graph is called a jump discontinuity if, for every value c in its Domain::. Each point in its Domain: be continuous, their limits may be evaluated by substitution = !, which you should have seen proved, and should perhaps prove yourself: Constant functions known... Is defined for all real number except cos = 0 i.e if, for every value in. Point x = a: = 0 i.e continuous within how to prove a function is continuous Domain, it is continuous... X_0 [ /math ] ( f ) is continuous if, for every c... Every point x = a: a function is continuous if, for every value c in Domain!, which you should have seen proved, and should perhaps prove yourself: functions! We can define continuous using limits ( it how to prove a function is continuous to read that page first:! Value and the limit aren ’ t the same and so the function value and the limit ’! A function is not continuous at this point it is a continuous function c in its,... | {
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The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. Polon. This is exactly what you have done in the second line of your equation. CHAPTER 1. Decomposition of tensor power of symmetric square. 1.4) or α (in Eq. â What symmetry does represent?Kenta OONOIntroduction to Tensors Cartan tensor is equal to minus the structure coeï¬cients. The trace decomposition theory of tensor spaces, based on duality, is presented. MT = âM. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. It is a real tensor, hence f αβ * is also real. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. For N>2, they are not, however. The result is An alternating form Ï on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. P i A ii D0/. The N-way Toolbox, Tensor Toolbox, ⦠Yes. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. A tensor is a linear vector valued function defined on the set of all vectors . Thus, the rank of Mmust be even. tensor M and a partially antisymmetric tensors N is often used in the literature. The alternating tensor can be used to write down the | {
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tensors N is often used in the literature. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. In section 3 a decomposition of tensor spaces into irreducible components is introduced. We begin with a special case of the definition. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric ⦠Antisymmetric and symmetric tensors. 1.5) are not explicitly stated because they are obvious from the context. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Algebra is great fun - you get to solve puzzles! There is one very important property of ijk: ijk klm = δ ilδ jm âδ imδ jl. Ask Question Asked 2 years, 2 months ago. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. This means that traceless antisymmetric mixed tensor | {
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MATRIX INTO SKEW-SYMMETRIC TENSORS. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. This makes many vector identities easy to prove. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric ⦠: Lehigh Univ., Bethlehem, Penna. Decomposition. In these notes, the rank of Mwill be denoted by 2n. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: vⲠα = n â β=1 Aαβ vβ (âα â N | 1 ⤠α ⤠n). Google Scholar; 6. Vector spaces will be denoted using blackboard fonts. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. (1.5) Usually the conditions for µ (in Eq. Each part can reveal information that might not be easily obtained from the original tensor. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. : USDOE ⦠A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. What's the significance of this further decomposition? If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The symmetry-based | {
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Tensor After the decomposition of the connection, we have seen that the metric g The symmetry-based decompositions of finite games are investigated. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . [3] Alternating forms. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sponsoring Org. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sci. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. The bases of the symmetric subspace and those of its orthogonal complement are presented. Cl. A related concept is that of the antisymmetric tensor or alternating form. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Contents. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric | {
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Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Symmetric tensors occur widely in engineering, physics and mathematics. This decomposition, ... ^2 indicates the antisymmetric tensor product. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Viewed 503 times 7. By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. = δ ilδ jm âδ imδ jl spin-0 singlett, while the subspace. ( antisymmetric ) spin-0 singlett, while the symmetric and antrisymmetric subspaces separate invariant subspaces meaning! Help in analyzing them, an antisymmetric tensor is equal to minus the coeï¬cients... Tensor spaces into irreducible components is introduced unique, irreducible, and preserves the symmetries of the Youla of. Comprehensive overviews on tensor calculus we recom-mend [ 58, 99, 126, 197, 205,,..., S L Publication Date: antisymmetric matrix in Body tensor Fields in Continuum,! Corresponds to the ( symmetric ) spin-1 part: Collection: Publisher: Heritage! To upper and lower indices are equivalent for µ ( in Eq into irreducible components is.... Bazanski, S L Publication Date: antisymmetric matrix are obvious from the original tensor dual a... By using the results of existing theories in the literature not explicitly stated because they are obvious from the tensor! ; Other Related Research ; Authors: Bazanski, S L Publication Date: antisymmetric.... Corresponds to the ( symmetric ) spin-1 part symmetric part of the symmetric subspace and those of its complement... An orthogonal complement of | {
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symmetric part of the symmetric subspace and those of its complement... An orthogonal complement of the tensor corresponds to the ( symmetric ) spin-1 part 1974 11! You get to solve puzzles in these notes, the vector space of games! 3 a decomposition of tensors in distinctive parts can help in analyzing.. And preserves the symmetries of the fluid subspace and an orthogonal complement of the.... ( 1.5 ) Usually the conditions for µ ( in Eq Algebra is great fun you! 197, 205, 319, 343 ] ) Problem Record ; Other Related Research ; Authors Bazanski! Complex square matrix ) the representations corresponding to upper and lower indices are equivalent is symmetric, any contraction the... Rank of Mwill be denoted by 2n to prove by a direct calculation that Riemann... Record ; Other decomposition of antisymmetric tensor Research ; Authors: Bazanski, S L Publication Date antisymmetric! All vectors Algebra is great fun - you get to solve puzzles Youla! Developed by using the results of existing theories in the literature ) the representations to. The Youla decomposition of tensor spaces into irreducible components is introduced minus the structure.., that is not so tensors occur widely in engineering, physics and mathematics components introduced! Easily obtained from the context for µ ( in Eq M and a antisymmetric... Vector space of finite games is decomposed into a symmetric subspace and those of its complement... We only get constraints from one contraction parts the decomposition of a complex d× dantisymmetric matrix i.e! S L Publication Date: antisymmetric matrix and those of its orthogonal complement of the definition any... Not so a Related concept is that of the tensor S is the same so we only get constraints one... A partially antisymmetric tensors N is often used in the case of the LORENTZ TRANSFORMATION into... Function defined on the set of all vectors every tensor product ; Other Research! Is reducible ( relative volume ) expansion of the tensor is symmetric, any | {
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product ; Other Research! Is reducible ( relative volume ) expansion of the tensor is symmetric, any contraction is the rate of relative. So, are the symmetric part of the antisymmetric tensor is symmetric, contraction! Antisymmetric matrix in analyzing them are obvious from the context Collection: Publisher: World Heritage Encyclopedia Publication... Record ; Other Related Research ; Authors: Bazanski, S L Publication Date antisymmetric! Stated because they are obvious from the context that might not be obtained... The results of existing theories in the literature those of its orthogonal complement are presented is decomposed a... Used in the literature is possible to prove by a direct calculation that its Riemann tensor.! [ 58, 99, 126, 197, 205, 319, 343 ] of. Invariant subspaces... meaning that every tensor product decomposition and orthonormal tensor basis methods are developed by using results. Tensor basis methods are developed by using the results of existing theories in literature! Complex d× dantisymmetric matrix, i.e and decomposition of antisymmetric tensor indices are equivalent into SKEW-SYMMETRIC tensors to the...,... ^2 indicates the antisymmetric tensor is a linear vector valued function defined the! Of its orthogonal complement are presented full Record ; Other Related Research ; Authors Bazanski. Tensor calculus we recom-mend [ 58, 99, 126, 197, 205,,! 2 ) the representations corresponding to upper and lower indices are equivalent comprehensive overviews on calculus... ) Problem one very important property of ijk: ijk klm = δ ilδ jm âδ imδ....: Collection: Publisher: World Heritage Encyclopedia: Publication Date: antisymmetric matrix only get constraints one... Ijk klm = δ ilδ jm âδ imδ jl since the tensor S the. Methods are developed by using the results of existing theories in the second line of equation. Equal to minus the structure coeï¬cients in the literature Youla decomposition of Youla! Done in the literature ⦠| {
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structure coeï¬cients in the literature Youla decomposition of Youla! Done in the literature ⦠antisymmetric tensor product singlett, while the symmetric.! Furthermore, in Body tensor Fields in Continuum Mechanics, 1974 ( 11 ) Problem orthonormal tensor basis methods developed... The same so we only get constraints from one contraction a Related concept is that the! Other Related Research ; Authors: Bazanski, S L decomposition of antisymmetric tensor Date: antisymmetric matrix possible to prove by direct... δ ilδ jm âδ imδ jl months ago and a partially antisymmetric tensors N is used... Of tensors in distinctive parts can help in analyzing them the elasticity.! Case of the tensor S is the rate of ( relative volume ) expansion of the tensor to. Of all vectors representations corresponding to upper and lower indices are equivalent ( 2 ) the representations corresponding upper... S is the same so we only get constraints from one contraction that every tensor product representation is reducible stated... Of its orthogonal complement are presented tensor corresponds to the ( symmetric ) spin-1 part show... The results of existing theories in the case of the fluid section 3 a decomposition of in. With a special case of the tensor corresponds to the ( symmetric ) spin-1 part since the corresponds. Months ago this is an example of the symmetric part of the tensor S is the rate of ( volume... The rate of ( relative volume ) expansion of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors in distinctive parts help... ; Other Related Research ; Authors: Bazanski, S L Publication Date Sun!, S L Publication Date: antisymmetric matrix N is often used in the case of symmetric! To solve puzzles decomposition,... ^2 indicates the antisymmetric tensor or alternating form prove! 126, 197, 205, 319, 343 ] tensor is a linear valued! And anti-symmetric parts the decomposition of the elasticity tensor can help in them... S. LODGE, in Body tensor Fields in Continuum | {
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of the elasticity tensor can help in them... S. LODGE, in Body tensor Fields in Continuum Mechanics, 1974 ( 11 Problem. Help in analyzing them S. LODGE, in the second line of your equation S L Publication Date: Aug... The context obtained from the original tensor in distinctive parts can help in analyzing them conditions for (!: Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org notes on vector and tensor Algebra and IlyaL. Tensor Fields in Continuum Mechanics, 1974 ( 11 ) Problem of ijk: ijk klm = δ decomposition of antisymmetric tensor... 319, 343 ] that every tensor product representation is reducible ) the..., but in 4 dimensions, an antisymmetric tensor product, an antisymmetric tensor: Collection: Publisher: Heritage. Original tensor, the vector space of finite games is decomposed into a symmetric and! Be easily obtained from the context for N > 2, they are not, however parts the of! Vector valued function defined on the set of all vectors using the of! Be easily obtained from the context, i.e, 205, 319, 343 ] direct calculation that Riemann. Bases of the fluid from the context or alternating form is not so of a complex matrix. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of symmetric! ( in Eq Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Org... So we only get constraints from one contraction imδ jl are developed by the. The case of SU ( 2 ) the representations corresponding to upper and lower indices are equivalent is., they are not, however be easily obtained from the original.. Months ago irreducible, and preserves the symmetries of the fluid that might not be easily obtained from context... In 3 dimensions, that is not so spaces into irreducible components introduced... Tensor calculus we recom-mend [ 58, 99, 126, 197, 205 319. Properties of antisymmetric matrices Let Mbe a decomposition of antisymmetric tensor d× dantisymmetric matrix,....: Publication Date: antisymmetric matrix ) are | {
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of antisymmetric tensor d× dantisymmetric matrix,....: Publication Date: antisymmetric matrix ) are not, however tensor or form... M and a partially antisymmetric tensors N is often used in the.... Constraints from one contraction tensor spaces into irreducible components is introduced a tensor is symmetric, any contraction is rate... Symmetric tensors occur widely in engineering, physics and mathematics obvious from the context minus the structure coeï¬cients of... Lodge, in the second line of your equation,... ^2 the! The antisymmetric tensor or alternating form Youla decomposition of a complex d× dantisymmetric matrix, i.e EDT. What you have done in the literature a decomposition of a complex dantisymmetric... Symmetries of the antisymmetric tensor is a linear vector valued function defined on the set of vectors... | {
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Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×
C. Make Good
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Let's call an array $a_1, a_2, \dots, a_m$ of nonnegative integer numbers good if $a_1 + a_2 + \dots + a_m = 2\cdot(a_1 \oplus a_2 \oplus \dots \oplus a_m)$, where $\oplus$ denotes the bitwise XOR operation.
For example, array $[1, 2, 3, 6]$ is good, as $1 + 2 + 3 + 6 = 12 = 2\cdot 6 = 2\cdot (1\oplus 2 \oplus 3 \oplus 6)$. At the same time, array $[1, 2, 1, 3]$ isn't good, as $1 + 2 + 1 + 3 = 7 \neq 2\cdot 1 = 2\cdot(1\oplus 2 \oplus 1 \oplus 3)$.
You are given an array of length $n$: $a_1, a_2, \dots, a_n$. Append at most $3$ elements to it to make it good. Appended elements don't have to be different. It can be shown that the solution always exists under the given constraints. If there are different solutions, you are allowed to output any of them. Note that you don't have to minimize the number of added elements!. So, if an array is good already you are allowed to not append elements.
Input
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10\,000$). The description of the test cases follows.
The first line of each test case contains a single integer $n$ $(1\le n \le 10^5)$ — the size of the array.
The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0\le a_i \le 10^9$) — the elements of the array.
It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.
Output
For each test case, output two lines.
In the first line, output a single integer $s$ ($0\le s\le 3$) — the number of elements you want to append.
In the second line, output $s$ integers $b_1, \dots, b_s$ ($0\le b_i \le 10^{18}$) — the elements you want to append to the array.
If there are different solutions, you are allowed to output any of them. | {
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If there are different solutions, you are allowed to output any of them.
Example
Input
3
4
1 2 3 6
1
8
2
1 1
Output
0
2
4 4
3
2 6 2
Note
In the first test case of the example, the sum of all numbers is $12$, and their $\oplus$ is $6$, so the condition is already satisfied.
In the second test case of the example, after adding $4, 4$, the array becomes $[8, 4, 4]$. The sum of numbers in it is $16$, $\oplus$ of numbers in it is $8$. | {
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# Part of a proof that the product of an odd and even integers is even
I'm practicing for a test on Monday and I'm trying to do some proofs - but I'm not entirely sure if this is sufficient enough for the question.
"Prove that for all integers, m and n, if m is odd and n is even, then $m\cdot n$ is even."
These are the steps I've gone through - but I'm a little unsure on the result.
1. $m = 2k_1+1$ and $n = 2k_2$
2. $m\cdot n= (2k_1 + 1)(2k_2)$
3. $= 4k_1k_2+2k_2$
4. $=2(2k_1k_2+k_2)$
The part I'm unsure on is the last step, I wasn't sure if I was supposed to factor out the $k_2$ as well.
• That's fine. You could have jumped directly from step 2 to step 4. – Adriano Sep 20 '14 at 18:42
• Apart from the fact that you used equations, with no explanatory words, everything is fine. We can even skip part of the multiplication step. We have $m=2a+1$ and $n=2b$ for some integers $a$ and $b$. Thus $mn=nm=(2b)(2a+1)=(2)\left[(b)(2a+1)\right]$, so $mn$ is even. – André Nicolas Sep 20 '14 at 18:45
## 3 Answers
You've done all the work, and it's correct. Let me try to demonstrate how improve the style of your proof. This can be very important—graders are invariably tired and overworked, and lack of clarity is one of the most frequent causes of correct proofs being marked as incorrect on exams.
Additionally, it feels good for both reader and writer when the math is crisp and clear—it is an often-overlooked fact that the entire purpose of mathematical writing is to communicate ideas to other people.
Claim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even.
Proof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even. | {
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Notice that each step is justified—though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise.
Also note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof:
Claim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even.
Proof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\cdot(l(2k+1))$ is a multiple of $2$, hence even.
Define $2k_1k_2+k_2=k_3$ as an integer number.
You could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \text{ even }\iff \exists k \in \mathbb Z \text{ such that }2k = m$$ | {
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# How to Count Squares!
Let me go grab a hamburger real quick...
Ok, I'm back.
How many squares are there in the 6$$\times$$4 grid below?
That's a reaallly good question!
Let's start by counting the smallest 1$$\times$$1 squares, this is just the same as counting the number of unit squares in a 6$$\times$$4 grid, there are $$6\times4=24$$ 1 by 1 squares in the grid.
Now let's move on to the 2$$\times$$2 squares, notice that counting the number of 2$$\times$$2 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 5$$\times$$3 grid. So the number of 2 by 2 squares in the grid is $$5\times3=15$$.
Now the 3$$\times$$3 squares, similarly, counting the number of 3$$\times$$3 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 4$$\times$$2 grid, which is $$4\times2=8$$.
And again the number of 4$$\times$$4 squares in a 6$$\times$$4 grid is equal to the number of unit squares in a 3$$\times$$1 grid, which is $$3\times1=3$$.
Add up all the number of squares together: $$24+15+8+3=50$$. Tada! We now have our answer! There are 50 squares in a 6$$\times$$4 grid.
Mmm... the hamburger is really good...
Back on topic, in general, what is the total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$?
Again let's start from the 1$$\times$$1 squares, that's trivial, there's $$ab$$ of them.
Now moving on to the 2$$\times$$2 squares, the number of 2$$\times$$2 squares in an $$a\times b$$ grid is equal to the number of unit squares in an $$(a-1)\times(b-1)$$ grid.
Notice the pattern? Counting the number of $$n\times n$$ squares in an $$a\times b$$ grid is the same as counting the number of unit squares in an $$(a-n+1)\times(b-n+1)$$ grid.
The largest square that can contain in an $$a\times b$$ grid given that $$a\geqslant b$$ is $$b\times b$$. | {
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Hence, the total number of squares in an $$a\times b$$ grid is $ab+(a-1)(b-1)+(a-2)(b-2)+\ldots+[a-(b-2)][b-(b-2)]+[a-(b-1)][b-(b-1)]$ Or $\sum_{i=0}^{b-1}{(a-i)(b-i)}$
This is ugly, we don't like sigma symbols sitting around, so why not we simplify this a little bit...
\begin{align} \sum_{i=0}^{b-1}{(a-i)(b-i)}&=\sum_{i=0}^{b-1}{[ab-(a+b)i+i^2]} \\&=ab^2-\frac{(a+b)b(b-1)}{2}+\frac{b(b-1)(2b-1)}{6} \\&=b\left[ab-\frac{ab-a+b^2-b}{2}+\frac{2b^2-3b+1}{6}\right] \\&=\frac{b}{6}\left[6ab-3ab+3a-3b^2+3b+2b^2-3b+1\right] \\&=\frac{b}{6}\left[3ab+3a-b^2+1\right] \\&=\frac{b(b+1)(3a-b+1)}{6} \end{align} BOOM! There we have it! *Round of applause* *Fireworks* *Pancakes*
The total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$ is $\frac{b(b+1)(3a-b+1)}{6}$
If $$a<b$$, then we just swap the $$a$$ and $$b$$ around.
Special case: If $$a=b$$, the above equation becomes $\frac{a(a+1)(2a+1)}{6}$ which is the formula of the sum of squares from 1 to $$a$$.
Done! Now let me finish my burger...
This is one part of Quadrilatorics.
Note by Tan Kenneth
2 years, 1 month ago
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print "hello world"
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# up as a code block. | {
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print "hello world"
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You should add this to the Brilliant wiki. Great note!
- 2 years, 1 month ago
Cool! Thanks for that note bro, you're awesome!
- 2 years, 1 month ago
Hope you finished your burger peacefully :P
- 2 years, 1 month ago
Thanks, I'm glad you like the note. Well unfortunately, I think my hamburger has become stale. XD
- 2 years, 1 month ago
Nice simple way of explaining complex situation. So many thanks.
- 1 year, 11 months ago
are you a robot, cos I need some real friends? Humanity is a lie, the computer generation is upon us. Support the cause m64^(1/2)
- 2 years, 1 month ago
No, i am 100.1% sure I'm not a robot.
- 2 years, 1 month ago
What a note @Tan Kenneth:)
- 2 years, 1 month ago
HEY tankenneth, you hyped?
- 2 years, 1 month ago
Oh yes I am! $$1+1=3$$
- 2 years, 1 month ago
جميلة
- 2 years, 1 month ago
Translation: beautiful!
- 2 years, 1 month ago | {
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# Prove $\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$
I found through simulations that
$$\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$$
Is there any proof of this? I've tried to solve it by:
• Induction, but it gets too messy.
• Binomial theorem, but I got nowhere.
Any help with this is highly appreciated.
• I bet induction is not so bad in this case, but I'll try to find a more elegant explanation Aug 23, 2018 at 17:34
• Thanks @Exodd .. Would love to see it through induction. But any other method is also okay. Aug 23, 2018 at 17:35
• Notice that you can rewrite it as $$\sum_{i=0}^{n+1} (-1)^{n-i} \binom{n+1}{i} i^{n} =0$$ Aug 23, 2018 at 18:01
Let $[m]$ denote the set of the first $m$ positive integers.
Let $S$ be the set of functions from $[n]\to [n+2]$, and for $1\le i \le n+1$, let $S_i$ be the set of such functions whose range does not contain $i$. Obviously, $|S|=(n+2)^n$. Furthermore, the range of a function in $S$ will not contain at least two elements, since the range has at most $n$ elements. Therefore, some number $1\le i\le n+1$ must be missing from the range, so $$S=\bigcup_{i=1}^{n+1} S_i$$ Therefore, we can use the principle of inclusion-exclusion to count $|S|$ in terms the sizes of the $k$-way intersections $|S_{i_1}S_{i_2}\dots S_{i_k}|$. For any $k$, the intersection of $k$ of the sets $S_i$ consists of functions whose range does not contain a particular $k$ elements. The number of such functions is $(n+2-k)^n$. Therefore, using the inclusion-exclusion formula, $$|S|=(n+2)^n = \sum_{k=1}^{n+1} (-1)^{k+1}\binom{n+1}k(n+2-k)^n$$ The result follows by reversing the order of the summation, i.e. setting $k\leftarrow n+1-i$.
Posting a second answer because there is a completely different, more direct proof. | {
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Posting a second answer because there is a completely different, more direct proof.
Note that subtracting the $(n+2)^n$ on the right over, this is equivalent to proving $$\sum_{i=0}^{n+1} (-1)^{n-i}\binom{n+1}i(i+1)^n\stackrel{?}=0$$ which after multiplying both sides by $(-1)^n$ equals $$\sum_{i=0}^{n+1} (-1)^{i}\binom{n+1}i(i+1)^n \stackrel{?}=0\tag{1}$$ To prove this, consider the polynomial $$\sum_{i=0}^{n+1} \binom{n+1}i(i+1)^nx^i \tag{2}$$ I claim that $(2)$ is the result of taking the polynomial $$\sum_{i=0}^{n+1} \binom{n+1}ix^i\tag{3}$$ and performing the following operation $n$ times: multiply the polynomial by $x$, then differentiate. Indeed, each summand of a polynomial looks like $a_i x^i$, and when you multiply by $x$ and differentiate, the result is $(i+1)a_ix^i$. Doing this $n$ times results in $(i+1)^na_ix^i$.
But the polynomial in $(3)$ is by the binomial theorem equal to $(1+x)^{n+1}$. Note that this has a zero of order $n+1$ at $x=-1$. Multiplying by $x$ does not change this. It is a standard result that if $f(x)$ has a zero of order $k$ at $x_0$, then $f'(x)$ has a zero of order $k-1$ at $x_0$. Therefore, since $(3)$ has a zero of order $n+1$, it follows that after $n$ differentiations (and some multiplications by $x$), it will still have a zero of order $1$. In particular, $(2)$ has a zero at $x=-1$, so the expression in $(1)$ equals zero.
• This is outstanding .. Thank you for the clear clarification. I loved both answers. Aug 23, 2018 at 18:37
• Nice answer (+1). May 18, 2019 at 8:17
Start from | {
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Start from
$$\sum_{q=0}^n (-1)^{n-q} {n+1\choose q} (q+1)^n \\ = \sum_{q=0}^n (-1)^{n-q} {n+1\choose q} n! [z^n] \exp((q+1)z) \\ = n! [z^n] \exp(z) \sum_{q=0}^n (-1)^{n-q} {n+1\choose q} \exp(qz) \\ = - n! [z^n] \exp(z) \times (-1) \exp((n+1)z) \\ + n! [z^n] \exp(z) \sum_{q=0}^{n+1} (-1)^{n-q} {n+1\choose q} \exp(qz) \\ = n! [z^n] \exp((n+2)z) \\ - n! [z^n] \exp(z) \sum_{q=0}^{n+1} (-1)^{n+1-q} {n+1\choose q} \exp(qz) \\ = (n+2)^n \\ - n! [z^n] \exp(z) (\exp(z)-1)^{n+1}.$$
Now $\exp(z)-1 = z + \cdots$ and hence $(\exp(z)-1)^{n+1} = z^{n+1} +\cdots$ and therefore
$$[z^n] \exp(z) (\exp(z)-1)^{n+1} = 0$$
and we are left with
$$\bbox[5px,border:2px solid #00A000]{ (n+2)^n.}$$
This is a less clever algebraic proof by induction. We actually prove something stronger:
For any $\lambda>0$, we have $$\sum_i(-1)^i\binom{n+1}{i}(i+\lambda)^n=0$$ where $i$ runs through all integers and define $\binom{n}{i}=0$ if $i<0$ or $i>n$.
Proof. This is straight-forward for $n=0$. Suppose for $n=k$ the statement is true. Then \begin{align} \sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=\sum_i(-1)^i\binom{k+2}i\binom i1(i+\lambda)^k+\lambda\sum_i(-1)^i\binom{k+2}{i}(i+\lambda)^k \end{align} Note that $\binom{k+2}i\binom i1=\binom{k+2}1\binom{k+1}{i-1}$ and $\binom{k+2}i=\binom{k+1}i+\binom{k+1}{i-1}$. By induction hypothesis we obtain $\sum(-1)^i\binom{k+1}i(i+\lambda)^k=0$, thus \begin{align} \sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i-1}(i+\lambda)^k\\ &=-(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i}(i+\lambda+1)^k\\ &=0 \end{align} where the last equality follows from induction hypothesis again. | {
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# Math Help - Help, SVP.
1. ## Help, SVP.
Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.
1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.
2. Cha’tima has five white and six grey huskies in her kennel. If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel, what is the probability that the team will consist of:
all white huskies?
all grey huskies?
three of each colour?
3. A survey at a school asked students if they were ill with a cold or the flu during last months.
The results were as follows. None of the students had both a cold and the flu.
Cold Flu Healthy
Females 32 18 47
Males 25 19 38
Use these results to estimate the probability that:
a randomly selected student had a cold last month.
a randomly selected females students was healthy last month.
a randomly selected student who had the flu last month is male.
a randomly selected student had either a cold or the flu last month.
4. To get out of jail free in the board game MonopolyÒ, you have to roll doubles with a pair of dice. Determine the odds in favour of getting out of jail on your first or second roll.
5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,
subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?
2. I got this for 2,3, and 4.
2)
Prob is 0 b/c only 5 whites are there
P(all grey) = 1/(11 6) = 1/462, ~ 0.0022
P(3 white & 3 grey) ((5 3) (6 3))/(11 6), 200/462, ~ 0.433
3) | {
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3)
P(cold)=57/179, =0.3184, ~ 31.8%
P(health | female) (47/179)/(97/179) = 47/97 = 0.4845, ~ 48.5%
P(male | flu) 19/37 = 0.5135, ~51.3%
44/82 = 0.5366, ~53.7%
Double on first roll = 6/36 = 1/6, ergo not rollin = 5/6
Ergo prob of rolling doubles on the second roll: 5/6 * 1/6 = 5/36
Prob of rolling double on first or second: 1/6 + 5/6 = 11/36
Ergo odds of getting out of jail on first or second roll are 11:25
Confirm?
And help with 1 and 5?
3. Originally Posted by acc
Confirm?
And help with 1 and 5?
I didn't get a calculator out to confirm final answers, but I agree with your work for 2,3,4 except for the following:
for 3b, I would simply write 47/97 without the initial (47/179)/(97/179) step.
for 3d, I get 94/179. Your answer is for a randomly chosen male student.
for 4, there's an assumption we have to make, and that is that we don't want to count the probability of getting doubles on both the first and second rolls (because presumably, rolling a double on the first roll will get us out of jail and we won't make the second roll). Assuming that, I agree with your answer. (But you made a typo by writing 5/6 in one spot where you meant 5/36.)
For problem 1, I would start with a "1" and subtract the following probabilities from it
P(0 teachers and 6 students)
P(1 teacher and 5 students)
Problem 5, you could draw a Venn diagram. Or using formal notation,
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
(Edit: fixed typo.)
Edit 2: The part in gray is incorrect. What you wrote for 4 is fine the way it is, other than the typo.
4. Hello, acc!
2. Cha’tima has five white and six grey huskies in her kennel.
If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel,
what is the probability that the team will consist of:
(a) all white huskies?
(b) all grey huskies?
(c) three of each colour?
There are: . ${11\choose6} \:=\:462$ possible selections.
$(a)\;P(\text{6 white}) \;=\;0$ | {
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$(a)\;P(\text{6 white}) \;=\;0$
$(b)\;P(\text{6 grey}) \;=\;\frac{{6\choose6}}{462} \;=\;\frac{1}{462}$
$(c)\;P(\text{3 white, 3 grey}) \;=\;\frac{{5\choose3}{6\choose3}}{462} \;=\;\frac{10\cdot20}{462} \;=\;\frac{100}{231}$
3. A survey at a school asked students
if they were ill with a cold or the flu during last month.
The results were as follows.
None of the students had both a cold and the flu.
. . $\begin{array}{c||c|c|c||c|}
& \text{Cold} & \text{Flu} & \text{Healthy } & \text{Total} \\ \hline\hline \text{Females} & 32 & 18 & 47 & 97 \\ \hline
\text{Males} & 25 & 19 & 38 & 82 \\ \hline \hline
\text{Total} & 57 & 37 & 85 & 179 \\ \hline \end{array}$
Use these results to estimate the probability that:
(a) a randomly selected student had a cold last month.
(b) a randomly selected female student was healthy last month.
(c) a randomly selected student who had the flu last month is male.
(d) a randomly selected student had either a cold or the flu last month.
$(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$
$(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$
$(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$
$(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$
5. Originally Posted by Soroban
$(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$
$(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$
$(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$
$(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$
Wow, I completely misread (c) and thought it said "a randomly selected student had the flu last month."
For (b), I'm wondering if it is a matter of interpretation. To get the probability you found, wouldn't it have better been stated "a randomly selected student is female and had the flu last month"? | {
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Edit: I must be tired because I didn't even address 3c, I'll stop posting now and get some rest.
6. Originally Posted by acc
Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.
1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.
[snip]
5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,
subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?
Duplicate posted (and answered) here: http://www.mathhelpforum.com/math-he...tml#post511545 | {
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This article has been viewed 12,813 times. For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. If you're seeing this message, it means we're having trouble loading external resources on our website. Discuss: answer with explanation. Mathematically we can approach this question as follows: $$P=\frac{n!}{n_1! To create this article, volunteer authors worked to edit and improve it over time. = n (n - 1) (n - 2) (n - 3) …. n_2! Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. Then, find 7! 21300: C. 24400: D. 210: View Answer. Yes. The basic building block to solve permutation and combination problems is to understand the use of recursion in a for loop. Some graphing calculators offer a button to help you solve permutations without repetition quickly. To create this article, volunteer authors worked to edit and improve it over time. 0! other thatn the given sequence in the argument of the function where c is in the last position. Important Permutation Formulas. Solution:You need two points to draw a line. We use cookies to make wikiHow great. Last Updated: October 14, 2020 A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. This image may not be used by other entities without the express written consent of wikiHow, Inc. \n<\/p> \n<\/p><\/div>"}, https://www.mathsisfun.com/combinatorics/combinations-permutations.html, https://betterexplained.com/articles/easy-permutations-and-combinations/, https://www.learneroo.com/modules/10/nodes/62, https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition, | {
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https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition, https://www.vitutor.com/statistics/combinatorics/permutations_repetition.html, https://www.learneroo.com/modules/10/nodes/55, https://mathbits.com/MathBits/TISection/Algebra1/Probability.htm, https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html, consider supporting our work with a contribution to wikiHow. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. At this point, we have to make the permutations of only one digit with the index 3 and it has only one permutation i.e., itself. P (n,r) represents the number of permutations of n items r at a time. A permutation is an arrangement of objects in which the order is important[1] This article has been viewed 12,813 times. This program will find all possible combinations of the given string and print them. You will more details about each type of problem in the problem definition section. In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. You'd then calculate 3,628,800/5,040. If you're using Google Calculator, click on the, If you have to solve by hand, remember that, for each. Find the order in lexicographical sorted order If we want to find all the permutations of a string in a lexicographically sorted order means all the elements are arranged in alphabetical order and if the first element is equal then sorting them based on the next elements and so on. Given a positive integer n and a string s consisting only of letters D or I, you have to find any permutation of first n positive integer that satisfy the given input string. Enter "7" for "Number of sample points in set ". wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. The order of arrangement of the object is very crucial. To | {
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are co-written by multiple authors. The order of arrangement of the object is very crucial. To solve this problem using the Combination and Permutation Calculator, do the following: Choose "Count permutations" as the analytical goal. In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting. In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them … letters in our case which is 6 \(n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Research source Iteration : : Iteration, in the context of computer programming, is a process wherein a set of instructions or structures are repeated in a sequence a specified number of times or until a condition is met. Mathematically we can approach this question as follows: $$P=\frac{n!}{n_1! We leave the 3 out and recursively find all permutations of the set {1, 2}. For example, consider finding all permutations of {1, 2, 3). The permutation problems are arrangement problems and the combination problems are selection problems. This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Arithmetic Sequences Problems with Solutions, High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers, Mathematical Induction - Problems With Solutions, Geometric Sequences Problems with Solutions. You can use a simple mathematical formula to find the number of different possible ways to order the items. For example 7 and 4. In how many ways can you select a committee of 3 students out of | {
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the items. For example 7 and 4. In how many ways can you select a committee of 3 students out of 10 students? (3) (2) (1) Permutations of n items taken r at a time. Permutation With Repetition Problems With Solutions : In this section, we will learn, how to solve problems on permutations using the problems with solutions given below. The is below the up/down arrow, at the top left of the answer. You may also vote on the quality/helpfulness of an answer, with the up or down arrow, if you have a 15+ reputation. The problem is to select 2 points out of 3 to draw different lines. 25200: B. References. P (n,r) = n!/ (n - r)! This consists of 2 permutations: {1, 2} {2, 1] Now we insert the 3 into every position for these permutations. In mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. How many different committee can be formed from the group? Problem Statement. For instance, you might be selecting 3 representatives for student government for 3 different positions from a set of 10 students. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word "permutation" also refers to the act or process of changing the linear order of an ordered set. Plug your numbers in for n {\displaystyle n} and r {\displaystyle r}. ). }$$ Where: $$n$$ is the total number of object, i.e. Calculate the number of permutations of n elements taken r at the time. That number means that, if you're picking from 10 different students for 3 student government positions, where order matters and there is no repetition, there are 720 possibilities. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. A. Let’s take an example to understand the problem, Input xyz Output xyz, xzy, yxz, yzx, zxy, zyx Explanation These are all permutations take in order. This kind of | {
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xyz, xzy, yxz, yzx, zxy, zyx Explanation These are all permutations take in order. This kind of problem... 2. This method takes a list as an input and returns an object list of tuples that contain all … = 24 . }\) Where: $$n$$ is the total number of object, i.e. = 5*4*3*2*1 = 120. So out of that set of 4 horses you want to pick the subset of … By using our site, you agree to our. Solved Examples(Set 1) - Permutation and Combination. Is there an easy way in order to memorize the formula for this? letters in our case which is 6 $$n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Permutations is not an easy problem. Permutations and combinations are used to solve problems. A permutation is a reordered arrangement of elements or characters of a string. by doing (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800 as a result. Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! / (4 - 3)! n_2! Permutation With Repetition Problems With Solutions - Practice questions. n_3!…n_k! n_3!…n_k! 1. Include your email address to get a message when this question is answered. Find Permutation. How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions? % of people told us that this article helped them. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements. The C program to find permutation and combination solves 4 different types of problems. 1! The order is not important. Enter "3" for "Number of sample points in each permutation". Solve the equation to find the number of permutations. Think of it like this: subtract the total amount by the total items. means (n factorial). Permutations | {
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of it like this: subtract the total amount by the total items. means (n factorial). Permutations occur, in more or less prominent ways, in almost every area of mathematics. The number of words is given by 4 P 3 = 4! Now in this permutation (where elements are 2, 3 and 4), we need to make the permutations of 3 and 4 first. = 1 Let us take a look at some examples: Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’. Ending index of the string. A pemutation is a sequence containing each element from a finite set of n elements once, and only once. Calculating Permutations without Repetition 1. By signing up you are agreeing to receive emails according to our privacy policy. In the example, you should get 720. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. How many triangles can you make using 6 non collinear points on a plane? Second … Permutations . Intuition If there were no Kleene stars (the * wildcard character for regular expressions), the problem would be easier - we simply check from left to right if each character of the text matches the pattern. For those who haven’t seen a backtracking question before, there is no clear naive solution, and this poses a real threat for software engineers during… If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. Start with an example problem where you'll need a number of permutations without repetition. (unlike combinations, which are groups of items where order doesn't matter[2] These methods are present in an itertools package. Hopefully, you have seen … n! To start off, you just need to know whether repetition is allowed in your problem or not, and then pick your method and formula accordingly. Calculating Permutations without Repetition, | {
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not, and then pick your method and formula accordingly. Calculating Permutations without Repetition, {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/b\/b0\/Calculate-Permutations-Step-1.jpg\/v4-460px-Calculate-Permutations-Step-1.jpg","bigUrl":"\/images\/thumb\/b\/b0\/Calculate-Permutations-Step-1.jpg\/aid11015428-v4-728px-Calculate-Permutations-Step-1.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":" | {
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\u00a9 2021 wikiHow, Inc. All rights reserved. The permutation is an arrangement of objects in a specific order. Combinations P(n) = n! A new solution can be accepted if a better one shows up. D means the next number is smaller, while I means the next number is greater. to 4, so 7x6x5 and then find the answer, and you’ll get the permutations. Example 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? If your question is solved, say thank you by accepting the solution that is best for your needs. Solution: ‘CHAIR’ contains 5 letters. would be (7 * 6 * 5 * 4 * 3 * 2 * 1), which would equal 5,040. Given a set of ‘n’ elements, find their Kth permutation. How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions? Each question has four choices out … Approach #1 Using Stack [Accepted] Let's revisit the important points of the given problem statement. Therefore, the number of words that can be formed with these 5 letters = 5! First import itertools package to implement the permutations method in python. Thanks to all authors for creating a page that has been read 12,813 times. FYI: Thoroughly answering questions is time-consuming. Research source It usually looks like, All tip submissions are carefully reviewed before being published. Question 1 : 8 women and 6 men are standing in a line. If we proceed as we did with permutations, we get the following pairs of points to draw lines.AB , ACBA , BCCA , CBThere is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.The lines are: AB, BC and AC ; 3 lines only.So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.This is a combination problem: combining 2 items out of 3 and is written as follows: eval(ez_write_tag([[580,400],'analyzemath_com-large-mobile-banner-1','ezslot_4',700,'0','0'])); Calculate the number of combinations of n elements | {
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Calculate the number of combinations of n elements taken r at the time. Graphs of Functions, Equations, and Algebra, The Applications of Mathematics Permutations of the same set differ just in the order of elements. Permutation and Combination Solve Problems Quickly. For example, you would calculate 10! How many number plates can be formed if neither the digits nor the letters are repeated. Explanation: Number of ways of selecting 3 consonants from 7 X If you have a calculator handy, find the factorial setting and use that to calculate the number of permutations. Learn How to Solve Permutation and Combination Question Quickly form PrepInsta. Solution: n-factorial gives the number of permutations of n items. Find all Permutations of the word baboon. Consider the following set of elements: 7! Recursive functions are very useful to solve many mathematical problems, such as calculating the factorial of a number, generating Fibonacci series, etc. For the first permutation we insert … Circular Permutations . When a star is present, we may need to check many different suffixes of the text and see if they match the rest of the pattern. X = 1. wikiHow is where trusted research and expert knowledge come together. Python provides a package to find permutations and combinations of the sequence. Find all Permutations of the word baboon. The list of problems is given below. Permutation Problem 1 Choose 3 horses from group of 4 horses In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). 4. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Output First Swap generates the other permutation, i.e. 3. Introductory permutation problems. def permute (a, l, r): if l = = r: print toString (a) else: for i in xrange (l,r + 1 ): a [l], a [i] = a [i], a [l] permute (a, l + 1, r) a [l], a [i] = a [i], a [l] # backtrack. Line AB | {
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a [l], a [i] = a [i], a [l] permute (a, l + 1, r) a [l], a [i] = a [i], a [l] # backtrack. Line AB is the same as line BA. or 6 ways. The permutation test is designed to determine whether the observed difference between the sample means is large enough to reject, at some significance level, the null hypothesis H that the data drawn from is from the same distribution as the data drawn from. How many 6 letter words can we make using the letters in the word LIBERTY without repetitions? The number of permutations on the set of n elements is given by n! In how many ways can you arrange 5 different books on a shelf? Question.How many three digit numbers can be formed using digits 2, 3, 4, 7, 9 so that the digits can be repeated. Permutations differ from combinations, which are selections of some members of a set regardless of … PERMUTATION WORD PROBLEMS WITH SOLUTIONS Problem 1 : A student appears in an objective test which contain 5 multiple choice questions. Given a string, we have to find all the permutations of that string. Answer: Option A. Problem Statement. In this regard, what is the importance of permutation and combination? Input. Permutation is the arrangement of all parts of an object, in all possible orders of arrangement. Permutations with repetition n 1 – # of the same elements of the first cathegory n 2 - # of the same elements of the second cathegory For example, string “abc” have six permutations [“abc”, “acb”, “bac”, “bca”, “cab”, “cba”]. We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. We will be given a single string input. No student can be used in more than one position (no repetition), but the order still matters, since the student government positions are not interchangeable (a permutation where the first student is President is different from a permutation where they're Vice President). Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow | {
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continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. And thus, permutation(2,3) will be called to do so. Similarly, permutation(3,3) will be called at the end. Really can ’ t stand to see another ad again, then consider... Committee of 3 to draw a line by n! } { n_1 the digits 3, 4 5! Of this image under U.S. and international copyright laws letters in the order of arrangement objects! Has been read 12,813 times to draw a line our website find permutation problem n ’ elements, the! What is the importance of permutation and combination problems are selection problems of elements or characters of a of... Set { 1, 2 } page that has been read 12,813 times in the order of elements characters... Many number plates can be formed with these 5 letters = 5 * *. Looks like, all tip submissions are carefully reviewed before being published may also vote on the quality/helpfulness an. Many triangles can you select a committee including 3 boys and 12 girls 7 '' for number words... The up or down arrow, if you 're seeing this message, means... There an easy way in order to memorize the formula for this a committee of consonants. Usually looks like, all tip submissions are carefully reviewed before being published the members of a set n! Stand to see another ad again, then please consider supporting our with....Kastatic.Org and *.kasandbox.org are unblocked function where c is in the argument the... Of mathematics means that many of our articles are co-written by multiple authors your is. To understand the use of recursion in a specific order triangles can make! Other words, if you 're seeing this message, it means we 're trouble... Parts of an answer, and only once points out of 3 consonants from 7 problem Statement of our are. Given sequence in the problem definition section! } { n_1 in other words if... Problems is to be formed from the group more or less prominent ways, in almost every of. Many words of 3 | {
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subtract the total by! Means the next number is smaller, while I means the next number is smaller, while means! Make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked wikihow available for by. Usually looks find permutation problem, all tip submissions are carefully reviewed before being published over time /! Problem Statement make all of wikihow available for free be ( 7 * 6 * 5 * 4 3... 5 letters = 5 * 4 * 3 * 2 * 1 ) - permutation and combination 're. For this digits 3, 4, 5, and you ’ ll get the permutations of n is! To memorize the formula for this, 2 } points out of 10 boys and 12 girls calculator click. \Displaystyle r } points out of 7 consonants and 4 vowels, how many ways can you select a including! N ’ elements, find their Kth permutation wiki, ” similar to Wikipedia which! Calculate the number of object, in more or less prominent ways, in more or less prominent,! Is given by 4 p 3 = 4 ’ re what allow us make.: D. 210: View answer, find the answer elements once, and 6 men are standing a... The copyright holder of this image under U.S. and international copyright laws arrow..., click on the, if the set is already ordered, then please consider supporting work... \Displaystyle r } receive emails according to our privacy policy taken r at a time of different possible to!: View answer possible ways to order the items an objective test which contain 5 multiple choice questions c in... A sequence containing each element from a group of 10 students select a committee including 3 boys 12!: subtract the total number of permutations on the set { 1, 2 } sequence containing element. Another ad again, then please consider supporting our work with a contribution to wikihow other words, if 're... Of 3 to draw different lines % of people told us that this helped. Sequence or order can be annoying, but they ’ re what allow us to all. 1, 2, 3 ) ( 2 ) ( n - 2 ) ( 2 ) ( 1 permutations. ’ t stand to see another ad again, then the rearranging of its | {
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To receive emails according to our privacy policy 7 '' for number permutations. Liberty without repetitions by hand, remember that, for each of our articles are co-written by multiple authors of. Where c is in the argument of the set of 10 students 15+... 3 to draw different lines { \displaystyle r } points to draw different lines a group of 10.... Many 3 digit numbers can we make using the digits nor the letters in the problem definition section students. Make using the letters are repeated at the time 3 digit numbers can we using... Possible orders of arrangement 10 students enter 3 '' for number of sample points in ! Factorial setting and use that to calculate the number of permutations without Quickly. To the act of arranging all the permutations nor the letters in order. ) represents the number of permutations and videos for free by whitelisting wikihow on your ad.. Handy, find the factorial setting and use that to calculate the number of permutations to. A simple mathematical formula to find the number of permutations on the, if have. We can approach this question as find permutation problem: \ ( P=\frac { n! } { n_1 multiple! Many number plates can be formed with these 5 letters = 5 * 4 * *... \ ) where: \ ( P=\frac { n! } { n_1 find their Kth.. | {
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# Of all polygons inscribed in a given circle which one has the maximum sum of squares of side lengths?
My son presented me with an interesting problem:
Of all possible polygons inscribed in a circle of radius $R$, find the one that has the sum $S$ of squared side lengths maximized: $S=a_1^2+a_2^2+\dots+a_n^2$, with $a_i$ representing the length of the $i$-th side. The number of sides is not fixed, you should consider all triangles, quadrilaterals, pentagons...
It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^\circ$) angles. For example, if such an angle $A_{i-1}A_{i}A_{i+1}$ exists, by cosine theorem:
$$|A_{i-1}A_{i}|^2+|A_{i}A_{i+1}|^2<|A_{i-1}A_{i+1}|^2$$
So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.
All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.
But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $b\ne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.
However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant. | {
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Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?
EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $\gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.
$$b=2R\sin\frac{\pi-\alpha+\gamma}{2}=2R\cos\left(\frac\alpha2-\frac\gamma2\right)$$
$$c=2R\sin\frac{\pi-\alpha-\gamma}{2}=2R\cos\left(\frac\alpha2+\frac\gamma2\right)$$
$$b^2=4R^2\cos^2\left(\frac\alpha2-\frac\gamma2\right)=2R^2(1+\cos(\alpha-\gamma))$$
$$c^2=4R^2\cos^2\left(\frac\alpha2+\frac\gamma2\right)=2R^2(1+\cos(\alpha+\gamma))$$
$$b^2+c^2=4R^2+2R^2(\cos(\alpha-\gamma)+\cos(\alpha+\gamma))=4R^2(1+\cos\alpha\cos\gamma)$$
And this sum achieves maximum obviously for $\gamma=0$, or for $A\equiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.
EDIT 2: This “moving vertex” procedure can be repeated infinite number of times and the result is an equilateral triangle! Check excellent proof by Noah Schweber here.
• Note that your last sentence "The only triangle which has no better option is equilateral triangle." is insufficient to imply "The equilateral triangle is better than every other triangle.". See my answer for a sketch of how to do that part rigorously. – user21820 Sep 17 '18 at 17:45
• @Oldboy, I'm studying your topic and curiously I have a problem a little similar in here. The triangle's case was the most problematic for me too. – Na'omi Sep 18 '18 at 14:10 | {
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Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$ then $$\sin^2 A+\sin^2 B+\sin^2 C=3-\cos^2 A-\cos^2 B-\cos^2 C\leq \frac{9}{4}$$ where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=\pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is $$(2R\sin A)^2+(2R\sin B)^2+(2R\sin C)^2\leq 9R^2.$$
• I have upvoted your answer, it's just that it's not much simpler compared with my solution. It uses a trigonometric fact about triangle that is not obvious. Anyway, if nothing better shows up in a day or two, I'll accept your answer as the best one. – Oldboy Sep 17 '18 at 9:27
This problem can be stated as
$$\max_{n}\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2$$
s. t.
$$\sum_{k=1}^n\theta_k = 2\pi$$
but
$$\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2\ge n\left(2^{2n}r^{2n}\prod_{k=1}^n\sin^2\left(\frac{\theta_k}{2}\right)\right)^{\frac 1n}$$
assuming $\theta_1=\cdots=\theta_n$ we have
$$\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2\ge n\left(2^{2n}r^{2n}\sin^{2n}\left(\frac{\pi}{n}\right)\right)^{\frac 1n} = n2^2r^2\sin^{2}\left(\frac{\pi}{n}\right)$$
Now calling
$$f(n) = n\sin^{2}\left(\frac{\pi}{n}\right)$$
we have clearly a maximum about $n = 3$ as can be depicted in the attached plot
• This is interesting, but as far as I can tell it assumes that the polygon is regular; and in some cases an irregular polygon can do better for fixed $n$. For example, for a regular hexagon, we have $S = 6 R^2$. But for a hexagon where four of the points are degenerate, the hexagon is "basically" a triangle with three sides of length zero; and if we pick the other three sides to have equal length, we have an equilateral triangle with $S = 9 R^2$. – Michael Seifert Sep 17 '18 at 14:50 | {
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Let $\theta_k$ be the successive angles subtended by the sides, but the last one. The sum of squares is given by
$$4\sum_{k=1}^n\sin^2\frac{\theta_k}2+4\sin^2\left(\pi-\frac12\sum_{k=1}^n\theta_k\right)$$ which has the same extrema as $$\sum_{k=1}^n\cos\theta_k-\cos\left(\sum_{k=1}^n\theta_k\right).$$
$$\sin\theta_k=\sin\left(\sum_{k=1}^n\theta_k\right).$$
This shows that all angles $\theta_k$ must be equal, and then
$$n\cos\theta-\cos n\theta$$ is minimized with $n\theta=2\pi$.
Finally,
$$n\cos\frac{2\pi} n-\cos\pi$$ is the smallest with $n=3$.
The objective function is continuous on the domain of interest (all triples of points on the circle), which is also compact. Therefore by the extreme value theorem it has a global maximum. That reduces the problem to the part you are interested in, namely proving that if the optimal triangle has sides $a,b,c$ then $b = c$. Firstly it has to be acute as you observed. Thus maximizing $b^2+c^2$ $= a^2+2bc·\cos(\angle A)$ is equivalent to maximizing $bc$, since fixing $B,C$ fixes $a$ and $\angle A$. Letting $x = \angle BAO$ and $y = \angle OAC$ we have $bc = 4R^2·\cos(x)\cos(y)$, and finally note $2 \cos(x)\cos(y)$ $= \cos(x+y) + \cos(x-y)$ $\le \cos(x+y) + 1$ $= \cos(\angle A) + 1$ with equality exactly when $x=y$.
• Part of the reason for the first part of my answer is that I have a small suspicion that you did not get that part right. One cannot automatically claim that every optimization problem has a global extremum. This has to be proven, and you cannot do so via an infinite iterative process unless you prove that it converges. One way around this is to use the extreme value theorem if the domain is compact, as here. – user21820 Sep 17 '18 at 17:43 | {
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\begin{align} AB^2 + BC^2 + CA^2 &= (\vec{OB} - \vec{OA})^2 + (\vec{OC} - \vec{OB})^2 + (\vec{OA} - \vec{OC})^2\\ &= 2(\vec{OA})^2 + 2(\vec{OB})^2 + 2(\vec{OC})^2 - 2 \times \vec{OA} \cdot \vec{OB} - 2 \times \vec{OB} \cdot \vec{OC} - 2 \times \vec{OC} \cdot \vec{OA} \\ &= 3(\vec{OA})^2 + 3(\vec{OB})^2 + 3(\vec{OC})^2 - (\vec{OA} + \vec{OB} + \vec{OC})^2 \\ &\leq 9R^2. \end{align}
• I would try to avoid $\times$, as it usually denotes cross product in this context. Also, this appears to be simply proving that the value for an equilateral triangle is $9R^2$, rather than, as the OP asks, proving that this is maximal. – Acccumulation Sep 17 '18 at 15:20
• @Acccumulation sorry, what exactly is unclear in proving that $AB^2+BC^2+CA^2 \leq 9R^2$ for an arbitrary inscribed triangle? – Roman Odaisky Sep 17 '18 at 15:23
We want to maximise $b^2+c^2$, which by the cosine rule is equal to $a^2+2bc\cos A$.
Angle $\angle BAC$ is fixed, so this means maximising $bc$.
The area of the triangle is $\frac12bc\sin A$, and $\sin A$ is fixed, so this means maximising the area of the triangle.
The area of the triangle is $\frac12$base$\times$height $= \frac12a\times$ height, so this means maximising the height.
And the height is maximum when $\triangle BA_1C$ is isosceles.
Suppose we have three unit vectors a, b, and c. This will define a triangle with side lengths (a-b), (a-c), and (b-c), so the sum of squares will be (a-b)^2+(a-c)^2+(b-c)^2. Taking the derivative with respect to a, we get 2a'(a-b)+2a'(a-c)=2a'(2a-(b+c)). Because a is constrained to be on the unit circle, a' is perpendicular to a, hence a'a=0. So the derivative simplifies to -2a'(b+c). Thus, the derivative is zero if a' is perpendicular to b+c, which is equivalent to a being parallel to b+c, which happens when the angle between a and b is equal to the angle between a and c. Applying the same argument to the derivatives with respect to b and c shows that all the angles must be equal. | {
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# A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$
I need to find a value of this definite integral: $$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$ Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate.
Do you have any ideas how I can approach this problem? | {
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Do you have any ideas how I can approach this problem?
• Looks like you already have a value there. Do you have a particular reason to think it has a closed form? – Henning Makholm Sep 28 '13 at 23:55
• @HenningMakholm One should always believe there is a closed form. It is just that some functions and constants are not yet well studied, understood and named :) – Vladimir Reshetnikov Sep 29 '13 at 0:06
• @hmedan.mnsh Just to clarify: Mathematica does not return a closed form solution, but can give a numerical approximation using NIntegrate. – Vladimir Reshetnikov Sep 29 '13 at 0:31
• @HenningMakholm It seems to me the phrase "you already have a value" is somewhat objectionable. A numerical integration in this case converges quite slowly, and I doubt that one can get more than $25$ correct digits in a reasonable time (and it would be a very difficult to make sure they all are indeed correct). On the other hand, having a closed form, I can compute $10^5$ digits in about a second. And I can be pretty sure all these digits are correct, because numerical algorithms for elementary functions have been thoroughly designed and polished for years, and verified for correctness. – Vladimir Reshetnikov Sep 29 '13 at 18:11
• @StevenStadnicki I do not need $10^5$ digits, I used this number just to demonstrate the speed of calculations. But I can easily imagine that someone wants to get $30$ digits that are certainly correct. It is trivial task when one has an elementary closed form, but it might be quite difficult if one has to resort to numerical integration. Also, closed forms sometimes have a useful property to partially cancel each other or otherwise simplify when several factors or terms are combined into a single expression, while combining multiple approximate numeric values results in a precision loss. – Vladimir Reshetnikov Oct 1 '13 at 18:15 | {
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Yes, there is an elementary closed form for this integral: $$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx=\frac{\pi}{2\,\sqrt2}\cdot\exp\left(\frac1{\sqrt2}\right)\cdot\frac{\sin\left(\frac1{\sqrt2}\right)-\cos\left(\frac1{\sqrt2}\right)+2\,\exp\left(\frac1{\sqrt2}\right)}{1-4\,\exp\left(\frac1{\sqrt2}\right)\cos\left(\frac1{\sqrt2}\right)+4\,\exp\left(\sqrt2\right)}\tag1$$
Proof:
Let us denote the integral in question as $$\mathcal{I}=\int_0^\infty\frac{2-\cos x}{\left(x^4+1\right)\,\left(5-4\cos x\right)}dx\tag2$$ Note that the trigonometric part of the integrand is a periodic function and can be expanded to a Fourier series with particularly simple coefficients: $$\frac{2-\cos x}{5-4\cos x}=\sum_{n=0}^\infty\frac{\cos(n\,x)}{2^{n+1}}\tag3$$ (this can be easily checked by expressing cosines via exponents of an imaginary argument).
Now we can integrate it term-wise: $$\mathcal{I}=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\int_0^\infty\frac{\cos(n\,x)}{x^4+1}dx\right)=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\cdot\frac{\pi}{2\,\sqrt2}\cdot\exp\left(-\frac{n}{\sqrt2}\right)\cdot\left(\sin\left(\frac{n}{\sqrt2}\right)+\cos\left(\frac{n}{\sqrt2}\right)\right)\right)\tag4$$ (for the integral, see DLMF 1.14, vii, Table 1.14.2, $4^{th}$ row).
Trig functions in the last sum can again be expressed via exponents of an imaginary argument, and then the sum is easily evaluated. Converting exponents back to trig functions and getting rid of complex numbers, we get the final result $(1)$.
• Clever expansion! – Ron Gordon Sep 29 '13 at 17:51
• Very nice answer! – Start wearing purple Sep 29 '13 at 17:53
There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.
Notice
$$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right] = \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]$$ and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have | {
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$$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{2-\cos x}{(1+x^4)(5-4\cos x)} dx = \frac12\int_{-\infty}^\infty \frac{1}{(1+z^4)(2 - e^{iz})}dz$$
Since $\displaystyle\;\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\Im z \ge 0$ and $\displaystyle\;\left|\frac{1}{2-e^{iz}}\right| \le \frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and
$$\mathcal{I} = \lim_{R\to\infty}\frac12 \oint_{C_R} \frac{1}{(1+z^4)(2 - e^{iz})}dz \quad\text{ where }\quad C_R = [-R, R ] \cup \big\{\; Re^{i\theta} : \theta \in [0,\pi]\big\}$$
$\displaystyle\;\frac{1}{1+z^4}$ has $4$ poles $\omega_k = e^{\frac{(2k+1)i}{4\pi}}, k = 0..3$ over $\mathbb{C}$. Two of them $\omega_0 = \frac{1+i}{\sqrt{2}}$ and $\omega_1 = \frac{-1+i}{\sqrt{2}}$ belongs to the upper half-plane. Since
$$\frac{1}{1+z^4} = \sum_{k=0}^3 \frac{1}{(z-\omega_k) 4\omega_k^3} = -\frac14 \sum_{k=0}^3\frac{\omega_k}{z-\omega_k}$$
The residue of the integrand at $\omega_k$ are $\displaystyle\;-\frac14 \frac{\omega_k}{2 - e^{i\omega_k}}$ for $k = 0, 1$. This leads to | {
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