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I am preparing for my CS exams first year, I need help in understanding the below solution for given $$T(n)= \begin{cases} 3T(n-1),& \mbox{if }n>0\\ 1,& \mbox{otherwise} \end{cases}$$ equation.
solution: $$T(n) = 3T(n-1)$$ $$T(n) = 3 ( 3T(n-2)) = 3^2T(n-2)$$ $$T(n)= 3^2(3T(n-3))$$ how are we getting this $(n-3)$ from $(n-2)?$
We know that $T(n) = 3T(n-1)$. But this is true for all $n$. So it must be true for $n - 2$, as well. So $T(n-2) = 3T((n - 2) - 1)$. Simplifying inside the $T$, we have $T(n-2) = 3T(n-3)$. Since $T(n) = 3^2T(n-2)$, substituting $T(n-2) = 3T(n-3)$ gives us $T(n) = 3^2(3T(n-3))$.
We can simplify one step further, to get $T(n) = 3^3T(n-3)$.
• You might as well mention also that $\forall n \ge 0, T(n) = 3^n$. – Wildcard Jan 4 '17 at 2:35


text:
## Approximating the permanent of a matrix with nonnegative entries

03/30/2010 - 12:15pm
03/30/2010 - 1:10pm
Speaker:
Mark Huber (Claremont McKenna College)
Abstract:

Finding the determinant of a matrix is made easy by taking advantage of invariants arising from its geometric nature. The permanent slightly tweaks the formula for the determinant, but changes it into a combinatorial entity, making calculation far more difficult. Finding the permanent of a matrix, even when all the entries of the matrix are 0 or 1, is a #P complete problem. Still, all is not lost, as there are upper bounds like Minc's inequality, and lower bounds like Van der Waerden's Inequality. I will show how these two inequalities can be effectively used to construct a Monte Carlo algorithm for approximating the permanent. The permanent is related to counting perfect matchings in a bipartite graph, and this algorithm gives the first method for exact uniform simulation of perfect matchings for a nontrivial class of bipartite graphs.

Where:
Millikan 208 (Pomona College)

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Search: All articles in the CMB digital archive with keyword resonance

 Expand all        Collapse all Results 1 - 4 of 4

1. CMB 2014 (vol 57 pp. 658)

Thang, Nguyen Tat
 Admissibility of Local Systems for some Classes of Line Arrangements Let $\mathcal{A}$ be a line arrangement in the complex projective plane $\mathbb{P}^2$ and let $M$ be its complement. A rank one local system $\mathcal{L}$ on $M$ is admissible if roughly speaking the cohomology groups $H^m(M,\mathcal{L})$ can be computed directly from the cohomology algebra $H^{*}(M,\mathbb{C})$. In this work, we give a sufficient condition for the admissibility of all rank one local systems on $M$. As a result, we obtain some properties of the characteristic variety $\mathcal{V}_1(M)$ and the Resonance variety $\mathcal{R}_1(M)$. Keywords:admissible local system, line arrangement, characteristic variety, multinet, resonance varietyCategories:14F99, 32S22, 52C35, 05A18, 05C40, 14H50

2. CMB 2010 (vol 54 pp. 56)

Dinh, Thi Anh Thu
 Characteristic Varieties for a Class of Line Arrangements Let $\mathcal{A}$ be a line arrangement in the complex projective plane $\mathbb{P}^2$, having the points of multiplicity $\geq 3$ situated on two lines in $\mathcal{A}$, say $H_0$ and $H_{\infty}$. Then we show that the non-local irreducible components of the first resonance variety $\mathcal{R}_1(\mathcal{A})$ are 2-dimensional and correspond to parallelograms $\mathcal{P}$ in $\mathbb{C}^2=\mathbb{P}^2 \setminus H_{\infty}$ whose sides are in $\mathcal{A}$ and for which $H_0$ is a diagonal. Keywords:local system, line arrangement, characteristic variety, resonance varietyCategories:14C21, 14F99, 32S22, 14E05, 14H50

3. CMB 2009 (vol 53 pp. 347)

Papageorgiou, Nikolaos S.; Staicu, Vasile
 Multiple Nontrivial Solutions for Doubly Resonant Periodic Problems We consider semilinear periodic problems with the right-hand side nonlinearity satisfying a double resonance condition between two successive eigenvalues. Using a combination of variational and degree theoretic methods, we prove the existence of at least two nontrivial solutions. Keywords:double resonance, generalized LL-condition, Leray-Schauder degree, Cerami conditionCategory:34B15

4. CMB 2001 (vol 44 pp. 323)

Schuman, Bertrand
 Une classe d'hamiltoniens polynomiaux isochrones Soit $H_0 = \frac{x^2+y^2}{2}$ un hamiltonien isochrone du plan $\Rset^2$. On met en \'evidence une classe d'hamiltoniens isochrones qui sont des perturbations polynomiales de $H_0$. On obtient alors une condition n\'ecessaire d'isochronisme, et un crit\ere de choix pour les hamiltoniens isochrones. On voit ce r\'esultat comme \'etant une g\'en\'eralisation du caract\ere isochrone des perturbations hamiltoniennes homog\`enes consid\'er\'ees dans [L], [P], [S]. Let $H_0 = \frac{x^2+y^2}{2}$ be an isochronous Hamiltonian of the plane $\Rset^2$. We obtain a necessary condition for a system to be isochronous. We can think of this result as a generalization of the isochronous behaviour of the homogeneous polynomial perturbation of the Hamiltonian $H_0$ considered in [L], [P], [S]. Keywords:Hamiltonian system, normal forms, resonance, linearizationCategories:34C20, 58F05, 58F22, 58F30
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# If m is a positive integer and m^2 is divisible by 48, then

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If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 11:42
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If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

Can someone please explain this one? Knewton's explanation was not useful to me. Thx
OA
[Reveal] Spoiler:
D
[Reveal] Spoiler: OA
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Re: Division question [#permalink]  10 Oct 2011, 11:56
Well m^2 is divisible by 48

if we brake 48 we will get 4*4*3

now we know that m is an integer therefore, m^2 is completely divisible by 48 then to get the square number that will be divided by 48 we need to multiply 48 by 3
so that we will get 4*4*3*3 ....
square root of the above number will be 4*3 = 12
thus m = 12 and the greatest number that will divide 12 is 12 itself ...thus the answer.
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Re: Division question [#permalink]  10 Oct 2011, 12:05
arcanis2000 wrote:
Q: If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

Can someone please explain this one? Knewton's explanation was not useful to me. Thx
OA
[Reveal] Spoiler:
D

M^2 is divisible by 48 so M^2 must be multiple of 48.
If the value of M is Multiples of 12 then it will satisfy the condition. If we If M is 12 or 24 or 36 then it ans is D but if M = 48 then answer should be 16.

Is the question right? Or am i missing some thing?
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Re: Division question [#permalink]  10 Oct 2011, 12:07
vyassaptarashi wrote:
Well m^2 is divisible by 48

if we brake 48 we will get 4*4*3

now we know that m is an integer therefore, m^2 is completely divisible by 48 then to get the square number that will be divided by 48 we need to multiply 48 by 3
so that we will get 4*4*3*3 ....
square root of the above number will be 4*3 = 12
thus m = 12 and the greatest number that will divide 12 is 12 itself ...thus the answer.

This explanation looks fine but why M shouldn't be 48. There is no mention in the question about M's maximum limit.
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Re: Division question [#permalink]  10 Oct 2011, 12:22
1
KUDOS
Well, the largest integer, which will divide m, has asked in the question. This means there will be no other integer that will divide m is larger than what the answer is.
Also, It is not asked us that which of the following will be the largest integer that will divide m....

Thus m can not be 48 because the largest integer that divides 48 will be 48 ...which is not in the options, hence can not be correct. 16 is also not correct because it is the largest number in the options but larger that 16 divisors are also present.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 14:45
ok, I will try to explain this as best as possible.
the following is a tree diagram, or columns. hope it is easy to see it

m^2____m__ 2 2
____m__ 2 2 3

so, 2 * 2* 3= 12

In another words,
find out the prime factors of 48, ( 2,2,2,2,3)

, you know that M^2 is m repeated twice, so you

should distribute the the prime factors evenly

among the 2 m's, the 3 will be left over so you

should just put it out there under one of the m's.

Now, for every columns take one common number

which is 2 for the first columns, 2 for the second

column and 3 in the last column. multiplying the

prime factors ( 2*2*3) = 12...

best of luck
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 16:37
m^2 is divisible by 48

=> 48 is a factor of m^2

=> 2^4*3 is a factor of m^2

to make the above number a perfect square m^2 must have another 3 as a factor

=> (2^4)*(3^2) must be a factor of m^2

=> (2^2)*(3^1) = 12 must be a factor of m.

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 18:31
its mentioned indirectly by saying m has to be an integer.

if m^2 is not a perfect square then m will not be an integer.

for m to be in an integer, m^2 has to be a perfect square.

lets say m^2=48*3 = 144 (perfect square ) => m =12 (integer)

lets say m^2 = 48(not a perfect square) = > m= sqrt(48) = 2*sqrt(12) (not an integer)

arcanis2000 wrote:

@Spidy001: The perfect square is what Knewton gave as part of the answer. In what part of this question does it suggest that M^2 has to be a perfect square? What am I missing?

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 00:25
I dont Get it. Factors of 48 are= 2,2,2,2 and 3 and we know that 48 is a factor of m. Therefor 48 is also a factor of m^2.
So (2^4)*3 is a factor of m and m^2. But how do you come to the solution of 12? I don't see the link between (2^4)*3 and (2^2)*(3^2). Can someone explain please?

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 00:49
m^2 is a divisor of 48. M =/= divisor of 48.

Find the prime factors of 48, [2,2,2,2,3]. When you root square a number you double the amount of it's divisor, so if you root them number it should be half. So divide in half all the repeating multiples. You know there must be a 3, since 3 is a prime and no 2 integer forms a multiple of 3. Also half the 2s are gone so you have a two 2s left. 3*2*2 = 12.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:25
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M3tm4n wrote:
I dont Get it. Factors of 48 are= 2,2,2,2 and 3 and we know that 48 is a factor of m. Therefor 48 is also a factor of m^2.
So (2^4)*3 is a factor of m and m^2. But how do you come to the solution of 12? I don't see the link between (2^4)*3 and (2^2)*(3^2). Can someone explain please?

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If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?
(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

m^2 is a positive perfect square divisible by 48, the least such perfect square is 144=48*3=12^2. Thus the least value of m is 12, which means that m in any case must be divisible by 12.

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Hope it helps.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:36
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:43
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MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:46
yeah it is clear! Thanks, i wish i could explain it always like you did now.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 06:01
Bunnel thanks for the explanation
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  24 Aug 2014, 12:43
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  21 Jul 2015, 12:36
Bunuel wrote:
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.

Thanks alot for the help here. I have a quick doubt, the question is asking for the largest value of the largest value of m. Why are we finding the least value of m. What am I missing here.

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  21 Jul 2015, 13:08
neeraj609 wrote:
Bunuel wrote:
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.

Thanks alot for the help here. I have a quick doubt, the question is asking for the largest value of the largest value of m. Why are we finding the least value of m. What am I missing here.

The question is asking for the largest values from the given options. You are supposed to find which one of the 5 options gives you the correct answer. You are confusing between finding the "maximum possible value" when no options are given and "maximum possible value" out of the given options. For our use, the least value worked . If lets say the least did not work, we could have looked for a multiple of 12 (say 24 or 26 or 48 etc.).
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  21 Jul 2015, 16:18
Thanks alot for the response. I was also able find the explanation from on of the another post from Bunnel.
Re: If m is a positive integer and m^2 is divisible by 48, then   [#permalink] 21 Jul 2015, 16:18
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• SHUHRAH ALI S ALLAMI

Articles written in Bulletin of Materials Science

• Design and synthesis of organic dyes with various donor groups: promising dyes for dye-sensitized solar cells

The present work is involved in the computational and experimental studies of organic dyes and their applications as dye-sensitized solar cells (DSSCs). This comprised the study of three hydrazone-based sensitizers ($E$)-2-cyano-$N'$-((2-hydroxynaphthalen-1-yl)methylene)acetohydrazide (CHMA), ($E$)-2-cyano-$N'$-(4-(dimethylamino)benzylidene)acetohydrazide (CDBA), ($E$)-$N'$-(anthracen-9-ylmethylene)-2-cyanoacetohydrazide (AMCH) that have been prepared and confirmed by means of several analytical procedures like Fourier transform infrared, UV–visible and nuclear magnetic resonance techniques to investigate the best possible selection for DSSCs by computational and experimental techniques. The computational methods are applied to optimize the structures of prepared organic dyes via density functional theory (DFT) method at B3LYP/6-311G(p,d) level of theory. The time-dependent DFT (TD-B3LYP/6-311G**) was used with and without solvent to find out the absorption spectra and matched with the experimental data and the electro-optical and reorganization energies of prepared dyes were further investigated. The results revealed that the prepared dyes would be better sensitizers for DSSCs because of small highest-occupied molecular orbital–lowest unoccupied molecular orbital energy gap. Moreover, on the basis of the above results, we fabricated the devices via the doctor blade method to study the photovoltaic performance with the prepared dyes (CHMA, CDBA and AMCH). The dye AMCH exhibited the maximum efficiency with commercial TiO$_2$.

• # Bulletin of Materials Science

Volume 44, 2021
All articles
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• # Dr Shanti Swarup Bhatnagar for Science and Technology

Posted on October 12, 2020

Prof. Subi Jacob George — Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru
Chemical Sciences 2020


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# Synthetic Evidence of Benzene

Synthetic Evidence of Benzene

(1) Benzene is unreactive to addition reactions. Due to the presence of 03 alternate double bonds in benzene structure, the chemists of that time believe it to react easily with bromine in the dark at room temperature. But due to uniform delocalized electrons (aromaticity) between all C – C bonds, benzene regained unusual stability and does not react with bromine.

(2) The hydrogenation across double bond easily occurs.

Benzene contains 03 double bonds and should liberate three times the energy that of cyclohexene. i.e., (3x 119) = 357 kJ mol¹. But the actual value for benzene is -208 kJ mol¹. So benzene is (357-119) = 152 kJ mol¹ more stable than expected as it does not contain three ordinary alternate double bonds. All these three double bonds are resonating within the ring.

The benzene ring has thus delocalized electrons. Hence, all C-C bond lengths are the same.

Summary:

1. Benzene is a flat molecule with six carbons bonded in a planar ring. It has unusual stability.
2. Each carbon is covalently joined to two adjoining carbons and one hydrogen. The remaining outer electron of each carbon is shared with the electron of the next carbon in the ring. Thus, the outer electrons of all the six carbons are delocalized above and below the plane of the ring, giving the ring unusual stability.
3. The six carbon atoms form a perfectly regular hexagon. There are no distinct single or double bonds within the benzene. The delocalization of the ring makes each count as one and a half bonds between the carbons. Hence, all the carbon-carbon bonds have exactly the same length-somewhere between single and double bonds.
4. Benzene resists addition reactions because these reactions would involve breaking the delocalization and losing stability.
Make sure you also check our other amazing Article on : Orbital Picture of Benzene


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# [SOLVED] Completing the Square

• May 28th 2010, 03:54 PM
Zero17
[SOLVED] Completing the Square
Hi all, this is my first post here.

Ok, I'm using a math program in college (I forgot everything I learned in highschool math and am retaking courses to get up tp speed) for some algebra level math. I'm currently working on a section called "completing the square". Everything was going fine until this problem:

http://img294.imageshack.us/img294/2425/whatm.png

I wasn't sure how to go about doing the problem, and this is how the program responded. Normally I will pick up on how to do something after it shows me, but I'm not sure how it went about doing this one. First off, the problem started as:

X^2+(x+2)^2=10^2. Where did the 4X come from? The other part I'm a bit confused about is, a few lines down, the program goes from:

x^2+2x+2=50
to:
X^2+2x+?=48+?

It then changes it to a 1. What's up with that? I don't recall ever doing that in highschool. The program isn't explaining how it did any of this either, other than what you see here. Can someone enlighten me?
• May 28th 2010, 04:21 PM
11rdc11
Quote:

Originally Posted by Zero17
Hi all, this is my first post here.

Ok, I'm using a math program in college (I forgot everything I learned in highschool math and am retaking courses to get up tp speed) for some algebra level math. I'm currently working on a section called "completing the square". Everything was going fine until this problem:

http://img294.imageshack.us/img294/2425/whatm.png

I wasn't sure how to go about doing the problem, and this is how the program responded. Normally I will pick up on how to do something after it shows me, but I'm not sure how it went about doing this one. First off, the problem started as:

X^2+(x+2)^2=10^2. Where did the 4X come from? The other part I'm a bit confused about is, a few lines down, the program goes from:

x^2+2x+2=50
to:
X^2+2x+?=48+?

It then changes it to a 1. What's up with that? I don't recall ever doing that in highschool. The program isn't explaining how it did any of this either, other than what you see here. Can someone enlighten me?

$(x+2)^2 = (x+2)(x+2)$

Now just use foil and you get

$x^2 +4x +4$
• May 28th 2010, 04:21 PM
bigwave
Quote:

Originally Posted by Zero17
x^2+(x+2)^2=10^2. Where did the 4X come from? The other part I'm a bit confused about is, a few lines down, the program goes from:

x^2+2x+2=50
to:
X^2+2x+?=48+?

$
x^2+(x+2)^2=10^2
$

the $4x$ comes from expanding $(x+2)^2$
which become $x^2 + 4x + 4$

after simplifing you get $x^2 + 2x + 2 = 50$
however this is not in the format of $( )^2 = ()^2$so you subtract 1 from both sides.
$x^2 + 2x + 2 -1 = 50 -1$

now you have
$
x^2 + 2x +1 = 49 \rightarrow (x+1)^2 = 49 = 7^2
$
• May 28th 2010, 04:49 PM
Zero17
I see, now it makes sense. Thanks guys!


text:
# SONATA

System of nearrings and their applications

Version 2.9.1
Released 2018-10-07

This project is maintained by Erhard Aichinger, Peter Mayr

# GAP Package SONATA

The SONATA package provides methods for the construction and analysis of finite nearrings.

The current version of this package is version 2.9.1, released on 2018-10-07. For more information, please refer to the package manual. There is also a README file and a tutorial.

## Dependencies

This package requires GAP version >=4.9

The following additional GAP packages are not required, but suggested:

## Authors

Erhard Aichinger, Franz Binder, Jürgen Ecker, Peter Mayr, Christof Nöbauer.

## Introduction

SONATA stands for “systems of nearrings and their applications”. It provides methods for the construction and the analysis of finite nearrings. A left nearring is an algebra (N;+,.), where (N,+) is a (not necessarily abelian) group, (N,.) is a semigroup, and x(y+z) = xy + xz holds for all x,y,z in N.

As a typical example of a nearring, we may consider the set of all mappings from a group G into G, where the addition is the pointwise addition of mappings in G, and the multiplication is composition of functions. If functions are written on the right of their arguments, then the left distributive law holds, while the right distributive law is not satisfied for non-trivial G.

The SONATA package provides methods for the construction and analysis of finite nearrings.

• Methods for constructing all endomorphisms and all fixed-point-free automorphisms of a given group.
• Methods for constructing the following nearrings of functions on a group G: the nearring of polynomial functions of G (in the sense of Lausch-Nöbauer); the nearring of compatible functions of G; distributively generated nearrings such as I(G), A(G), E(G); centralizer nearrings.
• A library of all small nearrings (up to order 15) and all small nearrings with identity (up to order 31).
• Functions to obtain solvable fixed-point-free automorphism groups on abelian groups, nearfields, planar nearrings, as well as designs from those.
• Various functions to study the structure (size, ideals, N-groups, …) of nearrings, to determine properties of nearring elements, and to decide whether two nearrings are isomorphic.
• If the package XGAP is installed, the lattices of one- and two-sided ideals of a nearring can be studied interactively using a graphical representation.

## References

• J.R. Clay, Nearrings: Geneses and applications, Oxford University Press – Oxford, New York, Tokyo, 1992.
• C. Cotti Ferrero and G. Ferrero, Nearrings. some developments linked to semigroups and groups, Kluwer Academic Publishers, Dordrecht, Boston, London, 2002.
• H. Lausch and W. Nöbauer, Algebra of polynomials, North-Holland, Amsterdam, London; American Elsevier Publishing Company, New York, 1973.
• J.D.P. Meldrum, Near-rings and their links with groups, Pitman (Advanced Publishing Program), Boston, Mass., 1985.
• G.F. Pilz, Near-rings, 2nd ed., North-Holland Publishing Company – Amsterdam, New York, Oxford, 1983.
• H. Wähling, Theorie der Fastkörper, Thales-Verlag, Essen, 1987.

## Citing

You can get more info by typing Cite("SONATA"); in the gap prompt.