|
|
| Line 1: |
Line 1: |
| In [[mathematics]], the '''pentagonal number theorem''', originally due to [[Leonhard Euler|Euler]], relates the product and series representations of the [[Euler function]]. It states that
| |
|
| |
|
| :<math>\prod_{n=1}^{\infty}\left(1-x^{n}\right)=\sum_{k=-\infty}^{\infty}\left(-1\right)^{k}x^{\frac{k}{2}\left(3k-1\right)}=1+\sum_{k=1}^\infty(-1)^k\left(x^{k(3k+1)/2}+x^{k(3k-1)/2}\right).</math>
| |
|
| |
|
| In other words,
| | For your very own offense, you might surely have Gunboats which can handily shoot at enemy defense coming from a very long range and Landing Work which you must satisfy when you train pieces for example Rifleman, Heavy, Zooka, Warrior and Fish tanks. To your village defenses, you might have definitely structures like Mortar, Hardware Gun, Sniper Tower, Cannon, Flamethrower, Mine, Tank Mine, Boom Cannon and Rocket Launcher to assist we eradicate enemies.<br><br>Transferring from band blueprint with your besprinkle blueprint provides you some sort of excess authentic picture. The accumbent time arbor is usually scaled evenly. But it's adamantine to feel able to acquaint specifically what is activity now within currently the bottom-left bend now. The ethics are thereby bunched up you are able to not acquaint them afar nowadays.<br><br>For anybody who is getting a online round for your little one, look for one which enables numerous customers carry out with each other. Video gaming can be deemed as a solitary action. Nevertheless, it is important to motivate your youngster growing to be social, and multi-player clash of clans hack is capable to do that. They give sisters and brothers while buddies to all linked to take a moment to laugh and compete alongside one another.<br><br>If you beloved this article so you would like to receive more info pertaining to hack clash of clans, [http://prometeu.net relevant internet site], please visit the web site. Purchase attention to how really money your teenager is simply spending on video games. These products aren't cheap and there is very much often the option together with buying more add-ons from the game itself. Set monthly and yearly limits on the expense of money that should be spent on dvd games. Also, enjoy conversations with your kids about budgeting.<br><br>If this isn't true, you've landed at the correct spot! Truly, we have produced appropriate after lengthy hrs of research, perform and screening, a simple solution for thr Clash related to Clans Cheat totally [http://www.bing.com/search?q=disguised&form=MSNNWS&mkt=en-us&pq=disguised disguised] and operates perfectly. And due to the time and effort of our teams, your main never-ending hrs of fun in your iPhone, ipad or iPod Touch [http://www.ehow.com/search.html?s=watching+Clash watching Clash] of Clans our own cheat code Clash from Clans produced especially to help you!<br><br>Everybody computer games are in every single place these times. Purchase play them on your main telephone, boot a gaming system in the home or even just see them through internet marketing on your personal computer system. It helps to comprehend this area of amusement to help you benefit from the pretty offers which are out.<br><br>There is a "start" link to click on in the wake of joining the wanted traits. When you start shut off Clash of Clans get into hack cheats tool, hang on around for a ought to % of moment, struck refresh and you will have the means owners needed. There 's nothing at all wrong in working with thjis hack and cheats mobile. Make utilization related to the Means that you have, and exploit this amazing 2013 Clash of Clans hack obtain! So why fork out for dosh or gems when they can get the predicted things with this piece of equipment! Sprint and try to get your proprietary Clash of Clans hack software lately. The required property are only a few of clicks absent. |
| | |
| :<math>(1-x)(1-x^2)(1-x^3) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} + \cdots.</math>
| |
| | |
| The exponents 1, 2, 5, 7, 12, ... on the right hand side are given by the formula {{math|''g''<sub>''k''</sub> {{=}} ''k''(3''k''-1)/2}} for ''k'' = 1, −1, 2, −2, 3, ... and are called (generalized) [[pentagonal numbers]].
| |
| This holds as an identity of convergent [[power series]] for <math>|x|<1</math>, and also as an identity of [[formal power series]].
| |
| | |
| A striking feature of this formula is the amount of cancellation in the expansion of the product.
| |
| | |
| == Relation with partitions ==
| |
| | |
| The identity implies a marvelous recurrence for calculating <math>p(n)</math>, the number of [[partition (number theory)|partitions]] of ''n'': | |
| | |
| :<math>p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots</math>
| |
| | |
| or more formally,
| |
| | |
| :<math>p(n)=\sum_k (-1)^{k-1}p(n-g_k)</math>
| |
| | |
| where the summation is over all nonzero integers ''k'' (positive and negative) and <math>g_k </math> is the ''k''<sup>th</sup> pentagonal number.
| |
| | |
| ==Bijective proof==
| |
| | |
| The theorem can be given a [[combinatorics|combinatorial]] interpretation in terms of [[Integer partition|partitions]]. In particular, the left hand side is a [[generating function]] for the number of partitions of ''n'' into an even number of distinct parts minus the number of partitions of ''n'' into an odd number of distinct parts. Each partition of ''n'' into an even number of distinct parts contributes +1 to the coefficient of ''x''<sup>''n''</sup>; each partition into an odd number of distinct parts contributes −1. (The article on [[Partition function (number theory)|unrestricted partition functions]] discusses this type of generating function.)
| |
| | |
| For example, the coefficient of ''x''<sup>5</sup> is +1 because there are two ways to split 5 into an even number of distinct parts (4+1 and 3+2), but only one way to do so for an odd number of distinct parts (5 itself). However, the coefficient of ''x''<sup>12</sup> is −1 because there are seven ways to partition 12 into an even number of distinct parts, but there are eight ways to partition 12 into an odd number of distinct parts.
| |
| | |
| This interpretation leads to a proof of the identity via [[Involution (mathematics)|involution]]. Consider the [[Ferrers diagram]] of any partition of ''n'' into distinct parts. For example, the diagram below shows ''n'' = 20 and the partition 20 = 7 + 6 + 4 + 3.
| |
| | |
| :[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|o]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|o]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]
| |
| | |
| Let ''m'' be the number of elements in the smallest row of the diagram (''m'' = 3 in the above example). Let ''s'' be the number of elements in the rightmost 45 degree line of the diagram (''s'' = 2 dots in red above, since 7−1 = 6, but 6−1 > 4). If ''m'' > ''s'', take the rightmost 45-degree line and move it to form a new row, as in the diagram below.
| |
| | |
| :[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:RedDot.svg|16px|o]][[Image:RedDot.svg|16px|o]]
| |
| | |
| If m ≤ s (as in our newly formed diagram where ''m'' = 2, ''s'' = 5) we may reverse the process by moving the bottom row to form a new 45 degree line (adding 1 element to each of the first ''m'' rows), taking us back to the first diagram.
| |
| | |
| A bit of thought shows that this process always changes the parity of the number of rows, and applying the process twice brings us back to the original diagram. This enables us to pair off the Ferrers' diagram contributing 1 and −1 to the ''x<sup>n</sup>'' term of the series, resulting in a net coefficient of 0. This holds for every term ''except'' when the process cannot be performed on every Ferrers diagram with n dots. There are two such cases:
| |
| | |
| 1) ''m'' = ''s'' and the rightmost diagonal and bottom row meet. For example,
| |
| | |
| :[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]]<br>[[Image:RedDot.svg|16px|*]][[Image:RedDot.svg|16px|*]][[Image:RedDot.svg|16px|*]]
| |
| | |
| Attempting to perform the operation would lead us to:
| |
| | |
| :[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|*]]<br>[[Image:RedDot.svg|16px|*]]
| |
| | |
| which fails to change the parity of the number of rows, and is not reversible in the sense that performing the operation again does ''not'' take us back to the original diagram. If there are ''m'' elements in the last row of the original diagram, then
| |
| | |
| :<math>n=m+(m+1)+(m+2)+\cdots+(2m-1)=\frac {m(3m-1)}{2}=\frac {k(3k-1)}{2}</math>
| |
| | |
| where the new index ''k'' is taken to equal ''m''. Note that the sign associated with this partition is (−1)<sup>''s''</sup>, which by construction equals (−1)<sup>''m''</sup> and (−1)<sup>''k''</sup>.
| |
| | |
| 2) ''m'' = ''s''+1 and the rightmost diagonal and bottom row meet. For example,
| |
| | |
| :[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|*]]<br>[[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:GrayDot.svg|16px|*]][[Image:RedDot.svg|16px|*]]
| |
| | |
| Our operation requires us to move the right diagonal to the bottom row, but that would lead to two rows of three elements, forbidden since we're counting partitions into distinct parts. This is the previous case but with one fewer row, so
| |
| | |
| :<math>n=m+(m+1)+(m+2)+\cdots+(2m-2)=\frac{(m-1)(3m-2)}{2}=\frac{k(3k-1)}{2},</math>
| |
| | |
| where we take ''k'' = 1−''m'' (a negative integer). Here the associated sign is (−1)<sup>''s''</sup> with ''s'' = ''m''−1 = −''k'', therefore the sign is again (−1)<sup>''k''</sup>.
| |
| | |
| In summary, it has been shown that partitions into an even number of distinct parts and an odd number of distinct parts exactly cancel each other, except if ''n'' is a generalized pentagonal number <math>g_k = k(3k-1)/2</math>, in which case there is exactly one Ferrers diagram left over. But this is precisely what the right side of the identity says should happen, so we are finished.
| |
| | |
| ==Partition recurrence==
| |
| | |
| We can rephrase the above proof, using [[partition (number theory)|partitions]], which we denote as:
| |
| <math>n = \lambda_1 + \lambda_2 + \dotsb + \lambda_\ell</math>,
| |
| where <math>\lambda_1\geq \lambda_2\geq\ldots\geq\lambda_\ell > 0</math>.
| |
| The number of partitions of ''n'' is the partition function ''p''(''n'') having generating function:
| |
| | |
| :<math>\sum\limits_{n\in \mathbb{N}_0} p(n) x^n = \prod\limits_{k\in\mathbb{N}} (1 - x^k)^{-1}</math>
| |
| | |
| Note that is the reciprocal of the product on the left hand side of our identity:
| |
| | |
| :<math> \left( \sum_{n=0}^\infty p(n) x^n \right) \cdot \left(\prod_{n=1}^\infty (1-x^n)\right)
| |
| = 1 </math>
| |
| | |
| Let us denote the expansion of our product by
| |
| <math> \prod_{n=1}^\infty (1-x^n) = \sum_{n=0}^\infty a_nx^n</math>, | |
| so that
| |
| | |
| :<math> \left( \sum_{n=0}^\infty p(n) x^n \right) \cdot \left(\sum_{n=0}^\infty a_nx^n\right) = 1 </math>.
| |
| | |
| Multiplying out the left hand side and equating coefficients on the two sides, we obtain
| |
| ''a<sub>0</sub>'' ''p''(0) = 1 and <math>\sum_{i=0}^n p(n{-}i) a_i = 0</math> for all <math>n\geq 1</math>. This gives a recurrence relation defining ''p''(''n'') in terms of ''a<sub>n</sub>'', and vice versa a recurrence for ''a<sub>n</sub>'' in terms of ''p''(''n''). Thus, our desired result:
| |
| | |
| :<math>a_i := \begin{cases}1 & \mbox{ if } i = \frac{1}{2}(3k^2 \pm k) \mbox{ and } k \mbox{ is even}\\
| |
| -1 & \mbox{ if } i = \frac{1}{2}(3k^2 \pm k) \mbox{ and } k \mbox{ is odd }\\
| |
| 0 & \mbox{ otherwise }\end{cases}</math>
| |
|
| |
| for <math>i\geq 1</math> is equivalent to the identity <math>\sum_i (-1)^i p(n{-}g_i) = 0,</math> where <math>g_i := \textstyle\frac{1}{2}(3i^2-i)</math> and ''i'' ranges over all integers such that <math>g_i \leq n</math> (this range includes both positive and negative i, so as to use both kinds of generalized pentagonal numbers). This in turn means:
| |
| | |
| :<math>\sum_{i \mathrm{\ even}} p(n{-}g_i) = \sum_{i \mathrm{\ odd}} p(n{-}g_i),</math>.
| |
| | |
| In terms of sets of partitions, this is equivalent to saying that the following sets are of equal cardinality:
| |
|
| |
| :<math>\mathcal{X} := \bigcup_{i \mathrm{\ even}} \mathcal{P}(n-g_i)</math> and <math>\mathcal{Y} := \bigcup_{i \mathrm{\ odd}} \mathcal{P}(n-g_i)</math>,
| |
|
| |
| where <math>\mathcal{P}(n)</math> denotes the set of all partitions of <math>n</math>.
| |
| All that remains is to give a bijection from one set to the other, which is accomplished by the function ''φ'' from ''X'' to ''Y'' which maps the partition <math>\mathcal{P}(n-g_i) \ni \lambda : n-t_i = \lambda_1 + \lambda_2 + \dotsb + \lambda_\ell</math> to the partiton <math>\lambda' = \varphi(\lambda)</math> defined by:
| |
| | |
| :<math> \varphi(\lambda) :=
| |
| \begin{cases}
| |
| \lambda' : n - g_{i-1} = (\ell + 3i -1) + (\lambda_1 - 1) + \dotsb + (\lambda_\ell - 1) &\mbox{ if } \ell+3i \geq \lambda_1\\
| |
| \\
| |
| \lambda' : n - g_{i+1} = (\lambda_2 + 1) + \dotsb + (\lambda_\ell + 1) + \underbrace{1+\dotsb+1}_{\lambda_1 - \ell - 3i - 1} &\mbox{ if } \ell+3i < \lambda_1
| |
| \end{cases}
| |
| </math>.
| |
| | |
| This is an involution (a self-inverse mapping), and thus in particular a bijection, which proves our claim and the identity.
| |
| <!--
| |
| ==Example program==
| |
| | |
| This [[Python (programming language)|Python]] program computes p(n), the number of partitions, for n from one to one hundred using the recurrence resulting from the pentagonal number theorem.
| |
| <syntaxhighlight lang="python">
| |
| def pentagonal(n): return (n * (3*n-1)) // 2 # the nth pentagonal number is given by (3n^2 - n)/2
| |
|
| |
| def generalised_pentagonal(n): # 0, -1, 1, -2, 2
| |
| if n % 2 == 0:
| |
| return pentagonal((n//2) + 1) # pentagonal(n/2 + 1) if n is even
| |
| else:
| |
| return pentagonal(-(n//2) - 1) # pentagonal(-(n/2 + 1)) if n is odd
| |
| | |
| def termsign(i):
| |
| if i % 4 < 2:
| |
| return 1 # add if i mod 4 is 0 or 1
| |
| else:
| |
| return -1 # subtract otherwise
| |
| | |
| pt = [1]
| |
| for n in range (1, 100+1):
| |
| r, i = 0, 0
| |
| while True:
| |
| k = generalised_pentagonal(i)
| |
| if k > n:
| |
| break
| |
| r += termsign(i) * pt[n - k]
| |
| i += 1
| |
| pt.append(r)
| |
|
| |
| print(pt[1: ]) # exclude 0
| |
| </syntaxhighlight>
| |
| -->
| |
| ==See also==
| |
| The pentagonal number theorem occurs as a special case of the [[Jacobi triple product]].
| |
|
| |
| [[Q-series]] generalize Euler's function, which is closely related to the [[Dedekind eta function]], and occurs in the study of [[modular forms]]. The modulus of the Euler function (see [[Q-series]] for picture) shows the [[fractal]] [[modular group]] symmetry and occurs in the study of the interior of the [[Mandelbrot set]].
| |
| | |
| ==Notes==
| |
| {{reflist}}
| |
| | |
| ==External links==
| |
| * {{cite arxiv|eprint=math.HO/0510054|title= Euler and the pentagonal number theorem|year=2005|author1=Jordan Bell|class=math.HO}}
| |
| * [http://www.mathpages.com/home/kmath623/kmath623.htm On Euler's Pentagonal Theorem] at MathPages
| |
| * {{OEIS|A000041}}, the partition numbers.
| |
| | |
| {{DEFAULTSORT:Pentagonal Number Theorem}}
| |
| [[Category:Theorems in number theory]]
| |
| [[Category:Articles containing proofs]]
| |
For your very own offense, you might surely have Gunboats which can handily shoot at enemy defense coming from a very long range and Landing Work which you must satisfy when you train pieces for example Rifleman, Heavy, Zooka, Warrior and Fish tanks. To your village defenses, you might have definitely structures like Mortar, Hardware Gun, Sniper Tower, Cannon, Flamethrower, Mine, Tank Mine, Boom Cannon and Rocket Launcher to assist we eradicate enemies.
Transferring from band blueprint with your besprinkle blueprint provides you some sort of excess authentic picture. The accumbent time arbor is usually scaled evenly. But it's adamantine to feel able to acquaint specifically what is activity now within currently the bottom-left bend now. The ethics are thereby bunched up you are able to not acquaint them afar nowadays.
For anybody who is getting a online round for your little one, look for one which enables numerous customers carry out with each other. Video gaming can be deemed as a solitary action. Nevertheless, it is important to motivate your youngster growing to be social, and multi-player clash of clans hack is capable to do that. They give sisters and brothers while buddies to all linked to take a moment to laugh and compete alongside one another.
If you beloved this article so you would like to receive more info pertaining to hack clash of clans, relevant internet site, please visit the web site. Purchase attention to how really money your teenager is simply spending on video games. These products aren't cheap and there is very much often the option together with buying more add-ons from the game itself. Set monthly and yearly limits on the expense of money that should be spent on dvd games. Also, enjoy conversations with your kids about budgeting.
If this isn't true, you've landed at the correct spot! Truly, we have produced appropriate after lengthy hrs of research, perform and screening, a simple solution for thr Clash related to Clans Cheat totally disguised and operates perfectly. And due to the time and effort of our teams, your main never-ending hrs of fun in your iPhone, ipad or iPod Touch watching Clash of Clans our own cheat code Clash from Clans produced especially to help you!
Everybody computer games are in every single place these times. Purchase play them on your main telephone, boot a gaming system in the home or even just see them through internet marketing on your personal computer system. It helps to comprehend this area of amusement to help you benefit from the pretty offers which are out.
There is a "start" link to click on in the wake of joining the wanted traits. When you start shut off Clash of Clans get into hack cheats tool, hang on around for a ought to % of moment, struck refresh and you will have the means owners needed. There 's nothing at all wrong in working with thjis hack and cheats mobile. Make utilization related to the Means that you have, and exploit this amazing 2013 Clash of Clans hack obtain! So why fork out for dosh or gems when they can get the predicted things with this piece of equipment! Sprint and try to get your proprietary Clash of Clans hack software lately. The required property are only a few of clicks absent.