"점화식"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
Pythagoras0 (토론 | 기여) (→선형점화식) |
Pythagoras0 (토론 | 기여) (→메타데이터: 새 문단) |
||
| (같은 사용자의 중간 판 4개는 보이지 않습니다) | |||
| 7번째 줄: | 7번째 줄: | ||
| − | |||
| − | ==기본적인 점화식 | + | ==기본적인 점화식== |
* <math>a_{n+1} - a_n = c</math> : 등차수열 | * <math>a_{n+1} - a_n = c</math> : 등차수열 | ||
| 22번째 줄: | 21번째 줄: | ||
==기본 점화식의 응용== | ==기본 점화식의 응용== | ||
| − | * <math>a_{n+1} = ka_n + c</math | + | * <math>a_{n+1} = ka_n + c</math> |
** 양변에 적당한 상수를 더하면 <math>(a_{n+1} + p)= k(a_n + p)</math> 꼴로 만들 수 있다. | ** 양변에 적당한 상수를 더하면 <math>(a_{n+1} + p)= k(a_n + p)</math> 꼴로 만들 수 있다. | ||
** 일반항이 <math>(a_n + p)</math> 인 수열은 공비 <math>k</math> 인 등비수열, <math>a_n + p =(a_1 + p) k^{n-1}</math> | ** 일반항이 <math>(a_n + p)</math> 인 수열은 공비 <math>k</math> 인 등비수열, <math>a_n + p =(a_1 + p) k^{n-1}</math> | ||
| − | ** 적당한 상수 <math>p</math> 는 어떻게 찾냐고? 생각해 볼 것. | + | ** 적당한 상수 <math>p</math> 는 어떻게 찾냐고? 생각해 볼 것. |
| − | ** ex) <math>a_{n+1} = 2a_{n} + 3</math>, 초기항 1 | + | ** ex) <math>a_{n+1} = 2a_{n} + 3</math>, 초기항 1 양변에 3을 더하면 <math>(a_{n+1} +3 )= 2( a_{n} + 3)</math>, 적당한 상수 <math>k</math> 에 대하여 <math>a_{n} + 3 = k 2^n</math> 초항을 만족시키는 <math>k</math> 값은 2이므로, <math>a_n = 2^{n+1} - 3</math> |
| − | * <math>a_{n+1} = ra_n + b_n</math> 꼴의 점화식 | + | * <math>a_{n+1} = ra_n + b_n</math> 꼴의 점화식 |
** 양변을 <math>r^{n+1}</math> 로 나눈 후, <math>a_n / r^n</math> 에 대한 점화식을 푸는 것이 한 방법. <math>b_n</math> 이 등비수열인 경우 효과적이다. | ** 양변을 <math>r^{n+1}</math> 로 나눈 후, <math>a_n / r^n</math> 에 대한 점화식을 푸는 것이 한 방법. <math>b_n</math> 이 등비수열인 경우 효과적이다. | ||
| − | ** 양변에 적당히 <math>n</math> 에 대한 식을 더해서 공비 <math>r</math> 에 대한 등비수열 꼴로 만들 수 있는 경우가 많다. | + | ** 양변에 적당히 <math>n</math> 에 대한 식을 더해서 공비 <math>r</math> 에 대한 등비수열 꼴로 만들 수 있는 경우가 많다. |
| − | *** <math>b_n</math> 이 다항식인 경우, 양변에 <math>b_n</math> 과 같은 차수의 다항식을 (계수를 문자로 두고) 더해서, 등비수열 꼴로 만든 후에 계수 비교를 통해 문자를 찾는다. | + | *** <math>b_n</math> 이 다항식인 경우, 양변에 <math>b_n</math> 과 같은 차수의 다항식을 (계수를 문자로 두고) 더해서, 등비수열 꼴로 만든 후에 계수 비교를 통해 문자를 찾는다. |
| − | *** ex ) <math>a_{n+1} = 2a_n + 3n + 5</math> 인 경우, 양변에 <math>pn + q</math> 를 더하면:<math>a_{n+1} + pn + q = 2a_n + (3+p)n + (5+q)</math | + | *** ex ) <math>a_{n+1} = 2a_n + 3n + 5</math> 인 경우, 양변에 <math>pn + q</math> 를 더하면:<math>a_{n+1} + pn + q = 2a_n + (3+p)n + (5+q)</math> 우변이 <math>2(a_n+ pn + q)</math> 인 경우에 등비수열이 되니까, <math>3+p = 2p, \quad 5+q=2q</math> 이므로 <math>p=3, \quad q=5</math>. 그러므로:<math>a_n + 3n + 5 = k2^n</math>. 초기항이 주어진 경우 k 를 찾을 수 있다. |
| − | * 점화식에 덧셈 기호가 없을 때 | + | * 점화식에 덧셈 기호가 없을 때 |
| − | ** 로그를 취하면 도움이 됨. 로그의 밑은 계산이 간단하도록 적절히 선택하기. | + | ** 로그를 취하면 도움이 됨. 로그의 밑은 계산이 간단하도록 적절히 선택하기. |
| − | ** ex) <math>a_{n+1} = 4a_n^2</math> : 밑 2 인 로그를 취하면 <math>\log a_{n+1} = 2\log a_n + 2</math | + | ** ex) <math>a_{n+1} = 4a_n^2</math> : 밑 2 인 로그를 취하면 <math>\log a_{n+1} = 2\log a_n + 2</math> 이제 <math>\log a_n = b_n</math>에 대한 점화식을 풀면 됨. (양변에 2를 더해서 …) |
| − | * 점화식이 분수꼴일때 | + | * 점화식이 분수꼴일때 |
| − | ** 역수를 취하면 도움이 될 때가 많음. (만능은 아님) | + | ** 역수를 취하면 도움이 될 때가 많음. (만능은 아님) |
| − | ** ex) <math>{a_{n+1}} = \frac{2a_n}{3a_n+1}</math> : 역수를 취하면 <math>\frac{1}{a_{n+1}} = \frac{1}{2} \frac{1}{a_n} + \frac{3}{2}</math>. 이제 <math>b_n = \frac{1}{a_n}</math> 에 대한 점화식으로 보고 풀면 됨. | + | ** ex) <math>{a_{n+1}} = \frac{2a_n}{3a_n+1}</math> : 역수를 취하면 <math>\frac{1}{a_{n+1}} = \frac{1}{2} \frac{1}{a_n} + \frac{3}{2}</math>. 이제 <math>b_n = \frac{1}{a_n}</math> 에 대한 점화식으로 보고 풀면 됨. |
| − | * 점화식에 <math>a_n</math> 과 <math>S_n</math> 이 함께 나올 때 | + | * 점화식에 <math>a_n</math> 과 <math>S_n</math> 이 함께 나올 때 |
| − | ** <math>S_n - S_{n-1}=a_n</math><math>(n \ge 2)</math> , <math>S_1 = a_1 </math> 의 관계를 사용하면 <math>a_n</math> 만의 점화식으로 만들 수 있다. | + | ** <math>S_n - S_{n-1}=a_n</math><math>(n \ge 2)</math> , <math>S_1 = a_1 </math> 의 관계를 사용하면 <math>a_n</math> 만의 점화식으로 만들 수 있다. |
| − | ** ex) <math> a_n = 2 S_n + 3^n</math> 일 때, <math> a_1 = 2 a_1 + 3</math> 이므로 <math>a_1 = -3</math>:<math>a_{n+1} = 2 S_{n+1} + 3^{n+1}</math> 식과 <math> a_n = 2 S_n + 3^n</math> 식을 빼 주면 <math>a_{n+1} - a_n = 2 a_{n+1} + 2\cdot 3^{n}</math | + | ** ex) <math> a_n = 2 S_n + 3^n</math> 일 때, <math> a_1 = 2 a_1 + 3</math> 이므로 <math>a_1 = -3</math>:<math>a_{n+1} = 2 S_{n+1} + 3^{n+1}</math> 식과 <math> a_n = 2 S_n + 3^n</math> 식을 빼 주면 <math>a_{n+1} - a_n = 2 a_{n+1} + 2\cdot 3^{n}</math> 이제부터는 알아서 할 것. 부분합과 일반항이 함께 등장하는 점화식에서는 초기 조건에서 실수를 할 가능성이 높으므로, <math>a_1, a_2, a_3</math> 정도는 점화식으로부터 직접 구해 보는 것이 좋을 것이다. 그리고 구한 일반항 <math>a_n</math> 은 <math>(n \ge 2)</math> 에서만 성립하는 식일 가능성도 있다. |
| 56번째 줄: | 55번째 줄: | ||
* <math>pa_{n+2} + qa_{n+1} + ra_n = 0</math> 꼴의 점화식 | * <math>pa_{n+2} + qa_{n+1} + ra_n = 0</math> 꼴의 점화식 | ||
* <math>pa_{n+2} + qa_{n+1} + ra_n = b_n</math> 꼴의 점화식 | * <math>pa_{n+2} + qa_{n+1} + ra_n = b_n</math> 꼴의 점화식 | ||
| − | * [[선형점화식]] 항목을 참조 | + | * [[상수계수 선형점화식]] 항목을 참조 |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
==관련된 항목들== | ==관련된 항목들== | ||
| 75번째 줄: | 64번째 줄: | ||
* [[선형대수학]] | * [[선형대수학]] | ||
| − | |||
| − | |||
| − | == | + | ==사전 형태의 자료== |
| + | * http://ko.wikipedia.org/wiki/점화식 | ||
| + | * http://en.wikipedia.org/wiki/Recurrence_relation | ||
| + | [[분류:고교수학]] | ||
| + | [[분류:수열]] | ||
| + | |||
| + | == 메타데이터 == | ||
| − | + | ===위키데이터=== | |
| − | + | * ID : [https://www.wikidata.org/wiki/Q740970 Q740970] | |
| − | * | ||
| − | |||
| − | + | == 노트 == | |
| − | + | ===말뭉치=== | |
| + | # The procedure for finding the terms of a sequence in a recursive manner is called recurrence relation.<ref name="ref_f83c3639">[https://www.tutorialspoint.com/discrete_mathematics/discrete_mathematics_recurrence_relation.htm Discrete Mathematics]</ref> | ||
| + | # We study the theory of linear recurrence relations and their solutions.<ref name="ref_f83c3639" /> | ||
| + | # Doing so is called solving a recurrence relation .<ref name="ref_2fd665a6">[http://discrete.openmathbooks.org/dmoi2/sec_recurrence.html Solving Recurrence Relations]</ref> | ||
| + | # Recall that the recurrence relation is a recursive definition without the initial conditions.<ref name="ref_2fd665a6" /> | ||
| + | # , 485\ldots\text{.}\) Solution Finding the recurrence relation would be easier if we had some context for the problem (like the Tower of Hanoi, for example).<ref name="ref_2fd665a6" /> | ||
| + | # Remember, the recurrence relation tells you how to get from previous terms to future terms.<ref name="ref_2fd665a6" /> | ||
| + | # Recurrence relations are used to reduce complicated problems to an iterative process based on simpler versions of the problem.<ref name="ref_9f182e6c">[https://brilliant.org/wiki/recurrence-relations/ Brilliant Math & Science Wiki]</ref> | ||
| + | # Identifying a candidate problem to solve with a recurrence relation: The problem can be reduced to simpler cases.<ref name="ref_9f182e6c" /> | ||
| + | # The term difference equation sometimes (and for the purposes of this article) refers to a specific type of recurrence relation.<ref name="ref_eb26bcef">[https://en.wikipedia.org/wiki/Recurrence_relation Recurrence relation]</ref> | ||
| + | # A recurrence relation is an equation that expresses each element of a sequence as a function of the preceding ones.<ref name="ref_eb26bcef" /> | ||
| + | # This defines recurrence relation of first order.<ref name="ref_eb26bcef" /> | ||
| + | # The recurrence of order two satisfied by the Fibonacci numbers is the archetype of a homogeneous linear recurrence relation with constant coefficients (see below).<ref name="ref_eb26bcef" /> | ||
| + | # In this article, we will see how we can solve different types of recurrence relations using different approaches.<ref name="ref_5bf4f82a">[https://www.geeksforgeeks.org/different-types-recurrence-relations-solutions/ Different types of recurrence relations and their solutions]</ref> | ||
| + | # Therefore, we need to convert the recurrence relation into appropriate form before solving.<ref name="ref_5bf4f82a" /> | ||
| + | # There are a variety of methods for solving recurrence relations, with various advantages and disadvantages in particular cases.<ref name="ref_d4428fa4">[https://www.whitman.edu/mathematics/cgt_online/book/section03.04.html 3.4 Recurrence Relations]</ref> | ||
| + | # One method that works for some recurrence relations involves generating functions.<ref name="ref_d4428fa4" /> | ||
| + | # If we can find an explicit representation for the series for this function, we will have solved the recurrence relation.<ref name="ref_d4428fa4" /> | ||
| + | # We can often solve a recurrence relation in a manner analogous to solving a differential equations by multiplying by an integrating factor and then integrating.<ref name="ref_fc851a7a">[https://aofa.cs.princeton.edu/20recurrence/ Recurrence Relations]</ref> | ||
| + | # Wolfram|Alpha can solve various kinds of recurrences, find asymptotic bounds and find recurrence relations satisfied by given sequences.<ref name="ref_30bdd107">[https://www.wolframalpha.com/examples/mathematics/discrete-mathematics/recurrences/ Alpha Examples: Recurrences]</ref> | ||
| + | # This is called a recurrence relation.<ref name="ref_9cc2053b">[http://www.cs.cornell.edu/courses/cs312/2008sp/lectures/lec17.html Lecture 17: Recurrence relations]</ref> | ||
| + | # We somehow need to figure out how often the first versus the second branch of this recurrence relation will be taken.<ref name="ref_9cc2053b" /> | ||
| + | # We will first find a recurrence relation for the execution time.<ref name="ref_9cc2053b" /> | ||
| + | # These recurrence relations are shown to be closely related to each other.<ref name="ref_08c49046">[https://aip.scitation.org/doi/10.1063/1.1665327 Recurrence Relations for the Multiplicities in the Classical Groups]</ref> | ||
| + | # Racah has obtained a recurrence relation for the ``inner multiplicity'' (multiplicity of weights).<ref name="ref_08c49046" /> | ||
| + | # Moreover, Kostant's recurrence relation for the partition function as well as Racah's recurrence relation for the inner multiplicity have been generalized.<ref name="ref_08c49046" /> | ||
| + | # In this paper, the recurrence relation for negative moments along with negative factorial moments of some discrete distributions can be obtained.<ref name="ref_d56ddfd9">[https://www.hindawi.com/journals/jps/2015/972035/ Recurrence Relation and Accurate Value on Inverse Moment of Discrete Distributions]</ref> | ||
| + | # Is `D(4)` easier to calculate using the factorial formula or using the recurrence relation?<ref name="ref_37551064">[http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/LRGF.html Recurrence Relations and Generating Functions]</ref> | ||
| + | # Therefore the solution to the recurrence relation will have the form: a n =a2n+b18n.<ref name="ref_831ad1eb">[https://www.sanfoundry.com/discrete-mathematics-questions-answers-recurrence-relation/ Discrete Mathematics Questions and Answers]</ref> | ||
| + | # Therefore the solution to the recurrence relation will have the form: a n =a.6n+b.n.6n.<ref name="ref_831ad1eb" /> | ||
| + | # Explanation: Check for the left side of the equation with all the options into the recurrence relation.<ref name="ref_831ad1eb" /> | ||
| + | # Together with the initial conditions, the recurrence relation provides a recursive definition for the elements of the sequence.<ref name="ref_4c82613d">[https://link.springer.com/chapter/10.1007/978-1-4757-4878-9_6 Recurrence Relations]</ref> | ||
| + | # The solution techniques for these two classes of recurrence relation are discussed in detail in the Recurrence page.<ref name="ref_686afdb9">[https://web.cs.wpi.edu/~cs2223/d98/notes/recurrence/classify/classify.html cs2223 Classifying Recurrence Relations]</ref> | ||
| + | # We also justified Pascal's Formula as a recurrence relation.<ref name="ref_8ba8c9fd">[https://math.illinoisstate.edu/day/courses/old/305/contentrecursion.html Recurrence Relations]</ref> | ||
| + | # Let's write a recurrence relation for the sum of the elements in column 1 of the triangle.<ref name="ref_8ba8c9fd" /> | ||
| + | # Write a recurrence relation H(n) to describe the number of moves required when the left tower has n rings on it.<ref name="ref_8ba8c9fd" /> | ||
| + | # Write a recurrence relation b(n) to describe the balance due on the loan after n payments.<ref name="ref_8ba8c9fd" /> | ||
| + | # A recurrence relation is an equation which represents a sequence based on some rule.<ref name="ref_a6553953">[https://byjus.com/maths/recurrence-relation/ Recurrence Relation-Definition, Formula and Examples]</ref> | ||
| + | # Once the values of a 0 and a 1 are specified, the whole sequence {a i } i 0 is completely specified by the recurrence relation.<ref name="ref_50edb03b">[https://staff.cdms.westernsydney.edu.au/cgi-bin/cgiwrap/zhuhan/dmath/dm_readall.cgi?page=21 Discrete Mathematics]</ref> | ||
| + | # Observe that the difference of the highest index (n+1) and the lowest index (n-2) is exactly the order 6 of the recurrence relation.<ref name="ref_50edb03b" /> | ||
| + | # Since the highest index in the recurrence relation is n+1 while the lowest index is n-2, their difference (n+1)-(n-2)=3 must be the order m, i.e. m=3.<ref name="ref_50edb03b" /> | ||
| + | # For each fixed m, these two generating functions share the same denominator, hence the same recurrence relation .<ref name="ref_2e974bde">[https://encyclopedia2.thefreedictionary.com/recurrence+relation recurrence relation]</ref> | ||
| + | # After studying it carefully, it is found that the solutions cannot be given exactly due to the complicated three-term recurrence relation .<ref name="ref_2e974bde" /> | ||
| + | # Recurrence relations can also be used to calculate some sequences that are usually computed nonrecursively , e.g. via a closed-form formula .<ref name="ref_7a1d0504">[http://oeis.org/wiki/Recurrence Recurrence]</ref> | ||
| + | # Of course recurrence relations are not limited in application to sequences of integers.<ref name="ref_7a1d0504" /> | ||
| + | ===소스=== | ||
| + | <references /> | ||
| − | == | + | == 메타데이터 == |
| − | * | + | ===위키데이터=== |
| − | * [ | + | * ID : [https://www.wikidata.org/wiki/Q740970 Q740970] |
| − | * | + | ===Spacy 패턴 목록=== |
| − | [ | + | * [{'LOWER': 'recurrence'}, {'LEMMA': 'relation'}] |
| − | + | * [{'LOWER': 'recurrent'}, {'LEMMA': 'sequence'}] | |
2021년 1월 2일 (토) 06:31 기준 최신판
개요
- 점화식 : 수열의 여러항들이 만족시키는 관계
- 점화식이 만족하는 수열의 일반항을 알아 내는 문제 등
- 보통의 경우 초기항이 주어져야 완전한 답을 낼 수 있다.
기본적인 점화식
- \(a_{n+1} - a_n = c\) : 등차수열
- \(a_{n+1} / a_n = c\) : 등비수열
- \(a_{n+1} - a_n = b_n\) : 계차수열 참고.
- \(a_{n+1} / a_n = b_n\) : 계차수열을 통한 풀이에서, <모든 항을 더하>지 않고 <모든 항을 곱하>면 됨.
기본 점화식의 응용
- \(a_{n+1} = ka_n + c\)
- 양변에 적당한 상수를 더하면 \((a_{n+1} + p)= k(a_n + p)\) 꼴로 만들 수 있다.
- 일반항이 \((a_n + p)\) 인 수열은 공비 \(k\) 인 등비수열, \(a_n + p =(a_1 + p) k^{n-1}\)
- 적당한 상수 \(p\) 는 어떻게 찾냐고? 생각해 볼 것.
- ex) \(a_{n+1} = 2a_{n} + 3\), 초기항 1 양변에 3을 더하면 \((a_{n+1} +3 )= 2( a_{n} + 3)\), 적당한 상수 \(k\) 에 대하여 \(a_{n} + 3 = k 2^n\) 초항을 만족시키는 \(k\) 값은 2이므로, \(a_n = 2^{n+1} - 3\)
- \(a_{n+1} = ra_n + b_n\) 꼴의 점화식
- 양변을 \(r^{n+1}\) 로 나눈 후, \(a_n / r^n\) 에 대한 점화식을 푸는 것이 한 방법. \(b_n\) 이 등비수열인 경우 효과적이다.
- 양변에 적당히 \(n\) 에 대한 식을 더해서 공비 \(r\) 에 대한 등비수열 꼴로 만들 수 있는 경우가 많다.
- \(b_n\) 이 다항식인 경우, 양변에 \(b_n\) 과 같은 차수의 다항식을 (계수를 문자로 두고) 더해서, 등비수열 꼴로 만든 후에 계수 비교를 통해 문자를 찾는다.
- ex ) \(a_{n+1} = 2a_n + 3n + 5\) 인 경우, 양변에 \(pn + q\) 를 더하면\[a_{n+1} + pn + q = 2a_n + (3+p)n + (5+q)\] 우변이 \(2(a_n+ pn + q)\) 인 경우에 등비수열이 되니까, \(3+p = 2p, \quad 5+q=2q\) 이므로 \(p=3, \quad q=5\). 그러므로\[a_n + 3n + 5 = k2^n\]. 초기항이 주어진 경우 k 를 찾을 수 있다.
- 점화식에 덧셈 기호가 없을 때
- 로그를 취하면 도움이 됨. 로그의 밑은 계산이 간단하도록 적절히 선택하기.
- ex) \(a_{n+1} = 4a_n^2\) : 밑 2 인 로그를 취하면 \(\log a_{n+1} = 2\log a_n + 2\) 이제 \(\log a_n = b_n\)에 대한 점화식을 풀면 됨. (양변에 2를 더해서 …)
- 점화식이 분수꼴일때
- 역수를 취하면 도움이 될 때가 많음. (만능은 아님)
- ex) \({a_{n+1}} = \frac{2a_n}{3a_n+1}\) : 역수를 취하면 \(\frac{1}{a_{n+1}} = \frac{1}{2} \frac{1}{a_n} + \frac{3}{2}\). 이제 \(b_n = \frac{1}{a_n}\) 에 대한 점화식으로 보고 풀면 됨.
- 점화식에 \(a_n\) 과 \(S_n\) 이 함께 나올 때
- \(S_n - S_{n-1}=a_n\)\((n \ge 2)\) , \(S_1 = a_1 \) 의 관계를 사용하면 \(a_n\) 만의 점화식으로 만들 수 있다.
- ex) \( a_n = 2 S_n + 3^n\) 일 때, \( a_1 = 2 a_1 + 3\) 이므로 \(a_1 = -3\)\[a_{n+1} = 2 S_{n+1} + 3^{n+1}\] 식과 \( a_n = 2 S_n + 3^n\) 식을 빼 주면 \(a_{n+1} - a_n = 2 a_{n+1} + 2\cdot 3^{n}\) 이제부터는 알아서 할 것. 부분합과 일반항이 함께 등장하는 점화식에서는 초기 조건에서 실수를 할 가능성이 높으므로, \(a_1, a_2, a_3\) 정도는 점화식으로부터 직접 구해 보는 것이 좋을 것이다. 그리고 구한 일반항 \(a_n\) 은 \((n \ge 2)\) 에서만 성립하는 식일 가능성도 있다.
선형점화식
- \(pa_{n+2} + qa_{n+1} + ra_n = 0\) 꼴의 점화식
- \(pa_{n+2} + qa_{n+1} + ra_n = b_n\) 꼴의 점화식
- 상수계수 선형점화식 항목을 참조
관련된 항목들
사전 형태의 자료
메타데이터
위키데이터
- ID : Q740970
노트
말뭉치
- The procedure for finding the terms of a sequence in a recursive manner is called recurrence relation.[1]
- We study the theory of linear recurrence relations and their solutions.[1]
- Doing so is called solving a recurrence relation .[2]
- Recall that the recurrence relation is a recursive definition without the initial conditions.[2]
- , 485\ldots\text{.}\) Solution Finding the recurrence relation would be easier if we had some context for the problem (like the Tower of Hanoi, for example).[2]
- Remember, the recurrence relation tells you how to get from previous terms to future terms.[2]
- Recurrence relations are used to reduce complicated problems to an iterative process based on simpler versions of the problem.[3]
- Identifying a candidate problem to solve with a recurrence relation: The problem can be reduced to simpler cases.[3]
- The term difference equation sometimes (and for the purposes of this article) refers to a specific type of recurrence relation.[4]
- A recurrence relation is an equation that expresses each element of a sequence as a function of the preceding ones.[4]
- This defines recurrence relation of first order.[4]
- The recurrence of order two satisfied by the Fibonacci numbers is the archetype of a homogeneous linear recurrence relation with constant coefficients (see below).[4]
- In this article, we will see how we can solve different types of recurrence relations using different approaches.[5]
- Therefore, we need to convert the recurrence relation into appropriate form before solving.[5]
- There are a variety of methods for solving recurrence relations, with various advantages and disadvantages in particular cases.[6]
- One method that works for some recurrence relations involves generating functions.[6]
- If we can find an explicit representation for the series for this function, we will have solved the recurrence relation.[6]
- We can often solve a recurrence relation in a manner analogous to solving a differential equations by multiplying by an integrating factor and then integrating.[7]
- Wolfram|Alpha can solve various kinds of recurrences, find asymptotic bounds and find recurrence relations satisfied by given sequences.[8]
- This is called a recurrence relation.[9]
- We somehow need to figure out how often the first versus the second branch of this recurrence relation will be taken.[9]
- We will first find a recurrence relation for the execution time.[9]
- These recurrence relations are shown to be closely related to each other.[10]
- Racah has obtained a recurrence relation for the ``inner multiplicity (multiplicity of weights).[10]
- Moreover, Kostant's recurrence relation for the partition function as well as Racah's recurrence relation for the inner multiplicity have been generalized.[10]
- In this paper, the recurrence relation for negative moments along with negative factorial moments of some discrete distributions can be obtained.[11]
- Is `D(4)` easier to calculate using the factorial formula or using the recurrence relation?[12]
- Therefore the solution to the recurrence relation will have the form: a n =a2n+b18n.[13]
- Therefore the solution to the recurrence relation will have the form: a n =a.6n+b.n.6n.[13]
- Explanation: Check for the left side of the equation with all the options into the recurrence relation.[13]
- Together with the initial conditions, the recurrence relation provides a recursive definition for the elements of the sequence.[14]
- The solution techniques for these two classes of recurrence relation are discussed in detail in the Recurrence page.[15]
- We also justified Pascal's Formula as a recurrence relation.[16]
- Let's write a recurrence relation for the sum of the elements in column 1 of the triangle.[16]
- Write a recurrence relation H(n) to describe the number of moves required when the left tower has n rings on it.[16]
- Write a recurrence relation b(n) to describe the balance due on the loan after n payments.[16]
- A recurrence relation is an equation which represents a sequence based on some rule.[17]
- Once the values of a 0 and a 1 are specified, the whole sequence {a i } i 0 is completely specified by the recurrence relation.[18]
- Observe that the difference of the highest index (n+1) and the lowest index (n-2) is exactly the order 6 of the recurrence relation.[18]
- Since the highest index in the recurrence relation is n+1 while the lowest index is n-2, their difference (n+1)-(n-2)=3 must be the order m, i.e. m=3.[18]
- For each fixed m, these two generating functions share the same denominator, hence the same recurrence relation .[19]
- After studying it carefully, it is found that the solutions cannot be given exactly due to the complicated three-term recurrence relation .[19]
- Recurrence relations can also be used to calculate some sequences that are usually computed nonrecursively , e.g. via a closed-form formula .[20]
- Of course recurrence relations are not limited in application to sequences of integers.[20]
소스
- ↑ 1.0 1.1 Discrete Mathematics
- ↑ 2.0 2.1 2.2 2.3 Solving Recurrence Relations
- ↑ 3.0 3.1 Brilliant Math & Science Wiki
- ↑ 4.0 4.1 4.2 4.3 Recurrence relation
- ↑ 5.0 5.1 Different types of recurrence relations and their solutions
- ↑ 6.0 6.1 6.2 3.4 Recurrence Relations
- ↑ Recurrence Relations
- ↑ Alpha Examples: Recurrences
- ↑ 9.0 9.1 9.2 Lecture 17: Recurrence relations
- ↑ 10.0 10.1 10.2 Recurrence Relations for the Multiplicities in the Classical Groups
- ↑ Recurrence Relation and Accurate Value on Inverse Moment of Discrete Distributions
- ↑ Recurrence Relations and Generating Functions
- ↑ 13.0 13.1 13.2 Discrete Mathematics Questions and Answers
- ↑ Recurrence Relations
- ↑ cs2223 Classifying Recurrence Relations
- ↑ 16.0 16.1 16.2 16.3 Recurrence Relations
- ↑ Recurrence Relation-Definition, Formula and Examples
- ↑ 18.0 18.1 18.2 Discrete Mathematics
- ↑ 19.0 19.1 recurrence relation
- ↑ 20.0 20.1 Recurrence
메타데이터
위키데이터
- ID : Q740970
Spacy 패턴 목록
- [{'LOWER': 'recurrence'}, {'LEMMA': 'relation'}]
- [{'LOWER': 'recurrent'}, {'LEMMA': 'sequence'}]