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- ==Recursion== ...th>n \geq 3</math>. This is the simplest nontrivial example of a [[linear recursion]] with constant coefficients. There is also an explicit formula [[#Binet's7 KB (1,111 words) - 13:57, 24 June 2024
- == Solution 1 (Single Variable Recursion) == == Solution 2 (Multivariable Recursion by Algebra) ==19 KB (3,128 words) - 20:38, 23 July 2024
- The above uses the similarity between the Fibonacci recursion|recursive definition, <math>F_{n+2} - F_{n+1} - F_n = 0</math>, and the pol We rewrite the left side of <math>(\bigstar)</math> as a linear expression of <math>r:</math>10 KB (1,595 words) - 15:30, 24 August 2024
- ...ath>f(k)=ak^2+bk+c,</cmath> where <math>a,b,</math> and <math>c</math> are linear combinations of <math>x_1,x_2,x_3,x_4,x_5,x_6,</math> and <math>x_7.</math> == Solution 2 (Linear Combination) ==9 KB (1,312 words) - 11:10, 15 December 2024
- ...ill be extremely lengthy because I am assuming just very basic concepts of linear algebra. The concepts here extend to higher levels of mathematics, so feel ...st assess the kernel of this matrix (also known as the null space), or the linear subspace of the domain of <math>\hat{T}</math> where everything gets mapped15 KB (2,406 words) - 22:56, 23 November 2023
- ...all <math>n\ge 1</math>. To do this, we use the following matrix form of a linear recurrence relation ...r all <math>n\ge 1</math>. We conclude that <math>a_n</math> satisfies the linear recurrence <math>a_{n+1}=225a_n-a_{n-1}</math>.13 KB (2,214 words) - 16:39, 28 November 2024
- ==Proof using Recursion== ...on is <math>F_n = F_{n-1} + F_{n-2}.</math> This is a constant coefficient linear homogenous recurrence relation. We also know that <math>F_0 = 0</math> and6 KB (953 words) - 20:37, 30 May 2024
- The '''characteristic polynomial''' of a linear [[operator]] refers to the [[polynomial]] whose roots are the [[eigenvalue] ...often used to find closed forms for the solutions of [[#Linear recurrences|linear recurrences]].19 KB (3,412 words) - 13:57, 21 September 2022
- :<math>\text{Our plan is to view each } a_{n} \text{ as a linear combination of the weights } w_{1}\text{, ..., } w_{s} \text{ and track the ...text{the linear combinations possible in the recursion, in other words the recursion in the problem is phrased as}</math>4 KB (786 words) - 07:46, 12 March 2024
- .../math> for all <math>n\geq 2</math>. The characteristic polynomial of this linear recurrence is <math>x^2-x-2=0</math>, which has roots <math>2</math> and <m ...h>. Using the fact that <math>b_0=0, b_1=1,</math> we can solve a pair of linear equations for <math>k_1, k_2</math>:5 KB (956 words) - 23:43, 9 November 2024
- ...In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of <math>C</math> to see if there ...ath>n=3</math>, and <math>n=4</math> results in a system of <math>4</math> linear equations<math>\newline</math>10 KB (1,578 words) - 05:18, 15 December 2024
- ...ast three rolls are <math>123</math> for the game to end. Summing over the recursion, we obtain ...= 1-P_o</math> and <math>P_e(1) = 1 - P_o(1)</math>, we have a 3-equation linear system which is not hard to solve. The final answer is <math>Po=\frac{216}{11 KB (1,860 words) - 12:12, 24 January 2024
- ...possible roots is found, and in some cases the initial guess can cause the recursion to cycle or diverge instead of converging to a root. ...seek a root of <math>f(x)</math>. Since all nonconstant [[Linear function|linear functions]] have exactly one root, as long as <math>f'(x_i) \neq 0</math> w13 KB (2,298 words) - 22:34, 28 May 2023
- ...th> The values <math>x_0</math> and <math>x_1</math> used initially in the recursion are guesses. ...s <math>(x_{i-1}, f(x_{i-1}))</math> and <math>(x_i, f(x_i))</math> is a [[linear function]], it has exactly one root (unless <math>f(x_{i-1}) = f(x_i)</math2 KB (337 words) - 14:30, 9 May 2022
- == Solution 4 (Third-order Homogeneous Linear Recurrence Relation) == == Solution 5 (Polynomial + Recursion) ==5 KB (886 words) - 22:57, 1 November 2024
- ...r of remaining unrevealed cards of each color. Since [[expected value]] is linear, the expected value of the total number of correct card color guesses acros ==Solution 5 (Pseudo-recursion)==13 KB (2,010 words) - 23:33, 2 September 2024
- Now, you can bash through solving linear congruences, but there is a smarter way. Notice that <math>5|x-6,6|x-6</mat ===Solution 1 (Recursion)===64 KB (10,960 words) - 22:01, 13 December 2024