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- ...hould be wary: some authors give the Fibonacci sequence with the [[initial condition]]s <math>F_0 = F_1 = 1</math> (or equivalently <math>F_1 = 1, F_2 = 2</math7 KB (1,111 words) - 13:57, 24 June 2024
- ...=(0,i),(1,i-1),...,(i-1,1),(i,0)</math>, so <math>i+1</math> options. The initial corner can be any of the four legs, but each of the four permutations of th ...<math>L_2</math>, <math>L_3</math>, and <math>L_4</math> iff the problem's condition is met.7 KB (1,276 words) - 19:51, 6 January 2024
- ...<math>f(1) = 0.</math> In general for <math>f(x),</math> the bug has three initial edges to pick from. From there, since the bug cannot return to <math>A</mat Note that the condition is equivalent to that <math>b_i \in \{1, 2, 3\}</math> for all <math>1 \leq19 KB (3,128 words) - 20:38, 23 July 2024
- Suppose that <math>n = 40</math>, and that the terms of the initial sequence <math>r_1, r_2, \dots, r_{40}</math> are distinct from one another <math>1.</math> To make sure the first condition is met, we perform the following: We know that there are <math>k</math> ter8 KB (1,269 words) - 09:55, 26 June 2024
- Iterating the condition <math>f(3x) = 3f(x)</math>, we find that <math>f(x) = 3^kf\left(\frac{x}{3^ ...math>. Similarly, using <math>f(3x)=3f(x)</math>, we get a dilation of our initial figure by a factor of 3 for the next interval and so on.5 KB (779 words) - 14:53, 2 November 2024
- ...equence <math> a_0, a_1, a_2, \ldots </math> of real numbers satisfies the condition '''Lemma 1.''' If the two initial terms of the sequence are nonzero and distinct, then every term of the sequ8 KB (1,443 words) - 15:51, 28 November 2008
- Claim. The only positive integers <math>n</math> that satisfy the condition are perfect multiples of <math>3</math>. ...equiv 3 \pmod 9</math>. Clearly, no integers of this form will satisfy the condition because <math>2007</math> is a perfect multiple of <math>9</math>.15 KB (2,558 words) - 22:08, 28 July 2024
- ...grapes eaten can be computed as the sum of the arithmetic progression with initial term <math>n</math> (the number of grapes eaten by the child in <math>1</ma ...integers, <math>n\geq 2(c-1)</math>, and <math>n</math> is minimized. (The condition <math>n\geq 2(c-1)</math> states that even the last child had to eat a non-4 KB (762 words) - 17:40, 12 January 2024
- ...ath> or <math>2</math> and similarly for <math>f(2)</math>. By the second condition, we have ...h>f(n)=1</math> for all <math>n\ge 2</math>, but this does not satisfy the initial conditions. Indeed, we would have3 KB (571 words) - 07:18, 19 July 2016
- ...than or equal to all of its neighbors, is <math>>0</math>, which violates condition <math>(ii)</math>. There are <math>2^{mn}</math> possible configurations o ...ue by (i). And by (ii) the value would have to be 0, which contradicts our initial construction. Therefore, all the cells bordering a 1 have to be 2. Looks to3 KB (627 words) - 21:33, 9 April 2018
- ...the initial edge counter. Let Y be the other two counters that satisfy the condition. The line X should not have more than two counters on it. Player A should p4 KB (754 words) - 19:34, 3 June 2014
- If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of u Suppose our fraction is <math>\frac{a}{b}</math>. The given condition means <math>a+b=1000</math>. Now, if <math>a</math> and <math>b</math> shar6 KB (965 words) - 20:09, 2 December 2024
- a) Find all positive integers with initial digit <math>6</math> such that the integer formed by deleting <math>6</math We use the same notation as part a. Then, the condition can be represented as <math>m*10^{n-1} + x = 35x</math> where <math>m</math1 KB (237 words) - 23:16, 2 May 2020
- ...a sequence of <math>2015</math> distinct real numbers such that after one initial move is applied to the sequence -- no matter what move -- there is always a ...he total group, then all the members are reduced to the average. Under the condition that two arbitrary elements are chosen first, we need only <math>N\ge4</mat8 KB (1,405 words) - 19:13, 26 July 2022
- .../math> denotes the number of <math>n</math>-digit numbers that satisfy the condition. We claim <math>a_n=4\cdot 9^{n-1}-a_{n-1}</math>, with <math>a_1=4</math>. ...</math> denote the number of <math>n</math>-digit numbers that satisfy the condition. Clearly <math>a_1 = 4</math> and <math>a_2 = 32</math>. We claim that fo10 KB (1,799 words) - 10:37, 28 February 2023
- ...ince <math>x_n</math> is clearly positive for all <math>n</math>, from the initial definition), <math>x_{n+1} > 4</math> if and only if <math>x_{n} > 4</math> The condition where <math>x_m\leq 4+\frac{1}{2^{20}}</math> gives the motivation to make10 KB (1,631 words) - 10:06, 31 October 2024
- Plugging this condition into (1') to substitute <math>u</math>, we get ...)=(1+a)/2</math>, and when making the substitution in each equation of the initial set of equations, we obtain a new equation in the form of the sine addition15 KB (2,208 words) - 00:25, 1 February 2024
- satisfies the condition. label("Initial", (2.5,-1));15 KB (2,233 words) - 10:02, 3 November 2024
- ...fairly obvious that <math>n \geq 3</math>; so we may break up the initial condition into two sub-conditions.16 KB (2,240 words) - 22:16, 26 January 2024
- ...<math>k</math> since <math>a=2k+1.</math> One thing to keep in mind is the initial assumption <math>(a-1)^2 < n < a^2.</math> .../math>, while <math>g(n)</math> continues to <math>(a+1)^2</math>. If this condition is satisfied, we can figure out the value of <math>k</math> based on <math>16 KB (2,816 words) - 07:28, 24 November 2024