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- ==Problems== ([[2006 AMC 10A Problems/Problem 2|Source]])10 KB (1,761 words) - 02:16, 12 May 2023
- [[functional equation for the zeta function|functional equation]]: From the functional equation9 KB (1,547 words) - 02:04, 13 January 2021
- ...e, <math>f(x,y) = \mathrm{lcm}(x,y)</math> is a solution to the functional equation. [[Category:Intermediate Algebra Problems]]4 KB (538 words) - 12:24, 12 October 2021
- [[Category:Olympiad Algebra Problems]] [[Category:Functional Equation Problems]]5 KB (923 words) - 18:51, 21 January 2024
- ...but there do exist other solutions to this which are not solutions to the equation of this problem.) Hence all solutions to our equation are of the form <math>f(x) = kx</math>. It is easy to see that real value3 KB (490 words) - 21:09, 18 September 2018
- We first note that <math> \displaystyle f(x) = 0 </math> is a solution to the equation. Henceforth we shall consider other solutions to the equations, i.e., func ...satisfy the desired condition. Thus the only solutions to the functional equation are <math> \displaystyle f(x) = 0 </math> and <math> \displaystyle f(x) = x2 KB (339 words) - 22:46, 4 May 2007
- ...for all <math>x </math>. It is easy to see that this satisfies the given equation. Q.E.D. ...<math> \displaystyle y = \frac{-x}{f(x)-1} </math> is the solution to the equation <math> \displaystyle y = x + y\cdot f(x) </math>, which is positive when <7 KB (1,179 words) - 19:59, 28 March 2010
- ...</math>) that <math>f(xb)=bf(x)</math>. Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>. Repeating this process we find that <math [[Category:Olympiad Algebra Problems]]3 KB (664 words) - 07:14, 28 August 2022
- * [[2001 IMO Shortlist Problems]] [[Category:Olympiad Algebra Problems]]2 KB (404 words) - 11:39, 1 November 2020
- ...math> which satisfy the constraint <math>wx=yz</math> we get the following equation: So we get the quadratic equation:4 KB (661 words) - 00:10, 19 November 2023
- [[Category:Olympiad Algebra Problems]] [[Category:Functional Equation Problems]]3 KB (523 words) - 23:45, 8 July 2017
- [[Category:Olympiad Algebra Problems]] [[Category:Functional Equation Problems]]3 KB (523 words) - 16:03, 13 August 2023
- Substituing in the original equation we find <math>f(x)=0, \ \forall x \in \mathbb{R}</math> or <math>f(x)=a, \ ...0) \rfloor \neq 0</math>, and now <math>y=0</math> in the given functional equation yields <math>f(0) = f(x)\lfloor f(0) \rfloor</math> for all <math>x \in \ma7 KB (1,128 words) - 11:54, 3 April 2024
- [[Category:Olympiad Algebra Problems]] [[Category:Functional Equation Problems]]4 KB (763 words) - 12:11, 3 September 2024
- So, using this in equation <math>(2) </math>, we get ...x) = 1 - \frac{x^{2}}{2}} </cmath> as the only solution to this functional equation.1 KB (199 words) - 06:50, 24 June 2024
- Here is a list of '''Olympiad Books''' that have Olympiad-level problems used to train students for future [[mathematics]] competitions. *''Inequalities An Approach Through Problems - '''B. J. Venkatachala'''19 KB (2,581 words) - 13:25, 1 November 2024
- [[Category:Olympiad Algebra Problems]] [[Category:Functional Equation Problems]]1 KB (234 words) - 00:20, 19 November 2023
- ...e claim that <math>f(k)=1</math> for all <math>k\ge n</math>. Indeed, from Equation (1) we have <math>f(n+1)!\equiv 1 \mod n\cdot n!</math>, and this is only p ...2</math>. Fix <math>k > 1</math> and suppose that <math>f(k)=k</math>. By Equation (1) we have that3 KB (571 words) - 07:18, 19 July 2016
- ...s <math>x</math> such that, for some positive constant <math>a</math>, the equation [[Category:Olympiad Algebra Problems]]1 KB (216 words) - 12:33, 29 January 2021
- *[[IMO Problems and Solutions]] [[Category:Olympiad Algebra Problems]]4 KB (633 words) - 07:08, 20 November 2023