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wu :: forums    riddles    medium (Moderators: Icarus, william wu, SMQ, towr, Grimbal, ThudnBlunder, Eigenray)    Functions satisfying  2 {f(x)}^2 - f(2x) = 1 « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Functions satisfying  2 {f(x)}^2 - f(2x) = 1  (Read 468 times) Michael Dagg Senior Riddler     Gender: Posts: 500 Functions satisfying  2 {f(x)}^2 - f(2x) = 1   « on: Mar 2nd, 2008, 2:01pm » Quote Modify Determine the complex-valued functions  f(x)  which have power series   expansions that converge near  0  and which satisfy  2 {f(x)}^2 - f(2x) = 1   inside the circle of convergence. IP Logged Regards, Michael Dagg pex Uberpuzzler     Gender: Posts: 880 Re: Functions satisfying  2 {f(x)}^2 - f(2x)   « Reply #1 on: Mar 4th, 2008, 2:56pm » Quote Modify Amazing! If I haven't messed up too badly, hidden: either f(x) = -1/2 independent of x, f(x) = cos(kx) for some k, or f(x) = cosh(kx) for some k. The latter two solutions include the constant function f(x) = 1 as the special case k = 0. IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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