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y=2 starts out asymptotic to both of your functions, and I checked that a function based on y=−1 will kinda oscillate around it with an increasing amplitude for some time, after that I don’t know
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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f(x)=2cos(2xa) where a is an arbitrary parameter.
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Thanks! I find new one from this : f(x)=2⋅cosh(a⋅2x)
y=2 starts out asymptotic to both of your functions, and I checked that a function based on y=−1 will kinda oscillate around it with an increasing amplitude for some time, after that I don’t know
I think there are a lot of pathological functions satisfying the FE
@Yajat Shamji @Aryan Sanghi @Mahdi Raza @Jeff Giff @David Stiff @Vinayak Srivastava @Akela Chana @Siddharth Chakravarty
Not all functions but at the moment, f(x)=2 and f(x)=−1 are the boring constant functions which work
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Piecewise functions involving −ϕ and ϕ1 can be made