For particular case b=2a we have sin(2b sk)/(2b)= sin(b sk)/b which yields cos(b sk)=1 or sk=2k pi/b. It is true also if the sequence sk that approaches infinity as k approaches infinity is given by the formula sk=k .
If a=1 and b>1 is e.g. irrational you can always construct s_k \to \infty so that sin (bs_k)=b sin (s_k). Let f(x)=sin (bx) and g(x)=b sin (x) and consider all k such that f(2\pi k)>0. There are infinitely many such k since b is irrational. Then f(2\pi k)>g(2\pi k). Consider now the next zero of f after 2\pi k, say x_k. Since b>1 then x_k lies in the zone where g is positive, hence f(x_k)
Dear Dr. Metafune,
My actual problem contain 2 equations:
Sin((a-b)s_k)=(a-b) Sin(s_k) and (1-b)Sin(a s_k)=(1-b) Sin((1-b)s_k)
where real parameters a, b are distinct and not equal zero or one.
It is given that there exists a sequence s_k\to\infty such that the equations are true.
There are trivial solutions s_k=n pi and a, b are rational numbers.
How do you think is there a solution other than trivial?
You found a nice contr example to the simplified version that contains only one equation. I tried to use your idea to prove the existence of the solution of 2 equations other than the trivial, but could not.
It looks like the construction works in a general case.
Choosing b=1-a the second equation: (1-b)Sin(as)=a Sin((1-b)s)
is identity, so the first equation has some solutions with irrational a by Dr. Metafune's construction. Thanks again!
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