Ah, I missed that one. Well, I don't have that much time right now, so I'll try a quick argument: assume f (for any given alpha) is continuous on domain S. We see that any irrational for which f is continuous must have an epsilon-neighborhood not containing any rationals and vice versa. Also...
To me, there doesn't seem to be any \alpha > 0 that makes the function continuous (let alone differentiable). Choose any epsilon-neighborhood around a rational x = \frac p q, [a,b] = [x-\epsilon,x+\epsilon] . Since we can make (b-a)2k arbitrarily large, we can make the difference between \lceil...
The problem here is that the exponential function on the complex plane is not injective. That is, even if ea = eb, we cannot infer that a = b. Analogously, even if sin(0) = sin(\pi), we cannot infer that 0 = \pi.
Probably a typo there. a and b differ by a multiple of 2 i pi, i.e. a = b + 2k...
This one has already been answered, but here's an alternate solution:
\lfloor \frac a b \rfloor = \frac{a- a \mod b}{b}
We have that
10^{20000} \equiv (-3)^{200} = 9^{100} \pmod{10^{100}+3}
The units digit is then the last digit of
\frac{10^{20000}- 9^{100}}{10^{100}+3}
i.e...
\mathbb{R}^\mathbb{R} is the set of all functions on \mathbb{R}, and is a vector space under scalar multiplication and addition of functions. sin x and cos x are members of this vector space and are linearly independent. They therefore form the basis of a subspace V. Any function in V can be...
Mark44, T is not an operator on \mathbb{R}^2. It is an operator on the function space spanned by sin x and cos x. To be more precise, any vector in the space that T is operating on is of the form a sin x + b cos x for a,b reals. T is not a rotation operator on that function space. What would...
I have been re-reading some linear analysis the last week, and I have been playing around a bit with the linear subspace of \mathbb{R}^\mathbb{R} that you get with the basis {sin x, cos x}. It didn't take me long to realize that the family of transformations parametrized by theta T_\theta ...