Let
a) Show that has a unique solution.
b) Show that f(x) is not periodic.
My proof so far:
Now, I realize that f(x) = 2 iff x=0 in this case, but will that be enough for a proof?
And since f(x) = 2 only once, then it cannot be periodic.
Let
a) Show that has a unique solution.
b) Show that f(x) is not periodic.
My proof so far:
Now, I realize that f(x) = 2 iff x=0 in this case, but will that be enough for a proof?
And since f(x) = 2 only once, then it cannot be periodic.
well, that won't work as a proof!
use the sum-to-product formula for cosine:
...or maybe, you can continue with your line of thought, but add that since the range of cosine is [-1,1], the only way we can get two is if both terms are 1, since neither can be greater than 1, and if one of them is less than 1, the other would have to be greater than 1 to add up to 2 etc etc etc ...
This is not quite right. The two k's need not be the same, so we have:
and
for some , then eliminating between these we have:
,
but as is irrational this is imposible unless .
(PS it is obvious that the proof must hinge on being irrational so any
proof that does not use this fact must be deficient in some manner)
RonL
If the proof that I posted contained all the symbols that I had intended, then it would be correct. I do understand that we need two different possible values of K. I thought that I had used a K’ for the second. But several edits must have dropped some symbols; what was posted above is flawed.
The function is and as noted both of those terms must be 1 if the sum is two.
We get . Because it is the same x in the sum we get the following.
This means that and being irrational means or .