Hi ↑ stuart clark:
This might work:
Skrytý text:First of all there is
for
and
with nonzero
values on the interval boundaries, so we can look for the roots on the interval
and then double the count.
Then, for fixed
assuming
, for every
, there is for
So no matter what, in all points
the sign of the value of
is determined by expression
,
which is positive for even
and negative for odd
and it therefore alternates running through those
points.
By intermediate value theorem,
has at least
real zeros inside the interval
.
On the other hand,
can be transformed into
using transformation
on each term , where
is
Chebyshev polynomial of the first kind of degree
.
Note that on interval
where we work,
is injective function.
Thus
becomes ultimately polynomial of degree
; also visible in explicit expression
.
As such, it has at most
real zeros.
This leads to final result that
has
real roots on the interval
.