there exists a positive integer 
,
have at least one block of 
in their decimal representation.
be real numbers such that 
.
such that for every 
is satisfied? Explain the answer.
![\[\iint_{R} \sin(xy) \,dx\,dy, \quad R = \left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]\]](http://latex.artofproblemsolving.com/a/2/0/a205868dd99f4794de53ad1db50f34b17c70a923.png)
be a group, 
a subgroup of 
and 
an endomorphism with the following property:
such that for any 
there exist 
with 
and 
,
.
is abelian.
is a subgroup of 
are complex numbers with absolute value 
.
define 
.
converges. Show that,
.
such that there exists an integer polynomial 
with no integer roots and a constant 
such that
has at least 
distinct prime divisors for all positive integers 
is an integer, then there exist invertible matrices 
with non-zero entries such that:![\[A_1^{-1} + A_2^{-1} + \ldots + A_n^{-1} = (A_1 + A_2 + \ldots + A_n)^{-1}.\]](http://latex.artofproblemsolving.com/a/f/9/af9ff29771952abe37a00f3830c72e8c517e1cd6.png)
topic in College Math :coolspeak:
be a closed and countable set. Prove that if the analytic function 
is bounded, then 
is constant.
Over all real polynomials 
of degree 
what is the largest possible number of negative coefficients of 
fibonacci numbers 
)
and 
Y by
What is the smallest value of 
such that for any polynomial 
of degree 
with real coefficients, there exists a polynomial 
of degree at most with real coefficients such that the graphs of 
and 
intersect at exactly points?
such that for any polynomial 
of degree 
with real coefficients, there exists a polynomial 
of degree at most with real coefficients such that the graphs of 
and 
intersect at exactly points?This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
real roots counting multiplicity. So if 
the considering the 
-
gives that for any three given reals 
always has two real root if 
and 
lol).
works, first clean all the polynomial of degree 
using 
and from there many a suitable pick, in fact 
are not in our control, so that the polynomial has all real roots with no multiplicity and none of them are zero.
we can obviously shift the parabola by a constant acording to what is needed due to discriminant criteria so we should be fine here, due to continuity in fact you can just avoid a root being zero or the resulting parabola being tangent to the 
axis, for 
you have something unfree of the form 
and so you pick some 
on the free part where 
have freedom acording to needs like if 
and could just be suficiently small given we put 
to have two roots and that in order to not have repeated roots or the cubic being tangent to the 
you basically have control over a depressed cubic upon doing a similar setup as seen above just shift everything by a small constant at the end to satisfy all the given conditions needed and now we will show this is the kind of work you have to do.
then we prove it for 
,
so we just use the free term to focusing on a polynomial of the kind 
for 
,
is a root twice then we just shift by a very small constant as needed whether is above or below, one of them should work by continuity and IVT, since have 
as a root is only a one thing thing on a small interval we can also use continuity to avoid this and still satisfy the no double roots or more condition, hence by using the inductive hypothesis this is also completed and thus our claim is proven.
.
.
with degree less than 
has 100 roots for 
.
such that 
.
as the real roots of 
and 
is equal to 
and 
which is a contradiction. It follows that 
and 
with 
being the coefficient of 
being the leading coefficient. Obviously this is possible. From here we just alter the other coefficients so that each one of the coefficients becomes equal to the coefficients of the polynomial that has all these 
and so we are done. 
.
