AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a group, 
a subgroup of 
and 
an endomorphism with the following property:
such that for any 
there exist 
with 
and 
, for any 
.
is abelian.
is a subgroup of , prove that $H=G
be an abelian additive group such that all nonzero elements have infinite order and for each prime number 
we have the inequality 
, where 
, 
(where the sum has summands) and 
is the order of the quotient group 
(the index of the subgroup 
). of finite index is isomorphic to .
, 
.
.
are fixed positive integers.
are fixed positive real numbers.
be the set of all functions 
with the following properties: is defined for all real numbers and takes only real values.
the following equality holds: 
such that
,
.
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines 
and 
meet at a point . Let 
be the point of the line 
, different from 
, such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
for at least one 
;
for every positive integers 
and 
;
there are infinitely many positive integers such that 
for all 
.
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
balls to 
boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
Such That 
is even?
be defined by 
(a) Show that has exactly one point of discontinuity. at its point of discontinuity.
. We write 
if 
then 
.
we have 
, if 
we have 
and if 
we have 
.
But, actually, @alexheinis, this is from ISI Entrance Exam 2019, which is just a high school level exam. So, can you please post a little bit easier solution ?
. We write if then 
. we have , if we have and if we have .
, while for 
,
But, actually, @alexheinis, this is from ISI Entrance Exam 2019, which is just a high school level exam. So, can you please post a little bit easier solution ?
in Alexheinis's answer. But doing that during the exam would take away much time and I guess very few students could do something close to that. So although it is not something beyond the syllabus, but it is on the harder side with respect to the participants.
; 
,
Making cases like sir alexhienis did, we arrive at the answer.
and 
.
. We write if then . we have , if we have and if we have .
with another meaning, see #3 for the definition. So it's not about truncating series, but just about the quotient of two functions. How you prove them (#14) is a different matter: the first one is a derivative and for the second one you can use Taylor or de l'Hopital or 
, whatever you prefer. I liked this problem.
Mcq and solved only 2 Subjective problems correctly. Is there any chance of my qualification for the interview?
in the last-but-one term.
as well. Otherwise no equality.
Mcq and solved only 2 Subjective problems correctly. Is there any chance of my qualification for the interview?
. We write if then . we have , if we have and if we have .
Taking 
both sides, we have :
Let the part inside be 
. we write RHS as 
gives us 
. Multiplying numerator and denominator with and then 
, and putting appropriate limits, we are left with 
. Let this be 
.
which turns out to be 
for 
, 
for 
and 
for 
one point .
both sides, we have :
on both sides, you need to assure first that 
for all 
.
both sides, we have : on both sides, you need to assure first that for all .
![$$=\lim_{z\to 0+} \frac{\ln [1+(\cos(z^x)-1) ]}{\cos(z^x)-1} \cdot \frac{\cos(z^x)-1}{z^x} \cdot \frac{z^x}{z}$$](http://latex.artofproblemsolving.com/7/f/f/7ff4fd935ece04b18030ac3c327b7cffbe54b5e2.png)
Since all the limits exist,We have
.
is always positive.( Since 
)
Since all the limits exist,We have. is always positive.( Since )
Now you should calculate this limit carefully., what you did is: 
Be careful and see what it should be actually.)
. We will now explicitly find the function .
and for large enough , is obviously positive. So, taking natural logarithms both side and using the fact that 
is continuous over 
we get![\[g(x) \coloneqq \ln(f(x)) =
\lim_{n \to \infty} n \cdot \ln\left( \cos\left( \frac{1}{n^x} \right) \right)
= \lim_{n \to 0^+} \frac{\ln(\cos(n^x))}{n}.\]](http://latex.artofproblemsolving.com/8/1/2/812b0c65a0ef7296a512de19ba2264ae787414d7.png)
The limit is of the form 
, so applying L'Hôpital's Rule we get![\[g(x) = \lim_{n \to 0^+} \frac{1}{\cos(n^x)} \times -\sin(n^x) \times xn^{x-1}.\]](http://latex.artofproblemsolving.com/c/4/c/c4ce8fc1ce965a1ee92f72f5af3523e59f65c0e5.png)
Using approximations 
and 
around , we get
Taking exponential on both the sides, we get
From here, its clear that the only point of discontinuity is and we're done. 
be defined as:![\[
g(x) := \lim_{n \to \infty} \log \left( \cos^n \left( \frac{1}{n^x} \right) \right)
\]](http://latex.artofproblemsolving.com/d/2/e/d2ea59985a88d3e88dc1a356c1a7c16bac2b935b.png)
for all 
, since 
remains positive for all 
.
![\[
\lim_{n \to \infty} \log \frac{1 - 2 \sin^2 \left( \frac{1}{2n^x} \right)}{-2 \sin^2 \left( \frac{1}{n^x} \right)} = 1
\]](http://latex.artofproblemsolving.com/5/4/5/545388e64793a9044c0d01875cc086138e279b9a.png)
![\[
\lim_{n \to \infty} \left( -2 \sin^2 \left( \frac{1}{2n^x} \right) \right) = 0
\]](http://latex.artofproblemsolving.com/d/c/d/dcdd3db46fc03ee08bf1166559a77d529118628e.png)
![\[
\lim_{n \to \infty} \left( \frac{-2 \sin^2 \left( \frac{1}{2n^x} \right)}{\left( \frac{1}{2n^x} \right)^2} \cdot \left( \frac{1}{2n^x} \right)^2 \cdot n \right) =
\begin{cases}
0 & \text{if } x > \frac{1}{2} \\
-\frac{1}{2} & \text{if } x = \frac{1}{2} \\
-\infty & \text{if } x < \frac{1}{2}
\end{cases}
\]](http://latex.artofproblemsolving.com/1/b/b/1bb1737d19ffbdc4b18c7be1db366d2c8aa7ff9b.png)
can be described as:![\[
f(x) := e^{g(x)} = \lim_{n \to \infty} \cos^n \left( \frac{1}{n^x} \right)
\]](http://latex.artofproblemsolving.com/1/3/7/1374b416b41479d86cd7eee0f0159032250628e1.png)
:![\[
f\left( \frac{1}{2} \right) = \lim_{n \to \infty} \cos^n \left( \frac{1}{n^{\frac{1}{2}}} \right) = \frac{1}{\sqrt{e}}
\]](http://latex.artofproblemsolving.com/4/6/a/46af9d3a2da6b7759944ac8ee09caac32daa91f1.png)
:![\[
f(x) =
\begin{cases}
0 & \text{if } x \in \left( 0, \frac{1}{2} \right) \\
\frac{1}{\sqrt{e}} & \text{if } x = \frac{1}{2} \\
1 & \text{if } x > 1
\end{cases}
\]](http://latex.artofproblemsolving.com/0/1/b/01b97bf8cd975d9741f6159394abdfc8b618f47b.png)
Here we make use of the limits 
and 
(Derivatives of 
and 
at 
respectively).![\[
\lim_{n \to \infty} n \ln\left(\cos\left(\frac{1}{n^x}\right)\right) = \lim_{n \to \infty} -\frac{n^{1 - 2x}}{2}=
\begin{cases}
-\infty & \text{if } 0 < x < \frac{1}{2}, \\
-\frac{1}{2} & \text{if } x = \frac{1}{2}, \\
0 & \text{if } x > \frac{1}{2}.
\end{cases}
\]](http://latex.artofproblemsolving.com/9/8/8/9886158f32d1e05500c7c2c1abf17aaf42b3324c.png)
Lemma. If 
, then 
.
, for all 
there exists 
such that![\[
0 < |x - a| < \delta \implies f(x) < -M.
\]](http://latex.artofproblemsolving.com/e/4/2/e42b2abf8af3cb94b9e97c2a2c31c383459c096d.png)
Set 
, where 
. There exists 
such that![\[
f(x) < \ln \varepsilon \implies 0 < e^{f(x)} < \varepsilon.
\]](http://latex.artofproblemsolving.com/e/a/e/eae43756ae6077f3223979d61a5fd54a3235f757.png)
So 
, as desired. 
, for 
,![\[
f(x) = \lim_{n \to \infty} \cos^n\left(\frac{1}{n^x}\right) = \lim_{n \to \infty} e^{n \ln\left(\cos\left(\frac{1}{n^x}\right)\right)} = e^{\lim_{n \to \infty} n \ln\left(\cos\left(\frac{1}{n^x}\right)\right)}.
\]](http://latex.artofproblemsolving.com/f/a/1/fa17452f8bdec4dc2872966000bccdf3202de9f9.png)
And by the lemma, for 
,![\[
f(x) = \lim_{n \to \infty} \cos^n\left(\frac{1}{n^x}\right) = \lim_{n \to \infty} e^{n \ln\left(\cos\left(\frac{1}{n^x}\right)\right)} = 0.
\]](http://latex.artofproblemsolving.com/6/f/3/6f35eeae87d657fbda78b2c455a128bcc0d4cc0c.png)
![\[
f(x) =
\begin{cases}
0 & \text{if } 0 < x < \frac{1}{2}, \\
\frac{1}{\sqrt{e}} & \text{if } x = \frac{1}{2}, \\
1 & \text{if } x > \frac{1}{2}.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/9/c/b9c09e9082f8826cd6980cce2151a5ace3444e47.png)
It is clear that is discontinuous only at 
and that 
. 
such that:
Then 
But I am stuck. I came across this functional equation by trying to find a limit of a sequence, I can show that if it is helpful. 
has solutio
![$$f(x)=\lim_{n\to\infty} \cos^n\left(\dfrac{1}{n^x}\right)=\lim_{n\to\infty} \left[1+\left(\cos\left(n^{-x}\right)-1\right)\right]^n=\lim_{n\to\infty} \left(1+\dfrac{a_n}{n}\right)^n$$](http://latex.artofproblemsolving.com/3/7/e/37ed21e0666829b2a6b70e24bbc7ad3f9429bd84.png)
where 
. If we define, 
, then we have 
.
and the quantity in the parentheses goes to 
,
.
.
.
.
.
.
.
.
.
,
,is continus.
.
.
,
.
.
that is,
Where 
denotes the error function. Computing 
So 
.
where,![\[
g(x) = \lim_{n\to \infty} \ln\left(\cos^n\left(\frac 1{n^x}\right)\right).
\]](http://latex.artofproblemsolving.com/5/e/c/5ec013130a40057c74aae3b1561592aca3eccd3d.png)