AoPS Academy
be external direct sum of vector spaces 
and 
over a field 
.let 
and 
and 
is subspaces.
. Let 
be the external direct sum of and . show that is isomorphic to under the correspondence 
![$\{x_n\} \subseteq [0,1]$](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 
.
such that
diverges for 
(harmonic sequence) but converges for 
because 
, is there a value between and such that 
converges![$f(x)\in R[x]$](http://latex.artofproblemsolving.com/a/1/5/a15dc97c051e81ae9ff0c060eaa4367bc723ccc9.png)
show that 
has at least one root of multiplicity one
be a metric space and 
be a function such that 
.
Prove that for all 
, 
exists, where 
is 
.
Prove that the amount of the limit does not depend on choosing 
.
be a function, two times derivable on 
for which there exist 
such that![\[ \frac { f(b)-f(a) }{b-a} \neq f'(c) ,\]](http://latex.artofproblemsolving.com/a/3/1/a31861ef909a6e17ed8b495b9b9b16d887ba573f.png)
for all 
.
.
defined for 
being an interval (connected subset of the real line) and 
not being in this image, it follows that we may assume that 
for all . But then 
for all 
and so 
is a minimum point of 
. Of course, this can be written in 11-th grade vocabulary. 
has the same sign for all 
.Can you be give me more detailes please 
?The official solution is based on the fact that the function 
is injective ,hence monotonic.
such that is a connected subset of the plane and the function 
is continuous on this domain, thus its image is a connected subset of the line, thus an interval. I agree however that this is not a solution of an 11-th grade student, but that's how life is. I will not be amazed if in a few years I see complex analysis, Lebesgue integration and other such stuff at RMO. It's quite à pity, it gives huges advantages to some people. 
should be defined on the set of pairs such that 
to ensure that its domain is connected. Of course, all the rest works with this modification, I don't know how I could write such a stupid thing. 
, then 
is strictly concave up or down on some neighbourhood of the point , thus one can draw a close enough parallel to the line tangent at to the function's graph that intersects the graph in two points 
and 
. The slope of that tangent would be 
, equal to the slope of the tangent at , that is,.
is connected, but why is connected ? has clearly two connected components. However with this spirit we can also solve the problem, extend 
to whole plane defining 
and now 
is continuous on whole plane and do similar thing.
has the same sign for all .Can you be give me more detailes please ?The official solution is based on the fact that the function is injective ,hence monotonic.
there exists an 
with the desired property. So you get 
only for values in some set 
, which has to be dense in the reals, but need not be an interval, so the IVP you use later on in your proof does not give a contradiction. (See the solutions above for proofs that avoid this problem.)
is continuous on 
and that is true because 
is twice differentiable.
. In the first paragraph you show that for all 
has to contain some 
(which means that is a dense subset of the real numbers). This part of your proof is fine. But in order to make the proof of the green claim work you need to show that contains all real numbers. This is missing in your proof (if I interpret you proof correctly).
holds inequality 
![\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\frac{1 - e^{-2} \cos\left(2\left(u + \tan u\right)\right)}
{1 - 2e^{-2} \cos\left(2\left(u + \tan u\right)\right) + e^{-4}}
\, \mathrm{d}u
\]](http://latex.artofproblemsolving.com/9/1/1/91165e6b531233c56b799ed32ded982d57b6c87a.png)
for all 
and now make 
close to 