Exercise for young people

[Steven11]

I have always been out of this section, because I was not up to face the problems I read.
Actually I would like some foreing friends to visit us.
So I thought of proposing an exercise suitable for a student who has just terminated liceo.

Find the smallest real number $a>1$ that satisfies
$\frac{a+sinx}{a+siny}<=e^(y-x)\quad\quad\quad forall x<=y$

Bye, and let me know if my english was bad, I do not want to make a bad impression

[Paolo902]

"Steven":Yes, obviously correct.
You can look at the two solutions here
http://olimpiadi.ing.unipi.it/oliForum/ ... hp?t=10242
In the second post you can find a link in which you read the second solution, the one that uses Lagrange theorem.

All right, I've given a look at the links. Thank you very much indeed.

"Steven":
It was a problem of the school SNS, 1990.

Wow, what a good news! I can't believe I've solved it... I'm very satisfied!

"Steven":
Thank you for having paid attention to my problem.

Enjoy your evening

Thank you for having proposed it. Looking forward to hearing from you again very soon. Enjoy your evening too.

[Steven11]

Yes, obviously correct.
You can look at the two solutions here
http://olimpiadi.ing.unipi.it/oliForum/ ... hp?t=10242
In the second post you can find a link in which you read the second solution, the one that uses Lagrange theorem.
It was a problem of the school SNS, 1990.

Thank you for having paid attention to my problem.

Enjoy your evening

[Paolo902]

Well, if I'm not wrong another solution (I mean yours) can be easily found by using ... (:wink:) theorem.

In your hint, you suggested the use of the function $f(x): xsin(logx)$. Now, the this function is derivable and it is
$f'(x)=sinlogx+coslogx$

This may be written as well as (remembering the addition formulae for sine)
$f'(x)=sqrt2sin(pi/4+logx)$

Now, I have looked for a good interval over which the function is not only derivable but also continuos (these are the hypothesis of Lagrange's theorem). Indeed, if you consider $[e^x,e^y]$ then there exists a $z in ]e^x,e^y[$ for which
$\frac{e^ysinloge^y-e^xsinloge^x}{e^y-e^x}=sqrt2sin(pi/4+logz)$
$\frac{e^ysiny-e^xsinx}{e^y-e^x}=sqrt2sin(pi/4+logz)$

You can see that the RHS can reach at maximimum the value $sqrt2$ and at minimum the value $-sqrt2$.

Therefore
$-sqrt2<=\frac{e^ysiny-e^xsinx}{e^y-e^x}<=sqrt2$

Now, since the expression we had $\frac{a+sinx}{a+siny}<=e^(y-x)$ can be rewritten as
$e^x(a+sinx)<=e^y(a+siny)$ (as I had already done in my last post)
we can solve it for $a$; this gives
$a>=\frac{e^ysiny-e^xsinx}{e^y-e^x}$ (since $a>1$ and $x
Therefore, we can conclude again saying that $a=sqrt2$.

What have I said? What do you think? Is it correct? You said in your last post that there exists a place where there is the solution... would you be so kind to indicate me it, please (so I can learn something...)? Thanks a lot.

That's extremely kind of you.

Bye, my dear friend.

[Paolo902]

Hi Steven. Thanks for your corrections (I've done an edit in my previous message!). You were right, I just forgot... it was a mistake in writing . Thanks.

Well, your solution seems to be very interesting. Now I can't, I hope to be able to think about it this afternoon. I'll tell you later.

Thanks, see you!

[Steven11]

Good evening,
actually you forgot in two occasions $e^x$, because you only wrote $cosx$ instead of $e^xcosx$.
Anyway, it was only a mistake concerning wrinting, in fact the value you found is correct.

I cannot appreciate Camillo's equations because my knowledge about analysis is limitated to what I studied at school this year.
I did not study additional arguments of analysis at home.

Anyway I also read another solution: I can give the features, and if you want I will give you the link where you can find it.
You have to obtain your disequation in this form:
$a>=...$
after consider the following function
$f(z)=zsin(lnz)$
Can you go on?

Have a good night!

[Paolo902]

Hi Steven, good to hear from you also here in this section. So, you found exercises in English corner difficult, didn't you? I can't believe this... You are so good! And Camillo's equations are so beautiful! (I'm not joking, but I do want to thank Camillo for proposing his analysis exercises; I've been learning a lot since I began to write here. Thanks also to Kroldar who patiently explains me my mistakes...).
Anyway, I thank you for proposing this exercise. Let's see (just one thing: I have not terminated liceo yet... I try to do it, but I'm not sure...).

Since we know that $a>1$ we can write the expression as
$a+sinx<=e^(y-x)(a+siny)$
which means
$e^x(a+sinx)<=e^y(a+siny)$

Now, LHS and RHS looks very very similar. So, I would propose to consider the function $f(x): y=e^x(a+sinx)$. We could study this function, but we are only interested in the sign of $f'(x)$. Indeed, we are looking for an $a \in RR, a>1$ that let the function be non-decreasing. So, by differentiating we get $y'=ae^x+e^xsinx+e^xcosx=e^x(a+sinx)+e^xcosx$. Now, we want
$f'(x)>=0 => e^x(a+sinx)+e^xcosx>=0$
Let's divide for $e^x$; this yields to
$a>=-cosx-sinx$.

Now, the function $g(x)=-cosx-sinx$ as a maximum in $x=5/4pi$ (I've found it with differential calculus, but you can do it also without differentiating); so, the max point is $(5/4pi, sqrt2)$. So, it should be (you asked for the smallest!) $a=sqrt2$ (I've tried to substitute it into the equation, it works. I do hope the solution is correct).

Please, dear Steven, let me know. Thanks a lot. Enjoy your evening.

Paolo