A subtle discussion of a prob from regional 2019

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Mehrab4226
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A subtle discussion of a prob from regional 2019

Unread post by Mehrab4226 » Wed Mar 10, 2021 10:19 pm

The question was,
$f(x+y)=xf(x)+2f(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$

If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And why does our 2nd approach give us a different kinda a result?
Last edited by Mehrab4226 on Thu Mar 11, 2021 11:54 pm, edited 3 times in total.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Dustan
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Re: A subtle discussion of a prob from regional 2019

Unread post by Dustan » Wed Mar 10, 2021 10:25 pm

Correct your statement.

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Mehrab4226
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Re: A subtle discussion of a prob from regional 2019

Unread post by Mehrab4226 » Wed Mar 10, 2021 10:27 pm

But why does the function go weird when we use x=2019 and y=0?
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

~Aurn0b~
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Re: A subtle discussion of a prob from regional 2019

Unread post by ~Aurn0b~ » Thu Mar 11, 2021 10:53 pm

Mehrab4226 wrote:
Wed Mar 10, 2021 10:19 pm
The question was,
$f(x+y)=xf(x)+2(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$

If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And why does our 2nd approach give us a different kinda a result?
is that $2(y)$, or $2f(y)$?
and the question didnt give any info about domain range?

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: A subtle discussion of a prob from regional 2019

Unread post by ~Aurn0b~ » Thu Mar 11, 2021 11:04 pm

Mehrab4226 wrote:
Wed Mar 10, 2021 10:19 pm
The question was,
$f(x+y)=xf(x)+2(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$

If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And why does our 2nd approach give us a different kinda a result?
I think the question is wrong....
$P(0,x)\Rightarrow f(x)=x$, Plugging it into the main equation, we get that it doesnt satisfy, so there must be no solution to the functional equation.
but i dont think we can even say that because we dont know about the domain and range

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Mehrab4226
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Re: A subtle discussion of a prob from regional 2019

Unread post by Mehrab4226 » Thu Mar 11, 2021 11:56 pm

~Aurn0b~ wrote:
Thu Mar 11, 2021 11:04 pm
Mehrab4226 wrote:
Wed Mar 10, 2021 10:19 pm
The question was,
$f(x+y)=xf(x)+2(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$

If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And why does our 2nd approach give us a different kinda a result?
I think the question is wrong....
$P(0,x)\Rightarrow f(x)=x$, Plugging it into the main equation, we get that it doesnt satisfy, so there must be no solution to the functional equation.
but i dont think we can even say that because we dont know about the domain and range
Yes, it should be $2f(y)$. I am thinking the same. This came on BDMO 2019 regional. That is why I was just curious if the question was wrong.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

Re: A subtle discussion of a prob from regional 2019

Unread post by Dustan » Fri Mar 12, 2021 10:02 am

Mehrab4226 wrote:
Wed Mar 10, 2021 10:19 pm
The question was,
$f(x+y)=xf(x)+2f(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$

If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And why does our 2nd approach give us a different kinda a result?
i think it has other soln too. but as they said $f(2019)=a/b$ we should count the 2nd one.

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Mehrab4226
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Re: A subtle discussion of a prob from regional 2019

Unread post by Mehrab4226 » Fri Mar 12, 2021 12:15 pm

Dustan wrote:
Fri Mar 12, 2021 10:02 am
Mehrab4226 wrote:
Wed Mar 10, 2021 10:19 pm
The question was,
$f(x+y)=xf(x)+2f(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$

If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And why does our 2nd approach give us a different kinda a result?
i think it has other soln too. but as they said $f(2019)=a/b$ we should count the 2nd one.
That may be the logical thing to do.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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