A subtle discussion of a prob from regional 2019
- Mehrab4226
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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?
$f(x+y)=xf(x)+2f(y)-x-y$
$f(2019)=\frac{a}{b} ; a+b = ?$
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
-Henri Poincaré
Re: A subtle discussion of a prob from regional 2019
Correct your statement.
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: A subtle discussion of a prob from regional 2019
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
-Henri Poincaré
Re: A subtle discussion of a prob from regional 2019
is that $2(y)$, or $2f(y)$?Mehrab4226 wrote: ↑Wed Mar 10, 2021 10:19 pmThe 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?
and the question didnt give any info about domain range?
Re: A subtle discussion of a prob from regional 2019
I think the question is wrong....Mehrab4226 wrote: ↑Wed Mar 10, 2021 10:19 pmThe 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?
$P(0,x)\Rightarrow f(x)=x$, 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
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: A subtle discussion of a prob from regional 2019
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.~Aurn0b~ wrote: ↑Thu Mar 11, 2021 11:04 pmI think the question is wrong....Mehrab4226 wrote: ↑Wed Mar 10, 2021 10:19 pmThe 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?
$P(0,x)\Rightarrow f(x)=x$, 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
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
-Henri Poincaré
Re: A subtle discussion of a prob from regional 2019
i think it has other soln too. but as they said $f(2019)=a/b$ we should count the 2nd one.Mehrab4226 wrote: ↑Wed Mar 10, 2021 10:19 pmThe 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?
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: A subtle discussion of a prob from regional 2019
That may be the logical thing to do.Dustan wrote: ↑Fri Mar 12, 2021 10:02 ami think it has other soln too. but as they said $f(2019)=a/b$ we should count the 2nd one.Mehrab4226 wrote: ↑Wed Mar 10, 2021 10:19 pmThe 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?
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
-Henri Poincaré