2 variable Functional Equations

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Suppose a function $f : R->R$ satisfies the following conditions
$f(4xy) = 2y[f(x+y)+f(x-y)]$
$f(5)=3$
What is the value of $f(2015)$?
I am currently stuck after $x=y$ which gives out:
$f(4y^2) = 2y(f(2y))$
After getting this, I notice that
$4y^2 = (2y)^2$. So maybe the function also has this property
$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks
functions
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up vote
1
down vote
favorite
Suppose a function $f : R->R$ satisfies the following conditions
$f(4xy) = 2y[f(x+y)+f(x-y)]$
$f(5)=3$
What is the value of $f(2015)$?
I am currently stuck after $x=y$ which gives out:
$f(4y^2) = 2y(f(2y))$
After getting this, I notice that
$4y^2 = (2y)^2$. So maybe the function also has this property
$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks
functions
"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12
My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose a function $f : R->R$ satisfies the following conditions
$f(4xy) = 2y[f(x+y)+f(x-y)]$
$f(5)=3$
What is the value of $f(2015)$?
I am currently stuck after $x=y$ which gives out:
$f(4y^2) = 2y(f(2y))$
After getting this, I notice that
$4y^2 = (2y)^2$. So maybe the function also has this property
$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks
functions
Suppose a function $f : R->R$ satisfies the following conditions
$f(4xy) = 2y[f(x+y)+f(x-y)]$
$f(5)=3$
What is the value of $f(2015)$?
I am currently stuck after $x=y$ which gives out:
$f(4y^2) = 2y(f(2y))$
After getting this, I notice that
$4y^2 = (2y)^2$. So maybe the function also has this property
$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks
functions
edited Aug 15 at 13:18
asked Aug 15 at 12:43
SuperMage1
708210
708210
"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12
My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19
add a comment |Â
"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12
My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19
"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12
"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12
My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19
My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19
add a comment |Â
1 Answer
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If you look at the function $f(x) = frac35x$, you can see that
- $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$
- $f(5) = frac35 cdot 5 = 3$
For this function, $f(2015) = 1209$ indeed.
It remains to show that this is the only function that satisfies the above conditions.
It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you look at the function $f(x) = frac35x$, you can see that
- $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$
- $f(5) = frac35 cdot 5 = 3$
For this function, $f(2015) = 1209$ indeed.
It remains to show that this is the only function that satisfies the above conditions.
It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.
add a comment |Â
up vote
1
down vote
If you look at the function $f(x) = frac35x$, you can see that
- $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$
- $f(5) = frac35 cdot 5 = 3$
For this function, $f(2015) = 1209$ indeed.
It remains to show that this is the only function that satisfies the above conditions.
It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you look at the function $f(x) = frac35x$, you can see that
- $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$
- $f(5) = frac35 cdot 5 = 3$
For this function, $f(2015) = 1209$ indeed.
It remains to show that this is the only function that satisfies the above conditions.
It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.
If you look at the function $f(x) = frac35x$, you can see that
- $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$
- $f(5) = frac35 cdot 5 = 3$
For this function, $f(2015) = 1209$ indeed.
It remains to show that this is the only function that satisfies the above conditions.
It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.
answered Aug 15 at 15:05
Ronald
533410
533410
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"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12
My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19