Decompose maximum function?
If $a,b,k in mathbb{R}$, is there a way to write the expression $$max(a-kb,0)$$ as $$max(a-b,0)+kf(a,b)$$ for some function f?
I am looking to factor out $k$ from inside the max expression, but I haven't had any luck.
analysis
asked Dec 9 at 15:49
user128836
546
add a comment |
If $a,b,k in mathbb{R}$, is there a way to write the expression $$max(a-kb,0)$$ as $$max(a-b,0)+kf(a,b)$$ for some function f?
I am looking to factor out $k$ from inside the max expression, but I haven't had any luck.
analysis
asked Dec 9 at 15:49
user128836
546
add a comment |
If $a,b,k in mathbb{R}$, is there a way to write the expression $$max(a-kb,0)$$ as $$max(a-b,0)+kf(a,b)$$ for some function f?
I am looking to factor out $k$ from inside the max expression, but I haven't had any luck.
analysis
asked Dec 9 at 15:49
user128836
546
If $a,b,k in mathbb{R}$, is there a way to write the expression $$max(a-kb,0)$$ as $$max(a-b,0)+kf(a,b)$$ for some function f?
I am looking to factor out $k$ from inside the max expression, but I haven't had any luck.
analysis
analysis
asked Dec 9 at 15:49
user128836
546
asked Dec 9 at 15:49
user128836
546
asked Dec 9 at 15:49
user128836
546
asked Dec 9 at 15:49
user128836
546
asked Dec 9 at 15:49
user128836
546
546
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
No, that can't work, for several reasons.
First consider $k=0$, that means you want to have $max(a,0)=max(a-b,0)$, which may not be true.
But more generally, if $a,b$ are fixed, then $g(k)=max(a-b,0)+kf(a,b)$ is a linear function in $k$ (fixed term $max(a-b,0)$ and slope $f(a,b)$), while $h(k)=max(a-kb,0)$ is (generally) not, it's 2 pieces; one's a linear function, the other the constant function ($0$) function.
The only case when $h(k)$ is linear is when $b=0$, then $h(k)$ is a constant. Then you can choose $f(a,b)=0$. But in general this is not possible.
answered Dec 9 at 18:32
Ingix
3,214145
Thanks! That helped.
– user128836
Dec 9 at 21:28
add a comment |
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1 Answer
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No, that can't work, for several reasons.
First consider $k=0$, that means you want to have $max(a,0)=max(a-b,0)$, which may not be true.
But more generally, if $a,b$ are fixed, then $g(k)=max(a-b,0)+kf(a,b)$ is a linear function in $k$ (fixed term $max(a-b,0)$ and slope $f(a,b)$), while $h(k)=max(a-kb,0)$ is (generally) not, it's 2 pieces; one's a linear function, the other the constant function ($0$) function.
The only case when $h(k)$ is linear is when $b=0$, then $h(k)$ is a constant. Then you can choose $f(a,b)=0$. But in general this is not possible.
answered Dec 9 at 18:32
Ingix
3,214145
Thanks! That helped.
– user128836
Dec 9 at 21:28
add a comment |
No, that can't work, for several reasons.
First consider $k=0$, that means you want to have $max(a,0)=max(a-b,0)$, which may not be true.
But more generally, if $a,b$ are fixed, then $g(k)=max(a-b,0)+kf(a,b)$ is a linear function in $k$ (fixed term $max(a-b,0)$ and slope $f(a,b)$), while $h(k)=max(a-kb,0)$ is (generally) not, it's 2 pieces; one's a linear function, the other the constant function ($0$) function.
The only case when $h(k)$ is linear is when $b=0$, then $h(k)$ is a constant. Then you can choose $f(a,b)=0$. But in general this is not possible.
answered Dec 9 at 18:32
Ingix
3,214145
Thanks! That helped.
– user128836
Dec 9 at 21:28
add a comment |
No, that can't work, for several reasons.
First consider $k=0$, that means you want to have $max(a,0)=max(a-b,0)$, which may not be true.
But more generally, if $a,b$ are fixed, then $g(k)=max(a-b,0)+kf(a,b)$ is a linear function in $k$ (fixed term $max(a-b,0)$ and slope $f(a,b)$), while $h(k)=max(a-kb,0)$ is (generally) not, it's 2 pieces; one's a linear function, the other the constant function ($0$) function.
The only case when $h(k)$ is linear is when $b=0$, then $h(k)$ is a constant. Then you can choose $f(a,b)=0$. But in general this is not possible.
answered Dec 9 at 18:32
Ingix
3,214145
No, that can't work, for several reasons.
First consider $k=0$, that means you want to have $max(a,0)=max(a-b,0)$, which may not be true.
But more generally, if $a,b$ are fixed, then $g(k)=max(a-b,0)+kf(a,b)$ is a linear function in $k$ (fixed term $max(a-b,0)$ and slope $f(a,b)$), while $h(k)=max(a-kb,0)$ is (generally) not, it's 2 pieces; one's a linear function, the other the constant function ($0$) function.
The only case when $h(k)$ is linear is when $b=0$, then $h(k)$ is a constant. Then you can choose $f(a,b)=0$. But in general this is not possible.
answered Dec 9 at 18:32
Ingix
3,214145
answered Dec 9 at 18:32
Ingix
3,214145
answered Dec 9 at 18:32
Ingix
3,214145
answered Dec 9 at 18:32
Ingix
3,214145
3,214145
Thanks! That helped.
– user128836
Dec 9 at 21:28
add a comment |
Thanks! That helped.
– user128836
Dec 9 at 21:28
Thanks! That helped.
– user128836
Dec 9 at 21:28
Thanks! That helped.
– user128836
Dec 9 at 21:28
add a comment |
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