Derivation $f(tx) = t^{mu}f(x) text{ } forall x in mathbb{R}^n text{ {0}} text{ } forall t in mathbb{R}^+$
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Let $mu in mathbb{R}$ be a real number and $f:mathbb{R}^n$ {$0$} $to mathbb{R}$ a function that is positive homegenous with degree $mu$, which means:
$$f(tx) = t^{mu}f(x) text{ } forall x in mathbb{R}^n text{ {0}} text{ } forall t in mathbb{R}^+$$
Furthermore, $f$ is totally differentiable in $mathbb{R}^n$ {$0$}.
How can I show that for a fix $x in mathbb{R^n}$ {$0$} the function $mathbb{R^+} ni t to f(tx) in mathbb{R}$ is totally differentiable in every $t in mathbb{R}^+$ and calculate the function in the position $t$ with the multidimensional chain rule?
I know that the chain rule is
But I struggle with the function.
analysis functions derivatives partial-derivative chain-rule
edited Nov 16 at 9:05
the_candyman
8,56921944
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up vote
1
down vote
favorite
Let $mu in mathbb{R}$ be a real number and $f:mathbb{R}^n$ {$0$} $to mathbb{R}$ a function that is positive homegenous with degree $mu$, which means:
$$f(tx) = t^{mu}f(x) text{ } forall x in mathbb{R}^n text{ {0}} text{ } forall t in mathbb{R}^+$$
Furthermore, $f$ is totally differentiable in $mathbb{R}^n$ {$0$}.
How can I show that for a fix $x in mathbb{R^n}$ {$0$} the function $mathbb{R^+} ni t to f(tx) in mathbb{R}$ is totally differentiable in every $t in mathbb{R}^+$ and calculate the function in the position $t$ with the multidimensional chain rule?
I know that the chain rule is
But I struggle with the function.
analysis functions derivatives partial-derivative chain-rule
edited Nov 16 at 9:05
the_candyman
8,56921944
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $mu in mathbb{R}$ be a real number and $f:mathbb{R}^n$ {$0$} $to mathbb{R}$ a function that is positive homegenous with degree $mu$, which means:
$$f(tx) = t^{mu}f(x) text{ } forall x in mathbb{R}^n text{ {0}} text{ } forall t in mathbb{R}^+$$
Furthermore, $f$ is totally differentiable in $mathbb{R}^n$ {$0$}.
How can I show that for a fix $x in mathbb{R^n}$ {$0$} the function $mathbb{R^+} ni t to f(tx) in mathbb{R}$ is totally differentiable in every $t in mathbb{R}^+$ and calculate the function in the position $t$ with the multidimensional chain rule?
I know that the chain rule is
But I struggle with the function.
analysis functions derivatives partial-derivative chain-rule
edited Nov 16 at 9:05
the_candyman
8,56921944
Let $mu in mathbb{R}$ be a real number and $f:mathbb{R}^n$ {$0$} $to mathbb{R}$ a function that is positive homegenous with degree $mu$, which means:
$$f(tx) = t^{mu}f(x) text{ } forall x in mathbb{R}^n text{ {0}} text{ } forall t in mathbb{R}^+$$
Furthermore, $f$ is totally differentiable in $mathbb{R}^n$ {$0$}.
How can I show that for a fix $x in mathbb{R^n}$ {$0$} the function $mathbb{R^+} ni t to f(tx) in mathbb{R}$ is totally differentiable in every $t in mathbb{R}^+$ and calculate the function in the position $t$ with the multidimensional chain rule?
I know that the chain rule is
But I struggle with the function.
analysis functions derivatives partial-derivative chain-rule
analysis functions derivatives partial-derivative chain-rule
edited Nov 16 at 9:05
the_candyman
8,56921944
edited Nov 16 at 9:05
the_candyman
8,56921944
edited Nov 16 at 9:05
the_candyman
8,56921944
edited Nov 16 at 9:05
the_candyman
8,56921944
edited Nov 16 at 9:05
the_candyman
8,56921944
8,56921944
asked Nov 16 at 8:32
user616397
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1 Answer
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If $x ne 0$ is fixed , then put $a:=f(x)$ and $g_x(t):=f(tx)$.
We have that $g_x(t)=t^{mu}f(x)=a t^{mu}$.
Can you proceed ?
answered Nov 16 at 9:18
Fred
42.8k1643
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If $x ne 0$ is fixed , then put $a:=f(x)$ and $g_x(t):=f(tx)$.
We have that $g_x(t)=t^{mu}f(x)=a t^{mu}$.
Can you proceed ?
answered Nov 16 at 9:18
Fred
42.8k1643
add a comment |
up vote
0
down vote
accepted
If $x ne 0$ is fixed , then put $a:=f(x)$ and $g_x(t):=f(tx)$.
We have that $g_x(t)=t^{mu}f(x)=a t^{mu}$.
Can you proceed ?
answered Nov 16 at 9:18
Fred
42.8k1643
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If $x ne 0$ is fixed , then put $a:=f(x)$ and $g_x(t):=f(tx)$.
We have that $g_x(t)=t^{mu}f(x)=a t^{mu}$.
Can you proceed ?
answered Nov 16 at 9:18
Fred
42.8k1643
If $x ne 0$ is fixed , then put $a:=f(x)$ and $g_x(t):=f(tx)$.
We have that $g_x(t)=t^{mu}f(x)=a t^{mu}$.
Can you proceed ?
answered Nov 16 at 9:18
Fred
42.8k1643
answered Nov 16 at 9:18
Fred
42.8k1643
answered Nov 16 at 9:18
Fred
42.8k1643
answered Nov 16 at 9:18
Fred
42.8k1643
42.8k1643
add a comment |
add a comment |
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