How do I Solve $varphi(t)=(varphi(t /2^n))^{4^n}$?
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I am trying to solve the equation $varphi(t)=(varphi(t /2^n))^{4^n}$ for $varphi$. By guessing I see that $varphi(t)=e^{- sigma^2t^2/2}$ for all $sigma>0$ satisfies the equation above. How could I show it in a more precise way? The hint in the book is to compute the limit of the $R.H.S$ of the equation above as $n to infty$. I also know that $varphi$ is real valued and bounded above by $1$.
Any hints?
calculus
asked Nov 15 at 14:01
user3503589
1,1221721
add a comment |
up vote
1
down vote
favorite
I am trying to solve the equation $varphi(t)=(varphi(t /2^n))^{4^n}$ for $varphi$. By guessing I see that $varphi(t)=e^{- sigma^2t^2/2}$ for all $sigma>0$ satisfies the equation above. How could I show it in a more precise way? The hint in the book is to compute the limit of the $R.H.S$ of the equation above as $n to infty$. I also know that $varphi$ is real valued and bounded above by $1$.
Any hints?
calculus
asked Nov 15 at 14:01
user3503589
1,1221721
Do you have any info about the continuity of $varphi$? Also, $varphi=0$ is also a solution.
– 5xum
Nov 15 at 14:02
yes its continuous. Its the fourier transform of a probability measure.
– user3503589
Nov 15 at 14:10
As it's name is $varphi(t)$, I assume that we are considering a characteristic function :) @5xum hence I assume $varphi(0) =1$.
– Stockfish
Nov 15 at 14:18
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to solve the equation $varphi(t)=(varphi(t /2^n))^{4^n}$ for $varphi$. By guessing I see that $varphi(t)=e^{- sigma^2t^2/2}$ for all $sigma>0$ satisfies the equation above. How could I show it in a more precise way? The hint in the book is to compute the limit of the $R.H.S$ of the equation above as $n to infty$. I also know that $varphi$ is real valued and bounded above by $1$.
Any hints?
calculus
asked Nov 15 at 14:01
user3503589
1,1221721
I am trying to solve the equation $varphi(t)=(varphi(t /2^n))^{4^n}$ for $varphi$. By guessing I see that $varphi(t)=e^{- sigma^2t^2/2}$ for all $sigma>0$ satisfies the equation above. How could I show it in a more precise way? The hint in the book is to compute the limit of the $R.H.S$ of the equation above as $n to infty$. I also know that $varphi$ is real valued and bounded above by $1$.
Any hints?
calculus
calculus
asked Nov 15 at 14:01
user3503589
1,1221721
asked Nov 15 at 14:01
user3503589
1,1221721
asked Nov 15 at 14:01
user3503589
1,1221721
asked Nov 15 at 14:01
user3503589
1,1221721
asked Nov 15 at 14:01
user3503589
1,1221721
1,1221721
Do you have any info about the continuity of $varphi$? Also, $varphi=0$ is also a solution.
– 5xum
Nov 15 at 14:02
yes its continuous. Its the fourier transform of a probability measure.
– user3503589
Nov 15 at 14:10
As it's name is $varphi(t)$, I assume that we are considering a characteristic function :) @5xum hence I assume $varphi(0) =1$.
– Stockfish
Nov 15 at 14:18
add a comment |
Do you have any info about the continuity of $varphi$? Also, $varphi=0$ is also a solution.
– 5xum
Nov 15 at 14:02
yes its continuous. Its the fourier transform of a probability measure.
– user3503589
Nov 15 at 14:10
As it's name is $varphi(t)$, I assume that we are considering a characteristic function :) @5xum hence I assume $varphi(0) =1$.
– Stockfish
Nov 15 at 14:18
Do you have any info about the continuity of $varphi$? Also, $varphi=0$ is also a solution.
– 5xum
Nov 15 at 14:02
Do you have any info about the continuity of $varphi$? Also, $varphi=0$ is also a solution.
– 5xum
Nov 15 at 14:02
yes its continuous. Its the fourier transform of a probability measure.
– user3503589
Nov 15 at 14:10
yes its continuous. Its the fourier transform of a probability measure.
– user3503589
Nov 15 at 14:10
As it's name is $varphi(t)$, I assume that we are considering a characteristic function :) @5xum hence I assume $varphi(0) =1$.
– Stockfish
Nov 15 at 14:18
As it's name is $varphi(t)$, I assume that we are considering a characteristic function :) @5xum hence I assume $varphi(0) =1$.
– Stockfish
Nov 15 at 14:18
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
If $y(t) = ln varphi(t)$, the equation becomes
$$ y(t) = 4^n y(t/2^n) $$
Assume $y$ is twice differentiable at $0$.
By Taylor's theorem the
right side is $4^n (y(0) + 2^{-n} t y'(0) + 2^{-2n} t^2 y''(0)/2 + o(2^{-2n})$, so for the limit to exist we need $y(0)=0$ and $y'(0)=0$,
and then $y(t) = c t^2$ where $c = y''(0)/2$. Thus
$$ varphi(t) = exp(ct^2)$$
answered Nov 15 at 14:21
Robert Israel
313k23206452
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If $y(t) = ln varphi(t)$, the equation becomes
$$ y(t) = 4^n y(t/2^n) $$
Assume $y$ is twice differentiable at $0$.
By Taylor's theorem the
right side is $4^n (y(0) + 2^{-n} t y'(0) + 2^{-2n} t^2 y''(0)/2 + o(2^{-2n})$, so for the limit to exist we need $y(0)=0$ and $y'(0)=0$,
and then $y(t) = c t^2$ where $c = y''(0)/2$. Thus
$$ varphi(t) = exp(ct^2)$$
answered Nov 15 at 14:21
Robert Israel
313k23206452
add a comment |
up vote
4
down vote
accepted
If $y(t) = ln varphi(t)$, the equation becomes
$$ y(t) = 4^n y(t/2^n) $$
Assume $y$ is twice differentiable at $0$.
By Taylor's theorem the
right side is $4^n (y(0) + 2^{-n} t y'(0) + 2^{-2n} t^2 y''(0)/2 + o(2^{-2n})$, so for the limit to exist we need $y(0)=0$ and $y'(0)=0$,
and then $y(t) = c t^2$ where $c = y''(0)/2$. Thus
$$ varphi(t) = exp(ct^2)$$
answered Nov 15 at 14:21
Robert Israel
313k23206452
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If $y(t) = ln varphi(t)$, the equation becomes
$$ y(t) = 4^n y(t/2^n) $$
Assume $y$ is twice differentiable at $0$.
By Taylor's theorem the
right side is $4^n (y(0) + 2^{-n} t y'(0) + 2^{-2n} t^2 y''(0)/2 + o(2^{-2n})$, so for the limit to exist we need $y(0)=0$ and $y'(0)=0$,
and then $y(t) = c t^2$ where $c = y''(0)/2$. Thus
$$ varphi(t) = exp(ct^2)$$
answered Nov 15 at 14:21
Robert Israel
313k23206452
If $y(t) = ln varphi(t)$, the equation becomes
$$ y(t) = 4^n y(t/2^n) $$
Assume $y$ is twice differentiable at $0$.
By Taylor's theorem the
right side is $4^n (y(0) + 2^{-n} t y'(0) + 2^{-2n} t^2 y''(0)/2 + o(2^{-2n})$, so for the limit to exist we need $y(0)=0$ and $y'(0)=0$,
and then $y(t) = c t^2$ where $c = y''(0)/2$. Thus
$$ varphi(t) = exp(ct^2)$$
answered Nov 15 at 14:21
Robert Israel
313k23206452
answered Nov 15 at 14:21
Robert Israel
313k23206452
answered Nov 15 at 14:21
Robert Israel
313k23206452
answered Nov 15 at 14:21
Robert Israel
313k23206452
313k23206452
add a comment |
add a comment |
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Do you have any info about the continuity of $varphi$? Also, $varphi=0$ is also a solution.
– 5xum
Nov 15 at 14:02
yes its continuous. Its the fourier transform of a probability measure.
– user3503589
Nov 15 at 14:10
As it's name is $varphi(t)$, I assume that we are considering a characteristic function :) @5xum hence I assume $varphi(0) =1$.
– Stockfish
Nov 15 at 14:18