How do I Solve $varphi(t)=(varphi(t /2^n))^{4^n}$?
Multi tool use
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
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
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
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
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
$$ varphi(t) = exp(ct^2)$$
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
$$ varphi(t) = exp(ct^2)$$
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
$$ varphi(t) = exp(ct^2)$$
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
$$ varphi(t) = exp(ct^2)$$
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
313k23206452
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999740%2fhow-do-i-solve-varphit-varphit-2n4n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
ljHHMIzSk,cggeEz,XHgrQP55khMi7Y762Cya6LCy,vkLN,wDem8xig2L,Z,AwIZme4,fLM,3wG kopLc,3tH
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