Mixing
Applying Bessel's inequality to a particular orthogonal system
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I'm studying PDE with a book called "A first course in PDE with complex variables and transform methods", in this book the author presents Bessel's inequality as follows:
$$ sum_n=1^infty C_n^ 2int_a^b phi_n^ 2 dx leq int_a^b f^ 2 dx $$
where $$C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$ are the Fourier coefficients of $f(x)$ with respect to the orthogonal functions $ phi_n(x) $.
if we consider the functions $1, cos(x), sin(x),cos(2x), sin(2x),... $ that are orthogonal on the interval $[-pi,pi]$. Then the Fourier series of $f(x)$ is
$$ sum_n=0^infty a_n cos(nx) + sum_n=1^infty b_n sin(nx) $$
where $$ a_n = frac int_-pi^pi f(x)cos(nx) dx int_-pi^pi cos^2(nx) dx n=0,1,.. $$
and $$ b_n = frac int_-pi^pi f(x)sin(nx) dx int_-pi^pi sin^2(nx) dx n=1,2,.. $$
From now I don't know if what I do is correct
$$ sum_n=0^infty a_n^ 2 int_-pi^pi cos^2(nx) + sum_n=1^infty b_n^ 2 int_-pi^pi sin^2(nx) dx = sum_n=0^infty a_n^ 2 pi + sum_n=1^infty b_n^ 2 pi = pi left( sum_n=0^infty a_n^ 2+ sum_n=1^infty b_n^ 2 right)leq int_-pi^pi f^ 2 dx $$
but then $$ a_0^ 2 + sum_n=1^inftyleft( a_n^ 2 + b_n^ 2 right) leq frac1pi int_-pi^pi f^ 2 dx $$
Now if $f(-pi) = f(pi)$ the Fourier series of $f'(x)$ is
$$ sum_n=1^infty Big(nb_ncos(nx) - na_nsin(nx) Big) = sum_n=1^infty nb_ncos(nx) +sum_n=1^infty (-na_n)sin(nx) $$
And $$ sum_n=1^infty (nb_n)^2int_-pi^pi cos^2(nx) + sum_n=1^infty (-na_n)^2int_-pi^pi sin^2(nx) = sum_n=1^infty (nb_n)^2 pi + sum_n=1^infty (-na_n)^2 pi =
sum_n=1^infty pi n^2 Big( a_n^2 +b_n^2 Big) leq int_-pi^pi ( f' )^2 dx$$ and then
$$ sum_n=1^infty n^2 Big( a_n^2 +b_n^2 Big) leq frac1pi int_-pi^pi ( f' )^2 dx $$
What do you think?, Is it correct?
pde fourier-analysis fourier-series
asked Jul 29 at 5:17
tnt235711
406
add a comment |Â
up vote
0
down vote
favorite
I'm studying PDE with a book called "A first course in PDE with complex variables and transform methods", in this book the author presents Bessel's inequality as follows:
$$ sum_n=1^infty C_n^ 2int_a^b phi_n^ 2 dx leq int_a^b f^ 2 dx $$
where $$C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$ are the Fourier coefficients of $f(x)$ with respect to the orthogonal functions $ phi_n(x) $.
if we consider the functions $1, cos(x), sin(x),cos(2x), sin(2x),... $ that are orthogonal on the interval $[-pi,pi]$. Then the Fourier series of $f(x)$ is
$$ sum_n=0^infty a_n cos(nx) + sum_n=1^infty b_n sin(nx) $$
where $$ a_n = frac int_-pi^pi f(x)cos(nx) dx int_-pi^pi cos^2(nx) dx n=0,1,.. $$
and $$ b_n = frac int_-pi^pi f(x)sin(nx) dx int_-pi^pi sin^2(nx) dx n=1,2,.. $$
From now I don't know if what I do is correct
$$ sum_n=0^infty a_n^ 2 int_-pi^pi cos^2(nx) + sum_n=1^infty b_n^ 2 int_-pi^pi sin^2(nx) dx = sum_n=0^infty a_n^ 2 pi + sum_n=1^infty b_n^ 2 pi = pi left( sum_n=0^infty a_n^ 2+ sum_n=1^infty b_n^ 2 right)leq int_-pi^pi f^ 2 dx $$
but then $$ a_0^ 2 + sum_n=1^inftyleft( a_n^ 2 + b_n^ 2 right) leq frac1pi int_-pi^pi f^ 2 dx $$
Now if $f(-pi) = f(pi)$ the Fourier series of $f'(x)$ is
$$ sum_n=1^infty Big(nb_ncos(nx) - na_nsin(nx) Big) = sum_n=1^infty nb_ncos(nx) +sum_n=1^infty (-na_n)sin(nx) $$
And $$ sum_n=1^infty (nb_n)^2int_-pi^pi cos^2(nx) + sum_n=1^infty (-na_n)^2int_-pi^pi sin^2(nx) = sum_n=1^infty (nb_n)^2 pi + sum_n=1^infty (-na_n)^2 pi =
sum_n=1^infty pi n^2 Big( a_n^2 +b_n^2 Big) leq int_-pi^pi ( f' )^2 dx$$ and then
$$ sum_n=1^infty n^2 Big( a_n^2 +b_n^2 Big) leq frac1pi int_-pi^pi ( f' )^2 dx $$
What do you think?, Is it correct?
pde fourier-analysis fourier-series
asked Jul 29 at 5:17
tnt235711
406
I don't get what you are trying to accomplish: apply Bessel inequality to a particular orthogonal system (as it seems in the body of the question) or to prove Bessel inequality (as in the title of the question)?
– Bob
Jul 29 at 5:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm studying PDE with a book called "A first course in PDE with complex variables and transform methods", in this book the author presents Bessel's inequality as follows:
$$ sum_n=1^infty C_n^ 2int_a^b phi_n^ 2 dx leq int_a^b f^ 2 dx $$
where $$C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$ are the Fourier coefficients of $f(x)$ with respect to the orthogonal functions $ phi_n(x) $.
if we consider the functions $1, cos(x), sin(x),cos(2x), sin(2x),... $ that are orthogonal on the interval $[-pi,pi]$. Then the Fourier series of $f(x)$ is
$$ sum_n=0^infty a_n cos(nx) + sum_n=1^infty b_n sin(nx) $$
where $$ a_n = frac int_-pi^pi f(x)cos(nx) dx int_-pi^pi cos^2(nx) dx n=0,1,.. $$
and $$ b_n = frac int_-pi^pi f(x)sin(nx) dx int_-pi^pi sin^2(nx) dx n=1,2,.. $$
From now I don't know if what I do is correct
$$ sum_n=0^infty a_n^ 2 int_-pi^pi cos^2(nx) + sum_n=1^infty b_n^ 2 int_-pi^pi sin^2(nx) dx = sum_n=0^infty a_n^ 2 pi + sum_n=1^infty b_n^ 2 pi = pi left( sum_n=0^infty a_n^ 2+ sum_n=1^infty b_n^ 2 right)leq int_-pi^pi f^ 2 dx $$
but then $$ a_0^ 2 + sum_n=1^inftyleft( a_n^ 2 + b_n^ 2 right) leq frac1pi int_-pi^pi f^ 2 dx $$
Now if $f(-pi) = f(pi)$ the Fourier series of $f'(x)$ is
$$ sum_n=1^infty Big(nb_ncos(nx) - na_nsin(nx) Big) = sum_n=1^infty nb_ncos(nx) +sum_n=1^infty (-na_n)sin(nx) $$
And $$ sum_n=1^infty (nb_n)^2int_-pi^pi cos^2(nx) + sum_n=1^infty (-na_n)^2int_-pi^pi sin^2(nx) = sum_n=1^infty (nb_n)^2 pi + sum_n=1^infty (-na_n)^2 pi =
sum_n=1^infty pi n^2 Big( a_n^2 +b_n^2 Big) leq int_-pi^pi ( f' )^2 dx$$ and then
$$ sum_n=1^infty n^2 Big( a_n^2 +b_n^2 Big) leq frac1pi int_-pi^pi ( f' )^2 dx $$
What do you think?, Is it correct?
pde fourier-analysis fourier-series
asked Jul 29 at 5:17
tnt235711
406
I'm studying PDE with a book called "A first course in PDE with complex variables and transform methods", in this book the author presents Bessel's inequality as follows:
$$ sum_n=1^infty C_n^ 2int_a^b phi_n^ 2 dx leq int_a^b f^ 2 dx $$
where $$C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$ are the Fourier coefficients of $f(x)$ with respect to the orthogonal functions $ phi_n(x) $.
if we consider the functions $1, cos(x), sin(x),cos(2x), sin(2x),... $ that are orthogonal on the interval $[-pi,pi]$. Then the Fourier series of $f(x)$ is
$$ sum_n=0^infty a_n cos(nx) + sum_n=1^infty b_n sin(nx) $$
where $$ a_n = frac int_-pi^pi f(x)cos(nx) dx int_-pi^pi cos^2(nx) dx n=0,1,.. $$
and $$ b_n = frac int_-pi^pi f(x)sin(nx) dx int_-pi^pi sin^2(nx) dx n=1,2,.. $$
From now I don't know if what I do is correct
$$ sum_n=0^infty a_n^ 2 int_-pi^pi cos^2(nx) + sum_n=1^infty b_n^ 2 int_-pi^pi sin^2(nx) dx = sum_n=0^infty a_n^ 2 pi + sum_n=1^infty b_n^ 2 pi = pi left( sum_n=0^infty a_n^ 2+ sum_n=1^infty b_n^ 2 right)leq int_-pi^pi f^ 2 dx $$
but then $$ a_0^ 2 + sum_n=1^inftyleft( a_n^ 2 + b_n^ 2 right) leq frac1pi int_-pi^pi f^ 2 dx $$
Now if $f(-pi) = f(pi)$ the Fourier series of $f'(x)$ is
$$ sum_n=1^infty Big(nb_ncos(nx) - na_nsin(nx) Big) = sum_n=1^infty nb_ncos(nx) +sum_n=1^infty (-na_n)sin(nx) $$
And $$ sum_n=1^infty (nb_n)^2int_-pi^pi cos^2(nx) + sum_n=1^infty (-na_n)^2int_-pi^pi sin^2(nx) = sum_n=1^infty (nb_n)^2 pi + sum_n=1^infty (-na_n)^2 pi =
sum_n=1^infty pi n^2 Big( a_n^2 +b_n^2 Big) leq int_-pi^pi ( f' )^2 dx$$ and then
$$ sum_n=1^infty n^2 Big( a_n^2 +b_n^2 Big) leq frac1pi int_-pi^pi ( f' )^2 dx $$
What do you think?, Is it correct?
pde fourier-analysis fourier-series
asked Jul 29 at 5:17
tnt235711
406
edited Jul 30 at 3:08
asked Jul 29 at 5:17
tnt235711
406
asked Jul 29 at 5:17
tnt235711
406
asked Jul 29 at 5:17
tnt235711
406
406
I don't get what you are trying to accomplish: apply Bessel inequality to a particular orthogonal system (as it seems in the body of the question) or to prove Bessel inequality (as in the title of the question)?
– Bob
Jul 29 at 5:39
add a comment |Â
I don't get what you are trying to accomplish: apply Bessel inequality to a particular orthogonal system (as it seems in the body of the question) or to prove Bessel inequality (as in the title of the question)?
– Bob
Jul 29 at 5:39
I don't get what you are trying to accomplish: apply Bessel inequality to a particular orthogonal system (as it seems in the body of the question) or to prove Bessel inequality (as in the title of the question)?
– Bob
Jul 29 at 5:39
I don't get what you are trying to accomplish: apply Bessel inequality to a particular orthogonal system (as it seems in the body of the question) or to prove Bessel inequality (as in the title of the question)?
– Bob
Jul 29 at 5:39
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865788%2fapplying-bessels-inequality-to-a-particular-orthogonal-system%23new-answer', 'question_page');
);
Post as a guest
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
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
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
I don't get what you are trying to accomplish: apply Bessel inequality to a particular orthogonal system (as it seems in the body of the question) or to prove Bessel inequality (as in the title of the question)?
– Bob
Jul 29 at 5:39