If $int_{a}^{b} f(x) g(x) dx=0$ with $f(x)=sum_{i=0}^{infty} a_n x^n$, can I integrate term by term?
Suppose $int_{a}^{b} f(x,l) g(x) dx=0...(1)$ with Taylor series of $f(x)=sum_{i=0}^{infty} a_n x^n$, where $a_n$ depends upon parameter $l$. I want to construct non-zero bounded integrable function $g(x)$.
Since we are on finite interval, series $sum_{i=0}^{infty} a_n x^n g(x)$ converges to $f(x)g(x)$ uniformly(?), can I integrate term by term?
Then problem reduces to construction of $g(x)$ such that $int_{a}^{b}x^n g(x) dx=0$ for $n=0,1,2...$ which I can do. My problem is with integrating term by term, can I construct $g(x)$ in this way?
real-analysis integration fourier-analysis uniform-convergence
asked Jan 4 at 3:48
ershersh
1659
add a comment |
Suppose $int_{a}^{b} f(x,l) g(x) dx=0...(1)$ with Taylor series of $f(x)=sum_{i=0}^{infty} a_n x^n$, where $a_n$ depends upon parameter $l$. I want to construct non-zero bounded integrable function $g(x)$.
Since we are on finite interval, series $sum_{i=0}^{infty} a_n x^n g(x)$ converges to $f(x)g(x)$ uniformly(?), can I integrate term by term?
Then problem reduces to construction of $g(x)$ such that $int_{a}^{b}x^n g(x) dx=0$ for $n=0,1,2...$ which I can do. My problem is with integrating term by term, can I construct $g(x)$ in this way?
real-analysis integration fourier-analysis uniform-convergence
asked Jan 4 at 3:48
ershersh
1659
3
If you have uniform convergence, then yes. But in general, no.
– mathworker21
Jan 4 at 3:50
also, what does $l$ have to do with anything?
– mathworker21
Jan 4 at 3:51
1
$sum_{n} a_n x^n$ converges to $f(x)$ uniformly on the interval $[a,b]$ if $|a| < R$ and $|b|<R$, where $f$ is analytic in the disk ${x: |x| < R}$ (it may or may not converge uniformly if $|a| = R$ or $|b| = R$). Then, $g$ being integrable, $sum_n a_n int_a^b x^n g(x) ; dx = int_a^b f(x) g(x); dx$ by the Dominated Convergence Theorem.
– Robert Israel
Jan 4 at 4:03
@mathworker21, my coefficients $a_n$ depend upon parameter $l$. I thought it wouldn't be harmful to mention it here.
– ersh
Jan 4 at 4:14
@RobertIsrael Thanks!
– ersh
Jan 4 at 4:15
add a comment |
Suppose $int_{a}^{b} f(x,l) g(x) dx=0...(1)$ with Taylor series of $f(x)=sum_{i=0}^{infty} a_n x^n$, where $a_n$ depends upon parameter $l$. I want to construct non-zero bounded integrable function $g(x)$.
Since we are on finite interval, series $sum_{i=0}^{infty} a_n x^n g(x)$ converges to $f(x)g(x)$ uniformly(?), can I integrate term by term?
Then problem reduces to construction of $g(x)$ such that $int_{a}^{b}x^n g(x) dx=0$ for $n=0,1,2...$ which I can do. My problem is with integrating term by term, can I construct $g(x)$ in this way?
real-analysis integration fourier-analysis uniform-convergence
asked Jan 4 at 3:48
ershersh
1659
Suppose $int_{a}^{b} f(x,l) g(x) dx=0...(1)$ with Taylor series of $f(x)=sum_{i=0}^{infty} a_n x^n$, where $a_n$ depends upon parameter $l$. I want to construct non-zero bounded integrable function $g(x)$.
Since we are on finite interval, series $sum_{i=0}^{infty} a_n x^n g(x)$ converges to $f(x)g(x)$ uniformly(?), can I integrate term by term?
Then problem reduces to construction of $g(x)$ such that $int_{a}^{b}x^n g(x) dx=0$ for $n=0,1,2...$ which I can do. My problem is with integrating term by term, can I construct $g(x)$ in this way?
real-analysis integration fourier-analysis uniform-convergence
real-analysis integration fourier-analysis uniform-convergence
asked Jan 4 at 3:48
ershersh
1659
asked Jan 4 at 3:48
ershersh
1659
edited Jan 4 at 4:14
ersh
asked Jan 4 at 3:48
ershersh
1659
asked Jan 4 at 3:48
ershersh
1659
asked Jan 4 at 3:48
ershersh
1659
1659
3
If you have uniform convergence, then yes. But in general, no.
– mathworker21
Jan 4 at 3:50
also, what does $l$ have to do with anything?
– mathworker21
Jan 4 at 3:51
1
$sum_{n} a_n x^n$ converges to $f(x)$ uniformly on the interval $[a,b]$ if $|a| < R$ and $|b|<R$, where $f$ is analytic in the disk ${x: |x| < R}$ (it may or may not converge uniformly if $|a| = R$ or $|b| = R$). Then, $g$ being integrable, $sum_n a_n int_a^b x^n g(x) ; dx = int_a^b f(x) g(x); dx$ by the Dominated Convergence Theorem.
– Robert Israel
Jan 4 at 4:03
@mathworker21, my coefficients $a_n$ depend upon parameter $l$. I thought it wouldn't be harmful to mention it here.
– ersh
Jan 4 at 4:14
@RobertIsrael Thanks!
– ersh
Jan 4 at 4:15
add a comment |
3
If you have uniform convergence, then yes. But in general, no.
– mathworker21
Jan 4 at 3:50
also, what does $l$ have to do with anything?
– mathworker21
Jan 4 at 3:51
1
$sum_{n} a_n x^n$ converges to $f(x)$ uniformly on the interval $[a,b]$ if $|a| < R$ and $|b|<R$, where $f$ is analytic in the disk ${x: |x| < R}$ (it may or may not converge uniformly if $|a| = R$ or $|b| = R$). Then, $g$ being integrable, $sum_n a_n int_a^b x^n g(x) ; dx = int_a^b f(x) g(x); dx$ by the Dominated Convergence Theorem.
– Robert Israel
Jan 4 at 4:03
@mathworker21, my coefficients $a_n$ depend upon parameter $l$. I thought it wouldn't be harmful to mention it here.
– ersh
Jan 4 at 4:14
@RobertIsrael Thanks!
– ersh
Jan 4 at 4:15
3
3
If you have uniform convergence, then yes. But in general, no.
– mathworker21
Jan 4 at 3:50
If you have uniform convergence, then yes. But in general, no.
– mathworker21
Jan 4 at 3:50
also, what does $l$ have to do with anything?
– mathworker21
Jan 4 at 3:51
also, what does $l$ have to do with anything?
– mathworker21
Jan 4 at 3:51
1
1
$sum_{n} a_n x^n$ converges to $f(x)$ uniformly on the interval $[a,b]$ if $|a| < R$ and $|b|<R$, where $f$ is analytic in the disk ${x: |x| < R}$ (it may or may not converge uniformly if $|a| = R$ or $|b| = R$). Then, $g$ being integrable, $sum_n a_n int_a^b x^n g(x) ; dx = int_a^b f(x) g(x); dx$ by the Dominated Convergence Theorem.
– Robert Israel
Jan 4 at 4:03
$sum_{n} a_n x^n$ converges to $f(x)$ uniformly on the interval $[a,b]$ if $|a| < R$ and $|b|<R$, where $f$ is analytic in the disk ${x: |x| < R}$ (it may or may not converge uniformly if $|a| = R$ or $|b| = R$). Then, $g$ being integrable, $sum_n a_n int_a^b x^n g(x) ; dx = int_a^b f(x) g(x); dx$ by the Dominated Convergence Theorem.
– Robert Israel
Jan 4 at 4:03
@mathworker21, my coefficients $a_n$ depend upon parameter $l$. I thought it wouldn't be harmful to mention it here.
– ersh
Jan 4 at 4:14
@mathworker21, my coefficients $a_n$ depend upon parameter $l$. I thought it wouldn't be harmful to mention it here.
– ersh
Jan 4 at 4:14
@RobertIsrael Thanks!
– ersh
Jan 4 at 4:15
@RobertIsrael Thanks!
– ersh
Jan 4 at 4:15
add a comment |
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3
If you have uniform convergence, then yes. But in general, no.
– mathworker21
Jan 4 at 3:50
also, what does $l$ have to do with anything?
– mathworker21
Jan 4 at 3:51
1
$sum_{n} a_n x^n$ converges to $f(x)$ uniformly on the interval $[a,b]$ if $|a| < R$ and $|b|<R$, where $f$ is analytic in the disk ${x: |x| < R}$ (it may or may not converge uniformly if $|a| = R$ or $|b| = R$). Then, $g$ being integrable, $sum_n a_n int_a^b x^n g(x) ; dx = int_a^b f(x) g(x); dx$ by the Dominated Convergence Theorem.
– Robert Israel
Jan 4 at 4:03
@mathworker21, my coefficients $a_n$ depend upon parameter $l$. I thought it wouldn't be harmful to mention it here.
– ersh
Jan 4 at 4:14
@RobertIsrael Thanks!
– ersh
Jan 4 at 4:15