Is term by term integration possible in the following case?











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Suppose we are given that $$int_1^4 g(t)f(t) dt=0 tag{1}$$



Suppose I am able to write $g(t)$ as Taylor series $displaystyle g(t)= sum_{n=1}^{infty} a_n t^n$. Now plug in series of $g(t)$ in equation $(1)$. What should be the conditions on $f(t)$ for term by term integration?



I think it should be that $t^n f(t)$ should converge uniformly to $g(t)f(t)$ and that $t^n f(t)$ should be integrable for every $n$. Am I correct?






calculus real-analysis







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  • 1




    If the Taylor series of $g$ converges uniformly for $tin[1,4]$ and $g(t)f(t)$ is integrable.
    – Yadati Kiran
    Nov 22 at 17:21








  • 1




    If the Taylor series converges uniformly in the given interval to its function $g$, and of course $g(t)f(t)$ is integrable, then you can integrate term by term.
    – uniquesolution
    Nov 22 at 17:21












  • Thanks! If all the coefficients in series of $g(t)$ are non-zero, can I conclude that equation $(1)$ is true if and only if $int_1^4 t^n f(t) dt=0$ for every $n$?
    – ersh
    Nov 22 at 17:36












  • No, you cant! For example take $g(t) = t+t^2$ and $f = 4 cdot 1_{[0,1/2]} - 1_{[1/2,1]}$. Then $int_0^1 f(t) g(t) , d t = 0$, but $int_0^1 f(t) t , d t = 1/8$. If $f$ is a.e. non-negative, then you can conlude that $int_1^4 t^n f(t) , d t =0$.
    – p4sch
    Nov 22 at 17:51












  • @p4sch That is great! Actually my $f$ is a.e non-negative. How to prove this fact of yours? Any reference
    – ersh
    Nov 22 at 18:01

















up vote
1
down vote

favorite












Suppose we are given that $$int_1^4 g(t)f(t) dt=0 tag{1}$$



Suppose I am able to write $g(t)$ as Taylor series $displaystyle g(t)= sum_{n=1}^{infty} a_n t^n$. Now plug in series of $g(t)$ in equation $(1)$. What should be the conditions on $f(t)$ for term by term integration?



I think it should be that $t^n f(t)$ should converge uniformly to $g(t)f(t)$ and that $t^n f(t)$ should be integrable for every $n$. Am I correct?






calculus real-analysis







share|cite|improve this question




















  • 1




    If the Taylor series of $g$ converges uniformly for $tin[1,4]$ and $g(t)f(t)$ is integrable.
    – Yadati Kiran
    Nov 22 at 17:21








  • 1




    If the Taylor series converges uniformly in the given interval to its function $g$, and of course $g(t)f(t)$ is integrable, then you can integrate term by term.
    – uniquesolution
    Nov 22 at 17:21












  • Thanks! If all the coefficients in series of $g(t)$ are non-zero, can I conclude that equation $(1)$ is true if and only if $int_1^4 t^n f(t) dt=0$ for every $n$?
    – ersh
    Nov 22 at 17:36












  • No, you cant! For example take $g(t) = t+t^2$ and $f = 4 cdot 1_{[0,1/2]} - 1_{[1/2,1]}$. Then $int_0^1 f(t) g(t) , d t = 0$, but $int_0^1 f(t) t , d t = 1/8$. If $f$ is a.e. non-negative, then you can conlude that $int_1^4 t^n f(t) , d t =0$.
    – p4sch
    Nov 22 at 17:51












  • @p4sch That is great! Actually my $f$ is a.e non-negative. How to prove this fact of yours? Any reference
    – ersh
    Nov 22 at 18:01















up vote
1
down vote

favorite









up vote
1
down vote

favorite











Suppose we are given that $$int_1^4 g(t)f(t) dt=0 tag{1}$$



Suppose I am able to write $g(t)$ as Taylor series $displaystyle g(t)= sum_{n=1}^{infty} a_n t^n$. Now plug in series of $g(t)$ in equation $(1)$. What should be the conditions on $f(t)$ for term by term integration?



I think it should be that $t^n f(t)$ should converge uniformly to $g(t)f(t)$ and that $t^n f(t)$ should be integrable for every $n$. Am I correct?






calculus real-analysis







share|cite|improve this question















Suppose we are given that $$int_1^4 g(t)f(t) dt=0 tag{1}$$



Suppose I am able to write $g(t)$ as Taylor series $displaystyle g(t)= sum_{n=1}^{infty} a_n t^n$. Now plug in series of $g(t)$ in equation $(1)$. What should be the conditions on $f(t)$ for term by term integration?



I think it should be that $t^n f(t)$ should converge uniformly to $g(t)f(t)$ and that $t^n f(t)$ should be integrable for every $n$. Am I correct?






calculus real-analysis




calculus real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 17:56









Robert Howard

1,9181822




1,9181822










asked Nov 22 at 17:11









ersh

264




264








  • 1




    If the Taylor series of $g$ converges uniformly for $tin[1,4]$ and $g(t)f(t)$ is integrable.
    – Yadati Kiran
    Nov 22 at 17:21








  • 1




    If the Taylor series converges uniformly in the given interval to its function $g$, and of course $g(t)f(t)$ is integrable, then you can integrate term by term.
    – uniquesolution
    Nov 22 at 17:21












  • Thanks! If all the coefficients in series of $g(t)$ are non-zero, can I conclude that equation $(1)$ is true if and only if $int_1^4 t^n f(t) dt=0$ for every $n$?
    – ersh
    Nov 22 at 17:36












  • No, you cant! For example take $g(t) = t+t^2$ and $f = 4 cdot 1_{[0,1/2]} - 1_{[1/2,1]}$. Then $int_0^1 f(t) g(t) , d t = 0$, but $int_0^1 f(t) t , d t = 1/8$. If $f$ is a.e. non-negative, then you can conlude that $int_1^4 t^n f(t) , d t =0$.
    – p4sch
    Nov 22 at 17:51












  • @p4sch That is great! Actually my $f$ is a.e non-negative. How to prove this fact of yours? Any reference
    – ersh
    Nov 22 at 18:01
















  • 1




    If the Taylor series of $g$ converges uniformly for $tin[1,4]$ and $g(t)f(t)$ is integrable.
    – Yadati Kiran
    Nov 22 at 17:21








  • 1




    If the Taylor series converges uniformly in the given interval to its function $g$, and of course $g(t)f(t)$ is integrable, then you can integrate term by term.
    – uniquesolution
    Nov 22 at 17:21












  • Thanks! If all the coefficients in series of $g(t)$ are non-zero, can I conclude that equation $(1)$ is true if and only if $int_1^4 t^n f(t) dt=0$ for every $n$?
    – ersh
    Nov 22 at 17:36












  • No, you cant! For example take $g(t) = t+t^2$ and $f = 4 cdot 1_{[0,1/2]} - 1_{[1/2,1]}$. Then $int_0^1 f(t) g(t) , d t = 0$, but $int_0^1 f(t) t , d t = 1/8$. If $f$ is a.e. non-negative, then you can conlude that $int_1^4 t^n f(t) , d t =0$.
    – p4sch
    Nov 22 at 17:51












  • @p4sch That is great! Actually my $f$ is a.e non-negative. How to prove this fact of yours? Any reference
    – ersh
    Nov 22 at 18:01










1




1




If the Taylor series of $g$ converges uniformly for $tin[1,4]$ and $g(t)f(t)$ is integrable.
– Yadati Kiran
Nov 22 at 17:21






If the Taylor series of $g$ converges uniformly for $tin[1,4]$ and $g(t)f(t)$ is integrable.
– Yadati Kiran
Nov 22 at 17:21






1




1




If the Taylor series converges uniformly in the given interval to its function $g$, and of course $g(t)f(t)$ is integrable, then you can integrate term by term.
– uniquesolution
Nov 22 at 17:21






If the Taylor series converges uniformly in the given interval to its function $g$, and of course $g(t)f(t)$ is integrable, then you can integrate term by term.
– uniquesolution
Nov 22 at 17:21














Thanks! If all the coefficients in series of $g(t)$ are non-zero, can I conclude that equation $(1)$ is true if and only if $int_1^4 t^n f(t) dt=0$ for every $n$?
– ersh
Nov 22 at 17:36






Thanks! If all the coefficients in series of $g(t)$ are non-zero, can I conclude that equation $(1)$ is true if and only if $int_1^4 t^n f(t) dt=0$ for every $n$?
– ersh
Nov 22 at 17:36














No, you cant! For example take $g(t) = t+t^2$ and $f = 4 cdot 1_{[0,1/2]} - 1_{[1/2,1]}$. Then $int_0^1 f(t) g(t) , d t = 0$, but $int_0^1 f(t) t , d t = 1/8$. If $f$ is a.e. non-negative, then you can conlude that $int_1^4 t^n f(t) , d t =0$.
– p4sch
Nov 22 at 17:51






No, you cant! For example take $g(t) = t+t^2$ and $f = 4 cdot 1_{[0,1/2]} - 1_{[1/2,1]}$. Then $int_0^1 f(t) g(t) , d t = 0$, but $int_0^1 f(t) t , d t = 1/8$. If $f$ is a.e. non-negative, then you can conlude that $int_1^4 t^n f(t) , d t =0$.
– p4sch
Nov 22 at 17:51














@p4sch That is great! Actually my $f$ is a.e non-negative. How to prove this fact of yours? Any reference
– ersh
Nov 22 at 18:01






@p4sch That is great! Actually my $f$ is a.e non-negative. How to prove this fact of yours? Any reference
– ersh
Nov 22 at 18:01

















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