which one of the following hold for all continuous function $f : [-pi ,pi] rightarrow mathbb{C}$
which one of the following hold For all continuous function $f : [-pi ,pi] rightarrow mathbb{C}$
$a)$ If $f(-t) =-f(t)$ for each $t in [-pi,pi]$,then $int_{-pi}^{pi}f(t)dt=0$
$b)$ $int_{-pi}^{pi}f(-t)dt=- int_{-pi}^{pi}f(t)dt $
My attempt : I thinks option b) will hold
and option a) will not hold because $int_{-pi}^{pi}f(t)dt=2int_{0}^{pi}f(t)dt neq 0$
Is it correct ?
real-analysis
asked Nov 21 '18 at 16:54
Messi fifa
51611
add a comment |
which one of the following hold For all continuous function $f : [-pi ,pi] rightarrow mathbb{C}$
$a)$ If $f(-t) =-f(t)$ for each $t in [-pi,pi]$,then $int_{-pi}^{pi}f(t)dt=0$
$b)$ $int_{-pi}^{pi}f(-t)dt=- int_{-pi}^{pi}f(t)dt $
My attempt : I thinks option b) will hold
and option a) will not hold because $int_{-pi}^{pi}f(t)dt=2int_{0}^{pi}f(t)dt neq 0$
Is it correct ?
real-analysis
asked Nov 21 '18 at 16:54
Messi fifa
51611
add a comment |
which one of the following hold For all continuous function $f : [-pi ,pi] rightarrow mathbb{C}$
$a)$ If $f(-t) =-f(t)$ for each $t in [-pi,pi]$,then $int_{-pi}^{pi}f(t)dt=0$
$b)$ $int_{-pi}^{pi}f(-t)dt=- int_{-pi}^{pi}f(t)dt $
My attempt : I thinks option b) will hold
and option a) will not hold because $int_{-pi}^{pi}f(t)dt=2int_{0}^{pi}f(t)dt neq 0$
Is it correct ?
real-analysis
asked Nov 21 '18 at 16:54
Messi fifa
51611
which one of the following hold For all continuous function $f : [-pi ,pi] rightarrow mathbb{C}$
$a)$ If $f(-t) =-f(t)$ for each $t in [-pi,pi]$,then $int_{-pi}^{pi}f(t)dt=0$
$b)$ $int_{-pi}^{pi}f(-t)dt=- int_{-pi}^{pi}f(t)dt $
My attempt : I thinks option b) will hold
and option a) will not hold because $int_{-pi}^{pi}f(t)dt=2int_{0}^{pi}f(t)dt neq 0$
Is it correct ?
real-analysis
real-analysis
asked Nov 21 '18 at 16:54
Messi fifa
51611
asked Nov 21 '18 at 16:54
Messi fifa
51611
edited Nov 21 '18 at 17:10
asked Nov 21 '18 at 16:54
Messi fifa
51611
asked Nov 21 '18 at 16:54
Messi fifa
51611
asked Nov 21 '18 at 16:54
Messi fifa
51611
51611
add a comment |
add a comment |
3 Answers
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oldest
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Actually both option you have given are equivalent. (Check b if you have some typos)
And both are correct . As function is odd so if you integrate it you get even function and by subtracting you get 0
answered Nov 21 '18 at 17:03
Shubham
1,5951519
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
add a comment |
option b) is false
$displaystyle underset{[-1,1]}inf fint_{-1}^{1}dtleint_{-1}^{1}f(t)dtle underset{[-1,1]}sup fint_{-1}^{1}dtiff 2underset{[-1,1]}inf fleint_{-1}^{1}f(t)dtle 2underset{[-1,1]}sup f$
so $f: xto cos^2(x)$ is counterexample
$displaystyle int_{-1}^{1}cos^2(t)dt =int_{0}^{1}1+cos(2t)dt=bigg[t+dfrac{sin(2t)}{2}bigg]^1_0=1+dfrac{sin2 }{2}>1ge f(x)ge0,quad forall xinmathbb{R}$
answered Nov 21 '18 at 18:23
Messi fifa
51611
add a comment |
It looks on the face of it that the other answer (the first you've got - this is the second) is just plainly obviously correct ... but in the abstract theory of functions I have seen so many ingeniously constructed 'pathological' functions that somehow manage to defy & foil what appear on the face of it to be the plainest of truths that I wouldn't even venture say with absolute certainty that $x=-ximplies x=0$.
Update
You have to be really careful looking at these, don't you, as it's so easy to make a little slip.
As for (a), it's just saying that the function is odd ... and therefore that the integral is 0, because every $operatorname{f}(-t)dt$ is cancelled by a $operatorname{f}(+t)dt$
But then (b) is stated independently of (a): if you're not careful, you can presume that the condition in (a) carries over into (b). And then another point you have to be careful about is that $$int_{-pi}^pioperatorname{f}(-t)dt$$ is not $$int_{-pi}^pioperatorname{f}(t)dt$$ with $-t$ substituted for $t$, but simply the integral of the function $operatorname{g}(t)$ obtained by reflecting $operatorname{f}(t)$ about $t=0$ ... meaning that in doing the integral it is the same integral but just carried out in reverse order. So (b) does not hold in general, but only when (a) applies ... in which case it's true by reason only of the being zero of the integral.
All this of course precludes any consideration of the sort of pathological function I was talking about at first, that can spring allmanner of surprise. But that is probably precluded by the continuity requirement anyway.
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
add a comment |
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3 Answers
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3 Answers
3
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oldest
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active
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Actually both option you have given are equivalent. (Check b if you have some typos)
And both are correct . As function is odd so if you integrate it you get even function and by subtracting you get 0
answered Nov 21 '18 at 17:03
Shubham
1,5951519
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
add a comment |
Actually both option you have given are equivalent. (Check b if you have some typos)
And both are correct . As function is odd so if you integrate it you get even function and by subtracting you get 0
answered Nov 21 '18 at 17:03
Shubham
1,5951519
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
add a comment |
Actually both option you have given are equivalent. (Check b if you have some typos)
And both are correct . As function is odd so if you integrate it you get even function and by subtracting you get 0
answered Nov 21 '18 at 17:03
Shubham
1,5951519
Actually both option you have given are equivalent. (Check b if you have some typos)
And both are correct . As function is odd so if you integrate it you get even function and by subtracting you get 0
answered Nov 21 '18 at 17:03
Shubham
1,5951519
answered Nov 21 '18 at 17:03
Shubham
1,5951519
answered Nov 21 '18 at 17:03
Shubham
1,5951519
answered Nov 21 '18 at 17:03
Shubham
1,5951519
1,5951519
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
add a comment |
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
ya,,u r saying right i have edited its now,,,,see again
– Messi fifa
Nov 21 '18 at 17:11
add a comment |
option b) is false
$displaystyle underset{[-1,1]}inf fint_{-1}^{1}dtleint_{-1}^{1}f(t)dtle underset{[-1,1]}sup fint_{-1}^{1}dtiff 2underset{[-1,1]}inf fleint_{-1}^{1}f(t)dtle 2underset{[-1,1]}sup f$
so $f: xto cos^2(x)$ is counterexample
$displaystyle int_{-1}^{1}cos^2(t)dt =int_{0}^{1}1+cos(2t)dt=bigg[t+dfrac{sin(2t)}{2}bigg]^1_0=1+dfrac{sin2 }{2}>1ge f(x)ge0,quad forall xinmathbb{R}$
answered Nov 21 '18 at 18:23
Messi fifa
51611
add a comment |
option b) is false
$displaystyle underset{[-1,1]}inf fint_{-1}^{1}dtleint_{-1}^{1}f(t)dtle underset{[-1,1]}sup fint_{-1}^{1}dtiff 2underset{[-1,1]}inf fleint_{-1}^{1}f(t)dtle 2underset{[-1,1]}sup f$
so $f: xto cos^2(x)$ is counterexample
$displaystyle int_{-1}^{1}cos^2(t)dt =int_{0}^{1}1+cos(2t)dt=bigg[t+dfrac{sin(2t)}{2}bigg]^1_0=1+dfrac{sin2 }{2}>1ge f(x)ge0,quad forall xinmathbb{R}$
answered Nov 21 '18 at 18:23
Messi fifa
51611
add a comment |
option b) is false
$displaystyle underset{[-1,1]}inf fint_{-1}^{1}dtleint_{-1}^{1}f(t)dtle underset{[-1,1]}sup fint_{-1}^{1}dtiff 2underset{[-1,1]}inf fleint_{-1}^{1}f(t)dtle 2underset{[-1,1]}sup f$
so $f: xto cos^2(x)$ is counterexample
$displaystyle int_{-1}^{1}cos^2(t)dt =int_{0}^{1}1+cos(2t)dt=bigg[t+dfrac{sin(2t)}{2}bigg]^1_0=1+dfrac{sin2 }{2}>1ge f(x)ge0,quad forall xinmathbb{R}$
answered Nov 21 '18 at 18:23
Messi fifa
51611
option b) is false
$displaystyle underset{[-1,1]}inf fint_{-1}^{1}dtleint_{-1}^{1}f(t)dtle underset{[-1,1]}sup fint_{-1}^{1}dtiff 2underset{[-1,1]}inf fleint_{-1}^{1}f(t)dtle 2underset{[-1,1]}sup f$
so $f: xto cos^2(x)$ is counterexample
$displaystyle int_{-1}^{1}cos^2(t)dt =int_{0}^{1}1+cos(2t)dt=bigg[t+dfrac{sin(2t)}{2}bigg]^1_0=1+dfrac{sin2 }{2}>1ge f(x)ge0,quad forall xinmathbb{R}$
answered Nov 21 '18 at 18:23
Messi fifa
51611
answered Nov 21 '18 at 18:23
Messi fifa
51611
answered Nov 21 '18 at 18:23
Messi fifa
51611
answered Nov 21 '18 at 18:23
Messi fifa
51611
51611
add a comment |
add a comment |
It looks on the face of it that the other answer (the first you've got - this is the second) is just plainly obviously correct ... but in the abstract theory of functions I have seen so many ingeniously constructed 'pathological' functions that somehow manage to defy & foil what appear on the face of it to be the plainest of truths that I wouldn't even venture say with absolute certainty that $x=-ximplies x=0$.
Update
You have to be really careful looking at these, don't you, as it's so easy to make a little slip.
As for (a), it's just saying that the function is odd ... and therefore that the integral is 0, because every $operatorname{f}(-t)dt$ is cancelled by a $operatorname{f}(+t)dt$
But then (b) is stated independently of (a): if you're not careful, you can presume that the condition in (a) carries over into (b). And then another point you have to be careful about is that $$int_{-pi}^pioperatorname{f}(-t)dt$$ is not $$int_{-pi}^pioperatorname{f}(t)dt$$ with $-t$ substituted for $t$, but simply the integral of the function $operatorname{g}(t)$ obtained by reflecting $operatorname{f}(t)$ about $t=0$ ... meaning that in doing the integral it is the same integral but just carried out in reverse order. So (b) does not hold in general, but only when (a) applies ... in which case it's true by reason only of the being zero of the integral.
All this of course precludes any consideration of the sort of pathological function I was talking about at first, that can spring allmanner of surprise. But that is probably precluded by the continuity requirement anyway.
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
add a comment |
It looks on the face of it that the other answer (the first you've got - this is the second) is just plainly obviously correct ... but in the abstract theory of functions I have seen so many ingeniously constructed 'pathological' functions that somehow manage to defy & foil what appear on the face of it to be the plainest of truths that I wouldn't even venture say with absolute certainty that $x=-ximplies x=0$.
Update
You have to be really careful looking at these, don't you, as it's so easy to make a little slip.
As for (a), it's just saying that the function is odd ... and therefore that the integral is 0, because every $operatorname{f}(-t)dt$ is cancelled by a $operatorname{f}(+t)dt$
But then (b) is stated independently of (a): if you're not careful, you can presume that the condition in (a) carries over into (b). And then another point you have to be careful about is that $$int_{-pi}^pioperatorname{f}(-t)dt$$ is not $$int_{-pi}^pioperatorname{f}(t)dt$$ with $-t$ substituted for $t$, but simply the integral of the function $operatorname{g}(t)$ obtained by reflecting $operatorname{f}(t)$ about $t=0$ ... meaning that in doing the integral it is the same integral but just carried out in reverse order. So (b) does not hold in general, but only when (a) applies ... in which case it's true by reason only of the being zero of the integral.
All this of course precludes any consideration of the sort of pathological function I was talking about at first, that can spring allmanner of surprise. But that is probably precluded by the continuity requirement anyway.
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
add a comment |
It looks on the face of it that the other answer (the first you've got - this is the second) is just plainly obviously correct ... but in the abstract theory of functions I have seen so many ingeniously constructed 'pathological' functions that somehow manage to defy & foil what appear on the face of it to be the plainest of truths that I wouldn't even venture say with absolute certainty that $x=-ximplies x=0$.
Update
You have to be really careful looking at these, don't you, as it's so easy to make a little slip.
As for (a), it's just saying that the function is odd ... and therefore that the integral is 0, because every $operatorname{f}(-t)dt$ is cancelled by a $operatorname{f}(+t)dt$
But then (b) is stated independently of (a): if you're not careful, you can presume that the condition in (a) carries over into (b). And then another point you have to be careful about is that $$int_{-pi}^pioperatorname{f}(-t)dt$$ is not $$int_{-pi}^pioperatorname{f}(t)dt$$ with $-t$ substituted for $t$, but simply the integral of the function $operatorname{g}(t)$ obtained by reflecting $operatorname{f}(t)$ about $t=0$ ... meaning that in doing the integral it is the same integral but just carried out in reverse order. So (b) does not hold in general, but only when (a) applies ... in which case it's true by reason only of the being zero of the integral.
All this of course precludes any consideration of the sort of pathological function I was talking about at first, that can spring allmanner of surprise. But that is probably precluded by the continuity requirement anyway.
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
It looks on the face of it that the other answer (the first you've got - this is the second) is just plainly obviously correct ... but in the abstract theory of functions I have seen so many ingeniously constructed 'pathological' functions that somehow manage to defy & foil what appear on the face of it to be the plainest of truths that I wouldn't even venture say with absolute certainty that $x=-ximplies x=0$.
Update
You have to be really careful looking at these, don't you, as it's so easy to make a little slip.
As for (a), it's just saying that the function is odd ... and therefore that the integral is 0, because every $operatorname{f}(-t)dt$ is cancelled by a $operatorname{f}(+t)dt$
But then (b) is stated independently of (a): if you're not careful, you can presume that the condition in (a) carries over into (b). And then another point you have to be careful about is that $$int_{-pi}^pioperatorname{f}(-t)dt$$ is not $$int_{-pi}^pioperatorname{f}(t)dt$$ with $-t$ substituted for $t$, but simply the integral of the function $operatorname{g}(t)$ obtained by reflecting $operatorname{f}(t)$ about $t=0$ ... meaning that in doing the integral it is the same integral but just carried out in reverse order. So (b) does not hold in general, but only when (a) applies ... in which case it's true by reason only of the being zero of the integral.
All this of course precludes any consideration of the sort of pathological function I was talking about at first, that can spring allmanner of surprise. But that is probably precluded by the continuity requirement anyway.
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
edited Nov 22 '18 at 1:10
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
answered Nov 21 '18 at 17:14
AmbretteOrrisey
57410
57410
add a comment |
add a comment |
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