Given: $int_{a}^{b}f=int_{a}^{b}g$, Can we always find some point $theta$ in $[a,b]$ such that...
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Given: $int_{a}^{b}f=int_{a}^{b}g$, Can we always find some point $theta$ in $[a,b]$ such that $f(theta)=g(theta)$
The 2019 Stack Overflow Developer Survey Results Are InLebesgue Integration and Convergence TheoremsLine integral on unit circle over non-holomorphic functionAny arbitrary closed smooth curve bounds a orientable surface?Basic facts related to Haar measureHow can I prove that there exists $cin[0,1]$ such that $intlimits_0^1intlimits_0^tf''(x)dxdt=frac{1}{2}f''(c)$Is $int f$ on [0,1] always equal to $int f$ on [x,1] when take the limit of x to 0Find $f(x)$ such that $int_{-infty}^infty int_{-a}^a frac{|y+x|^p}{1+y^2} f(x) dx dy<infty$Prove $int_{R^{+}} frac{1}{1+x} a(x) dx > 0$ given $int_{R^{+}} frac{1}{x} a(x) dx > 0$ and $int_{R^{+}} a(x) dx = 0$?Under what conditions are expected value and integral commutable?Can some transformation of a function always be found such that it has a specified integral?
$begingroup$
I noted something today, I don't know whether the result holds in general setting or not.
Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_{a}^{b}f=int_{a}^{b}g$.
Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$
Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_{[0,1]} g=1/3$
Now we can easily find some points which satisfy $x^2=1/3$.
There are lot of examples that I checked and every time I found that resullt holds good.
Can we write a proof?
Thanks for reading and helping out.
real-analysis integration
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
add a comment |
$begingroup$
I noted something today, I don't know whether the result holds in general setting or not.
Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_{a}^{b}f=int_{a}^{b}g$.
Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$
Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_{[0,1]} g=1/3$
Now we can easily find some points which satisfy $x^2=1/3$.
There are lot of examples that I checked and every time I found that resullt holds good.
Can we write a proof?
Thanks for reading and helping out.
real-analysis integration
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
1
$begingroup$
You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:24
1
$begingroup$
It suffices to prove that if $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
$endgroup$
– dmtri
Mar 21 at 6:24
add a comment |
$begingroup$
I noted something today, I don't know whether the result holds in general setting or not.
Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_{a}^{b}f=int_{a}^{b}g$.
Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$
Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_{[0,1]} g=1/3$
Now we can easily find some points which satisfy $x^2=1/3$.
There are lot of examples that I checked and every time I found that resullt holds good.
Can we write a proof?
Thanks for reading and helping out.
real-analysis integration
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
I noted something today, I don't know whether the result holds in general setting or not.
Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_{a}^{b}f=int_{a}^{b}g$.
Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$
Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_{[0,1]} g=1/3$
Now we can easily find some points which satisfy $x^2=1/3$.
There are lot of examples that I checked and every time I found that resullt holds good.
Can we write a proof?
Thanks for reading and helping out.
real-analysis integration
real-analysis integration
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
asked Mar 21 at 6:17
StammeringMathematicianStammeringMathematician
2,7951324
2,7951324
1
$begingroup$
You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:24
1
$begingroup$
It suffices to prove that if $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
$endgroup$
– dmtri
Mar 21 at 6:24
add a comment |
1
$begingroup$
You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:24
1
$begingroup$
It suffices to prove that if $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
$endgroup$
– dmtri
Mar 21 at 6:24
1
1
$begingroup$
You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:24
$begingroup$
You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:24
1
1
$begingroup$
It suffices to prove that if $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
$endgroup$
– dmtri
Mar 21 at 6:24
$begingroup$
It suffices to prove that if $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
$endgroup$
– dmtri
Mar 21 at 6:24
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This is just Rolle's theorem disguised.
Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
2
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
1
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
add a comment |
$begingroup$
An equivalent formulation (via $h=f-g$) is
Suppose $h$ is a continuous function on $[a,b]$ such that
$int_{a}^{b}h(x) , dx=0$.
Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?
and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_{a}^{b}h(x) , dx ne 0$.
(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
$endgroup$
add a comment |
$begingroup$
Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:
$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$
answered Mar 21 at 6:24
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, we have
$$int_a^b{(f(x)-g(x))dx} = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is just Rolle's theorem disguised.
Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
2
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
1
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
add a comment |
$begingroup$
This is just Rolle's theorem disguised.
Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
2
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
1
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
add a comment |
$begingroup$
This is just Rolle's theorem disguised.
Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
This is just Rolle's theorem disguised.
Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
answered Mar 21 at 6:25
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
40.3k33678
2
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
1
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
add a comment |
2
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
1
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
2
2
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
$begingroup$
Olof.What a nice solution!
$endgroup$
– Peter Szilas
Mar 21 at 6:42
1
1
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
$begingroup$
Dear Peter, thank you for the compliment
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 7:41
add a comment |
$begingroup$
An equivalent formulation (via $h=f-g$) is
Suppose $h$ is a continuous function on $[a,b]$ such that
$int_{a}^{b}h(x) , dx=0$.
Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?
and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_{a}^{b}h(x) , dx ne 0$.
(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
$endgroup$
add a comment |
$begingroup$
An equivalent formulation (via $h=f-g$) is
Suppose $h$ is a continuous function on $[a,b]$ such that
$int_{a}^{b}h(x) , dx=0$.
Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?
and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_{a}^{b}h(x) , dx ne 0$.
(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
$endgroup$
add a comment |
$begingroup$
An equivalent formulation (via $h=f-g$) is
Suppose $h$ is a continuous function on $[a,b]$ such that
$int_{a}^{b}h(x) , dx=0$.
Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?
and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_{a}^{b}h(x) , dx ne 0$.
(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
$endgroup$
An equivalent formulation (via $h=f-g$) is
Suppose $h$ is a continuous function on $[a,b]$ such that
$int_{a}^{b}h(x) , dx=0$.
Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?
and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_{a}^{b}h(x) , dx ne 0$.
(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
edited Mar 21 at 7:37
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
answered Mar 21 at 6:23
Martin RMartin R
30.8k33561
30.8k33561
add a comment |
add a comment |
$begingroup$
Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:
$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$
answered Mar 21 at 6:24
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:
$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$
answered Mar 21 at 6:24
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:
$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$
answered Mar 21 at 6:24
FredFred
48.5k11849
$endgroup$
Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:
$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$
answered Mar 21 at 6:24
FredFred
48.5k11849
answered Mar 21 at 6:24
FredFred
48.5k11849
answered Mar 21 at 6:24
FredFred
48.5k11849
answered Mar 21 at 6:24
FredFred
48.5k11849
48.5k11849
add a comment |
add a comment |
$begingroup$
As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, we have
$$int_a^b{(f(x)-g(x))dx} = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
$endgroup$
add a comment |
$begingroup$
As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, we have
$$int_a^b{(f(x)-g(x))dx} = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
$endgroup$
add a comment |
$begingroup$
As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, we have
$$int_a^b{(f(x)-g(x))dx} = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
$endgroup$
As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, we have
$$int_a^b{(f(x)-g(x))dx} = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
answered Mar 21 at 6:24
HaydenHayden
13.8k12448
13.8k12448
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1
$begingroup$
You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:24
1
$begingroup$
It suffices to prove that if $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
$endgroup$
– dmtri
Mar 21 at 6:24