Solving functional equation $F(x) = F(x^2 + 1)$ to prove that $F(x)$ is differentiable
Show that there exists a non-zero continuous function $f$ satisfying: $int_a^{a^2 + 1} f(x) dx = 0$ for all real $a$.
By letting $F'(x) = f(x)$, I have got that $F(x) = F(x^2 + 1)$, which implies that $F(x)$ is even, so $f(x)$ is odd.
I am stuck when it comes to showing that $f(x)$ is continuous (or equivalently that $F(x)$ is differentiable) using the information I currently have.
Is it necessary to find explicit expressions for $F(x)$ and / or $f(x)$ or is it possible to solve the problem without doing so? [I have tried guessing functions for $F(x)$ and using a series solution, but neither approach has worked.]
integration functions
asked Dec 3 '18 at 8:53
1123581321
11918
|
show 4 more comments
Show that there exists a non-zero continuous function $f$ satisfying: $int_a^{a^2 + 1} f(x) dx = 0$ for all real $a$.
By letting $F'(x) = f(x)$, I have got that $F(x) = F(x^2 + 1)$, which implies that $F(x)$ is even, so $f(x)$ is odd.
I am stuck when it comes to showing that $f(x)$ is continuous (or equivalently that $F(x)$ is differentiable) using the information I currently have.
Is it necessary to find explicit expressions for $F(x)$ and / or $f(x)$ or is it possible to solve the problem without doing so? [I have tried guessing functions for $F(x)$ and using a series solution, but neither approach has worked.]
integration functions
asked Dec 3 '18 at 8:53
1123581321
11918
1
You can't prove $f(x)$ is non-zero, because the constant zero function satisfies this integral equation. You also can't show it's continuous - changing its value at one point will not violate the property.
– Wojowu
Dec 3 '18 at 9:02
Also $f$ being continuous is not equivalent to $F$ being differntiable. $F$ is always differentiable (by definition) if exists. Which also brings the question: why do you think that $f$ has an antiderivative? Without the assumption that $f$ is continuous this may not be true.
– freakish
Dec 3 '18 at 9:09
@Wojowu What I meant was that assuming that $f(x)$ is not the zero function, show that the function is continuous - I can see that $f(x) = 0$ would satisfy the given integral
– 1123581321
Dec 3 '18 at 9:17
@Wojowu Also I'm confused about why I can't show that $f(x)$ is continuous
– 1123581321
Dec 3 '18 at 9:17
Take $f(x)=0$ for $xneq 1$ and $f(1)=1$. This is a discontinuous function also satisfying the equation.
– Wojowu
Dec 3 '18 at 9:18
|
show 4 more comments
Show that there exists a non-zero continuous function $f$ satisfying: $int_a^{a^2 + 1} f(x) dx = 0$ for all real $a$.
By letting $F'(x) = f(x)$, I have got that $F(x) = F(x^2 + 1)$, which implies that $F(x)$ is even, so $f(x)$ is odd.
I am stuck when it comes to showing that $f(x)$ is continuous (or equivalently that $F(x)$ is differentiable) using the information I currently have.
Is it necessary to find explicit expressions for $F(x)$ and / or $f(x)$ or is it possible to solve the problem without doing so? [I have tried guessing functions for $F(x)$ and using a series solution, but neither approach has worked.]
integration functions
asked Dec 3 '18 at 8:53
1123581321
11918
Show that there exists a non-zero continuous function $f$ satisfying: $int_a^{a^2 + 1} f(x) dx = 0$ for all real $a$.
By letting $F'(x) = f(x)$, I have got that $F(x) = F(x^2 + 1)$, which implies that $F(x)$ is even, so $f(x)$ is odd.
I am stuck when it comes to showing that $f(x)$ is continuous (or equivalently that $F(x)$ is differentiable) using the information I currently have.
Is it necessary to find explicit expressions for $F(x)$ and / or $f(x)$ or is it possible to solve the problem without doing so? [I have tried guessing functions for $F(x)$ and using a series solution, but neither approach has worked.]
integration functions
integration functions
asked Dec 3 '18 at 8:53
1123581321
11918
asked Dec 3 '18 at 8:53
1123581321
11918
edited Dec 3 '18 at 9:22
asked Dec 3 '18 at 8:53
1123581321
11918
asked Dec 3 '18 at 8:53
1123581321
11918
asked Dec 3 '18 at 8:53
1123581321
11918
11918
1
You can't prove $f(x)$ is non-zero, because the constant zero function satisfies this integral equation. You also can't show it's continuous - changing its value at one point will not violate the property.
– Wojowu
Dec 3 '18 at 9:02
Also $f$ being continuous is not equivalent to $F$ being differntiable. $F$ is always differentiable (by definition) if exists. Which also brings the question: why do you think that $f$ has an antiderivative? Without the assumption that $f$ is continuous this may not be true.
– freakish
Dec 3 '18 at 9:09
@Wojowu What I meant was that assuming that $f(x)$ is not the zero function, show that the function is continuous - I can see that $f(x) = 0$ would satisfy the given integral
– 1123581321
Dec 3 '18 at 9:17
@Wojowu Also I'm confused about why I can't show that $f(x)$ is continuous
– 1123581321
Dec 3 '18 at 9:17
Take $f(x)=0$ for $xneq 1$ and $f(1)=1$. This is a discontinuous function also satisfying the equation.
– Wojowu
Dec 3 '18 at 9:18
|
show 4 more comments
1
You can't prove $f(x)$ is non-zero, because the constant zero function satisfies this integral equation. You also can't show it's continuous - changing its value at one point will not violate the property.
– Wojowu
Dec 3 '18 at 9:02
Also $f$ being continuous is not equivalent to $F$ being differntiable. $F$ is always differentiable (by definition) if exists. Which also brings the question: why do you think that $f$ has an antiderivative? Without the assumption that $f$ is continuous this may not be true.
– freakish
Dec 3 '18 at 9:09
@Wojowu What I meant was that assuming that $f(x)$ is not the zero function, show that the function is continuous - I can see that $f(x) = 0$ would satisfy the given integral
– 1123581321
Dec 3 '18 at 9:17
@Wojowu Also I'm confused about why I can't show that $f(x)$ is continuous
– 1123581321
Dec 3 '18 at 9:17
Take $f(x)=0$ for $xneq 1$ and $f(1)=1$. This is a discontinuous function also satisfying the equation.
– Wojowu
Dec 3 '18 at 9:18
1
1
You can't prove $f(x)$ is non-zero, because the constant zero function satisfies this integral equation. You also can't show it's continuous - changing its value at one point will not violate the property.
– Wojowu
Dec 3 '18 at 9:02
You can't prove $f(x)$ is non-zero, because the constant zero function satisfies this integral equation. You also can't show it's continuous - changing its value at one point will not violate the property.
– Wojowu
Dec 3 '18 at 9:02
Also $f$ being continuous is not equivalent to $F$ being differntiable. $F$ is always differentiable (by definition) if exists. Which also brings the question: why do you think that $f$ has an antiderivative? Without the assumption that $f$ is continuous this may not be true.
– freakish
Dec 3 '18 at 9:09
Also $f$ being continuous is not equivalent to $F$ being differntiable. $F$ is always differentiable (by definition) if exists. Which also brings the question: why do you think that $f$ has an antiderivative? Without the assumption that $f$ is continuous this may not be true.
– freakish
Dec 3 '18 at 9:09
@Wojowu What I meant was that assuming that $f(x)$ is not the zero function, show that the function is continuous - I can see that $f(x) = 0$ would satisfy the given integral
– 1123581321
Dec 3 '18 at 9:17
@Wojowu What I meant was that assuming that $f(x)$ is not the zero function, show that the function is continuous - I can see that $f(x) = 0$ would satisfy the given integral
– 1123581321
Dec 3 '18 at 9:17
@Wojowu Also I'm confused about why I can't show that $f(x)$ is continuous
– 1123581321
Dec 3 '18 at 9:17
@Wojowu Also I'm confused about why I can't show that $f(x)$ is continuous
– 1123581321
Dec 3 '18 at 9:17
Take $f(x)=0$ for $xneq 1$ and $f(1)=1$. This is a discontinuous function also satisfying the equation.
– Wojowu
Dec 3 '18 at 9:18
Take $f(x)=0$ for $xneq 1$ and $f(1)=1$. This is a discontinuous function also satisfying the equation.
– Wojowu
Dec 3 '18 at 9:18
|
show 4 more comments
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1
You can't prove $f(x)$ is non-zero, because the constant zero function satisfies this integral equation. You also can't show it's continuous - changing its value at one point will not violate the property.
– Wojowu
Dec 3 '18 at 9:02
Also $f$ being continuous is not equivalent to $F$ being differntiable. $F$ is always differentiable (by definition) if exists. Which also brings the question: why do you think that $f$ has an antiderivative? Without the assumption that $f$ is continuous this may not be true.
– freakish
Dec 3 '18 at 9:09
@Wojowu What I meant was that assuming that $f(x)$ is not the zero function, show that the function is continuous - I can see that $f(x) = 0$ would satisfy the given integral
– 1123581321
Dec 3 '18 at 9:17
@Wojowu Also I'm confused about why I can't show that $f(x)$ is continuous
– 1123581321
Dec 3 '18 at 9:17
Take $f(x)=0$ for $xneq 1$ and $f(1)=1$. This is a discontinuous function also satisfying the equation.
– Wojowu
Dec 3 '18 at 9:18