Mixing
Finding all continuous functions such that $int_0^xf(t)dt=(f(x))^2+C $
$begingroup$
$C$ is a constant. FTC shows that $f(x)^2+C$ must be differentiable, which means that $f(x)^2$ is differentiable. But we don't know that $f(x)$ is differentiable then, right? I had the idea that the only two solutions are $f(x)=0$ or $f(x)=frac{x}{2}$, but this is assuming that $f(x)$ is differentiable. I'm not sure how to show that they are the only solutions (if they even are the only solutions).
real-analysis integration
edited Jan 27 at 3:18
Andrei
13.2k21230
asked Jan 27 at 3:04
Riley HRiley H
1046
$endgroup$
add a comment |
$begingroup$
$C$ is a constant. FTC shows that $f(x)^2+C$ must be differentiable, which means that $f(x)^2$ is differentiable. But we don't know that $f(x)$ is differentiable then, right? I had the idea that the only two solutions are $f(x)=0$ or $f(x)=frac{x}{2}$, but this is assuming that $f(x)$ is differentiable. I'm not sure how to show that they are the only solutions (if they even are the only solutions).
real-analysis integration
edited Jan 27 at 3:18
Andrei
13.2k21230
asked Jan 27 at 3:04
Riley HRiley H
1046
$endgroup$
3
$begingroup$
What are the hypotheses about $f$ ? $$f(x)^2 differentiable Longrightarrow f(x) differentiable$$ only around points where $f(x)$ is continuous and $f(x)not=0$. Look at $1/4(x+|x|)$ which satisfies your equation.
$endgroup$
– Duchamp Gérard H. E.
Jan 27 at 3:36
add a comment |
$begingroup$
$C$ is a constant. FTC shows that $f(x)^2+C$ must be differentiable, which means that $f(x)^2$ is differentiable. But we don't know that $f(x)$ is differentiable then, right? I had the idea that the only two solutions are $f(x)=0$ or $f(x)=frac{x}{2}$, but this is assuming that $f(x)$ is differentiable. I'm not sure how to show that they are the only solutions (if they even are the only solutions).
real-analysis integration
edited Jan 27 at 3:18
Andrei
13.2k21230
asked Jan 27 at 3:04
Riley HRiley H
1046
$endgroup$
$C$ is a constant. FTC shows that $f(x)^2+C$ must be differentiable, which means that $f(x)^2$ is differentiable. But we don't know that $f(x)$ is differentiable then, right? I had the idea that the only two solutions are $f(x)=0$ or $f(x)=frac{x}{2}$, but this is assuming that $f(x)$ is differentiable. I'm not sure how to show that they are the only solutions (if they even are the only solutions).
real-analysis integration
real-analysis integration
edited Jan 27 at 3:18
Andrei
13.2k21230
asked Jan 27 at 3:04
Riley HRiley H
1046
edited Jan 27 at 3:18
Andrei
13.2k21230
asked Jan 27 at 3:04
Riley HRiley H
1046
edited Jan 27 at 3:18
Andrei
13.2k21230
edited Jan 27 at 3:18
Andrei
13.2k21230
edited Jan 27 at 3:18
Andrei
13.2k21230
13.2k21230
asked Jan 27 at 3:04
Riley HRiley H
1046
asked Jan 27 at 3:04
Riley HRiley H
1046
asked Jan 27 at 3:04
Riley HRiley H
1046
1046
3
$begingroup$
What are the hypotheses about $f$ ? $$f(x)^2 differentiable Longrightarrow f(x) differentiable$$ only around points where $f(x)$ is continuous and $f(x)not=0$. Look at $1/4(x+|x|)$ which satisfies your equation.
$endgroup$
– Duchamp Gérard H. E.
Jan 27 at 3:36
add a comment |
3
$begingroup$
What are the hypotheses about $f$ ? $$f(x)^2 differentiable Longrightarrow f(x) differentiable$$ only around points where $f(x)$ is continuous and $f(x)not=0$. Look at $1/4(x+|x|)$ which satisfies your equation.
$endgroup$
– Duchamp Gérard H. E.
Jan 27 at 3:36
3
3
$begingroup$
What are the hypotheses about $f$ ? $$f(x)^2 differentiable Longrightarrow f(x) differentiable$$ only around points where $f(x)$ is continuous and $f(x)not=0$. Look at $1/4(x+|x|)$ which satisfies your equation.
$endgroup$
– Duchamp Gérard H. E.
Jan 27 at 3:36
$begingroup$
What are the hypotheses about $f$ ? $$f(x)^2 differentiable Longrightarrow f(x) differentiable$$ only around points where $f(x)$ is continuous and $f(x)not=0$. Look at $1/4(x+|x|)$ which satisfies your equation.
$endgroup$
– Duchamp Gérard H. E.
Jan 27 at 3:36
add a comment |
2 Answers
2
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oldest
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$begingroup$
If the equation in the title is true, and $f$ is differentiable, we can take the derivative with respect to $x$ of both sides. $$frac d{dx}int_0^x f(t) dt=f(x)$$ and $$frac d{dx}(f^2(x)+C)=2f(x)f'(x)$$ Then:$$f(x)=2f(x)f'(x)$$ This has solutions $f(x)=0$ or $f'(x)=frac 12$. The second one you can integrate and get $f(x)=frac x2+c$. You just need to show if any $c$ is OK, or only $c=0$.
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
Let $g(x)=f(x)^2$. Then $g'(x)=f(x)$ and $g(x)=[g'(x)]^2$, i.e., $sqrt{y}=|frac{dy}{dx}|$. This differential equation can be solved by separation of variables. From $pm dx=frac{dy}{sqrt{y}}$ , we get $pm x=2sqrt{y} + A$, where $A in mathbb{R}$ is an arbitrary constant, and hence $g(x)=y=(frac{pm x-A}{2})^2$ or $g(x)=y=0$. Therefore $|f(x)|=|frac{pm x-A}{2}|$ or $f(x)=0$.
answered Jan 27 at 3:52
lzralbulzralbu
640512
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
If the equation in the title is true, and $f$ is differentiable, we can take the derivative with respect to $x$ of both sides. $$frac d{dx}int_0^x f(t) dt=f(x)$$ and $$frac d{dx}(f^2(x)+C)=2f(x)f'(x)$$ Then:$$f(x)=2f(x)f'(x)$$ This has solutions $f(x)=0$ or $f'(x)=frac 12$. The second one you can integrate and get $f(x)=frac x2+c$. You just need to show if any $c$ is OK, or only $c=0$.
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
If the equation in the title is true, and $f$ is differentiable, we can take the derivative with respect to $x$ of both sides. $$frac d{dx}int_0^x f(t) dt=f(x)$$ and $$frac d{dx}(f^2(x)+C)=2f(x)f'(x)$$ Then:$$f(x)=2f(x)f'(x)$$ This has solutions $f(x)=0$ or $f'(x)=frac 12$. The second one you can integrate and get $f(x)=frac x2+c$. You just need to show if any $c$ is OK, or only $c=0$.
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
If the equation in the title is true, and $f$ is differentiable, we can take the derivative with respect to $x$ of both sides. $$frac d{dx}int_0^x f(t) dt=f(x)$$ and $$frac d{dx}(f^2(x)+C)=2f(x)f'(x)$$ Then:$$f(x)=2f(x)f'(x)$$ This has solutions $f(x)=0$ or $f'(x)=frac 12$. The second one you can integrate and get $f(x)=frac x2+c$. You just need to show if any $c$ is OK, or only $c=0$.
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
$endgroup$
If the equation in the title is true, and $f$ is differentiable, we can take the derivative with respect to $x$ of both sides. $$frac d{dx}int_0^x f(t) dt=f(x)$$ and $$frac d{dx}(f^2(x)+C)=2f(x)f'(x)$$ Then:$$f(x)=2f(x)f'(x)$$ This has solutions $f(x)=0$ or $f'(x)=frac 12$. The second one you can integrate and get $f(x)=frac x2+c$. You just need to show if any $c$ is OK, or only $c=0$.
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
answered Jan 27 at 3:24
AndreiAndrei
13.2k21230
13.2k21230
add a comment |
add a comment |
$begingroup$
Let $g(x)=f(x)^2$. Then $g'(x)=f(x)$ and $g(x)=[g'(x)]^2$, i.e., $sqrt{y}=|frac{dy}{dx}|$. This differential equation can be solved by separation of variables. From $pm dx=frac{dy}{sqrt{y}}$ , we get $pm x=2sqrt{y} + A$, where $A in mathbb{R}$ is an arbitrary constant, and hence $g(x)=y=(frac{pm x-A}{2})^2$ or $g(x)=y=0$. Therefore $|f(x)|=|frac{pm x-A}{2}|$ or $f(x)=0$.
answered Jan 27 at 3:52
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Let $g(x)=f(x)^2$. Then $g'(x)=f(x)$ and $g(x)=[g'(x)]^2$, i.e., $sqrt{y}=|frac{dy}{dx}|$. This differential equation can be solved by separation of variables. From $pm dx=frac{dy}{sqrt{y}}$ , we get $pm x=2sqrt{y} + A$, where $A in mathbb{R}$ is an arbitrary constant, and hence $g(x)=y=(frac{pm x-A}{2})^2$ or $g(x)=y=0$. Therefore $|f(x)|=|frac{pm x-A}{2}|$ or $f(x)=0$.
answered Jan 27 at 3:52
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Let $g(x)=f(x)^2$. Then $g'(x)=f(x)$ and $g(x)=[g'(x)]^2$, i.e., $sqrt{y}=|frac{dy}{dx}|$. This differential equation can be solved by separation of variables. From $pm dx=frac{dy}{sqrt{y}}$ , we get $pm x=2sqrt{y} + A$, where $A in mathbb{R}$ is an arbitrary constant, and hence $g(x)=y=(frac{pm x-A}{2})^2$ or $g(x)=y=0$. Therefore $|f(x)|=|frac{pm x-A}{2}|$ or $f(x)=0$.
answered Jan 27 at 3:52
lzralbulzralbu
640512
$endgroup$
Let $g(x)=f(x)^2$. Then $g'(x)=f(x)$ and $g(x)=[g'(x)]^2$, i.e., $sqrt{y}=|frac{dy}{dx}|$. This differential equation can be solved by separation of variables. From $pm dx=frac{dy}{sqrt{y}}$ , we get $pm x=2sqrt{y} + A$, where $A in mathbb{R}$ is an arbitrary constant, and hence $g(x)=y=(frac{pm x-A}{2})^2$ or $g(x)=y=0$. Therefore $|f(x)|=|frac{pm x-A}{2}|$ or $f(x)=0$.
answered Jan 27 at 3:52
lzralbulzralbu
640512
edited Jan 27 at 15:40
answered Jan 27 at 3:52
lzralbulzralbu
640512
answered Jan 27 at 3:52
lzralbulzralbu
640512
answered Jan 27 at 3:52
lzralbulzralbu
640512
640512
add a comment |
add a comment |
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$begingroup$
What are the hypotheses about $f$ ? $$f(x)^2 differentiable Longrightarrow f(x) differentiable$$ only around points where $f(x)$ is continuous and $f(x)not=0$. Look at $1/4(x+|x|)$ which satisfies your equation.
$endgroup$
– Duchamp Gérard H. E.
Jan 27 at 3:36