Mixing
How many types of functions $f(x)$ satisfy $(f(x))^2$ + $(f'(x))^2 = constant$?
0
$begingroup$
I know of two types of $f(x)$ that works.
First one is if $f(x) = c$, of course.
Second one is if it's a trigonometric function: $f(x) = c * cos(x)$ or $f(x) = c * sin(x)$.
My question is, are there any other types of functions that satisfy this criteria?
analysis functions constants
edited Jan 15 at 19:25
Henrik
6,03792030
asked Jan 15 at 19:09
throw123throw123
32
$endgroup$
add a comment |
0
$begingroup$
I know of two types of $f(x)$ that works.
First one is if $f(x) = c$, of course.
Second one is if it's a trigonometric function: $f(x) = c * cos(x)$ or $f(x) = c * sin(x)$.
My question is, are there any other types of functions that satisfy this criteria?
analysis functions constants
edited Jan 15 at 19:25
Henrik
6,03792030
asked Jan 15 at 19:09
throw123throw123
32
$endgroup$
$begingroup$
Solve the separable differential equation $(y')^2=c-y^2$.
$endgroup$
– Qurultay
Jan 15 at 19:17
add a comment |
0
0
0
$begingroup$
I know of two types of $f(x)$ that works.
First one is if $f(x) = c$, of course.
Second one is if it's a trigonometric function: $f(x) = c * cos(x)$ or $f(x) = c * sin(x)$.
My question is, are there any other types of functions that satisfy this criteria?
analysis functions constants
edited Jan 15 at 19:25
Henrik
6,03792030
asked Jan 15 at 19:09
throw123throw123
32
$endgroup$
I know of two types of $f(x)$ that works.
First one is if $f(x) = c$, of course.
Second one is if it's a trigonometric function: $f(x) = c * cos(x)$ or $f(x) = c * sin(x)$.
My question is, are there any other types of functions that satisfy this criteria?
analysis functions constants
analysis functions constants
edited Jan 15 at 19:25
Henrik
6,03792030
asked Jan 15 at 19:09
throw123throw123
32
edited Jan 15 at 19:25
Henrik
6,03792030
asked Jan 15 at 19:09
throw123throw123
32
edited Jan 15 at 19:25
Henrik
6,03792030
edited Jan 15 at 19:25
Henrik
6,03792030
edited Jan 15 at 19:25
Henrik
6,03792030
6,03792030
asked Jan 15 at 19:09
throw123throw123
32
asked Jan 15 at 19:09
throw123throw123
32
asked Jan 15 at 19:09
throw123throw123
32
32
$begingroup$
Solve the separable differential equation $(y')^2=c-y^2$.
$endgroup$
– Qurultay
Jan 15 at 19:17
add a comment |
$begingroup$
Solve the separable differential equation $(y')^2=c-y^2$.
$endgroup$
– Qurultay
Jan 15 at 19:17
$begingroup$
Solve the separable differential equation $(y')^2=c-y^2$.
$endgroup$
– Qurultay
Jan 15 at 19:17
$begingroup$
Solve the separable differential equation $(y')^2=c-y^2$.
$endgroup$
– Qurultay
Jan 15 at 19:17
add a comment |
2 Answers
2
active
oldest
votes
4
$begingroup$
Differentiating both sides:
$$2ff'+2f'f''=0$$
$$2f'(f+f'')=0$$
So either $f'=0$, which gives $f(x)=c$, or $f+f''=0$, which gives $f(x)=Asin(x)+Bcos(x)$.
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
$endgroup$
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
2
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
add a comment |
0
$begingroup$
Hint.
$$
f'(x) = pmsqrt{c-f(x)^2}
$$
then
$$
dx =pmfrac{df}{sqrt{c-f^2}}
$$
giving
$$
x + C_0 = pmarctanleft(frac{f}{sqrt{ c-f^2}}right)
$$
etc.
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
Differentiating both sides:
$$2ff'+2f'f''=0$$
$$2f'(f+f'')=0$$
So either $f'=0$, which gives $f(x)=c$, or $f+f''=0$, which gives $f(x)=Asin(x)+Bcos(x)$.
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
$endgroup$
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
2
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
add a comment |
4
$begingroup$
Differentiating both sides:
$$2ff'+2f'f''=0$$
$$2f'(f+f'')=0$$
So either $f'=0$, which gives $f(x)=c$, or $f+f''=0$, which gives $f(x)=Asin(x)+Bcos(x)$.
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
$endgroup$
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
2
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
add a comment |
4
4
4
$begingroup$
Differentiating both sides:
$$2ff'+2f'f''=0$$
$$2f'(f+f'')=0$$
So either $f'=0$, which gives $f(x)=c$, or $f+f''=0$, which gives $f(x)=Asin(x)+Bcos(x)$.
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
$endgroup$
Differentiating both sides:
$$2ff'+2f'f''=0$$
$$2f'(f+f'')=0$$
So either $f'=0$, which gives $f(x)=c$, or $f+f''=0$, which gives $f(x)=Asin(x)+Bcos(x)$.
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
answered Jan 15 at 19:23
TonyKTonyK
42.7k355134
42.7k355134
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
2
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
add a comment |
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
2
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
So just all vectors in span of ${ sin x , cos x } $
$endgroup$
– Jimmy Sabater
Jan 15 at 19:24
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
$begingroup$
@JimmySabater: No. $f(x)=c$ is not in the span of ${sin x,cos x}$.
$endgroup$
– TonyK
Jan 15 at 19:27
2
2
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
Note that you can also concatenate these solutions (at the points where sine and cosine have zero derivative) in many different ways: for example, I can take $f=1$ for $x leq 0$ and a cosine for $x geq 0$. That won't be smooth though, but it's still $C^1$.
$endgroup$
– Thomas Bakx
Jan 15 at 19:30
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
@ThomasBakx: Yes, you are quite right! I started to edit my answer to address your comment, but then I realised that it would take me all night...
$endgroup$
– TonyK
Jan 15 at 19:36
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
$begingroup$
I think it should be mentioned that constants $A$ and $B$ are not independent: $A^2+B^2=c$. Probably it was even better to represent the solution as $sqrt{c}cos(x+phi)$.
$endgroup$
– user
Jan 15 at 21:25
add a comment |
0
$begingroup$
Hint.
$$
f'(x) = pmsqrt{c-f(x)^2}
$$
then
$$
dx =pmfrac{df}{sqrt{c-f^2}}
$$
giving
$$
x + C_0 = pmarctanleft(frac{f}{sqrt{ c-f^2}}right)
$$
etc.
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
$endgroup$
add a comment |
0
$begingroup$
Hint.
$$
f'(x) = pmsqrt{c-f(x)^2}
$$
then
$$
dx =pmfrac{df}{sqrt{c-f^2}}
$$
giving
$$
x + C_0 = pmarctanleft(frac{f}{sqrt{ c-f^2}}right)
$$
etc.
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
$endgroup$
add a comment |
0
0
0
$begingroup$
Hint.
$$
f'(x) = pmsqrt{c-f(x)^2}
$$
then
$$
dx =pmfrac{df}{sqrt{c-f^2}}
$$
giving
$$
x + C_0 = pmarctanleft(frac{f}{sqrt{ c-f^2}}right)
$$
etc.
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
$endgroup$
Hint.
$$
f'(x) = pmsqrt{c-f(x)^2}
$$
then
$$
dx =pmfrac{df}{sqrt{c-f^2}}
$$
giving
$$
x + C_0 = pmarctanleft(frac{f}{sqrt{ c-f^2}}right)
$$
etc.
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
edited Jan 15 at 19:39
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
answered Jan 15 at 19:28
CesareoCesareo
8,9073516
8,9073516
add a comment |
add a comment |
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$begingroup$
Solve the separable differential equation $(y')^2=c-y^2$.
$endgroup$
– Qurultay
Jan 15 at 19:17