Finding function given ratio of derivatives w.r. to different variablesHow to find a function that meets a given condition?Numerical integration of functions sampled with constant and non-constant ratesCavalieri's Method of IndivisiblesFinding the frequencies of vibration of a drum; PDEHow can the same equation have two different partial derivatives with respect to the same variable?Numerical methods for calculating derivative of gamma functionroot finding for a tangent-like functionThe maximum value of the two variables function given some constraints.calculating rational approximations of arbitrary radical expressionsReconstruct a function $f: mathbbR^3 to mathbbR$ from two identities on its partial derivatives
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Finding function given ratio of derivatives w.r. to different variables
How to find a function that meets a given condition?Numerical integration of functions sampled with constant and non-constant ratesCavalieri's Method of IndivisiblesFinding the frequencies of vibration of a drum; PDEHow can the same equation have two different partial derivatives with respect to the same variable?Numerical methods for calculating derivative of gamma functionroot finding for a tangent-like functionThe maximum value of the two variables function given some constraints.calculating rational approximations of arbitrary radical expressionsReconstruct a function $f: mathbbR^3 to mathbbR$ from two identities on its partial derivatives
$begingroup$
Is there any way to find a function $f(a,b)$ up to some factors and a constant, given the ratio $$fracpartial f / partial apartial f/ partial b$$? If not, is there any other useful information about $f$ one can get from this ratio or a good numerical method to approximate $f$?
Thanks for helping me out.
calculus integration numerical-methods partial-derivative
asked 2 days ago
math_dealermath_dealer
103
$endgroup$
add a comment |
$begingroup$
Is there any way to find a function $f(a,b)$ up to some factors and a constant, given the ratio $$fracpartial f / partial apartial f/ partial b$$? If not, is there any other useful information about $f$ one can get from this ratio or a good numerical method to approximate $f$?
Thanks for helping me out.
calculus integration numerical-methods partial-derivative
asked 2 days ago
math_dealermath_dealer
103
$endgroup$
$begingroup$
EDIT: The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b) = a^2b$?
$endgroup$
– math_dealer
2 days ago
add a comment |
$begingroup$
Is there any way to find a function $f(a,b)$ up to some factors and a constant, given the ratio $$fracpartial f / partial apartial f/ partial b$$? If not, is there any other useful information about $f$ one can get from this ratio or a good numerical method to approximate $f$?
Thanks for helping me out.
calculus integration numerical-methods partial-derivative
asked 2 days ago
math_dealermath_dealer
103
$endgroup$
Is there any way to find a function $f(a,b)$ up to some factors and a constant, given the ratio $$fracpartial f / partial apartial f/ partial b$$? If not, is there any other useful information about $f$ one can get from this ratio or a good numerical method to approximate $f$?
Thanks for helping me out.
calculus integration numerical-methods partial-derivative
calculus integration numerical-methods partial-derivative
asked 2 days ago
math_dealermath_dealer
103
asked 2 days ago
math_dealermath_dealer
103
asked 2 days ago
math_dealermath_dealer
103
asked 2 days ago
math_dealermath_dealer
103
asked 2 days ago
math_dealermath_dealer
103
103
$begingroup$
EDIT: The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b) = a^2b$?
$endgroup$
– math_dealer
2 days ago
add a comment |
$begingroup$
EDIT: The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b) = a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
EDIT: The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b) = a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
EDIT: The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b) = a^2b$?
$endgroup$
– math_dealer
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose we know
$$
fracf_af_b=r(a,b).
$$
Then at a point $(a,b)$, the directional derivative of $f$ in the direction $[1,-r(a,b)]$ is $0$. This implies that $f$ is constant along the solution curves to the ODE
$$
fracdbda=-r(a,b).
$$
In your example
$$
fracf_af_b=frac2ba,
$$
the ODE we get is
$$
fracdbda=-frac2ba,
$$
which is separable. The solution curves are of the form $a^2b=C$, where $C$ is a constant. If $g$ is any function of one variable, then $g(a^2b)$ will be constant along the curves $a^2b=C$, so will give a solution to your original equation.
answered 2 days ago
Julian RosenJulian Rosen
12k12349
$endgroup$
add a comment |
$begingroup$
Not sure this is what you were looking for but $f(a,b)=lambda a+b$ has $$fracfracpartial fpartial afracpartial fpartial b=frac lambda 1=lambda$$
answered 2 days ago
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
1
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose we know
$$
fracf_af_b=r(a,b).
$$
Then at a point $(a,b)$, the directional derivative of $f$ in the direction $[1,-r(a,b)]$ is $0$. This implies that $f$ is constant along the solution curves to the ODE
$$
fracdbda=-r(a,b).
$$
In your example
$$
fracf_af_b=frac2ba,
$$
the ODE we get is
$$
fracdbda=-frac2ba,
$$
which is separable. The solution curves are of the form $a^2b=C$, where $C$ is a constant. If $g$ is any function of one variable, then $g(a^2b)$ will be constant along the curves $a^2b=C$, so will give a solution to your original equation.
answered 2 days ago
Julian RosenJulian Rosen
12k12349
$endgroup$
add a comment |
$begingroup$
Suppose we know
$$
fracf_af_b=r(a,b).
$$
Then at a point $(a,b)$, the directional derivative of $f$ in the direction $[1,-r(a,b)]$ is $0$. This implies that $f$ is constant along the solution curves to the ODE
$$
fracdbda=-r(a,b).
$$
In your example
$$
fracf_af_b=frac2ba,
$$
the ODE we get is
$$
fracdbda=-frac2ba,
$$
which is separable. The solution curves are of the form $a^2b=C$, where $C$ is a constant. If $g$ is any function of one variable, then $g(a^2b)$ will be constant along the curves $a^2b=C$, so will give a solution to your original equation.
answered 2 days ago
Julian RosenJulian Rosen
12k12349
$endgroup$
add a comment |
$begingroup$
Suppose we know
$$
fracf_af_b=r(a,b).
$$
Then at a point $(a,b)$, the directional derivative of $f$ in the direction $[1,-r(a,b)]$ is $0$. This implies that $f$ is constant along the solution curves to the ODE
$$
fracdbda=-r(a,b).
$$
In your example
$$
fracf_af_b=frac2ba,
$$
the ODE we get is
$$
fracdbda=-frac2ba,
$$
which is separable. The solution curves are of the form $a^2b=C$, where $C$ is a constant. If $g$ is any function of one variable, then $g(a^2b)$ will be constant along the curves $a^2b=C$, so will give a solution to your original equation.
answered 2 days ago
Julian RosenJulian Rosen
12k12349
$endgroup$
Suppose we know
$$
fracf_af_b=r(a,b).
$$
Then at a point $(a,b)$, the directional derivative of $f$ in the direction $[1,-r(a,b)]$ is $0$. This implies that $f$ is constant along the solution curves to the ODE
$$
fracdbda=-r(a,b).
$$
In your example
$$
fracf_af_b=frac2ba,
$$
the ODE we get is
$$
fracdbda=-frac2ba,
$$
which is separable. The solution curves are of the form $a^2b=C$, where $C$ is a constant. If $g$ is any function of one variable, then $g(a^2b)$ will be constant along the curves $a^2b=C$, so will give a solution to your original equation.
answered 2 days ago
Julian RosenJulian Rosen
12k12349
answered 2 days ago
Julian RosenJulian Rosen
12k12349
answered 2 days ago
Julian RosenJulian Rosen
12k12349
answered 2 days ago
Julian RosenJulian Rosen
12k12349
12k12349
add a comment |
add a comment |
$begingroup$
Not sure this is what you were looking for but $f(a,b)=lambda a+b$ has $$fracfracpartial fpartial afracpartial fpartial b=frac lambda 1=lambda$$
answered 2 days ago
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
1
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
add a comment |
$begingroup$
Not sure this is what you were looking for but $f(a,b)=lambda a+b$ has $$fracfracpartial fpartial afracpartial fpartial b=frac lambda 1=lambda$$
answered 2 days ago
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
1
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
add a comment |
$begingroup$
Not sure this is what you were looking for but $f(a,b)=lambda a+b$ has $$fracfracpartial fpartial afracpartial fpartial b=frac lambda 1=lambda$$
answered 2 days ago
AndreiAndrei
13.1k21230
$endgroup$
Not sure this is what you were looking for but $f(a,b)=lambda a+b$ has $$fracfracpartial fpartial afracpartial fpartial b=frac lambda 1=lambda$$
answered 2 days ago
AndreiAndrei
13.1k21230
answered 2 days ago
AndreiAndrei
13.1k21230
answered 2 days ago
AndreiAndrei
13.1k21230
answered 2 days ago
AndreiAndrei
13.1k21230
13.1k21230
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
1
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
add a comment |
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
1
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b)=a^2b$?
$endgroup$
– math_dealer
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
So the ratio is not a constant? Can you separate it into some function of $a$ multiplied with a function of $b$?
$endgroup$
– Andrei
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
$begingroup$
Yes sorry, the ratio is a function of a and b itself. No, one has only the analytic expression, but it could contain something like $sin(ab)$ etc.
$endgroup$
– math_dealer
2 days ago
1
1
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
$begingroup$
I'm not sure it's solvable in general case. Maybe some separation of variables is possible.
$endgroup$
– Andrei
2 days ago
add a comment |
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$begingroup$
EDIT: The idea is to find THE specific function $f$ in the situation where one only knows this ratio. Say the ratio is $frac2ba$, how do i get $f(a,b) = a^2b$?
$endgroup$
– math_dealer
2 days ago