Find the form of functional
$begingroup$
I know the form of f(x) and that of g(x) and I would like to find an expression for the function H such that H[g(x)] = f(x).
f is a polynomial and g is more complex and involves some exp and cos.
Is there any procedure to find H?
More details:
$f(x) = Ax^6+Bx^{12}$
$g(x) = e^xleft(cos{x}+1right)$
I know (numerically) g(x) and f(x) and, I would like to find a function such that H[g(x)] = f(x) without having to evaluate it from x.
Thank you,
functional-equations
asked Jan 2 at 17:55
user3783569user3783569
11
$endgroup$
add a comment |
$begingroup$
I know the form of f(x) and that of g(x) and I would like to find an expression for the function H such that H[g(x)] = f(x).
f is a polynomial and g is more complex and involves some exp and cos.
Is there any procedure to find H?
More details:
$f(x) = Ax^6+Bx^{12}$
$g(x) = e^xleft(cos{x}+1right)$
I know (numerically) g(x) and f(x) and, I would like to find a function such that H[g(x)] = f(x) without having to evaluate it from x.
Thank you,
functional-equations
asked Jan 2 at 17:55
user3783569user3783569
11
$endgroup$
$begingroup$
There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$.
$endgroup$
– Ben W
Jan 2 at 17:57
1
$begingroup$
For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct.
$endgroup$
– lulu
Jan 2 at 18:09
add a comment |
$begingroup$
I know the form of f(x) and that of g(x) and I would like to find an expression for the function H such that H[g(x)] = f(x).
f is a polynomial and g is more complex and involves some exp and cos.
Is there any procedure to find H?
More details:
$f(x) = Ax^6+Bx^{12}$
$g(x) = e^xleft(cos{x}+1right)$
I know (numerically) g(x) and f(x) and, I would like to find a function such that H[g(x)] = f(x) without having to evaluate it from x.
Thank you,
functional-equations
asked Jan 2 at 17:55
user3783569user3783569
11
$endgroup$
I know the form of f(x) and that of g(x) and I would like to find an expression for the function H such that H[g(x)] = f(x).
f is a polynomial and g is more complex and involves some exp and cos.
Is there any procedure to find H?
More details:
$f(x) = Ax^6+Bx^{12}$
$g(x) = e^xleft(cos{x}+1right)$
I know (numerically) g(x) and f(x) and, I would like to find a function such that H[g(x)] = f(x) without having to evaluate it from x.
Thank you,
functional-equations
functional-equations
asked Jan 2 at 17:55
user3783569user3783569
11
asked Jan 2 at 17:55
user3783569user3783569
11
edited Jan 2 at 21:45
user3783569
asked Jan 2 at 17:55
user3783569user3783569
11
asked Jan 2 at 17:55
user3783569user3783569
11
asked Jan 2 at 17:55
user3783569user3783569
11
11
$begingroup$
There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$.
$endgroup$
– Ben W
Jan 2 at 17:57
1
$begingroup$
For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct.
$endgroup$
– lulu
Jan 2 at 18:09
add a comment |
$begingroup$
There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$.
$endgroup$
– Ben W
Jan 2 at 17:57
1
$begingroup$
For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct.
$endgroup$
– lulu
Jan 2 at 18:09
$begingroup$
There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$.
$endgroup$
– Ben W
Jan 2 at 17:57
$begingroup$
There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$.
$endgroup$
– Ben W
Jan 2 at 17:57
1
1
$begingroup$
For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct.
$endgroup$
– lulu
Jan 2 at 18:09
$begingroup$
For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct.
$endgroup$
– lulu
Jan 2 at 18:09
add a comment |
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$begingroup$
There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$.
$endgroup$
– Ben W
Jan 2 at 17:57
1
$begingroup$
For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct.
$endgroup$
– lulu
Jan 2 at 18:09