Mixing
Nonlinear functional equation with tangent
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Please help me with this. I need to find a non-trivial function $g(x)$ which satisfy the following functional equation
$$tan(g(x))+g(x)+g(x)tan^2(g(x))=0$$
functional-equations
asked Feb 1 at 8:29
DarekDarek
334
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add a comment |
$begingroup$
Please help me with this. I need to find a non-trivial function $g(x)$ which satisfy the following functional equation
$$tan(g(x))+g(x)+g(x)tan^2(g(x))=0$$
functional-equations
asked Feb 1 at 8:29
DarekDarek
334
$endgroup$
$begingroup$
This is not a functional equation.
$endgroup$
– Ivan Neretin
Feb 1 at 11:24
$begingroup$
I am afraid you are wrong. This is a functional equation.
$endgroup$
– Darek
Feb 1 at 13:49
$begingroup$
OK, this is a functional equation in name only. Imagine a simple algebraic equation, like $x^3-3x^2-3x+1=0$. It is definitely not a functional equation, for there is no unknown function of any sort, is it? Now suppose that we rewrite this equation and substitute every single instance of $x$ with $f(t)$. Does this make it a functional equation? Technically yes, but...
$endgroup$
– Ivan Neretin
Feb 1 at 13:58
add a comment |
$begingroup$
Please help me with this. I need to find a non-trivial function $g(x)$ which satisfy the following functional equation
$$tan(g(x))+g(x)+g(x)tan^2(g(x))=0$$
functional-equations
asked Feb 1 at 8:29
DarekDarek
334
$endgroup$
Please help me with this. I need to find a non-trivial function $g(x)$ which satisfy the following functional equation
$$tan(g(x))+g(x)+g(x)tan^2(g(x))=0$$
functional-equations
functional-equations
asked Feb 1 at 8:29
DarekDarek
334
asked Feb 1 at 8:29
DarekDarek
334
edited Feb 1 at 8:34
Darek
asked Feb 1 at 8:29
DarekDarek
334
asked Feb 1 at 8:29
DarekDarek
334
asked Feb 1 at 8:29
DarekDarek
334
334
$begingroup$
This is not a functional equation.
$endgroup$
– Ivan Neretin
Feb 1 at 11:24
$begingroup$
I am afraid you are wrong. This is a functional equation.
$endgroup$
– Darek
Feb 1 at 13:49
$begingroup$
OK, this is a functional equation in name only. Imagine a simple algebraic equation, like $x^3-3x^2-3x+1=0$. It is definitely not a functional equation, for there is no unknown function of any sort, is it? Now suppose that we rewrite this equation and substitute every single instance of $x$ with $f(t)$. Does this make it a functional equation? Technically yes, but...
$endgroup$
– Ivan Neretin
Feb 1 at 13:58
add a comment |
$begingroup$
This is not a functional equation.
$endgroup$
– Ivan Neretin
Feb 1 at 11:24
$begingroup$
I am afraid you are wrong. This is a functional equation.
$endgroup$
– Darek
Feb 1 at 13:49
$begingroup$
OK, this is a functional equation in name only. Imagine a simple algebraic equation, like $x^3-3x^2-3x+1=0$. It is definitely not a functional equation, for there is no unknown function of any sort, is it? Now suppose that we rewrite this equation and substitute every single instance of $x$ with $f(t)$. Does this make it a functional equation? Technically yes, but...
$endgroup$
– Ivan Neretin
Feb 1 at 13:58
$begingroup$
This is not a functional equation.
$endgroup$
– Ivan Neretin
Feb 1 at 11:24
$begingroup$
This is not a functional equation.
$endgroup$
– Ivan Neretin
Feb 1 at 11:24
$begingroup$
I am afraid you are wrong. This is a functional equation.
$endgroup$
– Darek
Feb 1 at 13:49
$begingroup$
I am afraid you are wrong. This is a functional equation.
$endgroup$
– Darek
Feb 1 at 13:49
$begingroup$
OK, this is a functional equation in name only. Imagine a simple algebraic equation, like $x^3-3x^2-3x+1=0$. It is definitely not a functional equation, for there is no unknown function of any sort, is it? Now suppose that we rewrite this equation and substitute every single instance of $x$ with $f(t)$. Does this make it a functional equation? Technically yes, but...
$endgroup$
– Ivan Neretin
Feb 1 at 13:58
$begingroup$
OK, this is a functional equation in name only. Imagine a simple algebraic equation, like $x^3-3x^2-3x+1=0$. It is definitely not a functional equation, for there is no unknown function of any sort, is it? Now suppose that we rewrite this equation and substitute every single instance of $x$ with $f(t)$. Does this make it a functional equation? Technically yes, but...
$endgroup$
– Ivan Neretin
Feb 1 at 13:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no non-trivial real (I assume you meant real as you didn't specify anything else) solution. We can rewrite your equation (using the usual trigonometric identities) as
$$g(x) = - frac{tan(g(x))}{1+tan^2(g(x))} = -frac{1}{2} sin(2 g(x))$$
If there is some $x_0 in mathbb R$ such that $a := g(x_0) neq 0$ then we have
$$a = - frac{1}{2} sin(2 a)$$
which is equivalent to $-2a = sin(2a)$ which has the only real solution $a=0$. Therefore $g(x_0)=0$ and therefore $g(x) = 0$ for all $x$.
answered Feb 1 at 8:47
flawrflawr
11.7k32546
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add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
There is no non-trivial real (I assume you meant real as you didn't specify anything else) solution. We can rewrite your equation (using the usual trigonometric identities) as
$$g(x) = - frac{tan(g(x))}{1+tan^2(g(x))} = -frac{1}{2} sin(2 g(x))$$
If there is some $x_0 in mathbb R$ such that $a := g(x_0) neq 0$ then we have
$$a = - frac{1}{2} sin(2 a)$$
which is equivalent to $-2a = sin(2a)$ which has the only real solution $a=0$. Therefore $g(x_0)=0$ and therefore $g(x) = 0$ for all $x$.
answered Feb 1 at 8:47
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
There is no non-trivial real (I assume you meant real as you didn't specify anything else) solution. We can rewrite your equation (using the usual trigonometric identities) as
$$g(x) = - frac{tan(g(x))}{1+tan^2(g(x))} = -frac{1}{2} sin(2 g(x))$$
If there is some $x_0 in mathbb R$ such that $a := g(x_0) neq 0$ then we have
$$a = - frac{1}{2} sin(2 a)$$
which is equivalent to $-2a = sin(2a)$ which has the only real solution $a=0$. Therefore $g(x_0)=0$ and therefore $g(x) = 0$ for all $x$.
answered Feb 1 at 8:47
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
There is no non-trivial real (I assume you meant real as you didn't specify anything else) solution. We can rewrite your equation (using the usual trigonometric identities) as
$$g(x) = - frac{tan(g(x))}{1+tan^2(g(x))} = -frac{1}{2} sin(2 g(x))$$
If there is some $x_0 in mathbb R$ such that $a := g(x_0) neq 0$ then we have
$$a = - frac{1}{2} sin(2 a)$$
which is equivalent to $-2a = sin(2a)$ which has the only real solution $a=0$. Therefore $g(x_0)=0$ and therefore $g(x) = 0$ for all $x$.
answered Feb 1 at 8:47
flawrflawr
11.7k32546
$endgroup$
There is no non-trivial real (I assume you meant real as you didn't specify anything else) solution. We can rewrite your equation (using the usual trigonometric identities) as
$$g(x) = - frac{tan(g(x))}{1+tan^2(g(x))} = -frac{1}{2} sin(2 g(x))$$
If there is some $x_0 in mathbb R$ such that $a := g(x_0) neq 0$ then we have
$$a = - frac{1}{2} sin(2 a)$$
which is equivalent to $-2a = sin(2a)$ which has the only real solution $a=0$. Therefore $g(x_0)=0$ and therefore $g(x) = 0$ for all $x$.
answered Feb 1 at 8:47
flawrflawr
11.7k32546
answered Feb 1 at 8:47
flawrflawr
11.7k32546
answered Feb 1 at 8:47
flawrflawr
11.7k32546
answered Feb 1 at 8:47
flawrflawr
11.7k32546
11.7k32546
add a comment |
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$begingroup$
This is not a functional equation.
$endgroup$
– Ivan Neretin
Feb 1 at 11:24
$begingroup$
I am afraid you are wrong. This is a functional equation.
$endgroup$
– Darek
Feb 1 at 13:49
$begingroup$
OK, this is a functional equation in name only. Imagine a simple algebraic equation, like $x^3-3x^2-3x+1=0$. It is definitely not a functional equation, for there is no unknown function of any sort, is it? Now suppose that we rewrite this equation and substitute every single instance of $x$ with $f(t)$. Does this make it a functional equation? Technically yes, but...
$endgroup$
– Ivan Neretin
Feb 1 at 13:58