Solving this polynomial equation
So I'm trying to maximize a log function and I'm getting stuck in solving for $x$.
The original function is
$$g(x) = x^{-3} log (1+x^2) + x^{-1} log(1+x^2) - x^{-1}$$
Note that plotting this in R, I can see a clear maximum close to $e^1$ (though slightly to the right).
The derivative is equal to
$$g'(x) = log(1+x^2)(-3 x^{-4}-x^{-2})+frac{2}{1+x^2}(x^{-2}+1)+x^{-2}$$
But setting equal to zero, I'm having trouble separating the logarithm to get a closed form solution of $x_max$.
Edit: My original line below here was incorrect. Corrected now
This leads to
$$1+x^2 = exp left( frac{3x^2}{3+x^2} right) $$
calculus
asked Nov 26 at 14:14
Xiaomi
1,016115
|
show 3 more comments
So I'm trying to maximize a log function and I'm getting stuck in solving for $x$.
The original function is
$$g(x) = x^{-3} log (1+x^2) + x^{-1} log(1+x^2) - x^{-1}$$
Note that plotting this in R, I can see a clear maximum close to $e^1$ (though slightly to the right).
The derivative is equal to
$$g'(x) = log(1+x^2)(-3 x^{-4}-x^{-2})+frac{2}{1+x^2}(x^{-2}+1)+x^{-2}$$
But setting equal to zero, I'm having trouble separating the logarithm to get a closed form solution of $x_max$.
Edit: My original line below here was incorrect. Corrected now
This leads to
$$1+x^2 = exp left( frac{3x^2}{3+x^2} right) $$
calculus
asked Nov 26 at 14:14
Xiaomi
1,016115
You can not algebraically solve these kinds of equations. You might try a numerical approach.
– Cornman
Nov 26 at 14:16
1
Are you sure the derivative is correct? WA provides www4c.wolframalpha.com/Calculate/MSP/…
– Mohammad Zuhair Khan
Nov 26 at 14:19
Really? Because the question I'm working on seems to suggest you can solve for it
– Xiaomi
Nov 26 at 14:19
@raptor the derivative is correct. They're just written in different forms. Double checked by plotting both graphs
– Xiaomi
Nov 26 at 14:28
@Xiaomi very well then but IMO perhaps the rearrangement may help? Like I get $1+x^2=exp left( dfrac {3x^2}{x^2+3} right)$ from the new one whereas you got $1+x^2 = exp left( dfrac{2+x^2}{3+x^2} right)$. Can you double check your last steps because otherwise seems like $2+x^2=3x^2 implies 2x^2=2 implies x^2=1 implies x=pm 1.$ which should not be possible.
– Mohammad Zuhair Khan
Nov 26 at 14:40
|
show 3 more comments
So I'm trying to maximize a log function and I'm getting stuck in solving for $x$.
The original function is
$$g(x) = x^{-3} log (1+x^2) + x^{-1} log(1+x^2) - x^{-1}$$
Note that plotting this in R, I can see a clear maximum close to $e^1$ (though slightly to the right).
The derivative is equal to
$$g'(x) = log(1+x^2)(-3 x^{-4}-x^{-2})+frac{2}{1+x^2}(x^{-2}+1)+x^{-2}$$
But setting equal to zero, I'm having trouble separating the logarithm to get a closed form solution of $x_max$.
Edit: My original line below here was incorrect. Corrected now
This leads to
$$1+x^2 = exp left( frac{3x^2}{3+x^2} right) $$
calculus
asked Nov 26 at 14:14
Xiaomi
1,016115
So I'm trying to maximize a log function and I'm getting stuck in solving for $x$.
The original function is
$$g(x) = x^{-3} log (1+x^2) + x^{-1} log(1+x^2) - x^{-1}$$
Note that plotting this in R, I can see a clear maximum close to $e^1$ (though slightly to the right).
The derivative is equal to
$$g'(x) = log(1+x^2)(-3 x^{-4}-x^{-2})+frac{2}{1+x^2}(x^{-2}+1)+x^{-2}$$
But setting equal to zero, I'm having trouble separating the logarithm to get a closed form solution of $x_max$.
Edit: My original line below here was incorrect. Corrected now
This leads to
$$1+x^2 = exp left( frac{3x^2}{3+x^2} right) $$
calculus
calculus
asked Nov 26 at 14:14
Xiaomi
1,016115
asked Nov 26 at 14:14
Xiaomi
1,016115
edited Nov 26 at 14:44
asked Nov 26 at 14:14
Xiaomi
1,016115
asked Nov 26 at 14:14
Xiaomi
1,016115
asked Nov 26 at 14:14
Xiaomi
1,016115
1,016115
You can not algebraically solve these kinds of equations. You might try a numerical approach.
– Cornman
Nov 26 at 14:16
1
Are you sure the derivative is correct? WA provides www4c.wolframalpha.com/Calculate/MSP/…
– Mohammad Zuhair Khan
Nov 26 at 14:19
Really? Because the question I'm working on seems to suggest you can solve for it
– Xiaomi
Nov 26 at 14:19
@raptor the derivative is correct. They're just written in different forms. Double checked by plotting both graphs
– Xiaomi
Nov 26 at 14:28
@Xiaomi very well then but IMO perhaps the rearrangement may help? Like I get $1+x^2=exp left( dfrac {3x^2}{x^2+3} right)$ from the new one whereas you got $1+x^2 = exp left( dfrac{2+x^2}{3+x^2} right)$. Can you double check your last steps because otherwise seems like $2+x^2=3x^2 implies 2x^2=2 implies x^2=1 implies x=pm 1.$ which should not be possible.
– Mohammad Zuhair Khan
Nov 26 at 14:40
|
show 3 more comments
You can not algebraically solve these kinds of equations. You might try a numerical approach.
– Cornman
Nov 26 at 14:16
1
Are you sure the derivative is correct? WA provides www4c.wolframalpha.com/Calculate/MSP/…
– Mohammad Zuhair Khan
Nov 26 at 14:19
Really? Because the question I'm working on seems to suggest you can solve for it
– Xiaomi
Nov 26 at 14:19
@raptor the derivative is correct. They're just written in different forms. Double checked by plotting both graphs
– Xiaomi
Nov 26 at 14:28
@Xiaomi very well then but IMO perhaps the rearrangement may help? Like I get $1+x^2=exp left( dfrac {3x^2}{x^2+3} right)$ from the new one whereas you got $1+x^2 = exp left( dfrac{2+x^2}{3+x^2} right)$. Can you double check your last steps because otherwise seems like $2+x^2=3x^2 implies 2x^2=2 implies x^2=1 implies x=pm 1.$ which should not be possible.
– Mohammad Zuhair Khan
Nov 26 at 14:40
You can not algebraically solve these kinds of equations. You might try a numerical approach.
– Cornman
Nov 26 at 14:16
You can not algebraically solve these kinds of equations. You might try a numerical approach.
– Cornman
Nov 26 at 14:16
1
1
Are you sure the derivative is correct? WA provides www4c.wolframalpha.com/Calculate/MSP/…
– Mohammad Zuhair Khan
Nov 26 at 14:19
Are you sure the derivative is correct? WA provides www4c.wolframalpha.com/Calculate/MSP/…
– Mohammad Zuhair Khan
Nov 26 at 14:19
Really? Because the question I'm working on seems to suggest you can solve for it
– Xiaomi
Nov 26 at 14:19
Really? Because the question I'm working on seems to suggest you can solve for it
– Xiaomi
Nov 26 at 14:19
@raptor the derivative is correct. They're just written in different forms. Double checked by plotting both graphs
– Xiaomi
Nov 26 at 14:28
@raptor the derivative is correct. They're just written in different forms. Double checked by plotting both graphs
– Xiaomi
Nov 26 at 14:28
@Xiaomi very well then but IMO perhaps the rearrangement may help? Like I get $1+x^2=exp left( dfrac {3x^2}{x^2+3} right)$ from the new one whereas you got $1+x^2 = exp left( dfrac{2+x^2}{3+x^2} right)$. Can you double check your last steps because otherwise seems like $2+x^2=3x^2 implies 2x^2=2 implies x^2=1 implies x=pm 1.$ which should not be possible.
– Mohammad Zuhair Khan
Nov 26 at 14:40
@Xiaomi very well then but IMO perhaps the rearrangement may help? Like I get $1+x^2=exp left( dfrac {3x^2}{x^2+3} right)$ from the new one whereas you got $1+x^2 = exp left( dfrac{2+x^2}{3+x^2} right)$. Can you double check your last steps because otherwise seems like $2+x^2=3x^2 implies 2x^2=2 implies x^2=1 implies x=pm 1.$ which should not be possible.
– Mohammad Zuhair Khan
Nov 26 at 14:40
|
show 3 more comments
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You can not algebraically solve these kinds of equations. You might try a numerical approach.
– Cornman
Nov 26 at 14:16
1
Are you sure the derivative is correct? WA provides www4c.wolframalpha.com/Calculate/MSP/…
– Mohammad Zuhair Khan
Nov 26 at 14:19
Really? Because the question I'm working on seems to suggest you can solve for it
– Xiaomi
Nov 26 at 14:19
@raptor the derivative is correct. They're just written in different forms. Double checked by plotting both graphs
– Xiaomi
Nov 26 at 14:28
@Xiaomi very well then but IMO perhaps the rearrangement may help? Like I get $1+x^2=exp left( dfrac {3x^2}{x^2+3} right)$ from the new one whereas you got $1+x^2 = exp left( dfrac{2+x^2}{3+x^2} right)$. Can you double check your last steps because otherwise seems like $2+x^2=3x^2 implies 2x^2=2 implies x^2=1 implies x=pm 1.$ which should not be possible.
– Mohammad Zuhair Khan
Nov 26 at 14:40