How to solve equations where the power of $x$ is a function of $x$?
I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?
Any help will be greatly appreciated.
polynomials exponential-function lambert-w
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
asked Dec 2 '18 at 5:29
user3430220
111
add a comment |
I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?
Any help will be greatly appreciated.
polynomials exponential-function lambert-w
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
asked Dec 2 '18 at 5:29
user3430220
111
add a comment |
I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?
Any help will be greatly appreciated.
polynomials exponential-function lambert-w
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
asked Dec 2 '18 at 5:29
user3430220
111
I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?
Any help will be greatly appreciated.
polynomials exponential-function lambert-w
polynomials exponential-function lambert-w
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
asked Dec 2 '18 at 5:29
user3430220
111
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
asked Dec 2 '18 at 5:29
user3430220
111
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
edited Dec 2 '18 at 5:32
Rócherz
2,7762721
2,7762721
asked Dec 2 '18 at 5:29
user3430220
111
asked Dec 2 '18 at 5:29
user3430220
111
asked Dec 2 '18 at 5:29
user3430220
111
111
add a comment |
add a comment |
1 Answer
1
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I do not think that explicit solutions could be obtained for the zero's of
$$f(x)=x^{p,x-c} - a$$
and you will, more than likely, use numerical methods.
In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
$$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
$x^x=a^{frac1p}$.
For $cneq 0$, consider instead the function
$$g(x)=(px-c)log(x)-log(a)$$
$$g'(x)=-frac{c}{x}+p log (x)+p$$
$$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
$$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
$$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
c}{p}right)right)$$ Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
I do not think that explicit solutions could be obtained for the zero's of
$$f(x)=x^{p,x-c} - a$$
and you will, more than likely, use numerical methods.
In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
$$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
$x^x=a^{frac1p}$.
For $cneq 0$, consider instead the function
$$g(x)=(px-c)log(x)-log(a)$$
$$g'(x)=-frac{c}{x}+p log (x)+p$$
$$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
$$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
$$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
c}{p}right)right)$$ Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
add a comment |
I do not think that explicit solutions could be obtained for the zero's of
$$f(x)=x^{p,x-c} - a$$
and you will, more than likely, use numerical methods.
In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
$$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
$x^x=a^{frac1p}$.
For $cneq 0$, consider instead the function
$$g(x)=(px-c)log(x)-log(a)$$
$$g'(x)=-frac{c}{x}+p log (x)+p$$
$$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
$$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
$$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
c}{p}right)right)$$ Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
add a comment |
I do not think that explicit solutions could be obtained for the zero's of
$$f(x)=x^{p,x-c} - a$$
and you will, more than likely, use numerical methods.
In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
$$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
$x^x=a^{frac1p}$.
For $cneq 0$, consider instead the function
$$g(x)=(px-c)log(x)-log(a)$$
$$g'(x)=-frac{c}{x}+p log (x)+p$$
$$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
$$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
$$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
c}{p}right)right)$$ Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
I do not think that explicit solutions could be obtained for the zero's of
$$f(x)=x^{p,x-c} - a$$
and you will, more than likely, use numerical methods.
In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
$$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
$x^x=a^{frac1p}$.
For $cneq 0$, consider instead the function
$$g(x)=(px-c)log(x)-log(a)$$
$$g'(x)=-frac{c}{x}+p log (x)+p$$
$$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
$$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
$$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
c}{p}right)right)$$ Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
answered Dec 2 '18 at 7:29
Claude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
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