How to solve equation with exponents multiplied with a polynomial?
up vote
4
down vote
favorite
I don't know how to tackle equation with exponents multiplied with a polynomial in general.
For example:
$100 =2^x(1+x)$
Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.
polynomials exponential-function
edited Nov 23 at 18:08
Anurag A
25.5k12249
asked Nov 23 at 18:03
Chuanyuan Liu
303
add a comment |
up vote
4
down vote
favorite
I don't know how to tackle equation with exponents multiplied with a polynomial in general.
For example:
$100 =2^x(1+x)$
Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.
polynomials exponential-function
edited Nov 23 at 18:08
Anurag A
25.5k12249
asked Nov 23 at 18:03
Chuanyuan Liu
303
2
You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07
What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14
@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I don't know how to tackle equation with exponents multiplied with a polynomial in general.
For example:
$100 =2^x(1+x)$
Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.
polynomials exponential-function
edited Nov 23 at 18:08
Anurag A
25.5k12249
asked Nov 23 at 18:03
Chuanyuan Liu
303
I don't know how to tackle equation with exponents multiplied with a polynomial in general.
For example:
$100 =2^x(1+x)$
Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.
polynomials exponential-function
polynomials exponential-function
edited Nov 23 at 18:08
Anurag A
25.5k12249
asked Nov 23 at 18:03
Chuanyuan Liu
303
edited Nov 23 at 18:08
Anurag A
25.5k12249
asked Nov 23 at 18:03
Chuanyuan Liu
303
edited Nov 23 at 18:08
Anurag A
25.5k12249
edited Nov 23 at 18:08
Anurag A
25.5k12249
edited Nov 23 at 18:08
Anurag A
25.5k12249
25.5k12249
asked Nov 23 at 18:03
Chuanyuan Liu
303
asked Nov 23 at 18:03
Chuanyuan Liu
303
asked Nov 23 at 18:03
Chuanyuan Liu
303
303
2
You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07
What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14
@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26
add a comment |
2
You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07
What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14
@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26
2
2
You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07
You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07
What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14
What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14
@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26
@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
You can solve this with the Lambert W function: Write you equation as
$$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
with $y=1+x$ you get
$$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
$$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
y = frac{W(200ln 2)}{ln2}$$
So finally
$$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$
As an alternative you can use a simple iteration scheme from the equation
$$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$
answered Nov 23 at 21:29
gammatester
16.6k21632
add a comment |
up vote
0
down vote
As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).
In this particular case, you can reduce to the form
$$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.
answered Nov 23 at 21:58
Yves Daoust
123k668219
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You can solve this with the Lambert W function: Write you equation as
$$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
with $y=1+x$ you get
$$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
$$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
y = frac{W(200ln 2)}{ln2}$$
So finally
$$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$
As an alternative you can use a simple iteration scheme from the equation
$$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$
answered Nov 23 at 21:29
gammatester
16.6k21632
add a comment |
up vote
3
down vote
accepted
You can solve this with the Lambert W function: Write you equation as
$$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
with $y=1+x$ you get
$$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
$$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
y = frac{W(200ln 2)}{ln2}$$
So finally
$$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$
As an alternative you can use a simple iteration scheme from the equation
$$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$
answered Nov 23 at 21:29
gammatester
16.6k21632
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You can solve this with the Lambert W function: Write you equation as
$$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
with $y=1+x$ you get
$$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
$$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
y = frac{W(200ln 2)}{ln2}$$
So finally
$$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$
As an alternative you can use a simple iteration scheme from the equation
$$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$
answered Nov 23 at 21:29
gammatester
16.6k21632
You can solve this with the Lambert W function: Write you equation as
$$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
with $y=1+x$ you get
$$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
$$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
y = frac{W(200ln 2)}{ln2}$$
So finally
$$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$
As an alternative you can use a simple iteration scheme from the equation
$$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$
answered Nov 23 at 21:29
gammatester
16.6k21632
edited Nov 23 at 21:44
answered Nov 23 at 21:29
gammatester
16.6k21632
answered Nov 23 at 21:29
gammatester
16.6k21632
answered Nov 23 at 21:29
gammatester
16.6k21632
16.6k21632
add a comment |
add a comment |
up vote
0
down vote
As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).
In this particular case, you can reduce to the form
$$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.
answered Nov 23 at 21:58
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).
In this particular case, you can reduce to the form
$$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.
answered Nov 23 at 21:58
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
up vote
0
down vote
As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).
In this particular case, you can reduce to the form
$$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.
answered Nov 23 at 21:58
Yves Daoust
123k668219
As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).
In this particular case, you can reduce to the form
$$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.
answered Nov 23 at 21:58
Yves Daoust
123k668219
answered Nov 23 at 21:58
Yves Daoust
123k668219
answered Nov 23 at 21:58
Yves Daoust
123k668219
answered Nov 23 at 21:58
Yves Daoust
123k668219
123k668219
add a comment |
add a comment |
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2
You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07
What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14
@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26