Solving a nonpolynomial equation
$begingroup$
How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)
$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$
I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.
algebra-precalculus numerical-methods
edited Jan 10 at 5:44
Dylan
14.2k31127
asked Jan 9 at 19:35
KenKen
43028
$endgroup$
add a comment |
$begingroup$
How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)
$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$
I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.
algebra-precalculus numerical-methods
edited Jan 10 at 5:44
Dylan
14.2k31127
asked Jan 9 at 19:35
KenKen
43028
$endgroup$
2
$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42
$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04
add a comment |
$begingroup$
How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)
$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$
I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.
algebra-precalculus numerical-methods
edited Jan 10 at 5:44
Dylan
14.2k31127
asked Jan 9 at 19:35
KenKen
43028
$endgroup$
How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)
$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$
I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.
algebra-precalculus numerical-methods
algebra-precalculus numerical-methods
edited Jan 10 at 5:44
Dylan
14.2k31127
asked Jan 9 at 19:35
KenKen
43028
edited Jan 10 at 5:44
Dylan
14.2k31127
asked Jan 9 at 19:35
KenKen
43028
edited Jan 10 at 5:44
Dylan
14.2k31127
edited Jan 10 at 5:44
Dylan
14.2k31127
edited Jan 10 at 5:44
Dylan
14.2k31127
14.2k31127
asked Jan 9 at 19:35
KenKen
43028
asked Jan 9 at 19:35
KenKen
43028
asked Jan 9 at 19:35
KenKen
43028
43028
2
$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42
$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04
add a comment |
2
$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42
$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04
2
2
$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42
$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42
$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04
$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Computer algebra gives the solution:
$$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$
which suggests that you'll not easily find this solution by pen and paper.
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
add a comment |
$begingroup$
Hint: Expanding all we obtain
$$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
this is a cubic equation in $x$
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
$endgroup$
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
1
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Computer algebra gives the solution:
$$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$
which suggests that you'll not easily find this solution by pen and paper.
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
add a comment |
$begingroup$
Computer algebra gives the solution:
$$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$
which suggests that you'll not easily find this solution by pen and paper.
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
add a comment |
$begingroup$
Computer algebra gives the solution:
$$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$
which suggests that you'll not easily find this solution by pen and paper.
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
Computer algebra gives the solution:
$$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$
which suggests that you'll not easily find this solution by pen and paper.
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
answered Jan 9 at 19:43
David G. StorkDavid G. Stork
11.9k41735
11.9k41735
add a comment |
add a comment |
$begingroup$
Hint: Expanding all we obtain
$$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
this is a cubic equation in $x$
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
$endgroup$
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
1
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
add a comment |
$begingroup$
Hint: Expanding all we obtain
$$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
this is a cubic equation in $x$
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
$endgroup$
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
1
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
add a comment |
$begingroup$
Hint: Expanding all we obtain
$$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
this is a cubic equation in $x$
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
$endgroup$
Hint: Expanding all we obtain
$$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
this is a cubic equation in $x$
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
answered Jan 9 at 19:43
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.5k42867
78.5k42867
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
1
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
add a comment |
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
1
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
$endgroup$
– Ken
Jan 9 at 19:46
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
$begingroup$
Yes you can, and good luck!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 19:48
1
1
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
This is not an answer. It would better be given as a comment on the question.
$endgroup$
– robjohn♦
Jan 9 at 19:58
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
$begingroup$
But the Computer Algebra solution is an answer? Senseless your coment!
$endgroup$
– Dr. Sonnhard Graubner
Jan 9 at 20:02
add a comment |
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2
$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42
$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04