Solve $x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0$ [closed]
$begingroup$
How would I find the roots of this function?
begin{equation}
x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?
logarithms roots
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closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How would I find the roots of this function?
begin{equation}
x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?
logarithms roots
$endgroup$
closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How would I find the roots of this function?
begin{equation}
x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?
logarithms roots
$endgroup$
How would I find the roots of this function?
begin{equation}
x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?
logarithms roots
logarithms roots
asked Jan 18 at 17:08
ViktorViktor
16310
16310
closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.
You can have approximations using series expansions.
Around $x=0$,
$$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
$$x=frac{3}4$$
Around $x=3$, the expansion would be
$$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$ and the the solution is close to
$$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
log (3)right)}approx 3.00039$$
Edit
With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
1 & frac{3}{4} & 0.750000 \
2 & frac{25}{33} & 0.757576 \
3 & frac{1247}{1643} & 0.758977 \
4 & frac{5179}{6822} & 0.759162 \
5 & frac{11947}{15737} & 0.759166 \
6 & frac{1339361}{1764260} & 0.759163
end{array}
right)$$
$endgroup$
add a comment |
$begingroup$
Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$
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1
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
1
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
add a comment |
$begingroup$
I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.
You can have approximations using series expansions.
Around $x=0$,
$$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
$$x=frac{3}4$$
Around $x=3$, the expansion would be
$$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$ and the the solution is close to
$$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
log (3)right)}approx 3.00039$$
Edit
With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
1 & frac{3}{4} & 0.750000 \
2 & frac{25}{33} & 0.757576 \
3 & frac{1247}{1643} & 0.758977 \
4 & frac{5179}{6822} & 0.759162 \
5 & frac{11947}{15737} & 0.759166 \
6 & frac{1339361}{1764260} & 0.759163
end{array}
right)$$
$endgroup$
add a comment |
$begingroup$
If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.
You can have approximations using series expansions.
Around $x=0$,
$$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
$$x=frac{3}4$$
Around $x=3$, the expansion would be
$$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$ and the the solution is close to
$$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
log (3)right)}approx 3.00039$$
Edit
With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
1 & frac{3}{4} & 0.750000 \
2 & frac{25}{33} & 0.757576 \
3 & frac{1247}{1643} & 0.758977 \
4 & frac{5179}{6822} & 0.759162 \
5 & frac{11947}{15737} & 0.759166 \
6 & frac{1339361}{1764260} & 0.759163
end{array}
right)$$
$endgroup$
add a comment |
$begingroup$
If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.
You can have approximations using series expansions.
Around $x=0$,
$$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
$$x=frac{3}4$$
Around $x=3$, the expansion would be
$$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$ and the the solution is close to
$$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
log (3)right)}approx 3.00039$$
Edit
With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
1 & frac{3}{4} & 0.750000 \
2 & frac{25}{33} & 0.757576 \
3 & frac{1247}{1643} & 0.758977 \
4 & frac{5179}{6822} & 0.759162 \
5 & frac{11947}{15737} & 0.759166 \
6 & frac{1339361}{1764260} & 0.759163
end{array}
right)$$
$endgroup$
If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.
You can have approximations using series expansions.
Around $x=0$,
$$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
$$x=frac{3}4$$
Around $x=3$, the expansion would be
$$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$ and the the solution is close to
$$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
log (3)right)}approx 3.00039$$
Edit
With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
1 & frac{3}{4} & 0.750000 \
2 & frac{25}{33} & 0.757576 \
3 & frac{1247}{1643} & 0.758977 \
4 & frac{5179}{6822} & 0.759162 \
5 & frac{11947}{15737} & 0.759166 \
6 & frac{1339361}{1764260} & 0.759163
end{array}
right)$$
edited Jan 19 at 8:40
answered Jan 19 at 7:21
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
add a comment |
add a comment |
$begingroup$
Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$
$endgroup$
1
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
1
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
add a comment |
$begingroup$
Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$
$endgroup$
1
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
1
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
add a comment |
$begingroup$
Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$
$endgroup$
Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$
answered Jan 18 at 17:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
79.2k42867
1
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
1
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
add a comment |
1
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
1
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
1
1
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
$begingroup$
I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
$endgroup$
– Viktor
Jan 18 at 17:16
1
1
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
$endgroup$
– Robert Israel
Jan 18 at 17:23
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
$begingroup$
I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
$endgroup$
– Viktor
Jan 18 at 17:56
add a comment |
$begingroup$
I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.
$endgroup$
add a comment |
$begingroup$
I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.
$endgroup$
add a comment |
$begingroup$
I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.
$endgroup$
I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.
answered Jan 18 at 19:28
Lorenz H MenkeLorenz H Menke
9611
9611
add a comment |
add a comment |