Mixing
Help needed with an equation
$begingroup$
Can someone help me solve this
$e^{-x}=ln(1/x)$
I think these two functions are inverses of each other and since they both meet the line $y=x$ they intersect
But I'm not able to simplify the equality above to find their point of intersection
Help me, please.
linear-algebra functions systems-of-equations
edited Jan 27 at 16:59
Cesareo
9,4713517
asked Jan 27 at 4:31
LeeLee
526
$endgroup$
add a comment |
$begingroup$
Can someone help me solve this
$e^{-x}=ln(1/x)$
I think these two functions are inverses of each other and since they both meet the line $y=x$ they intersect
But I'm not able to simplify the equality above to find their point of intersection
Help me, please.
linear-algebra functions systems-of-equations
edited Jan 27 at 16:59
Cesareo
9,4713517
asked Jan 27 at 4:31
LeeLee
526
$endgroup$
1
$begingroup$
Try using Lambert W function.
$endgroup$
– Sujit Bhattacharyya
Jan 27 at 4:37
$begingroup$
WolframAlpha solves it as $x = 0.56714329040978dots$. The Inverse Symbolic Calculator gives this as $x = W(1)$ with the Lambert W function and $W(1)$ as the Omega constant.
$endgroup$
– Tito Piezas III
Jan 27 at 5:52
add a comment |
$begingroup$
Can someone help me solve this
$e^{-x}=ln(1/x)$
I think these two functions are inverses of each other and since they both meet the line $y=x$ they intersect
But I'm not able to simplify the equality above to find their point of intersection
Help me, please.
linear-algebra functions systems-of-equations
edited Jan 27 at 16:59
Cesareo
9,4713517
asked Jan 27 at 4:31
LeeLee
526
$endgroup$
Can someone help me solve this
$e^{-x}=ln(1/x)$
I think these two functions are inverses of each other and since they both meet the line $y=x$ they intersect
But I'm not able to simplify the equality above to find their point of intersection
Help me, please.
linear-algebra functions systems-of-equations
linear-algebra functions systems-of-equations
edited Jan 27 at 16:59
Cesareo
9,4713517
asked Jan 27 at 4:31
LeeLee
526
edited Jan 27 at 16:59
Cesareo
9,4713517
asked Jan 27 at 4:31
LeeLee
526
edited Jan 27 at 16:59
Cesareo
9,4713517
edited Jan 27 at 16:59
Cesareo
9,4713517
edited Jan 27 at 16:59
Cesareo
9,4713517
9,4713517
asked Jan 27 at 4:31
LeeLee
526
asked Jan 27 at 4:31
LeeLee
526
asked Jan 27 at 4:31
LeeLee
526
526
1
$begingroup$
Try using Lambert W function.
$endgroup$
– Sujit Bhattacharyya
Jan 27 at 4:37
$begingroup$
WolframAlpha solves it as $x = 0.56714329040978dots$. The Inverse Symbolic Calculator gives this as $x = W(1)$ with the Lambert W function and $W(1)$ as the Omega constant.
$endgroup$
– Tito Piezas III
Jan 27 at 5:52
add a comment |
1
$begingroup$
Try using Lambert W function.
$endgroup$
– Sujit Bhattacharyya
Jan 27 at 4:37
$begingroup$
WolframAlpha solves it as $x = 0.56714329040978dots$. The Inverse Symbolic Calculator gives this as $x = W(1)$ with the Lambert W function and $W(1)$ as the Omega constant.
$endgroup$
– Tito Piezas III
Jan 27 at 5:52
1
1
$begingroup$
Try using Lambert W function.
$endgroup$
– Sujit Bhattacharyya
Jan 27 at 4:37
$begingroup$
Try using Lambert W function.
$endgroup$
– Sujit Bhattacharyya
Jan 27 at 4:37
$begingroup$
WolframAlpha solves it as $x = 0.56714329040978dots$. The Inverse Symbolic Calculator gives this as $x = W(1)$ with the Lambert W function and $W(1)$ as the Omega constant.
$endgroup$
– Tito Piezas III
Jan 27 at 5:52
$begingroup$
WolframAlpha solves it as $x = 0.56714329040978dots$. The Inverse Symbolic Calculator gives this as $x = W(1)$ with the Lambert W function and $W(1)$ as the Omega constant.
$endgroup$
– Tito Piezas III
Jan 27 at 5:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is no analytical solution to this equation even using special functions. Then you will need some numerical method for finding the zero of function
$$f(x)=e^{-x}+log(x)$$ Its derivative
$$f'(x)=-e^{-x}+frac 1x$$ cancels when $x=-W(-1)$ which is a complex number; so, $f'(x)>0 $and there is only one root.
By inspection, $f(1)=frac 1e >0$. So, using Newton method with $x_0=1$ will give the following iterates
$$left(
begin{array}{cc}
n & x_n \
0 & 1.0000000000 \
1 & 0.4180232931 \
2 & 0.5413727558 \
3 & 0.5664266698 \
4 & 0.5671427445 \
5 & 0.5671432904
end{array}
right)$$
We could have generated a better estimate of the starting point building first the simplest Padé approximant at $x=1$
$$f(x) simeq frac{frac{1}{e}+frac{(2 e-1) }{2 e}(x-1)}{1+frac{1}{2}(x-1)}$$ giving $x_0=frac{2 e-3}{2 e-1}approx 0.549201$.
Continuing with $[1,n]$ Padé approximants, it will be closer and closer to the solution as shown below
$$left(
begin{array}{ccc}
1 & frac{-3+2 e}{-1+2 e} & 0.5492006529 \
2 & frac{-8+34 e-36 e^2+12 e^3}{-2+16 e-24 e^2+12 e^3} & 0.5651368901 \
3 & frac{30-300 e+636 e^2-504 e^3+144 e^4}{6-108 e+348 e^2-360 e^3+144 e^4} & 0.5670287743 \
4 & frac{-144+3216 e-12480 e^2+18000 e^3-11520 e^4+2880 e^5}{-24+1056 e-5520 e^2+10800 e^3-8640 e^4+2880 e^5} & 0.5671134681
end{array}
right)$$
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
add a comment |
$begingroup$
Alpha gives a numeric solution $x approx 0.56714$ I don't think it surprising that it did not give an analytic result. This looks like an equation that will not yield to algebra.
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
1
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
add a comment |
$begingroup$
As you observed, the functions $f(x)=e^{-x}$ and $g(x)=log(1/x)$ are inverses. If $f(x)$ intersects $y=x$ at $a$ then we have $f(a)=a$. Applying $g$ we get $a=g(a)$. Then $f(a)=g(a)$ and $a$ is a solution for the equation. So we have $a = f(a) = e^{-a}$, which is equivalent to $a e^a = 1$. This is precisely the equation that is solved by the Lambert $W$ function. Apply $W$ to get $a = W(1)$
answered Jan 31 at 19:23
jjagmathjjagmath
3387
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089137%2fhelp-needed-with-an-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no analytical solution to this equation even using special functions. Then you will need some numerical method for finding the zero of function
$$f(x)=e^{-x}+log(x)$$ Its derivative
$$f'(x)=-e^{-x}+frac 1x$$ cancels when $x=-W(-1)$ which is a complex number; so, $f'(x)>0 $and there is only one root.
By inspection, $f(1)=frac 1e >0$. So, using Newton method with $x_0=1$ will give the following iterates
$$left(
begin{array}{cc}
n & x_n \
0 & 1.0000000000 \
1 & 0.4180232931 \
2 & 0.5413727558 \
3 & 0.5664266698 \
4 & 0.5671427445 \
5 & 0.5671432904
end{array}
right)$$
We could have generated a better estimate of the starting point building first the simplest Padé approximant at $x=1$
$$f(x) simeq frac{frac{1}{e}+frac{(2 e-1) }{2 e}(x-1)}{1+frac{1}{2}(x-1)}$$ giving $x_0=frac{2 e-3}{2 e-1}approx 0.549201$.
Continuing with $[1,n]$ Padé approximants, it will be closer and closer to the solution as shown below
$$left(
begin{array}{ccc}
1 & frac{-3+2 e}{-1+2 e} & 0.5492006529 \
2 & frac{-8+34 e-36 e^2+12 e^3}{-2+16 e-24 e^2+12 e^3} & 0.5651368901 \
3 & frac{30-300 e+636 e^2-504 e^3+144 e^4}{6-108 e+348 e^2-360 e^3+144 e^4} & 0.5670287743 \
4 & frac{-144+3216 e-12480 e^2+18000 e^3-11520 e^4+2880 e^5}{-24+1056 e-5520 e^2+10800 e^3-8640 e^4+2880 e^5} & 0.5671134681
end{array}
right)$$
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
add a comment |
$begingroup$
There is no analytical solution to this equation even using special functions. Then you will need some numerical method for finding the zero of function
$$f(x)=e^{-x}+log(x)$$ Its derivative
$$f'(x)=-e^{-x}+frac 1x$$ cancels when $x=-W(-1)$ which is a complex number; so, $f'(x)>0 $and there is only one root.
By inspection, $f(1)=frac 1e >0$. So, using Newton method with $x_0=1$ will give the following iterates
$$left(
begin{array}{cc}
n & x_n \
0 & 1.0000000000 \
1 & 0.4180232931 \
2 & 0.5413727558 \
3 & 0.5664266698 \
4 & 0.5671427445 \
5 & 0.5671432904
end{array}
right)$$
We could have generated a better estimate of the starting point building first the simplest Padé approximant at $x=1$
$$f(x) simeq frac{frac{1}{e}+frac{(2 e-1) }{2 e}(x-1)}{1+frac{1}{2}(x-1)}$$ giving $x_0=frac{2 e-3}{2 e-1}approx 0.549201$.
Continuing with $[1,n]$ Padé approximants, it will be closer and closer to the solution as shown below
$$left(
begin{array}{ccc}
1 & frac{-3+2 e}{-1+2 e} & 0.5492006529 \
2 & frac{-8+34 e-36 e^2+12 e^3}{-2+16 e-24 e^2+12 e^3} & 0.5651368901 \
3 & frac{30-300 e+636 e^2-504 e^3+144 e^4}{6-108 e+348 e^2-360 e^3+144 e^4} & 0.5670287743 \
4 & frac{-144+3216 e-12480 e^2+18000 e^3-11520 e^4+2880 e^5}{-24+1056 e-5520 e^2+10800 e^3-8640 e^4+2880 e^5} & 0.5671134681
end{array}
right)$$
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
add a comment |
$begingroup$
There is no analytical solution to this equation even using special functions. Then you will need some numerical method for finding the zero of function
$$f(x)=e^{-x}+log(x)$$ Its derivative
$$f'(x)=-e^{-x}+frac 1x$$ cancels when $x=-W(-1)$ which is a complex number; so, $f'(x)>0 $and there is only one root.
By inspection, $f(1)=frac 1e >0$. So, using Newton method with $x_0=1$ will give the following iterates
$$left(
begin{array}{cc}
n & x_n \
0 & 1.0000000000 \
1 & 0.4180232931 \
2 & 0.5413727558 \
3 & 0.5664266698 \
4 & 0.5671427445 \
5 & 0.5671432904
end{array}
right)$$
We could have generated a better estimate of the starting point building first the simplest Padé approximant at $x=1$
$$f(x) simeq frac{frac{1}{e}+frac{(2 e-1) }{2 e}(x-1)}{1+frac{1}{2}(x-1)}$$ giving $x_0=frac{2 e-3}{2 e-1}approx 0.549201$.
Continuing with $[1,n]$ Padé approximants, it will be closer and closer to the solution as shown below
$$left(
begin{array}{ccc}
1 & frac{-3+2 e}{-1+2 e} & 0.5492006529 \
2 & frac{-8+34 e-36 e^2+12 e^3}{-2+16 e-24 e^2+12 e^3} & 0.5651368901 \
3 & frac{30-300 e+636 e^2-504 e^3+144 e^4}{6-108 e+348 e^2-360 e^3+144 e^4} & 0.5670287743 \
4 & frac{-144+3216 e-12480 e^2+18000 e^3-11520 e^4+2880 e^5}{-24+1056 e-5520 e^2+10800 e^3-8640 e^4+2880 e^5} & 0.5671134681
end{array}
right)$$
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
There is no analytical solution to this equation even using special functions. Then you will need some numerical method for finding the zero of function
$$f(x)=e^{-x}+log(x)$$ Its derivative
$$f'(x)=-e^{-x}+frac 1x$$ cancels when $x=-W(-1)$ which is a complex number; so, $f'(x)>0 $and there is only one root.
By inspection, $f(1)=frac 1e >0$. So, using Newton method with $x_0=1$ will give the following iterates
$$left(
begin{array}{cc}
n & x_n \
0 & 1.0000000000 \
1 & 0.4180232931 \
2 & 0.5413727558 \
3 & 0.5664266698 \
4 & 0.5671427445 \
5 & 0.5671432904
end{array}
right)$$
We could have generated a better estimate of the starting point building first the simplest Padé approximant at $x=1$
$$f(x) simeq frac{frac{1}{e}+frac{(2 e-1) }{2 e}(x-1)}{1+frac{1}{2}(x-1)}$$ giving $x_0=frac{2 e-3}{2 e-1}approx 0.549201$.
Continuing with $[1,n]$ Padé approximants, it will be closer and closer to the solution as shown below
$$left(
begin{array}{ccc}
1 & frac{-3+2 e}{-1+2 e} & 0.5492006529 \
2 & frac{-8+34 e-36 e^2+12 e^3}{-2+16 e-24 e^2+12 e^3} & 0.5651368901 \
3 & frac{30-300 e+636 e^2-504 e^3+144 e^4}{6-108 e+348 e^2-360 e^3+144 e^4} & 0.5670287743 \
4 & frac{-144+3216 e-12480 e^2+18000 e^3-11520 e^4+2880 e^5}{-24+1056 e-5520 e^2+10800 e^3-8640 e^4+2880 e^5} & 0.5671134681
end{array}
right)$$
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 5:07
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
add a comment |
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
$begingroup$
@Lee. I am obviously wrong when I write There is no analytical solution to this equation even using special functions. As Robert Israel commented, the solution is the so-called $Omega$ constant. Have a look at en.wikipedia.org/wiki/Omega_constant
$endgroup$
– Claude Leibovici
Jan 27 at 5:50
add a comment |
$begingroup$
Alpha gives a numeric solution $x approx 0.56714$ I don't think it surprising that it did not give an analytic result. This looks like an equation that will not yield to algebra.
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
1
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
add a comment |
$begingroup$
Alpha gives a numeric solution $x approx 0.56714$ I don't think it surprising that it did not give an analytic result. This looks like an equation that will not yield to algebra.
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
1
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
add a comment |
$begingroup$
Alpha gives a numeric solution $x approx 0.56714$ I don't think it surprising that it did not give an analytic result. This looks like an equation that will not yield to algebra.
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
$endgroup$
Alpha gives a numeric solution $x approx 0.56714$ I don't think it surprising that it did not give an analytic result. This looks like an equation that will not yield to algebra.
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
answered Jan 27 at 4:40
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
1
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
add a comment |
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
1
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
$begingroup$
Mathway gave me $approx{0.36787894412}$ which is $frac{1}{e}$.
$endgroup$
– Gnumbertester
Jan 27 at 4:42
1
1
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
Perhaps surprisingly, Maple doesn't solve it symbolically either. But the solution is $W(1)$ where $W$ is the Lambert W function.
$endgroup$
– Robert Israel
Jan 27 at 4:43
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@Gnumbertester Probably you entered the wrong equation. $1/e$ is certainly not a solution.
$endgroup$
– Robert Israel
Jan 27 at 4:44
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
$begingroup$
@RobertIsreal That's interesting. It's easy to see that it is not a solution but Mathway still gives it. Here is what I entered it in.<mathway.com/Calculus>
$endgroup$
– Gnumbertester
Jan 27 at 4:48
add a comment |
$begingroup$
As you observed, the functions $f(x)=e^{-x}$ and $g(x)=log(1/x)$ are inverses. If $f(x)$ intersects $y=x$ at $a$ then we have $f(a)=a$. Applying $g$ we get $a=g(a)$. Then $f(a)=g(a)$ and $a$ is a solution for the equation. So we have $a = f(a) = e^{-a}$, which is equivalent to $a e^a = 1$. This is precisely the equation that is solved by the Lambert $W$ function. Apply $W$ to get $a = W(1)$
answered Jan 31 at 19:23
jjagmathjjagmath
3387
$endgroup$
add a comment |
$begingroup$
As you observed, the functions $f(x)=e^{-x}$ and $g(x)=log(1/x)$ are inverses. If $f(x)$ intersects $y=x$ at $a$ then we have $f(a)=a$. Applying $g$ we get $a=g(a)$. Then $f(a)=g(a)$ and $a$ is a solution for the equation. So we have $a = f(a) = e^{-a}$, which is equivalent to $a e^a = 1$. This is precisely the equation that is solved by the Lambert $W$ function. Apply $W$ to get $a = W(1)$
answered Jan 31 at 19:23
jjagmathjjagmath
3387
$endgroup$
add a comment |
$begingroup$
As you observed, the functions $f(x)=e^{-x}$ and $g(x)=log(1/x)$ are inverses. If $f(x)$ intersects $y=x$ at $a$ then we have $f(a)=a$. Applying $g$ we get $a=g(a)$. Then $f(a)=g(a)$ and $a$ is a solution for the equation. So we have $a = f(a) = e^{-a}$, which is equivalent to $a e^a = 1$. This is precisely the equation that is solved by the Lambert $W$ function. Apply $W$ to get $a = W(1)$
answered Jan 31 at 19:23
jjagmathjjagmath
3387
$endgroup$
As you observed, the functions $f(x)=e^{-x}$ and $g(x)=log(1/x)$ are inverses. If $f(x)$ intersects $y=x$ at $a$ then we have $f(a)=a$. Applying $g$ we get $a=g(a)$. Then $f(a)=g(a)$ and $a$ is a solution for the equation. So we have $a = f(a) = e^{-a}$, which is equivalent to $a e^a = 1$. This is precisely the equation that is solved by the Lambert $W$ function. Apply $W$ to get $a = W(1)$
answered Jan 31 at 19:23
jjagmathjjagmath
3387
answered Jan 31 at 19:23
jjagmathjjagmath
3387
answered Jan 31 at 19:23
jjagmathjjagmath
3387
answered Jan 31 at 19:23
jjagmathjjagmath
3387
3387
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089137%2fhelp-needed-with-an-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Try using Lambert W function.
$endgroup$
– Sujit Bhattacharyya
Jan 27 at 4:37
$begingroup$
WolframAlpha solves it as $x = 0.56714329040978dots$. The Inverse Symbolic Calculator gives this as $x = W(1)$ with the Lambert W function and $W(1)$ as the Omega constant.
$endgroup$
– Tito Piezas III
Jan 27 at 5:52