What is the condition that $e^x=ax+a-1$ has non-negative solution?
Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?
calculus systems-of-equations roots
asked Dec 28 '18 at 9:51
Math_Y
605
add a comment |
Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?
calculus systems-of-equations roots
asked Dec 28 '18 at 9:51
Math_Y
605
Can you please tell what methods have you tried to solve this question?
– Jaideep Khare
Dec 28 '18 at 9:53
1
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
– Kavi Rama Murthy
Dec 28 '18 at 9:58
add a comment |
Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?
calculus systems-of-equations roots
asked Dec 28 '18 at 9:51
Math_Y
605
Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?
calculus systems-of-equations roots
calculus systems-of-equations roots
asked Dec 28 '18 at 9:51
Math_Y
605
asked Dec 28 '18 at 9:51
Math_Y
605
asked Dec 28 '18 at 9:51
Math_Y
605
asked Dec 28 '18 at 9:51
Math_Y
605
asked Dec 28 '18 at 9:51
Math_Y
605
605
Can you please tell what methods have you tried to solve this question?
– Jaideep Khare
Dec 28 '18 at 9:53
1
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
– Kavi Rama Murthy
Dec 28 '18 at 9:58
add a comment |
Can you please tell what methods have you tried to solve this question?
– Jaideep Khare
Dec 28 '18 at 9:53
1
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
– Kavi Rama Murthy
Dec 28 '18 at 9:58
Can you please tell what methods have you tried to solve this question?
– Jaideep Khare
Dec 28 '18 at 9:53
Can you please tell what methods have you tried to solve this question?
– Jaideep Khare
Dec 28 '18 at 9:53
1
1
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
– Kavi Rama Murthy
Dec 28 '18 at 9:58
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
– Kavi Rama Murthy
Dec 28 '18 at 9:58
add a comment |
3 Answers
3
active
oldest
votes
Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
2
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
add a comment |
Hint :
Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$
The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$
Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
add a comment |
Hint:
The equation can be written
$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.
Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.
The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
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%2f3054737%2fwhat-is-the-condition-that-ex-axa-1-has-non-negative-solution%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
Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
2
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
add a comment |
Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
2
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
add a comment |
Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
edited Dec 28 '18 at 10:52
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
answered Dec 28 '18 at 10:36
Dr. Sonnhard Graubner
73.4k42865
73.4k42865
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
2
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
add a comment |
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
2
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
It is not related to amount of $a$?
– Math_Y
Dec 28 '18 at 10:48
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:51
2
2
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
IMO, this brings you farther from the solution than the original setting of the question.
– Yves Daoust
Dec 28 '18 at 12:45
add a comment |
Hint :
Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$
The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$
Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
add a comment |
Hint :
Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$
The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$
Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
add a comment |
Hint :
Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$
The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$
Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
Hint :
Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$
The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$
Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
edited Dec 28 '18 at 10:21
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
answered Dec 28 '18 at 10:03
Rebellos
14.5k31245
14.5k31245
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
add a comment |
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
you must also consider the case if $$a<0$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 10:08
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
@Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
– Rebellos
Dec 28 '18 at 10:10
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
Thank you very much. There exists a closed form relation for the solutions?
– Math_Y
Dec 28 '18 at 10:29
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
@Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
– Rory Daulton
Dec 28 '18 at 15:34
add a comment |
Hint:
The equation can be written
$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.
Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.
The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
add a comment |
Hint:
The equation can be written
$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.
Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.
The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
add a comment |
Hint:
The equation can be written
$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.
Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.
The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
Hint:
The equation can be written
$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.
Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.
The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
edited Dec 28 '18 at 12:54
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
answered Dec 28 '18 at 12:42
Yves Daoust
124k671222
124k671222
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3054737%2fwhat-is-the-condition-that-ex-axa-1-has-non-negative-solution%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
Can you please tell what methods have you tried to solve this question?
– Jaideep Khare
Dec 28 '18 at 9:53
1
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
– Kavi Rama Murthy
Dec 28 '18 at 9:58