Mixing
Number of roots of $f(x)=x^2-2^{x-frac{1}{x}}$
$begingroup$
How many roots does $f(x)=x^2-2^{x-frac{1}{x}}$ have in $(0,1]$?
Since this function is continuous, I plugged in a couple of values and looked at the signs of the function's values and I concluded that there is a root between $frac{1}{10}$ and $frac{1}{5}$.$1$ is also a root, so I believe there are two roots in $(0,1]$, but how to rigorously prove this?
real-analysis roots
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
$endgroup$
add a comment |
$begingroup$
How many roots does $f(x)=x^2-2^{x-frac{1}{x}}$ have in $(0,1]$?
Since this function is continuous, I plugged in a couple of values and looked at the signs of the function's values and I concluded that there is a root between $frac{1}{10}$ and $frac{1}{5}$.$1$ is also a root, so I believe there are two roots in $(0,1]$, but how to rigorously prove this?
real-analysis roots
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
$endgroup$
$begingroup$
Your function has two roots in the given interval.
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 15:53
$begingroup$
I believe it does, but how to prove this?
$endgroup$
– JustAnAmateur
Feb 2 at 15:56
$begingroup$
I'd suggest looking at it through Taylor series to prove there are only two roots. Plotting on Mathematica "confirms" your guess, but looking at the derivative directly seems to be rather complicated.
$endgroup$
– Clayton
Feb 2 at 16:54
add a comment |
$begingroup$
How many roots does $f(x)=x^2-2^{x-frac{1}{x}}$ have in $(0,1]$?
Since this function is continuous, I plugged in a couple of values and looked at the signs of the function's values and I concluded that there is a root between $frac{1}{10}$ and $frac{1}{5}$.$1$ is also a root, so I believe there are two roots in $(0,1]$, but how to rigorously prove this?
real-analysis roots
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
$endgroup$
How many roots does $f(x)=x^2-2^{x-frac{1}{x}}$ have in $(0,1]$?
Since this function is continuous, I plugged in a couple of values and looked at the signs of the function's values and I concluded that there is a root between $frac{1}{10}$ and $frac{1}{5}$.$1$ is also a root, so I believe there are two roots in $(0,1]$, but how to rigorously prove this?
real-analysis roots
real-analysis roots
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
asked Feb 2 at 15:46
JustAnAmateurJustAnAmateur
1096
1096
$begingroup$
Your function has two roots in the given interval.
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 15:53
$begingroup$
I believe it does, but how to prove this?
$endgroup$
– JustAnAmateur
Feb 2 at 15:56
$begingroup$
I'd suggest looking at it through Taylor series to prove there are only two roots. Plotting on Mathematica "confirms" your guess, but looking at the derivative directly seems to be rather complicated.
$endgroup$
– Clayton
Feb 2 at 16:54
add a comment |
$begingroup$
Your function has two roots in the given interval.
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 15:53
$begingroup$
I believe it does, but how to prove this?
$endgroup$
– JustAnAmateur
Feb 2 at 15:56
$begingroup$
I'd suggest looking at it through Taylor series to prove there are only two roots. Plotting on Mathematica "confirms" your guess, but looking at the derivative directly seems to be rather complicated.
$endgroup$
– Clayton
Feb 2 at 16:54
$begingroup$
Your function has two roots in the given interval.
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 15:53
$begingroup$
Your function has two roots in the given interval.
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 15:53
$begingroup$
I believe it does, but how to prove this?
$endgroup$
– JustAnAmateur
Feb 2 at 15:56
$begingroup$
I believe it does, but how to prove this?
$endgroup$
– JustAnAmateur
Feb 2 at 15:56
$begingroup$
I'd suggest looking at it through Taylor series to prove there are only two roots. Plotting on Mathematica "confirms" your guess, but looking at the derivative directly seems to be rather complicated.
$endgroup$
– Clayton
Feb 2 at 16:54
$begingroup$
I'd suggest looking at it through Taylor series to prove there are only two roots. Plotting on Mathematica "confirms" your guess, but looking at the derivative directly seems to be rather complicated.
$endgroup$
– Clayton
Feb 2 at 16:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Instead of finding the roots $x^2 = 2^{x - 1/x}$, let's consider the roots of $f(x) = 2 ln x - (x - 1/x) ln 2$. We have
$$f'(x) = -frac {ln 2} {x^2} left( x^2 - frac {2 x} {ln 2} + 1 right).$$
$f'(x)$ has two positive roots the product of which is $1$, therefore it has exactly one root $x_0$ on $(0, 1)$.
This means that $f$ decreases from $infty$ to $f(x_0)$ on $(0, x_0)$ and increases from $f(x_0)$ to $f(1) = 0$ on $(x_0, 1)$. Since $f(1) = 0$, we have $f(x_0) < 0$. Therefore $f$ has exactly two roots on $(0, 1]$.
answered Feb 2 at 18:56
MaximMaxim
6,2831221
$endgroup$
add a comment |
$begingroup$
Compute $$f(0.2),f(0.4)$$ and $$f(1)$$
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
add a comment |
Your Answer
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%2f3097419%2fnumber-of-roots-of-fx-x2-2x-frac1x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Instead of finding the roots $x^2 = 2^{x - 1/x}$, let's consider the roots of $f(x) = 2 ln x - (x - 1/x) ln 2$. We have
$$f'(x) = -frac {ln 2} {x^2} left( x^2 - frac {2 x} {ln 2} + 1 right).$$
$f'(x)$ has two positive roots the product of which is $1$, therefore it has exactly one root $x_0$ on $(0, 1)$.
This means that $f$ decreases from $infty$ to $f(x_0)$ on $(0, x_0)$ and increases from $f(x_0)$ to $f(1) = 0$ on $(x_0, 1)$. Since $f(1) = 0$, we have $f(x_0) < 0$. Therefore $f$ has exactly two roots on $(0, 1]$.
answered Feb 2 at 18:56
MaximMaxim
6,2831221
$endgroup$
add a comment |
$begingroup$
Instead of finding the roots $x^2 = 2^{x - 1/x}$, let's consider the roots of $f(x) = 2 ln x - (x - 1/x) ln 2$. We have
$$f'(x) = -frac {ln 2} {x^2} left( x^2 - frac {2 x} {ln 2} + 1 right).$$
$f'(x)$ has two positive roots the product of which is $1$, therefore it has exactly one root $x_0$ on $(0, 1)$.
This means that $f$ decreases from $infty$ to $f(x_0)$ on $(0, x_0)$ and increases from $f(x_0)$ to $f(1) = 0$ on $(x_0, 1)$. Since $f(1) = 0$, we have $f(x_0) < 0$. Therefore $f$ has exactly two roots on $(0, 1]$.
answered Feb 2 at 18:56
MaximMaxim
6,2831221
$endgroup$
add a comment |
$begingroup$
Instead of finding the roots $x^2 = 2^{x - 1/x}$, let's consider the roots of $f(x) = 2 ln x - (x - 1/x) ln 2$. We have
$$f'(x) = -frac {ln 2} {x^2} left( x^2 - frac {2 x} {ln 2} + 1 right).$$
$f'(x)$ has two positive roots the product of which is $1$, therefore it has exactly one root $x_0$ on $(0, 1)$.
This means that $f$ decreases from $infty$ to $f(x_0)$ on $(0, x_0)$ and increases from $f(x_0)$ to $f(1) = 0$ on $(x_0, 1)$. Since $f(1) = 0$, we have $f(x_0) < 0$. Therefore $f$ has exactly two roots on $(0, 1]$.
answered Feb 2 at 18:56
MaximMaxim
6,2831221
$endgroup$
Instead of finding the roots $x^2 = 2^{x - 1/x}$, let's consider the roots of $f(x) = 2 ln x - (x - 1/x) ln 2$. We have
$$f'(x) = -frac {ln 2} {x^2} left( x^2 - frac {2 x} {ln 2} + 1 right).$$
$f'(x)$ has two positive roots the product of which is $1$, therefore it has exactly one root $x_0$ on $(0, 1)$.
This means that $f$ decreases from $infty$ to $f(x_0)$ on $(0, x_0)$ and increases from $f(x_0)$ to $f(1) = 0$ on $(x_0, 1)$. Since $f(1) = 0$, we have $f(x_0) < 0$. Therefore $f$ has exactly two roots on $(0, 1]$.
answered Feb 2 at 18:56
MaximMaxim
6,2831221
answered Feb 2 at 18:56
MaximMaxim
6,2831221
answered Feb 2 at 18:56
MaximMaxim
6,2831221
answered Feb 2 at 18:56
MaximMaxim
6,2831221
6,2831221
add a comment |
add a comment |
$begingroup$
Compute $$f(0.2),f(0.4)$$ and $$f(1)$$
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
add a comment |
$begingroup$
Compute $$f(0.2),f(0.4)$$ and $$f(1)$$
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
add a comment |
$begingroup$
Compute $$f(0.2),f(0.4)$$ and $$f(1)$$
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
Compute $$f(0.2),f(0.4)$$ and $$f(1)$$
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Feb 2 at 16:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
79k42867
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
add a comment |
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
That is exactly what I did, this shows that there are $2$ roots, but it doesn't imply there is not a third one.
$endgroup$
– JustAnAmateur
Feb 2 at 16:21
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
Then show that the derivative between two points is decreasing
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:31
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
And how does this help? Is there any general result?
$endgroup$
– JustAnAmateur
Feb 2 at 16:43
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
$begingroup$
If it is so, then can the function cut the x-axes only one times since you have a sign change
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 16:48
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%2f3097419%2fnumber-of-roots-of-fx-x2-2x-frac1x%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
$begingroup$
Your function has two roots in the given interval.
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 15:53
$begingroup$
I believe it does, but how to prove this?
$endgroup$
– JustAnAmateur
Feb 2 at 15:56
$begingroup$
I'd suggest looking at it through Taylor series to prove there are only two roots. Plotting on Mathematica "confirms" your guess, but looking at the derivative directly seems to be rather complicated.
$endgroup$
– Clayton
Feb 2 at 16:54