Rigorous proof of a unique solution using Banach's Fixed Point Theorem The 2019 Stack Overflow Developer Survey Results Are InA corollary of Banach's fixed-point theoremUnique solution to $x^4 + 7x -1 = 0 $ on $[0,1]$ (Banach's fixed point theorem)Complete Condition in Banach Fixed Point TheoremGetting a root of a continuously differentiable function by Banach's Fixed Point Theorem.Proof of Banach's homeomorphism theorem without the contraction map principle.Problem in Banach Fixed Point Theorem for a functional equationBanach Fixed Point Theorem and the function $f(x)=x+e^-x$Completeness can be omitted from Banach Fixed Point Theorem?Banach's fixed point theoremcomplete metric space and unique fixed point
Where to refill my bottle in India?
Extreme, unacceptable situation and I can't attend work tomorrow morning
I see my dog run
Geography at the pixel level
Falsification in Math vs Science
The meaning of "à force"
Spanish for "widget"
What is the steepest angle that a canal can be traversable without locks?
What are the motivations for publishing new editions of an existing textbook, beyond new discoveries in a field?
Supports in 3d printing
is usb on wall sockets live all the time with out switches off
Is there a name of the flying bionic bird?
How long do I have to send payment?
Springs with some finite mass
Where does the "burst of radiance" from Holy Weapon originate?
Could JWST stay at L2 "forever"?
Access elements in std::string where positon of string is greater than its size
Is bread bad for ducks?
How are circuits which use complex ICs normally simulated?
Does light intensity oscillate really fast since it is a wave?
Typesetting a double Over Dot on top of a symbol
aging parents with no investments
Limit the amount of RAM Mathematica may access?
Why Did Howard Stark Use All The Vibranium They Had On A Prototype Shield?
Rigorous proof of a unique solution using Banach's Fixed Point Theorem
The 2019 Stack Overflow Developer Survey Results Are InA corollary of Banach's fixed-point theoremUnique solution to $x^4 + 7x -1 = 0 $ on $[0,1]$ (Banach's fixed point theorem)Complete Condition in Banach Fixed Point TheoremGetting a root of a continuously differentiable function by Banach's Fixed Point Theorem.Proof of Banach's homeomorphism theorem without the contraction map principle.Problem in Banach Fixed Point Theorem for a functional equationBanach Fixed Point Theorem and the function $f(x)=x+e^-x$Completeness can be omitted from Banach Fixed Point Theorem?Banach's fixed point theoremcomplete metric space and unique fixed point
$begingroup$
I would like to have feedback on the overall quality of the following proof.
Question: Prove that $x^5+7x-1=0$ has a unique solution in $[0,1]$.
Proof: Let $f(x)=frac1-x^57$ and note that any solution of $f(x)=x$ is in fact a solution of our polynomial equation. Since $[0,1]$ is closed in $mathbbR$, which is complete, it follows that $[0,1]$ is also complete. Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$. Together, $([0,1],d)$ is a complete metric space.
To satisfy Banach's fixed point theorem we require that $f$ is a contraction on $[0,1]$:
$d(f(x),f(y))=|f(x)-f(y)|=frac17|x^5-y^5|=frac17|x-y|cdot|x^4+x^3y+x^2y^2+xy^3+y^4|leq frac57|x-y|=frac57d(x,y)$
for all $x,yin[0,1]$. Thus $f$ is a contraction on $[0,1]$. Therefore, by Banach's fixed point theorem, $f$ must have a unique fixed point in $[0,1]$ and so the polynomial equation must have a unique solution in $[0,1]$.
All suggestions for improvements are welcome!
proof-verification metric-spaces fixed-point-theorems
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
$endgroup$
|
show 1 more comment
$begingroup$
I would like to have feedback on the overall quality of the following proof.
Question: Prove that $x^5+7x-1=0$ has a unique solution in $[0,1]$.
Proof: Let $f(x)=frac1-x^57$ and note that any solution of $f(x)=x$ is in fact a solution of our polynomial equation. Since $[0,1]$ is closed in $mathbbR$, which is complete, it follows that $[0,1]$ is also complete. Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$. Together, $([0,1],d)$ is a complete metric space.
To satisfy Banach's fixed point theorem we require that $f$ is a contraction on $[0,1]$:
$d(f(x),f(y))=|f(x)-f(y)|=frac17|x^5-y^5|=frac17|x-y|cdot|x^4+x^3y+x^2y^2+xy^3+y^4|leq frac57|x-y|=frac57d(x,y)$
for all $x,yin[0,1]$. Thus $f$ is a contraction on $[0,1]$. Therefore, by Banach's fixed point theorem, $f$ must have a unique fixed point in $[0,1]$ and so the polynomial equation must have a unique solution in $[0,1]$.
All suggestions for improvements are welcome!
proof-verification metric-spaces fixed-point-theorems
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
$endgroup$
$begingroup$
Why not just use IVT and MVT?
$endgroup$
– David Mitra
Mar 30 at 10:19
$begingroup$
I'm trying to improve my use of Banach's fixed point theorem and this seemed like a good question to start with
$endgroup$
– piece_and_love
Mar 30 at 10:24
1
$begingroup$
Ok! Looks good. You should say in the first line that $f(x)=x$ and $x^5+7x-1=0$ are equivalent equations (not merely that a solution of the former is a solution of the latter).
$endgroup$
– David Mitra
Mar 30 at 10:42
$begingroup$
I didn't quite understand the sentence "Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$". Once you start taking about completeness the metric should already be, implicitly or explicitly, fixed.
$endgroup$
– Anguepa
Mar 30 at 18:48
1
$begingroup$
I would state: Let $d$ denote the Euclidian metric, then $(mathbbR,d)$ is a complete metric space.
$endgroup$
– Carl Christian
Mar 30 at 19:07
|
show 1 more comment
$begingroup$
I would like to have feedback on the overall quality of the following proof.
Question: Prove that $x^5+7x-1=0$ has a unique solution in $[0,1]$.
Proof: Let $f(x)=frac1-x^57$ and note that any solution of $f(x)=x$ is in fact a solution of our polynomial equation. Since $[0,1]$ is closed in $mathbbR$, which is complete, it follows that $[0,1]$ is also complete. Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$. Together, $([0,1],d)$ is a complete metric space.
To satisfy Banach's fixed point theorem we require that $f$ is a contraction on $[0,1]$:
$d(f(x),f(y))=|f(x)-f(y)|=frac17|x^5-y^5|=frac17|x-y|cdot|x^4+x^3y+x^2y^2+xy^3+y^4|leq frac57|x-y|=frac57d(x,y)$
for all $x,yin[0,1]$. Thus $f$ is a contraction on $[0,1]$. Therefore, by Banach's fixed point theorem, $f$ must have a unique fixed point in $[0,1]$ and so the polynomial equation must have a unique solution in $[0,1]$.
All suggestions for improvements are welcome!
proof-verification metric-spaces fixed-point-theorems
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
$endgroup$
I would like to have feedback on the overall quality of the following proof.
Question: Prove that $x^5+7x-1=0$ has a unique solution in $[0,1]$.
Proof: Let $f(x)=frac1-x^57$ and note that any solution of $f(x)=x$ is in fact a solution of our polynomial equation. Since $[0,1]$ is closed in $mathbbR$, which is complete, it follows that $[0,1]$ is also complete. Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$. Together, $([0,1],d)$ is a complete metric space.
To satisfy Banach's fixed point theorem we require that $f$ is a contraction on $[0,1]$:
$d(f(x),f(y))=|f(x)-f(y)|=frac17|x^5-y^5|=frac17|x-y|cdot|x^4+x^3y+x^2y^2+xy^3+y^4|leq frac57|x-y|=frac57d(x,y)$
for all $x,yin[0,1]$. Thus $f$ is a contraction on $[0,1]$. Therefore, by Banach's fixed point theorem, $f$ must have a unique fixed point in $[0,1]$ and so the polynomial equation must have a unique solution in $[0,1]$.
All suggestions for improvements are welcome!
proof-verification metric-spaces fixed-point-theorems
proof-verification metric-spaces fixed-point-theorems
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
edited Mar 30 at 10:07
Thomas Shelby
4,7362727
4,7362727
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
asked Mar 30 at 10:03
piece_and_lovepiece_and_love
574
574
$begingroup$
Why not just use IVT and MVT?
$endgroup$
– David Mitra
Mar 30 at 10:19
$begingroup$
I'm trying to improve my use of Banach's fixed point theorem and this seemed like a good question to start with
$endgroup$
– piece_and_love
Mar 30 at 10:24
1
$begingroup$
Ok! Looks good. You should say in the first line that $f(x)=x$ and $x^5+7x-1=0$ are equivalent equations (not merely that a solution of the former is a solution of the latter).
$endgroup$
– David Mitra
Mar 30 at 10:42
$begingroup$
I didn't quite understand the sentence "Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$". Once you start taking about completeness the metric should already be, implicitly or explicitly, fixed.
$endgroup$
– Anguepa
Mar 30 at 18:48
1
$begingroup$
I would state: Let $d$ denote the Euclidian metric, then $(mathbbR,d)$ is a complete metric space.
$endgroup$
– Carl Christian
Mar 30 at 19:07
|
show 1 more comment
$begingroup$
Why not just use IVT and MVT?
$endgroup$
– David Mitra
Mar 30 at 10:19
$begingroup$
I'm trying to improve my use of Banach's fixed point theorem and this seemed like a good question to start with
$endgroup$
– piece_and_love
Mar 30 at 10:24
1
$begingroup$
Ok! Looks good. You should say in the first line that $f(x)=x$ and $x^5+7x-1=0$ are equivalent equations (not merely that a solution of the former is a solution of the latter).
$endgroup$
– David Mitra
Mar 30 at 10:42
$begingroup$
I didn't quite understand the sentence "Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$". Once you start taking about completeness the metric should already be, implicitly or explicitly, fixed.
$endgroup$
– Anguepa
Mar 30 at 18:48
1
$begingroup$
I would state: Let $d$ denote the Euclidian metric, then $(mathbbR,d)$ is a complete metric space.
$endgroup$
– Carl Christian
Mar 30 at 19:07
$begingroup$
Why not just use IVT and MVT?
$endgroup$
– David Mitra
Mar 30 at 10:19
$begingroup$
Why not just use IVT and MVT?
$endgroup$
– David Mitra
Mar 30 at 10:19
$begingroup$
I'm trying to improve my use of Banach's fixed point theorem and this seemed like a good question to start with
$endgroup$
– piece_and_love
Mar 30 at 10:24
$begingroup$
I'm trying to improve my use of Banach's fixed point theorem and this seemed like a good question to start with
$endgroup$
– piece_and_love
Mar 30 at 10:24
1
1
$begingroup$
Ok! Looks good. You should say in the first line that $f(x)=x$ and $x^5+7x-1=0$ are equivalent equations (not merely that a solution of the former is a solution of the latter).
$endgroup$
– David Mitra
Mar 30 at 10:42
$begingroup$
Ok! Looks good. You should say in the first line that $f(x)=x$ and $x^5+7x-1=0$ are equivalent equations (not merely that a solution of the former is a solution of the latter).
$endgroup$
– David Mitra
Mar 30 at 10:42
$begingroup$
I didn't quite understand the sentence "Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$". Once you start taking about completeness the metric should already be, implicitly or explicitly, fixed.
$endgroup$
– Anguepa
Mar 30 at 18:48
$begingroup$
I didn't quite understand the sentence "Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$". Once you start taking about completeness the metric should already be, implicitly or explicitly, fixed.
$endgroup$
– Anguepa
Mar 30 at 18:48
1
1
$begingroup$
I would state: Let $d$ denote the Euclidian metric, then $(mathbbR,d)$ is a complete metric space.
$endgroup$
– Carl Christian
Mar 30 at 19:07
$begingroup$
I would state: Let $d$ denote the Euclidian metric, then $(mathbbR,d)$ is a complete metric space.
$endgroup$
– Carl Christian
Mar 30 at 19:07
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
A hostile reviewer could raise the following issues:
- You do not explicitly define the function $p : [0,1] rightarrow mathbbR$ given by $p(x) = x^5 + 7x - 1$.
- You do not explicitly list the domain and co-domain of the function $f$ given by $f(x) = frac1-x^57$.
- You do not explicitly state that $p(x) = 0$ if and only if $f(x) = x$.
- You do not explicitly state that $f$ maps $[0,1]$ into itself.
- You would have chosen the Euclidian metric even when $f$ is not an algebraic function.
- The metric should be identified when you first mention completeness.
- Stictly speaking, $f$ does not satisfy Banach's fixed point theorem. The function $f$ satisfies the hypothesis of Banach's fixed point theorem.
The only really serious issue is point $4$. The function $g : [0,1] rightarrow mathbbR$ given by $g(x) =2$ is a contraction, but it has no fixed points. Nevertheless, it would serve you well to eliminate the remaining issues.
It is not necessary to apply this level of formalism in your daily work, but apply it when you need to remove any doubt about your ability to justify your reasoning.
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
$endgroup$
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
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%2f3168113%2frigorous-proof-of-a-unique-solution-using-banachs-fixed-point-theorem%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A hostile reviewer could raise the following issues:
- You do not explicitly define the function $p : [0,1] rightarrow mathbbR$ given by $p(x) = x^5 + 7x - 1$.
- You do not explicitly list the domain and co-domain of the function $f$ given by $f(x) = frac1-x^57$.
- You do not explicitly state that $p(x) = 0$ if and only if $f(x) = x$.
- You do not explicitly state that $f$ maps $[0,1]$ into itself.
- You would have chosen the Euclidian metric even when $f$ is not an algebraic function.
- The metric should be identified when you first mention completeness.
- Stictly speaking, $f$ does not satisfy Banach's fixed point theorem. The function $f$ satisfies the hypothesis of Banach's fixed point theorem.
The only really serious issue is point $4$. The function $g : [0,1] rightarrow mathbbR$ given by $g(x) =2$ is a contraction, but it has no fixed points. Nevertheless, it would serve you well to eliminate the remaining issues.
It is not necessary to apply this level of formalism in your daily work, but apply it when you need to remove any doubt about your ability to justify your reasoning.
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
$endgroup$
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
add a comment |
$begingroup$
A hostile reviewer could raise the following issues:
- You do not explicitly define the function $p : [0,1] rightarrow mathbbR$ given by $p(x) = x^5 + 7x - 1$.
- You do not explicitly list the domain and co-domain of the function $f$ given by $f(x) = frac1-x^57$.
- You do not explicitly state that $p(x) = 0$ if and only if $f(x) = x$.
- You do not explicitly state that $f$ maps $[0,1]$ into itself.
- You would have chosen the Euclidian metric even when $f$ is not an algebraic function.
- The metric should be identified when you first mention completeness.
- Stictly speaking, $f$ does not satisfy Banach's fixed point theorem. The function $f$ satisfies the hypothesis of Banach's fixed point theorem.
The only really serious issue is point $4$. The function $g : [0,1] rightarrow mathbbR$ given by $g(x) =2$ is a contraction, but it has no fixed points. Nevertheless, it would serve you well to eliminate the remaining issues.
It is not necessary to apply this level of formalism in your daily work, but apply it when you need to remove any doubt about your ability to justify your reasoning.
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
$endgroup$
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
add a comment |
$begingroup$
A hostile reviewer could raise the following issues:
- You do not explicitly define the function $p : [0,1] rightarrow mathbbR$ given by $p(x) = x^5 + 7x - 1$.
- You do not explicitly list the domain and co-domain of the function $f$ given by $f(x) = frac1-x^57$.
- You do not explicitly state that $p(x) = 0$ if and only if $f(x) = x$.
- You do not explicitly state that $f$ maps $[0,1]$ into itself.
- You would have chosen the Euclidian metric even when $f$ is not an algebraic function.
- The metric should be identified when you first mention completeness.
- Stictly speaking, $f$ does not satisfy Banach's fixed point theorem. The function $f$ satisfies the hypothesis of Banach's fixed point theorem.
The only really serious issue is point $4$. The function $g : [0,1] rightarrow mathbbR$ given by $g(x) =2$ is a contraction, but it has no fixed points. Nevertheless, it would serve you well to eliminate the remaining issues.
It is not necessary to apply this level of formalism in your daily work, but apply it when you need to remove any doubt about your ability to justify your reasoning.
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
$endgroup$
A hostile reviewer could raise the following issues:
- You do not explicitly define the function $p : [0,1] rightarrow mathbbR$ given by $p(x) = x^5 + 7x - 1$.
- You do not explicitly list the domain and co-domain of the function $f$ given by $f(x) = frac1-x^57$.
- You do not explicitly state that $p(x) = 0$ if and only if $f(x) = x$.
- You do not explicitly state that $f$ maps $[0,1]$ into itself.
- You would have chosen the Euclidian metric even when $f$ is not an algebraic function.
- The metric should be identified when you first mention completeness.
- Stictly speaking, $f$ does not satisfy Banach's fixed point theorem. The function $f$ satisfies the hypothesis of Banach's fixed point theorem.
The only really serious issue is point $4$. The function $g : [0,1] rightarrow mathbbR$ given by $g(x) =2$ is a contraction, but it has no fixed points. Nevertheless, it would serve you well to eliminate the remaining issues.
It is not necessary to apply this level of formalism in your daily work, but apply it when you need to remove any doubt about your ability to justify your reasoning.
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
answered Mar 30 at 18:33
Carl ChristianCarl Christian
6,0861723
6,0861723
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
add a comment |
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
$begingroup$
Thank you, this is exactly the kind of answer I was looking for. In the future I'll make sure to at least implement points 4 and 6.
$endgroup$
– piece_and_love
Mar 30 at 18:45
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%2f3168113%2frigorous-proof-of-a-unique-solution-using-banachs-fixed-point-theorem%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
$begingroup$
Why not just use IVT and MVT?
$endgroup$
– David Mitra
Mar 30 at 10:19
$begingroup$
I'm trying to improve my use of Banach's fixed point theorem and this seemed like a good question to start with
$endgroup$
– piece_and_love
Mar 30 at 10:24
1
$begingroup$
Ok! Looks good. You should say in the first line that $f(x)=x$ and $x^5+7x-1=0$ are equivalent equations (not merely that a solution of the former is a solution of the latter).
$endgroup$
– David Mitra
Mar 30 at 10:42
$begingroup$
I didn't quite understand the sentence "Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$". Once you start taking about completeness the metric should already be, implicitly or explicitly, fixed.
$endgroup$
– Anguepa
Mar 30 at 18:48
1
$begingroup$
I would state: Let $d$ denote the Euclidian metric, then $(mathbbR,d)$ is a complete metric space.
$endgroup$
– Carl Christian
Mar 30 at 19:07