Geometric transformations and limit (pi=4 revisited!!)
$begingroup$
Consider three geometric transformations.
1st: The geometric "proof" that hypotenuse of a right triangle is equal to the sum of squares of the other sides.
The link is here.
2nd: The famous "pi=4" "proof" by geometric transformation
3rd: Archimedes's method of calculation of pi
What is inherently different in these transformations that 1st 2nd yield wrong results while the 3rd doesn't? Is there any heuristic which could hint us when the transformation is valid for calculation of the limit?
I have read the discussion on mathstackexchange about pi=4, but still I couldn't get that intuition.
There was an interesting argument about the divergence of the derivatives of those two curves. But I still don't understand if it is a satisfactory condition to prove that Archimedes construction was a valid one.
If possible, please give 2 two explanations from intuitional and strictly mathematical point of view.
calculus real-analysis limits pi fake-proofs
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
asked Oct 12 '16 at 15:12
guserguser
18912
$endgroup$
add a comment |
$begingroup$
Consider three geometric transformations.
1st: The geometric "proof" that hypotenuse of a right triangle is equal to the sum of squares of the other sides.
The link is here.
2nd: The famous "pi=4" "proof" by geometric transformation
3rd: Archimedes's method of calculation of pi
What is inherently different in these transformations that 1st 2nd yield wrong results while the 3rd doesn't? Is there any heuristic which could hint us when the transformation is valid for calculation of the limit?
I have read the discussion on mathstackexchange about pi=4, but still I couldn't get that intuition.
There was an interesting argument about the divergence of the derivatives of those two curves. But I still don't understand if it is a satisfactory condition to prove that Archimedes construction was a valid one.
If possible, please give 2 two explanations from intuitional and strictly mathematical point of view.
calculus real-analysis limits pi fake-proofs
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
asked Oct 12 '16 at 15:12
guserguser
18912
$endgroup$
$begingroup$
This is nice food for thought...
$endgroup$
– imranfat
Oct 12 '16 at 15:18
add a comment |
$begingroup$
Consider three geometric transformations.
1st: The geometric "proof" that hypotenuse of a right triangle is equal to the sum of squares of the other sides.
The link is here.
2nd: The famous "pi=4" "proof" by geometric transformation
3rd: Archimedes's method of calculation of pi
What is inherently different in these transformations that 1st 2nd yield wrong results while the 3rd doesn't? Is there any heuristic which could hint us when the transformation is valid for calculation of the limit?
I have read the discussion on mathstackexchange about pi=4, but still I couldn't get that intuition.
There was an interesting argument about the divergence of the derivatives of those two curves. But I still don't understand if it is a satisfactory condition to prove that Archimedes construction was a valid one.
If possible, please give 2 two explanations from intuitional and strictly mathematical point of view.
calculus real-analysis limits pi fake-proofs
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
asked Oct 12 '16 at 15:12
guserguser
18912
$endgroup$
Consider three geometric transformations.
1st: The geometric "proof" that hypotenuse of a right triangle is equal to the sum of squares of the other sides.
The link is here.
2nd: The famous "pi=4" "proof" by geometric transformation
3rd: Archimedes's method of calculation of pi
What is inherently different in these transformations that 1st 2nd yield wrong results while the 3rd doesn't? Is there any heuristic which could hint us when the transformation is valid for calculation of the limit?
I have read the discussion on mathstackexchange about pi=4, but still I couldn't get that intuition.
There was an interesting argument about the divergence of the derivatives of those two curves. But I still don't understand if it is a satisfactory condition to prove that Archimedes construction was a valid one.
If possible, please give 2 two explanations from intuitional and strictly mathematical point of view.
calculus real-analysis limits pi fake-proofs
calculus real-analysis limits pi fake-proofs
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
asked Oct 12 '16 at 15:12
guserguser
18912
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
asked Oct 12 '16 at 15:12
guserguser
18912
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
edited Dec 24 '18 at 9:34
Glorfindel
3,41981830
3,41981830
asked Oct 12 '16 at 15:12
guserguser
18912
asked Oct 12 '16 at 15:12
guserguser
18912
asked Oct 12 '16 at 15:12
guserguser
18912
18912
$begingroup$
This is nice food for thought...
$endgroup$
– imranfat
Oct 12 '16 at 15:18
add a comment |
$begingroup$
This is nice food for thought...
$endgroup$
– imranfat
Oct 12 '16 at 15:18
$begingroup$
This is nice food for thought...
$endgroup$
– imranfat
Oct 12 '16 at 15:18
$begingroup$
This is nice food for thought...
$endgroup$
– imranfat
Oct 12 '16 at 15:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The key point is the lower semicontinuity (and in general the non-continuity) of the perimeter when you approximate (in some sense to be made more precise) a given region $R$ with some regions $R_n$ (for example polygons). So you can only say in general that
$$
textrm{Per}(R)leqliminf_n textrm{Per}(R_n).
$$
Nothing more.
If, for some reasons, you know a priori that, for any $n$, $textrm{Per}(R)geq textrm{Per}(R_n) $, then you can conclude
$$
textrm{Per}(R)= textrm{Per}(R_n).
$$
This is the case of the approximation of the circle with the inscribed polygons: you postulates that the chord is always less than the arc, i.e. the straight line is the minimum lenght curve, so $textrm{Per}(R)geq textrm{Per}(R_n) ;forall n$.
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation
$$
pi= textrm{Per}(R)<liminf_n textrm{Per}(R_n)=liminf_n 4=4.
$$
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
$endgroup$
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
add a comment |
$begingroup$
From the first figure pi<4. In each step n, the area of the polygon, a(n), is the same and equals 4. Hence lim a(n) = 4, as n tends to infinity and we still have pi < 4.
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
$endgroup$
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%2f1965355%2fgeometric-transformations-and-limit-pi-4-revisited%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
$begingroup$
The key point is the lower semicontinuity (and in general the non-continuity) of the perimeter when you approximate (in some sense to be made more precise) a given region $R$ with some regions $R_n$ (for example polygons). So you can only say in general that
$$
textrm{Per}(R)leqliminf_n textrm{Per}(R_n).
$$
Nothing more.
If, for some reasons, you know a priori that, for any $n$, $textrm{Per}(R)geq textrm{Per}(R_n) $, then you can conclude
$$
textrm{Per}(R)= textrm{Per}(R_n).
$$
This is the case of the approximation of the circle with the inscribed polygons: you postulates that the chord is always less than the arc, i.e. the straight line is the minimum lenght curve, so $textrm{Per}(R)geq textrm{Per}(R_n) ;forall n$.
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation
$$
pi= textrm{Per}(R)<liminf_n textrm{Per}(R_n)=liminf_n 4=4.
$$
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
$endgroup$
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
add a comment |
$begingroup$
The key point is the lower semicontinuity (and in general the non-continuity) of the perimeter when you approximate (in some sense to be made more precise) a given region $R$ with some regions $R_n$ (for example polygons). So you can only say in general that
$$
textrm{Per}(R)leqliminf_n textrm{Per}(R_n).
$$
Nothing more.
If, for some reasons, you know a priori that, for any $n$, $textrm{Per}(R)geq textrm{Per}(R_n) $, then you can conclude
$$
textrm{Per}(R)= textrm{Per}(R_n).
$$
This is the case of the approximation of the circle with the inscribed polygons: you postulates that the chord is always less than the arc, i.e. the straight line is the minimum lenght curve, so $textrm{Per}(R)geq textrm{Per}(R_n) ;forall n$.
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation
$$
pi= textrm{Per}(R)<liminf_n textrm{Per}(R_n)=liminf_n 4=4.
$$
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
$endgroup$
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
add a comment |
$begingroup$
The key point is the lower semicontinuity (and in general the non-continuity) of the perimeter when you approximate (in some sense to be made more precise) a given region $R$ with some regions $R_n$ (for example polygons). So you can only say in general that
$$
textrm{Per}(R)leqliminf_n textrm{Per}(R_n).
$$
Nothing more.
If, for some reasons, you know a priori that, for any $n$, $textrm{Per}(R)geq textrm{Per}(R_n) $, then you can conclude
$$
textrm{Per}(R)= textrm{Per}(R_n).
$$
This is the case of the approximation of the circle with the inscribed polygons: you postulates that the chord is always less than the arc, i.e. the straight line is the minimum lenght curve, so $textrm{Per}(R)geq textrm{Per}(R_n) ;forall n$.
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation
$$
pi= textrm{Per}(R)<liminf_n textrm{Per}(R_n)=liminf_n 4=4.
$$
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
$endgroup$
The key point is the lower semicontinuity (and in general the non-continuity) of the perimeter when you approximate (in some sense to be made more precise) a given region $R$ with some regions $R_n$ (for example polygons). So you can only say in general that
$$
textrm{Per}(R)leqliminf_n textrm{Per}(R_n).
$$
Nothing more.
If, for some reasons, you know a priori that, for any $n$, $textrm{Per}(R)geq textrm{Per}(R_n) $, then you can conclude
$$
textrm{Per}(R)= textrm{Per}(R_n).
$$
This is the case of the approximation of the circle with the inscribed polygons: you postulates that the chord is always less than the arc, i.e. the straight line is the minimum lenght curve, so $textrm{Per}(R)geq textrm{Per}(R_n) ;forall n$.
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation
$$
pi= textrm{Per}(R)<liminf_n textrm{Per}(R_n)=liminf_n 4=4.
$$
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
edited Oct 12 '16 at 15:41
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
answered Oct 12 '16 at 15:25
guestDiegoguestDiego
3,459417
3,459417
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
add a comment |
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
your external perimeter is always and always and always equal to 4, never less and never more.
$endgroup$
– hamam_Abdallah
Oct 12 '16 at 15:36
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Waht's the point of this comment?
$endgroup$
– guestDiego
Oct 12 '16 at 15:47
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
Of course this further property does't apply in your approximation with the external polygons and, in fact, for this approximation π=Per(R)<lim infnPer(Rn)=lim infn4=4. @guestDiego You can also approximate pi with circumscribed polygons. Naturally enough, they also converge to pi. So there is nothing connected with the fact of the polygons being circumscribed or inscribed to a curve or in this case a circle.
$endgroup$
– guser
Oct 12 '16 at 16:21
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
$begingroup$
@guser. Instead it is! The approximation with the external polygons works for circumscribed polygons but not for the polygons you showed in your construction. And the reason is exactly the one I explained.
$endgroup$
– guestDiego
Oct 12 '16 at 16:25
add a comment |
$begingroup$
From the first figure pi<4. In each step n, the area of the polygon, a(n), is the same and equals 4. Hence lim a(n) = 4, as n tends to infinity and we still have pi < 4.
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
$endgroup$
add a comment |
$begingroup$
From the first figure pi<4. In each step n, the area of the polygon, a(n), is the same and equals 4. Hence lim a(n) = 4, as n tends to infinity and we still have pi < 4.
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
$endgroup$
add a comment |
$begingroup$
From the first figure pi<4. In each step n, the area of the polygon, a(n), is the same and equals 4. Hence lim a(n) = 4, as n tends to infinity and we still have pi < 4.
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
$endgroup$
From the first figure pi<4. In each step n, the area of the polygon, a(n), is the same and equals 4. Hence lim a(n) = 4, as n tends to infinity and we still have pi < 4.
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
answered Sep 27 '17 at 11:47
frank trevorfrank trevor
1
1
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%2f1965355%2fgeometric-transformations-and-limit-pi-4-revisited%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
$begingroup$
This is nice food for thought...
$endgroup$
– imranfat
Oct 12 '16 at 15:18