Is this (1.11716..) a known/named constant?
$begingroup$
While looking at the Wikipedia entry for the Supergolden ratio aka Narayana's cow constant (cf. https://en.wikipedia.org/wiki/Supergolden_ratio), I noticed the following construction involving that constant and a 120-degree ($2pi/3$ radian) triangle on that page:
I then began to wonder about similar constructions with other fractional angles: $2pi/2$, $2pi/4$, ...
I found that an angle of $2pi/2$ gives a degenerate triangle based on the golden ratio:
$2pi/4$ gives a right triangle based on the square-root of the golden ratio:
And $2pi/6$ gives an equilateral triangle based on unity:
Nothing too surprising or new in the results so far, but an angle of $2pi/5$ (72-degrees) produced the following:
So, the positive solution to $-2 - x + sqrt 5 x - 2 x^2 + 2 x^4=0$, approximate value $1.11716..$, minimal polynomial $1
+ x + x^2 + x^3 - x^4 - x^5 - 2 x^6 + x^8$
Is this a known/named constant?
geometry constants
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
$endgroup$
add a comment |
$begingroup$
While looking at the Wikipedia entry for the Supergolden ratio aka Narayana's cow constant (cf. https://en.wikipedia.org/wiki/Supergolden_ratio), I noticed the following construction involving that constant and a 120-degree ($2pi/3$ radian) triangle on that page:
I then began to wonder about similar constructions with other fractional angles: $2pi/2$, $2pi/4$, ...
I found that an angle of $2pi/2$ gives a degenerate triangle based on the golden ratio:
$2pi/4$ gives a right triangle based on the square-root of the golden ratio:
And $2pi/6$ gives an equilateral triangle based on unity:
Nothing too surprising or new in the results so far, but an angle of $2pi/5$ (72-degrees) produced the following:
So, the positive solution to $-2 - x + sqrt 5 x - 2 x^2 + 2 x^4=0$, approximate value $1.11716..$, minimal polynomial $1
+ x + x^2 + x^3 - x^4 - x^5 - 2 x^6 + x^8$
Is this a known/named constant?
geometry constants
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
$endgroup$
$begingroup$
It is not a constant from this list, but this is only a small list...
$endgroup$
– Dietrich Burde
Jan 14 at 20:58
3
$begingroup$
The Inverse Symbolic Calculator doesn't have a match for the longer decimal expansion, $1.11715586807111ldots$.
$endgroup$
– Blue
Jan 14 at 21:05
$begingroup$
I was getting around to this question myself. Pi/4 is related to the square root of $1 - x^2 - 2 x^3 + x^4=0$. I can also relate Pi/3 to the square root of the root for $1 - x - x^2 - 2 x^3 + x^4=0$.
$endgroup$
– Ed Pegg
Jan 14 at 22:06
1
$begingroup$
The polynomial is known. lmfdb.org/NumberField/8.4.42903125.1
$endgroup$
– Ed Pegg
Jan 14 at 22:24
$begingroup$
The polynomials involved for Pi/7 and Pi/9 are not known at lmfdb.org.
$endgroup$
– Ed Pegg
Jan 14 at 23:11
add a comment |
$begingroup$
While looking at the Wikipedia entry for the Supergolden ratio aka Narayana's cow constant (cf. https://en.wikipedia.org/wiki/Supergolden_ratio), I noticed the following construction involving that constant and a 120-degree ($2pi/3$ radian) triangle on that page:
I then began to wonder about similar constructions with other fractional angles: $2pi/2$, $2pi/4$, ...
I found that an angle of $2pi/2$ gives a degenerate triangle based on the golden ratio:
$2pi/4$ gives a right triangle based on the square-root of the golden ratio:
And $2pi/6$ gives an equilateral triangle based on unity:
Nothing too surprising or new in the results so far, but an angle of $2pi/5$ (72-degrees) produced the following:
So, the positive solution to $-2 - x + sqrt 5 x - 2 x^2 + 2 x^4=0$, approximate value $1.11716..$, minimal polynomial $1
+ x + x^2 + x^3 - x^4 - x^5 - 2 x^6 + x^8$
Is this a known/named constant?
geometry constants
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
$endgroup$
While looking at the Wikipedia entry for the Supergolden ratio aka Narayana's cow constant (cf. https://en.wikipedia.org/wiki/Supergolden_ratio), I noticed the following construction involving that constant and a 120-degree ($2pi/3$ radian) triangle on that page:
I then began to wonder about similar constructions with other fractional angles: $2pi/2$, $2pi/4$, ...
I found that an angle of $2pi/2$ gives a degenerate triangle based on the golden ratio:
$2pi/4$ gives a right triangle based on the square-root of the golden ratio:
And $2pi/6$ gives an equilateral triangle based on unity:
Nothing too surprising or new in the results so far, but an angle of $2pi/5$ (72-degrees) produced the following:
So, the positive solution to $-2 - x + sqrt 5 x - 2 x^2 + 2 x^4=0$, approximate value $1.11716..$, minimal polynomial $1
+ x + x^2 + x^3 - x^4 - x^5 - 2 x^6 + x^8$
Is this a known/named constant?
geometry constants
geometry constants
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
edited Jan 14 at 22:15
WRSomsky
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
asked Jan 14 at 20:51
WRSomskyWRSomsky
45616
45616
$begingroup$
It is not a constant from this list, but this is only a small list...
$endgroup$
– Dietrich Burde
Jan 14 at 20:58
3
$begingroup$
The Inverse Symbolic Calculator doesn't have a match for the longer decimal expansion, $1.11715586807111ldots$.
$endgroup$
– Blue
Jan 14 at 21:05
$begingroup$
I was getting around to this question myself. Pi/4 is related to the square root of $1 - x^2 - 2 x^3 + x^4=0$. I can also relate Pi/3 to the square root of the root for $1 - x - x^2 - 2 x^3 + x^4=0$.
$endgroup$
– Ed Pegg
Jan 14 at 22:06
1
$begingroup$
The polynomial is known. lmfdb.org/NumberField/8.4.42903125.1
$endgroup$
– Ed Pegg
Jan 14 at 22:24
$begingroup$
The polynomials involved for Pi/7 and Pi/9 are not known at lmfdb.org.
$endgroup$
– Ed Pegg
Jan 14 at 23:11
add a comment |
$begingroup$
It is not a constant from this list, but this is only a small list...
$endgroup$
– Dietrich Burde
Jan 14 at 20:58
3
$begingroup$
The Inverse Symbolic Calculator doesn't have a match for the longer decimal expansion, $1.11715586807111ldots$.
$endgroup$
– Blue
Jan 14 at 21:05
$begingroup$
I was getting around to this question myself. Pi/4 is related to the square root of $1 - x^2 - 2 x^3 + x^4=0$. I can also relate Pi/3 to the square root of the root for $1 - x - x^2 - 2 x^3 + x^4=0$.
$endgroup$
– Ed Pegg
Jan 14 at 22:06
1
$begingroup$
The polynomial is known. lmfdb.org/NumberField/8.4.42903125.1
$endgroup$
– Ed Pegg
Jan 14 at 22:24
$begingroup$
The polynomials involved for Pi/7 and Pi/9 are not known at lmfdb.org.
$endgroup$
– Ed Pegg
Jan 14 at 23:11
$begingroup$
It is not a constant from this list, but this is only a small list...
$endgroup$
– Dietrich Burde
Jan 14 at 20:58
$begingroup$
It is not a constant from this list, but this is only a small list...
$endgroup$
– Dietrich Burde
Jan 14 at 20:58
3
3
$begingroup$
The Inverse Symbolic Calculator doesn't have a match for the longer decimal expansion, $1.11715586807111ldots$.
$endgroup$
– Blue
Jan 14 at 21:05
$begingroup$
The Inverse Symbolic Calculator doesn't have a match for the longer decimal expansion, $1.11715586807111ldots$.
$endgroup$
– Blue
Jan 14 at 21:05
$begingroup$
I was getting around to this question myself. Pi/4 is related to the square root of $1 - x^2 - 2 x^3 + x^4=0$. I can also relate Pi/3 to the square root of the root for $1 - x - x^2 - 2 x^3 + x^4=0$.
$endgroup$
– Ed Pegg
Jan 14 at 22:06
$begingroup$
I was getting around to this question myself. Pi/4 is related to the square root of $1 - x^2 - 2 x^3 + x^4=0$. I can also relate Pi/3 to the square root of the root for $1 - x - x^2 - 2 x^3 + x^4=0$.
$endgroup$
– Ed Pegg
Jan 14 at 22:06
1
1
$begingroup$
The polynomial is known. lmfdb.org/NumberField/8.4.42903125.1
$endgroup$
– Ed Pegg
Jan 14 at 22:24
$begingroup$
The polynomial is known. lmfdb.org/NumberField/8.4.42903125.1
$endgroup$
– Ed Pegg
Jan 14 at 22:24
$begingroup$
The polynomials involved for Pi/7 and Pi/9 are not known at lmfdb.org.
$endgroup$
– Ed Pegg
Jan 14 at 23:11
$begingroup$
The polynomials involved for Pi/7 and Pi/9 are not known at lmfdb.org.
$endgroup$
– Ed Pegg
Jan 14 at 23:11
add a comment |
0
active
oldest
votes
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%2f3073730%2fis-this-1-11716-a-known-named-constant%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3073730%2fis-this-1-11716-a-known-named-constant%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
$begingroup$
It is not a constant from this list, but this is only a small list...
$endgroup$
– Dietrich Burde
Jan 14 at 20:58
3
$begingroup$
The Inverse Symbolic Calculator doesn't have a match for the longer decimal expansion, $1.11715586807111ldots$.
$endgroup$
– Blue
Jan 14 at 21:05
$begingroup$
I was getting around to this question myself. Pi/4 is related to the square root of $1 - x^2 - 2 x^3 + x^4=0$. I can also relate Pi/3 to the square root of the root for $1 - x - x^2 - 2 x^3 + x^4=0$.
$endgroup$
– Ed Pegg
Jan 14 at 22:06
1
$begingroup$
The polynomial is known. lmfdb.org/NumberField/8.4.42903125.1
$endgroup$
– Ed Pegg
Jan 14 at 22:24
$begingroup$
The polynomials involved for Pi/7 and Pi/9 are not known at lmfdb.org.
$endgroup$
– Ed Pegg
Jan 14 at 23:11