Extending the Lindemann Weierstrass Theorem
up vote
3
down vote
favorite
What I want to do is tinker with the Lindemann Weierstrass Theorem so I can ask 'what is so special about the number $e$'? I'll state the theorem below.
Lindemann-Weierstrass Theorem (Baker's Reformulation)
Let $alpha_1, dots, alpha_n in bar{mathbb{Q}}$ be distinct and let $a_1, dots, a_n in bar{mathbb{Q}}$ be non-zero. Where $bar{mathbb{Q}}$ refers to the set of algebraic numbers. Then $$sum_{i=1}^n{a_i e^{alpha_i}} neq 0$$
Note that we really have a polynomial (of sorts) that we have evaluated at $e$.
So consider the set of polynomials evaluated at a variable $x in mathbb{R}$.
$$S_x=bigg {sum_{i=1}^n{a_i x^{alpha_i}}: alpha_i in bar{mathbb{Q}} text{ distinct}, a_iin bar{mathbb{Q}} text{ non-zero}, nin mathbb{N} bigg }$$
Now we can rewrite the theorem above as saying: $0notin S_e$ and actually with no effort at all we can extend the theorem for any algebraic number $beta$:
$0notin S_{e^beta}$. But $S_x$ is countably many polynomials (They are only kinda polynomials: the exponents are algebraic) and each have finitely many zeros. So the set $L={x in mathbb{R}: 0in S_x}$ is countable as well. That is to say that most complex numbers enjoy the property of not satisfying any polynomials of this type.
And $L$ the solutions of some generalized polynomial is a nontrivial(shown below) countable superset of the algebraic numbers.
Questions
Is there a name in the literature for $L$?
Are there any expectations on what $mathbb{R} backslash L $ must be? Is it likely that say $ln(2)in mathbb{R} backslash L $? Does there exist any polynomial with algebraic coefficients and exponents which $ln(2)$ satisfies?
We should comment that it can be seen immediately that $bar{Q} subseteq L$ which is why we should focus on transcendental numbers.
complex-analysis transcendence-theory
asked Nov 22 at 19:48
Mason
1,8881527
add a comment |
up vote
3
down vote
favorite
What I want to do is tinker with the Lindemann Weierstrass Theorem so I can ask 'what is so special about the number $e$'? I'll state the theorem below.
Lindemann-Weierstrass Theorem (Baker's Reformulation)
Let $alpha_1, dots, alpha_n in bar{mathbb{Q}}$ be distinct and let $a_1, dots, a_n in bar{mathbb{Q}}$ be non-zero. Where $bar{mathbb{Q}}$ refers to the set of algebraic numbers. Then $$sum_{i=1}^n{a_i e^{alpha_i}} neq 0$$
Note that we really have a polynomial (of sorts) that we have evaluated at $e$.
So consider the set of polynomials evaluated at a variable $x in mathbb{R}$.
$$S_x=bigg {sum_{i=1}^n{a_i x^{alpha_i}}: alpha_i in bar{mathbb{Q}} text{ distinct}, a_iin bar{mathbb{Q}} text{ non-zero}, nin mathbb{N} bigg }$$
Now we can rewrite the theorem above as saying: $0notin S_e$ and actually with no effort at all we can extend the theorem for any algebraic number $beta$:
$0notin S_{e^beta}$. But $S_x$ is countably many polynomials (They are only kinda polynomials: the exponents are algebraic) and each have finitely many zeros. So the set $L={x in mathbb{R}: 0in S_x}$ is countable as well. That is to say that most complex numbers enjoy the property of not satisfying any polynomials of this type.
And $L$ the solutions of some generalized polynomial is a nontrivial(shown below) countable superset of the algebraic numbers.
Questions
Is there a name in the literature for $L$?
Are there any expectations on what $mathbb{R} backslash L $ must be? Is it likely that say $ln(2)in mathbb{R} backslash L $? Does there exist any polynomial with algebraic coefficients and exponents which $ln(2)$ satisfies?
We should comment that it can be seen immediately that $bar{Q} subseteq L$ which is why we should focus on transcendental numbers.
complex-analysis transcendence-theory
asked Nov 22 at 19:48
Mason
1,8881527
The case with $x=ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $alpha$ such that $(ln 2)^alpha$ is algebraic?
– Matt F.
Dec 9 at 1:58
1
I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs.
– Alexandre Eremenko
Dec 12 at 14:48
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What I want to do is tinker with the Lindemann Weierstrass Theorem so I can ask 'what is so special about the number $e$'? I'll state the theorem below.
Lindemann-Weierstrass Theorem (Baker's Reformulation)
Let $alpha_1, dots, alpha_n in bar{mathbb{Q}}$ be distinct and let $a_1, dots, a_n in bar{mathbb{Q}}$ be non-zero. Where $bar{mathbb{Q}}$ refers to the set of algebraic numbers. Then $$sum_{i=1}^n{a_i e^{alpha_i}} neq 0$$
Note that we really have a polynomial (of sorts) that we have evaluated at $e$.
So consider the set of polynomials evaluated at a variable $x in mathbb{R}$.
$$S_x=bigg {sum_{i=1}^n{a_i x^{alpha_i}}: alpha_i in bar{mathbb{Q}} text{ distinct}, a_iin bar{mathbb{Q}} text{ non-zero}, nin mathbb{N} bigg }$$
Now we can rewrite the theorem above as saying: $0notin S_e$ and actually with no effort at all we can extend the theorem for any algebraic number $beta$:
$0notin S_{e^beta}$. But $S_x$ is countably many polynomials (They are only kinda polynomials: the exponents are algebraic) and each have finitely many zeros. So the set $L={x in mathbb{R}: 0in S_x}$ is countable as well. That is to say that most complex numbers enjoy the property of not satisfying any polynomials of this type.
And $L$ the solutions of some generalized polynomial is a nontrivial(shown below) countable superset of the algebraic numbers.
Questions
Is there a name in the literature for $L$?
Are there any expectations on what $mathbb{R} backslash L $ must be? Is it likely that say $ln(2)in mathbb{R} backslash L $? Does there exist any polynomial with algebraic coefficients and exponents which $ln(2)$ satisfies?
We should comment that it can be seen immediately that $bar{Q} subseteq L$ which is why we should focus on transcendental numbers.
complex-analysis transcendence-theory
asked Nov 22 at 19:48
Mason
1,8881527
What I want to do is tinker with the Lindemann Weierstrass Theorem so I can ask 'what is so special about the number $e$'? I'll state the theorem below.
Lindemann-Weierstrass Theorem (Baker's Reformulation)
Let $alpha_1, dots, alpha_n in bar{mathbb{Q}}$ be distinct and let $a_1, dots, a_n in bar{mathbb{Q}}$ be non-zero. Where $bar{mathbb{Q}}$ refers to the set of algebraic numbers. Then $$sum_{i=1}^n{a_i e^{alpha_i}} neq 0$$
Note that we really have a polynomial (of sorts) that we have evaluated at $e$.
So consider the set of polynomials evaluated at a variable $x in mathbb{R}$.
$$S_x=bigg {sum_{i=1}^n{a_i x^{alpha_i}}: alpha_i in bar{mathbb{Q}} text{ distinct}, a_iin bar{mathbb{Q}} text{ non-zero}, nin mathbb{N} bigg }$$
Now we can rewrite the theorem above as saying: $0notin S_e$ and actually with no effort at all we can extend the theorem for any algebraic number $beta$:
$0notin S_{e^beta}$. But $S_x$ is countably many polynomials (They are only kinda polynomials: the exponents are algebraic) and each have finitely many zeros. So the set $L={x in mathbb{R}: 0in S_x}$ is countable as well. That is to say that most complex numbers enjoy the property of not satisfying any polynomials of this type.
And $L$ the solutions of some generalized polynomial is a nontrivial(shown below) countable superset of the algebraic numbers.
Questions
Is there a name in the literature for $L$?
Are there any expectations on what $mathbb{R} backslash L $ must be? Is it likely that say $ln(2)in mathbb{R} backslash L $? Does there exist any polynomial with algebraic coefficients and exponents which $ln(2)$ satisfies?
We should comment that it can be seen immediately that $bar{Q} subseteq L$ which is why we should focus on transcendental numbers.
complex-analysis transcendence-theory
complex-analysis transcendence-theory
asked Nov 22 at 19:48
Mason
1,8881527
asked Nov 22 at 19:48
Mason
1,8881527
edited Dec 8 at 0:06
asked Nov 22 at 19:48
Mason
1,8881527
asked Nov 22 at 19:48
Mason
1,8881527
asked Nov 22 at 19:48
Mason
1,8881527
1,8881527
The case with $x=ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $alpha$ such that $(ln 2)^alpha$ is algebraic?
– Matt F.
Dec 9 at 1:58
1
I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs.
– Alexandre Eremenko
Dec 12 at 14:48
add a comment |
The case with $x=ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $alpha$ such that $(ln 2)^alpha$ is algebraic?
– Matt F.
Dec 9 at 1:58
1
I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs.
– Alexandre Eremenko
Dec 12 at 14:48
The case with $x=ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $alpha$ such that $(ln 2)^alpha$ is algebraic?
– Matt F.
Dec 9 at 1:58
The case with $x=ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $alpha$ such that $(ln 2)^alpha$ is algebraic?
– Matt F.
Dec 9 at 1:58
1
1
I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs.
– Alexandre Eremenko
Dec 12 at 14:48
I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs.
– Alexandre Eremenko
Dec 12 at 14:48
add a comment |
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',
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%2f3009566%2fextending-the-lindemann-weierstrass-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
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.
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%2f3009566%2fextending-the-lindemann-weierstrass-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
The case with $x=ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $alpha$ such that $(ln 2)^alpha$ is algebraic?
– Matt F.
Dec 9 at 1:58
1
I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs.
– Alexandre Eremenko
Dec 12 at 14:48