Mixing
Existence of Dirichlet Inverse.
$begingroup$
How do we show the existence of the Dirichlet Inverse of a function $f$? I know using induction that we can find a recursive form of $f^{-1}$. But I can't seem to show the existence. Any ideas?
Most sites I have searched assume the existence before giving a recursive formula.
analytic-number-theory
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
$endgroup$
add a comment |
$begingroup$
How do we show the existence of the Dirichlet Inverse of a function $f$? I know using induction that we can find a recursive form of $f^{-1}$. But I can't seem to show the existence. Any ideas?
Most sites I have searched assume the existence before giving a recursive formula.
analytic-number-theory
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
$endgroup$
$begingroup$
Iff $f(1) ne 1$ then $f^{-1}$ exists. Replace $f(n)$ by $f(n)/f(1)$ to obtain $f(1)=f^{-1}(1)= 1$. Then for $n ge 2, sum_{d | n} f^{-1}(n/d) f(d) = 0 implies f^{-1}(n) = -sum_{d |n, d > 1} f^{-1}(n/d) f(d)$ the latter being a well-defined induction formula. Equivalently you can look at $frac{1}{F(s)}= sum_{k=0}^infty (1-F(s))^k$ the latter being convergent in the ring of formal Dirichlet series.
$endgroup$
– reuns
Jan 23 at 16:33
add a comment |
$begingroup$
How do we show the existence of the Dirichlet Inverse of a function $f$? I know using induction that we can find a recursive form of $f^{-1}$. But I can't seem to show the existence. Any ideas?
Most sites I have searched assume the existence before giving a recursive formula.
analytic-number-theory
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
$endgroup$
How do we show the existence of the Dirichlet Inverse of a function $f$? I know using induction that we can find a recursive form of $f^{-1}$. But I can't seem to show the existence. Any ideas?
Most sites I have searched assume the existence before giving a recursive formula.
analytic-number-theory
analytic-number-theory
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
asked Jan 23 at 13:05
Jhon DoeJhon Doe
642314
642314
$begingroup$
Iff $f(1) ne 1$ then $f^{-1}$ exists. Replace $f(n)$ by $f(n)/f(1)$ to obtain $f(1)=f^{-1}(1)= 1$. Then for $n ge 2, sum_{d | n} f^{-1}(n/d) f(d) = 0 implies f^{-1}(n) = -sum_{d |n, d > 1} f^{-1}(n/d) f(d)$ the latter being a well-defined induction formula. Equivalently you can look at $frac{1}{F(s)}= sum_{k=0}^infty (1-F(s))^k$ the latter being convergent in the ring of formal Dirichlet series.
$endgroup$
– reuns
Jan 23 at 16:33
add a comment |
$begingroup$
Iff $f(1) ne 1$ then $f^{-1}$ exists. Replace $f(n)$ by $f(n)/f(1)$ to obtain $f(1)=f^{-1}(1)= 1$. Then for $n ge 2, sum_{d | n} f^{-1}(n/d) f(d) = 0 implies f^{-1}(n) = -sum_{d |n, d > 1} f^{-1}(n/d) f(d)$ the latter being a well-defined induction formula. Equivalently you can look at $frac{1}{F(s)}= sum_{k=0}^infty (1-F(s))^k$ the latter being convergent in the ring of formal Dirichlet series.
$endgroup$
– reuns
Jan 23 at 16:33
$begingroup$
Iff $f(1) ne 1$ then $f^{-1}$ exists. Replace $f(n)$ by $f(n)/f(1)$ to obtain $f(1)=f^{-1}(1)= 1$. Then for $n ge 2, sum_{d | n} f^{-1}(n/d) f(d) = 0 implies f^{-1}(n) = -sum_{d |n, d > 1} f^{-1}(n/d) f(d)$ the latter being a well-defined induction formula. Equivalently you can look at $frac{1}{F(s)}= sum_{k=0}^infty (1-F(s))^k$ the latter being convergent in the ring of formal Dirichlet series.
$endgroup$
– reuns
Jan 23 at 16:33
$begingroup$
Iff $f(1) ne 1$ then $f^{-1}$ exists. Replace $f(n)$ by $f(n)/f(1)$ to obtain $f(1)=f^{-1}(1)= 1$. Then for $n ge 2, sum_{d | n} f^{-1}(n/d) f(d) = 0 implies f^{-1}(n) = -sum_{d |n, d > 1} f^{-1}(n/d) f(d)$ the latter being a well-defined induction formula. Equivalently you can look at $frac{1}{F(s)}= sum_{k=0}^infty (1-F(s))^k$ the latter being convergent in the ring of formal Dirichlet series.
$endgroup$
– reuns
Jan 23 at 16:33
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%2f3084447%2fexistence-of-dirichlet-inverse%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%2f3084447%2fexistence-of-dirichlet-inverse%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
$begingroup$
Iff $f(1) ne 1$ then $f^{-1}$ exists. Replace $f(n)$ by $f(n)/f(1)$ to obtain $f(1)=f^{-1}(1)= 1$. Then for $n ge 2, sum_{d | n} f^{-1}(n/d) f(d) = 0 implies f^{-1}(n) = -sum_{d |n, d > 1} f^{-1}(n/d) f(d)$ the latter being a well-defined induction formula. Equivalently you can look at $frac{1}{F(s)}= sum_{k=0}^infty (1-F(s))^k$ the latter being convergent in the ring of formal Dirichlet series.
$endgroup$
– reuns
Jan 23 at 16:33