How to simplify $2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2)$
$begingroup$
Let $k$ be a positive integer satisfying $k equiv 1 pmod 4$. Let $x in mathbb{N}$. Let $q$ be a prime number.
If
$$sigma(x) = sum_{d mid x}{d}$$
is the classical sum-of-divisors function, then how do I simplify the following expression?
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2)$$
When $k=1$, this is just
$$2n^2 - sigma(n^2) = D(n^2)$$
where $D(x)=2x-sigma(x)$ is the deficiency of $x$.
How about when $k>1$? Does the expression have a nice form?
MY ATTEMPT
Since $k>1$ and $k equiv 1 pmod 4$, then $k geq 5$, which implies that
$$sigma(q^{frac{k-1}{2}}) = q^{frac{k-1}{2}} + ldots + q + 1 = q^{frac{k-1}{2}} + sigma(q^{frac{k-3}{2}})$$
so that we obtain
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - q^{frac{k-1}{2}} sigma(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
$$= q^{frac{k-1}{2}} D(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
Is my derivation correct?
Note that when $gcd(q,n)=1$, then this is just
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - sigmaleft(q^{frac{k-1}{2}} n^2right) = Dbigg(q^{frac{k-1}{2}} n^2bigg).$$
elementary-number-theory proof-verification modular-arithmetic divisor-sum arithmetic-functions
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
$endgroup$
add a comment |
$begingroup$
Let $k$ be a positive integer satisfying $k equiv 1 pmod 4$. Let $x in mathbb{N}$. Let $q$ be a prime number.
If
$$sigma(x) = sum_{d mid x}{d}$$
is the classical sum-of-divisors function, then how do I simplify the following expression?
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2)$$
When $k=1$, this is just
$$2n^2 - sigma(n^2) = D(n^2)$$
where $D(x)=2x-sigma(x)$ is the deficiency of $x$.
How about when $k>1$? Does the expression have a nice form?
MY ATTEMPT
Since $k>1$ and $k equiv 1 pmod 4$, then $k geq 5$, which implies that
$$sigma(q^{frac{k-1}{2}}) = q^{frac{k-1}{2}} + ldots + q + 1 = q^{frac{k-1}{2}} + sigma(q^{frac{k-3}{2}})$$
so that we obtain
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - q^{frac{k-1}{2}} sigma(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
$$= q^{frac{k-1}{2}} D(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
Is my derivation correct?
Note that when $gcd(q,n)=1$, then this is just
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - sigmaleft(q^{frac{k-1}{2}} n^2right) = Dbigg(q^{frac{k-1}{2}} n^2bigg).$$
elementary-number-theory proof-verification modular-arithmetic divisor-sum arithmetic-functions
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
$endgroup$
1
$begingroup$
Your derivation is correct.
$endgroup$
– Cardioid_Ass_22
Jan 14 at 14:17
$begingroup$
Thank you for your feedback, @Cardioid_Ass_22.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 14:48
add a comment |
$begingroup$
Let $k$ be a positive integer satisfying $k equiv 1 pmod 4$. Let $x in mathbb{N}$. Let $q$ be a prime number.
If
$$sigma(x) = sum_{d mid x}{d}$$
is the classical sum-of-divisors function, then how do I simplify the following expression?
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2)$$
When $k=1$, this is just
$$2n^2 - sigma(n^2) = D(n^2)$$
where $D(x)=2x-sigma(x)$ is the deficiency of $x$.
How about when $k>1$? Does the expression have a nice form?
MY ATTEMPT
Since $k>1$ and $k equiv 1 pmod 4$, then $k geq 5$, which implies that
$$sigma(q^{frac{k-1}{2}}) = q^{frac{k-1}{2}} + ldots + q + 1 = q^{frac{k-1}{2}} + sigma(q^{frac{k-3}{2}})$$
so that we obtain
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - q^{frac{k-1}{2}} sigma(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
$$= q^{frac{k-1}{2}} D(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
Is my derivation correct?
Note that when $gcd(q,n)=1$, then this is just
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - sigmaleft(q^{frac{k-1}{2}} n^2right) = Dbigg(q^{frac{k-1}{2}} n^2bigg).$$
elementary-number-theory proof-verification modular-arithmetic divisor-sum arithmetic-functions
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
$endgroup$
Let $k$ be a positive integer satisfying $k equiv 1 pmod 4$. Let $x in mathbb{N}$. Let $q$ be a prime number.
If
$$sigma(x) = sum_{d mid x}{d}$$
is the classical sum-of-divisors function, then how do I simplify the following expression?
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2)$$
When $k=1$, this is just
$$2n^2 - sigma(n^2) = D(n^2)$$
where $D(x)=2x-sigma(x)$ is the deficiency of $x$.
How about when $k>1$? Does the expression have a nice form?
MY ATTEMPT
Since $k>1$ and $k equiv 1 pmod 4$, then $k geq 5$, which implies that
$$sigma(q^{frac{k-1}{2}}) = q^{frac{k-1}{2}} + ldots + q + 1 = q^{frac{k-1}{2}} + sigma(q^{frac{k-3}{2}})$$
so that we obtain
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - q^{frac{k-1}{2}} sigma(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
$$= q^{frac{k-1}{2}} D(n^2) - sigma(q^{frac{k-3}{2}}) sigma(n^2)$$
Is my derivation correct?
Note that when $gcd(q,n)=1$, then this is just
$$2 q^{frac{k-1}{2}} n^2 - sigma(q^{frac{k-1}{2}})sigma(n^2) = 2 q^{frac{k-1}{2}} n^2 - sigmaleft(q^{frac{k-1}{2}} n^2right) = Dbigg(q^{frac{k-1}{2}} n^2bigg).$$
elementary-number-theory proof-verification modular-arithmetic divisor-sum arithmetic-functions
elementary-number-theory proof-verification modular-arithmetic divisor-sum arithmetic-functions
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
edited Jan 16 at 11:16
Jose Arnaldo Bebita Dris
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
asked Jan 14 at 7:04
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,50741945
5,50741945
1
$begingroup$
Your derivation is correct.
$endgroup$
– Cardioid_Ass_22
Jan 14 at 14:17
$begingroup$
Thank you for your feedback, @Cardioid_Ass_22.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 14:48
add a comment |
1
$begingroup$
Your derivation is correct.
$endgroup$
– Cardioid_Ass_22
Jan 14 at 14:17
$begingroup$
Thank you for your feedback, @Cardioid_Ass_22.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 14:48
1
1
$begingroup$
Your derivation is correct.
$endgroup$
– Cardioid_Ass_22
Jan 14 at 14:17
$begingroup$
Your derivation is correct.
$endgroup$
– Cardioid_Ass_22
Jan 14 at 14:17
$begingroup$
Thank you for your feedback, @Cardioid_Ass_22.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 14:48
$begingroup$
Thank you for your feedback, @Cardioid_Ass_22.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 14:48
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%2f3072931%2fhow-to-simplify-2-q-frack-12-n2-sigmaq-frack-12-sigman2%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%2f3072931%2fhow-to-simplify-2-q-frack-12-n2-sigmaq-frack-12-sigman2%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
1
$begingroup$
Your derivation is correct.
$endgroup$
– Cardioid_Ass_22
Jan 14 at 14:17
$begingroup$
Thank you for your feedback, @Cardioid_Ass_22.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 14:48