double sigma in complex numberIf complex no. ($z$) satisfying $frac12leq |z|leq 4;,$ Then Max. and Min. of $left|z+frac1zright|$complex no. $z$ such that $ |z|<frac13$ and $ sum^n_r=1a_rz^r = 1;,$ Where $|a_r|<2$Product of one minus the tenth roots of unityComplex number (cube roots of unity)Summation over roots of unityMaximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$.Proving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityEvaluating the complex summationvalue of $|b_1|+|b_2|+cdots cdots +|b_k|$ isProve that $|z_1+z_2|^2leq (1+c)|z_1|^2+bigg(1+frac1cbigg)|z_2|^2$
How strictly should I take "Candidates must be local"?
What's the "normal" opposite of flautando?
They call me Inspector Morse
Distinction between apt-cache and dpkg -l
What are some noteworthy "mic-drop" moments in math?
Does a warlock using the Darkness/Devil's Sight combo still have advantage on ranged attacks against a target outside the Darkness?
Conservation of Mass and Energy
Error during using callback start_page_number in lualatex
Intuition behind counterexample of Euler's sum of powers conjecture
Bash script should only kill those instances of another script's that it has launched
PTIJ: wiping amalek’s memory?
How did Alan Turing break the enigma code using the hint given by the lady in the bar?
In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?
Is "conspicuously missing" or "conspicuously" the subject of this sentence?
When traveling to Europe from North America, do I need to purchase a different power strip?
How can The Temple of Elementary Evil reliably protect itself against kinetic bombardment?
Recommendation letter by significant other if you worked with them professionally?
How do I express some one as a black person?
Accepted offer letter, position changed
Hotkey (or other quick way) to insert a keyframe for only one component of a vector-valued property?
Should I tell my boss the work he did was worthless
Are all players supposed to be able to see each others' character sheets?
In the late 1940’s to early 1950’s what technology was available that could melt a LOT of ice?
Declaring and defining template, and specialising them
double sigma in complex number
If complex no. ($z$) satisfying $frac12leq |z|leq 4;,$ Then Max. and Min. of $left|z+frac1zright|$complex no. $z$ such that $ |z|<frac13$ and $ sum^n_r=1a_rz^r = 1;,$ Where $|a_r|<2$Product of one minus the tenth roots of unityComplex number (cube roots of unity)Summation over roots of unityMaximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$.Proving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityEvaluating the complex summationvalue of $|b_1|+|b_2|+cdots cdots +|b_k|$ isProve that $|z_1+z_2|^2leq (1+c)|z_1|^2+bigg(1+frac1cbigg)|z_2|^2$
$begingroup$
if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is
what i try
assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots
$x=beta_1,beta_2,cdots,beta_100$
$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$
how do i solve it help me please
complex-numbers
asked Mar 7 at 14:36
jackyjacky
1,028715
$endgroup$
add a comment |
$begingroup$
if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is
what i try
assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots
$x=beta_1,beta_2,cdots,beta_100$
$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$
how do i solve it help me please
complex-numbers
asked Mar 7 at 14:36
jackyjacky
1,028715
$endgroup$
$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51
$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01
1
$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50
$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20
$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32
add a comment |
$begingroup$
if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is
what i try
assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots
$x=beta_1,beta_2,cdots,beta_100$
$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$
how do i solve it help me please
complex-numbers
asked Mar 7 at 14:36
jackyjacky
1,028715
$endgroup$
if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is
what i try
assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots
$x=beta_1,beta_2,cdots,beta_100$
$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$
how do i solve it help me please
complex-numbers
complex-numbers
asked Mar 7 at 14:36
jackyjacky
1,028715
asked Mar 7 at 14:36
jackyjacky
1,028715
asked Mar 7 at 14:36
jackyjacky
1,028715
asked Mar 7 at 14:36
jackyjacky
1,028715
asked Mar 7 at 14:36
jackyjacky
1,028715
1,028715
$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51
$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01
1
$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50
$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20
$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32
add a comment |
$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51
$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01
1
$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50
$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20
$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32
$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51
$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51
$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01
$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01
1
1
$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50
$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50
$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20
$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20
$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32
$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We obtain
beginalign*
colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
&=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
&=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
&=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
&=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
&=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
&,,colorblue=0
endalign*
Comment:
In (1) we rewrite the sum as double sum and factor out $beta_j$.
In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.
In (3) we apply the finite geometric series formula.
In (4) we factor out common terms and multiply out the sums.
In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.
Note:
Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
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%2f3138939%2fdouble-sigma-in-complex-number%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We obtain
beginalign*
colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
&=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
&=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
&=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
&=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
&=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
&,,colorblue=0
endalign*
Comment:
In (1) we rewrite the sum as double sum and factor out $beta_j$.
In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.
In (3) we apply the finite geometric series formula.
In (4) we factor out common terms and multiply out the sums.
In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.
Note:
Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
add a comment |
$begingroup$
We obtain
beginalign*
colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
&=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
&=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
&=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
&=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
&=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
&,,colorblue=0
endalign*
Comment:
In (1) we rewrite the sum as double sum and factor out $beta_j$.
In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.
In (3) we apply the finite geometric series formula.
In (4) we factor out common terms and multiply out the sums.
In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.
Note:
Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
add a comment |
$begingroup$
We obtain
beginalign*
colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
&=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
&=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
&=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
&=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
&=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
&,,colorblue=0
endalign*
Comment:
In (1) we rewrite the sum as double sum and factor out $beta_j$.
In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.
In (3) we apply the finite geometric series formula.
In (4) we factor out common terms and multiply out the sums.
In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.
Note:
Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
We obtain
beginalign*
colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
&=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
&=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
&=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
&=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
&=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
&,,colorblue=0
endalign*
Comment:
In (1) we rewrite the sum as double sum and factor out $beta_j$.
In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.
In (3) we apply the finite geometric series formula.
In (4) we factor out common terms and multiply out the sums.
In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.
Note:
Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
answered 2 days ago
Markus ScheuerMarkus Scheuer
62.6k459149
62.6k459149
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%2f3138939%2fdouble-sigma-in-complex-number%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 sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51
$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01
1
$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50
$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20
$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32