$sum q^nsin(nx)$ complex analysis
up vote
-1
down vote
favorite
How to solve this one?
$$sum_{k=1}^n q^k sin(kx) = S(q,n)$$ where $n$ is natural number and $q$ is a real number
I already got the formula $$S(q,n) = Imleft(frac{qexp(ix)(1-q^nexp(nix))}{1-qexp(ix)}right)$$
but I don't know how to simplify the $1-qexp(ix)$ using the formulas
$$ sin(φ)=frac{exp(iφ)-exp(-iφ)}{2i}$$
$$ cos(φ)=frac{exp(iφ)+exp(-iφ)}{2},.$$
complex-analysis complex-numbers
asked Nov 22 at 7:51
Sergei Volkov
112
add a comment |
up vote
-1
down vote
favorite
How to solve this one?
$$sum_{k=1}^n q^k sin(kx) = S(q,n)$$ where $n$ is natural number and $q$ is a real number
I already got the formula $$S(q,n) = Imleft(frac{qexp(ix)(1-q^nexp(nix))}{1-qexp(ix)}right)$$
but I don't know how to simplify the $1-qexp(ix)$ using the formulas
$$ sin(φ)=frac{exp(iφ)-exp(-iφ)}{2i}$$
$$ cos(φ)=frac{exp(iφ)+exp(-iφ)}{2},.$$
complex-analysis complex-numbers
asked Nov 22 at 7:51
Sergei Volkov
112
$e^{ix}=cos, x+isin, x$. Hyperbolic functions are not involved.
– Kavi Rama Murthy
Nov 22 at 8:00
@SergeiVolkov: You have assumed $|qe^{ix}|<1$, but why?
– Yadati Kiran
Nov 22 at 9:10
Hm... I don't understand why you have the factor of $q e^{ix}$ in the numerator of $S_n$. The summation starts from 0 so the first term is equal to 1
– Mikalai Parshutsich
Nov 22 at 9:32
@MikalaiParshutsich: The first term is $0$ since $sin kx$ is $0$ when $k=0$ and has rewritten the summation $displaystylesum_{k=0}^n q^k sin(kx)=:text{Imaginary part}:left(displaystylesum_{k=0}^n q^k e^{ikx}right)$.
– Yadati Kiran
Nov 22 at 9:34
@YadatiKiran... Oh yeah! My mistake =(
– Mikalai Parshutsich
Nov 22 at 9:36
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How to solve this one?
$$sum_{k=1}^n q^k sin(kx) = S(q,n)$$ where $n$ is natural number and $q$ is a real number
I already got the formula $$S(q,n) = Imleft(frac{qexp(ix)(1-q^nexp(nix))}{1-qexp(ix)}right)$$
but I don't know how to simplify the $1-qexp(ix)$ using the formulas
$$ sin(φ)=frac{exp(iφ)-exp(-iφ)}{2i}$$
$$ cos(φ)=frac{exp(iφ)+exp(-iφ)}{2},.$$
complex-analysis complex-numbers
asked Nov 22 at 7:51
Sergei Volkov
112
How to solve this one?
$$sum_{k=1}^n q^k sin(kx) = S(q,n)$$ where $n$ is natural number and $q$ is a real number
I already got the formula $$S(q,n) = Imleft(frac{qexp(ix)(1-q^nexp(nix))}{1-qexp(ix)}right)$$
but I don't know how to simplify the $1-qexp(ix)$ using the formulas
$$ sin(φ)=frac{exp(iφ)-exp(-iφ)}{2i}$$
$$ cos(φ)=frac{exp(iφ)+exp(-iφ)}{2},.$$
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 22 at 7:51
Sergei Volkov
112
asked Nov 22 at 7:51
Sergei Volkov
112
edited Nov 22 at 22:03
asked Nov 22 at 7:51
Sergei Volkov
112
asked Nov 22 at 7:51
Sergei Volkov
112
asked Nov 22 at 7:51
Sergei Volkov
112
112
$e^{ix}=cos, x+isin, x$. Hyperbolic functions are not involved.
– Kavi Rama Murthy
Nov 22 at 8:00
@SergeiVolkov: You have assumed $|qe^{ix}|<1$, but why?
– Yadati Kiran
Nov 22 at 9:10
Hm... I don't understand why you have the factor of $q e^{ix}$ in the numerator of $S_n$. The summation starts from 0 so the first term is equal to 1
– Mikalai Parshutsich
Nov 22 at 9:32
@MikalaiParshutsich: The first term is $0$ since $sin kx$ is $0$ when $k=0$ and has rewritten the summation $displaystylesum_{k=0}^n q^k sin(kx)=:text{Imaginary part}:left(displaystylesum_{k=0}^n q^k e^{ikx}right)$.
– Yadati Kiran
Nov 22 at 9:34
@YadatiKiran... Oh yeah! My mistake =(
– Mikalai Parshutsich
Nov 22 at 9:36
add a comment |
$e^{ix}=cos, x+isin, x$. Hyperbolic functions are not involved.
– Kavi Rama Murthy
Nov 22 at 8:00
@SergeiVolkov: You have assumed $|qe^{ix}|<1$, but why?
– Yadati Kiran
Nov 22 at 9:10
Hm... I don't understand why you have the factor of $q e^{ix}$ in the numerator of $S_n$. The summation starts from 0 so the first term is equal to 1
– Mikalai Parshutsich
Nov 22 at 9:32
@MikalaiParshutsich: The first term is $0$ since $sin kx$ is $0$ when $k=0$ and has rewritten the summation $displaystylesum_{k=0}^n q^k sin(kx)=:text{Imaginary part}:left(displaystylesum_{k=0}^n q^k e^{ikx}right)$.
– Yadati Kiran
Nov 22 at 9:34
@YadatiKiran... Oh yeah! My mistake =(
– Mikalai Parshutsich
Nov 22 at 9:36
$e^{ix}=cos, x+isin, x$. Hyperbolic functions are not involved.
– Kavi Rama Murthy
Nov 22 at 8:00
$e^{ix}=cos, x+isin, x$. Hyperbolic functions are not involved.
– Kavi Rama Murthy
Nov 22 at 8:00
@SergeiVolkov: You have assumed $|qe^{ix}|<1$, but why?
– Yadati Kiran
Nov 22 at 9:10
@SergeiVolkov: You have assumed $|qe^{ix}|<1$, but why?
– Yadati Kiran
Nov 22 at 9:10
Hm... I don't understand why you have the factor of $q e^{ix}$ in the numerator of $S_n$. The summation starts from 0 so the first term is equal to 1
– Mikalai Parshutsich
Nov 22 at 9:32
Hm... I don't understand why you have the factor of $q e^{ix}$ in the numerator of $S_n$. The summation starts from 0 so the first term is equal to 1
– Mikalai Parshutsich
Nov 22 at 9:32
@MikalaiParshutsich: The first term is $0$ since $sin kx$ is $0$ when $k=0$ and has rewritten the summation $displaystylesum_{k=0}^n q^k sin(kx)=:text{Imaginary part}:left(displaystylesum_{k=0}^n q^k e^{ikx}right)$.
– Yadati Kiran
Nov 22 at 9:34
@MikalaiParshutsich: The first term is $0$ since $sin kx$ is $0$ when $k=0$ and has rewritten the summation $displaystylesum_{k=0}^n q^k sin(kx)=:text{Imaginary part}:left(displaystylesum_{k=0}^n q^k e^{ikx}right)$.
– Yadati Kiran
Nov 22 at 9:34
@YadatiKiran... Oh yeah! My mistake =(
– Mikalai Parshutsich
Nov 22 at 9:36
@YadatiKiran... Oh yeah! My mistake =(
– Mikalai Parshutsich
Nov 22 at 9:36
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Let me recast your formula using $n-1$ instead of $n$, for convenience. First, we reduce the calculation to the sum of a geometric progression
$$
sum_{k=0}^{n-1} q^k sin(kx) = Im sum_{k=0}^{n-1}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}},,$$
where we have used
$$
sum_{k=0}^{n-1} s^k = frac{1-s^n}{1-s}
$$
(this is easily proved for $sneq 1$ multiplying both sides by $1-s$ and noting the telescopic sum on the left-hand side; for $sto1$, the right-hand side reduces to $n$).
Now,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2},,
$$
whose imaginary part is
$$
frac{qsin x(1-q^ncos(nx))-q^n sin(nx) (1-qcos x)}{(1-qcos x)^2+(q sin x)^2},.
$$
Simplifying slightly, by means of trigonometric formulas,
$$
sum_{k=0}^{n-1} q^k sin(kx) = frac{qsin x - q^n sin(nx)+q^{n+1}sin((n-1)x)}{1-2qcos x+q^2},.
$$
answered Nov 22 at 9:43
Brightsun
3,85711532
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
add a comment |
up vote
0
down vote
Well, I have this solution for the problem:
$$
sum_{k=1}^{n} q^k sin(kx) = Im sum_{k1}^{n}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} ,,
$$
as a geometric series (I don't know why $|q, e^{ix}|<1$ as you said).
Then,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}q, e^{ix}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2}q, (cosx+isinx),,
$$
and
then simplifying the last expressions just by opening the brackets and then using the formula for the $sin$ of the sum we get
$$
sum_{k=1}^{n} q^k sin(kx) = qfrac{q^{n+1}sin(nx)-q^{n}sin((n+1)x)+sin x}{q^2-2qcos x+1}, .
$$
Sorry, I firstly haven't caught that in the sum k is going not from the 0 but from 1, that is my mistake.
edited Nov 23 at 14:08
Brightsun
3,85711532
answered Nov 22 at 22:02
Sergei Volkov
112
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
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',
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%2f3008860%2fsum-qn-sinnx-complex-analysis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let me recast your formula using $n-1$ instead of $n$, for convenience. First, we reduce the calculation to the sum of a geometric progression
$$
sum_{k=0}^{n-1} q^k sin(kx) = Im sum_{k=0}^{n-1}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}},,$$
where we have used
$$
sum_{k=0}^{n-1} s^k = frac{1-s^n}{1-s}
$$
(this is easily proved for $sneq 1$ multiplying both sides by $1-s$ and noting the telescopic sum on the left-hand side; for $sto1$, the right-hand side reduces to $n$).
Now,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2},,
$$
whose imaginary part is
$$
frac{qsin x(1-q^ncos(nx))-q^n sin(nx) (1-qcos x)}{(1-qcos x)^2+(q sin x)^2},.
$$
Simplifying slightly, by means of trigonometric formulas,
$$
sum_{k=0}^{n-1} q^k sin(kx) = frac{qsin x - q^n sin(nx)+q^{n+1}sin((n-1)x)}{1-2qcos x+q^2},.
$$
answered Nov 22 at 9:43
Brightsun
3,85711532
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
add a comment |
up vote
1
down vote
Let me recast your formula using $n-1$ instead of $n$, for convenience. First, we reduce the calculation to the sum of a geometric progression
$$
sum_{k=0}^{n-1} q^k sin(kx) = Im sum_{k=0}^{n-1}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}},,$$
where we have used
$$
sum_{k=0}^{n-1} s^k = frac{1-s^n}{1-s}
$$
(this is easily proved for $sneq 1$ multiplying both sides by $1-s$ and noting the telescopic sum on the left-hand side; for $sto1$, the right-hand side reduces to $n$).
Now,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2},,
$$
whose imaginary part is
$$
frac{qsin x(1-q^ncos(nx))-q^n sin(nx) (1-qcos x)}{(1-qcos x)^2+(q sin x)^2},.
$$
Simplifying slightly, by means of trigonometric formulas,
$$
sum_{k=0}^{n-1} q^k sin(kx) = frac{qsin x - q^n sin(nx)+q^{n+1}sin((n-1)x)}{1-2qcos x+q^2},.
$$
answered Nov 22 at 9:43
Brightsun
3,85711532
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
add a comment |
up vote
1
down vote
up vote
1
down vote
Let me recast your formula using $n-1$ instead of $n$, for convenience. First, we reduce the calculation to the sum of a geometric progression
$$
sum_{k=0}^{n-1} q^k sin(kx) = Im sum_{k=0}^{n-1}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}},,$$
where we have used
$$
sum_{k=0}^{n-1} s^k = frac{1-s^n}{1-s}
$$
(this is easily proved for $sneq 1$ multiplying both sides by $1-s$ and noting the telescopic sum on the left-hand side; for $sto1$, the right-hand side reduces to $n$).
Now,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2},,
$$
whose imaginary part is
$$
frac{qsin x(1-q^ncos(nx))-q^n sin(nx) (1-qcos x)}{(1-qcos x)^2+(q sin x)^2},.
$$
Simplifying slightly, by means of trigonometric formulas,
$$
sum_{k=0}^{n-1} q^k sin(kx) = frac{qsin x - q^n sin(nx)+q^{n+1}sin((n-1)x)}{1-2qcos x+q^2},.
$$
answered Nov 22 at 9:43
Brightsun
3,85711532
Let me recast your formula using $n-1$ instead of $n$, for convenience. First, we reduce the calculation to the sum of a geometric progression
$$
sum_{k=0}^{n-1} q^k sin(kx) = Im sum_{k=0}^{n-1}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}},,$$
where we have used
$$
sum_{k=0}^{n-1} s^k = frac{1-s^n}{1-s}
$$
(this is easily proved for $sneq 1$ multiplying both sides by $1-s$ and noting the telescopic sum on the left-hand side; for $sto1$, the right-hand side reduces to $n$).
Now,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2},,
$$
whose imaginary part is
$$
frac{qsin x(1-q^ncos(nx))-q^n sin(nx) (1-qcos x)}{(1-qcos x)^2+(q sin x)^2},.
$$
Simplifying slightly, by means of trigonometric formulas,
$$
sum_{k=0}^{n-1} q^k sin(kx) = frac{qsin x - q^n sin(nx)+q^{n+1}sin((n-1)x)}{1-2qcos x+q^2},.
$$
answered Nov 22 at 9:43
Brightsun
3,85711532
answered Nov 22 at 9:43
Brightsun
3,85711532
answered Nov 22 at 9:43
Brightsun
3,85711532
answered Nov 22 at 9:43
Brightsun
3,85711532
3,85711532
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
add a comment |
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
thank you a lot
– Sergei Volkov
Nov 22 at 19:57
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
Hi @SergeiVolkov welcome to Stack Exchange. In case you find that an answer indeed fits the question you have asked, it is customary to accept it by clicking on the "tick" symbol on the left.
– Brightsun
Nov 23 at 14:05
add a comment |
up vote
0
down vote
Well, I have this solution for the problem:
$$
sum_{k=1}^{n} q^k sin(kx) = Im sum_{k1}^{n}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} ,,
$$
as a geometric series (I don't know why $|q, e^{ix}|<1$ as you said).
Then,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}q, e^{ix}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2}q, (cosx+isinx),,
$$
and
then simplifying the last expressions just by opening the brackets and then using the formula for the $sin$ of the sum we get
$$
sum_{k=1}^{n} q^k sin(kx) = qfrac{q^{n+1}sin(nx)-q^{n}sin((n+1)x)+sin x}{q^2-2qcos x+1}, .
$$
Sorry, I firstly haven't caught that in the sum k is going not from the 0 but from 1, that is my mistake.
edited Nov 23 at 14:08
Brightsun
3,85711532
answered Nov 22 at 22:02
Sergei Volkov
112
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
add a comment |
up vote
0
down vote
Well, I have this solution for the problem:
$$
sum_{k=1}^{n} q^k sin(kx) = Im sum_{k1}^{n}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} ,,
$$
as a geometric series (I don't know why $|q, e^{ix}|<1$ as you said).
Then,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}q, e^{ix}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2}q, (cosx+isinx),,
$$
and
then simplifying the last expressions just by opening the brackets and then using the formula for the $sin$ of the sum we get
$$
sum_{k=1}^{n} q^k sin(kx) = qfrac{q^{n+1}sin(nx)-q^{n}sin((n+1)x)+sin x}{q^2-2qcos x+1}, .
$$
Sorry, I firstly haven't caught that in the sum k is going not from the 0 but from 1, that is my mistake.
edited Nov 23 at 14:08
Brightsun
3,85711532
answered Nov 22 at 22:02
Sergei Volkov
112
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
add a comment |
up vote
0
down vote
up vote
0
down vote
Well, I have this solution for the problem:
$$
sum_{k=1}^{n} q^k sin(kx) = Im sum_{k1}^{n}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} ,,
$$
as a geometric series (I don't know why $|q, e^{ix}|<1$ as you said).
Then,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}q, e^{ix}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2}q, (cosx+isinx),,
$$
and
then simplifying the last expressions just by opening the brackets and then using the formula for the $sin$ of the sum we get
$$
sum_{k=1}^{n} q^k sin(kx) = qfrac{q^{n+1}sin(nx)-q^{n}sin((n+1)x)+sin x}{q^2-2qcos x+1}, .
$$
Sorry, I firstly haven't caught that in the sum k is going not from the 0 but from 1, that is my mistake.
edited Nov 23 at 14:08
Brightsun
3,85711532
answered Nov 22 at 22:02
Sergei Volkov
112
Well, I have this solution for the problem:
$$
sum_{k=1}^{n} q^k sin(kx) = Im sum_{k1}^{n}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} ,,
$$
as a geometric series (I don't know why $|q, e^{ix}|<1$ as you said).
Then,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}q, e^{ix}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2}q, (cosx+isinx),,
$$
and
then simplifying the last expressions just by opening the brackets and then using the formula for the $sin$ of the sum we get
$$
sum_{k=1}^{n} q^k sin(kx) = qfrac{q^{n+1}sin(nx)-q^{n}sin((n+1)x)+sin x}{q^2-2qcos x+1}, .
$$
Sorry, I firstly haven't caught that in the sum k is going not from the 0 but from 1, that is my mistake.
edited Nov 23 at 14:08
Brightsun
3,85711532
answered Nov 22 at 22:02
Sergei Volkov
112
edited Nov 23 at 14:08
Brightsun
3,85711532
edited Nov 23 at 14:08
Brightsun
3,85711532
edited Nov 23 at 14:08
Brightsun
3,85711532
3,85711532
answered Nov 22 at 22:02
Sergei Volkov
112
answered Nov 22 at 22:02
Sergei Volkov
112
answered Nov 22 at 22:02
Sergei Volkov
112
112
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
add a comment |
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
Even if you let the sum start from $k=1$ instead of $k=0$, what you are doing is nothing but subtracting 1 (a real number) and hence you do not alter the imaginary part and the result (if you will, the imaginary part of 1 is 0). Secondly, at no point there is an assumption on the absolute value of $q e^{ix}$, since you are considering a finite sum! The formula $sum_{j=0}^{n-1} a^j=(1-a^{n})/(1-a)$ is valid for any $aneq 1$ and the sum just gives $n$ as $ato1$.
– Brightsun
Nov 23 at 14:12
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
thank you for your explanation, I'll try to understand this thing
– Sergei Volkov
Nov 23 at 20:30
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.
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%2f3008860%2fsum-qn-sinnx-complex-analysis%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
$e^{ix}=cos, x+isin, x$. Hyperbolic functions are not involved.
– Kavi Rama Murthy
Nov 22 at 8:00
@SergeiVolkov: You have assumed $|qe^{ix}|<1$, but why?
– Yadati Kiran
Nov 22 at 9:10
Hm... I don't understand why you have the factor of $q e^{ix}$ in the numerator of $S_n$. The summation starts from 0 so the first term is equal to 1
– Mikalai Parshutsich
Nov 22 at 9:32
@MikalaiParshutsich: The first term is $0$ since $sin kx$ is $0$ when $k=0$ and has rewritten the summation $displaystylesum_{k=0}^n q^k sin(kx)=:text{Imaginary part}:left(displaystylesum_{k=0}^n q^k e^{ikx}right)$.
– Yadati Kiran
Nov 22 at 9:34
@YadatiKiran... Oh yeah! My mistake =(
– Mikalai Parshutsich
Nov 22 at 9:36