Sum of two infinite complex geometric sums
$begingroup$
let $$C=fraccostheta2-fraccos2theta4+fraccos3theta8+...$$
$$S=fracsintheta2-fracsin2theta4+fracsin3theta8+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.
so the sum of $C+iS$ is
$$C+iS=(fraccostheta2-fraccos2theta4+fraccos3theta8+...)+i(fracsintheta2-fracsin2theta4+fracsin3theta8+...)$$
$$=frac12(costheta+isintheta)-frac14(cos2theta+isin2theta)+frac18(cos3theta+isin3theta) + ...$$
$$=frac12(costheta+isintheta)-frac14(costheta+isintheta)^2+frac18(costheta+isintheta)^3 + ...$$
So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?
So I've continued and got the series to here:
$$frac12[e^itheta-frac12(e^itheta)^2+frac14(e^itheta)^3...]$$ But I'm still not sure where to continue.
sequences-and-series complex-numbers
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
$endgroup$
add a comment |
$begingroup$
let $$C=fraccostheta2-fraccos2theta4+fraccos3theta8+...$$
$$S=fracsintheta2-fracsin2theta4+fracsin3theta8+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.
so the sum of $C+iS$ is
$$C+iS=(fraccostheta2-fraccos2theta4+fraccos3theta8+...)+i(fracsintheta2-fracsin2theta4+fracsin3theta8+...)$$
$$=frac12(costheta+isintheta)-frac14(cos2theta+isin2theta)+frac18(cos3theta+isin3theta) + ...$$
$$=frac12(costheta+isintheta)-frac14(costheta+isintheta)^2+frac18(costheta+isintheta)^3 + ...$$
So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?
So I've continued and got the series to here:
$$frac12[e^itheta-frac12(e^itheta)^2+frac14(e^itheta)^3...]$$ But I'm still not sure where to continue.
sequences-and-series complex-numbers
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
$endgroup$
$begingroup$
Use $costheta+isintheta=e^itheta$ and factor out $frac12$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21
add a comment |
$begingroup$
let $$C=fraccostheta2-fraccos2theta4+fraccos3theta8+...$$
$$S=fracsintheta2-fracsin2theta4+fracsin3theta8+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.
so the sum of $C+iS$ is
$$C+iS=(fraccostheta2-fraccos2theta4+fraccos3theta8+...)+i(fracsintheta2-fracsin2theta4+fracsin3theta8+...)$$
$$=frac12(costheta+isintheta)-frac14(cos2theta+isin2theta)+frac18(cos3theta+isin3theta) + ...$$
$$=frac12(costheta+isintheta)-frac14(costheta+isintheta)^2+frac18(costheta+isintheta)^3 + ...$$
So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?
So I've continued and got the series to here:
$$frac12[e^itheta-frac12(e^itheta)^2+frac14(e^itheta)^3...]$$ But I'm still not sure where to continue.
sequences-and-series complex-numbers
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
$endgroup$
let $$C=fraccostheta2-fraccos2theta4+fraccos3theta8+...$$
$$S=fracsintheta2-fracsin2theta4+fracsin3theta8+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.
so the sum of $C+iS$ is
$$C+iS=(fraccostheta2-fraccos2theta4+fraccos3theta8+...)+i(fracsintheta2-fracsin2theta4+fracsin3theta8+...)$$
$$=frac12(costheta+isintheta)-frac14(cos2theta+isin2theta)+frac18(cos3theta+isin3theta) + ...$$
$$=frac12(costheta+isintheta)-frac14(costheta+isintheta)^2+frac18(costheta+isintheta)^3 + ...$$
So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?
So I've continued and got the series to here:
$$frac12[e^itheta-frac12(e^itheta)^2+frac14(e^itheta)^3...]$$ But I'm still not sure where to continue.
sequences-and-series complex-numbers
sequences-and-series complex-numbers
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
edited Nov 15 '18 at 15:39
H.Linkhorn
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
asked Nov 15 '18 at 11:14
H.LinkhornH.Linkhorn
537313
537313
$begingroup$
Use $costheta+isintheta=e^itheta$ and factor out $frac12$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21
add a comment |
$begingroup$
Use $costheta+isintheta=e^itheta$ and factor out $frac12$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21
$begingroup$
Use $costheta+isintheta=e^itheta$ and factor out $frac12$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21
$begingroup$
Use $costheta+isintheta=e^itheta$ and factor out $frac12$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT
We have that
$$C+iS=sum_k=1^inftyfrac(-1)^k+12^ke^left(ikthetaright)
=-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k$$
then refer to geometric series which holds also for $r$ complex $|r|<1$.
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
$endgroup$
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
add a comment |
$begingroup$
Hint:
Using Intuition behind euler's formula and the special case $e^ipi=-1$
$$dfrac(cos t+isin t)^n(-1)^n-12^n=-dfrace^int(e^ipi)^n2^n=-left(dfrace^i(t+pi)2right)^n$$
Now for the common ratio$(r)$ of Geometric Series,
$|r|=left|dfrace^i(t+pi)2right|=dfrac12<1$
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
We have that
$$C+iS=sum_k=1^inftyfrac(-1)^k+12^ke^left(ikthetaright)
=-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k$$
then refer to geometric series which holds also for $r$ complex $|r|<1$.
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
$endgroup$
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
add a comment |
$begingroup$
HINT
We have that
$$C+iS=sum_k=1^inftyfrac(-1)^k+12^ke^left(ikthetaright)
=-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k$$
then refer to geometric series which holds also for $r$ complex $|r|<1$.
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
$endgroup$
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
add a comment |
$begingroup$
HINT
We have that
$$C+iS=sum_k=1^inftyfrac(-1)^k+12^ke^left(ikthetaright)
=-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k$$
then refer to geometric series which holds also for $r$ complex $|r|<1$.
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
$endgroup$
HINT
We have that
$$C+iS=sum_k=1^inftyfrac(-1)^k+12^ke^left(ikthetaright)
=-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k$$
then refer to geometric series which holds also for $r$ complex $|r|<1$.
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
edited Nov 15 '18 at 11:28
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
answered Nov 15 '18 at 11:21
gimusigimusi
93k84594
93k84594
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
add a comment |
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $fracexp(itheta)+1exp(itheta)+2exp(-itheta)+3 $which isn't the correct answer. I was told i should be $frac2exp(itheta)+15+4costheta$
$endgroup$
– H.Linkhorn
Nov 17 '18 at 18:31
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
$begingroup$
@H.Linkhorn We have that $$-sum_k=1^inftyleft(-frace^left(ithetaright)2right)^k=-frac11+frace^itheta2+1=frace^itheta2+e^itheta$$ Then multiply by $2+e^-itheta$ denominator and numerator to obtain teh result.
$endgroup$
– gimusi
Nov 17 '18 at 21:36
add a comment |
$begingroup$
Hint:
Using Intuition behind euler's formula and the special case $e^ipi=-1$
$$dfrac(cos t+isin t)^n(-1)^n-12^n=-dfrace^int(e^ipi)^n2^n=-left(dfrace^i(t+pi)2right)^n$$
Now for the common ratio$(r)$ of Geometric Series,
$|r|=left|dfrace^i(t+pi)2right|=dfrac12<1$
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
add a comment |
$begingroup$
Hint:
Using Intuition behind euler's formula and the special case $e^ipi=-1$
$$dfrac(cos t+isin t)^n(-1)^n-12^n=-dfrace^int(e^ipi)^n2^n=-left(dfrace^i(t+pi)2right)^n$$
Now for the common ratio$(r)$ of Geometric Series,
$|r|=left|dfrace^i(t+pi)2right|=dfrac12<1$
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
add a comment |
$begingroup$
Hint:
Using Intuition behind euler's formula and the special case $e^ipi=-1$
$$dfrac(cos t+isin t)^n(-1)^n-12^n=-dfrace^int(e^ipi)^n2^n=-left(dfrace^i(t+pi)2right)^n$$
Now for the common ratio$(r)$ of Geometric Series,
$|r|=left|dfrace^i(t+pi)2right|=dfrac12<1$
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
Hint:
Using Intuition behind euler's formula and the special case $e^ipi=-1$
$$dfrac(cos t+isin t)^n(-1)^n-12^n=-dfrace^int(e^ipi)^n2^n=-left(dfrace^i(t+pi)2right)^n$$
Now for the common ratio$(r)$ of Geometric Series,
$|r|=left|dfrace^i(t+pi)2right|=dfrac12<1$
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
answered Nov 15 '18 at 11:21
lab bhattacharjeelab bhattacharjee
229k15159279
229k15159279
add a comment |
add a comment |
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$begingroup$
Use $costheta+isintheta=e^itheta$ and factor out $frac12$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21