Infinite summation $sum_limitsn=0^infty fracsin(ntheta)2^n$
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So one of my past exam papers has the question:
Show:
$$sum_limitsn=0^infty fracsin(ntheta)2^n = fracsin(theta)5-4cos(theta)$$
My working:
$e^intheta = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^intheta$
Thus we get: $$Imleft(sum_n=0^infty left(frace^itheta2right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac11-frace^itheta2$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac2sin(theta)5-4cos(theta)$$
Any help would be great, thanks!
complex-numbers summation
edited yesterday
Zacky
5,3611754
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
$endgroup$
add a comment |
$begingroup$
So one of my past exam papers has the question:
Show:
$$sum_limitsn=0^infty fracsin(ntheta)2^n = fracsin(theta)5-4cos(theta)$$
My working:
$e^intheta = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^intheta$
Thus we get: $$Imleft(sum_n=0^infty left(frace^itheta2right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac11-frace^itheta2$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac2sin(theta)5-4cos(theta)$$
Any help would be great, thanks!
complex-numbers summation
edited yesterday
Zacky
5,3611754
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
$endgroup$
add a comment |
$begingroup$
So one of my past exam papers has the question:
Show:
$$sum_limitsn=0^infty fracsin(ntheta)2^n = fracsin(theta)5-4cos(theta)$$
My working:
$e^intheta = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^intheta$
Thus we get: $$Imleft(sum_n=0^infty left(frace^itheta2right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac11-frace^itheta2$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac2sin(theta)5-4cos(theta)$$
Any help would be great, thanks!
complex-numbers summation
edited yesterday
Zacky
5,3611754
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
$endgroup$
So one of my past exam papers has the question:
Show:
$$sum_limitsn=0^infty fracsin(ntheta)2^n = fracsin(theta)5-4cos(theta)$$
My working:
$e^intheta = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^intheta$
Thus we get: $$Imleft(sum_n=0^infty left(frace^itheta2right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac11-frace^itheta2$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac2sin(theta)5-4cos(theta)$$
Any help would be great, thanks!
complex-numbers summation
complex-numbers summation
edited yesterday
Zacky
5,3611754
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
edited yesterday
Zacky
5,3611754
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
edited yesterday
Zacky
5,3611754
edited yesterday
Zacky
5,3611754
edited yesterday
Zacky
5,3611754
5,3611754
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
826
add a comment |
add a comment |
1 Answer
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Testing... if $theta=fracpi2$, the original series is $frac12^1-frac12^3+frac12^5-frac12^7+cdots =left(frac12-frac18right)left(1+frac12^4+frac12^8+cdotsright)$ which becomes $frac38cdotfrac1615=frac25$
The official answer claims $frac15-4cdot 0=frac15$. Yours claims $frac25-4cdot 0=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
3,592514
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Testing... if $theta=fracpi2$, the original series is $frac12^1-frac12^3+frac12^5-frac12^7+cdots =left(frac12-frac18right)left(1+frac12^4+frac12^8+cdotsright)$ which becomes $frac38cdotfrac1615=frac25$
The official answer claims $frac15-4cdot 0=frac15$. Yours claims $frac25-4cdot 0=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
3,592514
$endgroup$
add a comment |
$begingroup$
Testing... if $theta=fracpi2$, the original series is $frac12^1-frac12^3+frac12^5-frac12^7+cdots =left(frac12-frac18right)left(1+frac12^4+frac12^8+cdotsright)$ which becomes $frac38cdotfrac1615=frac25$
The official answer claims $frac15-4cdot 0=frac15$. Yours claims $frac25-4cdot 0=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
3,592514
$endgroup$
add a comment |
$begingroup$
Testing... if $theta=fracpi2$, the original series is $frac12^1-frac12^3+frac12^5-frac12^7+cdots =left(frac12-frac18right)left(1+frac12^4+frac12^8+cdotsright)$ which becomes $frac38cdotfrac1615=frac25$
The official answer claims $frac15-4cdot 0=frac15$. Yours claims $frac25-4cdot 0=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
3,592514
$endgroup$
Testing... if $theta=fracpi2$, the original series is $frac12^1-frac12^3+frac12^5-frac12^7+cdots =left(frac12-frac18right)left(1+frac12^4+frac12^8+cdotsright)$ which becomes $frac38cdotfrac1615=frac25$
The official answer claims $frac15-4cdot 0=frac15$. Yours claims $frac25-4cdot 0=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
3,592514
answered Jan 1 at 23:43
jmerryjmerry
3,592514
answered Jan 1 at 23:43
jmerryjmerry
3,592514
answered Jan 1 at 23:43
jmerryjmerry
3,592514
3,592514
add a comment |
add a comment |
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