Mixing
Infinite summation $sum_limits{n=0}^infty frac{sin(ntheta)}{2^n}$
$begingroup$
So one of my past exam papers has the question:
Show:
$$sum_limits{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$$
My working:
$e^{intheta} = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^{intheta}$
Thus we get: $$Imleft(sum_{n=0}^infty left(frac{e^{itheta}}{2}right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac{1}{1-frac{e^{itheta}}{2}}$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac{2sin(theta)}{5-4cos(theta)}$$
Any help would be great, thanks!
complex-numbers summation
edited Jan 13 at 12:38
Zacky
6,5701958
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
$endgroup$
add a comment |
$begingroup$
So one of my past exam papers has the question:
Show:
$$sum_limits{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$$
My working:
$e^{intheta} = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^{intheta}$
Thus we get: $$Imleft(sum_{n=0}^infty left(frac{e^{itheta}}{2}right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac{1}{1-frac{e^{itheta}}{2}}$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac{2sin(theta)}{5-4cos(theta)}$$
Any help would be great, thanks!
complex-numbers summation
edited Jan 13 at 12:38
Zacky
6,5701958
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
$endgroup$
add a comment |
$begingroup$
So one of my past exam papers has the question:
Show:
$$sum_limits{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$$
My working:
$e^{intheta} = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^{intheta}$
Thus we get: $$Imleft(sum_{n=0}^infty left(frac{e^{itheta}}{2}right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac{1}{1-frac{e^{itheta}}{2}}$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac{2sin(theta)}{5-4cos(theta)}$$
Any help would be great, thanks!
complex-numbers summation
edited Jan 13 at 12:38
Zacky
6,5701958
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
$endgroup$
So one of my past exam papers has the question:
Show:
$$sum_limits{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$$
My working:
$e^{intheta} = cos(ntheta) + isin(ntheta)$
So $sin(ntheta)$ is the imaginary part of $e^{intheta}$
Thus we get: $$Imleft(sum_{n=0}^infty left(frac{e^{itheta}}{2}right)^nright)$$
Using the summation formula for infinite geometric series gives:
$$frac{1}{1-frac{e^{itheta}}{2}}$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$frac{2sin(theta)}{5-4cos(theta)}$$
Any help would be great, thanks!
complex-numbers summation
complex-numbers summation
edited Jan 13 at 12:38
Zacky
6,5701958
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
edited Jan 13 at 12:38
Zacky
6,5701958
asked Jan 1 at 23:29
PolynomialCPolynomialC
826
edited Jan 13 at 12:38
Zacky
6,5701958
edited Jan 13 at 12:38
Zacky
6,5701958
edited Jan 13 at 12:38
Zacky
6,5701958
6,5701958
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
1
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$begingroup$
Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$
The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
$begingroup$
Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$
The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
$endgroup$
add a comment |
$begingroup$
Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$
The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
$endgroup$
add a comment |
$begingroup$
Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$
The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
$endgroup$
Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$
The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
answered Jan 1 at 23:43
jmerryjmerry
8,4981022
8,4981022
add a comment |
add a comment |
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