calculation of sum, (1/6)*( sum_{n=-2 to 3} (1/2)* e^{-j (pi/3) k} )

avishaib

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Hi can someone explain me this?



16\displaystyle \dfrac{1}{6}\,n=23\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, (12ejπ3k)n=12521.5(1)k3ejπ3k\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}



Thank you
 
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Hi can someone explain me this?



16\displaystyle \dfrac{1}{6}\,n=23\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, (12ejπ3k)n=12521.5(1)k3ejπ3k\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}
The left-hand side is six terms. What did you get when you expanded the expression on the left-hand side, converted everything to a common denominator, and then simplified into one fractional expression?

Please be complete, including what summation formulas you applied. Thank you! ;)
 
Hi can someone explain me this?



16\displaystyle \dfrac{1}{6}\,n=23\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, (12ejπ3k)n=12521.5(1)k3ejπ3k\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}
You need to explain some notation to us. Do we assume that j2=1\displaystyle j^2=-1 rather than what mathematicians use which is i\displaystyle i.
If so then you need to know that eix=cos(x)+isin(x)\displaystyle e^{ix}=\cos(x)+i\sin(x).
Also you told us nothing about what k\displaystyle k might be.
 
Last edited:
Hi can someone explain me this?



16\displaystyle \dfrac{1}{6}\,n=23\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, (12ejπ3k)n=12521.5(1)k3ejπ3k\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}



Thank you
Let
x = 12ejπ3k\displaystyle \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}
and S be the sum. Then you have
S = 16x2Σn=5n=0xn\displaystyle \dfrac{1}{6\, x^2} \underset{n=0}{\overset{n=5}{\Sigma}}\, x^n
or
S = 16x21x61x\displaystyle \dfrac{1}{6\, x^2} \dfrac{1\, -\, x^6}{1\, -\, x}

Now simplify, i.e
x6 = 126[cos(6kπ3)jsin(6kπ3)]\displaystyle \dfrac{1}{2^6}\, [ cos(\dfrac{6\, k\, \pi}{3})\, -\, j\, sin(\dfrac{6\, k\, \pi}{3})\, ]
= 126\displaystyle \dfrac{1}{2^6}
etc. ..
 
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