Series Riddle

Calculus Level pending

∣ ∑ n = 0 ∞ 2 − n i ∣ 2 = ? \large \left | \sum_{n=0}^\infty 2^{-ni} \right | ^2 = \, ?

Clarification : i = − 1 i=\sqrt{-1} .

1 2 − 2 cos ⁡ ln ⁡ 2 \frac{1}{2-2\cos\ln2} 0 1 1 2 + 2 sin ⁡ ln ⁡ 2 \frac{1}{2+2\sin\ln2} 1 2 − 2 sin ⁡ ln ⁡ 2 \frac{1}{2-2\sin\ln2} 1 2 + 2 cos ⁡ ln ⁡ 2 \frac{1}{2+2\cos\ln2}

Bar Hemo by Bar Hemo , 26, Israel

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1 solution

Bar Hemo
Mar 17, 2016

∣ ∑ n = 0 ∞ ( 2 ) − n j ∣ = ∣ 1 1 − 2 − j ∣ = ∣ 1 1 − e − j ln ⁡ 2 ∣ = ∣ 1 1 − cos ⁡ ln ⁡ 2 + j sin ⁡ ln ⁡ 2 ∣ = \left|\displaystyle\sum_{n=0}^{\infty} (2)^{-nj}\right| =\left|\dfrac{1}{1-2^{-j}}\right|= \left|\dfrac{1}{1-e^{-j\ln2}}\right|=\left|\dfrac{1}{1-\cos\ln2+j\sin\ln2}\right|=

1 ( 1 − cos ⁡ ln ⁡ 2 ) 2 + ( sin ⁡ ln ⁡ 2 ) 2 = 1 1 − 2 cos ⁡ ln ⁡ 2 + ( cos ⁡ ln ⁡ 2 ) 2 + ( sin ⁡ ln ⁡ 2 ) 2 = 1 2 − 2 cos ⁡ ln ⁡ 2 \dfrac{1}{(1-\cos\ln2)^2+(\sin\ln2)^2}=\dfrac{1}{1-2\cos\ln2+(\cos\ln2)^2+(\sin\ln2)^2}=\dfrac{1}{2-2\cos\ln2}

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Note that the absolute value of the complex number z = a + b i z = a + bi is a 2 + b 2 \sqrt{ a^2+b^2} . I have added a ^square to your problem.

Calvin Lin Staff - 5 years, 2 months ago

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