Revision as of 22:41, 26 November 2023 by Admin (Created page with "'''Solution: B''' <math>K a_{\overline{12 \mid}}=2 K a_{\overline{4} \mid}</math> so <math>a_{\overline{12 \mid}}=2 a_{\overline{4} \mid}</math> and <math>\frac{1-(1+i)^{-12}}{i}=2 \frac{1-(1+i)^{-4}}{i}</math>. Let <math>x=(1+i)^{-4}</math>. Then <math>1-x^3=2(1-x)</math> so <math>x^3-2 x+1=0</math>. One root is clearly <math>x=1</math> so <math>(x-1)</math> is a factor. Thus <math>(x-1)\left(x^2+x-1\right)=0</math>. By the quadratic formula, <math>x=\frac{-1 \pm \sqr...")
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Exercise


Nov 26'23

Answer

Solution: B

[math]K a_{\overline{12 \mid}}=2 K a_{\overline{4} \mid}[/math] so [math]a_{\overline{12 \mid}}=2 a_{\overline{4} \mid}[/math] and [math]\frac{1-(1+i)^{-12}}{i}=2 \frac{1-(1+i)^{-4}}{i}[/math]. Let [math]x=(1+i)^{-4}[/math]. Then [math]1-x^3=2(1-x)[/math] so [math]x^3-2 x+1=0[/math]. One root is clearly [math]x=1[/math] so [math](x-1)[/math] is a factor.

Thus [math](x-1)\left(x^2+x-1\right)=0[/math]. By the quadratic formula, [math]x=\frac{-1 \pm \sqrt{1+4}}{2}[/math]. But [math]x=(1+i)^{-4}[/math] lies between 0 and 1 . So [math](1+i)^{-4}=x=\frac{-1+\sqrt{5}}{2}=.618034[/math]. Thus [math]i=(.618034)^{-1 / 4}-1=.12784=12.784 \%[/math].

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

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