If p=13+1336+133659+⋯and p2+ap+c=0.Find a, c also |c|=2.

Question

Genesis Reilly · Accepted Answer

p=13+1⋅31⋅2⋅(13)2+1⋅3⋅51⋅2⋅3⋅(13)3+⋯So&amp;nbsp;p+1=1+13+1⋅31⋅2⋅(13)2+1⋅3⋅51⋅2⋅3⋅(13)3+⋯Now notice that(1+x)n=1+nx+n(n−1)2!x2+⋯We can assume that&amp;nbsp;nx=13and&amp;nbsp;n(n−1)2!x2=1⋅31⋅2⋅(13)2and we will get&amp;nbsp;n-1n=3⟹n=-12⟹x=-23Therefore we have1+p=11+x=3

Elliot Roy · Accepted Answer

Since&amp;nbsp;p=13∑m≥02m+1m6m&amp;nbsp;and&amp;nbsp;nk=∮|z|=1(1+z)ndz2πizk+1, simplifying a geometric-series integrand eventually givesp=−2∮|z|=11+z1−4z+z2&amp;nbsp;dz&amp;nbsp;2πi.The denominator&#039;s roots are&amp;nbsp;2±3, but only&amp;nbsp;2−3&amp;nbsp;is enclosed. Sop=−2limz→2−31+zz−2−3=3−1.(More generally, the above technique shows&amp;nbsp;∑m≥02m+1mxm=(1-4x)-1/2-12x&amp;nbsp;for&amp;nbsp;|x|&amp;lt;14.) If we assume a,&amp;nbsp;c∈Z, the sought quadratic has roots&amp;nbsp;−1±3&amp;nbsp;so that&amp;nbsp;a=2,c=−2&amp;nbsp;because the quadratic is&amp;nbsp;(p+1)2−3=0.

If p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots and p^2+ap+c=0. Find a, c also |c|=2.

Answered question

Answer & Explanation

New Questions in Algebra I