**#Problem 6**

The cubic equation x^3+2x-1 = 0 has exactly one real root r. Note that 0.4 < r < 0.5.

(a)Prove that an increasing sequence of positive integers a1,a2,a3,.....exists, such that 1/2=r^a1+r^a2+r^a3+.... is possible

(b) Prove that the sequence that you found in part (a) is the unique increasing sequence

with the above property.

with the above property.

**Solution Scheme and approach**

(a)

As r is a root then we may write

r^3+2r-1=0 =>1-r^3=2r=>r/(1-r^3)=1/2=>r*(1-r^3)^-1=1/2

By expanding

r*(1+r^3+r^6+r^9+....)=1/2

As |r|<1 the series converges

=>r+r^4+r^7+r^10+....=1/2 equating with r^a1+r^a2+r^a3+..=1/2

we can write {a1,a2,a3....}::{1,4,7....}

(b)Proof by contradiction

If possible let another for another series b1,b2,b3....

r^x1+r^x2+r^x3+...=r^y1+r^y2+r^y3+...=N [N>0]............(1)

without losing the generality we can consider y1>x1 =>r^y1>r^x1

r^x1<=r^x1+r^x2+r^x3+..=r^y1+r^y2+r^y3+....

Dividing both sides by r^x1

1<=r^p1+r^p2+r^p3+......[yi-xi=pi]...........(2)

As 0

1<=r^p1+r^p2+r^p3+...<=r+r^2+r^3+....=r/1-r=1/(1-r)-1<1

A contradiction.

Hence only one sequence exists.

## No comments:

## Post a Comment