Attempt at Challenge problems for Math 504

Attempt at Challenge problems for Math 504

Nasser M. Abbasi

June 16, 2014

1 Problem 1

Solution:

We start with the solution we already¹ obtained for $E(Yi)$ which is

i−1 E (Yi) = 1+ ∑ E (Yk)Pik k=1

Let $E (Yi) = xi$ hence the above can be written as

i− 1 x = 1+ x P i ∑k=1k ik

But $Pik = 1 i$ then the above becomes

1i−1 xi = 1+ i∑k=1xk

Multiply by $i$ the above becomes

i−1 i xi= i+ ∑ xk k=1

Therefore, we obtain the following equations for $i= 1⋅⋅⋅r$

For $i= 1$

x1 = 1

(1)

For $i= 2$

2x2 = 2+ x1

(2)

For $i= 3$

3x3 = 3+ x1+ x2

(3)

For $i= 4$

4x4 = 4+ x1+ x2+ x3

(4)

etc...

Now evaluate (2)-(1) and (3)-(2) and (4)-(3), etc... we obtain the following equations

(2)-(1) gives

pict

(3)-(2) gives

pict

etc... Hence we see that for the $rth$ term we obtain

1 + r x xr =------r−1 r

Hence

xr = 1-+ xr−1 r

(6)

Now replace $r$ by $r− 1$ in the above we obtain

1 xr−1 = r−-1 + xr−2

replace the above in (6) we obtain

( ) 1- -1--- xr = r + r− 1 + xr−2

(7)

And again, in the above, $xr−2 = r1−2 + xr−3$ hence (7) becomes

( ( )) 1- --1-- -1--- xr = r + r − 1 + r− 2 + xr−3

(8)

and so on, until we get to $x1 = 1$ , hence we obtain

|--------------------------| | 1 -1- 1-- | | xr = r + r−1 + r−2 + ⋅⋅⋅+ 1| ---------------------------

Hence

r 1 xr = ∑ -- k=1k

Which is the harmonic series. Now, it is know that²

|----------------------| | | | limr→ ∞xr = log(r)− γ | -----------------------

Where $γ$ is Euler Gamma constant given by