#ATTENDANCE QUIZ FOR LECTURE 16 of Dr. Z.'s Math454(02) Rutgers University

# Please Edit this .txt page with Answers

#Email  ShaloshBEkhad@gmail.com 
#Subject: p16
#with an attachment called
#p16FirstLast.txt
#(e.g. p16DoronZeilberger.txt)

#Right after finishing watching the lecture but no later than Oct. 30, 2020, 8:00pm


THE NUMBER OF ATTENDANCE QUESTIONS WERE: 9

PLEASE LIST ALL THE QUESTIONS FOLLOWED, AFTER EACH, BY THE ANSWER

QUESTION #1:

Convert 6f(n)+12f(n+1)+18f(n+3) = 0 into explicit format

ANSWER:

f(n+3) = 0*f(n+2) - 2/3*f(n+1) - 1/3*f(n)

QUESTION #2:

f(n)=3*f(n-1)-4*f(n-2), f(0)=1, f(1)=-2

f(2) = -10, f(3) = -22, f(4) = -26

ANSWER:

f(5) = 3*(-26) - 4*(-22) = 10
f(6) = 3*10 - 4*(-26) = 134

QUESTION #3:

What is F(10^5)? How long did it take?
What is f(10^5)? How long did it take?

F := proc(n) local i, L; L := [1, -2]; for i to n - 1 do L := [L[2], 3*L[2] - 4*L[1]]; end do; end proc
f := proc(n) option remember; if n = 0 then 1; elif n = 1 then -2; else 3*f(n - 1) - 4*f(n - 2); end if; end proc

ANSWER:
F(10^5) = 
237935101918497760621346389...(fills up screen)
It takes 4.640 seconds to run this command.

f(10^5) - Error: too many levels of recursion.

QUESTION #4:

Consider the sequence that is defined by the recurrence f(n+1000)=f(n+999)+5*f(n). What is the operator such that ope(N) f(n)=0?

ANSWER:

We have f(n+1000)-f(n+999)-5*f(n) = 0, which corresponds to ope(N) = N^1000 - N^999 - N^5


QUESTION #5:

Characterize the sequence(s) that satisfy a (Homog) recurrence of order zero. 

ANSWER:

For a homogeneous equation we require the recurrence to equal 0. Since the recurrence has degree 0, then we have 0 recurrence elements which leaves us with 0 = 0.


QUESTION #6:

Can you prove that d(n)/n! -> 1/e?

ANSWER:

We have that d(n) = (n-1)(d(n-1) + d(n-2)). 

We know that e^x = Sum(x^n/n!, n=0..infinity).
So e^-1 = 1/e = Sum((-1)^n/n!, n=0..infinity).

The explicit formula for d(n) is:
d(n) = n!*Sum((-1)^i/i!, i=0..infinity).

Notice that this is equal to n!*e^-1. 
Hence, as n approaches infinity, d(n)/n! -> 1/e.


QUESTION #7:

What is the OEIS A number of the sequence:
1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496, 35696, 140152, 568504, 2390480, ...

ANSWER:

A000085: Number of self-inverse permutations on n letters, also known as involutions


QUESTION #8:

Find the operators in N and N^{-1} annihilated by w(n).
w(n) = w(n-1) - (n-1)w(n-2)

ANSWER:

We have w(n+2) = w(n+1) - (n+1)w(n) -> (n+1)w(n) - w(n+1) + w(n+2) = 0.
So, ope(n,N) = (n+1) - N + N^2, and
ope(n,N^-1) = (n+1) - N^-1 + N^-2.


QUESTION #9:

Do ?SumTools[Hypergeometric][ZeilbergerRecurrence]. look up the syntax of ZeilbergerRecurrence.

ANSWER:

Syntax: ZeilbergerRecurrence(T, n, k, f, l..u)