Question

Topic: University - Discrete Mathematics
Anyone willing to check my homework for me? I was working on it till 5 this morning and would just like some fresh eyes to see if I followed everything correctly. its attached in reply. THANK YOU =)

Answer 1

What Grade?

Answer 2

Im a senior in college

Answer 3

Oh Ok I Havent even Graduated High School lol im sorry

Answer 4

if i knew it would help

Answer 5

im in 9th grade im 15 so sorry

Answer 6

np thanks for the thought =)

Answer 7

I gave u a medal anyway

Answer 8

Do you know how to solve the recurrence in the first question? If you can find the closed-form solution, it is almost trivial to prove it.

Answer 9

no i couldnt figure it out thats why i i did it induction

Answer 10

I don't think your answer to the second question is right. That looks like a list checking function. It checks all elements of the list recursively until it finds an element which equal to x. If there is an element equal to x, return 10. Return 0 otherwise.

Answer 11

Hmm, I would use more words :\
It's not wordy enough! lol
I highlighted some changes that I would make.

In problem 2 I highlighted an error.
\(\large\rm B\cup C\ne S\) because it doesn't contain {{a}}.

Answer 12

Sorry I mean problem 4*

Answer 13

I suppose L[i] denotes the i-th element of the list and the first element of the list is L[1].

Answer 14

g[L list, i, x] basically does the following:
If i>10, return 0 (false)
If i<=10, check if the i-th element of the list L is equal to x. (Elseif L[i]=x)
If L[i]=x, return 10 (true)
If L[i]!=x, run g[L, i+1, x] (Calling g with the same list and x but incrementing the list index)

Answer 15

thank you

Answer 16

Or put it in another way using pseudocode:

g[L list, i, x]:
For(i<=10,i++)
{
If L[i]=x then
{
return 10
end program
}
}
return 0

Answer 17

Same logic as this C program

http://www.c4learn.com/c-programs/searching-element-in-array.html

Answer 18

For solving recurrence relations:
Assuming T(n)=r^n is a solution to this equation.
\[
T(n)=2T(n-1)+T(n-2)\\
r^n=2r^{n-1}+r^{n-2}\\
r^n-2r^{n-1}-r^{n-2}=0\\
r^2-2r-1=0\\
\]
So there are two solutions, \(r_1^n\) and \(r_2^n\). Any linear combination of these two solutions (\(c_1r_1^n+c_2r_2^n\)) will also satisfy the equation. However, not all solutions will satisfy the given initial conditions, namely T(0)=1 and T(1)=2. Find \(c_1\) and \(c_2\) using the initial conditions.

Answer 19

Once you find r1, r2, c1, and c2 the answer is very obvious.