Recent content by jhudson1

n^2 definitely. I guess what threw me was the floor function. I wasn't sure if I could multiply through that or not and say n^2/2. Since it seems that you can, 1/2 is clearly scaling the order of the algorithm, which is n^2

Homework Statement Here is a pastebin link to my question, it contains an algorithm, my trace of the algorithm, and my question.Homework Equations I am interested in the theta notation for the runtime of an algorithm The Attempt at a Solution Here is the link http://mathb.in/1202 My thoughts...

Great. Thank you!

Homework Statement Working on a program that will calculate the results of a shootout between 3 people, each of whom have a different level of accuracy (one hits his mark 1/3 of the time, one hits his mark 1/2 of the time, one hits his mark 1/1 of the time). Homework Equations...

Just worked the problem again and had a similar issue. SammyS I just now see your response. I will try it once more tonight and if I can't figure it out I'll ask my professor tomorrow. It seems like k(k+1) = \frac{k(k+1)(k+2)}{3} means k(k+1)+(k+1)((k+1)+1) =...

Yes, I'll do that right now

Homework Statement Prove that (1*2) + (2*3) + (3*4) + ... + n(n+1) = \frac{n(n+1)(n+2)}{3} for every positive integer n **edit: using mathematical induction** Homework Equations In section 3 I have posted my work so far. the k^3 term is bugging me, I'm not sure how this is going to play out...

Thank you so much! Looking at it that way (as a tool used to manipulate an equation) makes so much sense.

Homework Statement Working on a computer program that will create an amortization table (a table that lists each payment on an amortizing loan). I am going to use a simple c++ array to store each row of data, and need to know the number of months required to pay off a given loan so that i...

Yes. That is very easy and very intuitive, at least in my mind. Thanks, by the way!

You're right, I can't decompose that fraction that way. Take a look at this: \lim_{n \to \infty} \frac{2^n}{1+2^n}=\lim_{n \to \infty} \frac{1}{1+2^n}-\lim_{n \to \infty} \frac{2^n}{1+2^n} \lim_{n \to \infty} \frac{1}{1+2^n}=0 Leaving us with -\lim_{n \to \infty} \frac{2^n}{1+2^n} =...

I think he meant -1

As n approaches infinity 2^n/2^n becomes larger and larger, but lim_{n\to\infty}2^n/2^n will always be 1 because they grow larger and larger together. 1/1 = 1 32/32=1 4096/4096=1 2^2,000,000/2^2,000,000=1 If you think of it as f(x)=2^x/2^x, no matter what we put in as x (assuming x>0)...

You're right. I was looking at 2^n/2^n and thinking that the result would be \infty/\infty, or undefined or zero. Perhaps it is inappropriate to think of an undefined term as 0? Either way I was wrong, because it isn't \infty/\infty but rather 2^{\infty}/2^{\infty} which is certainly one.

Sure. Using rules for limits... \lim_{n\to{\infty}} \frac{2^n(1/2^n-1)}{2^n(1/2^n+1)} = \lim_{n\to{\infty}} \frac{2^n}{2^n}\times\lim_{n\to{\infty}}\frac{(1/2)^n-1}{(1/2)^n+1} \\=\lim_{n\to{\infty}} \frac{2^n}{2^n}\times\left [\lim_{n\to{\infty}}\frac{(1/2)^n}{(1/2)^n} +...