Recurrence Relation Puns

I have watched several videos on this topic and every time I think I understand I become more confused. For this specific problem I have shown where I have gotten so far:

https://imgur.com/a/yJfKDzG,

but after this, I am lost. Any help would be awesome.

T(n) = 2T(n-1) + O(1)

I think it's O(2^n ) because if I draw out the recursion tree, I have a tree of height n with 2^n leaves. This tree has O(2^n ) nodes.

https://preview.redd.it/gzmtyztpxuy71.png?width=351&format=png&auto=webp&s=5c6d2a482cdc04a88614da2ef72c6dcbb1fd0243

Find the general solution of the recurrence relation:

T(n) = 5T(n/4) – 4T(n/16) + 3n

(b) What is the form of the particular solution of the linear non-homogeneous

recurrence relation an = -32an – 2 + 256an – 4 + F(n)

( characteristic equation when factorised is (x + 4)2(x – 4)2 )

(i) F(n) = (n3 – 4)4n (ii) F(n) = 4n + 1

In the following pic, how did they get the logn circled in red?

So this is a solution to a recurrence relation problem that was on one of my homeworks. Here they didn't show any work and just straight up listed the Faulhaber formula. So am I just supposed to automatically recognize to apply that here or is there more to this and they just didn't want to show the work done?

Hi folks, been playing this game for a while now and several weeks ago I hit 20 students (ee5,000) and decided to unlock this theory. But the progress has been painfully slow ever since. I reached all the theory milestones and have been publishing several times a day to increase the multiplier. The q and c upgrades have been crazy slow too. I can't seem to break ee1,500, which is crazy considering I was at ee5,000 several weeks ago. Should I have not unlocked the theory just yet? To unlock a new student I need to reach ee5,200, which at this pace I think I would have probably reached it faster without the theory.

Any insights would be helpful, thank you in advance!

Gallery showing the interface and some example images. Here is the GitHub project page.

I've been interested in a particular fractal-generation method for a while and just finished my master's thesis on the subject. I decided to make and share an interactive version of the picture-making program I used so I can spread my fascination. In short, these are all variations on how color-coding the odd and even entries of Pascal's triangle generates the Sierpinski triangle fractal. I have a bit more explanation on the GitHub page.

This was my first time programming a GUI and I'm completely self-taught when it comes to programming, so I welcome advice and feedback.

Define a recurrence relation as well as a set of initial conditions for An, the number of ways to draw n cards with replacement from a standard deck and not draw two hearts consecutively.

After getting the ee5000 graduation I had good progress with T1. It was moving pretty fast but now it's just inching along. At e108 but the next publish is at e111 I have one of each of the four milestones, should i have a different configuration? I hit ee5200 after a lot of grinding. Do i abandon theories for a bit to get to ee5400? Theories seem to be a lot more powerful than normal student upgrades though.

EDIT: I’m an idiot I had auto buy set to 10x

I am having trouble with this problem, even though I think I know how it works.

Ratio from a_0 to a_1 = 1

Ratio from a_1 to a_2 = 2

Ratio from a_2 to a_3 = 2 (2*1)

Ratio from a_3 to a_4 = 4 (2*2)

Ratio from a_4 to a_5 = 8 (2*4)

Ratio from a_5 to a_6 = 32 (4*8)

I put the pattern I noticed in the parentheses (the ratio is the product of the two previous ratios); I just can't figure out how to translate this into a recurrence relation equation.

So then I would assume initial conditions are a_0 = 1, a_1 = 1, and a_2 = 2.

Noticing that these were all exponents of 2, something that I tried was a_n = a_(n-1) * 2^(n-1) * 2^(n-2), but it doesn't work when I plug in the known values in the sequence so it's wrong. I would appreciate any hints to get me on the right path.

Question Answer

I'm not following the logic of the second part. Wouldn't the d_(k-1) derangements include ones where i was in the first slot? For example the derangement 3412 of {1, 2, 3, 4}, 1 goes to 3, and the remaining {2,3,4} is rearranged to 34-2, which would be counted in d_(4-1) = d_3. But wouldn't 3412 also be included in the d_(k-2) derangements of the first part?

What am I missing here?

Hello, I have a couple of question regarding linear recurrence relations.

Given:
T(n) = 3T(n-1)-2T(n-2)

I can solve this recurrence relation using the characteristic polynomial etc.
The closed form is:
T(n) = a+b*2^n
Now we can say that T(n) = Theta(2^n)

My question is:
- first of all what does this function even represent? How can an algorithm have a time complexity function of this form? It doesn't have a constant term, which is absurd given that a recursion call always has a cost. Moreover the second term is -2T(n-2), so the second call goes back in time?!??!

- Why isn't the time complexity O(1)? If I draw the recursion tree each node has cost 0 because there's no constant term. So what does the closed form represent? the number of nodes?

I don't really understand how homogeneous linear recurrence relations relate to Algorithms

I have this recurrence relation:

T(n) = 2T(n-10) + 3 for n > 11, T(n) = 5 for n <= 10

so, the relation is as follows

5 5 5 5 5 5 5 5 5 5 13 13 13 13 13 13 13 13 13 13 29 29 ...

Basically every 10 numbers are the same, and they are equal to two times the previous number, plus three. (right?)

Can anybody help me find the closed form equation for this? Thanks so much!

I'm trying to solve it using the substitution method but all of the examples on the internet only showed one T(n) in the equation but that problem's got two of them. I don't know what to do. Please help.

If I do T(2), T(3), T(4)... I can easily see that the recurrence is T(n) = H_n, but I'm not getting to this result solving it normally. Here is my work. The relation I found at the end works for n > 1, but T(n) = H_n works for n = 1 also. Is my answer wrong?

Given:

For n = 1, T(1) = 1

For n > 1, T(n) = 7 T(n/2) + n^2

Find the closed form of the recurrence relation via the substitution method.

I have tried to substitute n/2, n/4, n/8, etc. but I can't intuitively find the relation from seeing the iterations. I would greatly appreciate any help.

I have watched several videos on this topic and every time I think I understand I become more confused. For this specific problem I have shown where I have gotten so far:

https://imgur.com/a/yJfKDzG,

but after this, I am lost. Any help would be awesome.

Can I have a recurrence relation like T(n) = T(n/2) - n ? Is this "-" impossible ?

How can a function have a negative cost (space and time complexity)?