Simplifying using Big Theta expression
By : Shubhadeep
Date : September 16 2020, 08:00 AM

will help you Yes. But, the result is wrong! It should be \Theta(3^n), as 3^n is an exponential function and grows faster than a polynomial function such as n^{5000}. Also, you can think about the limit of the given function over 3^n when n goes to infinity. code :
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Big Oh Notation and Big Theta Notation Simplifying
By : HarderK
Date : March 29 2020, 07:55 AM
seems to work fine (There's a TLDR at the end specifically about your question, I thought I'd give you a little bit about bigO as well) Maybe something that's more palpable than formalism: this Onotation stuff is supposed to compare functions according to their growth. If you compare polynomials according to their growth, then all you have to look at is the highest exponent (i.e. their "degree"): if they have the same degree, then they grow about as quickly as each other. If one has a higher degree than the other, then the one with the higher degree grows faster than the other.

Proving the BigTheta and other asymptotic definitions (BigTheta, BigOmega, BigO, littletheta, littleomega)
By : Amit Jaiswal
Date : March 29 2020, 07:55 AM
may help you . So in a future assignment, I noticed some problems that requested us to just "use" these rules. I was wondering if any rules existed for littletheta, and littleomega as well (using limit as x approaches infinity of f(x)/g(x)). , Definitions which do not involve limits are provided here: Big O

Big Theta Notation  simplifying
By : Xtylîsh Hêrò
Date : March 29 2020, 07:55 AM
it should still fix some issue Yes. Your assumptions are correct. The first one is Θ(n2) and the second one is Θ(n3). When you are using Θ notation you only require the largest term. In case of your second recurrence consider the n = 1000, then n3 = 1000000000. Where as 100n2 is just 100000000. As the value of n increases, n3 becomes more and more predominant than 100n2.

Time complexity of the given C function theta(nlogn) or theta(n^2logn)?
By : Rash1
Date : March 29 2020, 07:55 AM
wish of those help I think your answer is right, it is theta(nlogn) indeed. However, your analysis seems somewhat wrong. The outer loop executes O(n/2) times.

how to plot on a graph using the hypothesis function by substituting the value of theta 0 and theta 1
By : reader
Date : March 29 2020, 07:55 AM
This might help you It is the same way that we graph the linear equations. Let us assume h(x) as y and θ as some constant and x as x. So we basically have a linear expression like this y = m + p * x (m,p are constants) . To even simplify it assume the function as y = 2 + 4x. To plot this we will just assume the values of x from a range (0,5) so now for each value of x we will have corresponding value of x. so our (x,y) set will look like this ([0, 1, 2, 3, 4], [2, 6, 10, 14, 18]). Now the graph can be plotted as we know both x and y coords.



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