How Do You Prove a Sum of Cubes is Theta of n^4 Using Inequalities?

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To prove that the sum of cubes, 1^3 + 2^3 + ... + n^3, is Theta(n^4), one must use the definition of Theta notation and inequalities, focusing on positive integer values of n. The lower bound can be established by considering the sum from the middle term to the last term, which involves identifying the middle of the series. Clarification on the term "middle" indicates it refers to the central value in the sequence of cubes, not the average. The discussion also highlights confusion regarding the nature of the series, clarifying that it is not a geometric progression. Understanding these concepts is essential for correctly applying the Theta notation in this context.
mr_coffee
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Hello everyone, have a problem. I'm not allowed to use the books instructions on how to do the problem but instead I must do the following:
Prove each statement, assuming n is a variable that takes positive integer values.
Ignore the book's instructions for 41. Instead, use the definition of Theta notation and work with inequalities. Since everything is positive, you don't need to use absolute values. Doing the exercise this way is a requirement, not a suggestion. (The following hint, however, is just a suggestion. For the lower bound, consider the sum of the series from the middle to the end, and find a lower bound on that.)

The problem is the following:
1^3+2^3+3^3+...+n^3 is Theta(n^4).

The book only gives examples of inequalities, but this isn't, i'm not sure how to transform this into an inequality, does anyone know how to do this or have a website that will explain the process?

Also he says find the lower bound, consider the sum of the series from the middle to the end, what is the middle? if the first term is 1 and the last is n^3 is he saying, use (1+n^3)/2? But that's just the average, not middle to the end so I'm also confused on that.
I see this is a geometric progression, with a ratio of 3...so i could find a formula for the sum if i use the following:
ratio: 3
first term: 1
mythical next term: 3 * n^3
hm...maybe this isn't a geometric progression because usually the mythical next term doesn't have an n in the base but on the exp. But clearly each term is mutliplied by a ratio.

Any ideas?

note:
definition of theta is the following:
Definition (big-theta): Let f and g be functions from the set of
integers (or the set of real numbers) to the set of real numbers. Then
f(x) is said to be Theta( g(x) ), which is read as f(x) is big-theta
of g(x), if f(x) is O( g(x) ), and Omega( g(x) ). We also say that
f(x) is of order g(x).

Thanks!
 
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mr_coffee said:
i'm not sure how to transform this into an inequality
Invoke the definition of theta. (and then of O and of Omega)


what is the middle?
Your sum consists of n terms:

1³, 2³, 3³, ..., (n-1)³, n³

"middle" refers to the middle of this list.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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