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Assignment 2 Question 2(iii)

Posted by liang 
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Assignment 2 Question 2(iii)
May 14, 2008 01:03PM
This question should read as follows: "...for all positive integers n > 0".
Anonymous User
Re: Assignment 2 Question 2(iii)
May 14, 2008 01:37PM
Oops, too late, just give us free marks. we have already submitted.
Anonymous User
Re: Assignment 2 Question 2(iii)
May 15, 2008 06:49AM
I've also summitted, but here's the question...


I could be wrong on this one, but if 0 is not part of the set then I can't see how it will work.

RHS: 2 - 1/(2^4) = 1.9375 (or 2 - 1/16)

for (n > 0)

LHS: 1/(2^1) + 1/(2^2) + 1(2^3) + 1/(2^4) = 1/2 + 1/4 + 1/8 + 1/16 = 0.9375

you need the missing 1 (i.e. 1/(2^0))

I accept I'm the student here, but I cannot see that is will be possible to prove this quetion if n cannot be 0

A casual observation of the RHS shows that 2 - (something smaller than 1) will be greater than 1
Re: Assignment 2 Question 2(iii)
May 15, 2008 09:32AM
This will not have any impact on your marks for assignment 2 (i.e. you will not be penalized for this.)
Re: Assignment 2 Question 2(iii)
May 15, 2008 09:46AM
1/(2^0) is still included in the LHR calculation, i.e. LHS = 1/(2^0)+1/(2^1)+...+1/(2^4). The number from 0..4 is the value for i, not the value for n.
Re: Assignment 2 Question 2(iii)
May 15, 2008 03:02PM
In response to Justin, the identity is true whether or not 0 or 1 is your base element.

It's like being asked to prove that 1 + 2 + 3...+ n = n * (n+1) / 2 for n > 93. That relationship is true for n > 93 but it's also true if n > 52 or n > 9 or, in fact, for n >= 0. Essentially all that you're doing is changing the base element in your inductive proof.
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