Thread: exponent math, calculus?

Hey everyone. I need a little exponent help.

This is the problem that I am currently working on:
***Note: Since I don't know if I can use {sup}{/sup} and html on here, I will simply type my exponents in graphing calculator format.. like 5 ^ 3 = (5*5*5)=125 ***

"Write an interactive C++ program to calculate the following summation using an iteration,

Sum = 1 + 1/2 + 1/4 + ... + 1/2n (note: the n is an exponent for the denomenator, as 1/(2^n) )
Your program should prompt for the value of n. Please show your results for n = 10.
The series above has a closed form analytical solution, namely,

Sum(n) = 2 - 1/2n = 1 + 1/2 + 1/4 + ... + 1/2n (note: the n is an exponent for the denomenator, as 1/(2^n) )


In the output for your C++ program, compare the iterative solution to the analytical solution. "

I'm not exactly sure what to do...While poking around google, I found solutions to base/exponent problems. If I follow these, I think I may need <math.h> or <cmath> in order to do exponents with the "pow" command. Upon reviewing my problem, however, I began to wonder if maybe I need a loop or something. My teacher was explaining this problem on the board, and told us that the ... in "1 + 1/2 + 1/4 + ... + 1/2n " meant something like 1/(2^0) + 1/(2^1) + 1/(2^2) + ... + 1/(2^n) . Have any thoughts/tips/suggestions?
As always, any help would be greatly appreaciated.

Ok...I sort of see...

however, I am having trouble understanding a basic principle of this problem. You have this little set of things you need to do..in this case, the "1/(2^0) + 1/(2^1) + 1/(2^2) + ... + 1/(2^n) "

Because I am a noob/can't understand directions, I have no idea as to how I would implement this into code. How would you go about doing this? If you wish, you can make up a problem/solution that is somewhat similar.

Thank you for sharing your expertise.

I'm just wondering what type of class this is, and how much math have you had? Anyway this is not a very hard problem (unless you have not covered loops).

Code:

double sum = 0;

for(int i = 0, n = 10; i <=n; i++)
{
     sum += 1 / pow(2,n);
}

Hey, if they don't want to learn it fine.

Actually so far I'd have to say the OP is doing a good job. Granted the request for code is a little much but they at least stated the question / problem clearly. If they had kept on their current course I'm sure we could have guided them into a proper solution.

Hey, if they don't want to learn it fine.

If you insist on that attitude I suggest you leave.

Originally Posted by phosphorousgobu

Are you not allowed to ask people to show you how to do something on these forums?

Asking for examples doesn't seem to be bad. I think it's if someone came on and said "do my homework for me" that things get ugly.

And congradulations on understanding.

Code:

#include <cmath>
#include <iostream>

int main()

{
	double sum = 0;
	int n;
	std::cout << "Enter value for n: \n";
	std::cin >> n;
	for(int i = 0; i <=n; i++)
	sum += 1 / pow(2,n);
	std::cout << "Answer is " << sum << "\n";	
	return (0);
}

If I execute this (in Visual Studio 6.0) and enter "2" as my n value, I get a return of "0.75". Is this correct?

I'm still not sure what the heck I'm doing. From what I gleaned from the problem, you are starting off with 1. You keep dividing the variables by two until you reach n, which is actually 1/2^n. Let's look at what I just did. If I enter "2" as my n variable, what theoretically happens? Is it...
1 + 1/2 + 1/2^n ?
Since I put "2" as my n, I am understanding that it will go from the beginning up to where 2^2 (or 4) would be
1 + 1/2 + 1/4 = 1.75 (on a graphing calc).

There are either 1 of two things wrong...1). I don't understand what I'm doing and this is why I don't get the results, or 2). The code simply needs to be tweaked.


Thank you all very much for your patience and help.

Code:

	for(int i = 0; i <=n; i++)
	sum += 1 / pow(2,n);

should be

Code:

	for(int i = 0; i <=n; i++)
	    sum += 1 / pow(2,i);