# Using method to compute the sum of n numbers, not getting the return value

• 02-28-2013, 01:56 AM
abi
Using method to compute the sum of n numbers, not getting the return value
m(i) = i / ( i + 1) where i goes from 1 to 20

The program is supposed to show a table with i and the corresponding value m(i) listed.
i.e.

i m(i)
_____________
1 0.5000
2 1.1667
.
.
.
.
20 .....

I got my code to compile but I am getting the same value for all values of i (1 to 10).

Code:

public class HW5_13New
{
public static double sum(int i)
{
double add = 0;

for ( i = 1; i <= 20; i++ )
{
add = (double)i / ( i + 1);

}

}

public static void main(String[] args)
{

System.out.println("i          m(i)");
System.out.println("____________________");

for ( int i = 1; i <= 20; i++)
{
double ans = sum (i);

System.out.print("\n" + i + " " +"        " + ans);
}
}

}

• 02-28-2013, 02:12 AM
pbrockway2
Re: Using method to compute the sum of n numbers, not getting the return value
Code:

public static int sum(int i)
You have declared sum() to return an int. But in fact you intend it to return a floating point number (a double). So it should be declared to return a double.

In code like this you need to give some thought to what sort of thing (int or double) you intend each variable be.

Quote:

I am having trouble taking the value from the Sum method into the main method.
It would help to describe the problem. Does what you posted compile? Does it compile and run but print strange values at runtime? In the latter case, observe that i/(i+1) is always zero when i is an int. You can get a double answer by using a cast: (double)i/(i+1)
• 02-28-2013, 02:22 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
I had been completely oblivious of that. Just didn't occur to me. Rookie mistakes I guess. So I made that correction all of my variables are doubles now.

I guess I didn't describe my problem adequately enough. Yes it does compile, that is what i meant when I said it displays. After I made the variable type correction it shows 0.5 for all the i (1 - 20).

Thanks man.
• 02-28-2013, 03:29 AM
jim829
Re: Using method to compute the sum of n numbers, not getting the return value
Looks like you simply keep assigning a new value to add. So the only value saved is the last one in the loop.

Jim
• 02-28-2013, 04:30 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
I don't see a way to get around that. Could you tell me how I could store the value of each i to add?
• 02-28-2013, 04:38 AM
jim829
Re: Using method to compute the sum of n numbers, not getting the return value
Here is your sum method:

Code:

public static double sum(int i)  {
double add = 0;

for ( i = 1; i <= 20; i++ )
{
add = (double)i / ( i + 1);

}

}

First, you pass a value to i but then you don't use it because you set i to 1 in the for loop.
To update add, do the following:

Code:

Regards,
Jim
• 02-28-2013, 04:56 AM
pbrockway2
Re: Using method to compute the sum of n numbers, not getting the return value
I'm glad you've got the double/int thing sorted out.

Jim's right, you've got to "accumulate" the result. You can also use += for that.

Is there some reason you're not doing the whole thing in main()? I'm a fan of using separate methods, but here you are calculating the same thing over and over again.
• 02-28-2013, 05:18 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
I am learning to use separate methods and the assignment requires that I use it.
Wouldn't that result in a total sum.

I want this:
say it first takes i = 1 and comes out with 0.5
so add = 0.5
it displays 1 0.5
then
now it goes to i = 2 and gets 0.6667
so wouldnt add = 0.5 + 0.0667
and so on until it reaches 20
• 02-28-2013, 05:20 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
Couldn't that result in a total sum.
I want this:
say it first takes i = 1 and comes out with 0.5
so add = 0.5
it displays 1 0.5
then
now it goes to i = 2 and gets 0.6667
so wouldnt add = 0.5 + 0.0667
and so on until it reaches 20
• 02-28-2013, 05:28 AM
jim829
Re: Using method to compute the sum of n numbers, not getting the return value
Ok, perhaps you don't want to update "add." I reread the first thread. The loop in the sum method confused me. You don't need it.

Just simply return the value i/(i+1) from the sum method. The reason your method is messing up is because the i in the for loop is clobbering your
method argument. And since you aren't really summing anything you might want to change the name (but it isn't necessary for operation).

Regards,
Jim
• 02-28-2013, 05:43 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
Code:

public class HW5_13New
{
public static double sum(int i)
{

double add += (double)i / ( i + 1);

}

}

public static void main(String[] args)
{

System.out.println("i          m(i)");
System.out.println("____________________");

for ( int i = 1; i <= 20; i++)
{
double ans = sum (i);

System.out.print("\n" + i + " " +"        " + ans);
}
}

This results in the program giving me individual values for the values of i
sample
i m(i)
1 0.5
2 0.6667 <---- 0.5 + 0.6667
.
.
so on

i need
1 0.5
2 1.667
...

I tried a few things with this updated code but it isnt happening.
• 02-28-2013, 05:54 AM
jim829
Re: Using method to compute the sum of n numbers, not getting the return value
OK, let me try and understand. You want to create a list of values as follows:

i i/(i+1)

1 1/2
2 2/3
3 4/5
...
...
19 19/20
20 20/21

Am I correct? If so check out my post #10
Jim
• 02-28-2013, 06:00 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
Sorry for being unclear.

No that is not correct but that is what I am getting currently.

What I need is for it to add up as it goes from 1 - 20 and display the sum as it goes along

i i/(i+1)

1 1/2
2 (1/2)+ (2/3)
3 (1/2)+ (2/3) + (3/4)
...
• 02-28-2013, 06:18 AM
jim829
Re: Using method to compute the sum of n numbers, not getting the return value
OK, got it. What you need to to is sum of the values in add as we have told you. But you want to increase the number of terms you are summing for each call. This is controlled by the termination for value n which is now passed to sum. If you want to include the nth iteration in sum change to from < n to <= n.

Code:

public static double sum(int n)  {
double add = 0;

for (int i = 1; i < n; i++) {
add += (double)i / ( i + 1);
}
}

Next, in the other for loop, you pass the index to sum. Thus, each time sum is called in will calculate one more element.
Now you can play with it to fine tune it to the proper output.

Regards,
Jim
• 02-28-2013, 07:09 AM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
Ok I got that and came up with this. The only problem now is that it displays
1 0.0
2 0.5
3 1.16
...
do you see the trend?the values on the right have been pushed down per say.I poked at it but i can't see how to get it to start at 1 - 0.5.

Code:

public class HW5_13New
{
public static double sum(int n)
{
double add = 0;

for (int i = 1; i < n; i++)
{
add += (double)i / ( i + 1);
}

}

public static void main(String[] args)
{

System.out.println("i          m(i)");
System.out.println("____________________");

for ( int n = 1; n <= 20; n++)
{
sum (n);

System.out.print("\n" + n + " " +"        " + sum(n));
}
}

}

• 02-28-2013, 04:07 PM
jim829
Re: Using method to compute the sum of n numbers, not getting the return value
Try changing the loop comparison in the sum method from "<" to "<=".

Jim
• 02-28-2013, 08:59 PM
abi
Re: Using method to compute the sum of n numbers, not getting the return value
Turns out we didn't need to use different methods so I did it within the main method which was pretty easy.
Thanks