Compute pi

[ilovejava]

i have to write a program that displays the value of pi for values for i = 10000,20000.....,and 100000
i have to use the following series

pi = 4(1-(1/3)+(1/5)-(1/7)+(1/9)-(1/11_+...+(1/(2i +1))-(1/(2i +1))
any ideas

i have to do a program with looping i cant use arrays

can someone give me a hint and a strategy on how to do this problem

[rubberman]

This your school exercise. Make an effort - describe the processes needed, write some pseudo-code (or real code if you prefer), and we will be happy to comment on it. However, YOU need to make a good faith first effort at solving the problem.

[NormR1]

Look at the terms in the equation. Are there any repeating or stepped values that you can use in a loop?
Write a list of half a dozen terms with one term per line. Look for a pattern

[ilovejava]

hmmm ok

public class computingPi {
	public static void main(String args[]){
		double i, add, minus;
		i=0;
		add = (1/(2*i+1));
		for(int a=0;i<=10000 ; i++){
			double ans = 4*((1/(2*i+1)-(1/(2*i+1);
		}
	}

}

[ilovejava]

alright i have no idea what am i doing i no i have loop it but i dont think so thats correct

[ilovejava]

i see some stuff repeating
i know i have to do a for loop but im horrible and new to this

[NormR1]

What is the pattern you saw when you listed the first half dozen terms in the equation?

You need to figure out the algorithm BEFORE you write the code.

[ilovejava]

4((1/(2i +1))-(1/(2i +1)+..........) is this the pattern i is always incrementing by 1

[NormR1]

That is the equation. Extract out the terms of the equation. I'll give you the first one:
4

Add the next 5 terms to the list, one per line

[ilovejava]

1-(1/3)+(1/5)-(1/7)+(1/9)-(1/11)+(1/13)-(1/15)+(1/17)-(1/19)
so after u plugged 4 which is 1/9 i did 5 6 7 8 9 i extracted the eqn in paper and plugged numbers, yes i see a pattern

[NormR1]

I was trying to get you to put one term per line so you could see the pattern as you went from the value on one line to the next line.
Perhaps you can see it with the terms on one line.
To use a loop you need to see what changes from one term to the next.
Can you see that?

[ilovejava]

I was trying to get you to put one term per line so you could see the pattern as you went from the value on one line to the next line. Perhaps you can see it with the terms on one line. To use a loop you need to see what changes from one term to the next. Can you see that?

yes i see the pattern (1/(2(1)-1) - (1/(2(2)-1) + (1/(2(3)-1) - (1/(2(3)-1) + (1/(2(4)-1)

is this the patern you wnat me to look at

[NormR1]

Are you sure those are the first 5 terms? Where is the 4?

What things change from one term to the next?

[ilovejava]

Are you sure those are the first 5 terms? Where is the 4?

What things change from one term to the next?

4*(1/(2(1)-1) - 1/(2(2)-1) + 1/(2(3)-1) - 1/(2(3)-1) + 1/(2(4)-1))

sorry forgot to put the four in and arnt these the first 5 i plugged in 1 2 3 4 5

[NormR1]

What things change from one term to the next?

Why do you keep writing that equation? I'm asking for a list of the terms, not the equation.
Here are the first two terms:
+4/1
-4/3

[ilovejava]

What things change from one term to the next, starting with the second term?

Why do you keep writing that equation? I'm asking for a list of the terms, not the equation. Here are the first two terms: 4 -4/3

it adds then subtracts
it adds(i=2) subtracts when(i=3) adds when(i=4) subtracts when (i=5)
i see for a odd number it subtracts for a even number it adds

but yeah its something like x-x+x-x+x-x

[NormR1]

If the terms are:
+4/1
-4/3
+4/5
-4/7
I see that the sign alternates from one term to the next
and the value of the denominator increases by 2 for each term

What is a simple way to change the sign of a number?

[ilovejava]

If the terms are: +4/1 -4/3 +4/5 -4/7 I see that the sign alternates from one term to the next and the value of the denominator increases by 2 for each term

What is a simple way to change the sign of a number?

multiplying it by -1

[NormR1]

Ok, you've got most of what you need. Now write the expression that adds the next term to the current sum and changes the sign for each term. Put it in a loop and see what you get.

[ilovejava]

Ok, you've got most of what you need. Now write the expression that adds the next term to the current sum and changes the sign for each term. Put it in a loop and see what you get.

do {
			ans = 1/(2i-1)*(-1)
			
			
			
		} 
		while (ans<=10000);