Java Issue

k1ll3r_x · 09-13-2005, 10:43 AM

i am totally lost on this

Code:

public class Class1{
public static void main(String args[]) {
double a[]={3.0,2.0,3.0};
int i;
double h;
                for (i=0;i<3;i++){
                    if (i %2 ==0 ){
                   h=Math.ceil(Math.exp(Math.log(a[i])*2));//this is where im lost
                                           }
                else{
                   h=Math.ceil(Math.exp(Math.log(a[i])*2));
                                            }
 
System.out.print(h);
System.out.print(" ");
} }
}

I dont know why, but i work it out on paper and i get 6.0 4.0 6.0 and i really dont know why, as far as i know i%2 is equal to 0 because i is initiated to 0.
if anyone can help me out on this, i think the math.ceil has to do something on it too, but i just cant find out what, id really appreciate
thanx

edit* [i forgot about the if statement, plz excuse the issue

Nylex · 09-13-2005, 10:51 AM

Your calculation isn't based on the value of i, it's based on the value of a[i]. So going through the loop,

i = 0, a[i] = 3.0
i = 1, a[i] = 2.0
i = 2, a[i] = 3.0,

as declared in your array.

jlliagre · 09-13-2005, 10:53 AM

I see no "i%2" in your code.

Nylex · 09-13-2005, 12:39 PM

Also, does that code compile? You have an else statement with no if..

k1ll3r_x · 09-13-2005, 03:20 PM

yea sorry i edited it now, i did miss the if statement

jlliagre · 09-13-2005, 07:34 PM

What is this "if" statement for, as the instructions are the same in both cases ?

k1ll3r_x · 09-14-2005, 10:25 AM

some test stuff that i got, but we go through every single problem, and this is one that i just get, this is not the first problem i post, its been a few. But what i want to know is how it works and why i get the answer i get, and where it is that im lost

jlliagre · 09-14-2005, 11:59 AM

Can you clarify what piece of this code you do not understand and what is the problem you are trying to fix, if any.

k1ll3r_x · 09-14-2005, 03:24 PM

well i just dont understand how the first number comming out is a 9.0, not a 6.0

Nylex · 09-14-2005, 03:37 PM

Go through your code :/.

The first time through the loop, i = 0. So, a[i] = 3. Then you take the natural log (ln) of a[i] and multiply by 2:

(Math.log(a[i])*2)

After that, you're finding e^x, where x = (Math.log(a[i])*2):

Math.exp((Math.log(a[i])*2)).

Now e^ab = (e^a)^b, so that is the same as if you had written

(Math.exp(Math.log(a[i]))^2

and since e^x and ln x are inverse functions, Math.exp(Math.log(a[i])) is just the same as writing

a[i] (inverse functions "cancel each other out", so to speak).

Since i = 0, you know a[i] = 3, so therefore

h = Math.ceil((a[i]^2);

and that is just 9.0.

k1ll3r_x · 09-14-2005, 03:41 PM

ok you just got me even more lost, all i want to know is why im getting a 9.0 as the first number when i compile, i am getting a 6.0, and i am not suppossed to, i compile and run, and my results are [9.0 ,4.0, 9.0] and if i do it on paper, i get 6.0 4.0 and 6.0, why is it that i get that on paper, and why is it that the correct answer it 9.0, 4.0, and 9.0?

spooon · 09-14-2005, 04:07 PM

How do you get 6 when you "do it on paper"? As Nylex pointed out, the math is:
e ^ ( log(3) * 2 ) = ( e ^ log(3) ) ^ 2 = 3 ^ 2 = 9

Do it on a calculator if you don't believe us.

k1ll3r_x · 09-15-2005, 10:30 AM

yea i got it, my bad, it was just i hadnt really noticed the Math.log, although i didnt know, but know i get, so thanx for realz, i really appreciate the help.
-Jerry V.