Expontional Function in java and the number of vowels

[YellowJ]

Hii , every body .. how are you ?

    public static double CalculateTheExp(double x , double n)
    {
        double sum = 0;
        x = CalculatPower(2 , 5 );
        n = CalculatFact(5);
        if(x==0)
            return 1 ;
        else if (x>0)
        {
            return x+(x/(CalculateTheExp(x,n)));
        }
        else
        System.out.println("WRONG CHOICE!");
        return -1;
    }

    // calculate factorial.
    public static double CalculatFact(double n )
    {
     if(n == 0)
        return 1;
    else
      return n * CalculatFact(n-1);
    }

    // calculate Power.
    public static double CalculatPower(double n , double x )
    {
            if(n==0) // base case
            return 1;
        else if(n>0) // recursive case
            return x*CalculatPower(n-1,x);
                else
                    return -1;
    }

I WANNA FOUND e^x when e^x = 1+(x/1!)+(x^2)/2!+ ... x^n/n!

power and factorial methods worked ! but the exponential it does not work at all !!

the error !

at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)
at labassigmnetsix .Main.Calculate TheExp(Main.jav a:34)


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[Oralloy]

First off, these look like homework questions. So, if you have specific technique questions or whatever, then we'll answer them for you. We are not, however, in the business of doing folks homework. That's your job.

So, in your mathematic recursion problem, the error is in lines 4 and 5 of the code you posted.

As for your second problem, you have code that I'm not sure will even compile. Are you clear on what you're trying to achieve?

Luck!
Oralloy

[YellowJ]

the second one i already solved by myself and it's worked ! so forget it and i'll removed now!

but the first one even if it's homework , i did not
say i need someone to solve it to me i almost solved it
but .. i have problem with exp. function !!
here where i got the error & and i know i made
a mistake .. but i do not know what it is !!
return x+(x/(CalculateTheEx p(x,n)));

[Oralloy]

YellowJ,

The problem is that you are overloading x and n.

Look very carefully at lines 4 and 5 for a moment.

Then, if you still don't see it, change the function prototype to this:

public static double CalculateTheExp(final double x , final double n)

I'm trying to help here, without throwing the answer at you.

Remember that your calculation must promote the power series and factorial series in parallel. Ponder that, and write down what a few phases of the recursion should look like. Once you do that, I think you'll have a better idea of how to proceed.

If need be, step very carefully through each step of the exponental recursion. I think you'll find the error very quickly on the first pass.

Luck!

[YellowJ]

yup! but i do not need to change any thing in fact & pow methods ? right !

.

exp function !
basic case (if x = 0) --> return one and stop !
but if (x+1) .. 1 .. , return (x-1)+(x^n/n!) ! am a right !! because each index will call the previous one
to get the answer ! is that what do you mean !!

do i have change my code to :

return x-1+(x/(CalculateTheEx p(x,n)));

I'll try it now ,,

[Oralloy]

Good, you realized that you are clobbering the values of x and n. That alone renders your exponential algorithm an infinite loop.

You recognize the recurance term, yes?

Start simple, don't recurse. How is your exponental function initially called? What value should the first iteration of your function compute?

Then, take it only one step deeper. How is the function called, and what should it return?

Does that help?

[YellowJ]

actually am still confuse !
but if i think in that way .. e = 2.718
so do i have say (should x be >1)!!

[Oralloy]

Sorry, my spelling error - it should be "recurrence term"

What is the general term in the sum used to calculate the exponental?

Start from there....

[YellowJ]

e^x=1+ x/1!+x^2/2!+ …….+x^n/n!
-----

THAT LAST THING MY MIND FOUND !

Sum = 0;
Num = 0;
if(x==0)
return one ; 
else if(x>0)
Num = CalculatPower(x-1,n)+CalculateTheExp(CalculatPower(x,n),(CalculatPower(x,n)/(CalculatFact(n))));
Sum = Sum + Num;
return sum ; 
else 
return 
-1;

[Oralloy]

Actually that's one direct approach to the problem, but the recursive term is wrong. I never considered propagating the one from the bottom of the recursion. That is clever. Kudos!

I'm not certain if you observe that:

1 = x^0/0!

Still, if you leave the first term out, or make it a special case, you have the general term in the series, the recurrence term, which is:

x^n/n!

Which, using your code, you can easily calculate, correct?

x^n/n! = CalcPow(x, n) / CalcFact(n)

The trick with recursion is to use each iteration to calculate one term (or other logical segment) of the problem.

Where this problem is possibly fooling you is that you have several conditions that you are trying to look at in the formula:

e^x = 1 + x/1! + x^2/2! + x^3/3! + ...

Breaking down,

term(0) = 1

and

term(1) = x/1!

and

term(n) = x^n/n!

Now, if you let

sum(n) = sum(n..infinity , term(n))

Which is to say

sum(n) = x^n/n! + x^(n+1)/(n+1)! + x^(n+2)/(n+2)! + ...

And, specifically,

sum(2) = x^2/2! + x^3/3! + x^4/4! + ...

Then

e^x = 1 + x/1! + sum(2)

And

e^x = 1 + x/1! + x^2/2! + sum(3)

I know it's a little long winded. Does that help?

[YellowJ]

        if(x==0)
            return 1 ;
        else if (x>0)
            return sum+=CalculateTheExp (CalculatPower(x,n),CalculatPower(x,n-1)/(CalculatFact(n-1)));
        else
        return sum;
    }

I study the case :
tell me if am wrong !
base case is when x=0 , return 1 .

but look !
power has its own base case to stop ?
and factorial also it has its own base case to stop ?
how i can mange this

[YellowJ]

sorry i posted my last comment w/o reading your last post

[YellowJ]

yeah i understand now , because the n which is increase
one each step !

so i created method for add 1 for each n !

        public static double sum(double n)
    {
         if(n==0)
             return 1;
         else if(n>0)
             return sum(n+1);
         else 
             return -1;
    }

    public static double CalculateTheExp(double x , double n)
    {
        double sum=0;
        double num =0;
        if(x==0)
            return 1 ;
        else if (x>0)
            return num=CalculateTheExp (1,(CalculatPower(x,sum(n)))/(CalculatFact(sum(n))));
        else
        return sum = sum+num;
    }

but wait if sum(n) = add all series together, so i think i don't have
to use it ,, why i can not +1 directly in the parameter.
I don't understand the error !! i think am gonna give up >.<"

[Oralloy]

YellowJ,

No worries.

As for the terminating condition, each term of the calculation is computed as follows:

term(n) = x^n/n!

Has your instructor done any analysis of the terms of the formula for the exponental?

In a nutshell, for any x, let

term(n) = f(n)/g(n)

where

f(n) = x^n

and

g(n) = n!

So, we want to know what happens to term(n) as n goes to infinity.

Let's start with f(n) for the easy case:

For -1<x<1, f(n) converges on zero as n aproaches infinity

next we'll take the positive one case:

For x=1, f(n)=1 for all n

and negative one:

For x=-1, f(n)=1 for n even, f(n)=-1 for n odd

and now the positive numbers greater than one:

for x>1, f(n)=x^n, increasing without bound as n approaches infinity

and now, the negative numbers less than negative-one:

for x<-1, f(n)=x^n, increasing without bound with alternating sign, as n approaches infinity

The trick is with g(n), which is simply analyzed as

g(n)=n!, increasing without bound as n approaches infinity

so, what happens to f(n)/g(n) as n goes to infinity? Well, it turns out that we can rearrange the equation for terms as follows:

term(n) = f(n)/g(n) = (x/1)*(x/2)*(x/3)*...*(x/n)

From which we can observe that the iteration number n will eventually swamp x and force term(n) to converge on zero.

Needless to say, once the term is below the floating point minimum of your machine, then it will be approximated by zero, and be inconsequential in the computation of the sum.

Luck!

[Oralloy]

removed due to giving an answer improperly. Self edit.