Urgent Help me!!!!

[KingOcean34]

#include <iostream>
using namespace std;

int main (int argc, char *argv[])
{
int x,n;
cout<<"Please enter values for x : ";
cin>>x;
cout<<"Please enter values for n : ";
cin>>n;
float product=2.0;
float sum = n;
float fact=2.0;
float c = 1.0;

for (int i=2; i<=n ; i++)
{
sum = 2.0/(x-n);
product= product *= i;
fact=fact *=i/(x-2);

if(i%2==0) //every two times
sum+=(product/fact);
else
sum-=(product/fact);
}



cout<<"The answer is: "<<sum<<endl;
system("pause");
return 0;
}

WHY AM I GETTING inf as result????

[Ganado]

Hi, it would help us help you, if you put [code] and [/code] tags around your code, to let it be properly formatted. You can still edit your post and add them now.

What input are you entering for x and n?

And can you explain specifically what instructions you're trying to tell the computer to do on these two lines?

        product= product *= i;
        fact=fact *=i/(x-2);

You should turn on warnings for your compiler, your code has undefined behavior

19:30: warning: operation on 'product' may be undefined [-Wsequence-point]
20:28: warning: operation on 'fact' may be undefined [-Wsequence-point]
14:11: warning: unused variable 'c' [-Wunused-variable]

[KingOcean34]

i dont understand. where should i use that tags?

[Ganado]

Edit your post and add [code] {your code here} [/code]

It will look like this:
{your code here}

See my edit. Right now you have undefined or unspecified behavior (depending on C++14 vs C++17 rules, either way it should be avoided).

Did you mean to do

        product *= i;
        fact *=i/(x-2);
        

instead?

Also, you never answer my question, What input are you entering for x and n?

[keskiverto]

If x==n or x==2, then you have division by 0.

[KingOcean34]

x value must always be higher than n value.

i tried to do factorial in it.
Normally what does inf means?

Feel free to edit the code...
line 14 removed, no problem.

[Ganado]

Assuming the change in my previous post was applied (I don't know if that's actually what you intend):

Let's say x = 5, n = 4.
i =2
sum = 2.0 / (5 - 4) = 2.0 / 1 = 2.0

fact *= i/(x-2); --> fact = fact * (2 / 3)
2 / 3 is integer division, so you're doing fact = fact * 0 essentially.
Then you divide by fact... and boom -- you get infinity.

If you still want to do division, but not integer division, then do
fact *= static_cast<double>(i)/(x-2);
Again... not sure if that's actually what you intend.

You are also re-assigning "sum" each iteration within the loop. Probably not what you want.

______________________________________


What would really, really help is if you gave us some test cases. Because you haven't made it clear what you're trying to do. Because a normal factorial isn't a function of two variables.
Given: x = 5, n = 4, what should sum be?
Given: x = 6, n = 4, what should sum be?
Given: x = 6, n = 5, what should sum be?
etc.

[KingOcean34]

Write a program to prompt the user to key in x and n value to calculate the result for the following formula:

Result = 1+ 2!/(x-2) - 3!/(x-3) + 4!/(x-4) - ....... n!/(x-n)

Note: x value must be greater than n value.

this is the question..
sorry for not posting it earlier.
you got any other way to solve this question?

[keskiverto]

A loop, yes, but yours differs quite much from the expected.

See, what happens:

#include <iostream>
using namespace std;

int main (int argc, char *argv[])
{
  int x = 6;
  int n = 4;
  float product = 2.0;
  float sum = n;
  float fact=2.0;

  for ( int i=2; i<=n ; i++ ) {
    sum = 2.0/(x-n);
    product *= i;
    fact *= static_cast<float>(i)/(x-2);

    if(i%2==0) //every two times
      sum+=(product/fact);
    else
      sum-=(product/fact);

    std::cout << "Term " << i << " factorial " << product << " divider " << fact << " sum " << sum << '\n';
  }
  cout<<"The answer is: "<<sum<<endl;
}

Term 2 factorial 4 divider 1 sum 5
Term 3 factorial 12 divider 0.75 sum -15
Term 4 factorial 48 divider 0.75 sum 65
The answer is: 65

But the expectation is:

+2!/(x-2) =  2/4 = +0.5
-3!/(x-3) = -6/3 = -2
+4!/(x-4) = 24/2 = +12
Final = 11.5

[KingOcean34]

i dont get it !!
i need it urgently!!
please help me

[keskiverto]

Read again, what you have to add to a sum and compare that to the things the following program shows:

#include <iostream>
#include <iomanip>

int main()
{
  int n = 9;
  int x = 17;
  unsigned long long factorial = 1ULL;
  int sign = 1;
  std::cout << "Term 1 # 1\n";
  for ( int i=2; i<=n ; ++i ) {
    factorial *= i;
    std::cout << "Term " << i << " sign " << std::setw(2) << sign << " # " << i;
    std::cout << "! == " << std::setw(7) << factorial;
    std::cout << " # (x-" << i << ") == " << x - i << '\n';
    sign = -sign;
  }
}

Term 1 # 1
Term 2 sign  1 # 2! ==       2 # (x-2) == 15
Term 3 sign -1 # 3! ==       6 # (x-3) == 14
Term 4 sign  1 # 4! ==      24 # (x-4) == 13
Term 5 sign -1 # 5! ==     120 # (x-5) == 12
Term 6 sign  1 # 6! ==     720 # (x-6) == 11
Term 7 sign -1 # 7! ==    5040 # (x-7) == 10
Term 8 sign  1 # 8! ==   40320 # (x-8) == 9
Term 9 sign -1 # 9! ==  362880 # (x-9) == 8

[KingOcean34]

but still im not getting the answer correct...

i want the code to ask user to input value x and value n.
then it shows the answer.
thank you for helping.

[jonnin]

factorials get too big to fit into integers very quickly. Does your code work for a small value of N, like n < 10 ?

[lastchance]

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
   double x;
   int n;
   cout << "x: ";   cin >> x;
   cout << "n: ";   cin >> n;
   double answer = 1;
   for ( int i = 2, s = 1; i <= n; i++, s = -s) answer += s * tgamma( i + 1.0 ) / ( x - i );
   cout << "Answer: " << answer << '\n';
}

x: 17
n: 9
Answer: -41326