Thread: Newbie here...needs help

What bit are you having trouble with? post the code you've written so far, and someone will guide you.

Please read this before you post the code though

There's plenty of tutorials around on the web, including one on this site. Also, get yourself a good book (do a forum search to find recommendations).

Once you have written some code, people here will be more than willing to help you with it.

Here's a starter

Code:

#include <iostream>
using namespace std; // Read this

int main()
{
  int myValue;
  myValue = 11;
  cout <<myValue <<endl;
}

Don't worry if it doesn't all make sense yet, just stick with it, and it will get better

Consider the problem you are working on. Try to break it down a bit. You know what the overall task is (to compute a series approximation to e), but try to break it into smaller more obtainable tasks.

I think I have the answer but it maybe wrong, your mathematical constant had to be e=1+1/1+!2/1+!3/1+!4 thats what I came up with, since I continued getting errors when I attempted your problem without the correction. I would post the code, but please show effort.

Another thing my formula is correct, you would get errors division by 0 since 1!+2, therefore you would use my formula e=1+1/1+!2... This is all the help I'm going to give. The rest should be explanitory.

No, his original formula was correct.

e=1+ 1/(1!)+1/(2!)+1/(3!)+1/(4!)+ ... +1/(infinity!)

For both the loop in your program that calculates e, and the loop in your factorial function, I would suggest you look into something called a "for loop" (Sorry if you already knew that... you seem lost)

for (num=1;num<=i;num++)

This defeats the purpose of getting input for 'num'.

Anyway I don't even understand the problem. What in the hell is that 1!+1 stuff anyway? What exactly is it that you are calculating?

Code:

#ifndef IOSTREAM
#include <iostream>
#endif

using namespace std;

int main()
{
  int num = -1;

  while(num<0)
 {
   cout<<"Please enter a positive number: " << endl;
  
   cin>>num;
    
      if(num<0 || !cin.good()) // edit
   {
    cout << "number must be non-negative." << endl;
   }

 }
 
 // no clue here...
 
 double e=1;
 
   for (int i=1;i<=10;i++)
 {
   double fact=1;
  
      for (;num<=i;num++)
     {
        fact=fact*num;
        
        e=e+(1/fact);
     }
 // ... //
}

  cout<<"The value of e is: "<<e<<endl;

 return e;
}

I think the idea here may be to create a program that allows you to run a certain number of iterations to get an approximation of the value of e. You would have to choose how many iterations to run. The more you run, the slower it is, but the more accurate it is. I would create a function first that finds the factorial of a given number. That shouldn't be too hard. Then I would create a loop that ascends through positive integers, running the formula with each factorial. Here's an example:

Code:

double factorial(int num)
{
    double count;
    count =1;

    for (int i=1;i<=num;i++)
    {
        count*=i;
    }
     return count;
}

int main()
{
    int nIterations=20;
    double tempE=1;

    for (int i=1;i<=nIterations;i++)
    {
          tempE+=1/factorial(i);
    }

    cout << tempE << endl;

    return 0;
}

I'm not sure about the math really, but that should give you an approximation that's a bit less than the actual value. There may be some issues with factorial sizes, they can get quite big, but I think it's all good.

An improvement could be to keep going through iterations and displaying the result until the user presses a key, that would allow them to stop when accurate enough.

I hope this helps.

Oops, just reread the thread and realized the problem was already solved. Good work Diceman. Something else we could have all done better was to save time with the factorials. Why calculate 5! from the beginning if we just did 4! ? Just keep the factorial variable from one iteration to the next, and multiply by the next value. Eg:

Code:

double factorial;

for (...)
{
factorial*=i;
e+=1/factorial;
}

This would save a lot of time, allowing for more accuracy.