for regular series caluclation

we could use this

{
   double summand = 1;
   int n = 1;
   double sum = 1;

   do
   {
      summand = summand * x / n;
      sum = sum + summand;
      n--;
   }
   while (summand > 0.00001);

   return sum;
}

i.e for e^x= 1+x+(x^2)/2! + (x^3)/3! ....... so on

how would u change the above code if

e^x= 1-x+(x^2)/2! - (x^3)/3! +(x^4)/4! - (x^5)/5! ....... so on

in which the sum is alternated ... any help would be great ...

cheers

Keep a flag variable which will decide whether the operation should be addition or subtraction.

{
   double summand = 1;
   int n = 1;
   double sum = 1;
   bool flag = true;
   do
   {
      summand = summand * x / n;

[B]      if (flag)
           sum += summand;
      else
           sum -= summand;
      flag = !flag;[/B]

      n--;
   }
   while (summand > 0.00001);

   return sum;
}

how would u alternate the flag variable everytime ??

is it by using the loop

Maybe you need to look a bit harder at the code I posted. It takes care of changing the flag variable. Look out for the part posted in bold...

Comments
Rep for a hard-working (and patient!) moderator... --Joe(y)

yea but u have to alternate the ++ and -- everytime ... i tried the code but does not give correct output

how would u do it for

Approximate sine of θ = θ – (θ3/3!) + (θ5/5!) - (θ7/7!) + (θ9/9!)

The alteration is happening alright. Your logic is wrong. Look at it once again. Are you sure you are doing what the equation demands ?

wt would i need to do for the sin equation i mean what summand equation would i need

Maybe something like this:

double factorial (double value)
{
    double result = 1.0;
    while (fabs (value) > 0)
    {
        result *= value--;
    }
    return result;
}

int main ()
{
    double summand = 0.0, sum = 0.0;
    int value = 0, current = 0;
    bool flag = true;

    cout << "Enter the power of e whose answer you want: " ;
    cin >> value;
    getchar ();

    do
    {
        summand = pow ((double)value, (double)current) / factorial (current);
        if (flag)
            sum += summand;
        else
            sum -= summand;

        flag = !flag;
        ++current;
    }
    while (summand > 0.00001);
    cout << "\nAnswer is : " << sum;

    getchar ();
    return 0;
}

Just replace the ++current in my previous code with current += 2 and make the data type of the variable value as double and you should be good to go.

hey uoit fella..lol..can't we just use pow(theta,n)?? i think we can..

good luck! about 7 hours to go and we still didnt figure it out! lol

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