The exponential function, EXP(x), is defined to be the sum of the following infinite series:
Write a program that reads in a REAL value and computes
EXP() of that value using the series until the absolute value of
a term is less than a tolerance value, say 0.00001.
Click here to download this program.! --------------------------------------------------------- ! This program computes exp(x) for an input x using the ! infinite series of exp(x). This program adds the ! terms together until a term is less than a specified ! tolerance value. Thus, two values are required: ! the value for x and a tolerance value. In this program, ! he tolerance value is set to 0.00001 using PARAMETER. ! --------------------------------------------------------- PROGRAM Exponential IMPLICIT NONE INTEGER :: Count ! # of terms used REAL :: Term ! a term REAL :: Sum ! the sum of series REAL :: X ! the input x REAL, PARAMETER :: Tolerance = 0.00001 ! tolerance READ(*,*) X ! read in x Count = 1 ! the first term is 1 and counted Sum = 1.0 ! thus, the sum starts with 1 Term = X ! the second term is x DO ! for each term IF (ABS(Term) < Tolerance) EXIT ! if too small, exit Sum = Sum + Term ! otherwise, add to sum Count = Count + 1 ! count indicates the next term Term = Term * (X / Count) ! compute the value of next term END DO WRITE(*,*) 'After ', Count, ' iterations:' WRITE(*,*) ' Exp(', X, ') = ', Sum WRITE(*,*) ' From EXP() = ', EXP(X) WRITE(*,*) ' Abs(Error) = ', ABS(Sum - EXP(X)) END PROGRAM Exponential
After 35 iterations: Exp(10.) = 22026.4648 From EXP() = 22026.4648 Abs(Error) = 0.E+0
If the input is -5.0, it takes 21 iterations to reach EXP(-5.0)=6.744734943E-3. The value from using Fortran intrinsic function is 6.737946998E-3 and the absolute error is 6.787944585E-6.
After 21 iterations: Exp(-5.) = 6.744734943E-3 From EXP() = 6.737946998E-3 Abs(Error) = 6.787944585E-6
Therefore, the (i+1)-th term is equal to the product of the i-th term and x/(i+1). In the program, variable Term is used to save the value of the current term and is updated with
where Count is the value of i+1.Term = Term * (X / Count)