Building a library of mathematical functions.

Question

Q:

The exponential $f (x) = e^{x}$ is being defined by a power series:

Equation 1. Power series definition of

f (x) = e^{x}

\begin{matrix} e^{x} = 1 + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... \\ = \sum_{i = 0}^{\infty} \frac{x^{i}}{i!} \end{matrix}

Implement a class method double exp(double) inside a class Math choosing a package of your choice. The name clash with the Java standard class java.lang.Math is intended: You'll learn how to resolve naming conflicts.

Regarding practical calculations we replace the above infinite series by a limited one. So the number of terms to be considered will become a parameter which shall be configurable. Continue the implementation of the following skeleton:

Figure 288. An implementation sketch for the exponential

public class Math {
 ...

  /**
   * @param seriesLimit The last term's index of a power series to be included,
   */
  public static void setSeriesLimit(int seriesLimit) {...}

  /**
   * Approximating the natural exponential function by a finite
   * number of terms from the corresponding power series.
   *
   * A power series implementation has to be finite since an
   * infinite number of terms requires infinite execution time.
   *
   * The number of terms to be considered can be set by {@link #setSeriesLimit(int)}}
   *
   * @param x
   * @return
   */
  public static double exp(double x) {...}
}

Compare your results using seriesLimit=8 terms and the corresponding values from the professional implementation java.lang.Math.exp by calculating $e^{-3}$ , $e^{-2}$ , $e^{1}$ , $e^{2}$ and $e^{3}$ . What do you observe? Can you explain this result?

Do not forget to provide suitable Javadoc comments and check the generated HTML documentation for correctness.

Hints:

You should only use basic arithmetic operations like +, - * and /. Do not use Math.pow(double a, double b) and friends! Their implementations use power series expansions as well and are designed for a different purpose like having fractional exponent values.
The power series' elements may be obtained in a recursive fashion like e.g.:

$\frac{x^{3}}{3!} = \frac{x}{3} \frac{x^{2}}{2!}$

Answer 1

A:

Maven module source code available at P/Sd1/math/V1.
See hints regarding import.

Regarding the finite number of terms we provide a class variable ❶ having default value of 5 corresponding to just the first 1 + 5 = 6 terms:

e^{x} \approx 1 + \frac{x^{1}}{1!} + ... + \frac{x^{5}}{5!}

We also provide a corresponding setter method ❷ enabling users of our class to choose a different value:

public class Math {

  static int seriesLimit = 5; ❶

  /**
   *
   * @param seriesLimit The last term's index of a power series to be included,
   *   {@link }}.
   */
  public static void setSeriesLimit(int seriesLimit) { ❷
    Math.seriesLimit = seriesLimit;
  } ...

Calculating values by a finite series requires a loop:

public static double exp(double x) {
  double currentTerm = 1.,  // the first (i == 0) term x^0/0!
         sum = currentTerm;    // initialize to the power series' first term

  for (int i = 1; i <= seriesLimit; i++) {  // i = 0 has already been completed.
    currentTerm *= x / i;
    sum += currentTerm;
  }
  return sum;
}

You may also view the Javadoc and the implementation of double Math.exp(double). We may use the subsequent code snippet for testing and comparing our implementation:

Math.setSeriesLimit(6);

double byPowerSeries = Math.exp(1.);
System.out.println("e^1=" + byPowerSeries + ", difference=" +
                 (byPowerSeries - java.lang.Math.exp(1.)));

byPowerSeries = Math.exp(2.);
System.out.println("e^2=" + byPowerSeries + ", difference=" +
                 (byPowerSeries - java.lang.Math.exp(2.)));

byPowerSeries = Math.exp(3.);
System.out.println("e^3=" + byPowerSeries + ", difference=" +
                 (byPowerSeries - java.lang.Math.exp(3.)));

In comparison with a professional implementation we have the following results:

e^1=2.7180555555555554, difference=-2.262729034896438E-4
e^2=7.355555555555555, difference=-0.033500543375095226
e^3=19.412499999999998, difference=-0.67303692318767

Our implementation based on just 6 terms is quite good for small values of x. Larger values however exhibit growing differences.

This is due to the fact that our approximation is in fact just a polynomial of degree 6:

Figure 289. Comparing exponential and approximation

\begin{matrix} e^{x} & and \\ p_{6} (x) & = & 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} \\ + \frac{x^{4}}{4!} + \frac{x^{5}}{5!} + \frac{x^{6}}{6!} \end{matrix}

The plots show a particularly bad approximation for negative values. But for larger values outside the above scope the exponential will quickly diverge from its polynomial approximation as well.

Prev	Up	Next
Binomials, the recursive way	Home	Adding sine

Building a library of mathematical functions.

Implementing exponentials.

The exponential f ⁡ x = e x

The exponential $f (x) = e^{x}$