Recursively calculate e^x with Java using MacLaurin Series

问题

I need to write a recursive java method that will compute e^x called e(x,n) with the signature,

public static double eThree(double x, long n)

and it must use the MacLaurin series to compute e^x, which is the following observation:

1 + x(1 + x/2( 1 + x/3(1 + x/4(1 + ... ))))
e(x,0)= 1 A call to this would return a result of 1

With some help earlier, I was able to make one that didn't require this format, but I'm not sure what I would code in order to use the format above. Thanks guys, really appreciate any help that will be given!

回答1:

Does this work for you? I believe it implements the algorithm but it does not produce x^n.

public static double eThree(double x, long n) {
    return eThree(x, n, 1);
}

private static double eThree(double x, long n, int div) {
    double d = 1;
    if (n > 0) {
        d = d + (x / div) * (eThree(x, n - 1, div + 1));
    }
    return d;
}

it seems to think that:

2^0 = 1.0
2^1 = 3.0
2^2 = 5.0
2^3 = 6.333333333333333
2^4 = 7.0
2^5 = 7.266666666666667
2^6 = 7.355555555555555
2^7 = 7.3809523809523805
2^8 = 7.387301587301588
2^9 = 7.388712522045855

回答2:

public static double eThree(double x, long n) {
  return eThreeRec(x, n, 1);
}

public static double eThreeRec(double x, long n, long i){
  if(i==1) return 1;
  else{
    return 1 + (x/i)*eThreeRec(x, n, i+1);
  }
}

Can't test it now though.

回答3:

If I understand correctly, n here should represent the precision of the McLaurin series (i.e. up to how many terms we calculate). In that case, you should decrement the n counter every time you invoke the method recursively. You could write something like:

public static double eThree(double x, long n) {
  return eThreeRec(x, n, 1);
}

public static double eThreeRec(double x, long n, long i) {
  if( i >= n ) {
    return 1.0;
  } else {
    return 1.0 + (x/i) * eThreeRec(x, n, i + 1);
  }
}

来源：https://stackoverflow.com/questions/22065279/recursively-calculate-ex-with-java-using-maclaurin-series

标签

java

recursion