Natural Logs

Key Questions

  • The Answer is 1.

    ln(e) is the same thing as log_e(e)

    Because e^1 = e, log_e(e) = 1

  • Answer:

    You can approximate ln x by approximating int_1^x 1/t dt using Riemann sums with the trapezoidal rule or better with Simpson's rule.

    Explanation:

    For example, to approximate ln(7), split the interval [1, 7] into a number of strips of equal width, and sum the areas of the trapezoids with vertices:

    (x_n, 0), (x_n, 1/x_n), (x_(n+1), 0), (x_(n+1), 1/(x_(n+1)))

    If we use strips of width 1, then we get six trapezoids with average heights: 3/4, 5/12, 7/24, 9/40, 11/60, 13/84.

    If you add these, you find:

    3/4+5/12+7/24+9/40+11/60+13/84 ~~ 2.02

    If we use strips of width 1/2, then we get twelve trapezoids with average heights: 5/6, 7/12, 9/20,...,25/156,27/182

    Then the total area is:

    1/2 xx (5/6+7/12+9/20+...+25/156+27/182) ~~ 1.97

    Actually ln(7) ~~ 1.94591, so these are not particularly accurate approximations.

    Simpson's rule approximates the area under a curve using a quadratic approximation. For a given h > 0, the area under the curve of f(x) between x_0 and x_0 + 2h is given by:

    h/3(f(x_0) + 4f(x_0+h) + f(x_0+2h))

    Try improving the approximation using Simpson's rule, using h = 1/2 then we get six approximate areas to sum:

    h/3(25/6+73/30+145/84+241/180+361/330+505/546)~~1.947

    That's somewhat better.

  • ln(x) is the same as saying log_e(x). e is the base of the natural log.

Questions