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How to get the approximation of $\ln 2$ and prove results using the knowledge of senior high school(China)See in the textbook.(the figure accurate to the third decimal place)? Above the question,everything we are supposed to use(without proving):

  1. derivative and its main formulas to find a function's derivative
  2. definite integral
  3. Newton-Leibniz formula

In fact,there's an orignal question about it.See it in the picture.

So,I find a method to calculate it.But it is slowly to use actually.And the official answer is so ingenious to construct a fit inequality. I use the definite integral.

Since$$lnx=\int\frac{1}xdx$$ then according to the Newton-Leibniz formula:$$ln2=ln2-ln1=\int_1^2 \frac{1}xdx$$ Then the interval 1 to 2 is then divided into ten parts and approximated using the connection of the points of $\frac{1}x$ at the endpoints of the interval.It's hard to calculate and its approximation only accurate to the second decimal place.

icase233
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    It's the unique $x$ for which $e^x=2.$ First since $e^0=1<2<e^1=2.718...,$ one knows it lies in interval $[0,1].$ Then do bisection method enough times to trap it to the accuracy needed. – coffeemath Apr 22 '23 at 02:44
  • One pretty easy approach amounts to "range reduction" $\ln 2 = \ln(e(2/e)$ as covered in this previous Question. If a high precision result were needed, then continued fractions are better than the well known power series, which for $\ln 2$ involves the infamously slowly converging alternating harmonic series. See the last part of this older Answer of mine for details. – hardmath Apr 22 '23 at 03:08

1 Answers1

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You can use Newton's Method for approximating roots of an equation. Consider:

$f(x) = e^x - 2$

When $x = \ln(2)$, $f(x) = 0$, so:

Make a guess, say $x_0=1$, then Newton's Method says:

$x_{n+1}=x_n-f(x_n)/f'(x_n)$

$x_1 = x_0 - f(x_0)/f'(x_0)$ $= 1 - (e-2)/e$

Then:

$x_2 = x_1 - f(x_1)/f'(x_1)$

When you get to $x_3$ it's accurate to 6 decimal places ($x_2$ is accurate to 2).

Alternatively, you can use the power series representation of $\ln(1+x)$:

$\ln(1+x) = x-x^2/2+x^3/3-x^4/4+...$

Evaluating at $x=1$ gives you a rational approximation, but it's only accurate to three decimal places after 500 terms

wyboo
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