Question

How do you evaluate `ln 5`?

Answer 1

You can use Newton's method to help a little.

Explanation:

`ln 5` is the zero of the function `f(x) = e^x-5`

If you can calculate `e^x` then you can calculate approximations for `ln 5` using Newton's method...

`f'(x) = e^x`

Starting with an approximation `a_0`, iterate to get better approximations using the formula:

`a_(i+1) = a_i - (f(x))/(f'(x))=a_i - (e^x-5)/e^x=a_i - 1 + 5/e^x`

As for `e^x` you can use the Maclaurin series, which converges for any value of `x`:

`e^x = sum_(k=0)^oo x^k/(k!)`

I would be interested to know how well this method works using a truncated version of this series for `e^x`. How many terms for `e^x` does it require to get a `5` decimal place approximation for `ln 5`?

I may dig a little deeper...

Using `sum_(k=0)^9 x^k/(k!)` as an approximation for `e^x` and Newton's method, the approximation for `ln 5` I found was:

`ln 5 ~~ 1.6094454239`

Using `sum_(k=0)^10 x^k/(k!)` I found:

`ln 5 ~~ 1.6094389965`

Compare those with the proper value:

`ln 5 ~~ 1.6094379124`

If you do want to evaluate:

`sum_(k=0)^10 x^k/(k!)`

then it may be easier to use the form:

`sum_(k=0)^10 x^k/(k!) = 1+x(1+x/2(1+x/3(1+x/4(1+x/5(1+`

`x/6(1+x/7(1+x/8(1+x/9(1+x/10)))))))))`