## Calculating Logarithms

### December 18, 2009

Our solution is based on a description by Ed Sandifer. We define `epsilon`

as the desired accuracy of the final result:

`(define epsilon 1e-7)`

Newton’s method uses an initial estimate of 1.0, then calculates a new estimate using the formula for the derivative:

`(define (newton x2)`

(let loop ((x 1.0))

(if (< (abs (- (* x x) x2)) epsilon) x

(loop (- x (/ (- (* x x) x2) (+ x x)))))))

The implementation of Euler’s method merges two loops into one. The first pesudo-loop, at the `(< hi n)`

condition, calculates the smallest power of `base`

greater than `n`

. The second pseudo-loop performs Euler’s algorithm; `next`

is the geometric mean, `log-next`

is the arithmetic mean, and the `if`

chooses the sub-interval of the bisection:

`(define (euler base n)`

(let loop ((lo 1.0) (log-lo 0.0) (hi 1.0) (log-hi 0.0))

(cond ((< hi n) (loop lo log-lo (* hi base) (+ log-hi 1)))

((< (abs (- (/ log-lo log-hi) 1)) epsilon)

(/ (+ log-lo log-hi) 2))

(else (let* ((next (newton (* lo hi)))

(log-next (/ (+ log-lo log-hi) 2)))

(if (< next n)

(loop next log-next hi log-hi)

(loop lo log-lo next log-next)))))))

Here are some sample calculations:

`> (newton 2)`

1.4142135623746899

> (euler 10 612)

2.7867514193058014

> (expt 10 2.7867514193058014)

611.9999959982614

You can run the code at http://programmingpraxis.codepad.org/oOdXhS7g.

[…] Praxis – Calculating Logarithms By Remco Niemeijer In today’s Programming Praxis problem we have to implement Euler’s method for calculating logarithms […]

My Haskell solution (see http://bonsaicode.wordpress.com/2009/12/18/programming-praxis-calculating-logarithms/ for a version with comments):

Mine, in Python

[…] calculated the value of pi, and logarithms to seven digits, in two previous exercises. We continue that thread in the current exercise with a function that calculates […]

I note that the approved solution calculates (- (* x x) x2)) twice. Is there some reason you don’t use a temporary?

(define (newton x2)

(let loop ((x 1.0))

(let ((t (- (* x x) x2)))

(if (< (abs t) epsilon) x

(loop (- x (/ t (+ x x))))))))

In Erlang.

Interactive Erlang session