Hello World

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Here’s some LaTeX:

a0+a0a1+a0a1a2+⋯+a0a1a2⋯an=a01−a11+a1−a21+a2−⋱⋱an−11+an−1−an1+an a_{0} + a_{0}a_{1} + a_{0}a_{1}a_{2} + \cdots + a_{0}a_{1}a_{2} \cdots a_{n} = \cfrac{a_{0}}{1 - \cfrac{a_{1}}{1 + a_{1} - \cfrac{a_{2}}{1 + a_{2} - \cfrac{\ddots}{\ddots \cfrac{a_{n-1}}{1 + a_{n-1} - \cfrac{a_{n}}{1 + a_{n}}}}}}} 
(defun cont-frac (n d k)
  (labels ((iter (i result)
             (if (= i 0)
                 result
                 (iter (- i 1) (/ (funcall n i) (+ (funcall d i) result))))))
    (iter k 0)))

(format t "~a" (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 10))

What if we also wanted some in-line LateX? We could write an inline equation: E=mc2E = mc^2. That’s possible with \( delimiters. Or perhaps some inline code? Well, that is possible too, by using single backticks. Consider the above Lisp magic. We can explain how it works and use backticks to highlight its parts.

This Lisp code defines a function cont-frac, which computes a continued fraction. The function uses the helper iter, defined inside cont-frac using labels for local recursion to iterate and calculate the fraction.

We write a lambda function to handle the numerator and denominator values, which are then passed back as functions rather than values. The iter function is where the magic happens. It keeps calling itself with a reduced iteration count (i - 1), counting down until i has reached 0. And at each step, it computes a fraction.

The program terminates and outputs our final value when the lamdba’s index hits 0. Basically the Lisp interpreter is being told, “hey, simulate 10 iterations of 1 / (1 + 1 / (1 + 1 /...))” by the lambda function.

Image
Saint Catherine of Siena

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