3 ms·

The J language makes trivial to solve this problem without a tetration operator: (^~^:_1)5 2.12937

• H8crilA 7y ago
  How's that implemented? Newton's method like in the OP? I.e. if there's at least one solution there's an okay chance to find at most one solution?
  • xelxebar 7y ago
    Essentially, yes. > (^~^:_1) 5 First, let me break this down, in the interest of sharing the joys of J. The thing in parenthesis is our inverse operation of x^x; we then evaluate that at 5. The first ^ is exponentiation and the following ~ makes binary operators into unary ones by evaluating on the diagonal. In other words it turns f(x,y) into f(x) = f(x,x), which for x^y corresponds to x^x. Next we have the ^: modifier. This is function iteration, i.e. exponentiation over function composition. The function is on the left (here ^~) and the count on the right (here _1). So f ^: 2 means f(f(x)). In fact, this function exponentiation is extended to negative integers as well! In J, negative numbers are prefixed with _ (underscore), and f^:_1 actually gives us the inverse of f (if it exists)! Anyway, we can see inspect the definition for an operation's inverse using the b. verb. From the jconsole: ^~ b. _1 3 : '(- -&b@(*^.) % >:@^.)^:_ ]1>.b=.^.y'"0 :.(^~) The part we are interested in is in single quotes. Pulling this one apart would involve introducing the various operator combinators in J, so suffice it to say that the ^:_ part numerically approximates a fixed point of the part in parenthesis, which essentially is a straightforward transcription of the recursive relation in Newton's method.