Scheme "application: not a procedure;" when calculating a derivative

Question

I am brand new to Scheme and am working on an assignment to implement stochastic gradient descent. So far I believe I have the structure of the program correct however my procedure that takes the derivative of a function f(x) is giving me some trouble. In my "try" loop at the bottom of the code I recursively call (try (func-eval guess)) where (func-eval guess) calculates the next guess of my function's local minimal with the formula *x - alpha*f'(x)* where alpha = 0.1.

I seem to be getting an error when calculating the derivative... I am using Dr.Racket IDE and it has highlighted this following line as being problematic: (f (+ x dx)) ... which is the second line in my local derivative procedure:

(define (local-minimal first-guess)

;A way to check a guess
(define(good-enough? val1 val2)
(<(abs(- val1 val2)) 0.00001))

; x_new = x_old - alpha*f'(x)    /// f(x)=(x+1)^2+2  //// alpha = 0.1
(define (func-eval x)
  (- x (* 0.1((derivative (+ 2(expt (+ x 1) 2)) 0.00001)x))))

  (define (derivative f dx)
  (lambda (x)
    (/ (- (f (+ x dx)) (f x))
       dx)))

; trys the guess
(define (try guess)
  (if (good-enough? guess -1)
      guess
      (try (func-eval guess))))
(try first-guess)) 

I am getting an error saying:

application: not a procedure;
 expected a procedure that can be applied to arguments
  given: 3
  arguments...:
   -1.99999

Is this a syntax error? I thought that I would be able to say f(x+dx) by using (f (+ x dx)) .... does this mean that I need to put an operator before the f in those parenthesis?

Answer 1

When calling derivative, the first argument has to be a function. In the following procedure call, the expression in the first argument gets evaluated to a number, not a function:

(derivative (+ 2 (expt (+ x 1) 2)) 0.00001)

To fix it, pack the expression inside a lambda, which makes it an actual function:

(derivative (lambda (x) (+ 2 (expt (+ x 1) 2))) 0.00001)