How to solve the following equation using accumulate (Scheme)

scheme

I'm trying to do the following problem (there is a formula so I print-screened and uploaded it)

Formula http://img248.imageshack.us/img248/6558/problemh.jpg
(http://img248.imageshack.us/img248/6558/problemh.jpg)

Using accumulate:

(define (accumulate combiner null-value term a next b) (if (> a b) null-value (combiner (term a) (accumulate combiner null-value term (next a) next b))))

but I have pretty much no idea how to do it. It should look something like
(define (sum-func f k)...?
I'm not sure how I should define f(x+ j*h) so that to put the changing number j (changes from 0 to the given number k) inside of it… In other words, I'm pretty lost.

Thank you for the fast respond. I'm studying for my midterms and I'm trying to do the midterms from the previous years, but I just got stuck here. I know how to use the accumulate and everything in it, but I just don't know how to do the f(x + jh). I'm trying this:
(define (sum-func term a b h x) (define (next a) (+ a 1)) (define (term a) (term (+ x (* h a)))) (accumulate + 0 (term a) a (next a) b))
But it's not working…
In other words I don't know how to use the 'next' (changing part) inside the term..
(sorry if I'm explaining very clearly – English is not my native)

Best Answer

I am not sure if this is a homework problem or not but in any case I will provide you with some ideas on how to solve it. For the sake of simplicity let us assume our problem is to find the sum of squares of all numbers between j and k. You can use your accumulate function above to solve this simple problem. All you will have to do is define your combiner, term and next functions for this problem along with defining the final null-value.

Term function here should calculate the squares for each j. Therefore term will take j as a argument & will return the square of j as the result. (define (term j) (sqr j))

Next function should get the next j in the sequence. (define (next j) (+ j 1)

Combiner function should combine two terms together. (define (combiner t1 t2) (+ t1 t2))

Finally, null-value signifies the stop condition, the last value that should be passed to combiner when we have accumulated all the values from j to k. All we have to do is just define it as a zero in this case. (define null-value 0)

For your problem, most of these functions are same except for the term function. In this problem, the term function was a very simple one, it just found the square of the supplied number. However, in your case, it is not as simple as finding the square of j since you have several other constants defined.

Hope this will help you solve the problem.

If you need even more insight into this, you may want to look at the accumulate problems in SICP chapter 2.

Related Solutions

Using Scheme code to solve a quadratic equation

Well, one reason is that it's all on one line. You can make it more readable using something called pretty-printing, where you break it up into multiple lines and use whitespace:

(define (solve-quadratic-equation a b c)
  (define disc (sqrt (- (* b b)
                        (* 4.0 a c))))
  (/ (+ (- b) disc)
     (* 2.0 a)))

This way you can more clearly see the structure of the expressions.

And here's a quote from SICP:

There is no limit (in principle) to the depth of such nesting and to the overall complexity of the expressions that the Lisp interpreter can evaluate. It is we humans who get confused by still relatively simple expressions such as
(+ (* 3 (+ (* 2 4) (+ 3 5))) (+ (- 10 7) 6))
which the interpreter would readily evaluate to be 57. We can help ourselves by writing such an expression in the form
(+ (* 3
      (+ (* 2 4)
         (+ 3 5)))
   (+ (- 10 7)
      6))
following a formatting convention known as pretty-printing, in which each long combination is written so that the operands are aligned vertically. The resulting indentations display clearly the structure of the expression.

Lisp: How to write a Higher Order Function

This is a great way to learn how to fish. Much better than being given a fish.

Until then, here's how to approach the problem. Write a function which would do what (accumulate + 0) would do. Don't use the accumulate function; just write a defun which which does what your homework asks. Next, write a function which would do what (accumulate * 1) would do. What are the similarities, what are the differences between the two functions. For the most part, they should be identical except for the occurrence of the + and * operators.

Next, note that the accumulate function is to return a function which will look a lot like the two functions you wrote earlier. Now, using the insight that two functions you wrote are very similar, think how to apply that to the function which (defun accumulate ...) is to return.

Best Answer

Related Solutions

Using Scheme code to solve a quadratic equation

Lisp: How to write a Higher Order Function

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