[plt-scheme] HtDP Exercise 23.4.1
Hi Matthias --
David, it is entirely safe to navigate around the mathematics sections in
the book
Yes, but I can't resist the challenge of tackling and hopefully solving an
explicitly "mathematical" problem which I always considered beyond my reach
:) (which is why I'm up at 5:30 am taking a whack at this thing!).
I looked at the link sent to me by Thomas Chust and came up with this
version of integrate-kepler (before reading your comments below, which I
will):
(define (integrate-kepler f left right)
(local ((define (area-of-trap a b1 b2)
(* a ( / ( + (f b1) (f b2)) 2)))
(define midpoint ( / (+ right left) 2))
(define alt-left (- midpoint left))
(define alt-right (- right midpoint)))
(+ (area-of-trap alt-left left midpoint)
(area-of-trap alt-right midpoint right))))
Am I in the ballpark?
Thanks!
Dave
On Tue, Feb 10, 2009 at 6:40 AM, Matthias Felleisen <matthias at ccs.neu.edu>wrote:
>
>
> David, it is entirely safe to navigate around the mathematics sections in
> the book. As someone pointed out, they exist because it's so easy to use
> Scheme to bring across rather complex mathematical concepts. Indeed, you
> won't believe it but by reaching part VI, you have mastered a significant
> chunk of mathematics that you would have never considered within reach had I
> told you at the outset of your studies that you'd understand rudimentary
> concepts from "higher" algebra (advanced college material).
>
> If you want to figure out some basics, try this.
>
> Can you draw the curve for (define (f x) x), (define (g x) 10), and
> (define (h x) (- 10 x)) between 0 and 10?
>
> Can you determine the area that is enclosed by f, g, and h, the X and Y
> axis between 0 and 10?
>
> If so, how did you compute those?
>
> These are special cases for the exercise that you are tackling. If your
> solution comes close to your answers, you know it works for special cases.
> That's good. For somewhat more advanced uses, you would consider more
> sophisticated cases, such as sin, cos, expt, etc. In all those cases,
> mathematics tells you how to come up with a close-enough answer through a
> few calculations.
>
> And that's the whole point of testing. You hope that it works for the rest
> of it.
>
> Give it a try. -- Matthias
>
>
>
>
>
>
>
>
> On Feb 9, 2009, at 9:47 PM, David Yrueta wrote:
>
> Hi Thomas --
>>
>> Thank you for the link. I'm not familiar with the mathematical notation
>> used to represent the trapezoid rule function, but it resembles Scheme
>> enough to give me some clue on how to go forward.
>>
>> Here's my best effort at translation:
>>
>> For this problem, argument "a" signifies the x-axis coordinate for the
>> "left interval" and argument "b" signifies the x-axis coordinate for "right
>> interval."
>>
>> The length of each interval is equal to it's y-coordinate value, which can
>> be derived by applying function f to it's x-coordinate. The length of each
>> interval also equals the two base lengths of a trapezoid. The altitude of
>> the trapezoid is equal to the difference between the x-coordinate values of
>> a and b. So, if the area of a trapezoid = altitude * ((base1+ base2) / 2),
>> then the formula for the area under the curve equals [(b - a) * ((f a) + (f
>> b) / 2)].
>>
>> To calculate the areas for two trapezoids, one needs only to divide the
>> interval between a and b by 2, apply f to that value, and then use the
>> result as base2 for the trapezoid with (f a) as base1, and base1 for the
>> trapezoid with (f b) as base2.
>>
>> Does that sound about right? If so, I'll translate it to Scheme. Perhaps
>> that would make my thoughts easier to evaluate.
>>
>> Thanks for the time!
>>
>> Thanks,
>> Dave
>>
>> On Mon, Feb 9, 2009 at 5:33 PM, Thomas Chust <chust at web.de> wrote:
>> 2009/2/10 David Yrueta <dyrueta at gmail.com>:
>> > [...] Still don't really understand how this relates to the question of
>> computing the
>> > "area of each trapezoid," though. [...]
>>
>> Hello,
>>
>> http://en.wikipedia.org/wiki/Trapezoidal_rule has some figures that
>> may make the idea behind this integration method more clear: The graph
>> of the function is approximated by a polyline -- in the special case
>> of Kepler's rule a very coarse one with only two edges; the
>> approximation of the integral is the area of the polygon bounded by
>> this polyline, the two integration interval boundaries and the
>> coordinate axis of the integration variable.
>>
>> cu,
>> Thomas
>>
>>
>> --
>> When C++ is your hammer, every problem looks like your thumb.
>>
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>
>
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