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Topic: Functional equation (addendum)
Replies: 1   Last Post: May 28, 2005 10:46 PM

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Vincent Johns

Posts: 89
Registered: 1/25/05
Re: Functional equation (addendum)
Posted: May 28, 2005 10:46 PM
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In my previous posting, I forgot to mention something...

Vincent Johns wrote:
> Will wrote:
>

>> William Elliot wrote:
>>
>>

>>> Subject : Re: Functional equation
>>>
>>> On Mon, 23 May 2005, Will wrote:
>>>
>>>

>>>> A function f(x) has the following properties:
>>>> 1) f(1) = 1
>>>> 2) f(2x) = 4f(x) + 6
>>>> 3) f(x+2) = f(x) + 12x + 12
>>>>

>>>
>>> f(2x + 2) = 4f(x + 1) + 6
>>> f(2x + 2) = f(2x) + 12*2x + 12 = 4f(x) + 24x + 18
>>> f(x + 1) = f(x) + 6x + 3
>>>
>>> f(x + 1) - f(x) = 6x + 3
>>> f(x) = 3x(x - 1) + 3x + c = 3x^2 + c

>>
>> ...
>>
>> I understand your argument up to and including the line:
>> "f(x + 1) - f(x) = 6x + 3"
>>
>> But how do you conclude that:
>> "f(x) = 3x(x - 1) + 3x + c = 3x^2 + c" ?
>>

[...]
>
> First differences:
> Delta(x) := f(x + 1) - f(x) = 6x + 3
>
> Second differences:
> Delta^2(x) := Delta(x + 1) - Delta(x)
> = (6*(x + 1) + 3) - (6*x + 3)
> = (6*x + 9) - (6*x + 3)
> = 6
> and since the 2nd difference is a constant value, we know that the
> coefficient of the x^2 term is this divided by 2!, in other words = 6/2!
> = 6/2 = 3, so the polynomial fitting the points has the form
> p(x) := 3*x^2 + b*x + c

[...]

(Incidentally, I used some sloppy notation -- instead of "Delta(x)" I
should have written "Delta(f(x))", since I'm interested in the
differences of the values of the function, not of x. Sorry.)

From here, instead of using Newton's polynomial-fitting formula as I
suggested in my earlier post, you could (knowing that 3*x^2 is the
highest term) just subtract off the 3*x^2 values, leaving something of
the form

q(x) := b*x + c

which you can analyze the same way.

First, subtract 3*x^2 from each term, then compute another table of
differences:

x 1 2 3 ...
f(x) 1 10 25 ...
3*x^2 3 12 27 ...
q(x) :=
f(x) - 3*x^2 -2 -2 -2 ...
Delta(q(x)) 0 0 ...

Aha! (It's probably kind of obvious how this function behaves, but I'll
show the steps for the sake of illustration).

The coefficient of the linear term (value of b in p(x) = 3*x^2 + b*x +
c) is the constant value of Delta(q(x)) divided by 1!, so in this case

b = 0/1! = 0/1 = 0

and if it weren't zero, for the next step we could subtract b*x from the
values of (f(x) - 3*x^2) to find the constant term, c. But, since
subtracting zero won't change the values much, it's probably apparent
that c = -2.

So the (unique) polynomial fitting these values is

p(x) = a*x^2 + b*x + c
= 3*x^2 + 0*x + (-2)
= 3*x^2 - 2

Usually, Newton's formula will give you an answer more directly, but in
this case, I guess just subtracting the 3*x^2 makes the result obvious
pretty quickly.

-- Vincent Johns <vjohns@alumni.caltech.edu>
Please feel free to quote anything I say here.



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