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Whats the use of this?

  1. May 20, 2007 #1
    Hello everybody,

    I wanted to know the uses of such a function definition;

    f(x) = 3*f(x-1) - 3*f(x-2) + f(x-3) This works in linear and quadratic functions.

    I'll thankfull for your replies :)
  2. jcsd
  3. May 20, 2007 #2

    matt grime

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    In what way does that 'work in linear and quadratic functions'?
  4. May 20, 2007 #3
    As an example; say x^2 & assuming x = 4

    f(4) = 3*f(3) - 3*f(2) + f(1) which is true: 16 = 3*9 - 3*4 +1

    this is what I meant...

    Do you know any intresting uses of such a thing?
    Thanks for contributing :)
  5. May 20, 2007 #4


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    Let me explain how you can derive an equation like that one and why it works.

    Say you started with a very simple linear recurrence relation,

    1. f(x) = f(x-1) : This works for f(x) = 1.

    Now lets look for one that works for f(x) = x and also for f(x) = 1.

    Say f(x) = c f(x-1) + d f(x-2). Subst f(x)=1 and you find that "c+d=1". Next subst f(x)=x and you find that in addition to "c+d=1" we must also have "c + 2d = 0". Solving these give c=2 and d=-1. So our second recurance relation is,

    2. f(x) = 2 f(x-1) - f(x-2) : This works with f(x) = 1 and with f(x) = x

    Similar working with three simultaneous equations you can find the coefficients of a third order linear equation that works for f(x)=1 and for f(x)=x and for f(x)=x^2. This one turns out to be the one that you started this thread with. That is,

    3. f(x) = 3f(x-1) - 3f(x-2) + f(x-3) : This works with f(x)=1 and with f(x)=x and with f(x)=x^2

    Note that these are all linear mappings of the form, f(x) = L{ f(x-1), f(x-2), ...}

    It is important to realize that these are all linear in "f", whether or not f is linear in x is irrelevant to this point.

    By linearity L{ a f_1 + b f_2 + ... } = a L{f_1} + b L{f_2} + ...

    So you can see for example that since the second relation above "works" for f_1(x)=1 and f_2(x)=x then it automatically must work for all linear functions f(x) = ax+b.

    For the same reason the third relation must hold for all quadratic funtions and so on.
    Last edited: May 20, 2007
  6. May 21, 2007 #5
    Thankyou very much Uart, but I think I don't have enough knowledge of mathmatics to understand this derivation but I'll certainly keep it in a safe place and check it out once I've learned more maths... Thanks once again :)

    Actually I have derived this one by myself but using a very different technique which is given as follows;

    lets say we have 3 consecutive terms; f(5), f(4) & f(3) .

    f(5) - f(4) = rate of increase between these terms.

    f(4) - f(3) = rate of increase between these terms, which is sth like first derivative.

    Now ; f(5) - f(4) - ( f(4) - f(3) ) = rate of increase between the rates of increase which is sth like 2nd derivative.

    2nd derivative of quadratic and linear functions is always constant, using this fact and a little logic we'll have;

    f(5) - f(4) - ( f(4) - f(3) ) + ( f(5) - f(4) ) + f(5) = f(6)

    which means; f(6) = 3*f(5) - 3*f(4) + f(3)


    Thats what I did...

    So what are the uses of such function definitions?

    One use of this was solving projectile motion without being concerned about the equation of motion & having quick estimates for many functions without trying to evaluate the function's eq. especially when concerned with an irregular function.

    I'll be eagerly waiting for your reply :)
    Thanks once again.
    Last edited: May 21, 2007
  7. May 21, 2007 #6
    So the essentials of this method uart is doing is to think up of a recurrence relation and find let certain functions you want the recurrence relation to work with, to find out the exact recurrence relation? Relating the functions
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