## Whats the use of this?

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 :)

 Recognitions: Homework Help Science Advisor In what way does that 'work in linear and quadratic functions'?

 Quote by matt grime In what way does that 'work in linear and quadratic functions'?
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 :)

Recognitions:

## Whats the use of this?

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.

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

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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.