Simple..............or is it?

[cooldog182]

Hey there, despite my efforts I just can't see a function for :-

f(1) = 2
f(2) = 1
f(3) = 3

and also

g(1) = 1
g(2) = 3
g(3) = 3

these should be simple right?

 

[cooldog182]

can I make the assumption for :-

f(1) = 2
f(2) = 1
f(3) = 3

to be that f(n)= f(n-1) + f(n-2) ?

In this case it would force:-

f(0) = -1
f(-1) = 3

 

[galactus]

Try setting up a system of equations and solving.

Try a quadratic.

For instance \(\displaystyle ax^{2}+bx+c=2\)

Sub in x=1. Set the other two up accordingly by setting them equal to

1 and 3 and subbing in the x value given.

You'll have three equations with three unknowns: a,b,c

 

[cooldog182]

ah ok, i'll give it a go! thanks!

 

[daon]

\(\displaystyle f(n) = (2^{n-1}-1) + 2 \lfloor \frac{1}{n} \rfloor\)

 

[cooldog182]

Ah thanks, but that doesn't really work for n = 3

 

[cooldog182]

ah yes! i did them using 3 simultaneous equations! thank-yew!

 

[tkhunny]

can I make the assumption for

You can make ANY assumption you like. That is why there are infinitely many solutiuons. That is why many feel these silly problems a grave waste of time.