How do give the recursive form for a certain function?

[thename1000]

I only have the explanation for linear homogeneous ones. I don't know how to do these:

Give a recursive form (including bases) for the following functions:
a. f(n) = 2 + (-1)^n
b. f(n) = n(n+1)
c. f(n) = 2n - 2
d. f(n) = n^2 – n
e. f(n) = 3^n + n⋅3^n

If someone could explain one or two it would help alot.

Thanks

[honestrosewater]

Maybe someone could help if you gave more information. What are the domains and codomains supposed to be? What counts as a recursive form (what examples do you have)?

Without more info, my best guess is something like this. For (a), assuming n ranges over N, when n is odd, f(n) = 1; when n is even, f(n) = 3. So you could take n = 1 as your base:
f(1) = 1
and then use the fact that oddness and evenness alternate in N:
if f(n-1) = 1, then f(n) = 3; otherwise, f(n) = 1
But there are several ways that you could express that, and your teacher or book might want something else.

(b) can be expressed as the product of 2 and the sum of 1 to n:
f(1) = 2(1) = 2
f(2) = 2(1 + 2) = 6
f(3) = 2(1 + 2 + 3) = 12
...
So this definition involves another recursive function.

[HallsofIvy]

I only have the explanation for linear homogeneous ones. I don't know how to do these:

Give a recursive form (including bases) for the following functions:
a. f(n) = 2 + (-1)^n
b. f(n) = n(n+1)
c. f(n) = 2n - 2
d. f(n) = n^2 – n
e. f(n) = 3^n + n⋅3^n

If someone could explain one or two it would help alot.

Thanks
Step one, compute a few of the values to get a handle on exactly what it is.
for (a), f(1)= 2- 1= 1, f(2)= 2+ 1= 3, f(3)= 2-1= 1, etc. In other words, f(n)= 1 if n is odd, 3, if n is even.

Step two, try to find a relation between f(n) and f(n+1).
f(n+1)- f(n)= 2+ (-1)^(n+1)- 2- (-1)^n= (-1)(-1)^n- (-1)^n= (-2)(-1)^n.

Step three, check that this recursive formula gives the same thing you got in step one to reassure yourself.

f(n+1)= f(n)+ (-2)(-1)^n. That, together with f(1)= 1, is a recursive formula for this sequence.
Note that, according to this, f(2)= f(1)+ (-2)(-1)^1= 1+ 2= 3, f(3)= f(2)+ (-2)(-1)^2= 3- 2= 1, etc.

For b, f(n)= n(n+1), f(1)= 1(2)= 2, f(2)= 2(3)= 6, f(3)= 3(4)= 12, etc. f(n+1)- f(n)= (n+1)(n+2)- n(n+1). Factoring out (n+1), f(n+1)- f(n)= (n+1)(n+2-n)= 2(n+1). f(n+1)= f(n)+ 2(n+1), and f(1)= 2.

By that formula f(2)= f(1)+ 2(2)= 2+ 4= 6, f(3)= f(2)+ 2(3)= 6+ 6= 12, etc.

For (2), since there is no addition or subtraction (addition does not work well together with multiplication!), you could also consider the relation between f(n+1) and f(n) by dividing: f(n+1)/f(n)= (n+1)(n+2)/n(n+1)= (n+2)/n. f(n+1)= [(n+2)/n]f(n) with f(1)= 2, is another recurvsive formula for the same sequence. f(2)= [(1+2)/1]f(1)= 3(2)= 6, f(3)= [(2+2)/2]f(2)= 2(6)= 12, etc.