Question

Find f(1), f(2), and f(3) if f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, . . .

• f(n+1) = f(n) + 2

So, would it be f(n) = f(n+1) + 2? Or would I just keep it like the original and plug in 1, 2, 3. Thanks for any helpful replies.

Answer 1

As in algebra, put your known values on the right hand side, and the quantity to be evaluated on the left. Follow the definition of the function.

So if f(0)=1, then n=0, therefore
f(1)=f(0)+2
...

Answer 2

just add 2 each time
f(0) = 1
f(1) = 3
f(2) = 5
f(3) = 7 etc
That is an arithmetic sequence
f(n) = 1 + 2n

Answer 3

Ok thank for the responses, but there seems to be a contradiction between the two. Wouldn't f(1) = 1 + 2, which equals 3?

Answer 4

The f(n+1) is throwing me off what does that mean?

Answer 5

As Damon mentioned, it is an arithmetical sequence (or in general, any sequence).
So the zeroth term is denoted f(0), the 1st term f(1), etc.
So for
1,3,5,7,9,...
f(0)=1
f(1)=3
f(2)=5
...
the relation between them is therefore
f(n+1)=f(n)+2

This is called a recurrence relation, and you can only evaluate a certain term if the previous terms are know.

For the Fibonacci sequence, it would be
f(n)=f(n-1)+f(n-2)
f(0)=0
f(1)=1

so
f(2)=f(1)+f(0)=1+0=1
f(3)=f(2)+f(1)=1+1=2
f(4)=f(3)+f(2)=2+1=3
f(5)=f(4)+f(3)=3+2=5
....
to give the Fibonacci sequence
1,1,2,3,5,8,13,21,34....

Answer 6

OK example: f(n+1) = 3f(n)

f(1) = 3
f(2) = 6
f(3) = 9

Right?

Answer 7

Almost!
What you have done was for
f(n+1)=f(n)+3, and f(1)=3.
so f(2)=f(1)+3=3+3=6
f(3)=f(2)+3=6+3=9.

If you are solving f(n+1)=3f(n) with f(1)=3, then
f(2)=3*f(1)=3*3=9
f(3)=3*f(2)=3*9=27
f(4)=3*f(3)=3*27=81,
...
effectively a geometric sequence instead.

Answer 8

Oops. . .Sorry disregard previous post. . .

Answer 9

f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, . . .

Find f(1), f(2), f(3)

Answer 10

Is this the previous problem where f(n)=f(n-1)+3, or is this a new problem?

I don't see the recursive definition of f(n).

Answer 11

Yes, they both follow the same recursive definition. I was just trying the second part on my own to see if I understand. Sorry about the misunderstanding. . .