Asked by Francesca
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.
• 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.
Answers
Answered by
MathMate
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
...
So if f(0)=1, then n=0, therefore
f(1)=f(0)+2
...
Answered by
Damon
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
f(0) = 1
f(1) = 3
f(2) = 5
f(3) = 7 etc
That is an arithmetic sequence
f(n) = 1 + 2n
Answered by
Francesca
Ok thank for the responses, but there seems to be a contradiction between the two. Wouldn't f(1) = 1 + 2, which equals 3?
Answered by
Francesca
The f(n+1) is throwing me off what does that mean?
Answered by
MathMate
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....
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....
Answered by
Francesca
OK example: f(n+1) = 3f(n)
f(1) = 3
f(2) = 6
f(3) = 9
Right?
f(1) = 3
f(2) = 6
f(3) = 9
Right?
Answered by
MathMate
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.
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.
Answered by
Francesca
Oops. . .Sorry disregard previous post. . .
Answered by
Francesca
f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, . . .
Find f(1), f(2), f(3)
Find f(1), f(2), f(3)
Answered by
MathMate
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).
I don't see the recursive definition of f(n).
Answered by
Francesca
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. . .