SOLUTION: A function f is defined on the set of positive integers by f(1) = 1, f(3) = 3, f(2n) = n, f(4n + 1) = 2f(2n + 1) − f(n), and f(4n + 3) = 3f(2n + 1) − 2f(n) for all positive in

Question 1199565: A function f is defined on the set of positive integers by f(1) = 1, f(3) = 3, f(2n) = n, f(4n + 1) = 2f(2n + 1) − f(n), and f(4n + 3) =
3f(2n + 1) − 2f(n) for all positive integers n. Determine the summation of [f(4n + 1) + f(2n + 1) − f(4n + 3)] if i = 1 and n=10.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

I'm not sure how to read the summation you/we are supposed to find. So I'll leave that part of the problem to you; I will only show how to get the function values f(n) for each integer n.

(1) We are given f(1) = 1 and f(3) = 3; for all other integers n, the values of f(n) will be determined recursively from those two given function values.

(2) f(2n) = n

This defines the function values of the function when the function argument is even: f(2) = 1; f(4) = 2; f(6) = 3; etc.

(3) When the function argument is odd, it is either of the form 4n+1 or 4n+3. The other two pieces of given information tell us how to find the function values when the argument is odd (and not equal to 1 or 3).

The two recursive definitions for these function values are

f(4n + 1) = 2f(2n + 1) − f(n) [1]
f(4n + 3) = 3f(2n + 1) − 2f(n)[2]

Using these definitions....

f(5): 5 = 4(1)+1, so using [1], f(5) = 2(f(3))-f(1) = 2(3)-1 = 5
f(7): 7 = 4(1)+3, so using [2], f(7) = 3(f(3))-2(f(1)) = 3(3)-2(1) = 7
f(9): 9 = 4(2)+1, so using [1], f(9) = 2(f(5))-f(2) = 2(5)-1 = 9
f(11): 11 = 4(2)+3, so using [2], f(11) = 3(f(5))-2(f(2)) = 3(5)-2 = 13
f(13): 13 = 4(3)+1, so using [1], f(13) = 2(f(7))-f(3) = 14-3 = 11
f(15): 15 = 4(3)+3, so using [2], f(15) = 3(f(7))-2(f(3)) = 21-6 = 15
f(17): 17 = 4(4)+1, so using [1], f(17) = 2(f(9))-f(4) = 18-2 = 16
f(19): 19 = 4(4)+3, so using [2], f(19) = 3(f(9))-2(f(4)) = 27-4 = 23
...

I'm not seeing an easy pattern....

And, as I stated earlier, I'm not sure what summation we are supposed to be doing....

So I will end my response here.

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