Pure Recursive Function for f(n) = 6[SUP]n[/SUP] + 6n

[lpau001]

"Pure Recursive Function" for f(n) = 6ⁿ + 6n

Howdy! First of all I was going to explain a 'Pure Recursive Function' as how my professor defined it: A pure recursive function is stated by only using previous values of the function.. Like: F(n) = f(n-1) + f(n-2) Which means you can't have a stray 'n' in the function.. Hard to explain, but maybe it will make more sense if I continue.

Homework Statement

Find a pure recursive function of 6^{n} +6n, if it exists.

Homework Equations

I started off by writing down the first few values..
f(1) = 12
f(2) = 48
f(3) = 234
f(4) = 1320
f(5) = 7806
f(6) = 46692

The Attempt at a Solution

After a few minutes of messing around with the numbers, I came up with this recursive function.

f(n) = 6[f(n-1) - (n[5]-6)]

I was feeling great, because it works and it only took me like 10 minutes to figure out! woohoo! EXCEPT when I asked my professor about it, he said that's great, but it's not a 'Pure Recursive Definition' because of the (n[5]-6) that 'n' has nothing to do with previous values of the function, so it's not a 'pure' recursive function!

Anyways.. On to the REAL problem!

How would I find a pure recursive definition? I've noticed that the limit of the function as n approaches infinity of F(n+1) / F(n) = 6. So I was multiplying the 'last' value of the function by 6, and then subtracting a constant to get the current value. What's interesting is that the constant goes up by 30 each time! (starts out at 24 for f(2), then 54, 84 ...etc.) so like f(5) = 6f(4) - 114 ...Ok, so how do I relate that constant to a previous value? I can't think of a good way.. I also tried multiplying the previous value by 5, and adding that constant, but its way too large..

Anyways, I've been stuck on this for a few days. I'm wondering if I need to try some more complex algebra stuff instead of just simple division and addition / subtraction.

Thanks for any help!

 

[Curious3141]

Try this:

f(n) = 6^n + 6n ---eqn 1

f(n+1) = (6)(6^n) + 6n + 6 --eqn 2

Rearrange eqn 1 to get:

6^n = f(n) - 6n ---eqn 3

Sub eqn 3 into eqn 2:

f(n+1) = 6f(n) - 36n + 6n + 6 = 6f(n) - 30n + 6 ---eqn 4

Now find a similar expression for f(n-1) like in eqn 2, and again replace the 6ⁿ term with eqn 3. Multiply the resulting equation by a certain constant so that you can now add it to eqn 4 and cause the terms in n to cancel out. Just rearrange to get the form you want, and you're done.

 

[lpau001]

Try this:

f(n) = 6^n + 6n ---eqn 1

f(n+1) = (6)(6^n) + 6n + 6 --eqn 2

Rearrange eqn 1 to get:

6^n = f(n) - 6n ---eqn 3

Sub eqn 3 into eqn 2:

f(n+1) = 6f(n) - 36n + 6n + 6 = 6f(n) - 30n + 6 ---eqn 4

Now find a similar expression for f(n-1) like in eqn 2, and again replace the 6ⁿ term with eqn 3. Multiply the resulting equation by a certain constant so that you can now add it to eqn 4 and cause the terms in n to cancel out. Just rearrange to get the form you want, and you're done.

Wow.. Simply Amazing!

I really need to work more on using the function to help me out, instead of just trying to figure out how the function works!

You really made this problem look easy! Thank you very much!

 

[Curious3141]

Wow.. Simply Amazing!

I really need to work more on using the function to help me out, instead of just trying to figure out how the function works!

You really made this problem look easy! Thank you very much!

You're welcome.