Finding a closed form..

[daon]

I'm having difficulties with a problem in which we are given an initial value and a recursive formula for computing any other values, and we need to find a closed form of the sequence (that is, a single formula that will give the nth number in the sequence).

I am given:

\(\displaystyle \\
b(1) = 1\\
b(n>1) = \sum\limits_{i=1}^{n-1} b_{i}\)

Using this given information, I have found that:
\(\displaystyle b_1 = 1\\b_2=1\\b_3=2\\b_4=4\\b_5=8\\b_6=16\\b_7=32\\\)

Now, 2 things scream out to me. The two ones in the beginning tell me it may be a variation of the fibonacci sequence, of which I might be able to manipulate the closed form of into the above. The second thing is the 1,2,4,8,16,... screams out to me 2^n-1, but I can't figure out how to get the double 1's in the beginning.

Any help appreciated.
-Daon

 

[pka]

Use the ceiling function: \(\displaystyle \L
b_n = \left\lceil {2^{n - 2} } \right\rceil\).

 

[daon]

The least-integer function has not yet been introduced in this class. I wish it were that easy...

You see, finding the closed form is supposed to be the easy part. We then need to prove the closed form once we find it, and the concept of the ceiling function has not been discovered by the martians yet, as my professor would say.

 

[daon]

I have not yet discovered any function able to output this sequence and could still use assistance even though the assignment is past due. I still need to be able to do such problems on the midterm exam.

The function mentioned above is incorrect because when n=1 we have \(\displaystyle \L \lceil 2^{-1} \rceil = 0\) I believe...

But I found this: \(\displaystyle \L 2^{n-2} + \frac{3}{2} \lfloor \frac{1}{n} \rfloor\).

This seems to work, but I am still looking for a version without the use of any floor/ceiling functions. Thanks for all the help.

Daon

 

[pka]

The function mentioned above is incorrect because when n=1 we have \(\displaystyle \L \lceil 2^{-1} \rceil = 0\) I believe...Daon

You are absolutely wrong about that!
\(\displaystyle \L
\left\lceil {2^{ - 1} } \right\rceil = 1\)
The ceiling function returns the first integer not less than x.

 

[daon]

Ah yes, I was somehow creating a negative in my head that didn't exist

But isn't there another way to produce this sequence without these functions?