Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

The Method of Finite Differences, Extended -- and Shortened

Date: 12/22/2010 at 10:43:12
From: Abby
Subject: Write algrebraic formula for a series

I have read all your info about the method of finite differences, but the
series I am working with doesn't seem to fit any of the patterns, so I am
stuck.

The series is 

   1 16 63 160

I have listed all the differences, hoping to get to a common factor, but
no luck. I have looked at the diagonals of the differences to see a
pattern, and still no luck. It doesn't seem to be a polynomial. 

I can't seem to guess it or see any pattern.

How can I write an expression for the next term?



Date: 12/22/2010 at 12:20:57
From: Doctor Greenie
Subject: Re: Write algrebraic formula for a series

Hi, Abby --

The series is (or at least can be) described by a polynomial; and the
method of finite differences can lead you to that polynomial.

Here is the beginning of the work on the problem using the method of
finite differences:

    1    16    63    160
      15    47    97
         32    50
            18

With just these numbers, there is no row with a common difference. But you
can MAKE a row with a common difference simply by adding another "18" to
the row of 3rd differences. Then if you want, you can work backwards to
find what the next term in the original sequence would be. That can be
useful to check the formula you end up with: if the formula you end up
with produces that number for the 5th term in the sequence, then your
formula must be right.

    1    16    63    160   325
      15    47    97    165
         32    50    68
            18    18

At this point, you could continue with the usual method of finite
differences to find the polynomial formula for the sequence. But sometimes
the formula for the given sequence is relatively simple and you can find
it without going through the entire normal method of finite differences.
This example turns out to be a case like that.

To use this "shortcut" in the method of finite differences, you need to
know that if the n-th row of differences is constant, then that constant
difference is (n!) times the leading coefficient of the polynomial.

In your example, the constant 3rd difference is 18, so

   18 = (3!)a   where "a" is the leading coefficient
   18 = 6a
    3 = a

So the polynomial for your sequence is of the form

   t(n) = 3n^3 + .....

Because the polynomial for this particular example turns out to be
relatively simple, we can determine the rest of it by comparing each term
of the sequence with the value of the leading 3n^3 term:

   n   t(n)  3n^3  t(n) - n^3
  ---------------------------
   1    1      3       -2
   2   16     24       -8
   3   63     81      -18
   4  160    192      -32

The differences in the last column represent the "rest" of the polynomial,
which should be of degree 2 (or less). Examination of the numbers in the
last column leads rather easily to the final determination of the
polynomial that defines the sequence:

   n = 1:  -2 = -2(1)
   n = 2:  -8 = -2(4)
   n = 3: -18 = -2(9)
   n = 4: -32 = -2(16)

We see that the remaining part of the polynomial is simply

   -2(n^2)

And so the polynomial that defines this sequence is

   t(n) = 3n^3 - 2n^2

We can verify that this formula produces the given first four numbers of
the sequence, as well as the fifth number we predicted:

   t(n) = 3n^3 - 2n^2 
        = n^2(3n - 2)

   t(1) = 1^2(3(1) - 2) =   1(1) =   1
   t(2) = 2^2(3(2) - 2) =   4(4) =  16
   t(3) = 3^2(3(3) - 2) =   9(7) =  63
   t(4) = 4^2(3(4) - 2) = 16(10) = 160
   t(5) = 5^2(3(5) - 2) = 25(13) = 325

Of course, if you continue with the usual method of finite differences,
starting from the point above where we found (created) a constant 3rd
difference of 18, you should end up with this same polynomial formula.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
  


Date: 12/22/2010 at 21:44:08
From: Abby
Subject: Thank you (Write algrebraic formula for a series)

I learned more from your email than I've learned all year in my math class
so far. Thanks for being so complete and for showing me multiple ways to
work it out.

And thanks for being so fast. I was up for hours last night trying to
figure it out. I can sleep well tonight!!



Date: 12/23/2010 at 18:08:11
From: Doctor Greenie
Subject: Re: Thank you (Write algrebraic formula for a series)

Hi, Abby --

I'm glad I could help out.

It sounds as if you are a young student who really enjoys learning. It is
especially rewarding for me when I get that feeling from somebody I help
here at Dr. Math.

Thanks for your good work; and for your note of thanks.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/25/2010 at 00:18:53
From: Abby
Subject: Write algrebraic formula for a series

Dr. Greenie,

I have one more sequence I am working on. Can you check my thinking?

The series is

   4, 8, 14, 24

I have read all the stuff on Dr. Math, and it seems like a recursive
series.

I am not sure I know how to write it down on the computer, so let me write
it out:

   2 times the last term - 2^(n - 2)
   
So for example, the 2nd term is

   2*4 minus 0
   
And the third term would be

   2*8 minus 2

I am surprised my teacher would give a problem with a recursive series
(that's pretty hard core), so I was wondering if you saw a simpler
solution.

Looking forward to your view.

Thx, Abby



Date: 12/25/2010 at 02:56:41
From: Doctor Greenie
Subject: Re: Write algrebraic formula for a series

Hi, Abby --

Your recursive sequence formula doesn't work for going from the first term
(4) to the second one (8):

  2*4 - 2^0 = 8 - 1 = 7

You can get a polynomial formula for this sequence using exactly the same
strategy as for the earlier problem. But in this case the formula is not
as "nice"; and you probably have to use the formal method of finite
differences (instead of a shortcut) to find the formula for this sequence.

From the given sequence, we can see

    4   8   14    24
      4   6    10
        2    4
          2

We can simply assume that the row of 3rd differences is going to be a
constant "2"; using the method of finite differences will again give us a
polynomial formula of degree 3 that produces the given sequence.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/26/2010 at 21:09:18
From: Abby
Subject: Thank you (Write algrebraic formula for a series)

OK Dr. G -- I worked it out.

The fractional coefficient was a bit tricky but I was able to figure it
out. I forgot that anything to the power of 0 is 1; that's the mistake I
made in my first try.

Thanks again -- I really appreciate your help!!



Date: 12/26/2010 at 23:31:11
From: Doctor Greenie
Subject: Re: Thank you (Write algrebraic formula for a series)

Nice work, Abby --

It's been a pleasure to work with you on these problems.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/