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In a contest problem book, I found a reference to Newton's little formula that may be used to find the nth term of a numeric sequence. Specifically, it is a formula that is based on the differences between consecutive terms that is computed at each level until the differences match.

An example application of this formula for computing the nth term of the series (15, 55, 123, 225, 367, 555, 795, ....) involves computing the differences as shown below:

1) 1st Level difference is (40, 68, 102, 142, 188, 240)
2) 2nd Level difference is (28, 34, 40, 46, 52)
3) 3rd Level difference is (6, 6, 6, 6, 6) 

Now the nth term is $$15{n-1\choose 0} + 40{n-1\choose 1} + 28{n-1\choose 2} + 6{n-1\choose 3}$$ where the constant multipliers are the first term of the differences at each level in addition to the first term of the sequence itself.

I was not able to find any reference to this formula or a proof of it after searching on the web. Any explanation of this method is appreciated.

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    $\begingroup$ LaTeX formatting will make your question much easier to read. $\endgroup$ – The Count Dec 31 '16 at 20:04
  • $\begingroup$ Thank you for the comment. I have formatted the nth term expression. $\endgroup$ – erase.ego Dec 31 '16 at 20:23
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Suppose that we have a sequence:

$$a_0,a_1,...a_k$$

And we want to find the function of $n$ that defines $a_n$.

To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we call this operation on $a_n$ the forward difference. Then given $\Delta a_n$ we can find $a_n$. Sum both sides of the equation from $n=0$ to $x-1$, and note that we have a telescoping series:

$$\sum_{n=0}^{x-1} \Delta a_n=\sum_{n=0}^{x-1} (a_{n+1}-a_n)=a_{x}-a_{0}$$

Hence $a_n=a_0+\sum_{i=0}^{n-1} \Delta a_i$. Also $\Delta a_n=\Delta (0)+\sum_{i=0}^{n-1} \Delta^2 a_n$...and so forth. Using this we must have if the series converges:

$$a_n=a_0+\Delta (0) \sum_{x_0=0}^{n-1} 1+\Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1+\Delta \Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1+\cdots$$

Where $\Delta^i (0)$ denotes the first term ($n=0$) of the $i$ th difference sequence of $a_n$.

Through a combinational argument, If we take $\Delta^0 (0)=a_0$ and ${n \choose 0}=1$ we may get:

$$a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$$

If it is the case you want the sequence to start with $a_1$ we need to shift this result to the right one:

$$a_n=\sum_{i=0}^{\infty} \Delta^i(1) {n-1 \choose i}$$

Note when you re-define $a_0$ to be $a_1$ by shifting it's index to the right one, you're re-defining $a_1-a_0=\Delta^1(0)$ to be $a_2-a_1=a_{1+1}-a_{1}=\Delta^1(1)$.

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  • $\begingroup$ $a_1 = a_0 + \Delta a_0$, $a_2 = a_0 + 2\Delta a_0 + \Delta^2 a_0$ and $a_3 = a_0 + 3\Delta a_0 + 3\Delta^2 a_0 + \Delta^3 a_0$ are the first three terms. The co-efficients seem to correspond to those of the binomial expansion and it looks like the general result maybe proved using induction. $\endgroup$ – erase.ego Jan 3 '17 at 3:17
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    $\begingroup$ Take $D$ to be the operation mapping $f(x)$ to $f(x+1)$. And $I$ to be the operation mapping $f(x)$ to itself. $D,I$ are linear operators. So if we want to find $\Delta^n f(x)=(D-I)^n f(x)$ we may treat $(D-I)^n$ as a polynomial. Then use binomial theorem to expand and then operate on $f(x)$ with what we get for $(D-I)^n$ to get a closed form for the coefficients @erase.ego $\endgroup$ – Ahmed S. Attaalla Jan 3 '17 at 3:21
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Starting with the second differences we have

  • $28 = 28$,
  • $34 = 28 + 6$,
  • $40 = 28 + 6 + 6$
  • $46 = 28 + 6 + 6 + 6$$

Now for the first differences we have

  • $40 = 40$
  • $68 = 40 + 28$
  • $102 = 40 + 28 + 34 = 40 + 28 + (28 + 6)$
  • $142 = 40 + 28 + 34 + 50 = 40 + 28 + (28 + 6) + (28 + 6 + 6)$

So the original sequence is

  • $15 = 15$
  • $55 = 15 + 40$
  • $123 = 15 + 40 + 68 = 15 + 40 + [40 + 28]$
  • $225 = 15 + 40 + 68 + 102 = 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)]$
  • \begin{align} 367 &= 15 + 40 + 68 + 102 + 142 \\ &= 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)] + [40 + 28 + (28 + 6) + (28 + 6 + 6)] \end{align}

The patterns should be clear. So just count how many 15's, 40's, 28's, and 6's there are in each term.

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  • $\begingroup$ I am wondering if this argument can be generalized some how? $\endgroup$ – erase.ego Dec 31 '16 at 23:44
  • $\begingroup$ Well, it generalizes to any sequence given by a polynomial. I guess you could also generalize it to a sequence where the n-th differences aren't constant but have some nice behavior. $\endgroup$ – Daniel McLaury Dec 31 '16 at 23:46

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