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Playing with a pencil and paper notebook I noticed the following:

$x=1$

$x^3=1$


$x=2$

$x^3=8$


$x=3$

$x^3=27$


$x=4$

$x^3=64$


$64-27 = 37$

$27-8 = 19$

$8-1 = 7$


$19-7=12$

$37-19=18$


$18-12=6$


I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.

I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?

There is a similar less complicated pattern for computing $x^2$ values.

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    $\begingroup$ As a rule of thumb, casual mathematicians make no discoveries. You are using the finite-differences approach. For all $k$, $\Delta_kx^k=k!$. $\endgroup$
    – user65203
    Commented Aug 21, 2016 at 10:06
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    $\begingroup$ @YvesDaoust A re-discovery is a discovery just the same. $\endgroup$ Commented Aug 21, 2016 at 10:07
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    $\begingroup$ @6005: I wouldn't qualify a re-discovery as "new". $\endgroup$
    – user65203
    Commented Aug 21, 2016 at 10:09
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    $\begingroup$ I don't care if it's new or not, that's an awesome discovery! $\endgroup$
    – Vincent
    Commented Aug 21, 2016 at 11:25
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    $\begingroup$ At school I used to use this to draw parabolas. The teacher wanted us to make big tables of values, but that was too slow and too boring for me. When I noticed the pattern I thought I could derive something useful from it, I only got $(x+a)^2=x^2+2xa+a^2$, haha $\endgroup$
    – Oriol
    Commented Aug 21, 2016 at 15:39

4 Answers 4

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It's not something new, but for your discovery I applaud. This procedure is called the method of successive differences, and you can show that for every power the successive difference appears.

Let us say you have a sequence: $$ 1^3 \quad2^3\quad 3^3\quad 4^3\quad \ldots $$

Note that $x^3-(x-1)^3 = 3x^2-3x+1$. So we'll get a new sequence at the bottom: $$ 7 \quad 19\quad 37\quad 61\quad \ldots $$ Now, note that $3x^2-3x+1-(3(x-1)^2-3(x-1)+1) = 6(x-1)$. Hence, we'll get another series: $$ 0 \quad6\quad 12\quad 18\quad\ldots $$ Now, note that $6(x-1)-6((x-1)-1) = 6$! Now, the new sequence is: $$ 6\quad 6\quad 6\quad 6\quad 6\quad ... $$ So $6$ appears as the final difference! This shows the power of algebra. As an exercise, do this for $x^4$. See the pattern of the number at the end, and if you can say something for $x^n$.

The reason, as you can see, is that at each line above, the degree of the polynomial $f(x)-f(x-1)$ decreases by $1$. Hence, at the end of three lines, you are only going to get a constant polynomial.

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    $\begingroup$ This is exactly the same method Charles Babbage's Difference Engine used to calculate powers in Taylor series for trigonometric and other functions. $\endgroup$ Commented Aug 21, 2016 at 10:22
  • $\begingroup$ @ParclyTaxel Thank you for the information. Do you know about the final constant at the end for general $n$? (I don't) $\endgroup$ Commented Aug 21, 2016 at 10:39
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    $\begingroup$ The final constant is the factorial of the starting power. $\endgroup$ Commented Aug 21, 2016 at 10:44
  • $\begingroup$ Wow, that's nice. Thank you again. $\endgroup$ Commented Aug 21, 2016 at 10:47
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    $\begingroup$ +1 for applauding the OPs rediscovery. (It's one lots of budding mathematicians - including me years ago - make often when young.) $\endgroup$ Commented Aug 21, 2016 at 13:18
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What you have discovered is a finite difference calculation. For any function $f$, in this case the third-power function $$ f(n) = n^3 $$ we can define the forward difference, or forward discrete derivative: $$ \Delta f(n) = f(n+1) - f(n) = 3n^2 + 3n + 1 $$ Likewise, \begin{align*} \Delta \Delta f(n) = \Delta^2 f(n) &= 6n+ 6 \\ \Delta^3 f(n) &= 6 \\ \Delta^4 f(n) &= 0. \end{align*}

Your computation, $$ 5^3 = 64 + 37 + 18 + 6 $$ is the statement $$ f(5) = f(4) + \Delta f(3) + \Delta^2 f(2) + \Delta^3 f(1), $$ or more generally $$ f(n + 1) = f(n) + \Delta f(n-1) + \Delta^2 f(n-2) + \Delta^3 f(n-3). $$ This is one discrete analogue of Taylor series (the more common analogue is Newton's series). The reason it works is that, for $f(n) = n^3$, $\Delta^4$ and beyond are all zero. So the summation stops once we get to $\Delta^3$.


EDIT: Let me add that there is yet another identity here which resembles Taylor series, namely, $$ f(n-1) = f(n) - \Delta f(n) + \Delta^2 f(n) - \Delta^3 f(n). $$

Altogether, at least when $f$ is a polynomial, we thus have the following Taylor series-like formulas: \begin{align*} f(n+1) &= \sum_{i=0}^\infty [\Delta^i f](n - i) \\ f(n-1) &= \sum_{i=0}^\infty (-1)^i [\Delta^i f](n) \\ f(n+x) &= \sum_{i=0}^\infty \binom{x}{i} [\Delta^i f](n). \end{align*}

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For a little bit more, see the answer "General method for indefinite summation" which explains how exactly this representation using forward differences allows you to easily find the formula for indefinite summation of powers. Applied to your case you get:

0,  1,  8, 27
  1,  7, 19
    6, 12
      6

and hence:

$n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$.

which immediately gives: $ \newcommand\lfrac[2]{{\large\frac{#1}{#2}}} $

$\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = n\lfrac{n-1}{2}(1+\lfrac{n-2}{3}(6+\lfrac{n-3}{4}(6)))$

$\ = \lfrac{n^2 (n-1)^2}{4}$.

and then, if you prefer the indices to end at $n$:

$\sum_{k=1}^n k^3 = \lfrac{(n+1)^2 n^2}{4}$.

As you can see, hardly any computation was necessary to get this result!

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  • $\begingroup$ This is just beautiful. I had read a similar way to calculate the sum $\sum_{k=1}^n k^N$ for all positive numbers $N$. It involves integration. $\endgroup$ Commented Aug 22, 2016 at 23:34
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    $\begingroup$ @астонвіллаолофмэллбэрг: Thank you! If you count the number of operations you need, you will find that many other methods (including the integration-based method) are vastly more inefficient ($O(d^3)$ instead of $O(d^2)$ where $d$ is the degree), because they require first computing the formulae for all smaller $d$, each of which takes $O(d^2)$ time. The reason this method does not face that problem is that the binomial coefficients form the natural basis so that difference and summation are simply shift operations on the coefficients. $\endgroup$
    – user21820
    Commented Aug 23, 2016 at 6:57
  • $\begingroup$ Oh, that is wonderful. Thank you for this information, @user21820. $\endgroup$ Commented Aug 23, 2016 at 8:18
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To test your hypothesis you could work out the form of the differences from the first few cases. \begin{align*} 1^{3}-0^{3}&=1\\ 2^{3}-1^{3}&=7\\ 3^{3}-2^{3}&=19\\ 4^{3}-3^{3}&=37 \end{align*} For example rewrite out $(37-19)-(19-7)=18-6=6$ as: \begin{align*} \{(4^{3}-3^{3})-(3^{3}-2^{3})\}&-\{(3^{3}-2^{3})-(2^{3}-1^{3})\}\\ &=(4^{3}-2\cdot3^{3}+2^3)-(3^{3}-2\cdot2^{3}+1^{3})\\ &=4^{3}-3\cdot3^{3}+3\cdot2^3-1^{3}\qquad (\star)\\ &=6 \end{align*} So you have to find the difference of two differences to get to $6$ (this is called a finite difference pattern, and you have to iterate twice to get the result of $6$ for all such differences, any further iteration ending in a $0$). Now check that pattern $(\star)$ holds in general for some integer $k\ge3$: \begin{align*} k^{3}&-3\cdot(k-1)^{3}+3\cdot(k-2)^3-(k-3)^{3}\\ &=k^{3}-3(k^2-3k^2+3k-1) +3(k^3-2\cdot3k^2+2^2\cdot3k-2^3) -(k^3-3\cdot3k^2+3^2\cdot3k-3^3)\\ &=\ \ k^3\\ &\ -3k^3\ +\ 9k^2\ -\ 9k\ +\ 3\\ &\ +3k^3-18k^2+36k-24\\ &\ \ -k^3\ +\ \ 9k^2-27k+27\\ &=6 \end{align*}

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  • $\begingroup$ As an aside, from this you could deduce $\sum_{j=0}^{3}(-1)^{j}\binom{3}{j}(k-j)^3=6$. Similar binomial power sum identities can be found for the other finite difference calculations concerning $x^4$, $x^5$, and so on. $\endgroup$ Commented Aug 25, 2016 at 23:35

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