Is this a new method for finding powers?

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.

Answer

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.

Attribution
Source : Link , Question Author : blue-sky , Answer Author : Sarvesh Ravichandran Iyer

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