Prove polynomial identity $x^k-1=(x-1)(x^{k-1}+x^{k-2}+\ldots+1).$

Question

I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below:

$(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$

Answer 1

Here's another approach using roots of unity and the fundamental theorem of algebra.

The zeros of $x^k - 1$ are $\{e^{\frac{n2\pi i}{k}}\}$ for $0 \leq n < k$.

Therefore, we can rewrite $x^k - 1$ as:

$$x^k - 1 = M\prod_{n=0}^{k-1} (x-e^{\frac{n2\pi i}{k}})$$

for some $M$

Now, let's find the zeros of $P(x) = (x-1)(x^{k-1} + ... + 1)$

Clearly, $x = 1 = e^{\frac{0*2\pi i}{k}}$ is a zero.

What about $Q(x) = (x^{k-1} + ... + 1)$?

Let's substitute our $k-1$ roots of unity, $r_n$ where $n > 1$

$$Q(r_n) = \sum_{j=0}^{k-1} (r_n)^j = \sum_{j=0}^{k-1} (e^{\frac{n2\pi i}{k}})^j = \frac{1 - (e^{\frac{n2\pi i}{k}})^k}{1 - e^{\frac{n2\pi i}{k}}} = \frac{0}{1 - e^{\frac{n2\pi i}{k}}} = 0$$

Using the sum of a finite geometric series and the fact that: $(e^{\frac{n2\pi i}{k}})^k = e^{n2\pi i} = 1^n$

Therefore, we can rewrite $Q(x)$ as:

$$Q(x) = D' \prod_{n=1}^{k-1} (x-e^{\frac{n2\pi i}{k}})$$

and $P(x)$ as:

$$P(x) = D\prod_{n=0}^{k-1} (x-e^{\frac{n2\pi i}{k}})$$

for some constant $D$.

Now, we have that $P(x)$ agrees with $x^k - 1$ at $k$ points, which are the $k$ roots of unity.

However, we also have that $P(0) = -1$ and $0^k - 1 = -1$ We can sub this back into the factored expressions to see that $M = D$, or we can also conclude that because $(x-1)(x^{k-1} + ... + 1)$ and $x^k - 1$ agree at $k + 1$ distinct points, $(x-1)(x^{k-1} + ... + 1) = x^k - 1$ for all $x$.

Answer 2

It's like this: $(x-1)(x^{k-1}+x^{k-2}+\ldots+1)$

$=x(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)$

$=x^{k-1+1}+x^{k-2+1}+x^{k-3+1}+x^{k-4+1}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x-1$

$=x^{k}+x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x^2-x-1$

$=x^{k}+(x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x)-(x^{k-1}+x^{k-2}+x^{k-3}+...+x^2+x)-1$

$=x^k-1$

You can also prove it by reversing the solution above.

Answer 3

For fun:

Let $x$ be real, and consider the geometric series:

1)$\sum_{i=0}^{k-1} x^i = 1+x +x^2 +..+x^{k-1};$

2) $ x \sum_{i=0}^{k-1}x^i = \ \ x+x^2+...+x^{k-1} +x^k;$

Subtract : 2)-1):

$(x-1)\sum_{i=0}^{k-1} x^{i}= x^k-1$.

Answer 4

This one cries out for a simple proof by induction:

If $k = 1$ we evidently have

$x^1 - 1 = (x^1 - 1)(1), \tag 1$

and if $k = 2$:

$x^2 - 1 = (x -1)(x + 1) = (x - 1)\left ( \displaystyle \sum_0^1 x^i \right ); \tag 2$

if we now suppose that the formula holds for some positive $m \in \Bbb Z$,

$x^m - 1 = (x - 1) \left ( \displaystyle \sum_0^{m - 1} x^i \right ), \tag 3$

then

$x^{m + 1} - 1 = x^{m + 1} - x^m + x^m - 1$ $= x^m(x - 1) + (x - 1) \left ( \displaystyle \sum_0^{m - 1} x^i \right ) = (x - 1) \left ( x^m + \displaystyle \sum_0^{m - 1} x^i \right ) = (x - 1) \left ( \displaystyle \sum_0^m x^i \right ), \tag 4$

which shows that the formula

$x^k - 1 = (x - 1) \left ( \displaystyle \sum_0^{k -1} x^i \right ) \tag 5$

is valid for all positive integers $k$.