I am looking at a proof of the power rule and I see this:
$x^n - a^n = (x-a)(x^{n-1} + x^{n-2}*a + {...} + xa^{n-2} + a^{n-1})$
Where does this come from?
It's used in this proof which I'm trying to understand:
What is it called?
Asked
Active
Viewed 1,048 times
-1
-
For "what's it called", no particular name, you could say "factorisation of a difference of $n$th powers". For "where does it come from", try the history of maths SE. – David Jan 12 '18 at 03:43
-
Well, it's like the difference of perfect squares or difference of perfect cubes. You should know that $x^2-a^2 = (x-a)(x+a)$ and $x^3-a^3 = (x-a)(x^2+xa+a^2)$ – Landuros Jan 12 '18 at 03:43
-
Hint: $x^{n-1} + x^{n-2}*a + {...} + xa^{n-2} + a^{n-1}$ is the sum of a geometric progression. When I was at school, we studied those in the lead-up to limits and calculus. – PM 2Ring Jan 12 '18 at 05:10
-
Does this answer your question? Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ – tryst with freedom Jun 14 '21 at 16:19
2 Answers
2
Don't know that it has a name other than difference of powers identity. It follows for $\,z = \dfrac{x}{a}\,$ from the even simpler:
$$ \begin{align} z^n-1 &= (z^n\color{red}{-z^{n-1}})+ (\color{red}{z^{n-1}}-\color{blue}{z^{n-2}})+ \ldots +(\color{blue}{z}-1) \\ &= z^{n-1}(z-1) + z^{n-2}(z-1)+\ldots+ 1 \cdot (z-1) \\ &= (z-1)(z^{n-1}+z^{n-2}+\ldots+1) \end{align} $$
dxiv
- 76,497
1
$a $ is a root of $f (x)=x^n-a^n$... hence $(x-a)| f (x) $... this follows from the polynomial factor theorem
It remains to find the polynomial that is $\frac {x^n-a^n}{x-a} $
This can be done by synthetic division