- Thread starter
- #1

How can this be done?

Using the definition for (x - y), we have

$$

(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}

$$

but $n\notin\mathbb{Z}$.

- Thread starter dwsmith
- Start date

- Thread starter
- #1

How can this be done?

Using the definition for (x - y), we have

$$

(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}

$$

but $n\notin\mathbb{Z}$.

- Thread starter
- #2

It has to be the Taylor series.

How can this be done?

Using the definition for (x - y), we have

$$

(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}

$$

but $n\notin\mathbb{Z}$.

- Moderator
- #3

- Feb 7, 2012

- 2,770

It is the generalised ("Newton") binomial expansion $(1+x)^s = 1 + sx + \frac{s(s-1)}{2!}x^2 + \frac{s(s-1)(s-2)}{3!}x^3 + \ldots$, which is valid for any real number $s$, provided that $|x|<1$ (because it is an infinite series and you need that condition in order for it to converge). In this case you take $s = -\frac12$ and $x = -A^2u^2$.

How can this be done?

Using the definition for (x - y), we have

$$

(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}

$$

but $n\notin\mathbb{Z}$.