# A better approximation for $2(p+1)x^p - px - 2 = 0$

I would like to approximate the positive root of the following equation $$2(p+1)x^p - px - 2 = 0$$ where $p$ is an integer. We could use the formula $(1 - y)^p \approx 1 - py$ for $y$ small to get an approximation of root $x_0 \approx \frac{1}{2p+1}$. However, I believe that we can make a stronger approximation. Could you please suggest some ideas?

• Have you tried Newton's method ? (en.wikipedia.org/wiki/Newton%27s_method) – user10676 Jun 16 '11 at 8:16
• That approximation doesn't seem to be accurate for large $p$. Did you mean $x_0 \approx 1 - \frac{1}{2p+1}$? – Qiaochu Yuan Jun 16 '11 at 10:42
• If you are going to use this as part of a program, it could be a very good idea to use different "formulas" for the initial approximation for various ranges of values of $p$, say small, medium, and large. – André Nicolas Jun 16 '11 at 11:38

Write $x = 1 - \frac{r}{p}$. Then $x^p \approx e^{-r}$ and the equation becomes (approximately)

$$2(p+1) e^{-r} + r = p+2.$$

For large $p$ the rough approximation $r \approx 1$, substituted into the above, gives $r \approx \ln 2$. Letting $r = \ln 2 + s$ we get

$$(p+1) e^{-s} + \ln 2 + s = (p+2).$$

Since we know $s$ will be small for large $p$ we can further approximate this by

$$(p+1)(1 - s) + \ln 2 + s = (p+2)$$

giving $s = \frac{\ln 2 - 1}{p}$, hence

$$x \approx 1 - \frac{\ln 2}{p} + \frac{1 - \ln 2}{p^2}.$$

I am pretty sure this is accurate to slightly better than second-order:

• For $p = 10$ the positive root is about $0.936$ and the above gives about $0.934$.
• For $p = 100$ the positive root is about $0.993123$ and the above gives about $0.993099$.