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?
A better approximation for $2(p+1)x^p - px - 2 = 0$
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analysis
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1If 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. – 2011-06-16
1 Answers
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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 $0.993099$.