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Suppose $a \in \mathbb{R}$ is very large. Then there are two solutions to $x - \log x = a$. I was wondering what the asymptotic expansion of the larger solution to this equation is. The first term must be $a$ but what are the lower order terms?

user187437
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1 Answers1

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Let $x=a+b$, where $b\ll a$. Then

$$a=a+b-\log(a+b)=a+b-\log a-\log\left(1+\dfrac ba\right)\approx a+b-\log a-\dfrac ba$$

and

$$x\approx a+\frac{a\log a}{a-1}.$$

In blue, the exact curve.

enter image description here

You can also address the problem with good ol' Newton's iterations,

$$x\leftarrow x-\frac{x-\log x-a}{1-\dfrac1x}=x+\frac{x(\log x+a-x)}{x-1},$$

giving the approximations

$$a,\\ a+\frac{a\log a}{a-1},\\ a+\frac{a\log a}{a-1}+\frac{\left(a+\dfrac{a\log a}{a-1}\right)\log a}{a+\dfrac{a\log a}{a-1}-1},\\ \cdots$$

  • Damn that's a really good approximation. Testing with $a=30000$ nearly yields the exact value. Of course for very large $a$, the solution might as well be $x = a + \log(a)$ which makes sense since you'd have $a + \log(a) - \log(a + \log(a))$. $\log(a) \ll a$, so $\log(a + \log(a))$ will be pretty close to $\log(a)$ and the left hand side would be pretty close to $a$ after cancellation. This answer is a great example of how great this site is: despite years of studying math, I still have so much to learn - even simple stuff! – Cameron Williams Oct 18 '19 at 12:28
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    @CameronWilliams: you can easily obtain a better approximation by expending a second term of the logarithm and solving the quadratic equation. –  Oct 18 '19 at 12:42
  • The second order approximation is more accurate (but less elegant) : $x \approx a,\left( \sqrt{2\log{(a)}+{{\left( a-1\right) }^{2}}}-a+2\right) $ – leonbloy Oct 18 '19 at 12:52