How to solve this equation ?
$$\left(1-\frac{1}{x}\right)^x=0.01$$
Which expansion series can be used here ?
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It is not a polynomial equation and it only can be solved numerically. – ajotatxe Apr 24 '19 at 15:23
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@ajotaxte can you please explain a bit more that how to approach this problem ? – code_t Apr 24 '19 at 15:25
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$x=1.01060650\cdots$ – Dietrich Burde Apr 24 '19 at 15:34
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Numeric root finding is a large subject in numerical analysis. It is treated in any textbook. There are many methods. – Ross Millikan Apr 24 '19 at 15:36
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You can use newton method , to solve it numerically , see: http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html. – I0_0I Apr 24 '19 at 15:37
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@DietrichBurde can you please explain how you got this value ? – code_t Apr 24 '19 at 15:38
3 Answers
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It is not a polynomial equation. There are "closed-form" solutions, but they involve the Lambert W function:
$$ x = {\frac {2\;\ln \left( 10 \right) }{{\rm W} \left(- \ln \left( 10 \right)/50 \right)+2\,\ln \left( 10 \right) }}$$
where $W$ is a branch of the Lambert W function. Both the principal and $-1$ branches give real solutions.
EDIT: Actually, only the principal branch, as Peter Foreman pointed out. The numerical value is approximately $1.010606509247998$.
Slightly more generally, consider the equation $(1-1/x)^x = y$ where $0 < y < 1$, and we want $x > 1$. Writing $y = e^{-s}$ and taking log of both sides, $$ x \log(1-1/x) = -s $$ Let $x = s/(s-t)$ (where we want $0 < t < s$), so $1-1/x = t/s$, and the equation becomes $$ \log(t) - \log(s) = \log(t/s) = -s/x = t-s$$ i.e. $$ \log(t) - t = \log(s) - s $$ which we can write as $$ t e^{-t} = s e^{-s}$$ Of course this is true for $t=s$ (but we don't want that one). The other solution is $t = - W(-s e^{-s})$.
Robert Israel
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I'm actually quite interested as to how you rearrange the equation to get $f^{-1}(x)=\frac{\ln{(x)}}{\ln{(x)}-W(x\ln{(x)})}$ – Peter Foreman Apr 24 '19 at 15:38
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Hmm, when I plot it with Wolfram Alpha it doesn't look like it ought to have two solutions. What are the numerical value of the two real solutions you get? – hmakholm left over Monica Apr 24 '19 at 15:47
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1The $-1$st branch does not actually give a real solution in this case. It gives complex infinity. – Peter Foreman Apr 24 '19 at 15:52
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I think that a very quick and dirty way for this is binomial expansion and then converting it into a function and using Newton's method to find roots.
$$ \left ( 1 - \frac 1x \right)^x = 0.01 \\ \Rightarrow 1 + \left ( \frac {-1}{x} \right)x + \left ( \frac{x(x+1)}{2!} \right)\frac 1 {x^2} + \dots = 0.01$$
Now, use Newton's method to find the approximate roots.
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2Since $x$ is very close to $1$, the binomial series converges very slowly. It would be better to use Newton's method to compute Lambert W. – robjohn Apr 24 '19 at 17:28
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To answer the question of how to get the inversion formula:
$$
\begin{align}
\left(1-\frac1x\right)^x&=\frac1{100}\tag1\\
100\left(1-\frac1x\right)&=\exp\left(\log(100)-\frac{\log(100)}{x}\right)\tag2\\
-\log(100)\left(1-\frac1x\right)\exp\left(-\log(100)\left(1-\frac1x\right)\right)&=-\frac{\log(100)}{100}\tag3\\
-\log(100)\left(1-\frac1x\right)&=\mathrm{W}\!\left(-\frac{\log(100)}{100}\right)\tag4\\
x&=\frac{\log(100)}{\log(100)+\mathrm{W}\!\left(-\frac{\log(100)}{100}\right)}\tag5
\end{align}
$$
Explanation:
$(2)$: raise both sides to the $1/x$ power and multiply by $100$
$(3)$: multiply both sides by $-\frac{\log(100)}{100}\exp\left(-\log(100)\left(1-\frac1x\right)\right)$
$(4)$: apply Lambert W
$(5)$: solve for $x$
Use the branch of $\mathrm{W}\!\left(-\frac{\log(100)}{100}\right)\ne-\log(100)$ and we get $$ x\doteq1.0106065092479981249 $$
robjohn
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