Approximate value of k

Question

How do you solve $k$ in $\frac{(k-1)^{k-1}}{k^{k-2}}=n$ at least with a good approximation?

Is there tight approximation?

Answer 1

Our expression is equal to $$\left(1-\frac{1}{k}\right)^k \frac{k^2}{k-1}.$$ For $k$ reasonably large, the first term is fairly close to $e^{-1}$. How close?

Take the logarithm, and expand $\ln(1-1/k)$ in a Taylor series. The first two terms are given by $-\frac{1}{k}-\frac{1}{2k^2}$. Multiply by $k$, and exponentiate, approximately. We get $e^{-1}\left(1+\frac{1}{2k}\right)$.

That gives the approximation $e^{-1}\frac{k(2k+1)}{2(k-1)}$ for the left-hand side.

Set equal to $n$, and solve for $k$ using the quadratic formula. That should give a reasonably good approximation for largish $k$.

Answer 2

In the same spirit as André Nicolas's answer, considering large values of $k$ and $$A_k=\frac{(k-1)^{k-1}}{k^{k-2}}$$ we can write $$\log(A_k)=(k-1)\log(k-1)-(k-2)\log(k)$$ $$\log(A_k)=(k-1)\Big(\log(k)+\log(1-\frac 1k)\Big)-(k-2)\log(k)$$ $$\log(A_k)=\log(k)+(k-1)\log(1-\frac 1k)$$ Now, using Taylor series $$(k-1)\log(1-\frac 1k)=-1+\frac{1}{2 k}+\frac{1}{6 k^2}+O\left(\frac{1}{k^3}\right)$$ Taking the exponential $$\exp\Big((k-1)\log(1-\frac 1k)\Big)=\frac{1}{e}+\frac{1}{2 e k}+\frac{7}{24 e k^2}+O\left(\frac{1}{k^3}\right)$$ All of this makes $$A_k=\frac{k}{e}+\frac{1}{2 e}+\frac{7}{24 e k}+O\left(\frac{1}{k^2}\right)$$ So, ignoring high order terms, solving $A_k=n$ leads to the quadratic equation $$k^2+ \left(\frac{1}{2}-e n\right)k+\frac{7}{24}=0$$ the solution of which being $$k_2=\frac{e n}{2}+\frac{1}{12} \sqrt{36 e n (e n-1)-33}-\frac{1}{4}$$ If we had limited to first order, the solution would have been $$k_1=e n-\frac{1}{2}$$ For illustration purposes, using $n=10$, we would find $k_1=26.6828$, $k_2=26.6719$ while the "exact" solution would be $k=26.6716$.

If we consider that $n$ is large, Taylor expansion of $k_2$ would give as an estimate of the solution $$k_3=e n-\frac{1}{2}-\frac{7}{24 e n}$$