Maximum of nth roots of n (TOMATO Subjective 150)

Find the maximum among \(\mathbf { 1 , 2^{1/2} , 3^{1/3} , 4^{1/4} , … }\) .

Discussion

Consider the function \(\mathbf { f(x) = x^{1/x} }\) . We employ standard techniques to compute the maxima.

Take logarithm on both sides we have \(\mathbf { \log f(x) = \frac{1}{x} \log x }\) . Next find out the derivative:

\(\mathbf {\frac {1}{f(x)} f'(x) = \frac{-1}{x^2} \log x + \frac{1}{x}\cdot\frac{1}{x} implies f'(x) = f(x) \cdot \frac{1}{x^2} (1 – \log x) }\)

Since \(\mathbf { f(x) = x^{1/x} }\) is always positive for positive x and so is \(\mathbf {\frac{1}{x^2}}\) sign of the derivative depends only on (1-logx). Hence the derivative is 0 at x = e (2.71 approximately), positive before that and negative after that. Hence the function has a maxima at x = e.

We check the values at x=2 and x=3 and easy computations show that \(\mathbf { 3^{1/3} > 2^{1/2} }\). Hence \(\mathbf {3^{1/3} }\) is the largest value.

Special Note

One may ask for a non calculus proof of this problem. The basic idea is to understand that the inequality
\(\mathbf { n^{1/n} > (n+1)^{1/n+1}\Rightarrow n^{n+1} > (n+1)^n \Rightarrow n\cdot n^n > (n+1)^n \\ \Rightarrow n > \frac{(n+1)^n}{n^n}\Rightarrow n > (1+ \frac{1}{n})^n }\)

It is easy to show that the quantity \(\mathbf { (1+ \frac{1}{n})^n }\) lies within 2 and 3 for all values of n. Hence the inequality \(\mathbf {n > (1+ \frac{1}{n})^n }\) is true for n > 3. The result follows.

One Reply to “Maximum of nth roots of n (TOMATO Subjective 150)”

Leave a Reply

Your email address will not be published. Required fields are marked *