Studies and fun

TODAY'S PROBLEM

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
$\mathbf { n^{1/n} > (n+1)^{1/n+1} \implies n^{n+1} > (n+1)^n \implies n\cdot n^n > (n+1)^n \implies n > \frac{(n+1)^n}{n^n} \implies n > (1+ \frac{1}{n})^n }$

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.

Studies and fun

Saturday, 26 July 2014

No comments:

Post a Comment

Articles

About Me