### Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

### Little and Large

A point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?

### Approximating Pi

By inscribing a circle in a square and then a square in a circle find an approximation to pi. By using a hexagon, can you improve on the approximation?

# How Does Your Function Grow?

##### Stage: 5 Challenge Level:

Here are two solutions, one from Simon from Elizabeth College, Guernsey and one from Andrei from Tudor Vianu National College, Bucharest, Romania. First Simon's solution.

The functions are: \eqalign { A(n) &= \log(100n) \cr B(n) &= n^{100} \cr C(n) &= 100^n \cr D(n) &= n! .} Firstly I looked at Archimedes' choice, the logarithmic function. $$A(n) = \log(100n) = \log 100 + \log n.$$ I knew that when $n$ gets large $\log 100$ would become irrelevant. This meant that only $\log n$ would be important, and $\log n < n$ when $n > 1$. So $A(n)$ will be less than $n$ for large $n$. We find $n = \log(100)+ \log n$ for $n = 6.472775$ approximately and so $A(n) < n$ for all values of $n> 6.5$.\par The next task was establish whether $B(n)$ or $C(n)$ was greater for large numbers. I was certain of one thing: $B(n) = C(n) = 100^{100}$ for $n=100$.

Firstly I knew that $100^n$ would have $2n+1$ digits. I then established, where $x$ is the number of digits in $n$, that $n^{100}$ would have between $100(x-1)+1$ and $100x$ digits. So $C(n) = 100^n$ is the larger function for $n> 100$.

Therefore, so far, $C(n)$ is the largest function for large values of $n$.

The next task was to compare it to $D(n)$ or factorial $n$.

Using Stirling's formula: $$D(n) = n! = \sqrt{2\pi n} ({n\over e})^n.$$ As $n$ is a large positive number, $\sqrt {2\pi n}$ will be greater than 1. Therefore when ${n\over e}$ is greater than 100, $D(n) > C(n)$. The 'change over point' occurs for $n$ just less than $n = 100e$, that is $n \approx 270$.

Andrei considered the natural logarithms of the functions and plotted their graphs. For the factorial n function Andrei used Stirling's Approximation which is valid for large n.

I approximated the function for factorial n further by ignoring the term $\sqrt{2\pi n}$. \eqalign{ A(x)&= \log (n) + \log(100) \cr \log(B(n)) &= 100 \log(n)\cr \log C(n) &= n \log(100) \cr \log D(n)&\approx n\log(n) - n }. I know that $log(n)< n$ for the whole domain of definition of the logarithmic function, and I see that $A$ is the smallest. I have to compare the other functions. As $\log(100)\approx 4.60$, so: \eqalign { \log(B) &= 100 \log(n) \cr \log(C)&\approx 4.6n \cr \log(D) &\approx n (\log(n) -1).} In the limit for large $n$, $$\lim_{n\to \infty}{\log n\over n}\to 0$$ and so $B< C$.

For large $n$ we can neglect 1 in respect to $\log(n)$ in the approximate formula for $\log(D)$ and, as $4.6 < \log (n)$ we have $D> C> B> A$.

Here in the diagram, $A(x)$ is represented in red, $B(x)$ in green, $C(x)$ in blue and $D(x)$ in black. It can be observed that for large $n$ (here $n> 300$) the order of the magnitude of the functions is $D> C> B> A$.