How does your function grow?

Compares the size of functions f(n) for large values of n.

Age

16 to 18

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

Four enthusiastic mathematicians are asked to think of a function involving the number 100. The challenge is to think of the function which is biggest for big values of n

Archimedes chooses a logarithm function $A (n) = \log (100 n)$
Bernoulli decides to take 100th powers $B (n) = n^{100}$
Copernicus takes powers of 100 $C (n) = 100^{n}$
and, finally, de Moivre, who likes to be different, chooses the factorial function which he claims will be quite big enough without any reference to 100 at all $D (n) = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$

Which function is biggest for large values of n? Can you determine a value beyond which you know this function will be biggest?

[Extension: To find the exact switch-over value will be difficult and will require the clever use of a spreadsheet or computer.]

What could you say if the 100s were replaced by a million? billions? Create a convincing argument to prove your results to the mathematicians.

Student Solutions

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:

\begin{aligned} A (n) & = \log (100 n) \\ B (n) & = n^{100} \\ C (n) & = 100^{n} \\ D (n) & = n! . \end{aligned}

Firstly I looked at Archimedes' choice, the logarithmic function.

A (n) = \log (100 n) = \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 π n} (\frac{n}{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 $\sqrt{2\pi n}$.

\begin{aligned} A (x) & = \log (n) + \log (100) \\ \log (B (n)) & = 100 \log (n) \\ \log C (n) & = n \log (100) \\ \log D (n) & \approx n \log (n) - n \end{aligned} .

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:

\begin{aligned} \log (B) & = 100 \log (n) \\ \log (C) & \approx 4.6 n \\ \log (D) & \approx n (\log (n) - 1) . \end{aligned}

In the limit for large $n$,

lim_{n \to \infty} \frac{\log n}{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 $D> C> B> A$.

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