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Discrete Trends

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2
Congratulations to Curt from Reigate College for cracking this tough nut problem. Here is Curt's solution:

If $n = n^{1\over n}$, I first assert that for an $n> 1$, then $n^{1\over n}$ is larger than 1

If $n> 1$ then, raising both sides to the power of $1/n$ yields $n^{1\over n}> 1^{1\over n}$. So $n^{1\over n}$ is larger than 1 no matter what the value of $1/n$. Therefore, one could make $n^{1\over n}=1+R$. As $n^{1\over n}$ is always larger than 1, $R$ is always a positive real number. Now, raising both sides to the power $n$ one obtains:

$$n = 1 + nR + \textstyle{1\over 2}n(n-1)R^2 +\cdots.$$

(Incidentally, $n> nR$, therefore $R< 1$ and so $n^{1\over n}< 2$)

As n is an integer larger than one, at least the third term of expansion must exist. It is clear that:

$$\eqalign{ n &> \textstyle{1\over 2}n(n-1)R^2 \cr {2n\over n(n-1)}&> R^2\cr \sqrt{2\over n-1} &> R }.$$

So $R\rightarrow 0$ as $n\rightarrow \infty$ and

$$n^{1\over n} = 1 + R < 1 + \sqrt{2\over n-1}.$$

So $n^{1\over n}\rightarrow 1$ as $n\rightarrow \infty$.

As a corollary, I will show that $\sqrt{2/(n-1)}$ grows smaller for successive values of $n> 1$. Assume that for some positive $B$, $\sqrt{2/(n-1)}< \sqrt{2/(n+B-1)}$. Now if this is to be true $1/(n-1)< 1/(n+b-1)$, therefore $n-1> n+B-1$, therefore $B< 0$, contrary to our conditions. Therefore the assertion that for some positive B the value of the function increases is fallacious. This will prove useful later.

For $n= 1,\ 2,\ $ etc. we have $n= 1,\ 2^{1/2},\ 3^{1/3},\ 4^{1/4}=2^{1/2}\ ...$ and we see that the values increase to a maximum of $3^{1/3}$ and then start to decrease.

Now to prove that 3 is the value of $n$ pertaining to the maximum of this discrete function, I will find an $n$ such that $1+\sqrt{2/(n-1)}$ is less that the cube root of 3. From that point onwards, it is known that for successive values $\sqrt{2/(n-1)}$ decreases, therefore beyond this point the value of $n^{1\over n}$ decreases; more importantly, is less than $3^{1\over 3}$. For $n=19$ this holds true, as it will for subsequent values. The "in-between" values have been checked and are less than $3^{1\over 3}$. If $n\geq 19$ then

$$n^{1\over n} < 1 + \sqrt{2\over 18}= {4\over 3}< 3^{1\over 3}$$