Can you help me understand the properties and zeros of this function involving square roots terms?

Question

Consider this function ( I got this when I was solving an another problem)

f(n) = $$\left(\frac{1}{2}\right)\left[\left(\frac{n^2+\sqrt{n^4+16008n}}{2n}\right)+\sqrt{\left(\left(\frac{n^2+\sqrt{n^4+16008n}}{2n}\right)^2\right)-\frac{16008}{\frac{n^2+\sqrt{n^4+16008n}}{2n}}}\right]-n$$

When I searched "f(n)=0" on Wolfram alpha, it shows result: $$ n \geq \sqrt[3]{2001} $$ (That means f(n) has infinitely many zeroes ?)

Since,$$13\geq \sqrt[3]{2001}$$ f(13) should be zero , same for f(14), f(15) and so on... But when I evaluated f(13) on Wolfram alpha the result was $$5.32907\times10^{-15}$$

I also found that f(n)=0 whenever I plugged $$n \geq 60 $$ Can you help me with these questions.

1. If f(n) really has infinitely many zeroes ?

2. Is f(n) an algebraic function ?

3. Why Wolfram alpha behaves in this way ?

Answer 1

Bizarrely, this function is indeed identically equal to $0$ for all real numbers $n\ge\sqrt[3]{2001}$. It can be simplified to $$ \frac14\biggl( -3n + \sqrt{n^2+\frac{16008}n} + \sqrt{10n^2+\frac{16008}n-6n\sqrt{n^2+\frac{16008}n}} \biggr), $$ which further simplifies to $$ \frac14\biggl( -3n + \sqrt{n^2+\frac{16008}n} + \sqrt{\Bigl(3n-\sqrt{n^2+\frac{16008}n} \Bigr)^2} \biggr) $$ and therefore to $$ \frac14\biggl( -3n + \sqrt{n^2+\frac{16008}n} + \biggl| 3n-\sqrt{n^2+\frac{16008}n} \, \biggr| \biggr), $$ which finally equals $$ \begin{cases} \frac12\biggl( \sqrt{n^2+\frac{16008}n} -3n \biggr), & \text{if } \sqrt{n^2+\frac{16008}n} \ge 3n, \\ 0, & \text{if } \sqrt{n^2+\frac{16008}n} \le 3n. \end{cases} $$ This last inequality is the source of the changed behavior at $n=\sqrt[3]{2001}$.

Getting an answer of $5\times10^{-15}$ when plugging in $n=13$ is just a matter of machine-precision-level errors adding up to something that the computer thinks is nonzero.

Answer 2

To simplify the calculations, let $\,\frac{n^2 + \sqrt{n^4+16008n}}{2n} = a\,$ and $\,\frac{n^2 - \sqrt{n^4+16008n}}{2n} = b\,$ then $\,a+b = n\,$, $\,ab=-\frac{16008}{4n} = -\frac{16008}{4(a+b)} \iff 16008 = -4ab(a+b)\,$, and substituting these in $\,f(n)\,$:

$$ \require{cancel} \begin{align} 2\big(f(n) + n\big) = a + \sqrt{a^2 + \frac{4(a+b)\cancel{a}b}{\cancel{a}}} = a + \sqrt{(a+2b)^2} = a + |a+2b| \end{align} $$

When $\,a+2b \ge 0\,$ the above reduces to $\,2f(n) + \cancel{2n} = \cancel{2(a + b)} \implies f(n) = 0\,$.

The condition for $\,a+2b \ge 0\,$ is:

$$ 3n^2 \ge \sqrt{n^4+16008n} \;\;\iff\;\; 8 n^4 \ge 16008 n \;\;\iff\;\; n \ge \sqrt[3]{2001} $$