Why is this the nearest integer?

Question

I was trying to prove that $(N+\sqrt{N^2-1})^k$, where k is a positive integer, differs from the integer nearest to it by less than $(2N-\frac{1}{2})^{-k}$. Note: N is an integer greater than 1.

So, I tried to look for the answer of the question, which I have taken it from an exam paper.

It said that: We let T=$(N+\sqrt{N^2-1})^K +(N+\sqrt{N^2-1})^{-k}$= $(N+\sqrt{N^2-1})^k + (N-\sqrt{N^2-1})^k=2(N^k + kC2 N^{K-2}(N^2-1)+...$ which is clear it is an integer.

We know that: $(N-\frac{1}{2})^2 =N^2-N+\frac{1}{4}=\frac{5}{4}-N$ This is <0 when N > 1, since N is an integer.

So, $N-\frac{1}{2}<\sqrt{N^2-1}$.
$2N-\frac{1}{2} < N+\sqrt{N^2-1}$ $(2N-\frac{1}{2})^{-k}>(N+\sqrt{N^2-1}^{-k}$ Let $|T-(N+\sqrt{N^2-1})^{-k}$| we would be able to prove the question.

However, my question is that, we need to have the nearest integer, so, the integer that the answer used is $(N+\sqrt{N^2-1})^k + (N + \sqrt{N^2-1})^{-k}$.
Why is this the nearest integer?

Thank you so much for your reply.

Answer 1

Let $N$ be a fixed integer.

Using the expression you give,

$$\underbrace{(N+\sqrt{N^2-1})^k}_{A_k} + \underbrace{(N-\sqrt{N^2-1})^k}_{B_k>0}=\underbrace{2(N^k + \binom{k}{2} N^{k-2}(N^2-1)+...}_{C_k, \ \text{an integer}},\tag{1}$$

$C_k$ will be the closest integer to $A_k$ if their difference $B_k=C_k-A_k$ is proven to be $< 1/2$, whatever the values of $N>1$ and $k>0$ ;

Indeed :

$$B_k=N(1-\sqrt{1-1/N^2})^k < N(1-(1-1/N^2))^k=1/N^{2k-1} \leq 1/2^{2k-1}<1/2$$

due to inequality

$$\sqrt{1-x} < 1-x$$ valid for any $x$ such that $0 < x < 1$.

Remarks :

a) Let us show a numerical example in the case $N=2$ , with columns $k, A_k, B_k, C_k$ in that order :

$$\begin{array}{|r|r|r|r|} \hline 1&3.732050807 & 0.267949192 & 4\\ 2&13.928203230 & 0.071796769 & 14\\ 3&51.980762113 & 0.019237886 & 52\\ 4&193.994845223 & 0.005154776 & 194\\ 5&723.998618781 & 0.001381218 & 724\\ 6&2701.999629903 & 0.000370096 & 2702\\ \hline \end{array}$$

It is to be noted that numbers $C_k$ obey a second order recurrence relationship : $$C_{k+1}=4 C_k-C_{k-1}\tag{2}$$

(see https://oeis.org/A003500) and more generally :

$$C_{k+1}=2N C_k-C_{k-1}$$

(it comes from the fact that $A_k$ and $B_k$ can be considered as the two roots of quadratic characteristic equation associated with (2).

b) [in connection with (a)] : (1) can be considered for example as the analog of Binet formula for Fibonacci sequence defined by second order recurrence :

$$F_{k+1}=F_k+F_{k-1}$$

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section1)

Answer 2

By induction on $k$, we have $$ (N\pm \sqrt{N^2-1})^k=a\pm b\sqrt{N^2-1}$$ with $a,b\in\Bbb Z$. And of course $(N-\frac12)^2=N^2-N+\frac14<N^2-1 $ implies $$ 2N-\frac12<N+\sqrt{N^2-1}<2N.$$ Now from $$ (N+\sqrt{N^2-1})^k(N-\sqrt{N^2-1})^k=((N+\sqrt{N^2-1})(N-\sqrt{N^2-1}))^k=1^k=1,$$ we conclude that $(N+\sqrt{N^2-1})^k=a+b\sqrt{N^2-1}$ differs from the integer $2a$ by $$a-b\sqrt{N^2-1}=\frac1{(N+\sqrt{N^2-1})^k}<\frac1{(2N-\frac12)^k}.$$

Answer 3

Given that $\,N>1\,$ is an integer, define the real numbers $\, u := (N+\sqrt{N^2-1}),\,$ $\, v := (N-\sqrt{N^2-1}),\,$ and thus $\,u+v=2N, u\,v=1.\,$ Define the sequence $\, a(n) := u^n+v^n.\,$ This sequence satisfies $\,a(n) = a(-n)\,$ and a linear recurrence $\, a(n+1) = 2Na(n)-a(n-1) \,$ both for all integer $\,n\,$ with $\,a(0) = 2\,$ and $\,a(1) =2N.\,$ Thus all $\,a(n)\,$ are positive even integers. Read the Wikipedia article Lucas sequence for many details about such sequences.

Now notice that $\,0<v<\frac12\,$ and so $\,0<v^n<\frac12\,$ for all positive integer $\,n.\,$ Thus $\,a(n)\,$ is the closest integer to $\,u^n\,$ and $\,u^n=a(n)-v^n .\,$