Is a sum of equidistributed sequences $\sum \{n \alpha\}$ equidistributed?

We have two equidistributed sequences {n a} and {n b} (mod 1), in which a and b are irrational.

1) Is it true that the sum {na} + {nb} is equidistributed?

and

2) Is it true that {na} + {nb} = {n(a+b)} ?

For (1), n*a and n*b are both polynomials satisfying Weyl's criteria, and so the sum of these is I think also a polynomial satisfying the criteria (but this does not necessarily mean {na} + {nb} is equidistributed). For (2), well, I just tried a few calculations, so I am not sanguine about it but it seems true. Note: I do see that a + b might not be irrational, and in that case we would not satisfy Weyl's criteria in (2).

Thanks for any insights.

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"For (1), both expressions are polynomials...." No, $\lbrace na\rbrace$ is not a polynomial. –  Gerry Myerson Feb 1 '12 at 5:02
@GerryMyerson: hopefully the edit helped. –  daniel Feb 2 '12 at 19:43

(1): Not necessarily.

• We could take $a$ to be any irrational number in $(0,1)$ and $b=1-a$. In this case, for all $n>0$, $na+nb=n$ and $\{na\}+\{nb\}=1$.

• If 1, $a$ and $b$ are linearly independent over $\Bbb Q$, then by the Weyl equidistribution criterion, the sequence of points $(\{na\},\{nb\})$ will be equidistributed in the unit square. This means that $\{na\}+\{nb\}$ will not be equidistributed in the interval $[0,2]$ but will have a triangular probability density function, with maximum density at 1.

On the other hand, if $a=b$ and $a$ is irrational, then $\{na\}+\{nb\}=2\{na\}$ will be equidistributed in $[0,2]$, because $\{na\}$ is equidistributed in $[0,1]$. Also, it is possible for $\{na\}+\{nb\}$ to be equidistributed in a subinterval of $[0,2]$. For example, if $a=\sqrt{2}$, $b=-1/\sqrt{2}$, set $z:=\{nb\}$; then $na=-2nb$, so $$\{na\}+\{nb\}=\{-2z\}+z=\left\{ \begin{array}{l} 1-z, \qquad z\in(0,\frac12],\\ 2-z, \qquad z\in(\frac12,1]. \end{array} \right.$$ Since $z$ is equidistributed in $[0,1]$, it follows that $\{na\}+\{nb\}$ is equidistributed in $[\frac12,\frac32]$.

(2): Both $\{\{na\}+\{nb\}\}$ and $\{n(a+b)\}$ are integer translates of $na+nb$ which are in $[0,1)$, so $\{\{na\}+\{nb\}\}=\{n(a+b)\}$. However, $\{na\}+\{nb\}$ and $\{n(a+b)\}$ may differ by 1. For example, if you take $a=1/\sqrt{2}$, $b=1/\sqrt{2}$ and $n=1$, then $\{na\}+\{nb\}=2/\sqrt{2}=\sqrt{2}$, but $\{n(a+b)\}=\{\sqrt{2}\}=\sqrt{2}-1$.

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In fact, if $a+b=k/m$ is rational, $\{n(a+b)\}$ takes only $m$ different values, and since $\{na\} + \{nb\}$ is either $\{n(a+b)\}$ or $\{n(a+b)\}+1$, that takes at most $2m$ different values. –  Robert Israel Feb 1 '12 at 1:48