I am trying to show $$\sum_{d \leq x} \mu(d)\left\lfloor \frac{x}{d} \right\rfloor = 1 \;\;\;\; \forall \; x \in \mathbb{R}, \; x \geq 1 $$ I know that the sum over the divisors $d$ of $n$ is zero if $n \neq 1$. So we can rule out integers that are divisors of $x$ (if $x > 1$). I am not sure if I am on the right track to prove this. Any hints would be helpful. This is homework.
Asked
Active
Viewed 4,213 times
12
-
Similar: http://math.stackexchange.com/questions/8002 – Bart Michels Mar 15 '15 at 14:33
2 Answers
7
This is typical example in Möbius Inversion.
Use $\sum\limits_{d|n}\mu(d)=\epsilon(n)$ where $\epsilon(n)=1$ if $n=1$, and 0 otherwise. Now, $$\sum_{n\leq x}\sum_{d|n}\mu(d)=\sum_{n\leq x}\epsilon(n)=1$$ Changing the order of summation, we get $$\sum_{d\leq x}\mu(d)\sum_{d|n, n\leq x}1=1$$ Then the inside summation is $\left\lfloor\frac{x}{d}\right\rfloor$. So, we are done.
Sungjin Kim
- 20,102
-
-
The change of order works as follows: First, sum over $n$ inside instead of $d$. Then inside sum is over all multiples of $d$ but $\leq x$. That number is precisely $\lfloor\frac{x}{d}\rfloor$. – Sungjin Kim Mar 23 '13 at 15:49
-
Is it possible to deduce $\left| \sum_{d\leq x} \frac{\mu(d)}{d} \right| \leq 1$ – Tyler Hilton Mar 24 '13 at 23:21
-
Use $\lfloor\frac{x}{d}\rfloor=\frac{x}{d}+{\frac{x}{d}}$ where ${\frac{x}{d}}$ is the fractional part of $\frac{x}{d}$ – Sungjin Kim Mar 25 '13 at 02:35
-
Im not sure what you mean by fractional part. so [6/4] = 1 + 1/2? – Tyler Hilton Mar 25 '13 at 05:18
-
@TylerHilton see this answer: https://math.stackexchange.com/questions/2657156/prove-left-vert-sum-n-1n-frac-munn-right-vert-leqslant-1-where/2699866#2699866 – Sungjin Kim Mar 20 '18 at 05:41
5
It appears indeed to be an instance of the Moebius inversion formula, applied to $f(x) = \lfloor x \rfloor$ and $g(x) = 1$. We have $$ f(x)=\sum_{d \in \mathbf{Z}, 1 \le d \le x} g(x/d). $$ Then $$ g(x) = \sum_{d \in \mathbf{Z}, 1 \le d \le x} \mu(d) f(x/d) $$
Andreas Caranti
- 68,873
-
From both answers, I am confused on how the floor function is equivalent to the summation of the divisors – Tyler Hilton Mar 23 '13 at 00:22