Let $f:\Bbb{N}\to\Bbb{C}$ denotes the indicator function of squares. Express it in terms of Mobious function $\mu$.

Here $f(n)=\begin{cases} 1\ \text{if } n=m^2\text{ for some }m\in\Bbb{N}\\ 0\ \text{if otherwise} \end{cases}$

This is a multiplicative function. At first I define $g:\Bbb{N}\to\Bbb{C}$ be $g(n)$ to be number of square divisors of $n$.

Then it's easy to verify that $\mu\ast g=f$.

But later I realised that $g$ is nothing but $g= \mathbb{1}\ast f$ where $1$ denotes the constant function $1$. Then $\mu*g=(\mu*1)*f=\delta*f=f$ where $\delta(1)=1$ and $\delta(n)=0\ \forall n>1$.

So my solution i.e. the expression of $f$ is not giving anything special.

Can anyone help me in this regard? Thanks for help in advance.

Solution 1:

I'm not sure what your question is, but whether or not the expression $f=g\star 1$ is useful remains to be seen. We have \[ \sum _{n\leq x}f(n)=\sum _{d\leq x}g(d)\sum _{n\leq x\atop {d|n}}1=x\sum _{d\leq x}\frac {g(d)}{d}+\mathcal O\left (\sum _{d\leq x}|g(d)|\right ).\] If $g$ is weighted differently to $f$, this expression may be useful. If $g$ is weighted only on squares for example (where $f$ may not be), then the sum in the error is \[ \sum _{d^2\leq x}1=\sum _{d\leq \sqrt x}1\] so \[ \sum _{n\leq x}f(n)=x\sum _{d\leq x}\frac {g(d)}{d}+\mathcal O\left (\sqrt x\right ).\] Of course, the main term will simplify for the same reason, but the point I want to make is that the error term is small because the sum was of shorter length. Re-writing functions as certain convolutions so that the sums arising have different lengths has proven to be one of the most fruitful ideas in analytic number theory.

So you are right to notice that your convolution is nothing special, as long as sum length is nothing is special. But it is!

(And if you want a concrete example, then take $f$ to be the indicator of the squarefree numbers. Then you can find $g$ by the argument you make in your post. Now my argument above will give you the same error term and a main term \[ x\sum _{d\leq x}\frac {\mu (d)}{d^2}=x\sum _{d=1}^\infty \frac {\mu (d)}{d^2}+\mathcal O(1)=\frac {x}{\zeta (2)}+\mathcal O(1),\] so you've counted the squarefree numbers up to $x$.)

Solution 2:

I'm not quite sure if this is what you're looking for but it is interesting (and easy) to note that in your notation;

$f=\lambda*1 = \sum_{d \mid n} \lambda(d)$

where $\lambda$ is the Liouville function