# What is the solution of normalized harmonic series based on $4/3$ between one and two?

In music theory notes generated by the consequencing interval of $4/3$ generates harmonic series. Series can be normalized by multiplicating the fraction with a $2$ in power $n$.

What is a formula for $n$ depending on $m$ such that the ratio is always between $1$ and $2$?

I'm looking for integer solutions for n when m is a whole number:

$$1 \le (4/3)^m * 2^n \le 2$$

• Are $m$ and $n$ positive? – Julián Aguirre Aug 29 '18 at 11:02
• m and n can be either positive or negative. – MarkokraM Aug 29 '18 at 11:06

Take logarithms on both sides:

$$0\le m\log\frac43+n\log2\le\log2\;.$$

This is an area in the $(m,n)$ plane that lies between two parallel lines. Solving for $n$ yields

$$-m\log_2\frac43\le n\le1-m\log_2\frac43\;,$$

so

$$n=\left\lceil-m\log_2\frac43\right\rceil\approx\left\lceil-0.415m\right\rceil\;,$$

where $\lceil\cdot\rceil$ is the ceiling function.

The inequality can be rewritten as $$\Bigl(\frac{3}{4}\Bigr)^m\le 2^n\le2\,\Bigl(\frac{3}{4}\Bigr)^m.$$ Taging logarithms an dividing by $\log2>0$, we get $$\frac{\log(3/4)}{\log2}\,m\le n\le1+\frac{\log(3/4)}{\log2}\,m.$$ From here, it follows that $$n=\Bigl\lfloor\frac{\log(3/4)}{\log2}\,m\Bigr\rfloor,$$ where $\lfloor\ \rfloor$ is the floor function.

• I think you need the ceiling function instead of the floor function. – joriki Aug 29 '18 at 11:24
• Right. Floor gives normalization between 0 and 1, but ceil between 1 and 2. – MarkokraM Aug 29 '18 at 14:57
• @MarkokraM: I think you mean between $\frac12$ and $1$? – joriki Aug 31 '18 at 9:56
• @joriki To be more precise, yes, floor seems to normalize series between half and one. – MarkokraM Aug 31 '18 at 10:11