My question came from my musicial part: The note A4 on a piano is 440hz and the note A5 (one octave higher) is 880hz.

On the piano there are 12 notes between A4 and A5 (include).

Im trying to find three notes between A4 and A5 (hence on the scale of 440 to 880) Using the golden ratio: 1.618 (accuracy of three points is enough here).

More specific: There are 5 points on an axis from 440 to 880: A4=440 note1=?, note2=?, note3=?, A5=880

I want to find (note1, note2, note3) Where note2 is: “1.618 closer to note1 than note1 to A4”, note3 is “1.618 closer to note2 than note2 to note1” and A5 is “1.618 closer to note3 than note3 to note2.

Im not sure if one can prove that 3 points like this exist, but I guess we can prove that there are n points (n is natural) like that that exist, right?

How can I find these n’s? And when finding one, how can I find its n notes?

  • 2
    $\begingroup$ First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)? $\endgroup$ – Arthur Jul 30 '18 at 14:13
  • $\begingroup$ I meant for the first approach in your comment. $\endgroup$ – user1901968 Jul 30 '18 at 14:28

From what I'm understanding, you want the ratio between $d_2$ and $d_1$ to be 1.618, and so on for $d_3$ to $d_2$, and $d_4$ to $d_3$. You basically end up with a system of equations:


$$d_1=n_1-A_4$$ $$d_2=n_2-n_1$$ $$d_3=n_3-n_2$$ $$d_4=A_5-n_3$$

Where $n_k$ is the $k_{th}$ note and $A_4<n_1<n_2<n_3<n_4<A_5$ and $d_k$ is the distance between $n_k$ and $n_{k-1}$ and $n_0=A_4$ and $n_4=A_5$.

Then, by the golden ratio constraints you've asked for, we end up with:

$$n_3-n_2=(A_5-n_3)\cdot1.618$$ $$n_2-n_1=(n_3-n_2)\cdot1.618$$ $$n_1-A_4=(n_2-n_1)\cdot1.618$$

Solving these taking $A_4=440$ and $A_5=880$, we end up with

$$A_4=440$$ $$n_1=636.7701$$ $$n_2=758.3833$$ $$n_3=833.5459$$ $$A_5=880$$

All in Hz. Each of those are in the golden ratio to the preceding one. You can generalize this as long as you know your initial and final note.

  • 2
    $\begingroup$ This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for. $\endgroup$ – Ross Millikan Jul 30 '18 at 14:51
  • $\begingroup$ Exact solution for my specific question and generalization. $\endgroup$ – user1901968 Jul 30 '18 at 20:37

If what you want is the ratio of the frequencies to be $\varphi$ for each of the four ratios then this is impossible.

The interval on the ordinary $12$ tone scale is the twelfth root of $2$, which is $1.05946309436 \approx 1.06$. Four your five note scale you want $$ 2^{1/5} = 1.148698355 \approx 1.15 . $$ That's pretty far from $\varphi \approx 1.618$.

Even just a two note scale gives you a ratio pf $\sqrt{2} \approx 1.4 < \varphi$ .

  • $\begingroup$ As I read it, he doesn't want the ratio between the frequencies to be $\varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear. $\endgroup$ – Arthur Jul 30 '18 at 14:20
  • $\begingroup$ Exactly. And i meant the frequency interval $\endgroup$ – user1901968 Jul 30 '18 at 14:34
  • $\begingroup$ @user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio. $\endgroup$ – Ethan Bolker Jul 30 '18 at 15:38

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