7 ms·
>You might ask: if we have a baseless logarithm log(N), do we also have a “baseless exponential”? Sure we can, with some naive algebra. If we can take log(x,ba
by amavect 3mo ago
>You might ask: if we have a baseless logarithm log(N), do we also have a “baseless exponential”?
Sure we can, with some naive algebra. If we can take log(x,base) and drop the base, then we can also take pow(base,x) and drop the base. Since bits=log(2), then pow(bits)=2. You can probably connect it to the reverse of things, like integrals.
Also, for fun, I'll play with some notation tricks.
log(freq) = pitch
freq = pow(pitch)
octave = log(2)
400*Hz = 100*Hz*4 // the frequency 400 Hz equals 4 times 100 Hz
log(400*Hz) = log(100*Hz) + log(4)
log(400*Hz) = log(100*Hz) + 2*log(2)
log(400*Hz) = log(100*Hz) + 2*octave
log(400*Hz) = log(100*Hz) + 2*octave // the pitch of 400 Hz equals 2 octaves above the pitch of 100 Hz
cent = log(2)/1200
A4 = log(440*Hz)
B4 = A4 + 200*cent // the pitch B4 equals 200 cents above A4
B4 = log(440*Hz) + 200*log(2)/1200
B4 = log(440*Hz) + log(2^(2/12))
B4 = log(440*Hz * 2^(2/12))
pow(B4) = 493.883 Hz // the frequency of B4 equals 493.883 Hz
I like the intuition that baseless logarithm notation gives, and it also avoids needing to choose a specific reference point. I can also directly calculate by choosing an arbitrary base:
pow(log(440*Hz) + 200*log(2)/1200)
exp(ln(440) + 200*ln(2)/1200)
- ajkjk 3mo agoTrue, I guess you can just 'curry' exponentiation and say that's a baseless power. I couldn't find a clean notation for it so I gave up..
- amavect 3mo agoHah, I can use this to give decibels an actual unit. dB_P = log(10)/10 dB_F = log(10)/20 log(10*V) = log(V) + 20*dB_F // the level of 10 V equals 20 dB more than the power level of 1 V. SPL = 20*10^-6 * Pa hearing_damage = log(SPL) + 90*dB_F // hearing damage occurs over 90 dB_F above SPL (neglecting A-weighting) pow(hearing_damage) = pow(log(SPL) + 90*dB_F)) pow(hearing_damage) = pow(log(SPL) + 90*log(10)/20)) pow(hearing_damage) = SPL*pow(90*log(10)/20)) pow(hearing_damage) = SPL*31622.7766 // the pressure of hearing damage occurs above 31622 times SPL pow(hearing_damage) = 0.632455532 Pa // the pressure of hearing damage occurs above 0.632 Pa Very helpful!! Imagine combining the goofy list of decibel suffixes into a uniform notation. Write the logarithm first so the + or - stays in the same spot. log(reference_unit) + value*dB_F (or dB_P) log(reference_unit) - value*dB_F (or dB_P) https://en.wikipedia.org/wiki/Decibel#List_of_suffixes https://en.wikipedia.org/wiki/Decibel#List_of_suffixes
- Rotundo 3mo agoAmazing. Thank you for giving me a new mental tool.