Möbius Function - Identities and Mobidromes

[JohnnySegment86]

It turns out that last one's not too hard to prove: Take -log of LHS and expand log(1-x^n) in a power series. The coefficient of x^m is then (1/m)sum_{d|m} mu(d) where the sum is over the divisors of m. This equals 0 unless m=1 (random mobius function fact) so -log(LHS)=x. Tadaaa!! -_-

[DoubleDutchBust]

R E I N C A R N A T I O N S ANDO GOODS

MöbiusFunctionIdentitiesandMob­idromes

GeorgeHarrisonJohnLennonPaulMc­cartneyRingoStarMöbius FunctionRelativitynod

OnoYokoiNDIGOIdentitiesandMobi­drome

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GeorgeHarrisonJohnLennonPaulMc­cartn

MöbiusFunctionIdentitiesandMob­idromes

R E I N C A R N A T I O N S ANDO GOODS

3:13

[DoubleDutchBust]

Just as a special example for u.ok a song

MöbiusFunctionIdentitiesandMob­idromes

GeorgeHarrisonLaurelglenLagran­geLima

MöbiusFunctionIdentitiesandMob­idromes

Just as a special example for u.ok a song

3:13 a special place

[DoubleDutchBust]

Identities and MobidromesGMöbiusA

soreallynodlee yep! donyleeFunctioN

seangrayes really3:13 naes InverteD |x|

soreallynodlee yep! donyleeFunctioN

Identities and MobidromesGMöbiusA

[DoubleDutchBust]

There's a good reason these tables are numbered Honey, you just haven't thought of it yet - P!aTDMöbius Function - Identities andMobidromeskgrayinyangem

ThreehundredAndtwentythreelyno­donylee

3:13 |hat| 7:00

[DoubleDutchBust]

Mobidromeseangray

AgoodPlacedonylees

KalielgrayTwoplaces

Möbius Function 3:13

The first place |ok| uc

[NejiHyuuga1337]

unbelievable

[sacreebluee]

Is there a video following this one?

[donylee]

At the moment, no. Going deeper into the theorem requires mathematics which is too advance for me.

Thanks for your interest.