FIT3.2.2. Mobius Inversion Formula

Typo: At 14:36: Use variable other than k (say K) for the induction variable. Keep k as a generic exponent.

Field Theory/Number Theory: We state and prove the Mobius Inversion Formula. We apply the formula to several examples, including cyclotomic polynomials and the Euler totient function. (Reload: Badly synced audio after compression.)

[iTzSimple]

Seriously, thank you so much for posting all of your videos... I COULD NOT make it through my Number Theory class without your help! I appreciate it!

[dyntaos]

Very well presented! I like how you pause for a moment to allow things to sink in. Thank you!

[WEAVER]

IM IN AWE, TSUCH WONDERFUL ORCHESTRATION OF THE PROCESS. MY NEW GO TO MATH YOUTUBER

[joefagan]

Condensed but very well thought out and great presentation. I had it hit pause 20 times but I like it like that.

[chancemorgan]

Thanks so much for posting! You are a life saver!

[pauluk]

9:30-9:45 Didn't get this at first but finally spotted how this works (we are picking divisors 1 and p - p^0 and p^1 IOW, but any larger divisors necessarily contain squares giving '0' mu

11:45-12:10 It's impressive how this method from number theory connects to cyclotomic polynomial factoring. The condition |x| < 1 allows every N to give finite value and therefore gives us an NTF?

14:33-14:48 Thought the single use of 'k' for #primes confusing since k was used for prime multiplicity.

[krlsjke]

At 14:36, you state, "Suppose true for n with k prime factors, then consider f(n*p^k) ...." I believe you meant the p^k in the argument of f to be a general integer power of p. There is no reason the power of p must be limited to the integer k defining the base of the induction.

[anandafi]

I see this video after i saw avenger endgame😂😂😂

[MattSpaul]

I've been taught it as \sum{\mu(n) g(\frac{n}{d})} rather than \sum{\mu(\frac{n}{d}) g(n)}