a great first infinite product!

[@VeryLazyAngel]

"Not super interesting if it starts at n=1" I laughed at this for a bit too long

[@TheDeadOfNight37]

Hey, 0 is a cool number

[@goodplacetostop2973]

3:34 Michael and proof by induction, name a more iconic duo 8:04 Good Place To Stop and turn the lights off

[@p.abhijitsuhas26]

How do you always reply so quick? Genuinely impressed

[@goodplacetostop2973]

@@p.abhijitsuhas26 The daily video is uploaded at the same time, so it’s not that hard to login 1 minute before that time lol

[@MichaelPennMath]

How about those animations though??

[@goodplacetostop2973]

@@MichaelPennMath I like these animations actually, I was wondering yesterday how long does it take to draw them.

[@iooooooo1]

@@MichaelPennMath They're really good.

[@dragonite7780]

My first thought was maybe it'd be easier as an infinite sum so I used ln( ), it becomes the sum of ln(1-1/n) +ln(1+1/n) and I'd rather have a telescopic sum, so I noticed 1/(1-1/n) = 1+1/(n-1) , got the telescopic sum and then the answer was just e^ln(1/2)

[@matthieumoussiegt]

yes I prefer this proof because it does not involve induction

[@GiornoYoshikage]

Cool!

[@lexinwonderland5741]

100% same, i've almost never messed with the beast that is infinite products without taming them with a ln() first

[@rgqwerty63]

You can do this exact same proof without taking any logs. Just using the telescoping product 1-1/n^2 = (n-1)(n+1)/n^2

[@CarlosCoelhoLechner]

@@rgqwerty63 there is a typo in your comment ("(n+2)" instead of "(n+1)"), it's 1-1/n^2 = (n-1)(n+1)/n^2. : ) But yeah, the telescopic product is, by far, the most natural way to solve this. There was no need to bring an induction hypothesis into this. 🤷🏻‍♂️

[@blackkk07]

The partial product can be directly calculated. Notice that (1 - 1 / n^2) = (1 - 1 / n)(1 + 1 / n) = ((n - 1) / n)((n + 1) / n). So Π(1-1/n^2) = ((1/2)*(3/2)) * ((2/3)*(4/3)) * ((3/4)*(5/4)) ... (((N - 1)/N)*((N+1)/N)) = (almost all terms cancel, but remains the first and last term) = (1/2) * (N + 1) / N = (N + 1) / (2N).

[@GiornoYoshikage]

This is the way I solved it

[@mitchkovacs1396]

I used to mess around with products like this in undergrad. The product of 1 - 1/n^x for positive x from n=2 to infinity is a really interesting function. At x=1 it's infinitely differentiable but not analytic

[@lock_ray]

Seems cool... This is just from playing around in Desmos but it seems like exp(1-zeta(x)) is a really good fit for a potential closed form. Didn't try to prove it, but the graphs look similar 🤔

[@divisix024]

Attempt before watching: The roots of sin(πx)/((πx) (1-x^2)) are all the integers except 0, 1, and -1, so intuitively it should have factors (1-x/n) where n is an integer other than 0, 1, or -1. Bracketing the terms with n and -n together gives (1-x^2/n^2), so sin(πx)/((πx)(1-x^2))=Π (1-x^2/n^2) where the product is from n=2 to infinity. Taking the limit of the LHS as x tends to 1, the result is 1/2.

[@jackhandma1011]

The expression (N+1)/N really suggests it has something to do with telescoping (and it does).

[@o0ss_st]

Please explain..

[@Mryeo5354]

@@o0ss_st The operand can be simplified into ((n+1)/n) *((n-1)/n). The partial products of both factors can be simplified by converting them into telescoping products.

[@o0ss_st]

@@Mryeo5354 Thanks!

[@Skibidipurush]

i wish I had a teacher like him when I was prepping for INMO

[@PhiDelta]

I solved this one using a different approach: first I noticed that the product from n = 1 is a special case of Euler's product for sin(pi*x)/(pi*x) (i.e.: when x = 1), then I put sin(pi*x)/(pi*x*(1-x^2)) = our result as x approaches 1, then I used l'Hôpital's rule to solve the limit: everything cancels out so nicely that I couldn't resist posting my solution :)

[@shambosaha9727]

I did it the same way!!

[@jimskea224]

My method too!

[@littlekeegs8805]

I really like how you used the little animations, and then just cut back to having stuff already written on the board. It saves some time, but it's still perfectly easy to understand.

[@toasticide816]

I think starting at n=1 would make this a neat little 2 mark question in a test of sorts perhaps. Always important to be able to notice these things to save time. (And just to answer, the n=1 term forms a 0 term, making the whole product 0)

[@SteveSpieldberg360]

(1-(1/n^2)) can be rewritten as [(n-1)/n]*[(n+1)/n] via differences of squares (a^2-b^2). Taking the Kth multiple of [(n-1)/n] is the same as (k-1)!/k!, while the Kth multiple of [(k+1)/k] is [(k+1)!/(k!*2)], the 2 being in the denominator because the factorial started off at (2+1) rather than (2) or (2-1) and so it's there to compensate for the missing 2. Simplify [(k-1)!(k+1)!]/[k!k!2] to get (k+1)/2k. Take the limit as k approaches infinity and you get 1/2 as your answer.

[@MortezaSabzian-db1sl]

Nicely done

[@noway2831]

This can be done directly via telescoping series -- write (n^2-1)/n^2 = (n-1)/n*(n+1)/n, and using the definition of the product as the limit of partial products with commutivity of multiplication we can split it into two series multiplied, namely the product of (n-1)/n times the product of (n+1)/n with both products having n go from 2 to N. The limit of partial products will come out to be 1/N and (N+1)/2, giving (N+1)/(2N) when multiplied as conjectured in your IH. Very cool!

[@brendanward2991]

0:20 - "Post in the comments why it's not so super interesting if you start at n=1" - Even I can answer that one!

[@Lionfish5656]

It equals 0.

[@xyy2861]

At the 'consider' line on the right hand side of the equation: it should be (...) times P[from n=2 to k](1-1/n²). (n² not k²)

[@ZacharyKeithQ]

I wondered about this myself and was going to ask about it... Thanks for the clarifying comment!

[@weonlygoupfromhere7369]

Love the mini animation u did

[@stmmniko7836]

Everyone is talking about infinite sums. Infinite product: "Am I joke to you?"

[@chrisglosser7318]

Natural log: “Yes, you are a joke to me …”

[@stmmniko7836]

@@chrisglosser7318 XD

[@robertveith6383]

"Am I a joke to you?" is played out.

[@yeahyeah54]

Product of (1- 1/n^2) = e^ log(Product(..)) but log(Product(...))= sum log(1-1/n^2)= sum log(1-1/n) + log(1+1/n)= sum log[(n-1)/n] + log[(n+1)/n]=sum log[(n-1)/n] -log[n/(n+1)] , so it is a telescopic sum, hence Sum(...) = log(1/2) but e^log(1/2) = 1/2 QED

[@skylardeslypere9909]

It's actually not too difficult to show that the partial product is in fact (N+1)/(2N) by noticing (1-1/n²) = (n²-1)/n² = ((n+1)/n) * ((n-1)/n) The partial product of this will just result in a bunch of factorials, which cancel nicely to the desired result.

[@quangdienduong5805]

Use the following relation (1-1/n^2)=(n-1)/n . (n+1)/n. Product of (n-1)/n equals 1/N and product of (n+1)/n equals (N+1)/2.

[@26-dimesional_Cube]

If n starts at 1 instead of 2 then the result = 0 Becuase 1-1/(1^2) = 1-1/1 = 1-1=0 0*#the rest = 0 so the multiply = 0

[@nicepajuju3900]

Loving the animations! great work!!

[@markbracegirdle7110]

Factorising 1-1/n² as (1+1/n)(1-1/n) makes the product trivial, as most of the terms cancel out.

[@VSN1001]

I used the same method and replaced the upper limit with N as N approach infinity

[@hassanalihusseini1717]

That was quite nicely explained Prof. Penn. Thank you!

[@26-dimesional_Cube]

Summarize: PI(n=2->inf, 1-1/(n^2)) = lim(N->inf, PI(n=2->N, 1-1/(n^2))) Let a_n = 1-1/(n^2), P_n = PI(m=2->n, a_m) Make a table from n = 2 to n = 7 and calculate a_n, P_n to see a pattern. Note that for every n that is even, the pattern of the value of P_n is that the numerator is just n+1 and the denomenator is just the double of n (or 2n), so we may have a pattern like this (or function), P_n = (n+1)/(2n). But at odd n value then the pattern is no longer fit. But we can tweak a little bit that at every odd terms, we double the numerator and the denominator so that the value don't change. Pattern found! Note: There is a pattern P_n = (n+1)/(2n) that has been found! So we have a claim here: PI(n=2->N, 1-1/(n^2)) = (N+1)/(2N) But this has to be proof first to use it later *Proof (Using Induction): Part 1: Base case: n=2 PI(n=2->N, 1-1/(n^2)) = 1-1/(2^2) = 1-1/4 = 3/4 (N+1)/(2N) = (2+1)/(2*2) = 3/4 Because those have the same value so base case is check Part 2: Shows that for k >= 2 then we have PI(n=2->k, 1-1/(n^2)) = (k+1)/(2k) Consider: PI(n=2->(k+1), 1-1/(n^2)) = (1-1/(k+1)^2)*PI(n=2->k, 1-1/(n^2)) = (((k+1)^2 - 1)/(k+1)^2) * ((k+1)/(2k)) = (((k+1)^2 - 1)/(k+1) * (1/(2k)) = ((k^2 + 2k + 1 - 1)/(k+1) * (1/(2k)) = ((k^2 + 2k)/(k+1) * (1/(2k)) = (k^2 + 2k)/(2k(k+1)) = (k(k+2))/(2k(k+1)) = (k+2)/(2(k+1) It it the same as P_n function (pattern) evaluated at value k+1. So we have proof that this is true for any k>=2 Note here: The part (1-1/(k+1)^2) is the value a_n evaluated at k+1 (((k+1)^2 - 1)/(k+1)^2) is the common denominator of (1-1/(k+1)^2) (Remember we define a_n from the first place?) So we have proof the pattern, let's use that to finish the question PI(n=2->inf, 1-1/(n^2)) = lim(N->inf, PI(n=2->N, 1-1/(n^2))) = lim(N->inf, (N+1)/(2N)) = lim(N->inf, (1/2 + 1/(2N)) = 1/2 (As n goes to infinity then 2n goes to infinity and 1/2n is equal to 0) So PI(n=2->inf, 1-1/(n^2)) = 1/2 = 0.5 And that's a good place to stop

[@CM63_France]

Hi, I love these animations! But that would be a shame if you replace all by this, don't become a new Mathologer! You both have your own talent. These are the 3 technics for doing math vidéos : - Michale Penn : chalk and board, - BlackPenRedPen : marker pencils, - Mathologer : computer animations. When I saw the announcement of that video , I thought about a solution : use the development of sin pi x / pi x into an infinite product, divide by (1-x) and deal with the case x=1 (you have to calculate a limit)

[@DanielDH179]

4:52 that "k" was supposed to be "n".

[@manucitomx]

Thank you, professor.

[@ΧρήστοςΠαπαδημητρίου-μ6π]

Another way is to ignore the odd partial products entirely, and by ignore I mean not verifying that they follow the same pattern, because P_n is strictly decreasing and thus the limit of any subsequence is the same as the limit of the sequence as a whole.

[@JasonOvalles]

I like the new animations! The one at the 0:40 mark is a bit hard to read due to the color choice. Maybe consider using a color-blind friendly palette? The contrast between the colors in those palettes will help all users see the visualizations more clearly!

[@jimskea224]

This is a special case of Euler's infinte product formula for sin(x), applied at x=π. sin(x) = x Prod _{n=1}^∞ (1-x²/n²π²) => Prod _{n=2}^∞ (1-x²/n²π²) = sin(x)/x * π²/( π²-x²) and apply L'Hôpital's rule at x=π. Euler's formula can be easily obtained by considering an infinite polynomial P(x) with roots at 0, ±π, ±2π etc normalized such that P(x)/x=1 as x->0.

[@CzaroDziejCK]

0:43 cute animation :)

[@titwtomos]

The greatest distance at which a hunter can be certain of hitting a rabbit is 20 metres. If the rabbit can run away 20 metres in the time that it takes the hunter to re-load after missing then, given that the probability of hitting the rabbit is inversely proportional to the square of its distance, what is the probabilty that he will hit the rabbit in N shots if the rabbit is 40 metres away to begin with?

[@ΓιώργοςΚοτσάλης-σ1η]

Starts at n=2 since 1-\frac{1}{n} is 0 when n=1

[@kurax9115]

Very good beginners problem because its quite intuitive to solve

[@VSN1001]

I rewrote the product as N approach infinity ∏[(n+1)(n-1)]/n^2, lower bound n=2 upper bound n=N. A V shape cancellation would occur and the product telescopes to 1/2

[@seroujghazarian6343]

Funny how the sum series diverges but the product series converges

[@blake121666]

1 - 1/n^2 = (n^2 - 1)/n^2 = (n-1)(n+1)/n^2 Sum starting at 2 = 1*3/2*2 * 2*4/3*3 * 3*5/4*4 * 4*6/5*5 ... Notice how the n-1 in the numerator cancels one of the numbers in the previous denominator and the n+1 cancels a number in the following denominator. All but the one of the 2s in the first denominator therefore cancel out. And the n+1 denominator cancels the remaining number in the previous numerator, At stage k, therefore, you end up with 1/2 * (k+1)/k Product to infinity is therefore 1/2

[@yowut8075]

Just log the darn thing

[@knutthompson7879]

Induction. Is there anything you can’t do?

[@jorallan]

The surprise exam paradox :-)

[@lianggong7814]

This is related to Euler's technique in Basel problem.

[@pouzzler]

I ain't done math for 30 years and I found it without pen & paper :) 1-1/n^2 = (1-1/n)(1+1/n)=(n-1)/n . (n+1)/n 1-(1/(n+1)^2)=(1-1/(n+1))(1+1(n+1))=n/(n+1) . (n+2)/(n+1) The second term of the first product is the inverse of the first term of the second product. By recurrence, all terms multiply to 1 except the first term, 1/2 and the last term (○○+2)/(○○+1) This last term equals 1, and the product equals 1/2

[@angeloluisrocattojunior3425]

0:22 - Because it's equals zero.

[@alexrozenbom3430]

That's right. If n=1, the product element =0. So the entire product would equal 0.

[@kostaspapadopoulos1480]

And that's a good place to turn off the lights!

[@themushroom2130]

For the n+1/2n proof, couldnt you simplify the fractions as a product of n^2 - 1/n^2 and then factorise all the multiplicands using differences of squares and make all of it into factorials and cancel? i feel that takes less effort

[@adedamolayusuff5995]

If n started with one then wouldn't the product be 0?

[@tomkerruish2982]

Starting the product at n=1 reminds me of evaluating the product (x-a)(x-b)(x-c)...(x-z).

[@mrflibble5717]

Thanks Michael. Great explanation. Your work is very much appreciated. Clear explanation of the Infinite Product. The limit of the successive finite sums was nice, I look forward to othe infinite product solution methods. Michael, if you have time, would you go through the solution of the following Probability problem please? A purse contains 3 one cent coins, 4 two cent coins, two five cent coins, and one 10 cent coin. If a selection of coins are taken out, any number at a time, what is the Probability that the sum is 12 cents? Rgds, Rod Spencer

[@Thauanpolar22]

Use sin(x)/x = prod(k=1 to inf)(1-x²/(k*pi)²)

[@srittampanigrahi4810]

This is the Question from IIT JAM 2021 Mathematics

[@Deckname7]

I had to prove the closed form of this product via induction two weeks ago.

[@orenfivel6247]

product telescopes easily

[@srittampanigrahi4810]

Just factories it (1-{1/n}) *(1+(1/n)) then put value n=2 to n=N and you will see Some got cancelled and as n tends to infinite Value must be 1/2

[@benardolivier6624]

Math is actually easy: explore, find pattern, prove by induction.

[@seroujghazarian6343]

Also, if it starts at n=1, 1-1=0, and thus, the product would be 0

[@kingarth0r]

If n=1 the product is just 0

[@TIO540S1]

Thanks for this. I’ve always been slightly intimidated by infinite products. Can you speak on their applications? That is, not, say, physical applications but their broader scope in math.

[@meiwinspoi5080]

when n = 1 to infinity it is the easiest to calculate = Zero

[@tgx3529]

ln()=lim suma( ln(n-1)-ln n+ln(n+1)-ln n) where n=2...N, where N go till Infinity, there is -ln2,the origin result Is 1/2

[@abduvohidodiljonov9365]

a^2-b^2=(a-b)(a+b) , just need to use this.

[@joaquingutierrez3072]

My solution: It can easily be proved that if we change that infinity for a k, then the product evaluates to (1/2)(k+1)/k. Taking a limit as k goes to infinity we get 1/2, which is our answer.

[@arkitson]

That's exactly what he did.

[@joaquingutierrez3072]

Well, I it before looking at the video

[@Etothe2iPi]

4:52 1/n^2, not 1/k^2

[@rezabidar1042]

I would very much like to know an elementary proof of the Eulers formula Prod_1^infty(1-z^2/n^2)=sin(pi*z)/(pi*z), choose z=sqrt(-1) and we have Prod_1^infty(1+/n^2)=(e^pi-e^(-pi))/(2pi) Super Cool!!

[@elvisk6632]

Wow induction solves a lot of problems

[@1-Tryhxrd]

isn't the exponential form a much simpler way to solve this ?

[@krisbrandenberger544]

It would not be interesting if the product started at n=1, because that would make the first term, and hence, the entire product equal to zero.

[@yerel_nimni_avni]

On the "consider" part of the induction the last k should be n.

[@yerel_nimni_avni]

It's around 5:05 in the recording.

[@TheQEDRoom]

because the product will be 0 when we start at n=1

[@numoru]

Sin(x)

[@rdius4822]

n = 1 rather than 2 is maybe not interesting because 1-1/1^2 is 0 so the end product would also be 0. Nice problem!

[@pow3rofevil]

Ese producto es muy fácil, deberías calcular algunos con función gama 🤘🏻

[@lisandro73]

If it starts at. Zero doesn’t compensate the one case?

[@mw21016]

Today was one the best days on my life, i was capable to solve that excercise !!!! sí soy el mejor!!!

[@TurtleX21]

What does ii mean im bit confused with potetensial energy!.

[@tonyhaddad1394]

0:25 beacaus 0×y=0 1+1=2 1+2=3 i have a nice homework What 1+3 is ?

[@wyattstevens8574]

N=1 alone evaluates to 0.

[@gurkiratsingh7tha993]

I have sort of simplyfied riemann hypothesis and if you are able to solve the question given below then you just solved riemann hypothesis. Ques) prove that if a is not equal to 0.5 then (The improper integral from 1 to infinity of((summation from n=1 to infinity(e^((-n^2)*π*x)))*sin(b*ln(x)/2)*((x^((a-2)/2)) - (x^((-a-1)/2))) dx)) diverges Here a and b both are real numbers and a is between 0 and 1, e and π are the famous mathematical constants.

[@gavasiarobinssson5108]

Will there be more?

[@curiosityxxx4305]

may be cause 1 X 1 = 1

[@shenal8167]

If you start from n=1 then whole thing get 0

[@edmund3504]

If you start at n=1 the first term is 0, making the entire infinite product 0 as well.

[@Channel_Math_and_Physics]

I first factored it to (1-1/n)*(1+1/n) then separated the products. Prod of (1-1/n) from 1 to N is 1/N. (As (1/2)*(2/3)*(3/4)*...(N-1/N)=1/N) And prod of (1+1/n)=N+1/2.(As (3/2)*(4/3)...(N+1/N)=(N+1)/2 So, (1/N)*{(N+1)/2} =1/2*(N+1)/N =1/2*(1+1/N) Applying limit N>inf =1/2*(1+0) =1/2 Felt confident after doing it this way, then saw the solution, all the confidence went down the drain 😂

[@staswisniewski4101]

n=1 just makes everything 0 (cause 1-1/1=0)

[@anomalous5048]

1/2

[@terryendicott2939]

I have zero idea at what happens when n is one. pun intended

[@guidichris]

Zero is kinda boring which is what you get if you started with n=1.

[@lassekoch3573]

starting at 1: (1- 1/1) = 0, 0 * x = 0 so the whole series becomes = 0

[@wonka727h]

n=1 makes the first term becomes 0

[@samocali]

where n=1 the product would be 0 so this is pointless

[@lucaswilkins9217]

Michael, I really don't like this new intercalation of computer drawn screens. The changing parts makes it harder to follow: you can't read things at your own pace, and as you're not there writing is no indication of where any changes will be. Transitioning to the white screen is distracting and interrupts reading. The way lines are drawn is harder to read than the chalk board and the white background is hard on the eyes.

[@stephenbeck7222]

I disagree. I think you can follow it pretty easily and I like that it speeds up the simpler parts of the video. In one of the graphics he used a nice red highlighting technique to show what he was doing.

[@lucaswilkins9217]

@@stephenbeck7222 Fair enough. It's not like I couldn't follow it, it's just that I thought it was harder - maybe I'm getting old and got to that point where I dislike flashing lights and loud sounds, rather than finding them engaging. Personally, I found the red highlighting thing to be one of the worst aspects of it, I'd much prefer someone just pointing at it.

[@yanceyward3689]

n=1 is just zero.

[@SJ23982398]

Hey, a video I could follow too!

[@brandongroth4569]

Defines a_n and P_n, then never uses them again.

[@michaelempeigne3519]

if n = 1, then 1 - ( 1 / 1^2 ) = 1 - 1 = 0

[@palahnuk1]

Induction ;)

[@pastelrainbowsd5574]

I do not belong here

[@alexrozenbom3430]

ha ha

[@yahav897]

I lack formal notation, so I don't know how to prove using induction, or writing/thinking like Michael does. I have a few years until university; how could I improve my skills? I really want to learn maths formally. My idea was making 1 - (1/n^2) into (n-1)(n+1)/n^2. Plugging in values, you notice that many terms cancel out - and the ones that don't are 1 over 2, making the answer 1/2.

[@anasnaji9911]

fub

[@gordonstull1962]

I am not real sure, about the math behind this "infinite product", in my way of thinking, finite math would be closed-looped and infinite math would be open-loop. Quite obviously I am not a mathematician, particle-flow research has shown me time and time again, that action between empty-space and filled-space is that action that can either remove boundaries or produce boundaries. I know I should not assume anything and just go with that formula, but one thing comes to mind, being those exacting discrepancies between space & time and size & time coordinates, the infinite product would appear to be a force that pulls-straight?