I saw that as well, as I always try to check these formulae with the small values we already know the answers for. So for d= 1, 2, 3, 4, we know N(d) = 9, 9, 90, 90. The mistake he made was forgetting that the a_1=a_d pair wasn't included in the count of "10 choices", although at 8:17 he says "we're counting from a_2 to a_(d+1)/2" and promptly counts from a_1 instead. I have no clue how he managed to "take the 9 out and simplify the exponents" to get the expression shown at 10:36. His working should give a leading factor of 90√10, I think. If you use the correct formula for N(d), I think you should get: 9*√10*SUM/d=1...inf/(1/√10)^d which sums to (9*√10)*(1/√10)/(1 - 1/√10) = 9 / (1 - 1/√10) Finally, in the video, the sum of 9/√10*SUM/d=1...inf/(1/√10)^d would actually be (9/√10)*(1/√10)/(1 - 1/√10) = (9/10) / (1 - 1/(9*√10)*(1/√10)/(1 - 1/√10) but that is miscalculated also and the value of the correct sum magically appears in its place. It doesn't alter the convergence, just the accuracy of the means by which we reach upper bound for that limit, which I calculate to be about 13.2, and which seems reasonable to me.
I desperately want to know what this converges to. Lol. That seems to be a pretty hard problem though. I think I might just have to write some code to approximate the solution later. Edit: I wrote some code to do this, and it appears to be converging around ~3.37025
@@JorgeLuis-ts6qp How'd you get that expression? Did you just approximate the sum for a lot of bases, then use a regression to approximate a best fit curve?
Could be interesting to solve for different bases. I suppose a similar strategy can be used for different bases/encodings, but off-hand I'm not sure how it'd turn out with the more "exotic" ones (irrational, negative, complex bases).
If looking at natural numbers, the argument is really similar, your end result depending on the digits possible under that base (here 10 and 9 are prominent in a decimal base, where it would be 1 and 2 in binary). All those series will converge with a similar proof. As for an irrational base, that won't really work, since then you cannot express natural numbers in finite digit representations, if at all. If different bases interest you, look into p-adic numbers with p-adic metrics, these are really interesting for converging series that have no business converging in a traditional |a-b| metric space.
@@DrTaunu even in an irrational base, there's a finite number of palindromes of less than n digits right? You could then consider summing over all of them... Depending on how the sum converges (specifically, if it converges conditionally), the /order/ that we sum over them is relevant, and I don't think there's a single canonical ordering in this case...
Did you see the 9s and 10s at the end of the video? For any base, b, rational or not: sum(1/palindromes) < ( (b-1) / (sqrt(b) ) sum (d=1 to infinity) of ( (1/(sqrt(b))) ^d ) = (b-1) / (1 - 1/(sqrt(b))) For bases less than 1, the sum of palindromes is definitely ironic negative numbers.
My approach was to group palindromes of 2n+1 or 2n+2 digits together, since there's an equal amount of both. Then, the terms in the upper bound will be: 2*9*10^n/10^(2n) = 18*(1/10)^n, n>=0
My predict was that this sum is divergent, then trying to proof I realised that maybe isn’t the case and then I proved that it converge, very interesting exercise!
Although not relevant to the proof, I think he lost a √10 in the denominator when he gave the total for the final series. But as others have mentioned, N(d) should have been 9*10^floor((d-1)/2), which is 10 times less than the N(d) he gave. Given all the inequalities being thrown in I think that in addition to losing the floor in the exponent, I would have turned the 9 in N(d) into a 10 so that N(d)
To follow up on my own comment, I think I still have it wrong. It is clearer if you express everything using √10 so 10^(a/2) becomes √10^a and 10^b becomes √10^2b. Using the proper value for N(d) the upper-bound sum becomes (this is without using the 10-instead-of-9 change in N(d)): sum(d=1 to infinity){9*√10^(d-1)/√10^(2d-2)} which simplifies to sum(d=1 to infinity){9/√10^(d-1)} and then sum(d=0 to infinity){9/√10^d} by reindexing the sum and, by using the closed form for the series sum (a=9, r=1/√10) 9/(1-1/√10) and then multiplying numerator and denominator by √10 to give 9√10/(√10-1) which is about 13.162 This is the same as Michael's solution, so I think the d+1 instead of d-1 in his N(d) was exactly compensated when he lost some powers of 10 (through more than one error) during simplification.
Instead of bounding again with the floor((n+1)/2) (which I actually had as ceil(n/2)), we can split it up into n even vs. odd, and we eventually get that the sum is between 1.1 and 11. Evaluating this with Python gives : 10^6: 3.367469 10^7: 3.369772 10^8: 3.370002 so it seems that the sum is approximately 3.37, pretty snugly in the middle of our interval (assuming we take the middle as the geometric mean :P)
He did take it into account, since the geometric serie gave x/1-x, and the top x canceled out with the one on the denominator in front of the sum (x = square root of 10 here)
@@ludom.4652 actually x = 1/sqrt{10} (the absolute value of sqrt{10} isn't less than 1 so the geometric series doesn't converge if you take sqrt{10} as the common ratio), so it looks like he does account for the missing d=0 term (by using the formula x/(1-x)) but makes an algebra error? - the top x = 1/sqrt{10} should combine with the 1/sqrt{10} outside the sum to give 1/10. of course this really doesn't matter but I thought I'd point it out.
@Gabriel Flemming well, apart from that and the miscount at 8:32 and the rewrite blunder at 10:15... and even if i won't count the implied exclusion of 0... _Almost_ great work. >:-]
According to the OEIS (A118031), the first 1000 digits of this number are: 3.3702832594973733204921572985055311230714577794527784913350689259825197603494767589703011219269780898632350532421403168917126747933674147830127937467183247967252750087722634827550979318225637556877944404376796847493151313373647295531271637481054977957361205682332484460649774315808378576222780370382940370280769546170948671477666713704407509837747327681658124294157278347595895815147012803762301460078753895374219171398687823474425496450600364962569657596197751997355410808437695497820627954263483238465927514188721203312202864207253747796264903130226097564588742083842569397320030502643768192965449694458613706042979690968950583460074807869595979375862261937148481998657086618703808214697724773926778461748481209093452318737751966417981265923055052758889661866377594042994461127572237425467596287013687715350966157434717149952336086570739332966878807292450980178787018319902919155116505026319887004331982800424031530346620794654849578307161469569012827831247061491982966378324937801182371089307855265...
I was wondering already what the value would be. Can you get this value for base2, base4, base8, base16 also? This base10 certainly most interesting.. There could be a lot different versions of this numbers game.
Naively, palindromes grow roughly as fast as squares, and the sum of the reciprocals of the squares converges so I would expect this sequence to as well.
Is there a rigorous reason for saying they *grow* roughly as fast as squares? I only see a counting argument similar to the one in the video - i.e. the count of d-digit palindromes is roughly the same as d/2-digit numbers.
There is an exact 2:1 correspondence between the natural numbers and the palindromic natural numbers. For every natural number a formed as digits a_1 a_2 a_3 ... a_k you can form two palindromic numbers: One is the 2k-digit value using digits a_1 a_2 a_3 ... a_(k-1) a_k a_k a_(k-1) ... a_3 a_2 a_1 which is > a*10^k and < (a+1)*10^k The other is the (2k-1)-digit value using digits a_1 a_2 a_3 ... a_(k-1) a_k a_(k-1) ... a_3 a_2 a_1 which is > a*10^(k-1) and < (a+1)*10^(k-1) There is no other natural number which can form the same palindromic numbers in the same manner. This allows you to enumerate all the palindromic numbers so you could easily use a program to approximate the sum of the series. Since k=1+floor(log_10(a)) you could then write tighter upper and lower bounds for the original series, but I don't see any good way of evaluating these.
@@federicotedeschi3841 You're right, there is a special case for forming a 1-digit palindrome, in which case the lower bound a*10^(k-1) must include equality.
I predicted it would diverge because of its similarity to the harmonic series, turns out there are few enough of the palindromic numbers that it doesn't tend to infinity, I.E. the function N(d) increases fast enough, in which I don't think there is a single base where it diverges, since (1/sqrt(10)) would be replaced with 1/sqrt(base) which clearly is always less than one for non-trivial bases.
But almost all of them contain a 9, in the sense #{n in IN|n 1. But that means it must not be summable, just as the harmonic series (it suffices if the proportion does not approach 0), because if at least every D th element is there, the sum is >= infinity/D
10:47 I didn't trust that rewrite, so I did it myself. it should be 9*sqrt(1000)* sum( (1/sqrt(10))^d ). the final calculation will be 1/100 times the correct result. that being said, so long as it converges at ANY point, naturally, it converges, so you still answer the question correctly.
As usual great video, during the video I had a nice question in my head, which is : what is the natural density of the set of all palindromic numbers denoted by P, i.e. lim |P∩[1,x]|/x as x goes to +infinity ?? (if it exists of course)
That would be zero. If it was nonzero, then the series of reciprocals would diverge. (but we know it converges, so it must be zero) As a sketch of a proof (not a proof): If the density was 1/D, then that means roughly every Dth number is in P, so the series is roughly growing at the same rate as the series 1/(Dn), which diverges, since that's just a multiple of the harmonic series.
A related question is what's the density of palindromes in the set of all numbers with the same number of digits; this drops quickly: 100%, 9.89, 9.99, 1, 1, 0.01... If you then look at how this density changes across # digits, it oscillates back and forth between "10% of the last range's density", and "101% of the last range's density"
@@f5673-t1hYour reasoning is valid. I did a little math and it does indeed equal zero. Here is what I did: We first denote for every integer 𝑟≥0: 𝒥(𝑟) = [10^𝑟, 10^(𝑟+1)[ ∩ℕ (the set of integers with 𝑟+1 digits (not including 0)) Then we calculate a bound for |P∩𝒥(𝑟)|, using the same reasoning Michael did, we show that |P∩𝒥(𝑟)|≤10^⌊(𝑟+1)/2⌋≤10^((𝑟+1)/2)≤10*{10^(𝑟/2)} Now let 𝒙 be a real number ≥10, the number of digits in ⌊𝒙⌋ is : 𝑟 = ⌊log(⌊𝒙⌋)/log(10)⌋+1 Then we show that: P∩[1,𝒙] = (P∩𝒥(0))∪(P∩𝒥(1))∪( P∩𝒥(2))∪...∪(P∩𝒥(𝑟-1))∪(P∩[10^𝑟,𝒙]) Then: |P∩[1,𝒙]| ≤ |P∩𝒥(0)| + |P∩𝒥(1)| + ... + |P∩𝒥(𝑟-1)| + |P∩[10^𝑟,𝒙]| ...............≤ |P∩𝒥(0)| + |P∩𝒥(1)| + ... + |P∩𝒥(𝑟-1)| + |P∩𝒥(𝑟)| ...............≤ 10*{10^(0/2)} + ... + 10*{10^(𝑟/2)} ...............= 10*{sqrt(10)^0 + ... + sqrt(10)^𝑟} ...............= 10*[{sqrt(10)^(𝑟+1) - 1}/{sqrt(10) - 1}] ...............≤ 10*{sqrt(10)^(𝑟+1)} ...............= 10*{sqrt(10)^(⌊log(⌊𝒙⌋)/log(10)⌋+2)} ...............≤ 10*{sqrt(10)^(log(𝒙)/log(10)+2)} ...............= 100*sqrt(𝒙) Which gives us: 0≤ |P∩[1,𝒙]|/𝒙 ≤100/sqrt(𝒙) → 0 (as 𝒙 goes to +∞). (Correct me if I'm wrong).
It is of the order 1/sqrt(n) for numbers up to n. So the density drops to zero. That is easy to see because you can choose the first half of the digits and then the second half is forced (aside from details like requiring a non-zero starting digit, and being free to choose the middle digit if there is an odd number, but those don't change the order of the density). That also means the sum in the video converges because it is like a sum of terms n^(-3/2).
As this is zero, and someone else compared this sum to the sum of reciprocal of squares, I wonder: what's the density of palindromic numbers with respect to squares? Namely lim |P^[1,x]| / |S^[1,x]| when x goes to infinity (and ^ denotes intersection).
There’s a super minor technicality to watch out for, which doesn’t matter a whole lot for THIS series but can matter a LOT if you generalize this result too far: you can only guarantee that you preserve the associative-ness of an arbitrary number of terms in a convergent series if that series is absolutely convergent. In simple terms: you can only place parentheses in infinitely many places in a sum and guarantee that you preserve its sum if the sums of the absolute value of each term of the sun converges. This doesn’t matter with the reciprocal of palindromes series because all the terms are positive, so Dr. Penn could group terms without fear of affecting the final value of the sum. But consider the sum 1-1+1-1+1-1… As is, it diverges, but (1-1)+(1-1)+(1-1)+… = 0+0+0+…, which obviously converges to 0. The reason I’m splitting hairs to this degree is that it’s likely to matter to students who are interested in APPLYING infinite series. You can split hairs further with Banach spaces and other things that I have no experience with or knowledge of, but THIS particular difference matters even if you stop here with series, don’t delve into the theory any further, and only ever use them for real numbers.
I don't really see a reason to go that far. As you mentioned, that is not a problem in the video since the series consists of only positive terms and thus rearrangements preserve convergence. A simple mention of this, while avoiding technicalities, would more than suffice. Banach spaces are beyond what is necessary to fully grasp nearly everything related to series, muh less this example. Bringing more complicated topics up to solve a simple problem would only get viewers more confused
In a very weak sense, we expect that the number of palindromic numbers between 100^n and 100^(n+1) will be 10x the previous value given by n. This starts with n=0 -> 18 values, n=1 -> 180... forever. Always 18×10^n. Since the denominator is of the forn 100^n, we can underestimate all the denominators as being the power of 100 not greater than them, which gives the sum of 18/10^n as the overestimate, which converges to 2
@@DanTheManTerritorial so I’m guessing that it’s not related to any “known” constants such as pi or phi. I think it’s safe to assume that this is a transcendental number
According to the OEIS (A118031), the first 1000 digits of this number are: 3.3702832594973733204921572985055311230714577794527784913350689259825197603494767589703011219269780898632350532421403168917126747933674147830127937467183247967252750087722634827550979318225637556877944404376796847493151313373647295531271637481054977957361205682332484460649774315808378576222780370382940370280769546170948671477666713704407509837747327681658124294157278347595895815147012803762301460078753895374219171398687823474425496450600364962569657596197751997355410808437695497820627954263483238465927514188721203312202864207253747796264903130226097564588742083842569397320030502643768192965449694458613706042979690968950583460074807869595979375862261937148481998657086618703808214697724773926778461748481209093452318737751966417981265923055052758889661866377594042994461127572237425467596287013687715350966157434717149952336086570739332966878807292450980178787018319902919155116505026319887004331982800424031530346620794654849578307161469569012827831247061491982966378324937801182371089307855265...
Haven't watched the video yet, but wouldn't convergence/divergence be independent of base? I feel like it should be. Taking base 2 as given, any palindrome x of length n generates 2 palindromes of length n+2. Are these all of the possible palindromes? The fraction of palindromes will then double every time the number of digits increases by two. This means that the number of palindromes grows only logarithmically with N = 2^n, while 1/palindrome grows like 1/k.
Around 3:15, you state that ∑ (1-digit)^-1 is less than 1. In fact, 1/1 + ½ + ⅓ + … + 1/9 = 7129/2520 or about 2.829. Similarly, ∑(2-digit)^-1 = (∑(1-digit)^-1)/11 = 7129/27720 or about 0.2572. I don't think that will change the conclusion.
I just wrote up a quick python script to calculate this sum truncated at palindromes with a certain numbers of digits. Including palindromes up to 8 digits its value is 3.37, and graphing the cumulative sum up to that point it definitely looks convergent
There are 9 1-digit palindromic numbers, all less or equal to 1. Then there are 9 2-digit less than 1/10 and 9x10 3-digit such numbers less than 1/100, so the two sums combine to 18x1/10. Then, there are 9x10 4-digit less than 1/1000 and 9x10x10 5-digit such numbers less than 1/10000 and again the two sums combine to 18x1/100. So, if we add another copy of the 1-digit palindromic numbers we have a geometric series starting with 18 and a common ratio of 1/10, which adds up to 20 and subtracting the same copy we get that the sum is less than 11.
This approach of course has some issues because combining the 2k with the (2k+1)-digits palindromic numbers it's like assuming that the biggest palindromic numbers have an odd number of digits....
Should, yes. There are roughly about b^n palindromes in base b which have 2n digits (and the same number which have 2n-1 digits), so, as each reciprocal of a number with 2n digits is going to be at least b^(-(2n-1)) , etc. Uh... It will work out fine. Don’t feel like doing the calculation, and I should be finishing a project due in 6 hours not being on youtube
3.3702832594973733204921572985055311230714577794527784913350689259825197603494767589703011219269780898632350532421403168917126747933674147830127937467183247967252750087722634827550979318225637556877944404376796847493151313373647295531271637481054977957361205682332484460649774315808378576222780370382940370280769546170948671477666713704407509837747327681658124294157278347595895815147012803762301460078753895374219171398687823474425496450600364962569657596197751997355410808437695497820627954263483238465927514188721203312202864207253747796264903130226097564588742083842569397320030502643768192965449694458613706042979690968950583460074807869595979375862261937148481998657086618703808214697724773926778461748481209093452318737751966417981265923055052758889661866377594042994461127572237425467596287013687715350966157434717149952336086570739332966878807292450980178787018319902919155116505026319887004331982800424031530346620794654849578307161469569012827831247061491982966378324937801182371089307855265... According to entry A118031 in the OEIS.
Before watching the video: I believe the sequence will converge as it is strictly less than the harmonic sequence when grouping terms by the order of their denominator. As to what it converges to, I believe you'd need to work out the ratio of palindromic numbers to non-palindromic numbers of a given order, and take the limit of those partial sums.
Being strictly less than the harmonic series is no guarantor of convergence. The easy example is to remove the first term of the harmonic series: if you learned it as 1/1+1/2+1/3+1/4+…, then my strictly-less-than-and-also-doesn’t-converge sequence is 1/2+1/3+1/4+… You can also take out all the even or all the odd terms to get another strictly-less-than sequence that doesn’t converge: 1/1+1/3+1/5+1/7+… or 1/2+1/4+1/6+1/8+… So I agree that it’s less than the harmonic series, but you need to know how much less. In this case, it’s enough less that it converges to something like 3.7 if I remember correctly, but it need not have just based on being less than the harmonic series.
First draw the two graphs on the same plane. Note that the top curve of the area of the intersection is y=1-x² and the bottom curve is y=x². Then find their points of intersection by setting the two curves equal. { y=x² { y=1-x² x²=1-x² 2x²=1 x²=1/2 x=±1/√2 So the area of the intersection is equal to the integral ∫ Top curve - Bottom curve where the bounds of integration are ±1/√2 (from negative to positive). ∫ [ (1-x²) - x² ] dx from x=-1/√2 to x=1/√2 ∫ (1-2x²) dx from x=-1/√2 to x=1/√2 x - 2x³/3 evaluated from x=-1/√2 to x=1/√2 x(1-2x²/3) evaluated from x=-1/√2 to x=1/√2 Note that 1-2x²/3 is easy to compute at both x=±1/√2 x²=1/2, 2x²=1, 2x²/3=1/3, 1-2x²/3=1-1/3=2/3 So the answer is x(2/3) evacuated from x=-1/√2 to x=1/√2 [ (1/√2) - (-1/√2) ] × (2/3) (1/√2 + 1/√2) × (2/3) (2/√2) × (2/3) 4/(3√2) If you don't mind about having radicals in the denominator, that will be the final answer. Otherwise, 4/(3√2) = (4√2)/(3√2×√2) = (4√2)/(3×2) = (2√2)/3 .
Indeed, that's not the correct expression there. Using his miscounted d+1 it would've been 90√10 * sum (1/√10)^d, or using the correct d-1 in the denominator (see other comments on why that's the case) would turn into the proper 9√10 * sum (1/√10)^d. :-B
I tried to just calculate the sum in Mathematica but I couldn't figure it out. It went beyond its limits of precision after 10^40 terms. It looked like it was rather consistently diverging. The value of the partial sum of a set number of terms divided by the number of terms however seemed to converge to a number between 2 and 3. If the sum truly does converge then the value of the sum is absolutely enormous.
As far as I know, they divide calculus into three courses in unis in the US. Calulus 1 covers limits, derivatives, and their applications. Calculus 2 covers integration along with sequences and series. Calculus 3 covers vector and multivariable calculus.
In the US, Calculus 1 is limits, differentiation, and usually, definition of integration and the fundamental theorem of calculus (both parts). Calculus 2 typically is single-variable integration, sequences and series, and limit tests. Calculus 3 is multivariable calculus--vectors, differentiation of vector valued functions, and integration in multiple variables including path and surface integrals culminating in Stokes' Theorem in 3 variables (some teachers also include some linear algebra).
You did the sum of the Harmonic serie 1+1/2+1/3+1/4+...with orther positive series, but the Harmonic diverges, so I don't get it, really... For me you tried to solve something like x+3+"infinity" = 0 ;
It is verboten to mention here that the Online Encyclopedia of Integer Sequence holds the decimal expansion of the actual sum ? Having been sen cored twice, trying to give the reference, I'm giving up.
@@onecommunistboi Forbidden :-) From German (also in Dutch). "forbidden, especially by an authority" in English according to Oxford Language Dictionaries (online).
Regarding the "censoring" - please note that RU-vid does not allow you to put in any links to other websites. So if your previous posts contained links, that's why they were deleted.
When he talks about the last digit, he means the last digit you have a choice of which is the digit in the middle. The last digit of the number is fixed by the first digit so there is only 1 choice for it.
Whoa whoa whoa. You're supposed to be a math teacher aren't you? The heck are you doing at 2:04?! I agree you can separate the first 1 digit palindromes, and then the 2 digit palindromes, and then some, since you're just taking out a finite number of terms each time for a finite number of times. This is because the axiom of associativity implies you can re-associate a finite number of terms however you like. However, I disagree that you can express the entire infinite sequence like this. If you attempted this with Grandi's series, which infamously diverges, you could write 1-1+1-1+...=(1-1)+(1-1)+...=0+0+...=0. Thus, the convergence of a re-associated series has no implications on the convergence of the initial series. In this case you're lucky and it does work, since the sequence is positive, so the series either converges or approaches infinity. That is, the sequence of partial sums either converges or approaches infinity. Therefore any subsequence of the partial sums either converges (to the same value) or approaches infinity. Finally, you did not reorder the terms in your series at 2:04, so the partial sums of that expression will be a subsequence of the partial sums in the initial sequence. Nonetheless, this is the second most egregious error you can make with infinite series, the first being commuting terms. This sets a terrible example for students. I really expected more out of a math teacher with the gall to post this on youtube.
The order of the terms isn't defined in the first place tough. It is just the sum over all palindromes. For that to be well defined at all it needs to be the same result no matter the order. But I agree - even if I think that you put it a bit harshly - simply omitting (I don't think that he doesn't know this) the fact that you can't do this to any series you come across is a bit dangerous.
