My favourite fact about Pascal's Triangle is that the rows give information about the structure of simplices. The row with 1, 1 represents a point. The row 1, 2, 1 represents a line segment with its 2 end points. The row 1,3,3,1 represents a triangle with its 3 angles and 3 sides. The row 1,4,6,4,1 represents the tetrahedron with its 4 vertices, 6 edges and 4 faces. The row 1,5,10,10,5,1 represents the pentachoron with its 5 vertices, 10 edges, 10 faces and 5 cells. And so on and on.
Not only simplices, but also gnomon growth of n-cubes. The row 1, 2, 1 represents an original square, 2 rectangular gnomons at its farther sides, and a new square at the tip to enlarge the original square. The row 1,3,3,1 represents an original cube, 3 rectangles at farther sides, 3 more rectangles at the edges between the 3 new "side" rectangles, and a new square at the tip. The row 1,4,6,4,1 represents an original tesseract (4-cube), 4 "rectangular” tesseracts at its farther spaces, 6 "rectangular” tesseracts at the sides between the previous 4 "rectangular" tesseracts, 4 "rectangular” tesseracts at the edges between the previous 6 "rectangular" tesseracts, and one cubic tesseract at the tip. Etc. I wish I could draw it, but I can't.
My favourite are the first three diagonals. The first diagonal is just a series of 1's. The second diagonal is the sequence sum of the first diagonal (i.e. the counting numbers starting at 1) and the third diagonal is the sequential sum of the second diagonal which contains the counting numbers. From this can be derived useful summing formulas for a series of numbers provided you know the start number and the interval up to a given sequence number which was made famous by a young Gauss.
You get seirpinski's triangle when you colour in odd numbers, but you can also get similar interesting patterns by colouring in just numbers which aren't divisible by 3, or any other integer more than 2.
Labeling the first row as row 0, and the second row as row 1... if you consider the limit of the products of the successive differences in the number of elements in each row, you get wau!
Thank you for acknowledging it to be mount meru by an indian named pingal. He used it to find all possible combinations of long and short syllable (like 0 and 1) in a poem.he was brother of great grameterian Panini . But they date back to before Christ. You mentioned leibnitz series for pi but that was given by yet another indian mathematician Madhav 300 years prior to Lebnitz. He even exprezsed trignometric functions in series.
You can verfy if a number n is prime by looking in the n-th row and checking if every number (beside the 1´s) in that line is 0 modulo n. (the prove uses the binominal representation)
I think you can build a proof on the observation that you can generate row "n" of Pascal's Triangle as follows: Start with "1", then multiply it by n/1, then multiply that by (n-1)/2, then multiply that by (n-2)/3 . . . finally, multiply by 1/n to get the final "1." [Why? Can you prove that?] In all those steps, look at all the denominators. Before the final (1/n) multiplication, there will be a denominator that divides n, if and only if n is composite. And so, if n is prime, all the terms between the 'bookends' (the 1's) are divisible by n; if n is composite, then at least one of those terms won't be. This isn't an airtight argument, because sometimes a denominator divides its numerator, and so, doesn't get removed as a factor of the term being multiplied by the fraction. But a little fiddling with it, should reveal how to make it airtight. Fred
Holyshit that's tough to read here is an easier way. you notice that (p choose k)=p!/(p-k)!(k)!. Because p is prime, it doesn't divide any product of number all smaller than p (except 1 but it's not relevant). so (p choose k) = p * (p-1)!/(p-k)!(k)! = p*c. If it isn't prime (n choose k)=n!/(n-k)!(k)! will divide at least one number smaller than n and at least one of the (n choose k) will not be divisible by n. Because (n choose k)=(n choose n-k) you re gonna get at least 2.
If you alternate the signs of each term in each row and add the rows up, all will be zero except the first, leaving the value of the triangle at one. This means the sum of all non-convergent diagonals of the triangle (all of them) is also one.
For row n, the first number is 1, and every other number in the Pascal triangle can be obtained by multiplying the previous number by n-k+1, while dividing it by k. For example, take the sevenths row (n=7). The first number (k=0) is 1. Multiply 1 by 7-1+1=7 and divide it by 1 to get 7. Multiply 7 by 7-2+1=6 and divide it by 2 to get 21. Multiply 21 by 7-3+1=5 and divide by 3 to get 35. Multiply 35 by 7-4+1=4 and divide by 4 to get 35 (again). Multiply 35 by 7-5+1=3 and divide by 5 to get 21. Multiply 21 by 7-6+1=2 and divide by 6 to get 7. Multiply 7 by 7-7+1=1 and divide by 7 to get 1. Thus you get the 7th row of the triangle: 1 7 21 35 35 21 7 1 Also, you can get from the Pascal triangle to the Stirling numbers (actually, Stirling triangle) and to Catalan numbers.
Great video! Thanks for including e. For anyone interested, here are the papers that announced the discovery: H. J. Brothers, Finding e in Pascal’s triangle, Mathematics Magazine, Vol. 85, No. 1, 2012; page 51. H. J. Brothers, Pascal's triangle: The hidden stor-e, The Mathematical Gazette, Vol. 96, No. 535, 2012; pages 145-148.
As you showed, the sum of row n is 2ⁿ. If you normalize row n, C(n,k), by dividing by its sum, • then plot that, C(n,k)/2ⁿ, for constant n, as y-values against x = k-½n (which centers each row on its peak), • scaling the axes by shrinking x and expanding y by the factor √n, • the result approaches a perfect Gaussian curve ("Bell curve") as n→∞. Fred
Pasccal's triangle is of course very common in combinatorics. The triangular numbers, tetrahedral numbers, pentalope numbers and so forth can be read in the triangel. Those numbers are the coefficients of polynomials of different kinds. They can be used to solve certain kinds of probability problems.
Excellent video! I love that so, so many mathematical identities, sequences, and special sums are hidden in Pascal's Triangle. This video introduced me to a couple I wasn't previously aware of. I also appreciate that explanation was given to each observation, rather than just stating each as a "mysterious occurrence". One identity that wasn't mentioned here is that, if you left-justify the triangle, the first column is a sequence of 1's, the second column is the sequence of natural numbers, the third column is the sequence of triangular numbers, and the fourth column gives the sequence of tetrahedral numbers. This makes since in terms of binomial coefficients, since, for instance, the nth triangular number has the formula: (n(n+1))/2, which is the same as "n choose 2". I can't remember what type of number that the sequence of the fifth column gives. By the way, was anyone else slightly disturbed by the omission of a space on the bottom row of the slide at 3:01? It threw all the alignments off and frustrated my brain, haha!
Yes, when left-justified, the first column [column 0, because it's C(n,0) ], is all 1's; column 1 is the natural numbers; column 2 is the triangular numbers; column 3 is the tetrahedral numbers; column 4 is the pentatopal numbers; : column k is the k-dimensional simplex numbers. Each of those figures is a simplex; the k-dimensional analog of a triangle/pyramid. Each is the sum from 1 to k, of the numbers for 1 dimension lower. Fred
Hmm, simplex numbers, huh? I've never heard of those! I'll have to look into those more. Thanks for leaving a response! I did notice that each column contains basically partial sums of the column to the left of it, but I wasn't sure why. Now I know!
Actually, the partial sum property is just from the rule for generating Pascal's Triangle, along with the way we've rearranged it by left-justification. Also note that: n-trian = n(n+1)/2 = (n+1)-choose-2 n-tetrad = n(n+1)(n+2)/6 = (n+2)-choose-3 . . . etc. . . Fred
The i-th sum across all diagonal lines of constant slope (p,q) equals a function expressible as a constant-recursive sequence of order n=p+q which can be explicitly written as the n-th component of i-th power of an n*n matrix multiplied by a base vector of size n or equally as the solution to an ordinary linear differential equation of order n. Getting the specific matrix for the sum of a given diagonal slope is a little tricky but possible. Once you've figured out the specific matrix for the sum of a given diagonal slope (p,q), it's easy to compute the i-th power by converting it into an eigenvector problem by diagonalizing it using its singular value decomposition.
5:58 1. The first row containing a number other than 1, ("starting" with a 2,) going diagonally down in either direction is just adding 1 every time because of the 1s bordering the triangle. 2. Similarly, triangular numbers, numbers made by the sum of all natural numbers up to a specific one, can also be found in the third row going diagonally down from either 1 on that row (numbers like 1, 3, 6, 10, 15, 21, 28, etc.). 3. The eighth row, the one "starting" with a 7, only contains multiples of 7. This seems to be true for any row that starts with a prime number; all numbers in the row are divisible by the first non-1 number in the row. 4. Besides the first row, it seems the middle number in all rows "starting" with a square number is divisible by the square root of that number; in rows with no middle number, the two numbers next to each other with the same value are the middle number. I'll be honest, I can't 100% confirm facts 3 and 4, but they're still interesting in the first few rows of the triangle and apply in broader ways the less rows you consider. I still think all of these facts and the ones mentioned in the video are very interesting, whether true throughout the entire triangle or just the parts that the average mathematician sees in their lifetime.
if you go from the top of the triangle down a side, and add all the values you encounter on that side, the sum will equal the index(row number) of the row you stoped in. mind blowing!
Here's one. A point is 1. A line segment is 1 1. A square is 1 2 1 where 2 is the diagonal. In general, the pattern is you pick a point to start from and keep finding the set of points adjacent to the current set. What's awesome about this is you can immediately see how the powers of 2 make sense geometrically, and the set of points adjacent to any point on an n-dimensional cube is an n-1 dimensional simplex, explaining why the second diagonal of Pascal's triangle is called the simplex numbers. If you want to go deeper, ponder whether each number in the nth row of pascal's corresponds to a regular polytope.
If you use the Gamma-Function, then C(1, 0.5) = 4/pi and C(0, x) = sin(x pi)/(x pi) = sinc(x) the normalized sinus cardinalis. If P_n is the product of a row, then the limit n->∞ of P_n-1 x P_n+1 / P_n^2 = e.
Take every number in the triangle mod 2 and read each row as a number in binary (excluding row 0). Each line will either be a prime or a product of primes that have appeared as previous rows
the row where you take the reciprocals and do an alternating sum to give pi-2 is the triangular numbers, which are: 1 1+2=3 1+2+3=6 1+2+3+4=10 1+2+3+4+5=15 ... 1+2+3+...+n=1/2 n(n+1)
I don't know if I'm the first, but I've never heard that anyone else has noticed that Pascal's triangle has two lean-y sides that come together in a point at the top. Sorta like a triangle.
skoockum your not but good observation keep up the good work we're all learning something so cool and awesome like Fibonacci along with many others that have, and way past all of us so speaking of first you have to add the ppl back around 1553 I mean wat about 20 yrs before Internet let's work together not beat ppl to unending knowledge available for everyone
You forgot the Hockey Stick pattern. If you add up the numbers along a diagonal and then turn to go the other way diagonally, that number will equal the sum of all of the numbers on the previous diagonal.
Using (n+1) choose k where n equals a dimension, you can get the number of vertices, edges, and faces on a triangle in that dimension. (ex: 4th dimension. (4+1) choose 2 = 10. A 4 dimensional triangle has 10 edges) 5 choose (1 to 5) = 5 10 10 5 1 A 4th dimensional triangle: 5 - vertices 10 - edges 10 - faces (triangles) 5 - tetrahedra (3 dimensional triangle) 1 - 4-simplex (4 dimensional triangle) As far as I can tell it works starting with row 3 (3 3 1) showing a triangle (3 vertices, 3 edges, 1 face) and onward. n+1 choose 1 = vertices n+1 choose 2 = edges n+1 choose 3 = faces and so forth. It also works with squares. Instead of taking up more space to explain, I'll just leave the formula... (n choose k)*(2^(n-k))
Since the Fibonacci numbers show up and the limit of the difference between two consecutive terms tends to Phi. Therefore there is a connection to 0,1,pi,e and Phi.
In my arrogance and supposed knowledge of all things related to Pascal's Triangle (including it's Chinese origins), I said: "Right, show me what I don't know!" and duly proceeded to click on the 'play' button. I am both abashed and glad that you did indeed do so, for I was unaware of: the hexagonal properties within the triangle and the relationship of the triangle to both π and e. Now, to puff my chest out again (only for it to be once again deflated, as you'll soon learn), quite some years ago I was playing with geometric formal power series and their closed forms. On concentrating on the closed forms themselves and beginning with the function f(x) = 1/(1-x) (assuming convergence), I started a sequence of differentiating each function and multiplying the result by x. Thus if x*f_n (x) is x multiplied by the derivative of f_(n-1) (x), we get f_0 (x) = 1/(1 - x), then x*f_1 (x) = x/(1 - x)^2 x*f_2 (x) = (x + x^2) / (1 - x)^3 x*f_3 (x) = (x + 4x^2 + x) / (1 - x)^4 and so on. Apart from noticing that the denominator of f_n (x) is (1 - x)^(n+1), what interested me was that in the numerator of f_n (x) there were n terms of powers of x which summed to (n!). It was a short step to devising a "triangle" of which the first few terms were: 1) 1 2) 1 1 3) 1 4 1 4) 1 11 11 1 5) 1 26 66 26 1 6) 1 57 302 302 57 1 7) 1 120 1191 2416 1191 120 1 etc.... Having also derived a recurrence relation in the sequence, namely D(n, k) = (n-k+1)*D(n-1, k-1) + k*D(n-1, k) (where n is the row, k is the column and D is the function of these two variables), I was as pleased as punch (if not as good looking) that I may join the mathematical pantheon in having a sequence of numbers named after me. Alas, it was all to no avail, as I subsequently discovered some months later that some "obscure" 18th century mathematician by the name of Leonard Euler (you may not have heard of him:)), had already discovered this same sequence. Yes, they are called Euler numbers after the most prolific mathematician of all time who just could not leave one mere mortal one scrap of mathematical kudos! No, I'm not bitter (He said through grinding teeth). But one thing I do know that Herr Bloody Euler didn't: the second column "miraculously" conveys the number of bets involved in an n-horse 'yankee' which contains bets involving doubles, trebles, fourfolds, etc. So there! All being said, I am now both duly humbled and gladly educated by your most informative video. Thank you!
Tom Purcell you gets no points your nearly as smart as you boast and brag about your so smart where's your school where's your smartness towards something not already so easy to Google and Access come up with your own math problem that ppl havnt been doing way longer than you been able to drip snot from your nose
The rows of Pascal triangle look like nodes of the graphs of some kind of distributions. What kind of distributions are these? Gaussian? Cauchy? Student's t? Logistic? Any other? Why? What is the mechanism underneath? Are there any articles or books about it?
Pascals triangle shows up in the infinite nomial in its totality in the diagonal arrangement. Example Tetranomial 1 1 1 1 1 1 2 3 4 3 2 1 1 3 6 10 12 12 10 6 3 1 1 4 10 20 31 40 44 40 31 20 10 4 1 Do you see the 1,11,121? It goes on as you expand it. Tetranacci Sequence 1,1,2,4,8 by addition? Via prime factorization the tetranomial can be expressed as 11^n*101^n much as Pascals triangle can be expressed as 11^n. If you arrange the tetranomial in the classic mountain configuration and add the diagonal you get the tetranacci sequence. Once you realize that all n-nomials express n-nacci sequences and are related to palidromes you can deduce the prime factorizations for pentanomials, hexanomials, heptanomials, octanomials etc...
I have just discovered One more pattern.if you Represent the Numbers as Points,and Trace each line of How they Expanded,you will notice that It creates a Dimensional Pattern coincidentaly on The nth row,equal to what is the Reference of the Dimension.on Pascal's Two term(Binomial) Expansion(which is the Regular Pascal's triangle),a Square(2d) is Formed in 2nd row.
There's a cute approximation for that 'central spine' of the Triangle, C(2n,n) = (2n)! / [(n)!]² The simple form of the approximation is C(2n,n) ≈ 4ⁿ/√(πn) This form excludes n=0, because the approximation blows up there. A much better approximation is C(2n,n) ≈ 4ⁿ/ [√(πn)][1 + 1/(8n-½)] Fred
NormanicusDiabolicus I can disprove it. Claim: "People who died after birth are more likely to comment on this." (noone said "right after" birth.) Proof 1: (Disjunctive Syllogism) Suppose that: X ≝ ("left handed people are more likely to comment on this then people who died after birth.") Y ≝ ("People who died after birth are more likely to comment on this.") Everyone dies after birth, X∪Y, left handed or right handed people alike, (since you must be born to die right or left-handed.) If Y includes left-handed along with right-handed people, then left handed people are a proper subset of {lefties∪righties∪both} and {lefties∪righties∪both} ⊆ {people dead after birth} Justification for this is Modus Ponens: If "Y={lefties∪righties∪both}" then "{lefties} are a proper subset of {lefties∪righties∪both}." "Y={lefties∪righties∪both}" is affirmed therefore "{lefties}⊂{{lefties}∪{righties}∪both}, and {lefties∪righties∪both} ⊆{people dead after birth}. So, we have a Disjunctive Syllogism : (X∨Y), ~X (since X⊆Y) .·. Y▪ Proof 2 : Same Reasoning, using a Corresponding Conditional&Truth Table: XY: (( X∨Y )∧ (~X)) ⇒ Y TT T F F True T TF T F F True F FT T T T True T FF F F T True F so we have a Tautology: {T, F} = True, False respectively, and every row is Yrue under the main connective. This indicates that Y is True and not X, therefore Y is actually the case such that it is true that "people who died after birth are more likely to comment on this."▪
Pixel Bytes One person said "died after birth," which I disproved below, and you say "died AT birth" which I will prove here as requested by Normanicus. (Very different results) Proof by Disjunctive Syllogism: X =(Left handed people are more likely to comment then people who died at birth.) Y =(People who died AT birth, i.e. "dead babies" are more likely to comment!!) Disjunctive Syllogism: (X∨Y) ~Y (since Y is a dead baby) .·. X▪ Truth Table : XY: ((X∨Y)∧ (~Y)) ⇒ X TT. T F F True T TF. T T Τ True T FT. T F F True F FF. F F T True F is a Tautology, so "Left handed people are more likely to comment than" (dead babies,) or "people who died AT birth."▪ ~ most morbid proof ever ~
ONE thing is better than this triangle: catalan numbers. whatever the name your contry has.. they are everywhere! n-minos, n-nodes-trees, ways on a grid from one node to another, inverse coefficient of tan(x) polynomial development... and i'm forgetting the majority of the properties :D!
Here is a fun one brought to you by the number 9 and the Langlands project. You may not have seen this before. Les choses que tu n'as jamais vues. 121: 21-12=9 1331: 331-133=198=>1+9+8=18+>1+8=9 14641: 641-146=495+>4+9+5=18=>1+8=9 15101051: 1051-1510=-459. -4-5-9=> -18=-1-8= -9; alternately 11^5=161051=>1051-1610=-559=>-5-5-9= -18=>-1-8=-9 11 requires us to go outside the triangle a little: 11^1=11. 010 0110=>110-011=99=> 9+9=18=>1+8=9. Ce ne sont que des harmoniques.
So what should we actually name the triangle? Any suggestions are welcome. Basically everyone discovered this triangle independently and named it after themselves. That's unsettling. It looks like the kind of thing you would discover if you were bored in math class in elementary school... But then, calling it Pascal's Triangle is really unintuitive, and doesn't really make sense at all!
Well given that it doesn't really matter that much.. if someone ever does try to take it upon themselves to set a culture-neutral name, it would likely end up being something bland like "Binomial Triangle," which nobody would use because they're all used to the existing name(s.)
altrag I disagree. I'm familiar with it being Pascal's triangle, but presented with a better name, I am willing to reconsider. Knowing what I now know, I also wouldn't teach Pascal's triangle by its current name in a classroom, but only mention it contextually. It's an unnecessary ambiguity to students already struggling with learning disabilities.
+Unknown Entity > I am willing to reconsider. Most people would be "willing" to reconsider, but the problem is that the existing name is already well embedded in minds of established mathematicians and even with conscious effort, that's hard to change. A bigger issue still though is that its the name printed in millions upon millions of textbooks that would have to be scrapped and reprinted. Never mind if you wanted to rename everything in math and science that's been mis-attributed over the centuries. Babylonian Theorem anyone? > I also wouldn't teach Pascal's triangle by its current name in a classroom Why not? You have to call it something and the kids you're teaching wouldn't know the difference. And hopefully by the time they're old enough to understand (or even hear about) the historical and cultural implications of the name, they'd also be old enough to understand that things can have more than one name. I mean they know actors usually don't have the same name in different movies and they somehow get by. > to students already struggling with learning disabilities. Unless "extreme pedantry" has been classified as a learning disability, I'm not sure that this is relevant one way or the other. If a student is capable of learning about the triangle, then they'll learn it and if they aren't capable of learning it then they won't, no matter the name you give it.
altrag I understand you are opposed to the idea. I only minor in maths, so I'm not awfully familiar with expert opinions. But in my humble opinion most of those arguments are weak. Given that Pascal's triangle is in fact not called Pascal's triangle in most of the non-English speaking world (India and China are really fucking huge), what those mathematicians are accustomed to is actually already an exception to what they were taught. Besides, most textbooks are already bad, and anything until university level should be updated regularly anyway, as people change. It makes me think of an episode of Numberphile about τ (tau), where it was proposed that τ not replace π in the education system, but is used as a supplement in the introduction to irrational numbers, and then trigonometry, as it is initially more intuitive in high school maths. You don't sound like you're willing to try and understand how learning disabilities work. But in the off chance I'm wrong, I used to struggle with maths in school. I remember being thrown off at minor inconsistencies. In particular, Pascal's triangle (as it inconveniently is being called in Norway where I went to school) did not indicate to have any relation to either geometry or algebra (let alone probability as it was presented). For instance, I would overestimate the importance of chapters about concepts named after people, and miss the point of the learning objectives. After I started teaching, I learned that this is not completely unheard of. Some students, especially the Asperger's types are unable to organise new concepts that are not plainly algebraic and have descriptive names. Pascal's triangle is neither. Unfortunately, it took me until university to learn how to understand some things that I should have known before I started high school. The reason was that I had thought about it the wrong way, which was caused by confusion at the time of learning.
+Unknown Entity > I understand you are opposed to the idea. I'm not _opposed_ to the idea. I just don't see it being useful enough to warrant the effort needed to implement it. Certainly calling it the "Binomial Triangle" would give you an automatic tie-in to the binomial theorem, but unless you wanted a 400 word name, you're not going to be able to cover everything hidden in the triangle. At best you might get a vague reference to one or two ideas before the name becomes unwieldy, no matter what you choose. > not called Pascal's triangle in most of the non-English speaking world Sure, Indians and Chinese (and Italians and whoever else) will use their local names for the triangle but as long as they know what they're talking about, the name doesn't really matter to them either. The only time it would be an issue is if an English speaker tried to take a math course in India or something, and even then its all of a Google search away (or even less since you rarely find reference to the triangle's name without a corresponding diagram of some sort, and its a pretty recognizable structure.) > Numberphile about τ (tau) ... as it is initially more intuitive in high school maths. Not really. Tau gives you circumference without the factor of 2, and it makes it more obvious how to divide up the unit circle (1/4 tau is 1/4 of the circle!) But then when you look at area or arc length, you have to introduce a factor of 1/2. Tau and pi end up being fairly evenly distributed as the "easier" number in most of math and physics.. and factors of 2 are usually easier to work with than factors of 1/2, so pi kind of wins out there in my opinion. > did not indicate to have any relation to either geometry or algebra That comes back to the 400 word name. Pascal's triangle (or whatever you want to call it) and the Fibonacci sequence (also mis-named by the way) and several other fundamental numeric structures have applications in, or connections to, all sorts of weird places. There's no way you could name them in such a way that they make sense in all of those contexts. You have to pick something. As another example, does the phrase "prime number" sound terribly descriptive to you? > I would overestimate the importance of chapters about concepts named after people, and miss the point of the learning objectives. I don't know what to tell you there. Can you come up with a better name for Pascal's triangle that you believe would have made it easier for you? Did you have similar trouble with Pythagoras' theorem? Or the Fibonacci sequence? Or even the really "common" names like sine, cosine and tangent? I'm not trying to be glib here I'm really wondering if you have an answer to any of that. Its one thing to say "that didn't work for me" and quite another to say "but here's something that does."
It seems to be almost, but not quite, identical to Times, Times New Roman, Baskerville, or Big Caslon. None of my other fonts match any closer than those, although a few of them are sorta close.
Hmm if you use add and subtract pattern 1 1+1 1+2-1 1+3-3+1 1+4-6+4-1 youll likely get this 1 2 2 2 2 ... But when you use subtract and add 1 1-1 1-2+1 1-3+3-1 1-4+6-4+1 Youll get 1 0 0 0 0 0 ... Idk how.
I’m a bit late but yes, the modulus of a power set is 2^n because it is equivalent to the sum of every binomial co-efficient of a certain n (equivalent to the sum of every row is 2^n)
since the one at the top is the 0th row, this is just proof that (0 choose 0) = 1, i.e. there is only 1 way to arrange no objects: sucks to be you (the official maths lingo)
I think you've got an extra positive term in the middle but barring that.. why does that fact strike me as something that would relate to Riemann's Zeta function? Where on one side of zero everything is nice and orderly and then on the other side its.. well still orderly but a completely different (and messier-looking) kind of order. Someone up the comments mentioned a relation between Pascal's triangle and the primes, so I definitely wouldn't be surprised to see Riemann embedded in there somewhere as well. Edit: Nope no extra positive term. Just so many 1/2's in a row messing with my eyeballs.
Well what if we use the closed formula, C(n,m) = n!/[m!(n-m)!], and set n = -1? What we get is a whole string of "±∞/∞", because factorial of a negative integer is ±∞. But if we make some assumptions to try to tame the infinities, like using the factorial recursion, n! = n(n-1)!, then we get: ... -1 1 -1 1 1 -1 1 -1 ... which is just twice what you got. But this breaks the rule for summing adjacent entries to get the entry centered below them; 1+1 = 2 ≠ 1. OK, what if we treat each of these as a limit, where C(n,m) is replaced by the limit as ε→0 of C(n+ε, m+ε)? [Note that this calls on the "Pi" function, ∏(x) = Γ(x+1), which equals x! when x is an integer ≥ 0.] I haven't carried this out, but I think it can be done using l'Hôpital's Rule and the digamma function; which looks kinda hairy. But I suspect it will give your answer. For one thing, recall that the sum of row n is 2ⁿ, so the sum of row (-1) should be ½. Sure enough, although your sum doesn't converge in the usual sense, there is a looser sense in which it sums to ½, because of that 'extra' +½
...'yeah' but I'm thinking also of what physicists 'do' with infinite-cycle integrations, applying a centered weighting function and slowly lifting it (non-uniform pointwise convergence is 'unmathematical-scary-stuff' capable of spurious results)...
1:55 "If you find row n in Pascal's triangle, the second number corresponds to the number of points on the circle..." n=5 *But proceeds to show 6th row of PT* 💀
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My favourite fact about Pascal's Triangle is that the rows give information about the structure of simplices. The row with 1, 1 represents a point. The row 1, 2, 1 represents a line segment with its 2 end points. The row 1,3,3,1 represents a triangle with its 3 angles and 3 sides. The row 1,4,6,4,1 represents the tetrahedron with its 4 vertices, 6 edges and 4 faces. The row 1,5,10,10,5,1 represents the pentachoron with its 5 vertices, 10 edges, 10 faces and 5 cells. And so on and on.
Not only simplices, but also gnomon growth of n-cubes.
The row 1, 2, 1 represents an original square, 2 rectangular gnomons at its farther sides, and a new square at the tip to enlarge the original square.
The row 1,3,3,1 represents an original cube, 3 rectangles at farther sides, 3 more rectangles at the edges between the 3 new "side" rectangles, and a new square at the tip.
The row 1,4,6,4,1 represents an original tesseract (4-cube), 4 "rectangular” tesseracts at its farther spaces, 6 "rectangular” tesseracts at the sides between the previous 4 "rectangular" tesseracts, 4 "rectangular” tesseracts at the edges between the previous 6 "rectangular" tesseracts, and one cubic tesseract at the tip. Etc.
I wish I could draw it, but I can't.
My favourite are the first three diagonals. The first diagonal is just a series of 1's. The second diagonal is the sequence sum of the first diagonal (i.e. the counting numbers starting at 1) and the third diagonal is the sequential sum of the second diagonal which contains the counting numbers. From this can be derived useful summing formulas for a series of numbers provided you know the start number and the interval up to a given sequence number which was made famous by a young Gauss.
Already mentioned at 1:54
same
Whoever sponsored this is brilliant.
Whoa, brilliant pun.
@@abhilashsaha6098 Woah THAT was a brilliant pun
@@carsonwengert2854 No, THAT was a brilliant pun!
@@anawesomepet Are you sure about that? I think THAT was a brilliant pun
You get seirpinski's triangle when you colour in odd numbers, but you can also get similar interesting patterns by colouring in just numbers which aren't divisible by 3, or any other integer more than 2.
If you color in just the numbers that aren't divisible by 1, you get Pascal's Triangle! Amazing
Gold161803
LOL
If you're coloring them based on divisibility by n > 2, the interesting thing to do would be to have n colors, one for each possible residue mod n.
I think it's only for the primes
If you add ALL the numbers outside the triangle, you get 0, so cool!
Alternatively, e^πi + 1.
LOL
Truly remarkable!
In only if zero counts for anything except a power of a multiple of ten.
Labeling the first row as row 0, and the second row as row 1... if you consider the limit of the products of the
successive differences in the number of elements in each row, you get wau!
The Fibonacci sequence being there is the most amazing thing to me. The key to all life!
You SERIOUSLY deserve higher number of views!
Thank you very much to the creator of this video! It is really very beautiful Mathematics going on everywhere!
Thank you for acknowledging it to be mount meru by an indian named pingal. He used it to find all possible combinations of long and short syllable (like 0 and 1) in a poem.he was brother of great grameterian Panini . But they date back to before Christ. You mentioned leibnitz series for pi but that was given by yet another indian mathematician Madhav 300 years prior to Lebnitz. He even exprezsed trignometric functions in series.
You can verfy if a number n is prime by looking in the n-th row and checking if every number (beside the 1´s) in that line is 0 modulo n. (the prove uses the binominal representation)
Do you have a link to the proof? I'm interested.
I think you can build a proof on the observation that you can generate row "n" of Pascal's Triangle as follows:
Start with "1", then multiply it by n/1, then multiply that by (n-1)/2, then multiply that by (n-2)/3 . . . finally, multiply by 1/n to get the final "1." [Why? Can you prove that?]
In all those steps, look at all the denominators. Before the final (1/n) multiplication, there will be a denominator that divides n, if and only if n is composite.
And so, if n is prime, all the terms between the 'bookends' (the 1's) are divisible by n; if n is composite, then at least one of those terms won't be.
This isn't an airtight argument, because sometimes a denominator divides its numerator, and so, doesn't get removed as a factor of the term being multiplied by the fraction.
But a little fiddling with it, should reveal how to make it airtight.
Fred
Thanks! :)
Holyshit that's tough to read here is an easier way.
you notice that (p choose k)=p!/(p-k)!(k)!. Because p is prime, it doesn't divide any product of number all smaller than p (except 1 but it's not relevant). so (p choose k) = p * (p-1)!/(p-k)!(k)! = p*c.
If it isn't prime (n choose k)=n!/(n-k)!(k)! will divide at least one number smaller than n and at least one of the (n choose k) will not be divisible by n. Because (n choose k)=(n choose n-k) you re gonna get at least 2.
3zehnutters whao that is an interesting one... does it have a name?
the points on the circle i have never seen before, there's so many interesting facts about his triangle. great work!!
If you alternate the signs of each term in each row and add the rows up, all will be zero except the first, leaving the value of the triangle at one. This means the sum of all non-convergent diagonals of the triangle (all of them) is also one.
For row n, the first number is 1, and every other number in the Pascal triangle can be obtained by multiplying the previous number by n-k+1, while dividing it by k.
For example, take the sevenths row (n=7). The first number (k=0) is 1. Multiply 1 by 7-1+1=7 and divide it by 1 to get 7. Multiply 7 by 7-2+1=6 and divide it by 2 to get 21. Multiply 21 by 7-3+1=5 and divide by 3 to get 35. Multiply 35 by 7-4+1=4 and divide by 4 to get 35 (again). Multiply 35 by 7-5+1=3 and divide by 5 to get 21. Multiply 21 by 7-6+1=2 and divide by 6 to get 7. Multiply 7 by 7-7+1=1 and divide by 7 to get 1.
Thus you get the 7th row of the triangle: 1 7 21 35 35 21 7 1
Also, you can get from the Pascal triangle to the Stirling numbers (actually, Stirling triangle) and to Catalan numbers.
This comes from the fact that
nCr/nC(r-1) =( n-r+1) /r
Great video! Thanks for including e. For anyone interested, here are the papers that announced the discovery:
H. J. Brothers, Finding e in Pascal’s triangle, Mathematics Magazine, Vol. 85, No. 1, 2012; page 51.
H. J. Brothers, Pascal's triangle: The hidden stor-e, The Mathematical Gazette, Vol. 96, No. 535, 2012; pages 145-148.
Taking the nth derivative of the product of two functions, the coefficients also match the nth row of the triangle
Also each number represents number of ways to get to it by descenting from the first "1"
As you showed, the sum of row n is 2ⁿ. If you normalize row n, C(n,k), by dividing by its sum,
• then plot that, C(n,k)/2ⁿ, for constant n, as y-values against x = k-½n (which centers each row on its peak),
• scaling the axes by shrinking x and expanding y by the factor √n,
• the result approaches a perfect Gaussian curve ("Bell curve") as n→∞.
Fred
Holy shi
This follows from the fact that the binomial distribution approaches the normal distribution as n increases.
Um...for p = 1-p = ½, yes.
For other p values? Yes? No? I have't the time to work this right now.
Pasccal's triangle is of course very common in combinatorics. The triangular numbers, tetrahedral numbers, pentalope numbers and so forth can be read in the triangel. Those numbers are the coefficients of polynomials of different kinds. They can be used to solve certain kinds of probability problems.
Excellent video! I love that so, so many mathematical identities, sequences, and special sums are hidden in Pascal's Triangle. This video introduced me to a couple I wasn't previously aware of. I also appreciate that explanation was given to each observation, rather than just stating each as a "mysterious occurrence".
One identity that wasn't mentioned here is that, if you left-justify the triangle, the first column is a sequence of 1's, the second column is the sequence of natural numbers, the third column is the sequence of triangular numbers, and the fourth column gives the sequence of tetrahedral numbers. This makes since in terms of binomial coefficients, since, for instance, the nth triangular number has the formula:
(n(n+1))/2,
which is the same as "n choose 2". I can't remember what type of number that the sequence of the fifth column gives.
By the way, was anyone else slightly disturbed by the omission of a space on the bottom row of the slide at 3:01? It threw all the alignments off and frustrated my brain, haha!
Yes, when left-justified, the first column [column 0, because it's C(n,0) ], is all 1's;
column 1 is the natural numbers;
column 2 is the triangular numbers;
column 3 is the tetrahedral numbers;
column 4 is the pentatopal numbers;
:
column k is the k-dimensional simplex numbers.
Each of those figures is a simplex; the k-dimensional analog of a triangle/pyramid.
Each is the sum from 1 to k, of the numbers for 1 dimension lower.
Fred
Hmm, simplex numbers, huh? I've never heard of those! I'll have to look into those more. Thanks for leaving a response! I did notice that each column contains basically partial sums of the column to the left of it, but I wasn't sure why. Now I know!
Actually, the partial sum property is just from the rule for generating Pascal's Triangle, along with the way we've rearranged it by left-justification.
Also note that:
n-trian = n(n+1)/2 = (n+1)-choose-2
n-tetrad = n(n+1)(n+2)/6 = (n+2)-choose-3
. . . etc. . .
Fred
A n-simplex is a n-dimensional triangle
Expressing sine and cosine muliples as powers uses every second term in the multiple's row of the triangle
This is a great channel!!
Pascal Triangle is a very useful pattern of numbers since this can be use a lot of things and calculations about math, I love it
The i-th sum across all diagonal lines of constant slope (p,q) equals a function expressible as a constant-recursive sequence of order n=p+q which can be explicitly written as the n-th component of i-th power of an n*n matrix multiplied by a base vector of size n or equally as the solution to an ordinary linear differential equation of order n.
Getting the specific matrix for the sum of a given diagonal slope is a little tricky but possible.
Once you've figured out the specific matrix for the sum of a given diagonal slope (p,q), it's easy to compute the i-th power by converting it into an eigenvector problem by diagonalizing it using its singular value decomposition.
5:58
1. The first row containing a number other than 1, ("starting" with a 2,) going diagonally down in either direction is just adding 1 every time because of the 1s bordering the triangle.
2. Similarly, triangular numbers, numbers made by the sum of all natural numbers up to a specific one, can also be found in the third row going diagonally down from either 1 on that row (numbers like 1, 3, 6, 10, 15, 21, 28, etc.).
3. The eighth row, the one "starting" with a 7, only contains multiples of 7. This seems to be true for any row that starts with a prime number; all numbers in the row are divisible by the first non-1 number in the row.
4. Besides the first row, it seems the middle number in all rows "starting" with a square number is divisible by the square root of that number; in rows with no middle number, the two numbers next to each other with the same value are the middle number.
I'll be honest, I can't 100% confirm facts 3 and 4, but they're still interesting in the first few rows of the triangle and apply in broader ways the less rows you consider. I still think all of these facts and the ones mentioned in the video are very interesting, whether true throughout the entire triangle or just the parts that the average mathematician sees in their lifetime.
if you go from the top of the triangle down a side, and add all the values you encounter on that side, the sum will equal the index(row number) of the row you stoped in. mind blowing!
Here's one. A point is 1. A line segment is 1 1. A square is 1 2 1 where 2 is the diagonal. In general, the pattern is you pick a point to start from and keep finding the set of points adjacent to the current set. What's awesome about this is you can immediately see how the powers of 2 make sense geometrically, and the set of points adjacent to any point on an n-dimensional cube is an n-1 dimensional simplex, explaining why the second diagonal of Pascal's triangle is called the simplex numbers. If you want to go deeper, ponder whether each number in the nth row of pascal's corresponds to a regular polytope.
If you use the Gamma-Function, then C(1, 0.5) = 4/pi and C(0, x) = sin(x pi)/(x pi) = sinc(x) the normalized sinus cardinalis. If P_n is the product of a row, then the limit n->∞ of P_n-1 x P_n+1 / P_n^2 = e.
Take every number in the triangle mod 2 and read each row as a number in binary (excluding row 0). Each line will either be a prime or a product of primes that have appeared as previous rows
That's true for any number. It's either a prime, or a product of primes. What am I missing? The bit about "previously appeared" - what does that mean?
the row where you take the reciprocals and do an alternating sum to give pi-2 is the triangular numbers, which are:
1
1+2=3
1+2+3=6
1+2+3+4=10
1+2+3+4+5=15
...
1+2+3+...+n=1/2 n(n+1)
I don't know if I'm the first, but I've never heard that anyone else has noticed that Pascal's triangle has two lean-y sides that come together in a point at the top. Sorta like a triangle.
skoockum Are you serious? That's why they call it Pascal's TRIANGLE, my dude.
Am I serious? I'll have to check.
skoockum your not but good observation keep up the good work we're all learning something so cool and awesome like Fibonacci along with many others that have, and way past all of us so speaking of first you have to add the ppl back around 1553 I mean wat about 20 yrs before Internet let's work together not beat ppl to unending knowledge available for everyone
You forgot the Hockey Stick pattern. If you add up the numbers along a diagonal and then turn to go the other way diagonally, that number will equal the sum of all of the numbers on the previous diagonal.
U also have the hockey stick sum in Pascal's triangle..
I know one fact more. "Central" numbers in the triangle like (2n, n) minus neighbor number (2n, n-1) or (2n, n+1) will be Catalan numbers.
This is brilliant.
I love this channel!!!
Taking any number on the second 'column' and squaring it, gives the same as the sum of the right and down-right entry
Using (n+1) choose k where n equals a dimension, you can get the number of vertices, edges, and faces on a triangle in that dimension.
(ex: 4th dimension. (4+1) choose 2 = 10.
A 4 dimensional triangle has 10 edges)
5 choose (1 to 5) = 5 10 10 5 1
A 4th dimensional triangle:
5 - vertices
10 - edges
10 - faces (triangles)
5 - tetrahedra (3 dimensional triangle)
1 - 4-simplex (4 dimensional triangle)
As far as I can tell it works starting with row 3 (3 3 1) showing a triangle (3 vertices, 3 edges, 1 face) and onward.
n+1 choose 1 = vertices
n+1 choose 2 = edges
n+1 choose 3 = faces
and so forth.
It also works with squares. Instead of taking up more space to explain, I'll just leave the formula...
(n choose k)*(2^(n-k))
Since the Fibonacci numbers show up and the limit of the difference between two consecutive terms tends to Phi. Therefore there is a connection to 0,1,pi,e and Phi.
Lucas numbers can also be found. You can google for the evidence.
One fun fact about Pascal‘s triangle: If you close your eyes, all the odd numbers will disappear.
I tattooed pascal's triangle on the inside of my eyelid to prove you wrong.
I regret nothing
How did you fit the whole triangle on a finite surface?
Zacharie Etienne
You can make it fit by using a font half the size for the next row.
In my arrogance and supposed knowledge of all things related to Pascal's Triangle (including it's Chinese origins), I said: "Right, show me what I don't know!" and duly proceeded to click on the 'play' button.
I am both abashed and glad that you did indeed do so, for I was unaware of:
the hexagonal properties within the triangle and
the relationship of the triangle to both π and e.
Now, to puff my chest out again (only for it to be once again deflated, as you'll soon learn), quite some years ago I was playing with geometric formal power series and their closed forms. On concentrating on the closed forms themselves and beginning with the function
f(x) = 1/(1-x) (assuming convergence), I started a sequence of differentiating each function and multiplying the result by x. Thus if x*f_n (x) is x multiplied by the derivative of f_(n-1) (x), we get
f_0 (x) = 1/(1 - x), then
x*f_1 (x) = x/(1 - x)^2
x*f_2 (x) = (x + x^2) / (1 - x)^3
x*f_3 (x) = (x + 4x^2 + x) / (1 - x)^4
and so on.
Apart from noticing that the denominator of f_n (x) is (1 - x)^(n+1), what interested me was that in the numerator of f_n (x) there were n terms of powers of x which summed to (n!).
It was a short step to devising a "triangle" of which the first few terms were:
1) 1
2) 1 1
3) 1 4 1
4) 1 11 11 1
5) 1 26 66 26 1
6) 1 57 302 302 57 1
7) 1 120 1191 2416 1191 120 1
etc....
Having also derived a recurrence relation in the sequence, namely
D(n, k) = (n-k+1)*D(n-1, k-1) + k*D(n-1, k)
(where n is the row, k is the column and D is the function of these two variables),
I was as pleased as punch (if not as good looking) that I may join the mathematical pantheon in having a sequence of numbers named after me. Alas, it was all to no avail, as I subsequently discovered some months later that some "obscure" 18th century mathematician by the name of Leonard Euler (you may not have heard of him:)), had already discovered this same sequence. Yes, they are called Euler numbers after the most prolific mathematician of all time who just could not leave one mere mortal one scrap of mathematical kudos!
No, I'm not bitter (He said through grinding teeth).
But one thing I do know that Herr Bloody Euler didn't: the second column "miraculously" conveys the number of bets involved in an n-horse 'yankee' which contains bets involving doubles, trebles, fourfolds, etc. So there!
All being said, I am now both duly humbled and gladly educated by your most informative video.
Thank you!
Wow, i dont understand.
Stop humble bragging. It's annoying and doesn't fool anyone.
Tom Purcell you gets no points your nearly as smart as you boast and brag about your so smart where's your school where's your smartness towards something not already so easy to Google and Access come up with your own math problem that ppl havnt been doing way longer than you been able to drip snot from your nose
Yet, having had the wit to discover these things without the assistance of Herr Euler, you might bask in a little of his glory.
What is the music in this video? I really like it
Darude - Sandstorm
The Quartetto, TWV 43:D1 is a piece written by Georg Philipp Telemann for treble recorder, viola da gamba, strings, continuo.
Tipping Point Math thanks, there are no better pieces than ones written for the recorder
The worst part to me is how is it mountain of Peru on the Indian subcontinent. Bro, India and Meru are on different continents.
you can learn to do the nth root with the triangle and a formula
The out-most number on the other side of where the row starts with 1, is always 1. What a pattern!
I've been calling it Tartaglia's Triangle all my life, and I'm from Spain.
3 6 10 15 21 28 have patterns where there is +3,+4,+5,+6,+7(look at the diagonals
This is how energy fields interact expand to form matters anti matters and expand forever.
After layer 3 matters come into existence.
The ends are with a minus one and plus one.
The rows of Pascal triangle look like nodes of the graphs of some kind of distributions. What kind of distributions are these? Gaussian? Cauchy? Student's t? Logistic? Any other? Why? What is the mechanism underneath? Are there any articles or books about it?
See +ffggddss comment above (which might well have been meant as an answer to your question!)
Binomial it is.
If you keep adding the first non-zero digit of each row you will get ALL of the positive integers in sequence!
Thomas Joyce Wow, I didn't believe you until I tried it!
Jonathan Davey
You did better than I managed to.
I gave up, three quarters of the way to the end.
Actually you can see this by looking at the second diagonal
Pascals triangle shows up in the infinite nomial in its totality in the diagonal arrangement. Example Tetranomial
1
1 1 1 1
1 2 3 4 3 2 1
1 3 6 10 12 12 10 6 3 1
1 4 10 20 31 40 44 40 31 20 10 4 1
Do you see the 1,11,121? It goes on as you expand it. Tetranacci Sequence 1,1,2,4,8 by addition?
Via prime factorization the tetranomial can be expressed as 11^n*101^n much as Pascals triangle can be expressed as 11^n. If you arrange the tetranomial in the classic mountain configuration and add the diagonal you get the tetranacci sequence. Once you realize that all n-nomials express n-nacci sequences and are related to palidromes you can deduce the prime factorizations for pentanomials, hexanomials, heptanomials, octanomials etc...
I have just discovered One more pattern.if you Represent the Numbers as Points,and Trace each line of How they Expanded,you will notice that It creates a Dimensional Pattern coincidentaly on The nth row,equal to what is the Reference of the Dimension.on Pascal's Two term(Binomial) Expansion(which is the Regular Pascal's triangle),a Square(2d) is Formed in 2nd row.
I am going to use this for my math practicals presentations😅
There's a cute approximation for that 'central spine' of the Triangle,
C(2n,n) = (2n)! / [(n)!]²
The simple form of the approximation is
C(2n,n) ≈ 4ⁿ/√(πn)
This form excludes n=0, because the approximation blows up there. A much better approximation is
C(2n,n) ≈ 4ⁿ/ [√(πn)][1 + 1/(8n-½)]
Fred
The diagonals of the first row are natural numbers.
Fun fact! Did you know left-handed people are more likely to comment on this video than people who died at birth!!!!
And right-handed people are more likely to comment on this video than people who died after birth.
Pixel Bytes you forgot the question mark
But can you prove it ?
NormanicusDiabolicus
I can disprove it.
Claim: "People who died after birth are more likely to comment on this." (noone said "right after" birth.)
Proof 1: (Disjunctive Syllogism)
Suppose that:
X ≝ ("left handed people are more likely to comment on this then people who died after birth.")
Y ≝ ("People who died after birth are more likely to comment on this.")
Everyone dies after birth, X∪Y, left handed or right handed people alike, (since you must be born to die right or left-handed.) If Y includes left-handed along with right-handed people, then left handed people are a proper subset of
{lefties∪righties∪both} and {lefties∪righties∪both}
⊆ {people dead after birth}
Justification for this is Modus Ponens:
If "Y={lefties∪righties∪both}" then "{lefties} are a proper subset of {lefties∪righties∪both}."
"Y={lefties∪righties∪both}" is affirmed
therefore "{lefties}⊂{{lefties}∪{righties}∪both}, and {lefties∪righties∪both}
⊆{people dead after birth}.
So, we have a
Disjunctive Syllogism :
(X∨Y),
~X (since X⊆Y)
.·. Y▪
Proof 2 : Same Reasoning, using a
Corresponding Conditional&Truth Table:
XY: (( X∨Y )∧ (~X)) ⇒ Y
TT T F F True T
TF T F F True F
FT T T T True T
FF F F T True F
so we have a Tautology: {T, F} = True, False respectively, and every row is Yrue under the main connective. This indicates that Y is True and not X, therefore Y is actually the case such that it is true that "people who died after birth are more likely to comment on this."▪
Pixel Bytes
One person said "died after birth," which I disproved below, and you say "died AT birth" which I will prove here as requested by Normanicus. (Very different results)
Proof by Disjunctive Syllogism:
X =(Left handed people are more likely to comment then people who died at birth.)
Y =(People who died AT birth, i.e. "dead babies" are more likely to comment!!)
Disjunctive Syllogism:
(X∨Y)
~Y (since Y is a dead baby)
.·. X▪
Truth Table :
XY: ((X∨Y)∧ (~Y)) ⇒ X
TT. T F F True T
TF. T T Τ True T
FT. T F F True F
FF. F F T True F
is a Tautology, so "Left handed people are more likely to comment than" (dead babies,) or "people who died AT birth."▪
~ most morbid proof ever ~
u can find derivatives too from pascal s triangle😊
11^n begets Pascal's Triangle. (a+b)^n, etc. And on,for the powers of n.... and, thereby, the various roots proven.
If you add the numbers in each row then the sums of the numbers look like they are doubling
it really feels like there is all of math in Pascal's Triangle :D
The sum of x^(a+b) *(a,b) =-1/x for x bigger than -1 and smaller than zero.
Try The Silver Ratio
this is perfect to teach aliens our Number-system !!!
our?
there is also negative version of pascal's triangle
i used Pascal's triangle to generalise a formula for the sum of the sequence 1^k+2^k+3^k+4^k................(n terms)
4:57
I didn't understand this expression that how did the answer came as (22/7) after solving
1- 1/3+1/5-1/7
at 4:53 how did you get the second line, the factoring doesnt make sense???
Great video. The music is distracting
the triangular numbers on the third diagonal
ONE thing is better than this triangle: catalan numbers. whatever the name your contry has.. they are everywhere! n-minos, n-nodes-trees, ways on a grid from one node to another, inverse coefficient of tan(x) polynomial development... and i'm forgetting the majority of the properties :D!
dude where's my car (a denial of second thoughts), that is a significant one.
Omar Al Khayyam
I think it relates to basel problem.
Here is a fun one brought to you by the number 9 and the Langlands project. You may not have seen this before. Les choses que tu n'as jamais vues.
121: 21-12=9
1331: 331-133=198=>1+9+8=18+>1+8=9
14641: 641-146=495+>4+9+5=18=>1+8=9
15101051: 1051-1510=-459. -4-5-9=> -18=-1-8= -9; alternately 11^5=161051=>1051-1610=-559=>-5-5-9= -18=>-1-8=-9
11 requires us to go outside the triangle a little: 11^1=11.
010
0110=>110-011=99=> 9+9=18=>1+8=9.
Ce ne sont que des harmoniques.
So what should we actually name the triangle? Any suggestions are welcome.
Basically everyone discovered this triangle independently and named it after themselves. That's unsettling. It looks like the kind of thing you would discover if you were bored in math class in elementary school... But then, calling it Pascal's Triangle is really unintuitive, and doesn't really make sense at all!
Well given that it doesn't really matter that much.. if someone ever does try to take it upon themselves to set a culture-neutral name, it would likely end up being something bland like "Binomial Triangle," which nobody would use because they're all used to the existing name(s.)
altrag I disagree. I'm familiar with it being Pascal's triangle, but presented with a better name, I am willing to reconsider. Knowing what I now know, I also wouldn't teach Pascal's triangle by its current name in a classroom, but only mention it contextually. It's an unnecessary ambiguity to students already struggling with learning disabilities.
+Unknown Entity
> I am willing to reconsider.
Most people would be "willing" to reconsider, but the problem is that the existing name is already well embedded in minds of established mathematicians and even with conscious effort, that's hard to change. A bigger issue still though is that its the name printed in millions upon millions of textbooks that would have to be scrapped and reprinted. Never mind if you wanted to rename everything in math and science that's been mis-attributed over the centuries. Babylonian Theorem anyone?
> I also wouldn't teach Pascal's triangle by its current name in a classroom
Why not? You have to call it something and the kids you're teaching wouldn't know the difference. And hopefully by the time they're old enough to understand (or even hear about) the historical and cultural implications of the name, they'd also be old enough to understand that things can have more than one name. I mean they know actors usually don't have the same name in different movies and they somehow get by.
> to students already struggling with learning disabilities.
Unless "extreme pedantry" has been classified as a learning disability, I'm not sure that this is relevant one way or the other. If a student is capable of learning about the triangle, then they'll learn it and if they aren't capable of learning it then they won't, no matter the name you give it.
altrag I understand you are opposed to the idea. I only minor in maths, so I'm not awfully familiar with expert opinions. But in my humble opinion most of those arguments are weak.
Given that Pascal's triangle is in fact not called Pascal's triangle in most of the non-English speaking world (India and China are really fucking huge), what those mathematicians are accustomed to is actually already an exception to what they were taught. Besides, most textbooks are already bad, and anything until university level should be updated regularly anyway, as people change. It makes me think of an episode of Numberphile about τ (tau), where it was proposed that τ not replace π in the education system, but is used as a supplement in the introduction to irrational numbers, and then trigonometry, as it is initially more intuitive in high school maths.
You don't sound like you're willing to try and understand how learning disabilities work. But in the off chance I'm wrong, I used to struggle with maths in school. I remember being thrown off at minor inconsistencies. In particular, Pascal's triangle (as it inconveniently is being called in Norway where I went to school) did not indicate to have any relation to either geometry or algebra (let alone probability as it was presented). For instance, I would overestimate the importance of chapters about concepts named after people, and miss the point of the learning objectives. After I started teaching, I learned that this is not completely unheard of. Some students, especially the Asperger's types are unable to organise new concepts that are not plainly algebraic and have descriptive names. Pascal's triangle is neither.
Unfortunately, it took me until university to learn how to understand some things that I should have known before I started high school. The reason was that I had thought about it the wrong way, which was caused by confusion at the time of learning.
+Unknown Entity
> I understand you are opposed to the idea.
I'm not _opposed_ to the idea. I just don't see it being useful enough to warrant the effort needed to implement it. Certainly calling it the "Binomial Triangle" would give you an automatic tie-in to the binomial theorem, but unless you wanted a 400 word name, you're not going to be able to cover everything hidden in the triangle. At best you might get a vague reference to one or two ideas before the name becomes unwieldy, no matter what you choose.
> not called Pascal's triangle in most of the non-English speaking world
Sure, Indians and Chinese (and Italians and whoever else) will use their local names for the triangle but as long as they know what they're talking about, the name doesn't really matter to them either. The only time it would be an issue is if an English speaker tried to take a math course in India or something, and even then its all of a Google search away (or even less since you rarely find reference to the triangle's name without a corresponding diagram of some sort, and its a pretty recognizable structure.)
> Numberphile about τ (tau) ... as it is initially more intuitive in high school maths.
Not really. Tau gives you circumference without the factor of 2, and it makes it more obvious how to divide up the unit circle (1/4 tau is 1/4 of the circle!) But then when you look at area or arc length, you have to introduce a factor of 1/2. Tau and pi end up being fairly evenly distributed as the "easier" number in most of math and physics.. and factors of 2 are usually easier to work with than factors of 1/2, so pi kind of wins out there in my opinion.
> did not indicate to have any relation to either geometry or algebra
That comes back to the 400 word name. Pascal's triangle (or whatever you want to call it) and the Fibonacci sequence (also mis-named by the way) and several other fundamental numeric structures have applications in, or connections to, all sorts of weird places. There's no way you could name them in such a way that they make sense in all of those contexts. You have to pick something.
As another example, does the phrase "prime number" sound terribly descriptive to you?
> I would overestimate the importance of chapters about concepts named after people, and miss the point of the learning objectives.
I don't know what to tell you there. Can you come up with a better name for Pascal's triangle that you believe would have made it easier for you? Did you have similar trouble with Pythagoras' theorem? Or the Fibonacci sequence? Or even the really "common" names like sine, cosine and tangent? I'm not trying to be glib here I'm really wondering if you have an answer to any of that. Its one thing to say "that didn't work for me" and quite another to say "but here's something that does."
interesting fact 🤔
Sorry about the awkward question, but
What is the name of the font you use in this video? Feels mathematical, and I really want to know :3
It seems to be almost, but not quite, identical to Times, Times New Roman, Baskerville, or Big Caslon.
None of my other fonts match any closer than those, although a few of them are sorta close.
i feel like its times new roman with italicized letters for “x” and “y”
It's the same one used in Manim
I'm just going to call it number triangle, thanks
Lovely ❤😍
My college classmates nicknamed me Pascal. It was funny when the professors started calling me Pascal.
Good informations ,thanks but... very bad music
this is so cool
1:11 Why Crimea is colored in different color than the rest of Ukraine?
@@at7388 you spelled "orange" wrong
That is completely ridiculous indeed, legalizing theft!
Didn't notice
Hmm if you use add and subtract pattern
1
1+1
1+2-1
1+3-3+1
1+4-6+4-1
youll likely get this
1
2
2
2
2
...
But when you use subtract and add
1
1-1
1-2+1
1-3+3-1
1-4+6-4+1
Youll get
1
0
0
0
0
0
...
Idk how.
If you look at all numbers from the second diagonal you will notice that it goes from 1 to infinity starting from top to bottom
That was Brilliant. Thank you !!!!
Is there any relationship of the pascal's triangle with the concept of power sets??
I’m a bit late but yes, the modulus of a power set is 2^n because it is equivalent to the sum of every binomial co-efficient of a certain n (equivalent to the sum of every row is 2^n)
it counts from the two in the third column and just keeps counting
I can prove relationship of prime numbers and Riemann zeta zero using pascal triangle
since the one at the top is the 0th row, this is just proof that (0 choose 0) = 1, i.e. there is only 1 way to arrange no objects: sucks to be you (the official maths lingo)
*_...assuming symmetry in the line-before-the-first we get ...−½+½−½+½+½−½+½−½+..._*
I think you've got an extra positive term in the middle but barring that.. why does that fact strike me as something that would relate to Riemann's Zeta function? Where on one side of zero everything is nice and orderly and then on the other side its.. well still orderly but a completely different (and messier-looking) kind of order.
Someone up the comments mentioned a relation between Pascal's triangle and the primes, so I definitely wouldn't be surprised to see Riemann embedded in there somewhere as well.
Edit: Nope no extra positive term. Just so many 1/2's in a row messing with my eyeballs.
Well what if we use the closed formula, C(n,m) = n!/[m!(n-m)!], and set n = -1? What we get is a whole string of "±∞/∞", because factorial of a negative integer is ±∞.
But if we make some assumptions to try to tame the infinities, like using the factorial recursion, n! = n(n-1)!, then we get:
... -1 1 -1 1 1 -1 1 -1 ...
which is just twice what you got. But this breaks the rule for summing adjacent entries to get the entry centered below them; 1+1 = 2 ≠ 1.
OK, what if we treat each of these as a limit, where C(n,m) is replaced by the limit as ε→0 of C(n+ε, m+ε)?
[Note that this calls on the "Pi" function, ∏(x) = Γ(x+1), which equals x! when x is an integer ≥ 0.]
I haven't carried this out, but I think it can be done using l'Hôpital's Rule and the digamma function; which looks kinda hairy. But I suspect it will give your answer.
For one thing, recall that the sum of row n is 2ⁿ, so the sum of row (-1) should be ½.
Sure enough, although your sum doesn't converge in the usual sense, there is a looser sense in which it sums to ½, because of that 'extra' +½
...not an extra, +½, but missing, the implicit central, −½, due to ½-in-shifted wings...
"Extra" in the sense that, without it, the sequence is purely alternating.
...'yeah' but I'm thinking also of what physicists 'do' with infinite-cycle integrations, applying a centered weighting function and slowly lifting it (non-uniform pointwise convergence is 'unmathematical-scary-stuff' capable of spurious results)...
Cool
quality
im so confused about the triangle and i have to hand in a project but 8 30 in the morning lol wish me luck
Yang Hui (whey) saves writing huei,
gli is pronounced “yi” in italian. So Niccolò Tartayia. Not a big deal of course.
Multiply every numbers and get 0
1:55 "If you find row n in Pascal's triangle, the second number corresponds to the number of points on the circle..."
n=5
*But proceeds to show 6th row of PT*
💀
Hi. This is because the first row is conventionally referred to as "Row 0."
Still better than BRIGHT SIDE
that's
Khayyam Pascal
Pingala discovered this pyramid first