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Dr Barker
United Kingdom
Registrace 29. 12. 2020
Fun maths videos, covering lots of topics that I find interesting.
A Fun Exponential Equation
We solve the equation (9^x - 25^x)/15^{x -1} =16.
00:00 Getting started
03:45 Solving the quadratic
05:55 Complex solutions?
00:00 Getting started
03:45 Solving the quadratic
05:55 Complex solutions?
zhlédnutí: 3 309
Video
What Lies Above Pascal's Triangle?
zhlédnutí 141KPřed dnem
We explore how to extend Pascal's triangle upwards. Binomial series expansion proof: proofwiki.org/wiki/Binomial_Theorem#General_Binomial_Theorem en.wikipedia.org/wiki/Binomial_series 00:00 First considerations 01:44 Binomial coefficients 04:35 Extending further upwards 08:38 Binomial theorem 09:40 Proving the pattern continues 15:07 Non-integers 15:48 Alternative series expansion 20:45 Even mo...
An Amazing Floor & Square Root Identity
zhlédnutí 3,7KPřed 14 dny
We prove the identity ⌊sqrt{n} sqrt{n 1} sqrt{n 2}⌋ = ⌊sqrt{9n 8}⌋ for all natural numbers n. I came across this result as a problem posed here: F. David Hammer, Problem E3010, Amer. Math. Monthly 95 (1988), 133-134. 00:00 Sufficient conditions 01:31 Square roots of consecutive integers 05:37 Simplifying the inequalities 08:22 Bounds for sqrt{n(n 1)} 12:17 Bounds for sqrt{(n 1)(n 2)} 13:50 Boun...
The Best Diagram for Angle Sum Identities
zhlédnutí 3KPřed 21 dnem
Using a simple diagram, we derive the angle sum identities for sine and cosine from basic trigonometric results. 00:00 sin(θ ϕ) 03:04 cos(θ ϕ)
A Factor Tree Puzzle
zhlédnutí 1,9KPřed měsícem
We solve a problem involving finding lots of missing values across two prime factor trees. 00:00 Intro 00:41 Solution
Convergence of an Interesting Series
zhlédnutí 12KPřed měsícem
We show that the sum from 1 to infinity of sqrt{n^4 1} - n^2 is convergent. We also include a simple proof that the sum of 1/n^2 is convergent. 00:00 A trick to simplify 02:27 Convergence of 1/n^2 04:20 Method of differences 06:29 Considering negatives
A Tricky Roots of Cubics Problem
zhlédnutí 9KPřed měsícem
Given that α, β, γ are roots of the equation x^3 2x 1 = 0, we calculate α^5 β^5 γ^5, using some elegant algebraic techniques and shortcuts. 00:00 Simultaneous equations 02:46 Vieta's formulas 04:55 Sum of squares 06:14 Sum of cubes 11:48 Sum of fifth powers
A Satisfying Divisibility Proof
zhlédnutí 47KPřed měsícem
A Satisfying Divisibility Proof We prove that m^{17} n - m n^{17} is divisible by 10, for all integers m, n. 00:00 Divisibility by 2 01:55 Factorisation 03:25 Working modulo 5
Finding the Area of a Region
zhlédnutí 2,3KPřed 2 měsíci
We find the area of the region where | |x y| - y| ≤ |x - y| ≤ 3. 00:00 First inequality |x - y| ≤ 3 01:42 Second inequality x y ≥ 0 06:35 Second inequality x y ≤ 0 11:25 Finding the intersection
More Minimising Without Calculus
zhlédnutí 7KPřed 2 měsíci
We minimise a^2 b^2 c^2 d^2, subject to the constraint abcd = 9. 00:00 Intro 00:20 2 variable problem 03:56 Building on this case 05:00 Considering negatives 06:00 3 variable problem 10:03 4 variable problem
Finding the Range of a Function
zhlédnutí 3KPřed 2 měsíci
We determine the range of the function f(θ) = (1 - tanθ)/(1 cotθ). That is, the set of possible outputs, given real values of θ for which f(θ) is well-defined. 00:00 Domain 02:08 Substitution 03:34 Discriminant 05:34 Solving the inequality 06:41 Checking carefully: x = 0 08:03 Checking carefully: x = -1
A Short Number Theory Proof
zhlédnutí 8KPřed 2 měsíci
We explore positive integers with the same number of digits (in base 10) as distinct prime factors, and determine whether or not there are infinitely many such integers. 00:00 Examples & non-examples 02:14 Are there infinitely many? 03:53 Proof
A Floor & Ceiling Equation
zhlédnutí 7KPřed 2 měsíci
We solve the equation ⌊x^2 - x⌋ = ⌈1 - x^2⌉. 00:00 Sketching x^2 - x and 1 - x^2 01:53 Key values 05:26 Sketching ⌈1 - x^2⌉ 07:13 Sketching ⌊x^2 - x⌋
Minimising Without Calculus
zhlédnutí 6KPřed 3 měsíci
We minimise 5x^2 y^2 2z^2 - 2xy 4xz 6z without using calculus. Our approach relies on repeatedly completing the square. 00:00 Intro 00:50 Completing the square 02:43 Terms with x 04:11 Terms with z
Two Neat Properties of Cardioids
zhlédnutí 5KPřed 3 měsíci
We prove two properties of cardioids of the form r = a(1 cosθ). Firstly, we show that all chords passing through the origin have the same length. Secondly, we show that the tangents to the cardioid at either end of such a chord are perpendicular. 00:00 Property 1 02:36 Property 2 02:51 Differentiating 06:29 Perpendicularity 10:12 Simplifying
The Quadratic Formula No One Taught You
zhlédnutí 139KPřed 4 měsíci
The Quadratic Formula No One Taught You
Similar Triangles in the Complex Plane
zhlédnutí 2,7KPřed 4 měsíci
Similar Triangles in the Complex Plane
The Most Satisfying Proof: Product of 4 Consecutive Integers is not Square
zhlédnutí 24KPřed 6 měsíci
The Most Satisfying Proof: Product of 4 Consecutive Integers is not Square
shouldnt (1+x)^-1 = 1/(1+x)?
Out of bounds nonsense probability distributions. Glitches in quantum mechanics
Some 20 Years ago I played around with this Idea. You do not have to be a Mathematician to have Fun with it.
Sry but what a weird way of writing “x”?! 😅
@blackpenredpen - I’ve also expanded the standard triangle into higher dimensions. The problem I’m having is understanding Python enough to generalize the multinomial distribution.
I never knew what pascals triangle was! Now i know more than i am supposed to know about it 😅
Why would we continue the constant of 1's for the B but not the same constant of 1's for the A? Not seeing the connection.
Caught your video in recommended, didn't watch it in full but skipped ahead to 20:37 to get a sneak preview, I never thought about this, very interesting.
if u can move the ruler u only need the 0 and 1 cm marks to measure any integer distance
made me think that if the ancient mathematicians had a CZcams channel & modern dress, it wouldn't look too different from this
So I see some people choosing a=b=½, but you can have even more fun with any conjugate complex numbers with real part ½. Interestingly, the negative 3rd roots of unity have this property.
could we apply these principals to the complex plane? ie could a be (0.5 + 0.5i) and b could be (0.5-0.5i)? what would we see if we iterated on these values? 😊
This is only for (1+x)^n but what about (x+y)^n = 1x² + 2xy + 1y² (if n=2)
So in fact Pascal triangle is a Pascal half-plane.
Incredibly clean presentation! I love your style.
Awesome! Kp it up with these cool math videos!❤❤❤
Wow, I had no idea. Truly a new perspective.
I can't believe this, it's amazing how math can always shock you with new things with just simple things
1) Pascal triangle is about INTEGERS, more specifically- NATURAL NUMBERS. 2) What sense, say- geometrical sense it makes to extend it upwards?
My guess is, it's completely arbitrary
But what's above the rotated triangle now?
The existence of 3π/2 expanding Pascal's triangle and π/4 expanding Pascal's triangle implies the existence of 3π/4 expanding Pascal's triangle
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This is so fucking cool
square root of -0. I thought everyone knew this
As Below, So Above
lascaP triangle ?
Pascal would be rotating in his grave.😊
Care is needed here. There must be more to be un-earthed though.
Function keys. What they do depends on your operating system.
.. it's a reflection about the 0 wedge.....
but with flipping signs
it would look cool to design a visualization where the 0 void is at the top, the line of 1's is flat horizontal, and the two triangles expand to the two sides of the diagram
Handsome guy with interesting smart content! Subbed! 😂
This is really cool and all, but am I the only one distracted by the fact that when he writes an x on the board, because of the way he draws it it always looks like a cartoon butt with the cheeks facing each other? 😂
Wow, that is extremely cool and absolutely not what I expected.
Whats interesting is that if you add up all the numbers in the nth row in Pascal's Triangle, you get 2^n For example, 1 3 3 1 is the third row and 1+3+3+1 = 2^3 = 8. But at the -1st row at 4:18, you get zeros on the left and 1-1+1-1+... on the right, which is Grandi's series which "evaluates" to 1/2, which is 2^-1, and I can assume that the other negative rows "evaluate" to 2^n as well.
Is there a pretty way to define a and b to make it so when abs(x) > 1 then a would be 0 and b would be 1 and vice versa as in to fix the possible values of a and b depending only on x? Any of them can be defined like a = 1 - b (swapping them works just as well, in this case if we restrict b to be 1 or 0, it would work). But how would we calculate b like that using just the value of x? Probably it would be using limits to make sure things go to 0
I always wondered this and it's a super interesting result
👏
Excellent video! This is the first time I’ve seen the rest of Pascal’s Triangle explored. I got out the popcorn when I saw the 1, -2, 3, -4, …. appear. This gon b gud. 🤓
Well I'm amazed. What an incredible finish! A and B as coefficients that satisfy (a+b = 1), but for which X is undefined. Could there be complex values of X where there's a valid solution?
Very cool, thanks for sharing this!
Pascal's Trefoil
Did I miss something or did your first proof only works for X = 1?
I've investigated this myself, but it didn't occur to me that there could be infinitely many solutions. There is, however, as far as I can tell, only one symmetrical solution: −1½ ½ 0 0 ½ −1½ −1 ½ 0 ½ −1 ½ −½ ½ ½ −½ ½ 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 2 1 0 0 0 I have a related question. Have you ever come across variations of Pascal's triangle, such as one with 0, 1, 2, 0 in a row? 0 1 2 0 0 1 3 2 0 0 1 4 5 2 0 0 1 5 9 7 2 0 0 1 6 14 16 9 2 0 0 1 7 20 30 25 11 2 0 This particular variation has some fine properties, such as encompassing the Fibonacci numbers (descending shallow diagonals), the Lucas numbers (ascending shallow diagonals), the natural numbers, the odd numbers, the natural squares, the sums of the natural squares, and a host of other sequences. Perhaps more significantly, though, the individual elements of the shallow ascending diagonals provide the coefficients for polynomials that define L[2n], L[3n], L[4n], etc., in terms of L[n], the nth Lucas number. (Please excuse the formatting. Subscripts are fiddly and almost illegible in YT comments.) I'll give you a couple of illustrative examples: For even n, L[5n] = (L[n])^5 − 5(L[n])^3 + 5(L[n]). E.g., L[2] = 3; L[10] = 3^5 − 5 • 3^3 + 5 • 3 = 243 − 135 + 15 = 123. For odd n, change the minus to a plus. E.g.; L[3] = 4; L[15] = 4^5 + 5 • 4^3 + 5 • 4 = 1024 + 320 + 15 = 1364. Pascal's triangle can do similar things for Fibonacci numbers, but the polynomials are in terms of Lucas numbers, L[n], with all coefficients multiplied by F[n]. I won't bore you with the details.
Truly interesting! Also you speak so naturally, clearly and fluently. I'm just terrified of people who write x as an inverted c and c put together.
In short: another Pascal's triangle, but rotated 120 degrees counterclockwise, with the previously "left", now bottom chain of 1's alternating between 1 and -1, producing a Pascal triangle with alternating positive and negative values. And a whole bunch of zeroes
My mouth dropped when the triangle showed up again but rotated 😮
Fascinating! 😮
I've always extended Pascal's Triangle in my mind as if there was an extremely tiny, infinitesimal chance that there was a glitch in the summation. So you start with an infinite hexagonal grid of 0s, each 0 the sum of two other zeros. Except randomly, at some point in the infinite expanse...there's a "glitch" and 0+0=1. And from that, the entire Pascal's Triangle is generated. Similarly: the Fibonacci series.
From any number, the two numbers above are exactly half the original number. So what is above the 1? 0.5 and 0.5 What's above the 0 (next to the starting 1)? 0.5 and -0.5
I believe that provides the only symmetrical solution, but as the video shows, it is only one of infinitely many solutions.