![Dr Barker](/img/default-banner.jpg)
- Видео 186
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Dr Barker
Великобритания
Добавлен 29 дек 2020
Fun maths videos, covering lots of topics that I find interesting.
Gaussian Integral: Infinite Series Approach
We evaluate the Gaussian integral, using an approach based on series expansions.
Exchanging the order of summation and integration:
planetmath.org/criterionforinterchangingsummationandintegration
00:00 Setup
02:05 Infinite series
04:05 Integrating
08:03 Partial fractions
10:59 Simplifying the inner sum
14:15 Back to integration
17:45 Back to powers of e
21:23 Taking limits
Exchanging the order of summation and integration:
planetmath.org/criterionforinterchangingsummationandintegration
00:00 Setup
02:05 Infinite series
04:05 Integrating
08:03 Partial fractions
10:59 Simplifying the inner sum
14:15 Back to integration
17:45 Back to powers of e
21:23 Taking limits
Просмотров: 290
Видео
A Limit With A Cube Root
Просмотров 3,4 тыс.22 часа назад
We evaluate the limit as n tends to infinity of cbrt(n^3 n^2) - n. 00:00 Simplifying 03:12 Evaluating the limit
A Factorial Product Problem
Просмотров 2,7 тыс.14 дней назад
We show that the product 1! x 2! x ... x 1000! can be written in the form (a!)^2 b^3 c^4, where a, b, and c are positive integers. 00:00 Rewriting the product 01:59 Powers of 3 02:52 Powers of 4 05:05 Remaining terms
A Functional Equation Problem
Просмотров 4,4 тыс.21 день назад
Given a function which satisfies f(xy) = xf(y) yf(x) for all values of x and y, we find the value of f(72) given also that f(18) = 2, and f(108) = 36. We conclude with a systematic approach which can be used to to solve similar problems. 00:00 Generating new values 01:38 Making 72 04:21 More systematic approach
A Fun Exponential Equation
Просмотров 3,6 тыс.28 дней назад
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?
What Lies Above Pascal's Triangle?
Просмотров 196 тыс.Месяц назад
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
Просмотров 3,8 тыс.Месяц назад
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
Просмотров 3,1 тыс.Месяц назад
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
Просмотров 2 тыс.Месяц назад
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
Просмотров 12 тыс.2 месяца назад
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
Просмотров 9 тыс.2 месяца назад
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
Просмотров 47 тыс.2 месяца назад
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
Просмотров 2,4 тыс.2 месяца назад
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
Просмотров 7 тыс.2 месяца назад
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
Просмотров 3,1 тыс.3 месяца назад
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
The Quadratic Formula No One Taught You
Просмотров 141 тыс.4 месяца назад
The Quadratic Formula No One Taught You
Can You Solve Without Finding x & y?
Просмотров 4,1 тыс.5 месяцев назад
Can You Solve Without Finding x & y?
Similar Triangles in the Complex Plane
Просмотров 2,8 тыс.5 месяцев назад
Similar Triangles in the Complex Plane
I hadn't seen this one before but it is straightforward one you break it down as you have done. Paul Levy's original functional analysis treatise which Norbert Wiener used for his development of Brownian motion actually developed the Gaussian from volumes of n dimensional spheres. As usual you have done a very clear and succinct exposition.
80% of the math in one video!😀
Heavy duty!
More complicated than the usual approach, but on the other hand, one does not need to know polar coordinates... Very nice way! :) Did you come up with that on your own?
By the way, do you know the RUclips channel of Dr Peyam? He has a whole playlist on the Gaussian integral, with 12 different methods. But yours was not included there, so apparently it's really a new one?
Yes, I'm a fan of Dr Peyam! This approach is adapted from/related to something I remember seeing a few years ago - I can't seem to find a reference though for where I found it. I'll see if I can find it again.
Brilliant!!
And if we create a manifold where the absolute value is a polygon of size that number. A Saddleback?
Try convert 0 to other way, maybe i Try convert a or b number with i
Glorified hobby
Today I'm going to sleep earlier... I said to myself
I remember extending the Pascal triangle upwards on a whim in high school math, but I did not at all understand that the numbers (after arbitrarily going with 0 1 though I don't remember which direction I went in) correspond to these coefficients. It's really very beautiful.
👍👍👍👍
2^495562, 3^247263, 5^123124, 7^81774, 11^48697, 13^40508, 17^30240, 19^26789, 23^21757, 29^16939, 31^15704, 37^13041, 41^11724, 43^11155, 47^10164, 53^8955, 59^7992, 61^7720, 67^6979, 71^6559, 73^6370, 79^5850, 83^5538, 89^5137, 97^4675, 101^4464, 103^4374, 107^4194, 109^4104, 113^3940, 127^3451, 131^3339, 137^3171, 139^3115, 149^2877, 151^2835, 157^2709, 163^2583, 167^2500, 173^2410, 179^2320, 181^2290, 191^2140, 193^2110, 197^2050, 199^2020, 211^1894, 223^1774, 227^1734, 229^1714, 233^1674, 239^1614, 241^1594, 251^1497, 257^1461, 263^1425, 269^1389, 271^1377, 277^1341, 281^1317, 283^1305, 293^1245, 307^1161, 311^1137, 313^1125, 317^1101, 331^1017, 337^991, 347^961, 349^955, 353^943, 359^925, 367^901, 373^883, 379^865, 383^853, 389^835, 397^811, 401^799, 409^775, 419^745, 421^739, 431^709, 433^703, 439^685, 443^673, 449^655, 457^631, 461^619, 463^613, 467^601, 479^565, 487^541, 491^529, 499^505, 503^498, 509^492, 521^480, 523^478, 541^460, 547^454, 557^444, 563^438, 569^432, 571^430, 577^424, 587^414, 593^408, 599^402, 601^400, 607^394, 613^388, 617^384, 619^382, 631^370, 641^360, 643^358, 647^354, 653^348, 659^342, 661^340, 673^328, 677^324, 683^318, 691^310, 701^300, 709^292, 719^282, 727^274, 733^268, 739^262, 743^258, 751^250, 757^244, 761^240, 769^232, 773^228, 787^214, 797^204, 809^192, 811^190, 821^180, 823^178, 827^174, 829^172, 839^162, 853^148, 857^144, 859^142, 863^138, 877^124, 881^120, 883^118, 887^114, 907^94, 911^90, 919^82, 929^72, 937^64, 941^60, 947^54, 953^48, 967^34, 971^30, 977^24, 983^18, 991^10, 997^4 Calculated the prime powers Looks like there's 10^300 possible answers
Tried the substitution m = n+1/3. You now get the limit of (m^3 - m/3 + 4/27)^(1/3) - m + 1/3. Now the quadratic term within the cube root vanishes. To show that the linear and constant therms do not matter for the limit value of the cube root, is where I got stuck. But the 1/3 is already showing up.
Jones Jennifer Thompson Daniel Garcia Angela
I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.
Been a long time since my math undergrad, and this just reminded me why I fell in love with it. What a cool generalization
Can you make a video about Kruskals Count and how it’s different from the birthday paradox? They’re both pretty easy for laymen to understand with famous examples but I’m hoping you’ll show something not readily known
Так и не понял, что за заработок, т.е. это просто выгодня комиссия или что?
An offset perpendicular universe.
What I found interesting was how this extension of the triangle is entirely created by extending the line of 0 and 1 up and to the right.
22:53 I think that’s a brilliant maths quote: ‘bearing in mind that this isn’t actually going to be valid for any values’ but we’ll just keep going anyway cus it’s interesting!
With cbrt(n³ + n²) - n = n( cbrt(1 + 1/n) - 1) we can use cbrt(1 + x) = 1 + 1/3 * x - 1/9 * x² + ... if abs(x) <= 1 and so abs(1/x) >= 1. So we get n(1 + 1/3 * (1/n) - 1/9 * (1/n)² + ... - 1) = 1/3 - 1/9 * (1/n) + ... = 1/3 for n -> oo
Extensive use of zeros for extremums is one thing and exploiting the equal to sign with ratios to get equivalent identities is a physics thing 😅
lim qbrt(n^3+n^2)-n = lim n*(qbrt(1+1/n)-1) = lim n*(1+1/3n-1) = lim n*1/3n = 1/3
The 3n must be inside grouping symbols: lim n*[1 + 1/(3n) - 1] = lim n*1/(3n) = 1/3
@@robertveith6383 it's not 1/3*n 3n is grouped automatically
I personally love a more straight forward approach: ³√(n³+n²)-n =(³√(1+1/n)-1)/(1/n) L’hospital rule 0/0: =1/3*(1+1/n)^(-2/3)*(-1/n²)/(-1/n²) =1/3
This is how I did it
This is very interesting but probably harder to come up with. I would extract out the n^3 from the cube root we get n cube root(1+1/n) - n, and expand the cube root (1 + 1/n) to 2 terms to 1+1/3 1/n + o(1/n^2). This immediately gives 1/3.
That's how I did it!
I used the binomial expansion too but I didn't factor out the n^3 first which made it a little more interesting. I think I know what you mean - I wouldn't have thought that putting a sum of 1/3 and 2/3 power in the denominator would simplify things.
@@dugong369 the (1+epsilon)^n expansion is very well known at least in physics context.
@@ikarienator yes my way was less rigorous as well as a little more complicated. To be even sloppier, assume the limit is x. Then n^3+n^2 = (n+x)^3 in the limit. n^3+n^2 = n^3+3xn^2+3nx^2+x^3. In order for the 2 highest order terms to have matching coefficients, x=1/3.
Another option would be to proof: For all epsilon >0: For large enough n: (n+1÷3-epsilon)³<n³+n²<(n+1÷3)³ sandwich lemma implies forall epsilon (>0) 1÷3-epsilon<=L<=1÷3 implies L=1÷3
Optionally: (n^3 - n^2)^(1/3) = n(1 + 1/n)^(1/3) = n(1 + 1/3n + (1/3)(-2/3)(1/2n^2) + ... ) as a binomial expansion which converges when (1/n) < 1. Multiplying the series through by n, we get (n + 1/3 - 1/9n + ...). So the limit as n→∞ of ((n^3 - n^2)^1/3 - n) = Lim_n→∞ (n + 1/3 - 1/9n + ... - n ) = Lim_n→∞ (1/3 - 1/9n + ... ) = 1/3.
I have always understood it as squares are the sum of n odd natural nos so n² + 2n+1 = (n+1) ² so if 2n +1 can make such an impact n also can . Is this analogy even partially correct
1/3 I just did a Taylor's series of cbrt.
Did think that : n((1 + 1/n)^(1/3) - 1) might have been a first step followed by ((1 +1/n)^(1/3) - 1) /(1/n) but having seen the approach can see why this will fail.
It doesn't fail. Use a series expansion on n((1+1/n)^(1/3) - 1) and then cancel the 1's and muiltiply through by the n. You get (1/3 - terms in 1/n ...). The limit as n → ∞ is obviously 1/3.
@@RexxSchneider of course - a long day 🙂
I didn't know that Ronald Weasley is a good teacher.
Saar ! Do not redeem !
@@Regimeducamp I make bangers, not anthems, leave that to the artful dodgers.
This limit also turns out to be equal to f‘(0) where f(x) = (1+x)^(1/3). Just saying.
👌
You know I came up with the exact same method, I was reminded by a R.E. example lim(n to inf) ((n^2 +1)^(1/2) -n) where you rationalise the denominator, and then i remembered a video where you generalised rationalising the denominator for any order root.
I guess that this is the typical scenario when the limit cannot be distributed. BTW there is a criteria for this but I don't remember. Ploting the function in desmos shows that the solution holds also for n-> -inf. Great video.
This was very interesting. One would be wrong if assume that n^3 dominates at infinity and then get a limit of 1. It is a good lesson for students that a dominant term approximation is not always the correct choice. Another method (although fast-and-loose) is to note that (n^3+n^2)^(1/3) is not that much different at infinity from (n^3+n^2+n/3+1/27)^(1/3) = [(n+1/3)^3]^(1/3) = n + 1/3. Thus lim n-> inf (n^3+n^2)^(1/3) - n = lim n-> inf (n + 1/3) - n = 1/3.
@@brianstine2006 It's interesting that adding the n/3 + 1/27 contribution doesn't affect the limit, so while the n^3 term doesn't dominate at infinity, it seems like n^3 + n^2 does!
The explicit n^3 doesn't dominate at infinity because, being under the cube root sign, it only grows like n and is cancelled by the -n. Consider, e.g., pulling a factor of n out of the entire expression to get n((1+1/n)^{1/3} - 1), and looking at the taylor expansion of the binomial under the cube root, where the first term of the taylor expansion gets cancelled by the -1 at the end. A lot of confusing limits like this involve the first term(s) of some taylor expansions cancelling each other so only higher-order terms survive.
Generalizing, lim_{n to infinity} ( (n^m + n^{m-1})^{1/m} - n ) = 1/m.
I confused shouldnt the highest order term in the root match the order of the root?
Try checking that for m = 1 lim_{n to infinity} ( (n + n^2)^{1/1} - n ) = 1/1 => lim_{n to infinity} ( n + n^2 - n ) = 1 => lim_{n to infinity} ( n^2 ) = 1 ???
Sorry, typo, I've corrected it now.
Another method: =lim n*((1+1/n)^(1/3)-1), n-> inf = lim ((1+x)^(1/3)-1)/x, x->0 = lim (1+x/3-1)/x, x->0 =1/3
...or using Taylor expansion, which makes the result immediate and trivial
That's indeed what chatgpt came up with for when I tried the general case the other guy above mentioned
For the last step we can also note that, by definition, lim ((1+x)^(1/3)-1)/x, x->0 is the derivative of t^(1/3) at t = 1.
@@vladimir10 Use a binomial expansion and you don't need any calculus. (1 + 1/n)^m = (1 + m(1/n) + m(m-1)/2!(1/n)^2 ... ) Set m=1/3 for this case.
One thing worth mentioning is that with your definition, the infinite hypercube is disconnected, and the connected components are precisely the equivalence classes. So, the proof of bipartiteness splits up in two parts: 1) A graph is bipartite if all connected components are bipartite (here you need the axiom of choice to merge the partitions of the connected components) 2) Every connected component of the infinite dimensional hypercube is bipartite. Indeed, all connected components of the infinite dimensional hypercube are isomorphic, so it suffices to consider the connected component with only finitely many non-zero components. And for this connected component, the bipartiteness proof of the finite dimensional case carries over verbatim.
1 11 121 1331 14641 15(10)(10)51 16(15)(20)(15)61 17(21)(35)(35)(21)71 I'm am seeing a pattern
I found a different solution where a=996
Interesting! What values did you use for b and c?
@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.
@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.
@@DrBarker I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later.
@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996, b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497, c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125. We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.
But when a=0 it is no longer a quadratic function,it is a linear function
sorry to ask : what do you mean by : a functional equation ? apart from that : great content as usual !
Thank you! That basically means an equation which we might solve to find a function, like how f(xy) = f(x) + f(y) has a solution f(x) = log(x), rather than the usual solving an equation to find the value of a variable like 2x + 1 = 5.
А как насчёт ироцианальных чисел ) + дробей + не виданой мне штуковин + есть и другие виды чисел
The way he writes the letter X should be illegal
I would like to note that this generalization preserves in a certain way the property that the sum of the terms of the n'th row is 2^n (starting to count from the 0'th row). If you consider Ramanujan sumation, then this property actually holds for negative n.
Great man 👍🏻 love from India!
i'm not good at english so my explain might seems to be wierd a^2, b^2, c^2, d^2 are all greater than 0, so (a^2+b^2+c^2+d^2)/4 >= (a^2×b^2×c^2×d^2)^(1/4) because of relevence? of mathmatical mean? and geometrical mean? (idk what they are called in eng) and a^2+b^2+c^2+d^2 >= 4×(abcd)^(1/2) = 4×9^(1/2) = 4×3 =12 so a^2+b^2+c^2+d^2 always be equal or greater than 12 and a^2+b^2+c^2+d^2 = 12 when a^2 = b^2= c^2 = d^2 = 3 so a, b, c, d are 3^(1/2) but even number of them are negative
Can you use this knowledge to solve the Black ops 3 Kn-44 graph?
Bro I came here thinking the same thing 🤣.