phi^1=phi phi^2=phi+1 phi^3=2phi+1 phi^4=3phi+2 phi^5=5phi+3 phi^6=8phi+5 ... I think the Fibonacci sequence definetly is very much linked to phi actually...
Thanks, that is very interesting. BTW, the Lucas numbers also have a precise identity: L(n) = phi^n + (1-phi)^n and the Fibonacci numbers have a similar one: F(n) = (phi^n - (1-phi)^n) / sqrt(5) There is also a direct identity: L(n) = F(n+1) + F(n-1) Oh, and finally, from the closed form of F(n) above, follows: F(n) = round(phi^n / sqrt(5)) because the "error term" (1-phi)^n / sqrt 5 is always less than 1/2.
in fact if you use the golden ratio again to write phi^n you can find Fibonacci again. Nothing spectacular about Lucas Numbers: phi^1= phi + 0 phi^2= phi + 1 phi^3= 2phi + 1 phi^4= 3phi + 2 phi^5= 5phi + 3 ...
Same goes for any fibonacci style sequence, write out p subscript n for terms a and b(where a is the first term, like 1 in the fibonacci sequence, and b is the second term) and you find p subscript n=x (subscript n) *a+x (subscript n+1)*b when the starting series is a,b,p subscript 1, p subscript 2,... and x subscript n refers to the nth number of the fibonacci sequence
4:12 In defense of rounding: "People link mathematics with unnecessary precision ... with things that are pointless and over-the-top." What, like printing out a million places of pi? (ducks)
There was a reason for finding so many places of pi--they wanted to know whether or not the decimal sequence ever repeated, whether pi was truly irrational. And now we know that it is indeed irrational. That wasn't done just to nerdily rattle off as many decimal places as possible.
Dylan Rambow We can devise algorithms for numbers that could never fit into this universe. To determine if pi is transcendental, one must make a smart proof. No amount of computing power will create a mathematical proof.
I am a lowly stats 1 student, but given that we bounce from decimal place to decimal place depending on the problem, I like his essential notion that after a certain point, we just don't give a flying f---.
Call P=(1+sqrt(5))/2 and Q=(1-sqrt(5))/2, the two solutions of x^2-x-1=0. The Lucas sequence is simply P^n+Q^n, which happens to be all integers. Since Q is between -1 and 1, Q^n converges to 0, and that's why rounding works (for n>=2, that is).
I'm going to say it in English, rather than read it in English, realise its a french word, translate it and say it in French. Speaking words in your native language is definitely lazier.
I would argue against lucas numbers being superior, and here's why. If you take ɸ^1 you get ɸ+0, if you take ɸ^2 you get ɸ+1, now if you take ɸ^3, if you do some algebra given that ɸ^2 = ɸ+1, you get 2ɸ+1. ɸ^4 is 3ɸ+2, ɸ^5 is 5ɸ+3, and so on. It is very clear that ɸ^n = ɸF(n) + F(n-1). Is this something he hasn't come across, because this is mathematically exact, no rounding involved, and you get the fibonacci sequence from powers of the golden ratio, when you simplify it down.
Getting the numbers from rounding seems more magical, than making a decomposition. There is actually an exact decomposition of the lucas numbers to show that why the rounding gives these numbers: L_n= ɸ^n + (- ɸ)^(-n). But you're right, that doesn't mean the lucas numbers are superior. Since they both have a similar property and Matt actually didn't mention L_0 is approximated by ɸ^1, while L_1 equals ɸ^0.
ok then tell me what the 100th number in the Fibonacci sequence is, using your formula. Then tell me the 100th Lucas number is using the method from the video.
hitobite Although that is interesting stuff, I still think of ɸ^n giving fibonacci numbers is superior just because it's derived more simply. Fibonacci numbers seem more like the base line you go from for these types of series. I prefer Fibonacci numbers, but I get someone would prefer Lucas numbers.
Ow, my last comment was meant for Brandon Kekahuna. Bricks Of Awesome I would have to agree with Matt that the Fibonacci numbers are a bit overrated. I wouldn't say that the Lucas numbers are better.
Ah, but Matt, I can one-up you. Recall that φ possesses the curious property that φ^2 = φ + 1. With that in mind, let's look at some powers of φ. φ^1 = 1φ φ^2 = 1φ + 1 φ^3 = φ*φ^2 = φ(φ + 1) = φ^2 + φ = 2φ + 1 φ^4 = 3φ + 2 φ^5 = 5φ + 3 φ^6 = 8φ + 5 And so on. It is clear that φ^n = fib(n)φ + fib(n - 1). I think this is a much more intimate connection than the one Matt displayed with the Lucas numbers.
Hey Brady. I really like what you did with these newest three videos, how you broke them up and put a fun thumbnail with each. I could totally see all three of these as one long video, but I think that separating them allows for you to think about each concept/topic and process what you just watched before going onto the next one. Maybe do this in future videos. Anyway, keep making great videos.
It seems that Lucas numbers get closer and closer to Phi^n as they go.... 3 - phi^2 ~= 0.4 4 - phi^3 ~= 0.2 7 - phi^4 ~= 0.15 11 - phi^5 ~= 0.1 So I assume the "error" will get smaller and smaller.... Which means that after a few interactions, this sequence can estimate with incredibly good precision the "n" power of phi. It's amazing since the sequence itself is made by adding natural numbers. The sum of natural numbers is being used to calculate the n-power or an irrational number. ---------- Edit: On the other hand, that's probably just a coincidence. If what I said at the beginning of this comment is true, then phi^n must have a tendency to become a natural number as it goes to infinity. That doesn't seem true, but I haven't tried it out.
Johnny Luken I guess lack of self-esteem took a hold of me on that comment. Since this wasn't mentioned in the video I ended up assuming I was speaking non-sense. Thankfully, hitobite showed up and proved that I was right, which is nice.
hm hm. Let me clarify something. we have a function f(n) = f(n-1) + f(n-2). This function is defined by its first two values, any other value is dependent on those two alone. Mathematically spoken, this function has two degrees of freedom. Now, this function has special solutions in the form of a^n: the two roots of a^2 - a - 1 = 0 (as substitution in the definition tells us that a^(n+2) = a^(n+1) + a^n). But those two numbers are exactly phi and -1/phi (which equals 1-phi). This means that phi^n is a contestant for f, as is (1-phi)^n. More interestingly, any linear combination a*phi^n + b*(1-phi)^n will satisfy the conditions on f. And this combination has two degrees of freedom as well. Furthermore, you can always solve f(0) and f(1) to be your preferred starting numbers, like f(0) = 2308 and f(1) = 4261. This equation is always solvable, and the solution will always yield a function that satisfies both your initial condition as the function definition, so it is an explicit form of your function f. If we set f(0) = 2 and f(1) = 1, then a = 1 and b = -1, which proofs that: phi^n - (1-phi)^n are exactly the Lucas numbers. Now, as hitobite stated: 1. f(n) is an integer, by induction. 2. (1-phi)^n < 0.5 for n > 1 so phi^n rounded to the nearest integer indeed equals the n-th Lucas number. EDIT: this also proves that for any number q = a+sqrt(b), q^n 'has the tendency to become an integer when n goes to infinity', AND you can calculate these integers with recursion... if and only if abs(a - sqrt(b)) < 1 So you want to calculate q = 2+sqrt(7)? this number is a solution of x^2 - 4x - 3 = 0. Also, 2 - sqrt(7) > 2 - sqrt(9) = -1, so abs(q) < 1 Now we create the recursive function for which q^n satisfies the condition, which is: f(n) = 4f(n-1) + 3f(n-2) We can start with any two values we want, as the secondary term, (2-sqrt(7))^n, will go to 0. So we start with a decent approximation: f(0) = 1 f(1) = 4 so we get: 1, 4,19,88,409,1900,... And after a quick calculation, I estimate 2 + sqrt(7) to be 1900/409 = 4,645477 2 + sqrt(7) is actually equal to 4,645751, which means my answer is to 3 digits correct.
You are correct. Let phi be the golden ratio. One can show that x(n)=A*phi^n+B*(-1/phi)^n solves x(n) = x(n-1)+x(n-2) For Lucas sequence we have the system of equations: x(0)=A+B=2 x(1)=A*phi+B*(-1/phi)=1..................... Solve that as exercice :) The solution is A=B=1, so x(n)=phi^n+(1/phi)^n. As phi>1, when n goes to infinite, (-1/phi)^n goes to zero and x(n) goes to phi^n as you said
The amount that it misses by for the nth Lucas number (where 2 is the 0th one, of course), is exactly equal to the conjugate golden ratio ((1-sqrt(5))/2) to the nth power. So to exactly get the nth Lucas number, you add the nth powers of both ratios.
Phi, the golden ratio, has other interesting properties. It can be used to calculate the exact value of pi, and pi can be used to calculate the exact value of e. pi = 5 * arccos (phi / 2) e = (-1) ^ (1 / [pi * i]) To calculate e, you must use a calculator that can deal with imaginary numbers, because e is the "pi i" root of negative one, where 3.1415927... is the coefficient of i. Just search "(-1) ^ (1 / [pi * i])" on Google. You'll get e exactly. Google's search engine is a mighty fine calculator itself. You can prove that these equations are true by checking to see if you get zero after taking the calculated value of pi and subtracting the "real" pi from it. The same process can be done with e. The Google calculator is precise to the 308th digit, so it's pretty safe to say that these equations are 100% accurate.
+Jordan Fischer But Pi = 6 * arcsin(1/2) and arcsin and arccos are more easily calculated for inputs near zero. Tenth root of e = 1+1/10+1/(100*2!)+1/(1000*3!)+1/(10000*4!)+1/(100000*5!)+.....1/(10^n*n!) so n begins at 0. Converges faster than the usual factorial sum, then raise to the tenth power.
No calculator can calculate those numbers exactly because calculators generate rational numbers, while those numbers are all irrational. Your idea of a proof is also awful, because it doesn't matter how accurate the approximation is, it's still just an approximation. You need to use calculus which doesn't convert the numbers into rational approximations in order to actually prove the formulas.
You know what he meant, calculators can get within 0.000000001% of the exact answer. Given the infinite series I mentioned, though, I wonder why he chose to use complex numbers to derive e when he didn't need to.
Cooper Gates Because the formula using complex numbers is a simply expressed, finite formula. It's not an approximation, it's an identity. It's better to use complex numbers than to use an infinite series for those reasons.
I know this destroys the beautifulness of the rounding and the Lucas numbers, but here it is anyway. Using the quadratic equation used to define phi (which I'll write as p to be readable), you get that p² = p+1. Multiplying this by p^(n-2), a real number, you will get that for all n, p^n = p^(n-1)+p^(n-2). Therefore, starting a sequence with 1 and p, you will get the sequence 1, p, p², p³, ... The ratio of these numbers is always the golden ratio, and so you don't even need to round the powers of phi to get a sequence that satisfies the criterion that the ratio converges to infinity. In fact, it can be seen as the "best series" because it converges to the golden ratio instantly. I guess rounding makes the numbers more everyday and down-to-earth, but in my opinion, the unrounded sequence is still the better one...
I'm going to invent my own sequence so I can be known for something. How about you start with 1. And then you repeat the number 1. So start with 1 and then the second number is 1 and the third number is another 1 and so on and you just keep repeating that for infinity. What do I win?
Also that sequence is ζ(0) so it's sum is technically 1/2. Yes, 1/2=1+1+1+1+1+1+1+1... Cheeky, though objectionable proof. (Better ones exist) c=1+1+1+1+1+1... 2c= 2 +2 +2... -c=1-1+1-1+1-1... -c= 1-1+1-1+1... -2c=1 c=-1/2
I am curious: do different Fibonacci-type series approach phi at different rates? If so, what's the fastest known? Is it possible to construct any other kind of series that will do it faster?
0:18 - it just occured to me: Fibonacci was really named Bonacci, so his nickname is Fi-Bonaccci, you can spell it Phi-Bonacci or even Φ-Bonacci. He has Golden Ratio in his name!
I'm not going to defend the Fibonacci numbers as 'better' than the Lucas numbers, but I was curious as to why they appear in nature, and whether you could derive them from the golden ratio. So I set up a spreadsheet to try and find the best rational approximations to the golden ratio. On each line I increased the denominator by one, then found the 'best' numerator (by rounding) and compared the resulting ratio with the golden ratio. Then I went down the list (as far as the denominator being 60) and picked out any ratio which was more accurate than the previous best. This is what I found: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55
Although it could be considered "cheating" since it uses real numbers instead of integers, but the sequence starting with (1, phi) is so beautiful. 1 + phi = phi^2, so the next number is phi^2. Then, you have the same thing, just multiplied by phi, so you get phi^3 and so on... In the end, you have 1, phi, phi^2, phi^3, ... and the Nth number is phi^N (considering 1 is the "0th" number). It's like Lucas numbers, but without rounding !
By extension, if you start the sequence 1, Φ... you get the exact powers of Φ: 1, Φ, Φ^2, Φ^3, Φ^4 And this goes backward, if you start with any two consecutive powers of Φ in order (e.g. Φ^-10, Φ^-9) it will generate all the powers exactly.
Nebch12 Not sure what you're asking... pi is easily rounded off - practically every real world engineering application of pi uses an approximation. Same with zero - if you've lost your job and the mortgage is due, you may well say "I don't have any money" when there's actually 73 cents in your checking account... i is a little different. i would be difficult to round off to, or approximate. In that respect, it's more a number property than an actual quantity, isn't it? Taken out of its context as a factor of a negative square, what distinguishes one of them as an imaginary number? A seven could be a diameter or a circumference or a hypotenuse, but without the context to define its purpose, one couldn't tell which one a given instance of seven was...
A more compelling argument for the Lucas numbers, rather than rounding (which seems hand-wavey), is to consider that as the exponent increases, the difference between the result and the corresponding Lucas number integer approaches zero. phi^13=521.0019, for instance, and each successive power is even closer to a Lucas number.
FUN FACT: If you take the roots of the quadratic equation x^2=x+1 and call them a and b. Then, *nth Lucas number = a^n + b^n* For example, 0th Lucas number = a^0+b^0 = 1+1=2 1st Lucas number = a^1+b^1 = 1 2nd Lucas number = a^2+b^2 = 3 and so on... Also, one of the roots of that equation is infact the golden ratio. EDIT: I just realised that I accidentally proved why each Lucas number rounds off to phi^n As one of the roots is phi, let a=phi. So, b= 1-phi which is approx -0.6 So, if we want to find each Lucas number, say 4th Lucas number = phi^4 + (-0.6)^4 So, phi^4 = 4th Lucas number - (-0.6)^4 But, as (-0.6)^4 is definitely less than 0.5 and as the 4th Lucas number is definitely a natural number, removing (-0.6)^4 from the 4th Lucas number is not going to decrease it below 0.5 less than the number itself. Meaning, that if we happen to round up this difference, we will end up with the 4th Lucas Number. This means that phi^4 rounded up gives the 4th Lucas Number. Infact, phi^n will round off to the nth Lucas Number on one condition: (-0.6)^n must be less than 0.5, which will be true for all n>1. So, phi^n will have to round off to the nth Lucas Number. Also, the proof of nth Lucas Number = a^n + b^n is pretty easy, using Induction. All you have to do is find a+b=1 and a^2+b^2=3, then, show that as x^2=x+1, x^(n+2) = x^(n+1) + x^n [Multiplying by n] As a and b satisfy this equation, we get: a^(n+2) + b^(n+2) = a^(n+1)+b^(n+1) + a^n+b^n This is the exact way how we generate Lucas numbers; by adding consecutive terms to get the next term. So, this sequence is clearly the Lucas numbers.
I think it's also interesting to note that both the Fibonacci and Lucas numbers can be expressed exactly in terms of the Golden ratio without any need for rounding.
As much as it sticks in my craw to agree with old Matt that the Lucas numbers relate more to the Golden Ratio than the Fibonacci numbers do, surely an old pro like him can do better than rounding up/down to fit to an already known sequence. So lets help him out! Let A=GR and B=GR-1 or 1/GR ( they are both the same). Then A^1-B^1=1 : A^2+B^2=3 : A^3-B^3=4 : A^4+B^4=7 : A^5-B^5=11 : A^6+B^6=18 & so on & so forth. Voila the Lucas numbers which are the difference of the ODD powers and the SUM of the EVEN ones. What about a link between the triangular & the Fibonacci/Lucas numbers.
Hello once again. Since you've taken the trouble to ask I'm more than pleased to tell you. Take the two consecutive triangular numbers 21 and 28. Expand them as a Fibonacci series to give 49, 77, 126, 203, .... . Divide by 7 to give 3, 4, 7, 11, 18, 29, .... the Lucas Numders alias a Fibonacci derivative. Is it of any great significance? I doubt it but presumably the two curves are simultaneous at the points 21 & 28.
Ah, because 21 and 28 happen to be multiples of 7. Another consecutive pair would be 36 and 45, both divisible by 9, and dividing up gives the members 4 and 5 in 3, 1, 4, 5, 9, 14, 23, ... (less precise for phi approximation than Lucas or Fibonacci) Moving further into the triangular numbers yields lower ratios between consecutive terms unless some are skipped, same goes for square numbers. It just so happens that 5, 2, 7, 9, 16, 25, 41, 66, 107, ... contains three consecutive square numbers (hint: Use Pythagorean triples if you want more cases), and that is probably the such sequence with the smallest starting terms. It also gives a fun joke because 5, 2, 7, ... eventually reaches 999801, 1617712. Is phi roughly 1.617712? Not really, much closer to 1617712 / 999801. Shift the decimal point and divide.
+Cooper Gates. Actually old Matt is misguided. The golden ratio relates equally to both the Fibonacci & Lucas numbers, the powers of the GR being calculated in terms of quantities defined by two consecutive terms of the fibonnaci sequence.
Since phi = (1 + Sqrt(5)) / 2, it doesn't have to be derived from such a sequence, and to make phi's powers tend toward the Fibonaccis instead of Lucas numbers, multiply each by (5 + Sqrt(5)) / 10.
I think one answer to Brady's "isn't rounding kind of imprecise" complaint is to think to the previous video and remember that the "nth term/(n-1)th term" definition of the golden ratio we started from is already the _limit as n approaches infinity_ and that if you divide any two finite adjacent terms you will at best get an approximation of φ (one that gets better the bigger those two terms are); similarly the powers of φ more and more closely approximate the lucas numbers as the powers get higher (and start pretty dang close)...
The Fibonacci sequence actually has another link to the golden ratio: you can calculate the Fibonacci numbers using this formula: F(n)=(φ^n-(-φ)^(-n))/sqrt(5) where F(n) is the nth Fibonacci number and φ is the "major" golden ratio (1.6180...). This formula will work so that for every integer value of n, Fn will be an integer (even for negative numbers). When n increases, (-φ)^(-n) will converge to 0, which also leads to the fact that the ratio F(n)/F(n-1) converges to φ.
+gewoonWILLEM Every sequence of this form also has a similar formula. a(n+1) = ( ( c + b(φ)^1 )*φ^n + ( c - bφ )*(-φ)^n ) / sqrt(5) Where b and c are the starting values of your sequence. In fact, using this you find that the formula for the Lucas numbers is even simpler than that for the fibonacci: a(n+1) = φ^n + (-φ)^(-n).
You can get similar rounding results if you generalize the rule to X_n = A * X_n-1 + X_n-2 for some integer A. Using the same technique as in the proof video, the generalized golden ratio values will be of the form (A + sqrt(A^2 + 4)) / 2. The familiar golden ratio comes from the special case A = 1. But if you try A = 2 for example, you get a ratio of 1 + sqrt(2). The powers of this value round to 2, 6, 14, 34, 82, 198, 478, ..., which are the values of a Lucas-like sequence in which double the previous term is added to the term before it. (The ratios of consecutive terms do converge to 1 + sqrt(2) as expected.) For A = 3, you get a ratio of (3 + sqrt(13)) / 2. Once again, the powers of this value round to 3, 11, 36, 119, 393, 1298, 4287, ... , another valid sequence (tripling this time before adding the term before it). In fact, the first two terms are always A, A^2 + 2. (Except the rounding doesn't work for the first term when A = 1. Hmm.) Anyway, I don't know why this happens, I'm just playing around in Excel.
The golden ratio, written as a continued fraction (see wiki) reads ϕ = [1;1,1,1,1,1,1,1,1,1,1,1,…]. Its rational approximations are 1/1, 3/2, 5/3, 8/5, ... which consist of the Fibonacci numbers. So the two ARE closely related. That's why both appear in nature.
The Golden Fibonacci Numbers: The exact sequence for φ^n would be starting with the numbers 1,φ: 1, φ, (1+φ), (1+2φ), (2+3φ), (3+5φ), (5+8φ), (8+13φ), ..., (ƒ + [ƒ+2]φ) And when one does this the coefficients (ƒ) are the Fibonacci sequence.
Love your channel! It seems to me however that these numbers are not "superior" because of the power property, as the power property of the Lucas numbers is a necessary consequence of the fact that the quotient between two subsequent numbers in any sequence in the "Fibonacci family" (don't what is the real name) converges to the golden ratio as n goes to infinity. You can see it already in the video, where the errors are decreasing as n increases. Therefore, the nth term of a Fibonacci sequence with any starting terms can eventually be written pretty accurately as c*phi^n (where phi is the Golden Ratio and provided n is big enough). In case of the classic Fibonacci sequence, this c is equal to about 0.7243606... (according to my excel sheet)! It's just convenient beauty that the Lucas numbers have a c equal to 1. That is to say - they are both equally beautiful! No discrimination in mathematics madafakkaaa. Unless you're solving quadratic equations of course.
It's well known that you get from the Fibonacci to the Lucas sequence by swapping round 1 and 2 in the Fibonacci 1 2 3 5 8 13 . . . so you get the Lucas 2 1 3 4 7 11. . . if you keep to the same addition rule. What's not so obvious is that the 4 in this new, Lucas, sequence, doesn't just coincidentally equal 2(1x2). Try swapping round another consecutive Fibonacci pair, say 2 and 3 to get 3 2 5 7 12 19 . . . and note that 2(3x2) = 12. Note too that 12 is four along from 3 in the new sequence, and 3 is at index 4 in the original Fibonacci, (and the other way round for 2). Swap Fibonaccis 5 and 3 round and see you get 2(5x3) = 30 in the new sequence, and so on. Can anyone express this identity elegantly?
4 is the 4th Lucas number, but 12 is the 5th number in the next sequence you used. You also squared the 1st Lucas number but squared the second term in your second sequence (the 2, not the 3). What about 3, 1, 4, 5, 9, 14, 23, 37, 60, ...? It just so happens that 3*4*5 = 60. The next two terms are prime (97 and 157).
Shall we call this the Gates sequence? In this case it isn't a pair of immediately consecutive Fibonacci terms that are swapped round but a pair of terms, 3 and 1, separated by another, and which are then brought together as a consecutive pair to be added. This produces some interesting results too, especially if we bear in mind that the term to the left is -2. We then have two terms, -2 and 3, whose squares occur later on in the sequence, namely 4 and 9. Do this next-but-one-swap with other terms and we get the same result, for example the numbers 5 and 13, usually separated by 8 in the Fibonacci sequence. We get -8 13 5 18 23 41 64 105 169 379 . . . , in which -8 and 13 lead to their squares.
pi in base 706245 is (using the same principal as hexadecimal) is 3."99999" . allow "99999" to be equivalent to a single digit that equals that number in that base system. this is a wonderfully accurate representation of pi with minimal memorization
If phi is the golden ratio and phiconj = (1-sqrt(5))/2 = (1-phi) =-1/phi, The sequence is given exactly with no rounding by: L_n = phi^(n-1) + phiconj^(n-1) In fact any sequence that is generated by U_{n+1}= U_{n} +U_{n-1} can be expressed as U_n = a*phi^n + b*phiconj^n; where a and be are constants determined by the first two terms in the sequence. In this case a=1/phi and b = phi.
+Kyle W Or just zero-index the sequence. In other words, L_0 = 2, and then L_1 = 1 and L_2 = 3 will follow naturally. That closed-form formula also shows why the rounding thing works, as the phiconj^n part will vanish as n gets large (since |phiconj| < 1).
This video made me think about that "Any numbers when summed give you the golden ratio" And I tried to find the "magic / logic" behind it. Here is the result, //SPOILERS// it isn't surprising //SPOILERS END// U(1) = a U(2) = b U(3) = a + b U(4) = a + 2b U(5) = 2a + 3b U(6) = 3a + 5b … U(11) = 34a + 55b U(12) = 55a + 89b So, when you do the algebra (or "four basic operations"), somehow, anynumbers will give you Fibaonacci sequence back, so you will have golden ratio again. (Yes, it works with also Lucas Numbers, too.)
Yeah, that's the direction I went, too, but then I saw the Matt's actual proof, which doesn't involve any of that at all. It just uses X(n) = X(n-1) + X(n-2). Golden Proof - Numberphile
So I have the "Math Metal" song about phi that Phil Moriarty did with Dave from Boyinaband on my iPod and I quite enjoy it. I never really pay much attention to Phi, but when Matt started writing the digits I instantly found that in my head I could list them along without trying. So there we go, music does make things easier to remember.
I have discovered a cool explanation of why the Lucas series begins 2,1 , it will sound slightly strange to begin but I'm sure it makes sense))). The 2 means that in this series each number is the sum of the 2 previous numbers, And the 1 is 1 less than 2 to the power of 1. I have discovered some other Ratios associated with series' where each number is the sum of its 3,4,5 or 6 predecessors, when we take the powers of these ratios which extend towards 2, we have an equivalent Lucas series for each ratio. So the powers of the third ratio (1.839286755) round towards the 3,1,3 series, the 3 means each number is the sum of its 3 predecessors and the 1 and 3 are 1 less than 2 to the 1 and 2 squared. The 4th Ratio rounds to the 4,1,3,7, series then 5,1,3,7,15 and 6,1,3,7,15,31.My spreadsheet wont allow me any further so my progress has slowed somewhat, I'm not sure if the tenth powers rounds to the series beginning 1 or 10. This is another reason why the Lucas series is King as it ties these other ratios together with phi.
Love you answer at the end because really mother earth & the multiversal laws, seasons, cycles, planet orbit etc. etc. aren't exactly persice hence ovals over circles. At least thats our understanding. Man likes things to be to linear with a beginning and an end when we live in a cyclical world where the end is the beginning and the beginning is the end the its a spectrum and an almosty fuzzy point, kinda like rounding has to be chosen! THANK YOU for this video we have only recently discovered a Lucas Sequence and more accurately the Lucas Numbers and are wanting to learn more on the comparisons and or maybe better worded how these and the Fibonacci and be used together. Blessings
If you look at the successively higher powers of phi, you notice that the difference between the value and the nearest integer grows smaller, and in fact the difference is a power of the inverse of phi. Lucas numbers follow the formula phi^n + (-phi^(-n)).
Speaking of series, how about a video explaining the property of the addition of consecutive odd integers equaling squares (2^2=4=1+3, 3^2=1+3+5=9, 4^2=1+3+5+7=16 etc.) Never understood why this behaves like this.
Another interesting observation - the decimals for the powers are the same as the decimals for each multiple-1. ie. look at the decimals for 2phi and phi^3, 3phi and phi^4, and so on
I noticed, that as we raise phi to higher powers, the outcome is getting closer and closer to whole numbers, while other numbers multiplied by themselves seem to give pretty random decimal expansions. Why's that? Is phi the only number behaving this way?
Well, if you want Xn = φ^n, then choose X0 = 1, and X1 = φ. Since φ² = φ + 1, then φ^n = φ^(n-1) + φ^(n-2), and this way we indeed have Xn = X(n-1) + X(n-2), and Xn = φ^n.
I wouldn't call Lucas numbers "vastly superior", because if you divide by Sqrt[5] (e.g. Round[phi^n/Sqrt[5]] ) you get the Fibonacci numbers out instead. Though it may be not as elegant "vastly", I think, is a bit harsh.
rounding is just a new flavor of mathematics! it would be nice to see more videos on the 'laziness' of mathematics and the interesting things that pop out! (using the term 'laziness' loosely)!
Anyway, we round to get phi in Fibonacci too. 1/1 is 1, 2/1 is 2, 3/2 is 1.5, 5/3 is 1.666... (and it starts to look closer to phi), and so on. Only with n that tends to infinity we can say that Fn/Fn-1 = phi...
Two things: 1. The Fibonacci numbers are the nearest numbers to the golden ratio, because every fraction of two consecutive Fibonacci numbers is nearest to the golden ratio in form 1+1/1; 1+1/(1 + 1/1); 1+1/(1 + 1/(1+ 1/1)); ... 2. Not the special Lucas numbers 2, 1, 3, 4, 7, 11, 18, ... are the very interesting. Much more interesting are the common Lucas sequences U(P,Q) and V(P,Q) where U_n(1,-1) are the Fibonacci numbers and V_n(1,-1) are the special Lucas numbers. By the way, the Lucas Sequences U_n(a+1,a) represent the sequence a^n - 1 and V_n(a+1,a) represent the sequence a^n + 1. The last is intereseting, because if n is prime then V_n(a+1,a) - V_1 is divisible by n, which is the same as if n is prime, then a^n - a is divisble by n, which leads to a generalization: if n is a prime, then V_n(P,Q) - V_1(P,Q) is divisible by n.
Interesting comment on exactness in mathematics. I certainly prefer nice, exact quantites (the square root of 2 is just that; the number that, when multiplied by itself, equals 2, not some fractional estimate), but I can see there are things to be gained from looking at things in a more estimate-y way...maybe.
pi in base 706245 is (using the same principal as hexadecimal) is 3."99999" . allow "99999" to be equivalent to a single digit that equals that number in that base system
The more you feature Matt, the more I've gotten comfortable with his stye (which is cynical but often amusing). He's a good asset to the Numberphile team.
So, if the Lucas numbers are 1, 3, 4, 7, etc, when he says that φ^2 rounds up to 3, φ^3 rounds up to 4, φ^3 rounds up to 7, etc to conclude that the powers of φ are (rounded) the same as the Lucas numbers, he ignores the exception of φ itself. φ=φ^1=1.61... which would have to be rounded up to 2, which is not equal to 1, the first Lucas number. He should just have mentioned that this rule only doesn't apply for the first Lucas number. For the rest, I find this amazing, especially the fact that φ^n approaches the n-th Lucas number, more and more closely as n gets larger. When you think of this as follows: φ=1/2+1/2(sqrt(5)). Then φ^n, which has a lot of terms containing sqrt(5), becomes an integer, when n goes to infinity. Isn't that beautiful?
Why not having the Lucas Sequence start with "1, 3" instead of "2, 1"? I think that would make it nicer, because then it were monotonically increasing PLUS had the property "L(n) = round(phi^(n))" ("L(n)" being the n-th Lucas Number).
There is a little mistake in the video, as it shouldn't say that Lucas numbers are an approximation to ( ( 1 + SQRT( 5 ) ) / 2 )^n. Instead, it should have said that there is an equivalent way to build them which is, exactly: Ln = ( ( 1 + SQRT( 5 ) ) / 2 )^n + ( ( 1 - SQRT( 5 ) ) / 2 )^n. The second term, by the way, is the second solution to the golden equation: phi + 1 = phi ^ 2. I guess that solves the discussion about the unnecessary rounding.
Actually, I have found some rather interesting connection between the Fibonacci numbers and Phi. I am too lazy to write out how I got to this, so I'll just write the general form: (Phi^n = Fn*Phi + F(n-1 For any positive integer n, where Fn represents the nth Fibonacci number, and F(n-1) represents the Fibonacci number before the nth. And, well, I mean, someone most likelyfound that a long ago, but whatever.
Rounding powers of Phi [the golden ratio] to get the Lucas numbers, as shown in this video, must work for any base, like binary or tertiary representation of numbers, and not just for numbers represented as decimals (base 10). Perhaps a simpler statement of this result would be: The integer closest to (Phi)^N is the N'th Lucas number.
It is a well known fact that Fibonacci numbers arise from continious fraction decomposition of golden ratio thus their ratios are the best possible approximations of golden ratio. (any other rational number with denominator less or equal to corresponding Fibonacci number is further from golden ration)
Loved the video(s)! I was wondering if you guys could do a video on the Collatz Conjecture (3n+1 conjecture). I read about it online but it didn't give too much information in the article.
Yeah to me thats the best one. All over sequences are getting closer and closer to the golden ratio but this one already got it and keeps it forever. ok technically there are infinite amounts of this sequence too (just multiply the first and second number with anything) but 1 is a classy start.
i wrote the Lucas numbers on accident from (x) phibonatchi - (x-4) phibonatchi. where x is the how far along the sequence. and counting back. another nice one i found is 6,3,9,12,21,33,54,87,141,228,369 witch is (x) phibonatchi - (x-8)
The most important part of this video starts at 4:20....this is proof that nature "rounds"....I hadn't considered this before but it is an eye opener...
Hi, I enjoyed those three videos. It would have been good to mention that these calculations are amazingly easy to carry out using a spreadsheet using the 'copy down' function. Excel can also do the rounding as well. Keep up the good work! Mark Leach
A thought about the "rounding" issue. Couldn't you say that phi^n and the lucas sequence are each the 'limit' of the other? As phi^n continues, the rounding adjustment gets smaller and smaller approaching 0, The ratio of the lucas sequence, as it continues, aproaches phi with increasing accuracy. (adjustment approaches 0) I don't have time to calculate it right now, but it would be intersting to compare the degree of accuracy for each process at each step. A) is the adjustment the same at each step? B) if not, is there a formula that generalizes the difference? heading off to work... please have answer by 5:30pm EST ;-)
Hi folks, I have a question for you mathematically literate people. Inspired by videos like this I started looking at numbers that are exponents of natural numbers which are also "palindromic" (the same backward as forward). The first ones that caught my eye were these four powers of 11. 11^1 = 11 11^2 = 121 11^3 = 1331 11^4 = 14641 This particular pattern does not continue, as 11^5 = 161051. However there are a few more which are easy enough to find: 7^3 = 343 22^2 = 484 26^2 = 676 Are these types of numbers a thing, and if so, what are they called and where could I find out more about them?
Now meet the Lunar numbers, so called since they bear the same relationship to the sequence Narayana's Cows (OEIS 00930) as the Lucas numbers do to the Fibonacci, which is presented in the vid. The NC sequence goes 1 1 1 2 3 4 6 9 13 19 28 . . . and satisfies the recurrence formula a(n) = a(n-1) + a(n-3), meaning that you don't add two successive terms to get the next as in Fibonacci but add two terms that are one apart, eg 3+6=9, 4+9=13, etc. In the absence of any precise defining formula that I know of, the best approximation for the ratio constant between each term in this sequence and its immediate predecessor (call it the Bovine Ratio or by the Greek letter Moo) that I've been able to come up with is 1.465571232. Moo^2 = 2.147899036, rounding to 2. Moo^3 = 3.147899036, rounding to 3, (note in passing how (Moo^2) + 1 = Moo^3). Continuing this raising to successive powers and rounding we get the sequence 1 2 3 5 7 10 15 21 31 46 67 . . . which just about obeys the same NC recurrence formula stated above. (I say just about since 7+15 doesn't equal 21, maybe this anomaly is due to some imprecision in the value I got for the Bovine Ratio). This parallels the way Lucas numbers relate to the Fibonacci numbers and to the Golden Ratio as explained in the video.
+Chris G Lucas got his famous numbers by tweaking the Fibonacci sequence. Simply by reversing a pair, 1 2, to get 2 1 and then continuing 3 4 7 11 18 29 47 etc, he got an improved sequence (according to the video) which maps more closely one-to-one onto a sequence of Phi^2 Phi^3 Phi^4 etc. I was looking for an equivalent way to do this for the Narayana's Cows sequence, which originally goes 1 1 1 2 3 4 6 9 13 19 28 41 etc. How can we tweak that to correspond more closely to successive powers of what I've termed the Bovine Ratio? I think I've found the answer: start it with 3 1 1 and continue according to the same recurrence a(n) = a(n-1)+a(n-3): 4 5 6 10 15 21 31 46 67 98 etc. In general I conjecture we can "Lucasate" any sequence with a recurrence of the form a(n)=a(n-1)+a(n-m) by restarting it with m followed by (m-1) 1's. Thus where m=4, we give it the Lucas treatment by restarting it with 4 1 1 1. A kind of snipping out and reverse transcription.
Original Fibonacci sequence also has a unique property - if we take convergents of phi's continued fraction, we would get exactly the ratios of Fibonacci sequence terms.
I know this is kinda old, but as you raise Phi to higher powers, it gets extremely close to whole numbers. like by the time you raise it to the 25th power or something it's about within 0.000003 of a whole number
There is relation of phi and fibonacci number Just use the same formaula as Lucas number and divide by sqrt(5) before round it and you will get fibonacci
4:31 brings back memories of when I did maths in school and would add as many digits to endless decimal numbers as would fit on my calculator's screen, my flawless logic being "more numbers = more maths = superior maths; Q.E.D. I am the best at maths."
I always round to two decimal places it feels right. Not so imprecise to lose its value, but not so precise that it takes up too much space. I never round on certain in-between steps, like converting grams to moles, I round at the end. It’s like if π was rounded to 3.1, diameters greater than 5 units in length would lose precision, but 3.141592 is just unnecessary.
Great series of videos, and I would love to see a proof of this. Do the powers of phi more closely approach the corresponding terms of the Lucas sequence as it gets higher, or do they diverge?
The exact formula for Lucas numbers is phi^n + psi^n where psi = 1 - phi = -1/phi. The "psi^n" term will tend to 0 as n increases, therefore phi^n will be closer and closer to a whole Lucas number. To derive the formula you can start with the definition: f(n) = f(n-1) + f(n-2) and get it from there with methods I barely understand how to use, and definitely not enough to explain them.
In defense of rounding: At the low end of these sequences (whether Lucas or Fibonacci numbers), their relationships are NEAR the golden ratio. You have to round to get TO the golden ratio. (Because they APPROACH the golden ratio over time, not straight away.) So to go backwards FROM the golden ratio TO whole numbers, it will OF COURSE require rounding at the lowest numbers. But the rounding will be less and less necessary as the numbers get larger. I find the Lucas numbers to be a fascinating alternative. Not at all a Parker Square of a video.
![Megan Thee Stallion - Bigger In Texas [Official Video]](http://i.ytimg.com/vi/QxTkZTLWdo4/mqdefault.jpg)
-Unnecessary precision.
-Prints a mile of pi.
Prints a mile of pi.
Thus illustrating necessary precision. :-)
martinshoosterman Don’t forget “pointless and over the top”
The rounding thing comes from an exact relation -- look up Binet formulas.
??
So, to get a Lucas number, you have to take the Parker Square of φ. Interesting.
142857 is a parker 1/7th.
142857x7=999999
"There's some amazing things that drop out of rounding" like .14159265 ...
this comment is underrated.
Klobi for President I’m confused but u want to understand the joke
"roundind", like a circle, and pi goes 3.14159265
Parker Maths
@@cularre9544thank you...... now I might remember thoose last three numbers lol
phi^1=phi
phi^2=phi+1
phi^3=2phi+1
phi^4=3phi+2
phi^5=5phi+3
phi^6=8phi+5
...
I think the Fibonacci sequence definetly is very much linked to phi actually...
hahahahahahahahaha would you dare say so? XD
Wow, this is a great finding actually
shrdlu But it was already discovered?
He did a new video about this property after someone pointed it to him on Reddit.
Thanks, that is very interesting. BTW, the Lucas numbers also have a precise identity:
L(n) = phi^n + (1-phi)^n
and the Fibonacci numbers have a similar one:
F(n) = (phi^n - (1-phi)^n) / sqrt(5)
There is also a direct identity:
L(n) = F(n+1) + F(n-1)
Oh, and finally, from the closed form of F(n) above, follows: F(n) = round(phi^n / sqrt(5)) because the "error term" (1-phi)^n / sqrt 5 is always less than 1/2.
Side comment: how funny to see that the visual mark of the channel is the type of paper used. Rather unique I have to say.
Brown paper™
Right lol
Matt, you've done it again
I've fallen in love with yet another series of numbers
Look at how satisfied he is when he says "Golden Ratio". This guy loves maths.
no, he said the fibonacci sequence was over rated Martin.
he says he's a massive sceptic of the golden ratio, watch it again.
@@martinshoosterman u
??
in fact if you use the golden ratio again to write phi^n you can find Fibonacci again. Nothing spectacular about Lucas Numbers:
phi^1= phi + 0
phi^2= phi + 1
phi^3= 2phi + 1
phi^4= 3phi + 2
phi^5= 5phi + 3
...
Flewn not bad bro
Same goes for any fibonacci style sequence, write out p subscript n for terms a and b(where a is the first term, like 1 in the fibonacci sequence, and b is the second term) and you find p subscript n=x (subscript n) *a+x (subscript n+1)*b when the starting series is a,b,p subscript 1, p subscript 2,... and x subscript n refers to the nth number of the fibonacci sequence
I too noticed the same thing ; was just about to comment bout it
Phi^6 = 8phi + 5
4:12 In defense of rounding: "People link mathematics with unnecessary precision ... with things that are pointless and over-the-top."
What, like printing out a million places of pi?
(ducks)
Actually 5 trillion digits
Well you gotta have some fun from time to time.
There was a reason for finding so many places of pi--they wanted to know whether or not the decimal sequence ever repeated, whether pi was truly irrational. And now we know that it is indeed irrational. That wasn't done just to nerdily rattle off as many decimal places as possible.
Dylan Rambow
We can devise algorithms for numbers that could never fit into this universe. To determine if pi is transcendental, one must make a smart proof. No amount of computing power will create a mathematical proof.
I am a lowly stats 1 student, but given that we bounce from decimal place to decimal place depending on the problem, I like his essential notion that after a certain point, we just don't give a flying f---.
Call P=(1+sqrt(5))/2 and Q=(1-sqrt(5))/2, the two solutions of x^2-x-1=0. The Lucas sequence is simply P^n+Q^n, which happens to be all integers. Since Q is between -1 and 1, Q^n converges to 0, and that's why rounding works (for n>=2, that is).
I thought the Lucas numbers were: 4, 5, 6, 1, 2, 3, 7 !?
What?
+Nicolino Will .Star Wars!
4, 5, 6, 7
What? Is that a factorial in the end?
Martin Josip Kocijan 8 and 9 haven't been made yet.
"I'm going to say L-oo-cas instead of L-oo-ca because I'm lazy" - Matt logic
Nice to see you Gauss.
I'm going to say it in English, rather than read it in English, realise its a french word, translate it and say it in French.
Speaking words in your native language is definitely lazier.
I wonder how Matt chooses to say the loanword in English from French "foyer" as well, lol...
Immediately calls him l-oo-ca after saying that.
@@DeathBringer769 longer ago
I would argue against lucas numbers being superior, and here's why.
If you take ɸ^1 you get ɸ+0, if you take ɸ^2 you get ɸ+1, now if you take ɸ^3, if you do some algebra given that ɸ^2 = ɸ+1, you get 2ɸ+1. ɸ^4 is 3ɸ+2, ɸ^5 is 5ɸ+3, and so on. It is very clear that ɸ^n = ɸF(n) + F(n-1). Is this something he hasn't come across, because this is mathematically exact, no rounding involved, and you get the fibonacci sequence from powers of the golden ratio, when you simplify it down.
Getting the numbers from rounding seems more magical, than making a decomposition.
There is actually an exact decomposition of the lucas numbers to show that why the rounding gives these numbers: L_n= ɸ^n + (- ɸ)^(-n).
But you're right, that doesn't mean the lucas numbers are superior. Since they both have a similar property and Matt actually didn't mention L_0 is approximated by ɸ^1, while L_1 equals ɸ^0.
ok then tell me what the 100th number in the Fibonacci sequence is, using your formula.
Then tell me the 100th Lucas number is using the method from the video.
Bricks Of Awesome To be fair, the Lucas number show up in the decomposition of (2ɸ+1)*ɸ^n, while rounding ɸ^n/sqrt(5) gives the Fibonacci numbers.
hitobite Although that is interesting stuff, I still think of ɸ^n giving fibonacci numbers is superior just because it's derived more simply.
Fibonacci numbers seem more like the base line you go from for these types of series. I prefer Fibonacci numbers, but I get someone would prefer Lucas numbers.
Ow, my last comment was meant for Brandon Kekahuna.
Bricks Of Awesome I would have to agree with Matt that the Fibonacci numbers are a bit overrated. I wouldn't say that the Lucas numbers are better.
Ah, but Matt, I can one-up you.
Recall that φ possesses the curious property that φ^2 = φ + 1. With that in mind, let's look at some powers of φ.
φ^1 = 1φ
φ^2 = 1φ + 1
φ^3 = φ*φ^2 = φ(φ + 1) = φ^2 + φ = 2φ + 1
φ^4 = 3φ + 2
φ^5 = 5φ + 3
φ^6 = 8φ + 5
And so on. It is clear that φ^n = fib(n)φ + fib(n - 1). I think this is a much more intimate connection than the one Matt displayed with the Lucas numbers.
Hey Brady. I really like what you did with these newest three videos, how you broke them up and put a fun thumbnail with each. I could totally see all three of these as one long video, but I think that separating them allows for you to think about each concept/topic and process what you just watched before going onto the next one. Maybe do this in future videos. Anyway, keep making great videos.
It seems that Lucas numbers get closer and closer to Phi^n as they go....
3 - phi^2 ~= 0.4
4 - phi^3 ~= 0.2
7 - phi^4 ~= 0.15
11 - phi^5 ~= 0.1
So I assume the "error" will get smaller and smaller.... Which means that after a few interactions, this sequence can estimate with incredibly good precision the "n" power of phi.
It's amazing since the sequence itself is made by adding natural numbers.
The sum of natural numbers is being used to calculate the n-power or an irrational number.
----------
Edit:
On the other hand, that's probably just a coincidence.
If what I said at the beginning of this comment is true, then phi^n must have a tendency to become a natural number as it goes to infinity.
That doesn't seem true, but I haven't tried it out.
L_n = phi^n + (-phi)^(-n), the last term goes to zero.
So yes, phi^n goes closer and closer to a natural number.
Johnny Luken
I guess lack of self-esteem took a hold of me on that comment.
Since this wasn't mentioned in the video I ended up assuming I was speaking non-sense.
Thankfully, hitobite showed up and proved that I was right, which is nice.
hm hm. Let me clarify something.
we have a function f(n) = f(n-1) + f(n-2). This function is defined by its first two values, any other value is dependent on those two alone. Mathematically spoken, this function has two degrees of freedom.
Now, this function has special solutions in the form of a^n: the two roots of a^2 - a - 1 = 0 (as substitution in the definition tells us that a^(n+2) = a^(n+1) + a^n). But those two numbers are exactly phi and -1/phi (which equals 1-phi).
This means that phi^n is a contestant for f, as is (1-phi)^n. More interestingly, any linear combination a*phi^n + b*(1-phi)^n will satisfy the conditions on f. And this combination has two degrees of freedom as well.
Furthermore, you can always solve f(0) and f(1) to be your preferred starting numbers, like f(0) = 2308 and f(1) = 4261. This equation is always solvable, and the solution will always yield a function that satisfies both your initial condition as the function definition, so it is an explicit form of your function f.
If we set f(0) = 2 and f(1) = 1, then a = 1 and b = -1, which proofs that:
phi^n - (1-phi)^n are exactly the Lucas numbers.
Now, as hitobite stated:
1. f(n) is an integer, by induction.
2. (1-phi)^n < 0.5 for n > 1
so phi^n rounded to the nearest integer indeed equals the n-th Lucas number.
EDIT: this also proves that for any number q = a+sqrt(b), q^n 'has the tendency to become an integer when n goes to infinity', AND you can calculate these integers with recursion...
if and only if abs(a - sqrt(b)) < 1
So you want to calculate q = 2+sqrt(7)? this number is a solution of x^2 - 4x - 3 = 0.
Also, 2 - sqrt(7) > 2 - sqrt(9) = -1, so abs(q) < 1
Now we create the recursive function for which q^n satisfies the condition, which is:
f(n) = 4f(n-1) + 3f(n-2)
We can start with any two values we want, as the secondary term, (2-sqrt(7))^n, will go to 0. So we start with a decent approximation:
f(0) = 1
f(1) = 4
so we get:
1, 4,19,88,409,1900,...
And after a quick calculation, I estimate 2 + sqrt(7) to be 1900/409 = 4,645477
2 + sqrt(7) is actually equal to 4,645751, which means my answer is to 3 digits correct.
The reason is that the n=1th lucas number is given by φ^n + (-φ)^-n and as n approaches infinity (-φ)^-n approaches 0 so φ^n approaches a lucas number
You are correct.
Let phi be the golden ratio. One can show that x(n)=A*phi^n+B*(-1/phi)^n solves x(n) = x(n-1)+x(n-2)
For Lucas sequence we have the system of equations:
x(0)=A+B=2
x(1)=A*phi+B*(-1/phi)=1..................... Solve that as exercice :)
The solution is A=B=1, so x(n)=phi^n+(1/phi)^n. As phi>1, when n goes to infinite, (-1/phi)^n goes to zero and x(n) goes to phi^n as you said
Check out my Scratch programs! scratch.mit.edu/users/dinosaurwill08/
The amount that it misses by for the nth Lucas number (where 2 is the 0th one, of course), is exactly equal to the conjugate golden ratio ((1-sqrt(5))/2) to the nth power. So to exactly get the nth Lucas number, you add the nth powers of both ratios.
Phi, the golden ratio, has other interesting properties. It can be used to calculate the exact value of pi, and pi can be used to calculate the exact value of e.
pi = 5 * arccos (phi / 2)
e = (-1) ^ (1 / [pi * i])
To calculate e, you must use a calculator that can deal with imaginary numbers, because e is the "pi i" root of negative one, where 3.1415927... is the coefficient of i.
Just search "(-1) ^ (1 / [pi * i])" on Google. You'll get e exactly. Google's search engine is a mighty fine calculator itself.
You can prove that these equations are true by checking to see if you get zero after taking the calculated value of pi and subtracting the "real" pi from it. The same process can be done with e. The Google calculator is precise to the 308th digit, so it's pretty safe to say that these equations are 100% accurate.
+Jordan Fischer But Pi = 6 * arcsin(1/2) and arcsin and arccos are more easily calculated for inputs near zero.
Tenth root of e = 1+1/10+1/(100*2!)+1/(1000*3!)+1/(10000*4!)+1/(100000*5!)+.....1/(10^n*n!) so n begins at 0. Converges faster than the usual factorial sum, then raise to the tenth power.
That last sentence though...
No calculator can calculate those numbers exactly because calculators generate rational numbers, while those numbers are all irrational. Your idea of a proof is also awful, because it doesn't matter how accurate the approximation is, it's still just an approximation. You need to use calculus which doesn't convert the numbers into rational approximations in order to actually prove the formulas.
You know what he meant, calculators can get within 0.000000001% of the exact answer.
Given the infinite series I mentioned, though, I wonder why
he chose to use complex numbers to derive e when he didn't need to.
Cooper Gates Because the formula using complex numbers is a simply expressed, finite formula. It's not an approximation, it's an identity. It's better to use complex numbers than to use an infinite series for those reasons.
I know this destroys the beautifulness of the rounding and the Lucas numbers, but here it is anyway. Using the quadratic equation used to define phi (which I'll write as p to be readable), you get that p² = p+1. Multiplying this by p^(n-2), a real number, you will get that for all n, p^n = p^(n-1)+p^(n-2). Therefore, starting a sequence with 1 and p, you will get the sequence 1, p, p², p³, ... The ratio of these numbers is always the golden ratio, and so you don't even need to round the powers of phi to get a sequence that satisfies the criterion that the ratio converges to infinity. In fact, it can be seen as the "best series" because it converges to the golden ratio instantly. I guess rounding makes the numbers more everyday and down-to-earth, but in my opinion, the unrounded sequence is still the better one...
I'm going to invent my own sequence so I can be known for something. How about you start with 1. And then you repeat the number 1. So start with 1 and then the second number is 1 and the third number is another 1 and so on and you just keep repeating that for infinity. What do I win?
No research has been done - not that it has to be done - to get that 'sequence', so no, Kevin. You are not getting anything.
Nooooooooooooooo!
7 minute abs
someone beat you to it -- that's sequence A000012 in the OEIS :(
Also that sequence is ζ(0) so it's sum is technically 1/2.
Yes, 1/2=1+1+1+1+1+1+1+1...
Cheeky, though objectionable proof.
(Better ones exist)
c=1+1+1+1+1+1...
2c= 2 +2 +2...
-c=1-1+1-1+1-1...
-c= 1-1+1-1+1...
-2c=1
c=-1/2
i love matt, think he's by far the best numberphile 'guest'
I like that if Phi is the Golden ratio, then the sequence starting:
1, 1 + Phi, 2 + Phi, 3 + 2 Phi, ...
Is just a sequence of the powers of Phi.
I am curious: do different Fibonacci-type series approach phi at different rates? If so, what's the fastest known? Is it possible to construct any other kind of series that will do it faster?
0:18 - it just occured to me: Fibonacci was really named Bonacci, so his nickname is Fi-Bonaccci, you can spell it Phi-Bonacci or even Φ-Bonacci. He has Golden Ratio in his name!
I'm not going to defend the Fibonacci numbers as 'better' than the Lucas numbers, but I was curious as to why they appear in nature, and whether you could derive them from the golden ratio.
So I set up a spreadsheet to try and find the best rational approximations to the golden ratio. On each line I increased the denominator by one, then found the 'best' numerator (by rounding) and compared the resulting ratio with the golden ratio. Then I went down the list (as far as the denominator being 60) and picked out any ratio which was more accurate than the previous best. This is what I found:
2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55
This is an awesome video! Golden Ratio is one of my favorite areas of mathematics.
Although it could be considered "cheating" since it uses real numbers instead of integers, but the sequence starting with (1, phi) is so beautiful. 1 + phi = phi^2, so the next number is phi^2. Then, you have the same thing, just multiplied by phi, so you get phi^3 and so on...
In the end, you have 1, phi, phi^2, phi^3, ... and the Nth number is phi^N (considering 1 is the "0th" number).
It's like Lucas numbers, but without rounding !
By extension, if you start the sequence 1, Φ... you get the exact powers of Φ:
1, Φ, Φ^2, Φ^3, Φ^4
And this goes backward, if you start with any two consecutive powers of Φ in order (e.g. Φ^-10, Φ^-9) it will generate all the powers exactly.
An Engineer knows that EVERY number has already been rounded...
What about 0? What about pi? What about i?
Nebch12
Not sure what you're asking... pi is easily rounded off - practically every real world engineering application of pi uses an approximation. Same with zero - if you've lost your job and the mortgage is due, you may well say "I don't have any money" when there's actually 73 cents in your checking account...
i is a little different. i would be difficult to round off to, or approximate. In that respect, it's more a number property than an actual quantity, isn't it? Taken out of its context as a factor of a negative square, what distinguishes one of them as an imaginary number? A seven could be a diameter or a circumference or a hypotenuse, but without the context to define its purpose, one couldn't tell which one a given instance of seven was...
ThePeaceableKingdom I'm not an engineer and yet I got the joke better than most guys in this comment section^^
TheTck90
Thanks!
mwbhome e
Thanks! I've never heard that one!
A more compelling argument for the Lucas numbers, rather than rounding (which seems hand-wavey), is to consider that as the exponent increases, the difference between the result and the corresponding Lucas number integer approaches zero. phi^13=521.0019, for instance, and each successive power is even closer to a Lucas number.
FUN FACT:
If you take the roots of the quadratic equation
x^2=x+1
and call them a and b.
Then,
*nth Lucas number = a^n + b^n*
For example,
0th Lucas number = a^0+b^0 = 1+1=2
1st Lucas number = a^1+b^1 = 1
2nd Lucas number = a^2+b^2 = 3 and so on...
Also, one of the roots of that equation is infact the golden ratio.
EDIT:
I just realised that I accidentally proved why each Lucas number rounds off to phi^n
As one of the roots is phi, let a=phi.
So, b= 1-phi which is approx -0.6
So, if we want to find each Lucas number,
say 4th Lucas number = phi^4 + (-0.6)^4
So, phi^4 = 4th Lucas number - (-0.6)^4
But, as (-0.6)^4 is definitely less than 0.5 and as the 4th Lucas number is definitely a natural number, removing (-0.6)^4 from the 4th Lucas number is not going to decrease it below 0.5 less than the number itself. Meaning, that if we happen to round up this difference, we will end up with the 4th Lucas Number.
This means that phi^4 rounded up gives the 4th Lucas Number.
Infact, phi^n will round off to the nth Lucas Number on one condition: (-0.6)^n must be less than 0.5, which will be true for all n>1.
So, phi^n will have to round off to the nth Lucas Number.
Also, the proof of nth Lucas Number = a^n + b^n is pretty easy, using Induction.
All you have to do is find a+b=1 and a^2+b^2=3, then,
show that as x^2=x+1,
x^(n+2) = x^(n+1) + x^n [Multiplying by n]
As a and b satisfy this equation, we get:
a^(n+2) + b^(n+2) = a^(n+1)+b^(n+1) + a^n+b^n
This is the exact way how we generate Lucas numbers; by adding consecutive terms to get the next term. So, this sequence is clearly the Lucas numbers.
"almosty" is pretty much my favorite word now. Thanks Brady.
I think it's also interesting to note that both the Fibonacci and Lucas numbers can be expressed exactly in terms of the Golden ratio without any need for rounding.
As much as it sticks in my craw to agree with old Matt that the Lucas numbers relate more to the Golden Ratio than the Fibonacci numbers do, surely an old pro like him can do better than rounding up/down to fit to an already known sequence. So lets help him out! Let A=GR and B=GR-1 or 1/GR ( they are both the same). Then A^1-B^1=1 : A^2+B^2=3 : A^3-B^3=4 : A^4+B^4=7 : A^5-B^5=11 : A^6+B^6=18 & so on & so forth. Voila the Lucas numbers which are the difference of the ODD powers and the SUM of the EVEN ones. What about a link between the triangular & the Fibonacci/Lucas numbers.
Is the last sentence a tease, since triangular numbers
grow quadratically and the golden ratio sequences
exponentially?
Hello once again. Since you've taken the trouble to ask I'm more than pleased to tell you. Take the two consecutive triangular numbers 21 and 28. Expand them as a Fibonacci series to give 49, 77, 126, 203, .... . Divide by 7 to give 3, 4, 7, 11, 18, 29, .... the Lucas Numders alias a Fibonacci derivative. Is it of any great significance? I doubt it but presumably the two curves are simultaneous at the points 21 & 28.
Ah, because 21 and 28 happen to be multiples of 7. Another consecutive pair would be 36 and 45, both divisible by 9, and dividing up gives the members 4 and 5 in
3, 1, 4, 5, 9, 14, 23, ... (less precise for phi approximation than Lucas
or Fibonacci)
Moving further into the triangular numbers yields lower ratios between
consecutive terms unless some are skipped, same goes for square numbers.
It just so happens that 5, 2, 7, 9, 16, 25, 41, 66, 107, ... contains
three consecutive square numbers (hint: Use Pythagorean triples
if you want more cases), and that is probably the such sequence
with the smallest starting terms.
It also gives a fun joke because 5, 2, 7, ... eventually reaches
999801, 1617712.
Is phi roughly 1.617712? Not really, much closer to
1617712 / 999801. Shift the decimal point and divide.
+Cooper Gates. Actually old Matt is misguided. The golden ratio relates equally to both the Fibonacci & Lucas numbers, the powers of the GR being calculated in terms of quantities defined by two consecutive terms of the fibonnaci sequence.
Since phi = (1 + Sqrt(5)) / 2, it doesn't have to be derived from such a sequence, and to make phi's powers tend toward the Fibonaccis instead of Lucas numbers, multiply each by (5 + Sqrt(5)) / 10.
I think one answer to Brady's "isn't rounding kind of imprecise" complaint is to think to the previous video and remember that the "nth term/(n-1)th term" definition of the golden ratio we started from is already the _limit as n approaches infinity_ and that if you divide any two finite adjacent terms you will at best get an approximation of φ (one that gets better the bigger those two terms are); similarly the powers of φ more and more closely approximate the lucas numbers as the powers get higher (and start pretty dang close)...
The Fibonacci sequence actually has another link to the golden ratio: you can calculate the Fibonacci numbers using this formula:
F(n)=(φ^n-(-φ)^(-n))/sqrt(5)
where F(n) is the nth Fibonacci number and φ is the "major" golden ratio (1.6180...). This formula will work so that for every integer value of n, Fn will be an integer (even for negative numbers).
When n increases, (-φ)^(-n) will converge to 0, which also leads to the fact that the ratio F(n)/F(n-1) converges to φ.
+gewoonWILLEM
Every sequence of this form also has a similar formula.
a(n+1) = ( ( c + b(φ)^1 )*φ^n + ( c - bφ )*(-φ)^n ) / sqrt(5)
Where b and c are the starting values of your sequence. In fact, using this you find that the formula for the Lucas numbers is even simpler than that for the fibonacci:
a(n+1) = φ^n + (-φ)^(-n).
You can get similar rounding results if you generalize the rule to X_n = A * X_n-1 + X_n-2 for some integer A.
Using the same technique as in the proof video, the generalized golden ratio values will be of the form (A + sqrt(A^2 + 4)) / 2. The familiar golden ratio comes from the special case A = 1.
But if you try A = 2 for example, you get a ratio of 1 + sqrt(2). The powers of this value round to 2, 6, 14, 34, 82, 198, 478, ..., which are the values of a Lucas-like sequence in which double the previous term is added to the term before it. (The ratios of consecutive terms do converge to 1 + sqrt(2) as expected.)
For A = 3, you get a ratio of (3 + sqrt(13)) / 2. Once again, the powers of this value round to 3, 11, 36, 119, 393, 1298, 4287, ... , another valid sequence (tripling this time before adding the term before it).
In fact, the first two terms are always A, A^2 + 2. (Except the rounding doesn't work for the first term when A = 1. Hmm.)
Anyway, I don't know why this happens, I'm just playing around in Excel.
The golden ratio, written as a continued fraction (see wiki) reads ϕ = [1;1,1,1,1,1,1,1,1,1,1,1,…].
Its rational approximations are 1/1, 3/2, 5/3, 8/5, ... which consist of the Fibonacci numbers. So the two ARE closely related. That's why both appear in nature.
Great series of videos, definitely worth doing a series of videos on rounding!
The Golden Fibonacci Numbers:
The exact sequence for φ^n would be starting with the numbers 1,φ:
1, φ, (1+φ), (1+2φ), (2+3φ), (3+5φ), (5+8φ), (8+13φ), ..., (ƒ + [ƒ+2]φ)
And when one does this the coefficients (ƒ) are the Fibonacci sequence.
Or, start the sequence with 1 and Φ. 1 + Φ = Φ^2. Φ + Φ^2 = Φ^3...
Love your channel! It seems to me however that these numbers are not "superior" because of the power property, as the power property of the Lucas numbers is a necessary consequence of the fact that the quotient between two subsequent numbers in any sequence in the "Fibonacci family" (don't what is the real name) converges to the golden ratio as n goes to infinity. You can see it already in the video, where the errors are decreasing as n increases. Therefore, the nth term of a Fibonacci sequence with any starting terms can eventually be written pretty accurately as c*phi^n (where phi is the Golden Ratio and provided n is big enough). In case of the classic Fibonacci sequence, this c is equal to about 0.7243606... (according to my excel sheet)! It's just convenient beauty that the Lucas numbers have a c equal to 1.
That is to say - they are both equally beautiful! No discrimination in mathematics madafakkaaa. Unless you're solving quadratic equations of course.
I love his completely off-the-cuff defense of imprecision through rounding.
It's well known that you get from the Fibonacci to the Lucas sequence by swapping round 1 and 2 in the Fibonacci 1 2 3 5 8 13 . . . so you get the Lucas 2 1 3 4 7 11. . . if you keep to the same addition rule. What's not so obvious is that the 4 in this new, Lucas, sequence, doesn't just coincidentally equal 2(1x2). Try swapping round another consecutive Fibonacci pair, say 2 and 3 to get 3 2 5 7 12 19 . . . and note that 2(3x2) = 12. Note too that 12 is four along from 3 in the new sequence, and 3 is at index 4 in the original Fibonacci, (and the other way round for 2). Swap Fibonaccis 5 and 3 round and see you get 2(5x3) = 30 in the new sequence, and so on. Can anyone express this identity elegantly?
4 is the 4th Lucas number, but 12 is the 5th number in the next sequence you used. You also squared the 1st Lucas number but squared the second term
in your second sequence (the 2, not the 3).
What about 3, 1, 4, 5, 9, 14, 23, 37, 60, ...? It just so happens that 3*4*5 = 60.
The next two terms are prime (97 and 157).
Shall we call this the Gates sequence? In this case it isn't a pair of immediately consecutive Fibonacci terms that are swapped round but a pair of terms, 3 and 1, separated by another, and which are then brought together as a consecutive pair to be added. This produces some interesting results too, especially if we bear in mind that the term to the left is -2. We then have two terms, -2 and 3, whose squares occur later on in the sequence, namely 4 and 9. Do this next-but-one-swap with other terms and we get the same result, for example the numbers 5 and 13, usually separated by 8 in the Fibonacci sequence. We get -8 13 5 18 23 41 64 105 169 379 . . . , in which -8 and 13 lead to their squares.
pi in base 706245 is (using the same principal as hexadecimal) is 3."99999" . allow "99999" to be equivalent to a single digit that equals that number in that base system. this is a wonderfully accurate representation of pi with minimal memorization
If phi is the golden ratio and phiconj = (1-sqrt(5))/2 = (1-phi) =-1/phi,
The sequence is given exactly with no rounding by:
L_n = phi^(n-1) + phiconj^(n-1)
In fact any sequence that is generated by U_{n+1}= U_{n} +U_{n-1} can be expressed as U_n = a*phi^n + b*phiconj^n; where a and be are constants determined by the first two terms in the sequence.
In this case a=1/phi and b = phi.
If we start the series with 1, 3, then L_n = phi^n + phiconj^n which to me seems like a nicer result
+Kyle W Or just zero-index the sequence. In other words, L_0 = 2, and then L_1 = 1 and L_2 = 3 will follow naturally.
That closed-form formula also shows why the rounding thing works, as the phiconj^n part will vanish as n gets large (since |phiconj| < 1).
This video made me think about that "Any numbers when summed give you the golden ratio"
And I tried to find the "magic / logic" behind it.
Here is the result, //SPOILERS// it isn't surprising //SPOILERS END//
U(1) = a
U(2) = b
U(3) = a + b
U(4) = a + 2b
U(5) = 2a + 3b
U(6) = 3a + 5b
…
U(11) = 34a + 55b
U(12) = 55a + 89b
So, when you do the algebra (or "four basic operations"), somehow, anynumbers will give you Fibaonacci sequence back, so you will have golden ratio again. (Yes, it works with also Lucas Numbers, too.)
Yeah, that's the direction I went, too, but then I saw the Matt's actual proof, which doesn't involve any of that at all. It just uses X(n) = X(n-1) + X(n-2).
Golden Proof - Numberphile
This
ZipplyZane I know, what they shown as X there is what I shown here as U. The other part is just an algebraic example of the explanation. :)
Thanks for sharing! I knew the Lucas numbers, but that they link to Phi powers was new for me. Amazing properties...
So I have the "Math Metal" song about phi that Phil Moriarty did with Dave from Boyinaband on my iPod and I quite enjoy it. I never really pay much attention to Phi, but when Matt started writing the digits I instantly found that in my head I could list them along without trying. So there we go, music does make things easier to remember.
vids like these are why i love this channel!!!
I have discovered a cool explanation of why the Lucas series begins 2,1 , it will sound slightly strange to begin but I'm sure it makes sense))). The 2 means that in this series each number is the sum of the 2 previous numbers, And the 1 is 1 less than 2 to the power of 1. I have discovered some other Ratios associated with series' where each number is the sum of its 3,4,5 or 6 predecessors, when we take the powers of these ratios which extend towards 2, we have an equivalent Lucas series for each ratio. So the powers of the third ratio (1.839286755) round towards the 3,1,3 series, the 3 means each number is the sum of its 3 predecessors and the 1 and 3 are 1 less than 2 to the 1 and 2 squared. The 4th Ratio rounds to the 4,1,3,7, series then 5,1,3,7,15 and 6,1,3,7,15,31.My spreadsheet wont allow me any further so my progress has slowed somewhat, I'm not sure if the tenth powers rounds to the series beginning 1 or 10. This is another reason why the Lucas series is King as it ties these other ratios together with phi.
Love you answer at the end because really mother earth & the multiversal laws, seasons, cycles, planet orbit etc. etc. aren't exactly persice hence ovals over circles. At least thats our understanding. Man likes things to be to linear with a beginning and an end when we live in a cyclical world where the end is the beginning and the beginning is the end the its a spectrum and an almosty fuzzy point, kinda like rounding has to be chosen! THANK YOU for this video we have only recently discovered a Lucas Sequence and more accurately the Lucas Numbers and are wanting to learn more on the comparisons and or maybe better worded how these and the Fibonacci and be used together. Blessings
If you look at the successively higher powers of phi, you notice that the difference between the value and the nearest integer grows smaller, and in fact the difference is a power of the inverse of phi. Lucas numbers follow the formula phi^n + (-phi^(-n)).
Speaking of series, how about a video explaining the property of the addition of consecutive odd integers equaling squares (2^2=4=1+3, 3^2=1+3+5=9, 4^2=1+3+5+7=16 etc.) Never understood why this behaves like this.
Another interesting observation - the decimals for the powers are the same as the decimals for each multiple-1. ie. look at the decimals for 2phi and phi^3, 3phi and phi^4, and so on
I noticed, that as we raise phi to higher powers, the outcome is getting closer and closer to whole numbers, while other numbers multiplied by themselves seem to give pretty random decimal expansions. Why's that? Is phi the only number behaving this way?
It’s fitting that Matt’s Golden Ratio Trilogy shown in the end screen has 4 videos. 😊
The Lucas Numbers are quite interesting. My first book on the Fibonacci Numbers was actually a booklet on The Fibonacci and Lucas Numbers.
Well, if you want Xn = φ^n, then choose X0 = 1, and X1 = φ. Since φ² = φ + 1, then φ^n = φ^(n-1) + φ^(n-2), and this way we indeed have Xn = X(n-1) + X(n-2), and Xn = φ^n.
I wouldn't call Lucas numbers "vastly superior", because if you divide by Sqrt[5] (e.g. Round[phi^n/Sqrt[5]] ) you get the Fibonacci numbers out instead. Though it may be not as elegant "vastly", I think, is a bit harsh.
rounding is just a new flavor of mathematics!
it would be nice to see more videos on the 'laziness' of mathematics and the interesting things that pop out!
(using the term 'laziness' loosely)!
Anyway, we round to get phi in Fibonacci too. 1/1 is 1, 2/1 is 2, 3/2 is 1.5, 5/3 is 1.666... (and it starts to look closer to phi), and so on. Only with n that tends to infinity we can say that Fn/Fn-1 = phi...
Awesome series of vids! Clear, insightful and entertaining.
Two things:
1. The Fibonacci numbers are the nearest numbers to the golden ratio, because every fraction of two consecutive Fibonacci numbers is nearest to the golden ratio in form 1+1/1; 1+1/(1 + 1/1); 1+1/(1 + 1/(1+ 1/1)); ...
2. Not the special Lucas numbers 2, 1, 3, 4, 7, 11, 18, ... are the very interesting. Much more interesting are the common Lucas sequences U(P,Q) and V(P,Q) where U_n(1,-1) are the Fibonacci numbers and V_n(1,-1) are the special Lucas numbers.
By the way, the Lucas Sequences U_n(a+1,a) represent the sequence a^n - 1 and V_n(a+1,a) represent the sequence a^n + 1. The last is intereseting, because if n is prime then V_n(a+1,a) - V_1 is divisible by n, which is the same as if n is prime, then a^n - a is divisble by n, which leads to a generalization: if n is a prime, then V_n(P,Q) - V_1(P,Q) is divisible by n.
Interesting comment on exactness in mathematics. I certainly prefer nice, exact quantites (the square root of 2 is just that; the number that, when multiplied by itself, equals 2, not some fractional estimate), but I can see there are things to be gained from looking at things in a more estimate-y way...maybe.
pi in base 706245 is (using the same principal as hexadecimal) is 3."99999" . allow "99999" to be equivalent to a single digit that equals that number in that base system
The more you feature Matt, the more I've gotten comfortable with his stye (which is cynical but often amusing). He's a good asset to the Numberphile team.
So, if the Lucas numbers are 1, 3, 4, 7, etc, when he says that φ^2 rounds up to 3, φ^3 rounds up to 4, φ^3 rounds up to 7, etc to conclude that the powers of φ are (rounded) the same as the Lucas numbers, he ignores the exception of φ itself. φ=φ^1=1.61... which would have to be rounded up to 2, which is not equal to 1, the first Lucas number. He should just have mentioned that this rule only doesn't apply for the first Lucas number. For the rest, I find this amazing, especially the fact that φ^n approaches the n-th Lucas number, more and more closely as n gets larger. When you think of this as follows: φ=1/2+1/2(sqrt(5)). Then φ^n, which has a lot of terms containing sqrt(5), becomes an integer, when n goes to infinity. Isn't that beautiful?
Why not having the Lucas Sequence start with "1, 3" instead of "2, 1"? I think that would make it nicer, because then it were monotonically increasing PLUS had the property "L(n) = round(phi^(n))" ("L(n)" being the n-th Lucas Number).
(the latter applying for all n > 1)
There is a little mistake in the video, as it shouldn't say that Lucas numbers are an approximation to ( ( 1 + SQRT( 5 ) ) / 2 )^n. Instead, it should have said that there is an equivalent way to build them which is, exactly:
Ln = ( ( 1 + SQRT( 5 ) ) / 2 )^n + ( ( 1 - SQRT( 5 ) ) / 2 )^n.
The second term, by the way, is the second solution to the golden equation: phi + 1 = phi ^ 2. I guess that solves the discussion about the unnecessary rounding.
Actually, I have found some rather interesting connection between the Fibonacci numbers and Phi. I am too lazy to write out how I got to this, so I'll just write the general form:
(Phi^n = Fn*Phi + F(n-1
For any positive integer n, where Fn represents the nth Fibonacci number, and F(n-1) represents the Fibonacci number before the nth.
And, well, I mean, someone most likelyfound that a long ago, but whatever.
oh man "Lucas" numbers? In the last video I thought he said "Ludacris" numbers!
But I gotta, kn-kn-kn-know what-what's your fan-ta-ta-sy
Rounding powers of Phi [the golden ratio] to get the Lucas numbers, as shown in this video, must work for any base, like binary or tertiary representation of numbers, and not just for numbers represented as decimals (base 10).
Perhaps a simpler statement of this result would be: The integer closest to (Phi)^N is the N'th Lucas number.
It is a well known fact that Fibonacci numbers arise from continious fraction decomposition of golden ratio thus their ratios are the best possible approximations of golden ratio. (any other rational number with denominator less or equal to corresponding Fibonacci number is further from golden ration)
Loved the video(s)! I was wondering if you guys could do a video on the Collatz Conjecture (3n+1 conjecture). I read about it online but it didn't give too much information in the article.
Someone has probably already said this, but surely the "best" sequence is:
1, 1.618034...., 2.618034, 4.236068, 6.854102...
which matches perfectly?
Yeah to me thats the best one. All over sequences are getting closer and closer to the golden ratio but this one already got it and keeps it forever.
ok technically there are infinite amounts of this sequence too (just multiply the first and second number with anything) but 1 is a classy start.
i wrote the Lucas numbers on accident from (x) phibonatchi - (x-4) phibonatchi. where x is the how far along the sequence. and counting back. another nice one i found is 6,3,9,12,21,33,54,87,141,228,369 witch is (x) phibonatchi - (x-8)
The most important part of this video starts at 4:20....this is proof that nature "rounds"....I hadn't considered this before but it is an eye opener...
People link mathematics with unnecessary precision, says Matt : and that is exact!
Hi, I enjoyed those three videos. It would have been good to mention that these calculations are amazingly easy to carry out using a spreadsheet using the 'copy down' function. Excel can also do the rounding as well. Keep up the good work! Mark Leach
this is the most amazing thing i have ever seen
Consistently getting within rounding range of every power of the Golden Ratio is still pretty impressive.
A thought about the "rounding" issue.
Couldn't you say that phi^n and the lucas sequence are each the 'limit' of the other?
As phi^n continues, the rounding adjustment gets smaller and smaller approaching 0,
The ratio of the lucas sequence, as it continues, aproaches phi with increasing accuracy. (adjustment approaches 0)
I don't have time to calculate it right now, but it would be intersting to compare the degree of accuracy for each process at each step.
A) is the adjustment the same at each step?
B) if not, is there a formula that generalizes the difference?
heading off to work... please have answer by 5:30pm EST
;-)
Perhaps you should mention that it only works if you start at the first power, since phi^0 = 1, not 2 :P
Love you all numberphilers
Hi folks, I have a question for you mathematically literate people. Inspired by videos like this I started looking at numbers that are exponents of natural numbers which are also "palindromic" (the same backward as forward). The first ones that caught my eye were these four powers of 11.
11^1 = 11
11^2 = 121
11^3 = 1331
11^4 = 14641
This particular pattern does not continue, as 11^5 = 161051. However there are a few more which are easy enough to find:
7^3 = 343
22^2 = 484
26^2 = 676
Are these types of numbers a thing, and if so, what are they called and where could I find out more about them?
Thanks Smitty!
bw0n6 No problem. :)
Honestly I find rounding very useful
I made a formula for finding a phytagoras triple using rounding and other techniques usually considered useless
Rickson Geometry Dash
That sounds interesting! Care to elaborate?
rounding is rather important in Chemistry. Use significant figures to prevent artificial look of high precision.
Now meet the Lunar numbers, so called since they bear the same relationship to the sequence Narayana's Cows (OEIS 00930) as the Lucas numbers do to the Fibonacci, which is presented in the vid. The NC sequence goes 1 1 1 2 3 4 6 9 13 19 28 . . . and satisfies the recurrence formula a(n) = a(n-1) + a(n-3), meaning that you don't add two successive terms to get the next as in Fibonacci but add two terms that are one apart, eg 3+6=9, 4+9=13, etc. In the absence of any precise defining formula that I know of, the best approximation for the ratio constant between each term in this sequence and its immediate predecessor (call it the Bovine Ratio or by the Greek letter Moo) that I've been able to come up with is 1.465571232. Moo^2 = 2.147899036, rounding to 2. Moo^3 = 3.147899036, rounding to 3, (note in passing how (Moo^2) + 1 = Moo^3). Continuing this raising to successive powers and rounding we get the sequence 1 2 3 5 7 10 15 21 31 46 67 . . . which just about obeys the same NC recurrence formula stated above. (I say just about since 7+15 doesn't equal 21, maybe this anomaly is due to some imprecision in the value I got for the Bovine Ratio). This parallels the way Lucas numbers relate to the Fibonacci numbers and to the Golden Ratio as explained in the video.
+Chris G Lucas got his famous numbers by tweaking the Fibonacci sequence. Simply by reversing a pair, 1 2, to get 2 1 and then continuing 3 4 7 11 18 29 47 etc, he got an improved sequence (according to the video) which maps more closely one-to-one onto a sequence of Phi^2 Phi^3 Phi^4 etc. I was looking for an equivalent way to do this for the Narayana's Cows sequence, which originally goes 1 1 1 2 3 4 6 9 13 19 28 41 etc. How can we tweak that to correspond more closely to successive powers of what I've termed the Bovine Ratio? I think I've found the answer: start it with 3 1 1 and continue according to the same recurrence a(n) = a(n-1)+a(n-3): 4 5 6 10 15 21 31 46 67 98 etc. In general I conjecture we can "Lucasate" any sequence with a recurrence of the form a(n)=a(n-1)+a(n-m) by restarting it with m followed by (m-1) 1's. Thus where m=4, we give it the Lucas treatment by restarting it with 4 1 1 1. A kind of snipping out and reverse transcription.
I write the unnecessary decimal places, because I enjoy having that amount of accuracy and I also like to see how big the impact of rounding is.
Original Fibonacci sequence also has a unique property - if we take convergents of phi's continued fraction, we would get exactly the ratios of Fibonacci sequence terms.
I know this is kinda old, but as you raise Phi to higher powers, it gets extremely close to whole numbers. like by the time you raise it to the 25th power or something it's about within 0.000003 of a whole number
The rounding is what makes it applicable to Nature, and perhaps why Nature *decided* to use it.
There is relation of phi and fibonacci number
Just use the same formaula as Lucas number and divide by sqrt(5) before round it and you will get fibonacci
4:31 brings back memories of when I did maths in school and would add as many digits to endless decimal numbers as would fit on my calculator's screen, my flawless logic being "more numbers = more maths = superior maths; Q.E.D. I am the best at maths."
Lucas is pronounced "loo-cah" in French? Did he live on the second floor? Or the "
deuxième étage" in French?
Please note that this rule in Lucas Numbers stops working with 67 of the term. Take it to the top, I calculated in vain(
I always round to two decimal places it feels right. Not so imprecise to lose its value, but not so precise that it takes up too much space. I never round on certain in-between steps, like converting grams to moles, I round at the end. It’s like if π was rounded to 3.1, diameters greater than 5 units in length would lose precision, but 3.141592 is just unnecessary.
Great series of videos, and I would love to see a proof of this. Do the powers of phi more closely approach the corresponding terms of the Lucas sequence as it gets higher, or do they diverge?
The exact formula for Lucas numbers is phi^n + psi^n where psi = 1 - phi = -1/phi. The "psi^n" term will tend to 0 as n increases, therefore phi^n will be closer and closer to a whole Lucas number. To derive the formula you can start with the definition: f(n) = f(n-1) + f(n-2) and get it from there with methods I barely understand how to use, and definitely not enough to explain them.
In defense of rounding:
At the low end of these sequences (whether Lucas or Fibonacci numbers), their relationships are NEAR the golden ratio. You have to round to get TO the golden ratio. (Because they APPROACH the golden ratio over time, not straight away.) So to go backwards FROM the golden ratio TO whole numbers, it will OF COURSE require rounding at the lowest numbers. But the rounding will be less and less necessary as the numbers get larger. I find the Lucas numbers to be a fascinating alternative. Not at all a Parker Square of a video.