Can You Simplify? What do you think about this problem? If you're reading this ❤️. Hello My Friend ! Welcome to my channel. I really appreciate it! @higher_mathematics #maths #math
This is the simplest solution also derived independently
@buffalobilly6046Ай бұрын
Yes anytime you work with the “magic number” you need to know how to simplify
@oahuhawaii214129 күн бұрын
But you missed the coefficient of 2 simplifies things immensely: x³ = 2*x + 1 = √5 + 2 So, just square that twice, without having to deal with x in a symbolic manner, and substituting and simplifying yet again. x⁶ = 9 + 4*√5 x¹² = 161 + 72*√5
@ranjithamarakoon884224 күн бұрын
Nice work. ❤
@ayushdwivedi115417 күн бұрын
@@buffalobilly6046 can you tell me what is magic number?
@jpl569Ай бұрын
Another proof : noticing that (1+ √5)/2 = φ, and φ^2 = 1 + φ, we have the property φ^n = Un-2 + φ Un-1 for all n ≥ 2, where (Un) is the Fibonacci sequence. Then φ^12 = 89 + 144 φ = 161 + 72 √5. If you do not like finding the Fibonacci elements until n = 11, just take φ^6 = 5 + 8 φ, and square φ^6 : φ^12 = (φ^6)^2 = (5 + 8 φ)^2 = 89 + 144 φ = 161 + 72 √5. Thank you for your videos !
@EscviitashАй бұрын
I assume that you by Un-2 mean the (n-2)th Fibonacci number, or as it is usually written "F(n-2)" F(0) = 0 and F(1) = 1, and from there you get the rest of the Fibonacci numbers by F(n) = F(n-1) + F(n-2), and you can even go to in the negative direction by F(n) = F(n+2) - F(n+1). You will the get that F(11) = 89 and F(12) = 144. Compare this to the soltution and you will get that φ^x = F(n-1) + F(n)*φ, which you obviosusly got confused with the formula for finding the Nth Fibonacci number. A couple of examples where n is 0 or negative: Using F(n) = F(n+2) - F(n+1) you will get that F(-1) = 1, F(-2) = -1, F(-3) = 2, F(4) = -3 and so on, which is the same as the Fibonacci numbers in the positve direction, with the exception that they are negative if n is even. So if n = 0 the you get: φ^0 = F(-1) + F(0)*φ = 1 + 0*φ = 1 + 0 = 1 If n = -1 you get: φ^-1 (the reciprocal of φ) = F(-2) + F(-1)*φ = -1 + 1*φ = -1 + 1.618.. = 0.618... , which is indead the reciprocal of φ If n = -2 you get: φ^-2 (the reciprocal of φ^2) = F(-3) + F(-2)*φ = 2 + -1*φ = 2 - 1.618.. = 0.381... , which is indead the reciprocal of φ^2 This will hold true for any integer power of φ, not just for power of 2 or higher.
@jpl569Ай бұрын
@@Escviitash OK... I didn't use the usual notation for Fibonacci numbers, in order to achieve a nice and easy solution with positive integers. Thanks for your remark !
@robertveith6383Ай бұрын
@@jpl569 -- In your first post, n - 2 and n - 1 needed to be inside grouping symbols, respectively.
@jpl569Ай бұрын
@@robertveith6383 Yes, for sure, thanks !
@ffggddssАй бұрын
All good, except the index is off by one from the usual definition in parts of that. Yes, F(0)=0, F(1)=1; from which you eventually get F(11)=89, F(12)=144; but the rule for powers of phi is: φⁿ = F(n)φ + F(n-1) Fred
@46swaАй бұрын
This solution is far too cumbersome. (((Term^2)^2)^3 is much faster
@RexxSchneiderАй бұрын
You might alternately consider evaluating Term^4 = ((Term)^2)^2 as an intermediate step, then Term^8 = (Term^4)^2 and finally Term^12 = Term^8 * Term^4.
@DandoPorsaco-ho1zsАй бұрын
@@RexxSchneider That's how I did it.
@billweihmillerjr9481Ай бұрын
Or x^12= (x^4)(x^4 )(x^4). Solve x^4 once, pull the power down to 1. Square, pull power down to one, then uae the third, power down ... I liked his deliberatw approach, skipping magic ratio, phi, and avoiding the temptation to go complex when algebra,does just fine.
@billweihmillerjr9481Ай бұрын
Good problem to be VERY careful with.
@oahuhawaii214129 күн бұрын
Nope, cubing simplifies the calculation by eliminating the denominator, so 2 successive squares can be done easily. ((√5+1)/2)³ = (√5+1)³/8 = (5*√5 + 3*5 + 3*√5 + 1)/8 = (8*√5 + 16)/8 = √5 + 2 Then, just square that twice.
@jamesaitken4070Ай бұрын
I think it’s simpler just to expand out as ((((1+sqrt 5)/2)^2)^2)^3 as the expression squares out pretty simply each time
@milan.matejkaАй бұрын
Well, probably just as complicated to calculate, but you definitely don't risk a dead end.
@46swaАй бұрын
I agree with you. It's much quicker this way.
@user-gr5tx6rd4hАй бұрын
I did so and got the exact same answer.
@BotaTamas85Ай бұрын
In this particular case yes, but seeing how introducing an X for a part of the expression can simplify the problem is pretty valuable.
@benjamingrunbaum3601Ай бұрын
I also did it this way. I wouldn't have even thought of the "X"
@robertbox5399Ай бұрын
I was screaming, 'GET ON WITH IT!', Lol.
@AtanuKDey12 күн бұрын
I downvoted the video because of that. The man takes many steps to explain why 1 plus 4 is 5 by adding 1 to 1 four times to get 5. Three minutes gone.
@redotto100Ай бұрын
There’s another elegant way to solve this problem by recognizing that (1+sqrt(5))/2 is the eigenvalue of the matrix A = [0,1; 1, 1] with largest magnitude. Taking this matrix to powers produces matrices with Fibonacci numbers as entries A^p = [f(p-1), f(p); f(p), f(p+1)], so A^12 = [89, 144; 144, 233]. The largest eigenvalue of A^12 is ((1+sqrt(5))/2)^12. Simplifying the characteristic equation for A^12 gives 0 = x^2 - 322*x + 1 and computing the larger root gives the answer.
@wisdomokoro225511 күн бұрын
Just Brilliant!
@CGoldthorpe5 күн бұрын
Clearly this is more obscure, less intuitive, and less "pure"
@EscviitashАй бұрын
φ^z = F(z-1) + F(z)*φ holds true for any integer power of φ, negative, zero or positive. F(z) is the Fibonacci numbers with index z. To find the Fibonacci numbers with negative index you can reverse the formula to F(z) = F(z+2) - F(z+1) to get the sequence F(-1)=1, F(-2)=-1, F(-3)=2, F(-4)=-3, F(-5)=5, F(-6)=-8 and so on, i.e the same as the positve indexed Fibonanacci numbers, but negative if the index is even. Plug in 12 for z and you will get: φ^12 = F(12-1) + F(12)*φ = F(12-1) + F(12)*φ = F(11) + F(12)*φ = 89 + 144*φ which can be simplified to F(z-1)+F(z)/2 + (F(z)/2)*sqrt(5) if F(z) is even.
@pietergeerkens6324Ай бұрын
Such a great opportunity missed, to note in passing that [ (1 + √5) / 2 ]³ = (1 + 3√5 + 3*5 + 5√5) / 8 = (16 + 8√5) / 8 = 2 + √5; and thus give students insight into this instance (and others similar) of the Pell equation. Now the entire expression simplifies as [ (1 + √5) / 2 ]¹² = (2 + √5)⁴ = (4 + 4√5 + 5)² = 81 + 72√5 + 80 = 161 + 72√5
@robertveith6383Ай бұрын
No, it does *not* simplify to that. Go backward to some steps: [4 + 4sqrt(5) + 5]^2 = [9 + 4sqrt(5)]^2 = 81 + 72sqrt(5) + 80 = 161 + 72sqrt(5) *Answer*
@pietergeerkens6324Ай бұрын
@@robertveith6383 Oops! Thank you. I Copy-Pasted "²" twice instead of "√5". Now corrected.
@aspenrebelАй бұрын
Now that a good way to do it, start. Term to 3rd = 2 x sqrt of 5. Then sq, then sq =322
@aspenrebelАй бұрын
@@robertveith6383 that's what he did???
@oahuhawaii214129 күн бұрын
@aspenrebel: OP made an error, which someone pointed out, so OP edited his first comment to fix the error. You saw the first comment after it was corrected.
@thomasrebotier1741Ай бұрын
It's the golden ratio. That said, knowing the golden ratio and the afferent properties (x=1+1/x) is more a matter of general math culture, and olympiads are for teenagers. some of them will know and some won't at all, this is NOT the spirit of the olympiads. Those problems should be based on problem-solving ability, not on random general culture.
@koenth2359Ай бұрын
We only have to calculate φ^2, φ^4, φ^8 by consecutive squaring and then φ^12 =φ^4×φ^8. It's just four lines.
@panchostanza871213 күн бұрын
Here's what everyone is missing: the solution demonstrated would be doable by anyone who has done basic algebra
@ronaldnoll3247Ай бұрын
It is not clear to me what is simpler, the original equation or the calculated result.
@PixiltationАй бұрын
i don't think this video should be 11 minutes long
@billyoung8118Ай бұрын
I suspect his ornate way of writing "x" accounted for about 1/3 of the video length
@Philip-hv2kc29 күн бұрын
I don't see why he needs an X . You could simply do the basic arithmetic. But now I see that √5 if multiplied by itself 12 times is going to give different answers depending on original number of decimal points .
@souptikdam842415 күн бұрын
2x speed.
@xl0006 күн бұрын
this could have been a tweet... But this wouldn't be on KZbin
@golastname7686Күн бұрын
Right. Also, in a timed test, students ought to be trained to initially skip questions that take 11 minutes to answer (or that cause drowsiness). Unanswered questions might not mean the question is hard, but that students strategize time elsewhere.
@florianbasierАй бұрын
You complexified everything my friend. My idea was that we could detect a nice generalisation for Sn=x^n that would not force us to write Newton's formula with a coeff of 12. So I started looking at S1=(1+sqrt(5))/2, S2=(3+sqrt(5))/2, S3=2+sqrt(5). And now my idea of a nice formula vanished but S3 is so simple that it can be used to compute easily S6=S3²=9+4sqrt(5) and then S12=S6²=161+72sqrt(5)
@tungyeeso3637Ай бұрын
Nice and neat.
@kianooshboroojeni2551Ай бұрын
Want to have a shortcut? Fib(12) = 144. Considering the explicit form of Fib sequence, φ^12=144*sqrt(5) + ((1-sqrt(5))/2)^12 which is approximately equal to 144*sqrt(5) or 322.
@Ctrl_Alt_Sup5 күн бұрын
φ = (1+√5)/2 where φ is the golden ratio φⁿ=Fₙ₋₁+Fₙφ where (Fn) denotes the Fibonacci sequence φ¹²=F₁₁+F₁₂φ with F₁₁=89 and F₁₂=144 φ¹² =89+144((1+√5)/2)=161+72√5
@johnpool11112 күн бұрын
z = ½(1 + √5) is a root of x²-x-1, so z²=z+1. Now it is easy to calculate z^12 stepwise and while doing so replace each occurence of z² by z+1. This gives z^12=144z+89. Substituting z = ½(1 + √5) gives 161 + 72√5.
@JTBettencourtАй бұрын
A good teacher doesn’t just show a series of steps. A good teacher explains the goal and explains why each step works toward the goal. Otherwise this is just rote learning and not knowledge.
@ciprianteasca7823Ай бұрын
So, what's your point...?!
@RexxSchneiderАй бұрын
@@ciprianteasca7823 The point is that his variable 'x' is actually the golden ratio, φ. You should know that it has the property φ^2 = φ+1. That means that any power of φ can be reduced to a linear expression in φ. The simplification that provides is the key to finding an easy to evaluate expression for any power of φ, rather than the slog of evaluating a large binomial expansion with increasing powers of a variable. You can then evaluate φ^12 as ((φ^2 * φ)^2)^2 or ((φ^2)^2)^2 * (φ^2)^2, etc. as a linear expression in φ.
@echandlerАй бұрын
@@RexxSchneider *Any* simple radical expression for x can be cast as the root of some quadratic equation. φ is particularly nice!
@michaeledwards2251Ай бұрын
The point of the question is to determine whether students are aware of the significance of reducing the power demanding evaluation. This is allows programming with minimal error growth. (In this instance the Harmonic ratio and Fibonacci sequence are useful, and may be part of a bonus, but the most significant factor is spotting ways to minimize error growth in numerical calculations. )
@julianocgАй бұрын
In this case should use any other number, not the golden ratio.
@michaeledwards2251Ай бұрын
@@julianocg Why not use the golden ratio : it gives an opportunity for bonus insights.
@michaeledwards2251Ай бұрын
@@julianocg An additional factor is the real world doesn't follow neat and tidy rules : numerical evaluation is an artform whose purpose is to achieve a resolution with minimum error. It is forgotten how important numerical evaluation is to the design of cars, houses, buildings, roads and much more.
@RexxSchneiderАй бұрын
@@julianocg The whole point of picking the golden ration is that φ^2 = φ+1, which means that any power of φ can be successively reduced to a linear expression in φ. The key is that the student recognises that simplification provides a significant short-cut in this particular question.
@julianocgАй бұрын
@@RexxSchneider Just as I did.
@whycantiremainanonymous8091Ай бұрын
Without watching: (1+5^½)/2 is the constant phi (sorry, no Greek keyboard here). Phi ^ n = Fn×phi + Fn-1, where Fn is the nth number in the Fibonacci series. Moreover, this is approximately equal to Fn+1 + Fn-1. So, for power 12, this would be 233+89=322. The actual number is more like 319.99.
@literallydeadpoolАй бұрын
amazing, what a relation that is
@Ringcaat28 күн бұрын
Oh, impressive! Maybe some students knew that, but I doubt most did. I checked and it comes out right.
@jontallen331925 күн бұрын
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
@chriscurtain181629 күн бұрын
Surely the 'answer' here is simply a different way of writing the question?
@YoutubeHandlesSuckBallsКүн бұрын
Indeed.
@ald6980Ай бұрын
phi = x=(1+sqrt(5))/2; phi^n=F(n)phi+F(n-1), where F(n) - the n-th Fibonacci number: 1,1,2,3,5,8,13,21,..... F(n)=F(n-1)+F(n-2). The same another way: x^0 = 1 = 1+0*sqrt(5) = (1;0) x^1 =(0.5;0.5) x^n=(a(n);b(n)) = (a(n-1);b(n-1))+ (a(n-2);b(n-2)).
@ArtemKo___25 күн бұрын
½(1 + √5) = x x^12 = (x^2)^6 x^2 = t x^12 = (t^2)^3 and just calculate
@pythagorasaurusrex98534 күн бұрын
When you know some advanced geometry with the golden ratio, it is simple. (1+sqrt(5))/2 = phi, so phi^2=phi+1 square it: phi^4=phi^2+2phi+1=3phi+2 square again: phi^8=21phi+13 multiply last two results together: phi^12=phi^8*phi^4=(21phi+13)(3phi+2)=144phi+89=72*sqrt(5)+161. If you don't know the basic property of the golden ratio, then one will not get the first idea in this video. If you do, you can immediately start with step 2: x^2=x+1
@cecilponsaing2749Ай бұрын
A beautiful solution. I did it without x at first, carefully, and got it right.After that I could finally understand your simpler approach , and I then did it that way too. Thank you.
@paulortega5317Ай бұрын
Two Fibonacci series, each with different f(1) and f(2) values. 1,3 and 1,1. The latter can also be derived from (1/sqrt(5))*[((1+sqrt(5)/2)^n - ((1-sqrt(5)/2)^n]
@jontallen331925 күн бұрын
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
@oahuhawaii214129 күн бұрын
By factoring 12 to 3*2*2, raising to the power of 12 can be done by cubing, followed by squaring twice. Let x = √5 + 1 We have x³ = 5*√5 + 3*5 + 3*√5 + 1 = 8*√5 + 16 Thus, ((√5 + 1)/2)¹² = (((x/2)³)²)² = ((x³/8)²)² = (((8*√5 + 16)/8)²)² = ((√5 + 2)²)² = (4*√5 + 9)² = 72*√5 + 161
@samwong134910 күн бұрын
Hard because the question is incomplete. Some sort of result format specification needed. Decimals? Rational number format etc
@TakuT.28 күн бұрын
(1-√5)/2との和と積から解と係数の関係の逆からx^2-x-1=0は出せる。
@mariakoutsoulieri952528 күн бұрын
Needs calculations but is straightforward: ((1+ √5)/2)^12 = ((1+√5)x1/2))^12 = ((1+√5)x1/2))^2x6 = [(1+√5)^2)x (1/4)]^6. then we apply the (a+b)^2 formula = a^2 + 2ab + b^2... given that (a+b)^6 = (a+b)^(2x3) = ((a+b)^2)^3... we come to a point where we have... [1+√5(2+√5)]^3 x 1/(4^6).... we apply the rule (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3... the rest is pure calculations... Many calculations but no tricks!!!! straightforward answer. Finaly we multiply with 1/(4^6), basically division.
@virtual-vikingАй бұрын
It's much better to firstly calculate x^3=2+√5 because it immediately gets rid of the /2. Then find the solution from (2+√5)^4 via two successive squarings.
@oahuhawaii214129 күн бұрын
That's what I did.
@johnstanley5692Ай бұрын
Alternative? extract even and odd powers of expansion (1+x)^12 = p1(y) +x*p2(y) here x=sqrt(5), and y=x^2=5 then divide by 2^12. in this case p1(y)=y^6 + 66*y^5 + 495*y^4 + 924*y^3 + 495*y^2 + 66*y + 1 = 659456 -> 161. p2(y)=12*y^5 + 220*y^4 + 792*y^3 + 792*y^2 + 220*y + 12 =294912 -> 72=> (p1(y)+x*p2y)/2^12 = 161+72*sqrt(5)
@Bill_Woo29 күн бұрын
Small deduction for ambiguous "1" vs. "7" display. Yet a miracle occurs, that "c" and its inversion somehow produce "x", and it's clear and unambiguous. Every time I saw that I both winced and smiled.
@largepeep8710Ай бұрын
I bet the actual exam had only 5 lines to do working in lmao
@alexisgs8800Ай бұрын
I love how every overinflated ego posts a different solution using more and more advanced mathematics, criticizing his approach, but no one cared to ask who this video was made for. You don't teach how to solve such a problem the same way when your target audience is high school students as you would if it were made for people pursuing a university degree in mathematics. Some of his videos have "Entrance exam" in the title, for different universities, which implies that this may be for people who don't have all the techniques some of you are talking about.
@MiloTsukroffАй бұрын
A very elegant solution. (That is, if the year is before 1970 and no scientific calculators are available.) 161 + 72 * sqrt( 5 ) does indeed equal ( ( 1 + sqrt( 5 ) ) / 2 ) ^ 12
@Leeroy496 күн бұрын
Moivre-Binet formula for implicit Fibonacci solves :)
@JosSalinas3 күн бұрын
0:00 if you solve for the terms within the parentheses you get the golden ratio 🤩
@hippophileАй бұрын
Trying to keep it simple (to be safe) I just cubed the expression then squared it twice. Cubing gets rid of the denominator so it works out pretty nicely.
@pietergeerkens6324Ай бұрын
And a passing reference that this happens to other Pell Equation solutions as well, such as [ ( 3 + √13) / 2 ]³ = ( 27 + 3*9√13 + 3*3*13 + 13√13 ) / 8 = ( 27 + 27√13 + 117 + 13√13 ) / 8 = ( 144 + 40√13 ) / 8 = 18 + 5√13 and [ ( 5 + √21 ) / 2 ]³ = 55 + 12√21 , might bee in order, as well as that this can only (but doesn't always) happen for √(4n + 1).
@GlomgoldFlintheart13 күн бұрын
I did the same. If only 8% of students got it right, that must be bad students.
@braziliangentleman514824 күн бұрын
(1 + √5)/2 is the golden ratio, and it has some properties related to the fibonacci number. I will call this number g(for golden ration). First of all, we have g²=g+1; therefore, if we multiply g both sides, we get g³ = g² + g, and on we go. Now if we replace g², we end up with g³ = g + 1 + g, that being g³ = 2g + 1. If we do it recursively, you will notice that this expression will go like: gⁿ = Fibo(n)g + Fibo(n-1) being Fibo(k) the k-th number of the fibonacci sequence. For those who never saw this before, the fibonacci sequence is a sequence where the next term is the sum of the last 2 terms. It starts at 1. To get to the next term, we sum 1+0, because we have nothing before the starting term, then we get a 1. For the 3rd term, we sum 1+1 and get 2, being 1 and 1 the last 2 terms. For the next, we sum 1+2 and get 3, being 1 and 2 the last 2 terms, and so on. The Fibonacci sequence should look like this: Fibonacci {1, 1, 2, 3, 5, 8, 13, 21, 34, 55}. What is so important about this sequence is that if we get the ratio of 2 consecutive terms, we approach the golden ratio(named g before). now if we want to get the value of g¹², we need to get the value of Fibo(12) and Fibo(11), those values being 144 and 89. We end up with g¹² = 144g + 89. replacing (1 + √5)/2 in g, we get 161 + 72g
I' m ok with his 1's But the b are really weird. It's like a combination of two cursive styles
@johnblackledge4009Ай бұрын
8% of students got this right. 98% of the rest of humanity thought, "Why do we pay these people for this pointless gibberish?"
@DeadlyBlazeАй бұрын
98% of the rest of humanity can keep have that thought while working 996 unskilled labour till they're 70.
@donaldasayersАй бұрын
Revelling in his ignorance whilst posting using machines designed by people who understand mathematics, using mathematics.
@echandlerАй бұрын
The 8% get to respond when the 98% ask "Do you want fries with that?" The problem itself is good. This explanation dwells too long on routine algebra that someone encountering this problem should have already mastered. The good part is that the operation of raising a number to a power can be reduced in several ways. First is by nesting it as a sequence of squaring and cubing operations. Second is by using a recurrence relation. Third is that replacing x squared to a linear expression in x using *recursion*.
@stevealexander80106 күн бұрын
@@DeadlyBlaze Sorry dude - I have a math degree and there are any number of ppl who could never follow this vid yet who can buy & sell us. Bezos, Zuckerberg and Musk may be relative 'illiterati', but not "unskilled".
@amarjyoti7783Ай бұрын
Use the binomial theorem
@arekkrolak6320Ай бұрын
X is two straight lines crossing. What is this atrocity you are drawing? :)
@blacksmith67Ай бұрын
In most mathematics, x is like a sine wave from zero to 2πr with a diagonal line through it. I have seen his style of x before and it bugs me too.
@josepeixoto338429 күн бұрын
C'mon, it's so that you do not MIX it up with the x on 2x3=6
@justliberty407227 күн бұрын
@@josepeixoto3384 but nobody doing algebra ever writes 2x3...
You overcomplicated this poor little thing. Once you wrote that x²=x+1 you have: x^3=x².x=(x+1).x=x²+x=2x+1 x^6=(2x+1)²=4x²+4x+1=4.(x+1)+4x+1=8x+5 x^12=(8x+5)²=64x²+80x+25=64.(x+1)+80x+25=144x+89 So the value we are looking for is 72.(1+sqrt(5))+89=161+72.sqrt(5)
@TgfYTorseanmartinezs17 күн бұрын
Thanks for helping learn math. discover the math of reality. I stopped because I already know the answer.
@JulesMoyaert_photo28 күн бұрын
Thank you for showing me how to reach the result with all the substitutions.
@tanmaygawali-ri2nu20 күн бұрын
Smart guys unattempt question and saves time😎
@Valdagast27 күн бұрын
I don't know if I'm more impressed by the maths or the neat handwriting.
@julianocgАй бұрын
If you consider that (1+(5)^.5)/2=Phi, the golden ratio, and the relation Phi^n=Phi[n]*Phi+Phi[n-1], where Phi[n] is the n-th number of the Fibonaci sequence. Thus, Phi^12=Phi[12]*Phi+Phi[11], and finally, Phi^12=144*Phi+89.
If you happen to recognize ϕ, and know this one little trick, along with the first dozen Fibonacci numbers, you can do this one in seconds in your head: ϕⁿ = F(n)ϕ + F(n-1) In the given problem, n = 12, so we can write (½[1 + √5])¹² = ϕ¹² = F(12)ϕ + F(11) = 144ϕ + 89 = 72[1 + √5] + 89 = 161 + 72√5 And if you don't know the first dozen Fibonacci numbers, you can quickly produce them by the Fibonacci recursion rule, along with F(0) = 0, F(1) = 1: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 Fred
@jontallen331925 күн бұрын
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
@fatroberto301214 күн бұрын
Modern students will not understand why the final solution is any simpler than the initial problem. This is understandable, because the point of solving problems without a calculator is debatable. If only 8% are going to get the right answer, then why do it? They are more likely to have a calculator than a book of Mathematical Tables or a slide rule, the use of which was the object of these exercises when I was at school fifty years ago.
@jeffpudewell5884Ай бұрын
Awesome, thanks. Very helpful.
@jagjeetmule227124 күн бұрын
Isn't it simple ✓5 = 2.2360 then +1 and whole ÷ 2 and then 1.618 to the power 12 = approx 322
@Hatifnote7 күн бұрын
easy for anyone knowed or can see that x²=x+1.. Many examples in calculus are tributary to some anterior knowledge. Utility write any number as solution of some equation than try to simplify it by this equation Particular case :if there is sq root it will be root of second equation Ex: x=1-√5 (1-√5)¹⁰? (1-√5)²=1²-2*√5+5=6-2√5=4+2(1-√5) =4+2x x²=4+2x x⁴=16+16x+4x²=16+16x+4(4+2x) =32+24x x⁸=(32+24x)² =32²+2*32*24x+576 =1600+1536x x¹⁰=(1600+1536x)*(4+2x) =6400+3200x+6144+3072x =12544+6272x =12544+6272*(1-√5) =6272(2+1-√5)) =6272(1-√5) Finally x¹⁰=6272x
@knotwilg35966 күн бұрын
We know for a = phi: a²=a+1 so a³=a²+a=2a+1 = 2+ V5 a^12 = (2+ V5)^4 = (9+4V5)² = 161 + 72V5
@colinpountney3339 күн бұрын
A more elegant approach would be to iterate. After noting the expression to be raised to the power of 12 is a root of x^2 - x - 1, it follows that x^2 = x + 1. So x^3 = x^2 + x = 2x + 1. And therefore by iteration any power of x can be expressed in terms of ax + b. If x^n = ax + b then x^(n+1) = (a+b)x + a. You can do the rest in your head.
@prakharsahu764620 күн бұрын
Bro I think that You have to use binomial expression to solve this question. In India this math problem are simple . In India High school student slove this simple problem.
@elgb567119 сағат бұрын
You mean middle school, right?
@Stan_14415 күн бұрын
The worst part is: your pen is not working !
@PrypakАй бұрын
Just know that phi²=phi+1 and boom, ez solution, just develop (phi+1)², replacing phi² by phi+1, and then do it with (a*phi+b)³ and you've got it
@albertomontori2863Ай бұрын
After you arrived at (3x + 2)^3 , It would be more obvious to just do the cubic binomial expansion (a+b)^3= a^3 + b^3 + 3a^2b + 3ab^2; also the Tartaglia triangle was a viable method to do that binomial expansion!
It's perhaps easier to figure out first the relation phi^n = F_n . phi + F_{n-1} where F_n is the n-th Fibonacci number; then compute phi^12 directly.
@user-py8kl5gh2q29 күн бұрын
I just did this by repeated application of x^2, x^4, x^8 ; then x^4 * x^8, using x^2 = 1+x φ^2 = 1 + φ φ^4 = (φ^2)^2 = (1 + φ)^2 = 1 + 2φ + φ^2 = 1 + 2φ + (1 + φ) = 3φ + 2 φ^8 = (φ^4)^2 = (3φ + 2)^2 = 9φ^2 + 12φ + 4 = 9(1+φ) + 12φ + 4 = 21φ+13 φ^12 = (21φ + 13)(3φ+2) = 89 + 144φ
@amitvikramsinghvii-c79192 күн бұрын
Sir, I would like to take your views on solving (24x+13)(3x+2) Here, 24x*3= 72x By the way mind blowing lecture❤
@mehdimarashi173628 күн бұрын
Only 8% of 6th graders I assume?
@user-pt7py6hn3m13 күн бұрын
Why the equal numbers to X. Isn't it 20^6
@DanDirindon15 күн бұрын
3 sol. with x=(1+r5)/2 Common x2=(1+2r5+5)/4=(6+2r5)/4=(3+r5)/2 x4=x2*x2=(9+6r5+5)/4=(14+6r5)/4=(7+3r5)/2 x8=(49+42r5+45)/4=(94+42r5)/4=(47+21r5)/2 x12=x8*x4=(329+141r5+147r5+315)/4=(644+288r5)/4=161+72r5 Better x2=(1+2r5+5)/4=(6+2r5)/4=(3+r5)/2 x4=x2*x2=(9+6r5+5)/4=(14+6r5)/4=(7+3r5)/2 x6=x4*x2=(21+7r5+9r5+15)/4=(36+16r5)/4=9+4r5 x12=x6*x6=81+72r5+80=161+72r5 Easier x2=(1+2r5+5)/4=(6+2r5)/4=(3+r5)/2 x3=x2*x=(3+3r5+r5+5)/4=(8+4r5)/4=2+r5 x6=x3*x3=4+4r5+5=9+4r5 x12=x6*x6=81+72r5+80=161+72r5
@DanDirindon15 күн бұрын
Considering x2=x+1 does not simplify enough. x4=(x+1)2=x2+2x+1=x+1+2x+1=3x+2 x6=x4*x2=(3x+2)(x+1)=3x2+3x+2x+2=3x+3+5x+2=8x+5 x12=x6*x6=(8x+5)*2=64x2+80x+25=64x+64+80x+25=144x+89=72*(1+r5)+89=161+72r5
@661cyclist28 күн бұрын
I was always told that the number was *tau* - the Golden ratio - not phi . Often encountered in nature. But either Greek letter will do, I suppose.
@robertlunderwood20 күн бұрын
I immediately set the golden ratio to x and jumped to x^2 - x - 1 = 0 => x^2 = x + 1. Multiply both sides by x, simplify, square, simplify, square again, simplify, substitute, and solve, abusing x^2 = x + 1 whenever possible.
@Ridelto28 күн бұрын
No need to use x. Just split the numerator and denominator, then square the numerator three times getting the common factor out every time simplifying with the denominator, and finally multiply one more time the numerator by 7+3sqrt(5) to get 644/4 and (288/4)sqr(5) or 161 + 72sqr(5). That simple.
@lunsteeАй бұрын
You break down the 12th power into 2*2*3, taking x, squaring it twice, and then taking the cube. I would have taken the cube up front; x^3 = x^2*x = (x+1)*x = x^2+x = 2x+1, and then squared that twice instead. Or, observing that your x is the golden ratio φ, I would recognize that φ^n = (L(n) + F(n)*sqrt(5))/2 where L(n) and F(n) are the Lucas and Fibonacci series respectively. One can work out the series manually from the recursive definition, which for n=12 is feasible to do directly. For larger n, the methods for skipping ahead in these series are fundamentally the same as the manipulations shown in this video (multiplying two exponents).
@oahuhawaii214129 күн бұрын
Note that the cube is 2*x + 1, which eliminates the denominator, so squaring twice is easy. No need to track all the coefficients symbolically as they get large. x³ = 2*x + 1 = √5 + 2 x⁶ = 9 + 4*√5 x¹² = 161 + 72*√5
@coolbits2235Ай бұрын
I solved by factoring out 2 in step 1, this gives 2^-12 as a factor and (1+sqrt(5)) power 12 Now in each step you will get more factors that you can take out and power. Its almost the same amount of work that way.
@oahuhawaii214129 күн бұрын
The easiest way to hand simplify the expression is to cube it, then square twice. Try it to see how neat that is. You'll laugh at the narrator for taking the long and messy route.
@justinatgmail8 күн бұрын
Square it. Square that (^4). Do it one more time (^8) Multiply the 4th power and 8th power results. A little simplification and you get to the answer in about 1/4 the time.
@bogdanagrigoroaie584017 күн бұрын
(1+sqrt5)^12=(((1+sqrt5)^3)^2)^2 =((1+3sqrt5+15+5sqrt5)^2)^2 = ((16+8sqrt5)^2)^2= ((8(2+sqrt5))^2)^2 = 8^4 x(4+4sqrt5+5)^2 = (2^3)^4 x(9+4sqrt5)^2 = 2^12 x(81+72sqrt5+80) = 2^12 x(161+72sqrt5), and 2^12/2^12 = 1, and the result is 161+72sqrt5
@perakojot6524Ай бұрын
((1+sqrt(5))/2)^3 = 2+sqrt(5) (2+sqrt(5))^2=9+4*sqrt(5) (9+4*sqrt(5))^2=161+72*sqrt(5) This took me 2mins to compute.
@oahuhawaii214129 күн бұрын
Yeah, I did the same thing.
@PASHKULI10 күн бұрын
this expression is φ (The Golden ratio) raised to the 12 power. Then, φ² = 1 + φ.
@zanti4132Ай бұрын
Once you have established x² = x + 1 with x≠ 0, just multiply by x to get an equation in terms of x³: x³ = x² + x = (x + 1) + x = 2x + 1 From here keep going to get equations for each integer exponent: x⁴ = x³ + x² = (2x + 1) + (x + 1) = 3x + 2 x⁵ = x⁴ + x³ = (3x + 2) + (2x + 1) = 5x + 3 x⁶ = x⁵ + x⁴ = (5x + 3) + (3x + 2) = 8x + 5 x⁷ = x⁶ + x⁵ = (8x + 5) + (5x + 3) = 13x + 8 x⁸ = x⁷ + x⁶ = (13x + 8) + (8x + 5) = 21x + 13 etc. The reason for the Fibonacci sequence coming into play that other posters have pointed out becomes apparent.
@oahuhawaii214129 күн бұрын
But that's still a long and drawn out process. If you noticed the cube removes the denominator, then things get easy: x³ = 2*x + 1 = √5 + 2 Just square that twice!
@mariomilani7913 күн бұрын
It took me 2 minutes after rewriting (1+√5)/2 in four groups to the power of 3. By using once the formula of the cube, once the formula of the square and then rearranging the resulting number, you get the result.
@sloomsy9494Ай бұрын
Actually it's hard but I brove it by binomial theorem ☝️🤓
@u700731721 күн бұрын
A rule for solving this sort of problem: If it's too much like hard work, you've missed a trick... It's easy to see that... x^2 = x+1, so x^4 = 3x+2, so x^6 = 8x+5, so, finally x^12 = 144x+89 Then the result is easy to compute.
@the-boy-who-lived29 күн бұрын
This questions is lot easier if you recognize this is φ
@musicsubicandcebu1774Ай бұрын
Square by hand to obtain φ squared Multiply again by φ to obtain 3rd power Square the 2nd power to obtain 4 power Cube the 4th power
@user-yb5cn3np5q6 күн бұрын
I almost went into a rant about how arithmetic over Z[phi] could be hard, but... you know, when 8% can solve it, it's a very simple problem, because 95% usually didn't even try.
@markharder3676Ай бұрын
You could get (3x + 1)^3 via the binomial expansion. I.e. Pascal's triangle.
@minchomilev7 күн бұрын
(½(1 + √5))¹² = (¼(1 + 2√5 + 5))⁶ = (¼(6 + 2√5))⁶ = (½(3 + √5))⁶ = (¼(9 + 6√5 + 5))³ = (½(7 + 3√5))³ = ⅛(343 + 441√5 + 945 + 125√5) = 161 + 72√5 I can't get it why you made it so complicated?
@RealEverythingComputers16 күн бұрын
Interesting question. simple answer. thx for the vid
@X0000037014 күн бұрын
I thought your approach required minimal Algebra knowledge, so I thought it was an excellent approach. Some of the other solutions below were shorter but required more math knowledge to understand.
@ymarko15 күн бұрын
I'm not a mathematician. But after your many manipulations, what is the actual result? Because your final line is not much simple from the original line. If you do not simplified, what was the point to do all these manipulations? You can just solve the original one on a calculator.
@MrUtubePete17 күн бұрын
He took so long I forgot the original problem
@jerrygunning14494 күн бұрын
I was a software engineer for forty years. If this guy was working for me I would have fire him. I know this "guy" - he is far more interested in proving how smart is than actually getting any work done.
@spicymickfool25 күн бұрын
Decent presentation of the basics, but its a great opportunity to demonstate tge interconnectedness of ideas here. Once you have tgat expression fir x^2, yiu van get one from x^3. Then by breaking down non kinear terms into linear exoressions, you can save time. So one cannedplicitky introduce an iterative method. Isnt this closely tier to the fibonacci series?
@claireli88Ай бұрын
Since x²=x+1, then x⁴=(x²)² =(x+1)²=x²+2x+1 =x+1+2x+1=3x+2 x⁸=(x⁴)² =(3x+2)²=9x²+12x+4 =9(x+1)+12x+4=21x+13 x¹²=(x⁴)(x⁸) =(3x+2)(21x+13) =63x²+81x+26=63(x+1)+81x+26 =144x+89 Since x=(1+√5)/2, then x¹²=144(1+√5)/2+89 72+72√5+89=161+72√5
@oahuhawaii214129 күн бұрын
By factoring 12 to 3*2*2, the power of 12 can be cubing, followed by squaring twice. Let x = √5 + 1 We have x³ = 5*√5 + 3*5 + 3*√5 + 1 = 8*√5 + 16 Thus, ((√5 + 1)/2)¹² = (x/2)¹² = (((x/2)³)²)² = ((x³/8)²)² = (((8*√5 + 16)/8)²)² = ((√5 + 2)²)² = (4*√5 + 9)² = 72*√5 + 161
@justliberty407227 күн бұрын
What cultures write x with two curves rather that lines that cross?
@jeannieheard146519 күн бұрын
DQ Culture
@werner1348975 күн бұрын
So from memory from 30 yrs ago when I studied mathematics, isn’t this about a group modulo an ideal or something?
@black_eagleАй бұрын
You can go faster through the trivial algebra steps to save time. I think most people watching can handle it.