Hardest Exam Question | Only 8% of students got this math question correct
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- เผยแพร่เมื่อ 21 ส.ค. 2024
- Can You Simplify? What do you think about this problem?
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_x = ½(1 + √5) = ½( -(-1) + √( (-1)² - 4(1)(-1) )_
∴ _x² + (-1)x + (-1) = 0_
⇒ _x² = x + 1_
Multiply through by _x:_
_x³ = x² + x = (x + 1) + x = 2x + 1_
⇒ _x⁶ = (2x + 1)² = 4x² + 4x + 1 = 4(x + 1) + 4x + 1 = 8x + 5_
⇒ _x¹² = (8x + 5)² = 64x² + 80x + 25 = 64(x + 1) + 80x + 25 _
_= 144x + 89 = 72(1 + √5) + 89_
∴ *_x¹² = 161 + 72√5_*
This is the simplest solution also derived independently
Yes anytime you work with the “magic number” you need to know how to simplify
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
Nice work. ❤
@@buffalobilly6046 can you tell me what is magic number?
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 !
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.
@@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 !
@@jpl569 -- In your first post, n - 2 and n - 1 needed to be inside grouping symbols, respectively.
@@robertveith6383 Yes, for sure, thanks !
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
This solution is far too cumbersome. (((Term^2)^2)^3 is much faster
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.
@@RexxSchneider That's how I did it.
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.
Good problem to be VERY careful with.
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.
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
Well, probably just as complicated to calculate, but you definitely don't risk a dead end.
I agree with you. It's much quicker this way.
I did so and got the exact same answer.
In this particular case yes, but seeing how introducing an X for a part of the expression can simplify the problem is pretty valuable.
I also did it this way. I wouldn't have even thought of the "X"
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.
Just Brilliant!
Clearly this is more obscure, less intuitive, and less "pure"
I was screaming, 'GET ON WITH IT!', Lol.
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.
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.
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.
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
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*
@@robertveith6383 Oops! Thank you. I Copy-Pasted "²" twice instead of "√5". Now corrected.
Now that a good way to do it, start. Term to 3rd = 2 x sqrt of 5. Then sq, then sq =322
@@robertveith6383 that's what he did???
@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.
i don't think this video should be 11 minutes long
I suspect his ornate way of writing "x" accounted for about 1/3 of the video length
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 .
2x speed.
this could have been a tweet... But this wouldn't be on TH-cam
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.
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. )
In this case should use any other number, not the golden ratio.
@@julianocg
Why not use the golden ratio : it gives an opportunity for bonus insights.
@@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.
@@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.
@@RexxSchneider Just as I did.
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.
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)
Nice and neat.
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.
amazing, what a relation that is
Oh, impressive! Maybe some students knew that, but I doubt most did. I checked and it comes out right.
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
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.
So, what's your point...?!
@@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.
@@RexxSchneider *Any* simple radical expression for x can be cast as the root of some quadratic equation. φ is particularly nice!
Prop. Phi = x ,
x ^n = F(n) x + F(n-1)
= F(n)/2 + F(n-1) +
+ F(n)/2 * 5^.5
F(1) = F(2) = 1
F(n) = F(n-1) + F(n-2)
F(12) = 144,
F(11) = 89
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]
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
It is not clear to me what is simpler, the original equation or the calculated result.
Surely the 'answer' here is simply a different way of writing the question?
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
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
((1+√5)/2)^12 = phi^12
we know phi^2 = 1 + phi
phi^12 = phi^(2×6) = (phi^2)^6
= (1+phi)^6 = ((1+phi)^2)^3
= (1+2phi + phi^2)^3
= (1+2phi+1+phi)^3
= (2+3phi)^3
= (2+3phi)^2 (2+3phi)
= (4+12phi+9phi^2) (2+3phi)
= (4+12phi+9(1+phi)) (2+3phi)
= (4+12phi+9+9phi)(2+3phi)
= (13+21phi) (2+3phi)
= 26 + 39phi + 42phi + 63phi^2
= 26 + 81phi + 63(1+phi)
= 26 + 81phi + 63 + 63phi
= 89 + 144phi
= 89 + 144((1+√5)/2)
= 89 + (144/2) (1+√5)
= 89 + 72(1+√5)
= 89 + 72 + 72√5
= 161 + 72√5
so,
((1+√5)/2)^12 = 161 + 72√5
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.
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.
That's what I did.
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.
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)
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.
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
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.
½(1 + √5) = x
x^12 = (x^2)^6
x^2 = t
x^12 = (t^2)^3 and just calculate
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.
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).
I did the same. If only 8% of students got it right, that must be bad students.
(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
Ce nombre est φ, le nombre d'or et on sait que φ est solution de l'équation x² - x - 1 = 0 donc: φ² - φ - 1 = 0
d'où: φ² = φ + 1
φ¹² = (φ²)⁶ = (φ + 1)⁶ = [ (φ + 1)²]³ = ( φ² + 2φ + 1)³ = (φ + 1 + 2φ + 1)³
= (3φ + 2)³ = (3φ + 2)² (3φ + 2) = (9φ² + 12 φ + 4) (3φ + 2) = (9φ + 9 + 12 φ + 4) (3φ + 2)
= (21φ +13) (3φ + 2) = 63φ² + 42φ +39φ +26 = 63φ + 63 + 42φ +39φ +26
= 144φ + 89
= 144 (1 + √5)/2 + 89
= 72 (1 + √5) + 89
= 72 + 72√5 + 89
= 161 + 72√5
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
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
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!
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)).
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.
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
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.
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.
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.
Moivre-Binet formula for implicit Fibonacci solves :)
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
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
Use the binomial theorem
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).
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
Thank you for showing me how to reach the result with all the substitutions.
(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
F = PHI
F = ( sqrx( 5 ) + 1 ) ÷ 2
F^2 = F + 1
F^6 = ( F^2 )^3 = ( F + 1 )^3
..........
F^6 = 8F + 5
F^12 = ( F^6 )^2 = ( 8F + 5 )^2
.............
F^12 = 144F + 89
= 72 sqrx( 5 ) + 161
👍😁🤪👋
A = (1+√5)/2
A^2 = (3 + √5)/2
A^3 = [(1+√5)/2][(3+√5)/2]
= [3+4√5 + 5]/4
= 2 + √5
A^6 = 9 + 4√5
A^12 = 81 + 80 + 72√5
= 161 + 72√5
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.
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.
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
Smart guys unattempt question and saves time😎
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
Simple way: do match by square, squaring and cubing or cubing, squaring, squaring without 1/2 to avoid bother with fractions. Divide when it is convenient and keep count of number of 2s used.
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.
Awesome, thanks. Very helpful.
((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.
Yeah, I did the same thing.
Fibonacci sequence and the golden ratio ϕ= (1+ √5)/2, as already mentioned by others earlier. ϕ^12=89 + 144ϕ
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
You could get (3x + 1)^3 via the binomial expansion. I.e. Pascal's triangle.
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.
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.
(1 +sqrt(5))^12 = ((1+sqrt(5))^3)^4 = (1+3sqrt(5) +3*5+5sqrt(5))^4= (16+8*sqrt(5))^4=8^4*(2+sqrt(5))^4= 8^4*((2+sqrt(5))^2)^2
=8^4*(4+5+4*sqrt(5))^2= 8^4*(9+4*sqrt(5))2= 8^4*(81+80 + 72sqrt(5))= 8^4*(161+72*sqrt(5)). And 2^12 = 8^4 , =>
((1+sqrt(5))/2) = 8^4*(161+72*sqrt(5))/(8^4) = 161 +72*sqrt(5)
This is phi to the power twelve and phi squared equal phi + 1 , simplier this way
the golden rario?
@@synriser6742 Is indeed the golden ratio.
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?
this expression is φ (The Golden ratio) raised to the 12 power. Then, φ² = 1 + φ.
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.
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.
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!
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.
That was a lot of work vs just remembering phi^2 = phi + 1. Reducing it down to a cubic and using Pascal's triangle to expand out the coefficients.
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.
For this problem (x^12) direct expansion is probably much easier....but suppose the problem were x^67 or x^83, the method shown here would be much easier. x^67 could be written as (x^64)(x^2)(x)...just keep expanding out replacing x^2 with x+1. Same with x^83...write as (x^64)(x^16)(x^2)(x)
Well, for minimal resources, we can keep just the original numbers and toss in an accumulator. For exponentiation, convert the exponent to binary, and use simple squaring the accumulator and multiplying by the base when a '1' bit is sensed, as you read off the bits of the exponent from MSB to LSB. The accumulator is initialized to 1. [Must check to avoid 0⁰.] This is much like how multiplying is simply doubling the accumulator (shift left in binary) and adding the original value when a '1' bit is sensed. The accumulator is initialized to 0.
For exponent 12, we can do it symbolically without dealing with big numbers. We have 12 = 1100b:
x² = x + 1 [2 = 10b]
x²*x = x³ = (x + 1)*x = x² + x = 2*x + 1 [3 = 11b]
x⁶ = 4*x² + 4*x + 1 = 8*x + 5 [6 = 110b]
x¹² = 64*x² + 80*x + 25 = 144*x + 89 [12 = 1100b]
For the exponents you chose, the process gets cumbersome quickly. But the algorithm is simple: Preload accumulator to 1; thereafter, only square and if '1' bit, multiply by x.
For exponent 67:
x⁰: 1, x¹: x¹ᵇ, x²: x¹⁰ᵇ, x⁴: x¹⁰⁰ᵇ, x⁸: x¹⁰⁰⁰ᵇ, x¹⁶: x¹⁰⁰⁰⁰ᵇ, x³²: x¹⁰⁰⁰⁰⁰ᵇ, x³³: x¹⁰⁰⁰⁰¹ᵇ, x⁶⁶: x¹⁰⁰⁰⁰¹⁰ᵇ, x⁶⁷: x¹⁰⁰⁰⁰¹¹ᵇ
For exponent 83:
x⁰: 1, x¹: x¹ᵇ, x²: x¹⁰ᵇ, x⁴: x¹⁰⁰ᵇ, x⁵: x¹⁰¹ᵇ, x¹⁰: x¹⁰¹⁰ᵇ x²⁰: x¹⁰¹⁰⁰ᵇ, x⁴⁰: x¹⁰¹⁰⁰⁰ᵇ, x⁴¹: x¹⁰¹⁰⁰¹ᵇ, x⁸²: x¹⁰¹⁰⁰¹⁰ᵇ, x⁸³: x¹⁰¹⁰⁰¹¹ᵇ
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
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
I don't know if I'm more impressed by the maths or the neat handwriting.
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)
Simpler: once we get the relation x² = x + 1, we can obtain a recurrence relation for x^n.
Isn't it simple ✓5 = 2.2360 then +1 and whole ÷ 2 and then 1.618 to the power 12 = approx 322
Golden ratio to the power twelve. Powers of the golden ratio always approximate integers, in this case 322. If you need more accuracy, 322 minus 1/322 might well be good enough.
3:31 this is an old way to write multiple pairs of parenthesis and not go insane trying to not make a mistake in closing them and if there was another one it would have been { }
This is how to make a trivial task look hard and difficult by a long over-complicated solution.
This questions is lot easier if you recognize this is φ
X is two straight lines crossing. What is this atrocity you are drawing? :)
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.
C'mon, it's so that you do not MIX it up with the x on 2x3=6
@@josepeixoto3384 but nobody doing algebra ever writes 2x3...
(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?
Hasn’t no one tried simply using 1 and (1+sqrt5)/2 as the first two terms of a Fibonacci sequence, and calculating up to the 13th term of the progression to find the solution?
It’s as simple as adding up terms.
I bet the actual exam had only 5 lines to do working in lmao
Hard because the question is incomplete. Some sort of result format specification needed. Decimals? Rational number format etc
with a tiny bit of mathematical knowledge you recognise immediately the golden ratio and you know it is a solution of x=x+1
My old HP49 was able to quickly evaluate it to 161+72*sqrt(5) lol
My initial intuition was that maybe you could just go at it numerically and it would be faster than slinging algebra for 11+ minutes, ...and maybe get a closed-form answer numerically.
Assuming Olympiad answers prefer round numbers you could also test integer solutions in the form a + b*sqrt(c) lol
the way he wrote 1 is weird
Why ?? In what part of the world are you ?
I' m ok with his 1's
But the b are really weird.
It's like a combination of two cursive styles
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.
Thanks for helping learn math. discover the math of reality. I stopped because I already know the answer.
If the appearance of 144 and 89 in the third to last line seems disappointingly inauspicious for a golden ratio power, take heart-they are Fibonacci numbers. This is a reassuring sign that we are indeed working with the golden ratio.
jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci
Here's what everyone is missing:
the solution demonstrated would be doable by anyone who has done basic algebra
[(1+sqrt(5))/2]^3 = 2+sqrt(5)
[2+sqrt(5)]^2=9+4sqrt(5)
[9+4sqrt(5)]^2=161+72sqrt(5)=[(1+sqrt(5))/2]^12
Exactly how I solved it, too. You can do it in your head.
You can go faster through the trivial algebra steps to save time. I think most people watching can handle it.
Only 8% of 6th graders I assume?
Excellent !
But, since minute 8, several confusions between = and =>
Being lazy and using brute force, A = ½(1 + √5) squared is ½(3 + √5) - this takes 5 seconds.
½(3 + √5) squared is ½(7 + 3√5) - this also takes 5 seconds.
½(7 + 3√5) squared is ½(47 + 21√5) - this takes maybe 30 seconds.
Multiply A^4 x A^8 = ½(7 + 3√5) x ½(47 + 21√5) - this might take a minute or a bit more.
The answer is ¼(644 + 288√5) = 161+72√5
The whole thing takes 2 minutes, without any fancy technique of any sort.
Cube it first, as that eliminates the denominator. Then square twice. Super easy and fast. Try it and laugh at the narrator for taking the long and messy route.
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
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