Hardest Exam Question | Only 8% of students got this math question correct
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- čas přidán 28. 08. 2024
- Can You Simplify? What do you think about this problem?
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@higher_mathematics
#maths #math
_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"
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.
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.
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 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.
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.
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!
And almost all students will forget it an hour later, or right after the test.
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.
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.
Isn't it simple ✓5 = 2.2360 then +1 and whole ÷ 2 and then 1.618 to the power 12 = approx 322
Duh. Exactly. Some people are so removed from the real world they would look at a micrometer and think it must be some sort of giant earring or something. We didn't land men on the moon by answering mathematical problems with more problems. We did it by calculating real world solutions and applying them.
Right, it could have been solved in 30 seconds instead of partly solved in 11 minutes.
Actually, it can be calculated in 3 steps: Since x^12={[(x^3)^2}^2, we first obtain [(1 + √5)/2]^3 = 2 + √5. Then (2 + √5)^2 = 9 + 4√5 and finally, (9 + 4√5)^2 =81 + 4*2*9√5 +80 = 161+72√5. Even no algebra is needed!
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 CZcams
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.
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
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.
Let a be the number in parentheses. a is known as the golden ratio, and as such, it satisfies the equation x^2 - x - 1 = 0. Thus, a^2 = a + 1.
Given this, a^12 = (a^2)^6 = (a + 1)^6 = ((a + 1)^2)^3 = (a^2 + 2a + 1)^3 = (3a + 2)^3 = (3a + 2)(9a^2 + 12a + 4) = (3a + 2)(21a + 13) = 63a^2 + 81a + 26 = 144a + 89 = 72(1 + sqrt(5)) + 89 = 72sqrt(5) + 161 as desired.
((1+sqrt5)/2)^2=(3+sqrt5)/2, so ((1+sqrt5)/2)^4=(7+3sqrt5)/2, so ((1+sqrt5)/2)^6=(3+sqrt5)(7+3sqrt5)/4=(21+15+9sqrt5+7sqrt5)/4=9+4sqrt5, so ((1+sqrt5)/2)^12=81+16*5+72sqrt5=161+72sqrt5. Straight forward. Quicker.
x²-x-1=0
x²-x+1=2
x³+1=2x+2
x³=2x+1=2+sqrt5
x⁶=9+4sqrt5
x¹²=161+72sqrt5
수능 15번보다 쉬운듯
It is not clear to me what is simpler, the original equation or the calculated result.
Use the property of golden ratio: φⁿ = Fibonacci(n)φ + Fibonacci(n-1). So here, φ¹² = Fib(12)φ + Fib(11) = 144φ + 89 ≈ 321.997. This is the same as the final answer in the video. (Note that Fib(n) is zero based.)
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.
I would not just fire the guy who wasted 20 minutes answering the question with another problem. I would also fire the guy who came up with the question in the first place. When have you ever needed to raise anything to the 12th power in a real world situation? It takes about 30 seconds to type it into a python session or key it into a calculator, and get an actual usable numeric answer.
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.
Cut to the chase. Calculate the solution. 322. The presumption is that there is a practical need for the solution, and there it is, now it is ready to be used for whatever purpose the question was conceived.
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
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.
Right, so as I see it, it wasn't really solved, while several people here seemed to solve it well, faster, and more simply.
Hard because the question is incomplete. Some sort of result format specification needed. Decimals? Rational number format etc
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
(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?
8% of students got this right.
98% of the rest of humanity thought, "Why do we pay these people for this pointless gibberish?"
98% of the rest of humanity can keep have that thought while working 996 unskilled labour till they're 70.
Revelling in his ignorance whilst posting using machines designed by people who understand mathematics, using mathematics.
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*.
@@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".
@@echandler The "problem" is made up out of thin air with no basis in the real world. Why would you ever need to know the solution? If it were based on an actual real world problem, a terminal window with python called up or else a $4.99 calculator are all that is needed. We don't have to re-invent mathematics every time we have a reason to solve something. Obviously the answer is calculated at approximately 322. Done. Now, get back to work. Play time is over.
½(1 + √5) = x
x^12 = (x^2)^6
x^2 = t
x^12 = (t^2)^3 and just calculate
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.
Here's what everyone is missing:
the solution demonstrated would be doable by anyone who has done basic algebra
Surely the 'answer' here is simply a different way of writing the question?
Indeed.
Right, after 11 minutes he still did not solve for x. I think he would have gotten it marked wrong on the test, if he did not run out of time.
(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。
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)).
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.
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.
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
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.
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
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.
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
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
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
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)
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.
0:04 Is this suppose to be done without a calculator? If no, I did [(√5/2)+1]¹²=8150.7036132812426871779905887804. I used proper PEMDAS/PEDMSA.
Now, improper PEMDAS gives us
[(√5+1)/2]¹²=321.99689437998485814146050414865.
Now, I don't know a third way to calculate this because the equation is very specific. If it were written √5+1/2¹²=X, then there's many more ways to get it wrong.
0:00 if you solve for the terms within the parentheses you get the golden ratio 🤩
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
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
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 :)
((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.
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
((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
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!
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
(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 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
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.
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 don't know if I'm more impressed by the maths or the neat handwriting.
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
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
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...
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.
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.
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.
this expression is φ (The Golden ratio) raised to the 12 power. Then, φ² = 1 + φ.
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
👍😁🤪👋
(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
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.
The Golden Ratio is a very cool number - with one of its properties being that any whole power of it is equal to a Fibonacci Number times the Golden Ratio plus the next lower Fibonacci Number.
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.
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.
You mean middle school, right?
(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)
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)
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
You could get (3x + 1)^3 via the binomial expansion. I.e. Pascal's triangle.
Thank you for showing me how to reach the result with all the substitutions.
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
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
Use the binomial theorem
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.
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!
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
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
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.
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.
The worst part is: your pen is not working !
Simpler: once we get the relation x² = x + 1, we can obtain a recurrence relation for x^n.
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.
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.
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's been over 40 years since I was in a college algebra class.but to me the original equation looks simpler. You still have to find the square root of 5.
The problem only needs to be simplified to the sum of a few simple terms to get (A*√5 + B)/C, where the letter values are integers. The solution is 72*√5 + 161 .
The comments are way more informative and interesting than the video.
Thanks for helping learn math. discover the math of reality. I stopped because I already know the answer.