(1+sqrt(5))/2 is the golden ratio (phi). It satisfies that x²=x+1, thus x³ = x(x+1) = x² + x = 2x+1. x⁴ = x(2x+1) = 2x² + x = 2(x+1) +x = 3x +2 x^5 = x(3x+2) = 3x² +2x = 3(x+1) +2x = 5x + 3 It's easy to see the fibonacci series in there. Following the series, we will have that: x^12 will be 144x + 89. In fact, if we use this notation (ax+b), the result of phi^n will have a=n and b=(n-1). Please, note that 89 is the 11th and 144 is the 12th terms in the fibonacci series. Now all you have to do is turn x into phi. 144x + 89 = 144(1+sqrt (5)/2) + 89 = 72 + 72sqrt(5) + 89 = 161 + 72 sqrt (5) You can also see a pattern here. In the form (q + z*sqrt(5)), q (the rational part) will be (n/2 + (n-1)) = 3/2*n - 1, and the z (the irrational part) will be n/2. With this info, and as demonstration, we can easily calculate Phi^15, knowing that the 15th term of the fibonacci series is 610 and the 14th term is 377. solution: 682 + 305sqrt(5) ≈ 1364.00
There's quite a simple solution if we note that x = φ, where φ is golden ratio, the fundamental of which is φ²=φ+1 So we may subsequently substitute φ² while calculating φ¹² = (φ²)⁶=(φ+1)⁶=((φ+1)²)³= = (φ²+2φ+1)³=(3φ+2)³= (3φ+2)²(3φ+2)= (9φ²+12φ+4)(3φ+2)= (21φ+13)(3φ+2)= 63φ²+39φ+42φ+26= 144φ+89 Finally we substitute φ=(1+√5)/2 and get the answer 161+72√5
Let φ=(1+sqrt(5))/2 which is the golden ratio, n is an integer and F(n) is the n-th term of the Fibonacci series, then φ^n=F(n-1)+φ*F(n). For example, φ^(-3)=F(-4)+φ*F(-3)=-3+2φ.
Even faster. Recognize that the value being raised to the twelfth power is the value of the Golden Ratio, R, that satisfies the equation R^2 = 1+ R. Each successive power, R^n, is equal to (can be reduced) to a +b*R, where a is the (n-1)th integer in the Fibbonacci sequence, and be is the nth one. So for R^2, a = 1, b=1. For R^3, a=1, b=2. For R^4, a = 2, b =3. For R^5, a=3, b=5.......For R^12, a=89, b=144. So the answer is 89 + 144*R. R=(1 +sqrt(5))/2. So 89+144*R = 89 + 72 +72*sqrt(5) = 161 + 72*sqrt(5)/
This is actually not hard if you know the property that powers of phi have a connection to Fibonacci numbers. phi^n = F(n) + F(n-1)*phi. So in this case 144+89phi.
X^n represents just one real number, for every power of X. The X variable has been isolated from the number values. We need only complete the simplification of X^12. I'm assuming that both exact and numerical approximation are requested. Exact include the radical 5, while approximations compute the answer according to your required precision. Yeah, definitely, *after mastering* these you might want to turn to infinite sequences in whatever books you use. It's basically the way to examine number relationships and turn these into a single algorithm. Figuring them requires understanding the proofs used to construct them. Infinite and geometric number sequences can be evaluated, if existing, to create a general formula. For example, the equation (N*(N+1))/2 computes the sum of all positive integers, from 1 to N, inclusive. Ex1: N=4, (4*(5))/2 = 10 = (1+2+3+4) This problem is simple enough to walk the power ladder: 2,3,6,12. Identity: (A+B)^2 = A^2+B^2+2AB Identity: (A/B)^n = (A^n) / (B^n) Using the power division rule, and if you've done much base 2 math, especially in computing, you'll almost hear the 2^12 denom scream 4096. That's the size of most computer architecture real and virtual page sizes. Anyway, we could just do the numerators and divide that solution by 4096. But I want you to watch the rolling iterations. I haven't done all of the math steps but do list the procedures and intermediate and final answers. Do one or two steps to see if the answers match. X1= (1+sqrt(5)) / 2 X2= (X1 * X1)= 3+sqrt(5)) / 2 X3= (X1 * X2)= 2+sqrt(5) X6= (X3 * X3) = 9+4*sqrt(5) Note that the denominator cancels out nicely at X3, not to return in this set of values. X12=(X6)^2 =161+72(sqrt(5))
A much simpler solution is to take the inverse proportion of delta 5 to the tertiary coherent function divided by the cotangent of {5 + 63} and predetect the log of the variable operator with a squared coefficient minus infinity. And there you have it.
Actually, it should be extremely close to a Lucas number, and if you round the powers of phi, you will get the pure Lucas numbers, except obviously the first term, as that’s bigger than the second term, and that’s not how exponents of positive integer powers work
first cube it, then take 8 common and cancel it withdenominator . you get (2+sqrt(5))^4 square two times. first time, get 9+4sqrt(5) , squaring this gives 161+72sqrt(5) simple way. isn't it?
Fibonnacci Un+2=un+1 + un Solution = a GN^n + b o(GN)^n Hence GN^n matches fibonnacci with u0=1 u1= GN u2 = 1+ GN u3 = 1 + 2GN u4= 2 + 3 GN … u12 = 89 + 144 GN
……OR……plug it into a calculator. 🤔 You’re going to plug the final simplification into a calculator for a concise numerical answer anyway. Therefore, just plug the initial problem into the calculator. And save a lot of time. *cough*
@@trucid2 Doesn’t change the fact you’re still going to use a calculator in the end to get the most simplified value. This exercise would have much more bearing if one could simplify ball the way to a result in pure numerical format without roots. Otherwise, it’s just a game of “go as far as you can until you have to use a calculator”.
@@ltrizzle12 You can't write down the exact answer numerically. You have to express it in terms of roots. The point of the exercise is to learn how to simplify expressions, not to get a numerical answer.
@@trucid2 You could progress one step forward and technically you’ve “simplified” it. The greatest simplification is producing a value as accurate as feasible. Having a problem with a root and leaving in terms of a root is sophomoric & futile when a calculator is required to get some semblance of numerical value. If you entered 6*pi into a calculator, would you be okay with the calculator producing back “6*pi”? Of course not. You’d want to see approx. 18.85 to make practical sense out of it. It’s irrelevant when you can TRULY simplify for a largely accurate value. Irrational value or not, even taking it out to 2 or 3 decimal places is sufficient for most intents & purposes of human enterprise. I’m more of a pragmatist; don’t leave a root in the answer when one can reasonably round an irrational value to several decimal places with the assistance of a calculator. Would you leave a root in the denominator of an expression you were attempting to simplify? Of course not, because of social convention. In that same context, I’d never leave an answer with any visible root when one small further step of using a calculator generates a sensible, overt, numerical value that is plainly discernible by all who read it. OR just use the dang calculator from the onset of the problem-like I initially recommended-for an instant numerical solution. See? Took a wee bit, but we came full circle. 😎
At 3:50, instead of setting x^12 = (x^2)^6, we could rather consider ((x^3)^2)^2. We therefore find x^3 and then we square two times. It is very easy to find x^3: x^3 = x^2.x = (x+1)x = x^2 + x = (x +1) + x = 2x + 1. So we now square this two times. The first square: (2x + 1)^2 = 4x^2 + 4x +1 = 4(x+1) + 4x + 1 = 8x + 5 . We square again: (8x + 5)^2 = 64x^2 + 80x + 25 = 64(x+1) + 80x + 25 = 144x + 89.
@@master_sergik You apparently missed that x is not an 'unknown' but a very specific number. It is the number that is raised to the 12th power. Maybe the choice of the letter was not the right one, and this is what confused you. Probably φ would be more appropriate, as (1+sqrt(5))/2 is the so called "golden ratio". But the name of the variable is irrelevant. In any case, if the specific number (symbolized by x, or φ, or whatever) is squared, we get the number plus one. If we cube it, we get it two times plus one. And so on...
Esto es demasiado facil... Yo recrro a mis derivadas y difirenciales,a mis teoremas de exponentes, a raices explicadas como 1/?según la raiz.. Ponme algo mas diicil.como derivadas infinitas
Is it BECAUSE of the Golden Ratio that the surd part of the solution (=160.99689438) which is almost equal to the rational part of the solution (161). Do I need to go back to studying the Golden Ratio?
Also ((1+sqrt5)2)^6=9+4sqrt5 and 9 ~ 4sqrt5 so I suspect I've answered my own question and we're tending towards an equivalence here. Please enlighten me.
(1+sqrt(5))/2 is the golden ratio (phi). It satisfies that x²=x+1, thus x³ = x(x+1) = x² + x = 2x+1.
x⁴ = x(2x+1)
= 2x² + x
= 2(x+1) +x
= 3x +2
x^5 = x(3x+2)
= 3x² +2x
= 3(x+1) +2x
= 5x + 3
It's easy to see the fibonacci series in there. Following the series, we will have that:
x^12 will be 144x + 89.
In fact, if we use this notation (ax+b), the result of phi^n will have a=n and b=(n-1).
Please, note that 89 is the 11th and 144 is the 12th terms in the fibonacci series. Now all you have to do is turn x into phi.
144x + 89 = 144(1+sqrt (5)/2) + 89
= 72 + 72sqrt(5) + 89
= 161 + 72 sqrt (5)
You can also see a pattern here.
In the form (q + z*sqrt(5)), q (the rational part) will be (n/2 + (n-1)) = 3/2*n - 1, and the z (the irrational part) will be n/2.
With this info, and as demonstration, we can easily calculate
Phi^15, knowing that the 15th term of the fibonacci series is 610 and the 14th term is 377.
solution: 682 + 305sqrt(5) ≈ 1364.00
Also, with every iteration, the decimal part of the phi power will approach 0.
With phi^35, the decimal part will be 000000048...
Perfect. Much better than the solution in the video 😊
this is a much better solution.
Exploit x^2 = x + 1 each time to always keep simplifying the intermediate expressions as the form ax + b.
That keeps the numbers small and simple:
x^3 = x(x^2) = x(x+1) = x^2 + x = (x +1) + x = 2x + 1
x^6 = (x^3)^2 = (2x + 1)^2 = 4x^2 + 4x + 1 = 4(x+1) + 4x + 1 = 8x + 5
x^12 = (X^6)^2 = (8x + 5)^2 = 64 x^2 + 80 x + 25 = 64 (x+1) + 80 x + 25 = 144 x + 89
There's quite a simple solution if we note that x = φ, where φ is golden ratio, the fundamental of which is φ²=φ+1
So we may subsequently substitute φ² while calculating φ¹² = (φ²)⁶=(φ+1)⁶=((φ+1)²)³=
= (φ²+2φ+1)³=(3φ+2)³=
(3φ+2)²(3φ+2)=
(9φ²+12φ+4)(3φ+2)=
(21φ+13)(3φ+2)=
63φ²+39φ+42φ+26=
144φ+89
Finally we substitute
φ=(1+√5)/2 and get the answer
161+72√5
I think if x is cubed and then squared twice we get to the result faster.
Let φ=(1+sqrt(5))/2 which is the golden ratio, n is an integer and F(n) is the n-th term of the Fibonacci series, then
φ^n=F(n-1)+φ*F(n).
For example, φ^(-3)=F(-4)+φ*F(-3)=-3+2φ.
Even faster. Recognize that the value being raised to the twelfth power is the value of the Golden Ratio, R, that satisfies the equation R^2 = 1+ R. Each successive power, R^n, is equal to (can be reduced) to a +b*R, where a is the (n-1)th integer in the Fibbonacci sequence, and be is the nth one. So for R^2, a = 1, b=1. For R^3, a=1, b=2. For R^4, a = 2, b =3. For R^5, a=3, b=5.......For R^12, a=89, b=144. So the answer is 89 + 144*R. R=(1 +sqrt(5))/2. So 89+144*R = 89 + 72 +72*sqrt(5) = 161 + 72*sqrt(5)/
Nie każdy zna pojęcie złotego podziału,ciągu Fibonacciego itp.a wzory skróconego mnożenia większość. Zgrabne i przejrzyste rozwiązanie for every one.
Gostei muito professor! Realmente parece difícil, mas o senhor descomplicou. 👏🏻👏🏻👏🏻👏🏻👏🏻👏🏻
This is actually not hard if you know the property that powers of phi have a connection to Fibonacci numbers. phi^n = F(n) + F(n-1)*phi. So in this case 144+89phi.
amen, someone saw it
It's a good demo for students to learn solution
X^n represents just one real number, for every power of X. The X variable has been isolated from the number values. We need only complete the simplification of X^12.
I'm assuming that both exact and numerical approximation are requested. Exact include the radical 5, while approximations compute the answer according to your required precision.
Yeah, definitely, *after mastering* these you might want to turn to infinite sequences in whatever books you use. It's basically the way to examine number relationships and turn these into a single algorithm. Figuring them requires understanding the proofs used to construct them.
Infinite and geometric number sequences can be evaluated, if existing, to create a general formula. For example, the equation (N*(N+1))/2 computes the sum of all positive integers, from 1 to N, inclusive.
Ex1: N=4,
(4*(5))/2 = 10 = (1+2+3+4)
This problem is simple enough to walk the power ladder: 2,3,6,12.
Identity: (A+B)^2 = A^2+B^2+2AB
Identity: (A/B)^n = (A^n) / (B^n)
Using the power division rule, and if you've done much base 2 math, especially in computing, you'll almost hear the 2^12 denom scream 4096. That's the size of most computer architecture real and virtual page sizes. Anyway, we could just do the numerators and divide that solution by 4096. But I want you to watch the rolling iterations.
I haven't done all of the math steps but do list the procedures and intermediate and final answers.
Do one or two steps to see if the answers match.
X1= (1+sqrt(5)) / 2
X2= (X1 * X1)= 3+sqrt(5)) / 2
X3= (X1 * X2)= 2+sqrt(5)
X6= (X3 * X3) = 9+4*sqrt(5)
Note that the denominator cancels out nicely at X3, not to return in this set of values.
X12=(X6)^2
=161+72(sqrt(5))
Every solution in the comments is better than the one in the video 😊
Indeed!
A much simpler solution is to take the inverse proportion of delta 5 to the tertiary coherent function divided by the cotangent of {5 + 63} and predetect the log of the variable operator with a squared coefficient minus infinity. And there you have it.
Took the words outta my mouth!
Brilliant!
Nobody in their right mind would simplify that. It's the golden ratio to the 12th power. I'm positive that would result in bad karma.
Yes, equal to PHI^12...
Actually, it should be extremely close to a Lucas number, and if you round the powers of phi, you will get the pure Lucas numbers, except obviously the first term, as that’s bigger than the second term, and that’s not how exponents of positive integer powers work
😆
first cube it, then take 8 common and cancel it withdenominator . you get (2+sqrt(5))^4
square two times. first time, get 9+4sqrt(5) , squaring this gives 161+72sqrt(5)
simple way. isn't it?
Brilliant, the easiest solution.
所以到最后还是硬算出来的，你不如直接拆为2*2*3次方，直接算得了，由于2次的规律一致，还能节约点算法。
可以用重根的公式，其實不難，普遍中國、台灣高中生都會
Por que não usou binômio Newton / triângulo de Pascal ??
When the expression assumed as x, solutions is there.
Nice Exponents Math Simplification: [(1 + √5)/2]¹² = ?
Let: a = (1 + √5)/2; [(1 + √5)/2]¹² = a¹²
a² = [(1 + √5)/2]² = (6 + 2√5)/4 = (3 + √5)/2 = 1 + (1 + √5)/2 = 1 + a
a⁴ = (a²)² = (1 + a)² = 1 + 2a + a² = 1 + 2a + (1 + a) = 2 + 3a
a⁶ = (a²)(a⁴) = (1 + a)(2 + 3a) = 2 + 5a + 3a² = 2 + 5a + 3(1 + a) = 5 + 8a
a¹² = (a⁶)² = (5 + 8a)² = 25 + 80a + 64a²) = 25 + 80a + 64(1 + a) = 89 + 144a
= 89 + 144[(1 + √5)/2] = 89 + 72(1 + √5) = 161 + 72√5
Answer check:
[(1 + √5)/2]¹² = 322, 161 + 72√5 = 322; Confirmed
Math calculation was achieved on a smartphone with a standard calculator app
Final answer:
[(1 + √5)/2]¹² = 322 = 161 + 72√5
I did it in my mind. The trick is to calculate the cube first and the do the 4th power
No me extraña que las Matemáticas sean unas de las asignaturas más odiadas.
Bravo!!! 👏👏👏
Fibonnacci
Un+2=un+1 + un
Solution = a GN^n + b o(GN)^n
Hence GN^n matches fibonnacci with
u0=1
u1= GN
u2 = 1+ GN
u3 = 1 + 2GN
u4= 2 + 3 GN
…
u12 = 89 + 144 GN
Good solution
1.618 в 12 степени. 322 будет. Неужели сложно калькулятором посчитать?
Без замены и используя формулу куба сумы двух чисел решается быстрее.
I use a simple method,guickly get answer.
(1+√5):2=1,618 (Fibonacci)
Por qué no usar Binomio de Newton? Saludos.
Totally magnificent an optimum at finest😂❤
Ф := (√5+1)/2
Ф^n = fn * Ф + fn-1, where fn is the n-th Fibonacci number.
Hence Ф^12 = 144Ф + 89, now you only need to expand it, 72√5 + 161
a smart way NOT to use a calculator!
SPEAK LOUD, EXPLAIN, DEDUCE, REASONS, !!! DON'T NEED MUSICAL INSTRUMENTAL, I'D BRTTER SINTONIZE AN FM RADIO!.
Hermosa resolución!
简单问题复杂化，没完没了兜圈子。印度思维方式。😂
1. You're making it too complicated. x^2 ok, x^4 ok, x^8 ok, x^12=x^4*x^8.
2.The final equal sign (=) should not be written as =>, please!
PERO HABLA, EXPLICA , DEDUCE!. RAZONAMIENTO, !!! PARA MUSICA SINTONIZO UNA RADIO FM.
Conjugate it to get
2 to power 12 as answer
That is 32x32x2=32×64=2048
Great method. Well k own to some of us...but always good to see it on youtube for others.
It's not Simplify coz you're using more time algebra formula. This is complex for beginners
Muito interessante !!
Without binomial theorem! nice
Bravo!!!
вот извращение на потерю времени,
Impressive, but too long, mathematician enjoy short elegant solutions not tedious ones and striving to find them.
……OR……plug it into a calculator. 🤔
You’re going to plug the final simplification into a calculator for a concise numerical answer anyway.
Therefore, just plug the initial problem into the calculator. And save a lot of time.
*cough*
And learn nothing.
@@trucid2 Doesn’t change the fact you’re still going to use a calculator in the end to get the most simplified value. This exercise would have much more bearing if one could simplify ball the way to a result in pure numerical format without roots.
Otherwise, it’s just a game of “go as far as you can until you have to use a calculator”.
@@ltrizzle12 You can't write down the exact answer numerically. You have to express it in terms of roots. The point of the exercise is to learn how to simplify expressions, not to get a numerical answer.
@@trucid2 You could progress one step forward and technically you’ve “simplified” it. The greatest simplification is producing a value as accurate as feasible. Having a problem with a root and leaving in terms of a root is sophomoric & futile when a calculator is required to get some semblance of numerical value.
If you entered 6*pi into a calculator, would you be okay with the calculator producing back “6*pi”? Of course not. You’d want to see approx. 18.85 to make practical sense out of it. It’s irrelevant when you can TRULY simplify for a largely accurate value. Irrational value or not, even taking it out to 2 or 3 decimal places is sufficient for most intents & purposes of human enterprise.
I’m more of a pragmatist; don’t leave a root in the answer when one can reasonably round an irrational value to several decimal places with the assistance of a calculator.
Would you leave a root in the denominator of an expression you were attempting to simplify? Of course not, because of social convention. In that same context, I’d never leave an answer with any visible root when one small further step of using a calculator generates a sensible, overt, numerical value that is plainly discernible by all who read it.
OR just use the dang calculator from the onset of the problem-like I initially recommended-for an instant numerical solution. See? Took a wee bit, but we came full circle. 😎
@@ltrizzle12 The problem asked to simplify, not give an approximate value. If you were to do that on a test, you would not get full marks, if at all.
😮
Where is exponent 12 in the end?
At 3:50, instead of setting x^12 = (x^2)^6, we could rather consider ((x^3)^2)^2. We therefore find x^3 and then we square two times.
It is very easy to find x^3:
x^3 = x^2.x = (x+1)x = x^2 + x = (x +1) + x = 2x + 1.
So we now square this two times. The first square: (2x + 1)^2 = 4x^2 + 4x +1 = 4(x+1) + 4x + 1 = 8x + 5 .
We square again: (8x + 5)^2 = 64x^2 + 80x + 25 = 64(x+1) + 80x + 25 = 144x + 89.
how x³ can be equal 2x + 1? 🙄
check it if x=2:
2³ = 2*2 +1
8 = 5
😒
x=3:
27 = 7
🙊
x = 10:
1000 = 21
🤦🏻♀️
@@master_sergik The x in the video is phi, which is (1 + sqrt(5))/2. It's not an arbitrary number.
@@master_sergik You apparently missed that x is not an 'unknown' but a very specific number. It is the number that is raised to the 12th power. Maybe the choice of the letter was not the right one, and this is what confused you. Probably φ would be more appropriate, as (1+sqrt(5))/2 is the so called "golden ratio". But the name of the variable is irrelevant.
In any case, if the specific number (symbolized by x, or φ, or whatever) is squared, we get the number plus one. If we cube it, we get it two times plus one. And so on...
thanks for explaining, it is new info for me 👍
Esto es demasiado facil...
Yo recrro a mis derivadas y difirenciales,a mis teoremas de exponentes, a raices explicadas como 1/?según la raiz..
Ponme algo mas diicil.como derivadas infinitas
Me, an intellectual: φ¹²
(72√5 + 161)
Gold number ^22...😉
Super
too lengthy procedure
Uyku getirdiniz
Yorumlarda herkes çözdü size yetmez diye kağıt gönderiyorum
لماذا كل هذا التعب من البداية نعمل على التمرين بدون فرض x
Ну и чо? Как было так и осталось, с небольшими изменениями...
U make it longer
φ¹²=89+144φ,
φ¹²=161+72√5.
Painful.
答案就是
m + n SQRT(5)
m , n .不漂亮無聊 ！
323
I calculated it in my mind faster
懐かしい
On my watch you use a calculator. It’s 321.9969.
Devide and conquer... ()^12=((()^2)^2))^3 is really easy to get the final answer.
Shake Mohamed bin thogaluk Mathematics burude mela, mamde mela, tharadu mela, thale mela, =
황금비율인걸 알아도 좋고
그냥 차수줄이기로 풀어도 되는
고1수학 중상난이도 문제로도 있는
평범한 문제
Дурачится. Сразу видно. что не советское образование получил
Put it in a calculator
Faire une division euclidienne ou un Hörner aurait.ete plus simple
First of all you need to learn not to complicate things...
Instead of uploading video on CZcams asking to simplify it!!!😅😆🤣😁😂
Music is not golden to any degree
321.99?
knp gk x^2^2^2. x^2^2
Is it BECAUSE of the Golden Ratio that the surd part of the solution (=160.99689438) which is almost equal to the rational part of the solution (161).
Do I need to go back to studying the Golden Ratio?
Also ((1+sqrt5)2)^6=9+4sqrt5 and 9 ~ 4sqrt5 so I suspect I've answered my own question and we're tending towards an equivalence here. Please enlighten me.
Too long and innecesary solution.
Que perdida de tiempo.
Mejor tomo mi calculadora y lo resuelvo en 1 minuto
= ϕ¹²
Is this a simplification? Now count how many operations you performed to simplify? 🤣🤣🤣
Bro I got it in 20 seconds 😂
Dhur chata ki6ui Boja jaini
Y el triangulo de pascal????
~322
答案就是
m + n SQRT(5)
m , n .不漂亮無聊 ！
😮