The question is to find a solution to the differential equation, a" + b" =?. If it is understand that a and b can be functions then finding numerical values for a and b may be a part of the solution but will not determine what function a and function b are.
This is what I did. Find a and b, square them for a^2 and b^2, square them for a^4 and b^4, square them for a^8 and b^8, multiply the ^4 by the ^8 and divide by the ^1's gives you a^11 and b^11. All you ever have are binomials over a numeric denominator and the a's and b's are the same except one is + while the other is -. Really wasn't complicated at all.
There's a nice way using Viete's formulas - a,b are the roots of the equation x^2-(a+b)x + ab = 0, after finding ab=-1/2 you get 2x^2-2x-1 = 0 or x^(n+2)=x^(n+1) + (x^n)/2 - always true for roots a,b, giving the recursive relationship a^(n+2) + b^(n+2) = (a^(n+1) + b^(n+1)) + (a^n + b^n)/2.
Let a+b =S and ab=P. In the second equation a^2+b^2 =(a+b)^2-2ab=2 leads to ab=-1/2. If a and b are solutions of the quadratic equation X^2+PX+S=0, that give us a,b =(1±√3)/2.
Time taking and lengthy process. If you can find value of a-b by the formula (a-b)^2=(a+b)^2-4ab, after that find the value of a & b , finally you put the value get the answer
another solution: let d = (a+b-1=0), given g= (a^2+b^2-2=0) get remainder g/d = 2*b^2-2*b-1 =0 => b=1/2 +/- sqrt(3)/2. use synthetic division (a^11+b^11)/(a+b-1) and get remainder R(b): R(b)=11*b^10 - 55*b^9 + 165*b^8 - 330*b^7 + 462*b^6 - 462*b^5 + 330*b^4 - 165*b^3 + 55*b^2 - 11*b + 1. now use synthetic division R(b)/(b-(1/2+/-sqrt(3)/2) -> 989/32 => final result (a^11+b^11)=989/32)
I am always puzzled when I see solutions like that. Who the hell need it? You have two equations with two variables. Easily find both of them as (1+=sqrt(3)/2 and after any of their combination. How twisted mind should people have to post 16 minutes video with crazy computations for that task?
Now he got a result in Q where your result needs to be assumed to be in R ... unless you can get dont get rid of the "twisted mind" root in the term (x+sqrt(y))^11 + (x-sqrt(y))^11 ... whereas with binomial sum at least the odd exponents for sqrt(y) should cancel and the even are integer then 🤔
@@zyklos229Too complicated. In this particular case quadratic equation has 2 real roots. No other complex roots exists. To raise them in power 11 is not rocket science...
You could solve for a and b right after step 2. It would probably be easier (have to check). Also, this cannot be a math Olympiad question. This is something I was doing back in 9th grade or so for normal school exams.
alternative? 2*x^2-2*x-1 =0 & x^11+y^11= (2048*x^22-1)/(2048*x^11) . Use synthetic division (by 2*x^2-2*x-1) on both numerator and denominator to obtain remainders. i.e. x^11+y^11 -> (1129438*x+413402)/ (36544*x+13376) = 989/32 .
“Fantastic explanation! I love how you break down complex concepts. I do similar math tutorials and challenges on my channel. Check it out for a fresh perspective on math problems!”
You never said anything about 2ab, thus you changed the terms of the puzzle. Plus you misspelled "equation" early on, and your 1's look horrid. The answer is a=1.5 and b=-0.5. This fits the two terms of the puzzle that was stated. Oh, and your "music' is repetitive.
Childish maths problems are passed off as maths Olympiad problems for gullible viewers. Maths Olympiad, like the sporting Olympiads are tough and beyond the abilities of normal humans.
Besides, to multiply equations, you bring all terms to one side (e.g. a+b-1)=0 and then multiply them. You don't just multiply left terms with left terms and right terms with right terms. In other words, your solution is horse shit.
b=1-a substitute for b in a^2+b^2=2. You will get 2*a^2-2*a-1=0
you can find a and b
Questo è il modo migliore, non quello del video, che non so dove porti.
Yeah, but it takes literally hours to raise it to the power of 11 and calculate🤦♂️🤦♂️
The question is to find a solution to the differential equation, a" + b" =?. If it is understand that a and b can be functions then finding numerical values for a and b may be a part of the solution but will not determine what function a and function b are.
@@shingshing01It's a¹¹+b¹¹ not a''+b''.
This is what I did. Find a and b, square them for a^2 and b^2, square them for a^4 and b^4, square them for a^8 and b^8, multiply the ^4 by the ^8 and divide by the ^1's gives you a^11 and b^11. All you ever have are binomials over a numeric denominator and the a's and b's are the same except one is + while the other is -. Really wasn't complicated at all.
There's a nice way using Viete's formulas - a,b are the roots of the equation x^2-(a+b)x + ab = 0, after finding ab=-1/2 you get 2x^2-2x-1 = 0 or x^(n+2)=x^(n+1) + (x^n)/2 - always true for roots a,b, giving the recursive relationship a^(n+2) + b^(n+2) = (a^(n+1) + b^(n+1)) + (a^n + b^n)/2.
Nice problem, nice solution!
Let a+b =S and ab=P. In the second equation a^2+b^2 =(a+b)^2-2ab=2 leads to ab=-1/2. If a and b are solutions of the quadratic equation X^2+PX+S=0, that give us a,b =(1±√3)/2.
What I was thinking tbh
@@MegaSARITE nah ab is -1
@@thunderpokemon2456 recalcute and verify again!
@@MegaSARITE yeah sry it negative 1 by 2
2 equations, 2 parameters enough to solve.
(a-b)^2 = a^2 + b^2 - 2ab
(a-b)^2 = 2 - 2*(-1/2) = 3
a-b = sqrt(3)
a+b = 1
Add above
2a = 1+sqrt(3)
Hence a = (1+sqrt(3))/2 = 1.366
Hence b = 1-1.366 = -0.366
Therefore a^11 + b^11 = 31.
DUDE YOU NEED TO DRAW 1'S NOT LIKE STICKS CAUSE I THOUGHT YOU WERE TAKING THE DOUBLE DERIVATIVE OF A AND B!!!
Same😭
not for gymnasium students
YEAH! And in the answer it says g8g divided by 32. I never saw where the value of "g" was determined!!!
@@thomasharding1838are u stupid lol it’s 989
@@gilbertodeoliveirafrota5345hm? Not?
Time taking and lengthy process. If you can find value of a-b by the formula (a-b)^2=(a+b)^2-4ab, after that find the value of a & b , finally you put the value get the answer
another solution: let d = (a+b-1=0), given g= (a^2+b^2-2=0) get remainder g/d = 2*b^2-2*b-1 =0 => b=1/2 +/- sqrt(3)/2.
use synthetic division (a^11+b^11)/(a+b-1) and get remainder R(b):
R(b)=11*b^10 - 55*b^9 + 165*b^8 - 330*b^7 + 462*b^6 - 462*b^5 + 330*b^4 - 165*b^3 + 55*b^2 - 11*b + 1.
now use synthetic division R(b)/(b-(1/2+/-sqrt(3)/2) -> 989/32 => final result (a^11+b^11)=989/32)
Let Sn be a^n + b^n Then you can show Sn+1 = Sn + 1/2Sn-1
S11 can be calculated with primary school math starting from S1 and S2
And S_0 = a^0 + b^0 too!
a+b = 1
a²+b² = 2
Substitute a²+b² and rewrite it as (a+b)²-2ab = 2
(a+b)²-2ab = 2
[as (a+b) equals to 1]
=> (1)²-2ab = 2
=> 1-2ab = 2
=> -2ab = 1
(multiply both sides by -1)
so, 2ab = -1
ab = -1/2
S(n) = a^n + b^n
S(n) = (a+b)[S(n-1)] - abs[n-2]
S(n) = S[n-1] + 1/2[sn-2]
s(0) = a^0 + b^0 = 1+1 = 2
s(1) = 1 since a+b = 1
s(2) = 2 (given)
s(3) = s(2) + 1/2 s(1) = 2 + 1/2 = 2.5
s(4) = s(3) + 1/2 s(2) = 2.5 + 1 = 3.5
s(5) = s(4) + 1/2 s(3) = 3.5 + 1.25 = 4.75
s(6) = s(5) + 1/2 s(4) = 4.75 + 1.75 = 6.5
s(7) = s(6) + 1/2 s(5) = 6.5 + 2.375 = 8.875
s(8) = s(7) + 1/2 s(6) = 8.875 + 3.25 = 12.125
s(9) = s(8) + 1/2 s(7) = 12.125 + 4.4375 = 16.5625
s(10) = s(9) + 1/2 s(8) = 16.5625 + 6.0625 = 22.625
s(11) = s(10) + 1/2 s(9) = 22.625 + 1/2 (16.5625) = 8.28125 + 22.625 = 30.90625
so 30.90625 is our final answer.................
Perfect solution..👍. Here there is no other shortcut trick.. What is the point to find value of a and b?
The point is, that both solutions are equal ... technically each valid expression for "?" is a solution from math point of view.
nice
1:07 too boring and complicated
RHS value = power of a or b= 11
Don't think too much
Long proces
Brilliant mathematician 🎉 He rekindled my interest in mathematics at the age of 66 y
❤️❤️
I am always puzzled when I see solutions like that. Who the hell need it? You have two equations with two variables. Easily find both of them as (1+=sqrt(3)/2 and after any of their combination. How twisted mind should people have to post 16 minutes video with crazy computations for that task?
Now he got a result in Q where your result needs to be assumed to be in R ... unless you can get dont get rid of the "twisted mind" root in the term
(x+sqrt(y))^11 + (x-sqrt(y))^11 ... whereas with binomial sum at least the odd exponents for sqrt(y) should cancel and the even are integer then 🤔
@@zyklos229Too complicated. In this particular case quadratic equation has 2 real roots. No other complex roots exists. To raise them in power 11 is not rocket science...
How would you solve this question for an exam? 😁
I grew a beard watching this...but still enjoyed it.
🙏🙏
Music no need for this video. You can say) .
Think you!
it is very easy to find the value of a-b
But you are not trying to find values for a and b. You are trying to find what a" + b"equals.
You could solve for a and b right after step 2. It would probably be easier (have to check).
Also, this cannot be a math Olympiad question. This is something I was doing back in 9th grade or so for normal school exams.
a + b = 1 ← this is the sum (S)
(a + b)² = a² + 2ab + b²
(a + b)² = (a² + b²) + 2ab
1² = (2) + 2ab
1 = 2 + 2ab
2ab = - 1
ab = - 1/2 ← this is the product (P)
a & b are the solution of the equation:
x² - Sx + P = 0 → where S is the sum and where P is the product
x² - x - (1/2) = 0
x² - x + (1/4) - (1/4) - (1/2) = 0
x² - x + (1/4) = 3/4
[x - (1/2)]² = 3/4
x - (1/2) = ± (√3)/2
x = (1/2) ± (√3)/2 → the 2 solutions are:
a = (1 + √3)/2
b = (1 - √3)/2
a² = (1 + √3)²/2² = (1 + 2√3 + 3)/4 = (4 + 2√3)/4 = (2 + √3)/2
b² = (1 - √3)²/2² = (1 - 2√3 + 3)/4 = (4 - 2√3)/4 = (2 - √3)/2
a³ = a * a² = [(1 + √3)/2] * [(2 + √3)/2] = (2 + √3 + 2√3 + 3)/4 = (5 + 3√3)/4
b³ = b * b² = [(1 - √3)/2] * [(2 - √3)/2] = (2 - √3 - 2√3 + 3)/4 = (5 - 3√3)/4
a⁴ = (a²)² = (2 + √3)²/2² = (4 + 4√3 + 3)/4 = (7 + 4√3)/4
b⁴ = (b²)² = (2 - √3)²/2² = (4 - 4√3 + 3)/4 = (7 - 4√3)/4
a⁸ = (a⁴)² = (7 + 4√3)²/4² = (49 + 56√3 + 48)/16 = (97 + 56√3)/16
b⁸ = (a⁴)² = (7 - 4√3)²/4² = (49 - 56√3 + 48)/16 = (97 - 56√3)/16
a¹¹ = a³ * a⁸
a¹¹ = [(5 + 3√3)/4] * [(97 + 56√3)/16] = (485 + 280√3 + 291√3 + 504)/64 = (989 + 571√3)/64
b¹¹ = b³ * b⁸
b¹¹ = [(5 - 3√3)/4] * [(97 - 56√3)/16] = (485 - 280√3 - 291√3 + 504)/64 = (989 - 571√3)/64
a¹¹ + b¹¹ = [(989 + 571√3)/64)] + [(989 - 571√3)/64]
a¹¹ + b¹¹ = (989 + 571√3 + 989 - 571√3)/64
a¹¹ + b¹¹ = (2 * 989)/64
a¹¹ + b¹¹ = 989/32
Very nice ❤️❤️
i ❤ Mathematics
alternative? 2*x^2-2*x-1 =0 & x^11+y^11= (2048*x^22-1)/(2048*x^11) . Use synthetic division (by 2*x^2-2*x-1) on both numerator and denominator to obtain remainders. i.e. x^11+y^11 -> (1129438*x+413402)/ (36544*x+13376) = 989/32 .
ليه المرار الطافح ده ، استخدم نظرية ذات الحدين واخلص
You could use the a to the fifth equation to the same benefit.
What is 13multiplied by13,you did not check also
169
2 ans
Step 2 :::
(a+b)^2= (a-b)^2+4ab =1
(a-b)^2=(a+b)^2-4ab=1
a-b=1
Step 3::::
a+b=1
a-b=1
By adding
a=1
So b=0
Therefore,
a^11+b^11=1^11+0^11=1
😅😅bro u have mad the error in step-2 (a-b)= ±√3
A+B=1
(A+B)^2=1^2
A^2+B^2+2AB=1
A^2+B^2=-2AB+1
1=-2AB
AB=-1/2
A=-1/2b
A^11+B^11= -1/2B^11+B^11=2B^121-1/2B^11
❤️❤️
Change music ?
C’est vraiment passer derrière son genou pour se gratter la tête. 😂
2789/32 is result
1789/32.
smart way but ... too "dump". Easy to find a, b => done :-)
Bro wrong answer cause at the end you have made a calculation mistake 247/8 + 1/32 = 988/32 = 247/8
Bruh.
247/8 = 988/32.
988/32 + 1/32 = 989/32.
He's right.
And we never defined a or b. Interesting
((sin(23π/12)·√2)^11)+ ((cos(23π/12)·√2)^11)=989/32
2
Too long , attention span is 3 seconds max.🤣🤣🤣🧐🧐
Algebra hates it
“Fantastic explanation! I love how you break down complex concepts. I do similar math tutorials and challenges on my channel. Check it out for a fresh perspective on math problems!”
11 (??)
Solutions
a=(1+3^1/2)/2
b=(1-3^1/2)/2
If you swap a and b, that is also a solution.
The music is REALLY annoying
If a+b=1 then 2×(a+b)=2=(a²+b²) 🤔 not so hard from here on.... 2a+2b-a²-b²=0 ....etc😊
I would say 11😂
You never said anything about 2ab, thus you changed the terms of the puzzle.
Plus you misspelled "equation" early on, and your 1's look horrid.
The answer is a=1.5 and b=-0.5. This fits the two terms of the puzzle that was stated.
Oh, and your "music' is repetitive.
Thank you for your feedback 😊
@@mks-learning-easy You are welcome! The best of luck to you!
Childish maths problems are passed off as maths Olympiad problems for gullible viewers. Maths Olympiad, like the sporting Olympiads are tough and beyond the abilities of normal humans.
time wasted, too long, not the right way to solve this 😧
Ans=2. , as a=1. & b=1. a^11+b^11=1^11+1^11=1+1=2.
Wrong. That would give a+b=2. But here a+b=1.
Too much lengty.
Answer is 2
Besides, to multiply equations, you bring all terms to one side (e.g. a+b-1)=0 and then multiply them. You don't just multiply left terms with left terms and right terms with right terms. In other words, your solution is horse shit.
0
Music is horrible
Dumb way to solve this, there is a much easier way to solve it
Show it to us