there is a much simpler solution for this: the first two equations can be transformed into a second order polynom: separate x in the first equation x + y = 1 | - y x = 1 - y substitute x for 1 - y in the second equation (1 - y)² + y² = 2 apply second binominal formular (1 - 2y + y²) + y² = 2 simplify 2y² - 2y + 1 = 2 bring to polynom normal form 2y² - 2y + 1 = 2 | - 2 2y² - 2y - 1 = 0 | : 2 y² - y - 0.5 = 0 applying the pq-formula on that, we get two results y(1) = 0.366 y(2) = 1.366 Trying both values in the original equations, we see that y(1) is what we call 'scheinlösung' in germany, while y(2) actually adds up y = 1,366 x = 1 - 1.366 = -0.366 proof: x + y = 1 => 1.366 + ( - 0.366 ) = 1 x² + y² = 2 => 1.366² + (-0.366)² = 2 so then we can easily calculate the last equation and get the very same result as the insanely complex calculations in the video x^11 + y^11 = 1.366^11 + (-0.366)^11 = 30.9 = 989 / 32
this process also works, consider x^2+(1-x)^2=1 solving this you get 2x^2=2x+1 Multiply both sides by x 2x^3=2x^2+x 2x^3=2x+1+x 2x^3=3x+1 Now multiply again with x 2x^4=3x^2+x 2x^4=1.5(2x^2)+x 2x^4=1.5(2x+1)+x 2x^4=4x+1.5 Multiplying by x^3 on both sides 2x^7=4x^4+(1.5)x^3 2x^7=8x+3+(0.75)(2x^3) 2x^7=8x+3+(0.75)(3x+1) 2x^7=10.25x+3.75 Eventually you'll get x^11 in terms of x and y^11 in terms of y and the coefficients will be same, so you can add both of them to get the same answer.
Thank you Premath, These problems always intrigued me from school times, but I have NEVER known how to manage them... I'm still learning today with your lessons. Also, the precision in how you explain, the fact that you NEVER miss or "jump" a single step, even if it is evident, makes me awe, and thank you a lot!
My solution: Like you I first calculated that xy = -½. Next I noticed that since x + y = 1 we can get the general formula x^n + y^n = (x + y)(x^n + y^n) = x^(n+1) + y^(n+1) - ½(x^(n-1) + y^(n-1)) , which we can rewrite as x^(n+1) + y(n+1) = x^n + y^n - ½(x^(n-1) + y^(n-1)). If we now set A(n) = x^n + y^n we get, after relabeling, the recursion formula A(n) = A(n-1) + ½A(n-2), with A(1) = 1 and A(2) = 2. After some boring but simple arithmetics I arrived at the same answer X^11 + y^11 = A(11) = 989/32.
Thanks for the problem, I was a bit disappointed that we have to brute force it though, I was expecting a clever trick or variable substitution :). As many others in the chat I did resolve for x and y, but then calculate their square, 4th power and 6th power, which are quickly calculated to (26+/-sqrt(3))/8. x^11 is x^12/x which is (x^6.x^6)/x, and same for y. So you end up with the following formula x^11 + y^11 = -[(1-sqrt(3)).x^6.x^6 + (1+sqrt(3)).y^6.y^6], which leads to 989/32
Nice problem! Even more systematically: expand (x+y)*(x^n+y^n = x^(n+1)+y^(n+1) + x*y*(x^(n-1)+y^(n-1)). We know x+y=1, it's easy to show x*y=-1/2 so this gives a recurrence x^(n+1)+y^(n+1) = x^n+y^n + (x^(n-1)+y^(n-1))/2. The first 2 terms are given, recurrence gives the result for higher n.
I set up a recursive relationship as follows: Call A_n = x^n + y^n then A_n = (x+y) * B_n where B_n = x^(n-1) - x^ (n-2) * y + x^ (n-3) * y² - .... + y^ (n-1) we can rewrite B_n = x^(n-1) + y^ (n-1) + xy [x^(n-3) - x^ (n-4) * y + x^ (n-5) * y² - .... + y^ (n-3)] so B_n = A_(n-1) - xy A_(n-2) since x+y = 1, A_n = B_n and as you have also shown, from the first two equations follows that xy = -1/2 hence A_n = A_(n-1) + 1/2 A_(n-2) next you can calculate any A_n from A_1 = 1 and A_2 = 2
I did most of the math in the head (I only used a calculator for 11th power) Given: x+y= 1 , x² + y² = 2 First, isolate the “x” by subtracting the “y” over in the first equation. x+y= 1 x = 1-y Next, substitute “x” for the equation we found in step one. x² + y² = 2 (1-y)² + y² = 2 Distribute and simplify. (1-y)² +y² = 2 (1-y)(1-y) + y² = 2 (1-2y + y²) + y² = 2 2y² -2y + 1 = 2 Subtract the two over. 2y² -2y -1 = 0 Then, use the Quadratic formula, simplify, and solve for “y” (remember: quadratics have two answers). y= (2 ± √((-2)² -4(2)(-1))) / ((2)(2)) y= (2 ± √(12)) / (4) y = {1.366025404 , -0.3660254038} Luckily, the two answers from doing the Quadratic equation are both the value of “x” and “y”. “x” and “y” can be either value. For example, “y” would be “1.3660254038” if “x” is “-0.3660254038”. Now, do (x^11) + (y^11). (1.3660254038)^11 + (-0.3660254038)^11 = 30.90625
@@TheLibran38 this is the problem in this type of puzzle, you don't know what is expected from you really, your solution is not very helpful but technically correct. In this puzzle it is not important what is the value of x and y, but rather what is the value of xy
I did the UCCS Olympiad twice in the mid 1980s and was a judge for several years after that. I usually found that every question had an answer that was far easier than you'd originally expect (but required almost a mathematical artistry to get the answer). I first tried solving for (x,y) with the first 2 equations, and quickly realized that was the wrong way to solve because of the 11 exponent.
When you realize that x^2 = x+1/2 and y^2 = y+1/2, the exponentiation becomes easier. By doing (x^2)^2^2 * (x*x^2), and keep subsituting x^2 terms, you end up with 418/32 + 571(x+y)/32, and can insert x+y=1
I used the Binomial Theorem to find some of the powers of (x+y). 989/32 was the final result and this shows us the BEAUTIFUL CONSISTENCY of mathematics. Thank you for these problems.
What’s wrong with simply solving the quadratic equation from the first two and subbing in the last. Sure you end up with (1+_sqrt3)/2 to the power 11 for x and (1-+sqrt3)/2 to the power 11 for y and making use of binomial expansion and noting that odd powers cancel in the sum. For the even powers also you have many simplifications in the factorials. I got the same answer in less than it took the video to finish.
Yes, that's the most obvious solution. I solved it that way, too. But this method is more general and can be modified for solving similar problems where an explicit solution can't be obtained, e.g. with more variables.
@@viharsarok There is only one solution to a problem and the direct path is the easiest. Once a solution is found, we can fabricate an unlimited number of parallel solutions. This is waste of time when simple substitution method works.
Yes I also did it more quickly by first solving for x and y, then working out x^2, x^3, x^4, x^8, x^11 ... but yes there is also something worthwhile about solving the general problem. If the next question on the paper was very similar but with different (and more difficult) values for x+y, x^2+y^2, then you would be rewarded for having pursued the more general approach.
How about geometrical approach? Straight line x+y=1, equivalent of y=1-x, crosses a ring x^2+y^2=2. The x and y of the crossing points to the power eleven, added to each other would give the answer.
There's no i here, the resultant complex number wouldn't have magnitude 1, and it wouldn't rotate along the unit circle, If you have something else in mind please explain
@@lightspeed2014 (x-y)^2=3=>(x-y)=sqrt(3) or -(sqrt3), x & y has the same responsibility in the function so consider x>y => x-y=sqrt(3) =>x=1.336...; y=-0.336...
I did this slightly differently, but by heavens it was tedious. A big Pascal triangle with alternate terms being doubled or cancelled out, each multiplied by 3^n/2 where n is the entry in the Pascal triangle. So 2 x (1 x 3^0, 55 x 3^1, 330 x 3^2 + 462 x 3^3 + 165 x 3^4 + 11 x 3^5)/(2 ^11) which simplified to 989/32...
Please, can I suggest the following calculation which goes to same result? • x + y = 1 then y = -x + 1 • in x² + y² = 2 I replace y by -x + 1 and I get • x² + (-x + 1)² = 2 and I develop • x² + (x² - 2x + 1) = 2 • x² + x² - 2x + 1 - 2 = 0 • 2x² - 2x - 1 = 0 then I compute the discriminant D = b² - 4ac • D = (-2)² - (4*2*-1) = 4 - (-8) = 4 + 8 = 12 • D = 12 then D > 0 then x (case #1) = [-b + Sqrt(12)]/2a = [2 + Sqrt(12)]/4 = (1 + Sqrt(3))/2 = 1,366025 • D = 12 then D > 0 then x (case #2) = [-b - Sqrt(12)]/2a = [2 - Sqrt(12)]/4 = (1 - Sqrt(3))/2 = -0,366025 • I come back to x + y = 1 and I replace x by 1,366025 (case #1) and I get • 1,366025 + y = 1 then y = 1 - 1,366025 = -0,366025 • from x = 1,366025 and y = -0,366025 it's easy to calculate x^11 + y^11 • x^11 + y^11 = (1,366025)^11 + (-0,366025)^11 = 30,906 which is equal to your final answer: 989/32 • verification: 989/32 = 30,906 • note: with x (case #2) I should get the same result (of course): 30,906 (or 989/32) AND THANK YOU VERY MUCH FOR YOUR ALL VERY ATTRACTIVE AND VERY INSTRUCTIVE VIDEOS (I'm French, from Toulouse) !!!
@@HackedPC not really. I did it the same way as him and you could instead of calculating the numbers just put x and y (equal to 0,5 (+/-) sqrt(3/4)) to the power of 11 each. Wouldnt be as good as the solution since youd be multiplying it out to a very long term but it would give you a bracket in the end too ^^ So at least the approach is valid imo
x+y=1 x^2+y^2+2xy=1 2+2xy=1 2xy=-1 xy=-1/2 Now (x+y)(x^n-1+y^n-1) =x^n+y^n+xy(x^n-2+y^n-2) So x^n+y^n=(x+y)(x^n-1+y^n-1)-xy(x^n-2+y^n-2) Putting the value x+y & xy We get x^n+y^n=(x^n-1+y^n-1)+(1/2)*(x^n-2+y^n-2) Now it is a function like f(n)=f(n-1)+1/2*f(n-2) f(1)=x+y=1 f(2)=2 given f(3)=f(2)+1/2*f(1) =2+1/2=5/2 f(4)=5/2 + 1/2* 2=7/2 & so on
This problem can be solved using polynomial approach. Let us assume that x and y are zeroes of some quadratic polynomial f(X). So we have f(X) = (X-x)(X-y) = X² -(x+y)X+xy Now our task is to simply find those coefficients. We can easily calculate 2xy=(x+y)²-(x²+y²)=-1 =>xy=-1/2. Therefore, f(X)=X²-X-1/2. Since x and y are zeros of quadratic polynomial we have x²-x-1/2=0 and y²-y-1/2=0 => x²=x+1/2---(1) and y²=y+1/2---(2) Multiplying x both sides in eqn1 and y in eqn2 x³=x²+x/2 and y³=y²+y/2=> x³+y³=x²+y²+(x+y)/2=5/2. Similarly if we continue this we can see the pattern Xpow11+ypow11=xpow10+ypow10+1/2(xpow9+ ypow9). We can easily calculate xpow10 +ypow10 from xpow5+ypow5(This can be easily calculated from pattern) and xpow9+ypow9 from x³+y³.
We could have made a quadratic equation as we know the product and the sum of the two numbers. Ultimately, solving for x^11 + y^11. However, a good solution to appreciate problem-solving with the help of patterns.
Well i got the answer in a much simpler and shorter method. make x the subject in equation 1 resulting in x=1-y. then put this in equation 2 to get (1-y) all squared + ysquared = 2. from there you get a quadratic. use the almighty formula, getting y to be 1+root3/2. then use that to find x and subsequently the final answer
If x+y=1 and x^2+y^2=2 then by the formula (x+y)^2=x^2+y^2+2xy, we have xy=-1/2 . Now solving the eqns x+y=1 and xy=-1/2 we will get x=(1+√3)/2 and y =(1-√3)/2 using binomial expansion we will get x^11+y^11=31648/1024 =989/32
Once you calculate xy to be -1/2 and you the realize x11+y11 = (x+1)(x10-yx9+y2x8-y3x7+y4x6-y5x5+y6x4-y7x3+y8x2-y9x+y10) you can simplify very quickly :)
I did a similar concept, but used different equations that shrunk the process a little. I used the x+y and x2+y2 to get the x3+y3, the x2+x2 and x3+y3 to get x5+y5, the x3+y3 and x5+y5 to get x8+y8, and lastly, x3+y3 and x8+y8 to get the x11+y11.
I ever know you are a good professor, Sir, But now You suparse youself. If I have done (a+b)¹¹ I would have get my brain out! raising 11 is the same product from (a^5+b^5)(a^6+b^6), thanak you , sir!
Your videos are food form my mind. Your explanation is mind gym for me but I have one odd, blunt question: we have two unknows and three equations. Can't we just solve the first two equations and plug in the results in the third? Apologies again for my naivity.
От иррациональных корней не ожидаешь рациональных степенных выражений. Когда задача составляется от ответа, в этом есть лукавство автора. Но спасибо, необычное решение. Я начал решать в лоб и упёрся в тупик
Yes and doing by that tricks I am getting 2047 answer The value of X is coming 2 And y is -1 And after solving x11 +y 11 The answer is coming 2047. Try with yourself
If this could be solved with calculators, relatively quick to find x = (1 ± sqrt[3]) / 2 and y = (1 ∓ sqrt[3]) / 2, then calculate the desired answer directly. Granted far more an engineer's approach than a mathematician's.
Sir you teaches very good and your voice is also good
So nice of you Megha dear
Thank you! Cheers!
Keep rocking😀
Love and prayers from the USA!
You are the Bob Ross of Math 😉
I do agree..
@@PreMath 1oa1m1a
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
there is a much simpler solution for this:
the first two equations can be transformed into a second order polynom:
separate x in the first equation
x + y = 1 | - y
x = 1 - y
substitute x for 1 - y in the second equation
(1 - y)² + y² = 2
apply second binominal formular
(1 - 2y + y²) + y² = 2
simplify
2y² - 2y + 1 = 2
bring to polynom normal form
2y² - 2y + 1 = 2 | - 2
2y² - 2y - 1 = 0 | : 2
y² - y - 0.5 = 0
applying the pq-formula on that, we get two results
y(1) = 0.366
y(2) = 1.366
Trying both values in the original equations, we see that y(1) is what we call 'scheinlösung' in germany, while y(2) actually adds up
y = 1,366
x = 1 - 1.366 = -0.366
proof:
x + y = 1 => 1.366 + ( - 0.366 ) = 1
x² + y² = 2 => 1.366² + (-0.366)² = 2
so then we can easily calculate the last equation and get the very same result as the insanely complex calculations in the video
x^11 + y^11 = 1.366^11 + (-0.366)^11 = 30.9 = 989 / 32
Like my solve
this process also works,
consider x^2+(1-x)^2=1
solving this you get 2x^2=2x+1
Multiply both sides by x
2x^3=2x^2+x
2x^3=2x+1+x
2x^3=3x+1
Now multiply again with x
2x^4=3x^2+x
2x^4=1.5(2x^2)+x
2x^4=1.5(2x+1)+x
2x^4=4x+1.5
Multiplying by x^3 on both sides
2x^7=4x^4+(1.5)x^3
2x^7=8x+3+(0.75)(2x^3)
2x^7=8x+3+(0.75)(3x+1)
2x^7=10.25x+3.75
Eventually you'll get x^11 in terms of x and y^11 in terms of y and the coefficients will be same, so you can add both of them to get the same answer.
Хорош
Thank you Premath, These problems always intrigued me from school times, but I have NEVER known how to manage them... I'm still learning today with your lessons. Also, the precision in how you explain, the fact that you NEVER miss or "jump" a single step, even if it is evident, makes me awe, and thank you a lot!
My solution:
Like you I first calculated that xy = -½.
Next I noticed that since x + y = 1 we can get the general formula
x^n + y^n = (x + y)(x^n + y^n) = x^(n+1) + y^(n+1) - ½(x^(n-1) + y^(n-1)) , which we can rewrite as
x^(n+1) + y(n+1) = x^n + y^n - ½(x^(n-1) + y^(n-1)).
If we now set A(n) = x^n + y^n we get, after relabeling, the recursion formula
A(n) = A(n-1) + ½A(n-2), with A(1) = 1 and A(2) = 2.
After some boring but simple arithmetics I arrived at the same answer
X^11 + y^11 = A(11) = 989/32.
Beautifully complex, yet simple.
Thanks.
This question was one of the most interesting questions that I have come across, Thank you very much! Hope to see more!
i wish i had the teacher like you when i was at school!
i could easily win any olympiad around the world.
Your tutorial just is a simply genius!
Thanks for the problem, I was a bit disappointed that we have to brute force it though, I was expecting a clever trick or variable substitution :). As many others in the chat I did resolve for x and y, but then calculate their square, 4th power and 6th power, which are quickly calculated to (26+/-sqrt(3))/8. x^11 is x^12/x which is (x^6.x^6)/x, and same for y. So you end up with the following formula x^11 + y^11 = -[(1-sqrt(3)).x^6.x^6 + (1+sqrt(3)).y^6.y^6], which leads to 989/32
Thank you for a refresh of problems studied in my school. Your method of solution is simple and great.
Nice problem! Even more systematically: expand (x+y)*(x^n+y^n = x^(n+1)+y^(n+1) + x*y*(x^(n-1)+y^(n-1)). We know x+y=1, it's easy to show x*y=-1/2 so this gives a recurrence x^(n+1)+y^(n+1) = x^n+y^n + (x^(n-1)+y^(n-1))/2. The first 2 terms are given, recurrence gives the result for higher n.
Excellent
Thank you! Cheers!
Keep rocking😀
good method
@@user-jm1zo3ji6o Yeah!!
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
I set up a recursive relationship as follows:
Call A_n = x^n + y^n
then A_n = (x+y) * B_n
where B_n = x^(n-1) - x^ (n-2) * y + x^ (n-3) * y² - .... + y^ (n-1)
we can rewrite
B_n = x^(n-1) + y^ (n-1) + xy [x^(n-3) - x^ (n-4) * y + x^ (n-5) * y² - .... + y^ (n-3)]
so B_n = A_(n-1) - xy A_(n-2)
since x+y = 1, A_n = B_n
and as you have also shown, from the first two equations follows that xy = -1/2
hence A_n = A_(n-1) + 1/2 A_(n-2)
next you can calculate any A_n from A_1 = 1 and A_2 = 2
I really appreciate your content. Your problems are easy to understand and I find your solutions creative. Keep up the good work!
I appreciate that!
Thank you Nonsio! Cheers!
Keep rocking😀
Your teach is very very good but i was able to solve this math problem in a much simpler way
🇪🇬🇪🇬Greetings from Egypt
I did most of the math in the head (I only used a calculator for 11th power)
Given: x+y= 1 , x² + y² = 2
First, isolate the “x” by subtracting the “y” over in the first equation.
x+y= 1
x = 1-y
Next, substitute “x” for the equation we found in step one.
x² + y² = 2
(1-y)² + y² = 2
Distribute and simplify.
(1-y)² +y² = 2
(1-y)(1-y) + y² = 2
(1-2y + y²) + y² = 2
2y² -2y + 1 = 2
Subtract the two over.
2y² -2y -1 = 0
Then, use the Quadratic formula, simplify, and solve for “y” (remember: quadratics have two answers).
y= (2 ± √((-2)² -4(2)(-1))) / ((2)(2))
y= (2 ± √(12)) / (4)
y = {1.366025404 , -0.3660254038}
Luckily, the two answers from doing the Quadratic equation are both the value of “x” and “y”.
“x” and “y” can be either value. For example, “y” would be “1.3660254038” if “x” is “-0.3660254038”.
Now, do (x^11) + (y^11).
(1.3660254038)^11 + (-0.3660254038)^11
= 30.90625
I too did it like you. of course, the answer doesn't come out as a fraction of whole numbers...
@@TheLibran38 this is the problem in this type of puzzle, you don't know what is expected from you really, your solution is not very helpful but technically correct. In this puzzle it is not important what is the value of x and y, but rather what is the value of xy
yeah same for me
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
I did the UCCS Olympiad twice in the mid 1980s and was a judge for several years after that. I usually found that every question had an answer that was far easier than you'd originally expect (but required almost a mathematical artistry to get the answer). I first tried solving for (x,y) with the first 2 equations, and quickly realized that was the wrong way to solve because of the 11 exponent.
When you realize that x^2 = x+1/2 and y^2 = y+1/2, the exponentiation becomes easier. By doing (x^2)^2^2 * (x*x^2), and keep subsituting x^2 terms, you end up with 418/32 + 571(x+y)/32, and can insert x+y=1
I did not see a system of equations, but a pattern.
X^1+Y^1 =1
X^2+Y^2 =2
X^11+Y^11 =11.
Can be another way to look at it. 😂
@@menensa A way to look at it to get zero Marks!
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
I used the Binomial Theorem to find some of the powers of (x+y). 989/32 was the final result and this shows us the BEAUTIFUL CONSISTENCY of mathematics. Thank you for these problems.
What’s wrong with simply solving the quadratic equation from the first two and subbing in the last. Sure you end up with (1+_sqrt3)/2 to the power 11 for x and (1-+sqrt3)/2 to the power 11 for y and making use of binomial expansion and noting that odd powers cancel in the sum. For the even powers also you have many simplifications in the factorials. I got the same answer in less than it took the video to finish.
Yes, that's the most obvious solution. I solved it that way, too. But this method is more general and can be modified for solving similar problems where an explicit solution can't be obtained, e.g. with more variables.
@@viharsarok There is only one solution to a problem and the direct path is the easiest. Once a solution is found, we can fabricate an unlimited number of parallel solutions. This is waste of time when simple substitution method works.
@@librealgerien I don't agree. There is not "only one solution" and generalizations do matter.
Yes I also did it more quickly by first solving for x and y, then working out x^2, x^3, x^4, x^8, x^11 ... but yes there is also something worthwhile about solving the general problem. If the next question on the paper was very similar but with different (and more difficult) values for x+y, x^2+y^2, then you would be rewarded for having pursued the more general approach.
I don't think it's just about the solution, it's about the skill to look at things differently.
How about geometrical approach? Straight line x+y=1, equivalent of y=1-x, crosses a ring x^2+y^2=2. The x and y of the crossing points to the power eleven, added to each other would give the answer.
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
I think this’s unnecessarily complicated…you can just solve for x and y using a representation of re^(i*theta) and raise it to 11th power very easily…
There's no i here, the resultant complex number wouldn't have magnitude 1, and it wouldn't rotate along the unit circle, If you have something else in mind please explain
@@lightspeed2014 (x-y)^2=3=>(x-y)=sqrt(3) or -(sqrt3), x & y has the same responsibility in the function so consider x>y => x-y=sqrt(3) =>x=1.336...; y=-0.336...
Then how can you calculate the ^11 exponent without calculator?
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
@@namanarora2005 binomial expansion
I solved the first two equations for x and y, and then substituted those values into the third equation. Took about a minute.
This is my method too.
I hv also tried like this
@@rupamsaha4123 We probably would not get good marks for this method though!
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
@@PreservationEnthusiast Why?
Sir, I understood the even power substitutions, but not sure how you did the odd power substitutions.
Fascinating approach!
Thank you William! Cheers!
Keep rocking😀
You can also solve the math.by determining x-y & with that calculating x,y.....then from the value one can.easily solve the math
You can do it by Newton sums by by assuming X and Y as root of a quadratic Eqn
I did this slightly differently, but by heavens it was tedious. A big Pascal triangle with alternate terms being doubled or cancelled out, each multiplied by 3^n/2 where n is the entry in the Pascal triangle. So 2 x (1 x 3^0, 55 x 3^1, 330 x 3^2 + 462 x 3^3 + 165 x 3^4 + 11 x 3^5)/(2 ^11) which simplified to 989/32...
Why not simultaneous eqn x+y=1 and to the second eqn I get the answer right. Then after getting the answer we substitute on the third. ??
I like your videos. Thank you sir
Please, can I suggest the following calculation which goes to same result?
• x + y = 1 then y = -x + 1
• in x² + y² = 2 I replace y by -x + 1 and I get
• x² + (-x + 1)² = 2 and I develop
• x² + (x² - 2x + 1) = 2
• x² + x² - 2x + 1 - 2 = 0
• 2x² - 2x - 1 = 0 then I compute the discriminant D = b² - 4ac
• D = (-2)² - (4*2*-1) = 4 - (-8) = 4 + 8 = 12
• D = 12 then D > 0 then x (case #1) = [-b + Sqrt(12)]/2a = [2 + Sqrt(12)]/4 = (1 + Sqrt(3))/2 = 1,366025
• D = 12 then D > 0 then x (case #2) = [-b - Sqrt(12)]/2a = [2 - Sqrt(12)]/4 = (1 - Sqrt(3))/2 = -0,366025
• I come back to x + y = 1 and I replace x by 1,366025 (case #1) and I get
• 1,366025 + y = 1 then y = 1 - 1,366025 = -0,366025
• from x = 1,366025 and y = -0,366025 it's easy to calculate x^11 + y^11
• x^11 + y^11 = (1,366025)^11 + (-0,366025)^11 = 30,906 which is equal to your final answer: 989/32
• verification: 989/32 = 30,906
• note: with x (case #2) I should get the same result (of course): 30,906 (or 989/32)
AND THANK YOU VERY MUCH FOR YOUR ALL VERY ATTRACTIVE AND VERY INSTRUCTIVE VIDEOS (I'm French, from Toulouse) !!!
@@HackedPC not really. I did it the same way as him and you could instead of calculating the numbers just put x and y (equal to 0,5 (+/-) sqrt(3/4)) to the power of 11 each. Wouldnt be as good as the solution since youd be multiplying it out to a very long term but it would give you a bracket in the end too ^^
So at least the approach is valid imo
@@HackedPC ever heared somthing called pq formula
@@HackedPC OK i think you will learn it this or next year (at least your are in Germany)
@@HackedPC nobody
@@HackedPC i answered to the comment over the other (who said i am...)
Nice explanation, thanks
combine vietta and newton sum theorem
denote x and y are the root of a^2 - a - 1/2 = 0
P1 is x+y
P2 is x^2 + y^2
P3 is x^3 + y^3
.
.
P1 = 1
P2 = 2
P3 - P2 -(1/2)P1 = 0 --> P3 = 5/2
P_n = P_(n-1) + 1/2 P_(n-2)
P4 = P3 + 1/2 P2 = 7/2
P5 = P4 + 1/2 P3 = 19/4
P6 = P5 + 1/2 P4 = 26/4
P7 = 52/8 + 19/8 = 71/8
P8 = 71/8 + 26/8 = 97/8
P9 = 194/16 + 71/16 = 265/16
P10 = 265/16 + 97/16 = 362/16
P11 = 724/32 + 265/32 = 989/32
Very cool. Thanks a lot
f(n) = x^n + y^n; f(1) = 1; f(2) = 2; x.y = -1/2; => f(n) = f(n-1) + 1/2.f(n-2) and you can easily calculate a table to get the answer.
Nice!
f(1) = 1
f(2) = 2
f(3) = 2 + 1/2 = 5/2
f(4) = 5/2 + 1 = 7/2
f(5) = 7/2 + 5/4 = 19/4
f(6) = 19/4 + 7/4 = 26/4
f(7) = 26/4 + 19/8 = 71/8
f(8) = 71/8 + 26/4 = 97/8
f(9) = 97/8 + 71/16 = 265/16
f(10) = 265/16 + 97/16 = 362/16
f(11) = 362/16 + 265/32 = 989/32
Hello. How do you demonstrate: f(n) = f(n-1) + 1/2.f(n-2) ?
x+y=1
x^2+y^2+2xy=1
2+2xy=1
2xy=-1
xy=-1/2
Now
(x+y)(x^n-1+y^n-1)
=x^n+y^n+xy(x^n-2+y^n-2)
So
x^n+y^n=(x+y)(x^n-1+y^n-1)-xy(x^n-2+y^n-2)
Putting the value x+y & xy
We get
x^n+y^n=(x^n-1+y^n-1)+(1/2)*(x^n-2+y^n-2)
Now it is a function like
f(n)=f(n-1)+1/2*f(n-2)
f(1)=x+y=1
f(2)=2 given
f(3)=f(2)+1/2*f(1)
=2+1/2=5/2
f(4)=5/2 + 1/2* 2=7/2
& so on
@@francoism2232 (x+y)(x^(n-1)+y^(n-1)) = x^(n) + x.y^(n-1) +y.x^(n-1) + y^(n)
(x+y).f(n-1) = f(n) + x.y.f(n-2)
1.f(n-1) = f(n) - 1/2 . f(n-2)
f(n) = f(n-1) + 1/2 f(n-2)
Sorry for the messy notation.
This problem can be solved using polynomial approach.
Let us assume that x and y are zeroes of some quadratic polynomial f(X).
So we have f(X) = (X-x)(X-y) = X² -(x+y)X+xy
Now our task is to simply find those coefficients.
We can easily calculate 2xy=(x+y)²-(x²+y²)=-1 =>xy=-1/2.
Therefore,
f(X)=X²-X-1/2.
Since x and y are zeros of quadratic polynomial we have
x²-x-1/2=0 and y²-y-1/2=0 => x²=x+1/2---(1) and y²=y+1/2---(2)
Multiplying x both sides in eqn1 and y in eqn2
x³=x²+x/2 and y³=y²+y/2=> x³+y³=x²+y²+(x+y)/2=5/2.
Similarly if we continue this we can see the pattern
Xpow11+ypow11=xpow10+ypow10+1/2(xpow9+ ypow9).
We can easily calculate xpow10 +ypow10 from xpow5+ypow5(This can be easily calculated from pattern)
and xpow9+ypow9 from x³+y³.
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
You have to code for every instance of interaction for physics and it tends to go wrong . Just go with ideologies
GOOD morning, thank you for this important video ... could you tell me what program you use on your computer to make these videos?
If you are interested in olympiad-style algebra, check this video th-cam.com/video/MqJ6-1-6tzMa/w-d-xo.htmlnd several others in the channel
I doubt if a student would notice your method. Solve 1 and 2 as linear and quadratic for x and y and expand 3 by binomial theorem.
Excelente. Me encantó!!!
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
Simply the best
Thank you Yakoub! Cheers!
Keep rocking😀
Tedious but correct
Good Idea.
Sir you teaches very good and your voice is also good
So amazinggggg
Eq(1) and eq(2) gives one linear one quad and can be solved and then subed in eq(3). Why these long process?
x^2 = x+ 1/2 , y^2 = y + 1/2 makes it a bit easier
Beautiful sharing dear sir 🙏❤️🙏🙏
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
Es muy interesante.
Hello my friend. Nice job explaining this! Keep rocking, you are the best haha
Hey, thanks my friend!
Cheers!
Keep rocking😀
Analogy!!!
How did you find x4+y4 ? I'm getting 2 as answer. Please show me sir
We could have made a quadratic equation as we know the product and the sum of the two numbers. Ultimately, solving for x^11 + y^11. However, a good solution to appreciate problem-solving with the help of patterns.
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
Well i got the answer in a much simpler and shorter method. make x the subject in equation 1 resulting in x=1-y. then put this in equation 2 to get (1-y) all squared + ysquared = 2. from there you get a quadratic. use the almighty formula, getting y to be 1+root3/2. then use that to find x and subsequently the final answer
very well done bro, thanks for sharing this math olympiad question
My pleasure
Thank you! Cheers!
Keep rocking😀
If x+y=1 and x^2+y^2=2 then by the formula (x+y)^2=x^2+y^2+2xy, we have xy=-1/2 . Now solving the eqns x+y=1 and xy=-1/2 we will get x=(1+√3)/2 and y =(1-√3)/2 using binomial expansion we will get x^11+y^11=31648/1024 =989/32
I apply the third, fourth, fifth and eighth powers together with the given first and second powers to solve the above problem.
Once you calculate xy to be -1/2 and you the realize x11+y11 = (x+1)(x10-yx9+y2x8-y3x7+y4x6-y5x5+y6x4-y7x3+y8x2-y9x+y10) you can simplify very quickly :)
You can also get the values of x and y directly and calculate, right? Not very smart though.
Amazing amazing amazing
Thank you so much 😀
Glad you think so!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Stay blessed 😀
1st power is 1
2nd is 2
11 must be 11.
There ya go
lmao
@@dudono1744 ... 😀
I did a similar concept, but used different equations that shrunk the process a little. I used the x+y and x2+y2 to get the x3+y3, the x2+x2 and x3+y3 to get x5+y5, the x3+y3 and x5+y5 to get x8+y8, and lastly, x3+y3 and x8+y8 to get the x11+y11.
Awesome Steven
Thank you! Cheers!
Keep rocking😀
Amazing solution 👍thank you teacher🙏
You are welcome
Excellent
Thank you dear! Cheers!
Keep rocking😀
i took out the value off x and y and added 11 power
11. X= half . Or one by two . Y= half . So x+y = one whole . So 11 divided by half = 5 and half so 5 and half +5 and half = 11 .
I have a question. is another answer possible? (I got x=-0.5 and y=1.5, after x^11+y^11 i got 86+509/1024)
Thanks for your very complicated soulution. I soulve in just 2 min and every child (germany) at class 9 do that too🤝
very interesting trick to this question sir thanks keep it up!
Always welcome Nico dear
Thank you! Cheers!
Keep rocking😀
WOW!
Wow I have done my first Olympiad problem myself and also checked my answer thankyou sir for sharing this problem
thanks
I was able to solve. Cool problem!!!!
Excellent Pedro
Thank you! Cheers!
Keep rocking😀
You can easily solve it using quadratic equation And find value of x and y . But thanks for good question .
Very impressive and informative.
Glad you liked it!
Thank you John! Cheers!
Keep rocking😀
Another exercise that I've seen...
من المغرب شكرا لكم على المجهودات .
هxوy هما حلا المعادلةtمربع ناقص t ناقص2\1=0
نحسب بالمتتاليات المكونة بالحلين xوy ......
I ever know you are a good professor, Sir, But now You suparse youself. If I have done (a+b)¹¹ I would have get my brain out! raising 11 is the same product from (a^5+b^5)(a^6+b^6), thanak you , sir!
x+y=1=>(x+y)^2=1^2=1=> x^2+2xy+y^2=1=>2xy=1-(x^2+y^2)=
1-2=-1=>-1/2=xy=x(1-x)=x-x^2
Use Bhaskara to find x and x+y=1 to find y.
Nice
Your videos are food form my mind. Your explanation is mind gym for me but I have one odd, blunt question: we have two unknows and three equations. Can't we just solve the first two equations and plug in the results in the third? Apologies again for my naivity.
Try to do that and you will end up with the answer - much more knowledge needed in that case.
th-cam.com/video/Upvk1_M0B-k/w-d-xo.html
#binomial #polynomial
От иррациональных корней не ожидаешь рациональных степенных выражений. Когда задача составляется от ответа, в этом есть лукавство автора. Но спасибо, необычное решение. Я начал решать в лоб и упёрся в тупик
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html
You can have xy , then x-y then we will get value of x and y easily
Yes and doing by that tricks I am getting 2047 answer
The value of X is coming 2
And y is -1
And after solving x11 +y 11
The answer is coming 2047.
Try with yourself
XplusY=2 @ 1plus 1=2
Xsq plus Ysq also equals to 2 because 1sq plus 1sq=1plus 1=2
Similarly 1power 11 plus 1power11=1plus 1=2
Thank you , it is so cool.
Thank you too!
Cheers!
Keep rocking😀
Thank you
You're welcome Mahalakshmi
Thank you! Cheers!
Keep rocking😀
If this could be solved with calculators, relatively quick to find x = (1 ± sqrt[3]) / 2 and y = (1 ∓ sqrt[3]) / 2, then calculate the desired answer directly. Granted far more an engineer's approach than a mathematician's.
this is not a calc question
Why all this! We can get the values x and y just by using the two first equations and then substitute at the last equation
because calculating x^11 + y^11 when y=[1+root(3)]/2, x = [1-root(3)]/2, would be EXTREMELY annoying
@@Vertraic just put it into a calculator
what if we use trigonometry, like defining x as (sinx)^2 and y as (cosx)^2
you can't know if there is a z such that (sin z)^2 = x and (cos z)^2 = y
@@dudono1744 but i think we can find it out/
like in the second line: x^2 + y^2 = 2
=> (sin z)^4 + (cos z)^4 = 2
=> (0 + (sin z)^2 + (cos z)^2 + 0)^2 - 2 * (sin z)^2 * (cos z)^2 = 2
Watched in 2x speed but also voice is superb
You did not need x^4+y^4. Other than that well done!!!
a + bw + cw^2/aw + bw^2 + c + a + bw + cw^2/ aw^2 + b + cw^2 = 1
I will go for a coffee while someone solves the equation.
X^11 + y^11 = (x^10 + y^10)(y+x) = 729(√3)
Why not x¹¹+y¹¹=11? As we see x¹+y¹=2 & x²+y²=2, then we can come to a generalised equation which says
x^n+y^n=n
1.01.22.
You oughtta switch the signs for the next video. Can you find x^11+y^11 given x-y=1 and x^2-y^2=2?
Great suggestion!
Excellent
Thank you! Cheers!
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@@XJWill1
You gave answer in exponents
Give actual ans
I first find x2+y2, x4+y4, x8+y8. Then x3+y3. Intention was to do (x3+y3)(x8+y8). But I needed x5+y5. So I find that too. Got the same answer!
Excellent Garrick
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@@PreMath thank you! Now I look forward to solve ur daily qns
Well explained
Thank you Ravi dear! Cheers!
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But this math can simply be solved by determining the value of x and y using 4xy formula.
Yes we can ,but its deferent
Are calculators allowed in math Olympiads??
No doubt this problem can be solved in many ways but your method is easy to learn. So congratulations.
Thank you Jasbir dear! Cheers!
Keep rocking😀