Yes, and that P'(x)=(P'(x-1)+P'(x+1))/2 is only fulfilled for linear polynomials can also be seen easily (at least, if we know interpolation): Suppose, P' is a polynomial of degree n such that P'(x)=(P'(x-1)+P'(x+1))/2 2P'(x)-P'(x-1)=P'(x+1) holds for all x. Now, let a=P'(1) and b=P'(0). Then, we can recursively compute P'(2), P'(3), P'(4),...,P'(n). These are n+1 distinct points at which we prescribe the values, hence P' is the uniquely determined polynomial with degree at most n satisfying these prescriptions. Since the linear polynomial p(x)=(a-b)x+b also satisfies these conditions (because p(0)=b, p(1)=a and p(x+1)=(a-b)x+a-b+b=(a-b)x+a=2*((a-b)x+b)-((a-b)(x-1)+b))), P' must be equal to p and therefore be linear.
I wrote (2xi-3y)^3, expanded and then replaced using the original equations to get (2xi-3y)^3=8(1-i) which is easily solved by finding cubed roots of the RHS.
the first one can be done much more nicely, we start with : 36x²y- 27y^3 = 8 4x^3 - 27xy² = 4 By multiplying the second line by two and factoring the coefficients (27 = 3^3,36 = 3*3*2*2) we get : 3(2x)²(3y)-(3y)^3 = 8 (1) (2x)^3-3(2x)(3y)² = 8 (2) Lets define X = 2x and Y = 3y and see that (X+iY)^3 = X^3+3iX²Y-3XY²-iY^3 so that (2)+i(1) give (X+iY)^3 = 8(1+i) All that is left is to calculate the cube roots of 1+i and take the real and imaginary part. 1+i = sqrt(2)exp(i pi/4) so we need to calculate the real and imaginary parts of : exp(i pi/12), j exp(i pi/12)exp(i (pi/12+2pi/3)) = exp(3i pi/4) = (-1+i)/sqrt(2) and j² exp(i pi/12), with j = (-1+i sqrt(3))/2 hence X+iY = 2^(1+1/6-1/2)(-1+i)*w = (-1+i)*2w/cbrt(2) with w = 1, j or j² that is : X+iY = 2*(-1+i)/cbrt(2) or (-1+i)(-1+isqrt(3))/cbrt(2) or (-1+i)(-1-isqrt(3))/cbrt(2) X+iY = 2*(-1+i)/cbrt(2) or (-sqrt(3)+1-i(sqrt(3)+1))/cbrt(2) or (sqrt(3)+1+i(sqrt(3)-1))/cbrt(2) And finally : x,y = -2/2cbrt(2),2/3cbrt(2) x,y = -(sqrt(3)-1)/2cbrt(2),-(sqrt(3)+1)/3cbrt(2) x,y = (sqrt(3)+1)/2cbrt(2),(sqrt(3)-1)/3cbrt(2)
Q2) since the polynomials are homogeneous, setting x=t*y and comparing both equations gives us a cubic polynomial with an obvious solution at t=-2/3 (easily spotted if you simply notice the equation is almost a cube of something else). The rest is substitution. Q4) if we restrict x to natural numbers, then we end up with a difference equation a(n+1)=2a(n)-a(n)+2 whose solution is a(n) =n*a(0)-(n-1)*a(1)+n(n-1) P(x) = a(n) for all x in N, this means P(x) is defined by the same formula of a(n). So after rearranging terms, P(x) = x^2+x(a1-a0+1)+a0 = x^2+px+q
For Q4, you don't have to use binomial theorem at all. You can differentiate P for (n-2) times, you will get a quadratic polynomial, then prove the quadratic term is zero, and by induction, all n time differentials of P will be at most first order polynomials for n > 0, then you get P(x) = ax^2+bx+c. Put it back to get a = 1.
A more abstract solution to Q4 (turning it into a linear algebra problem): I start with the vector space of all polynomials of degree p(x+a) is a linear endomorphism on this vector space for any fixed a. In particular A^+: p(x)->p(x+1) and A^-:p(x)->p(x-1) are linear maps, as well as the identity id:p(x)->p(x). Writing the equation in terms of the linear operations, we need some vector p(x)=(a_0,...,a_n)^T, such that (1,0,...,0)^T+id*p(x)=(A^+*p(x)+A^-*p(x))/2. Rearranging and using linearity: (A^+ + A^- -2*id)/2p(x)=(1,0,...,0)^T. As soon as we know how the corresponding basis matrices for A^+ and A^- look like, we just need to solve a linear system of equations. It turns out (by applying the operations to the basis vectors) that for k>l a_{kl}^{+/-}=0 and for k>=l a_{k,l}^{+/-)=(+/-1)^{l-k}*binom(l,k). In particular, M=(A^+ +A^- -2*id)/2 has only zero entries below and on the diagonal starting at entry 1,2, while the diagonal starting at 1,3 has only nonzero entries with the entry 1,3 being 1. (stop here and explicitly write down A^+, A^- and M to see what I mean) This means that the matrix M is already in echelon form, so we can solve the equation Mp(x)=(1,0,...,0)^T by simple substitution starting at the bottom. But this means that a_n=0 => a_n-1=0 => ... => a_{3}=0 and a_{2}=1, while a_1 and a_0 are free parameters. Hence, independent of n, we get the 2-dimensional affine space of solutions +(0,0,1,0,...)^T, which translates back to x^2+ax+b. The same proof shows that for any fixed polynomial q, the set of all polynomials p satisfying q(x)+p(x)=(p(x+1)-p(x-1))/2 is a 2-dimensional affine vectorspace. (For the proof, n must be chosen bigger than deg(q)+2))
For the first question take y=kx, then divide both the equations. Simplify. You will get a cubic equations in k. Use synthetic division. You will get k= -2/3. Now solve the quadratic equation to get the other two roots. Now you have y in terms of x. Proceed to solve.
At 19:20 you have a(n-2) + 1 = a(0) + 1 = a(2) + a(0) This gives a(2) = 1, but you wrote down a(2) = 0, although you do go on to using 1 as the coefficient of x^2 (which is correct).
Since the order is the same for both E1 and E2 on the left hand side, what I did was just calculate E2/E1, let k=y/x, then it’s very easy to rewrite it as a function of k, now we have 1 variable and 1 equation so we can solve for k. I did some trial and error to find -2/3 is a root, so we can factorize it to find the other 2 roots. The rest is the same.
The first problem has an obvious symmetry between multipliers and exponents. That's why simplification 2x = u and 3y = t works well. Then assigning equations equal gives 3u^2t + 3ut^2 = u^3 + t^3, which is much easier to recognize as a cube of something. Knowing that (u + t)^3 = u^3 + 3u^2t + 3ut^2 + t^3 leads to an apparent try (u^2 - 2ut + t^2)(u + t) and when it fails a little bit to the correct (u + t)(u^2 - 4ut + t^2).
Here one very easy approach Substitute y=tx in bothe the eqn Divide both the eqn U will notice a perfect cube in terms of t forming (3t+2)^3=0 t=-2/3 Subst t value in any eqn u will get the correct answer
20:18 Yo, what’s up today? Here’s the homework... Your friend has chosen at random a card from a standard deck of 52 cards but keeps this card concealed. You have to guess what card it is. Before doing so, you can ask your friend either the question whether the chosen card is red or the question whether the card is the ace of spades. Your friend will answer truthfully. What question would you ask?
It doesnt matter which question we ask. The probability of geeting the right answer in each case is 1/26. Explanation: Case1 we ask if the card is red. the prob. of YES and NO is 1/2 each. after knowing the color, we have a 1/26 prob. of guessing the correct card in each case( Yes or No) so total prob= (1/2)*(1/26) + (1/2)*(1/26) = 1/26 Case2 we ask of the card is ace of spade. prob of YES = 1/52 prob of NO = 51/52 if yes: prob of guess =1(as we now know the exact card) if no: prob of guess= 1/51(1 card out of remaining 51 cards) total prob = (1/52)*(1) + (51/52)*(1/51) = 1/26
I think they’re the same. If you ask the first question the probability that you’re right is 1/26. For the second question, there’s a 1/52 chance you’ll have the answer right there, and if he says no (51/52) you’ll have a 1/51 chance of being right, meaning the second probability is (1/52)+(51/52)(1/51)=1/26.
I try to give quick sketch of proof. 1.Rewriting initial equation we get 2=-[P(x)-P(x-1)]+[P(x+1)-P(x)], it looks like in some way as a sum of two "consecutive" polynomials. Substituting W(x)=P(x+1)-P(x), we get: 2=-W(x-1)+W(x). *W(x) is polynomial as it is a linear combination of two polynomials. 2.Analyzing expression W(x)=W(x-1)+2 we can see, that in some way it is "shapely periodic function". I.e. if we consider W(x) in interval (x-1,x) and then W(x) in interval (x,x+1) then their looks, considering the shapes (considering the derivatives), are the same. The consecutive shapes are moved by vector [1,2]. But because W(x) is a polynomial, we can suspect whether W(x) is not linear. So we will try to prove this. We must show, that deg W(x)=1 3.In proving this fact we will use easy to prove theorem, that if A(x)=B(x) for x in R, then corresponding polynomial coefficients of A and B are the same. 4.Let n=deg W(x). Lets consider an(x-1)^2 in W(x-1). Expending this (using binomial coefficients) we can see, that the terms standing next to x^(n-1) in W(x) and W(x-1) are different. So the term standing next to x^(n-1) in W(x)-W(x-1) is non-zero. But because of W(x)-W(x-1)=2=2x^0 and point 3 we can see, that n-1=0, and thus def W(x)=1. So we have W(x)=a1x+a0. After substituting this to W(x)-W(x-1)=2, we can see, that W(x)=2x+a0, (a0 free). 5.Substituting above back to 2x+a0=W(x)=P(x+1)-P(x) we can see, that we play the same game as in the point 4, and we get, that P(x) is a quadratic function. And substituting back quadratic function P(x) to 2x+a0=P(x+1)-P(x) we can see, that: P(x)=x^2+alpha*x+beta, alpha, beta free.
Yes but even better with f(x)= P(x+1)-P(x) - 2x. Then f(x+1)= f(x) so since f is a polynomial f is constant. That is f’(x)=0. Then P’’(x+1) = P’’(x) so P’’(x) is constant which means P(x)= ax^2 + bx + c. By insertion a=1. Qed.
@@Aramil4 If you expand (2x-3iy)^3 you will see that (2x-3iy)^3=8+8i then, since x and y are real, you can solve the problem by matching real and imaginary parts.
Q2 is solvable in minutes if one knows the techniques of finite differences. I suspect this subject area is not taught anywhere these days. In days of old, whole books were written about it but, with the advent of computers, the knowledge has been “lost”.
Yes but homogenous equations also mean when everything has the same power, which he does have there. Pls don’t whoosh me, this is a math channel and I want to help.
For Q4, we can see that P(x+1)-P(x)-P(1)+P(0)=2x. So P(x+1)-P(x)=2x+P(1)-P(0). Add +1-1 on the RHS and rearranging, we get P(x+1)-P(x)=(2x+1)+(P(1)-P(0)-1). Using discrete calculus, and taking the "antiderivative", we get P(x)=x^2+(P(1)-P(0)-1)x+b. Substituting x=0, we get b=P(0). So P(x)=x^2+(P(1)-P(0)-1)x+P(0)
if you plot 4x^3 - 27x^2 = 4, then you get a figure which looks like a hyperbola on the right, but has two branches that are sort of like hyperbolas on the left. So maybe it could be called a "triperbola."
Man I was so close to getting the first problem, but I assumed adding the equations would give something of the form (Ax+By)^3=0 and that A and B would be + or - 2 and 3, but no combination would actually work turns out if that -18 at 5:55 were a +18, then (2x + 3y)^3 would = 0 so anything on the line y=(-3/2)x are solutions. I guess that'd be too easy lol
I solved it by differentiating both sides and comparing the coefficients of x^(n-3), obviously after assuming that n is larger than 2, and thus showing that a_n must be zero (contradiction).
Yes but even better with f(x)= P(x+1)-P(x) - 2x. Then f(x+1)= f(x) so since f is a polynomial f is constant. That is f’(x)=0. Then P’’(x+1) = P’’(x) so P’’(x) is constant which means P(x)= ax^2 + bx + c. By insertion a=1. Qed.
@@captainsnake8515 I’ve always found Olympiad geo to be largely just a long, LONG list of obscure theorems to memorize. Dunno if that would be best for Mr. Penn’s teaching style.
Because the first thing to try is a_n, which is the same on both sides. Then a_{n-1}, which again is the same on both sides. So you’re left with trying a_{n-2}.
(x+1)^2 + (x-1)^2 = 2x^2 + 2; yeah it's interesting how basically the ONLY thing that the x^2 terms do is to generate this random +1 on the left-hand side, but other than that the other terms bear no significance whatsoever.
I have watched many of your videos. I finally got fucked when i had no idea what "N choose 2" was. It must be some short hand notation for multiplication of powers in a polynomial. I'll have to sleep and think if I need to look that up.
geeeez that first problem was pretty miserable. very ugly and no clever insight in the solution except maybe the factoring part. cant help but feel like the problem designer(s) had a different method in mind that might have been nicer
So for the first one Start with 36ix^2y-27iy^3=8i and 8x^3-54xy^2=8 then adding you get 8x^3+36ix^2y-54xy^2-27iy^3=8+8i (2x+3iy)^3=8e^(ipi/4) 2x+3iy=2*2^(1/6)*(cos(pi/12)+isin(pi/12)), 2x+3iy=2*2^(1/6)*(-cos(pi/4)+isin(pi/4)), 2x+3iy=2*2^(1/6)*(-cos(5pi/12)-isin(5pi/12)) x=2^(1/6)cos(pi/12), y=2*2^(1/6)sin(pi/12)/3, x=-1/cubrt(2),y=2/3cubrt(2), x=-2^(1/6)cos(5pi/12),y=-2*2^(1/6)sin(5pi/12)
To pick a nit a bit, at 19:22 a2 = 1, not zero, right?
Yes. It has to be one. In the next step we got a monic quadratic polynomial. There a2 = 1.
Yep!
yup
For Q4 one can note that after differentiating both sides one gets P'(x)=(P'(x-1)+P'(x+1))/2 which is only satisfied by linear equations, so degP'
Wow!!! Mind blown🤯🤯
Yeah, I did it that way and everything kept cancelling out like crazy and I was left with a = 1.
Yes, and that P'(x)=(P'(x-1)+P'(x+1))/2 is only fulfilled for linear polynomials can also be seen easily (at least, if we know interpolation):
Suppose, P' is a polynomial of degree n such that P'(x)=(P'(x-1)+P'(x+1))/2 2P'(x)-P'(x-1)=P'(x+1) holds for all x.
Now, let a=P'(1) and b=P'(0). Then, we can recursively compute P'(2), P'(3), P'(4),...,P'(n). These are n+1 distinct points at which we prescribe the values, hence P' is the uniquely determined polynomial with degree at most n satisfying these prescriptions.
Since the linear polynomial p(x)=(a-b)x+b also satisfies these conditions (because p(0)=b, p(1)=a and p(x+1)=(a-b)x+a-b+b=(a-b)x+a=2*((a-b)x+b)-((a-b)(x-1)+b))), P' must be equal to p and therefore be linear.
I wrote (2xi-3y)^3, expanded and then replaced using the original equations to get (2xi-3y)^3=8(1-i) which is easily solved by finding cubed roots of the RHS.
I thought the same
But I am get in the form of cos and sin of pi/8 ,19pi/24, -13pi/24
the first one can be done much more nicely, we start with :
36x²y- 27y^3 = 8
4x^3 - 27xy² = 4
By multiplying the second line by two and factoring the coefficients (27 = 3^3,36 = 3*3*2*2) we get :
3(2x)²(3y)-(3y)^3 = 8 (1)
(2x)^3-3(2x)(3y)² = 8 (2)
Lets define X = 2x and Y = 3y and see that (X+iY)^3 = X^3+3iX²Y-3XY²-iY^3 so that (2)+i(1) give (X+iY)^3 = 8(1+i)
All that is left is to calculate the cube roots of 1+i and take the real and imaginary part.
1+i = sqrt(2)exp(i pi/4) so we need to calculate the real and imaginary parts of :
exp(i pi/12), j exp(i pi/12)exp(i (pi/12+2pi/3)) = exp(3i pi/4) = (-1+i)/sqrt(2) and j² exp(i pi/12), with j = (-1+i sqrt(3))/2
hence X+iY = 2^(1+1/6-1/2)(-1+i)*w = (-1+i)*2w/cbrt(2) with w = 1, j or j² that is :
X+iY = 2*(-1+i)/cbrt(2) or (-1+i)(-1+isqrt(3))/cbrt(2) or (-1+i)(-1-isqrt(3))/cbrt(2)
X+iY = 2*(-1+i)/cbrt(2) or (-sqrt(3)+1-i(sqrt(3)+1))/cbrt(2) or (sqrt(3)+1+i(sqrt(3)-1))/cbrt(2)
And finally :
x,y = -2/2cbrt(2),2/3cbrt(2)
x,y = -(sqrt(3)-1)/2cbrt(2),-(sqrt(3)+1)/3cbrt(2)
x,y = (sqrt(3)+1)/2cbrt(2),(sqrt(3)-1)/3cbrt(2)
Q2) since the polynomials are homogeneous, setting x=t*y and comparing both equations gives us a cubic polynomial with an obvious solution at t=-2/3 (easily spotted if you simply notice the equation is almost a cube of something else). The rest is substitution.
Q4) if we restrict x to natural numbers, then we end up with a difference equation a(n+1)=2a(n)-a(n)+2 whose solution is a(n) =n*a(0)-(n-1)*a(1)+n(n-1)
P(x) = a(n) for all x in N, this means P(x) is defined by the same formula of a(n).
So after rearranging terms, P(x) = x^2+x(a1-a0+1)+a0 = x^2+px+q
For Q4, you don't have to use binomial theorem at all. You can differentiate P for (n-2) times, you will get a quadratic polynomial, then prove the quadratic term is zero, and by induction, all n time differentials of P will be at most first order polynomials for n > 0, then you get P(x) = ax^2+bx+c. Put it back to get a = 1.
A more abstract solution to Q4 (turning it into a linear algebra problem):
I start with the vector space of all polynomials of degree p(x+a) is a linear endomorphism on this vector space for any fixed a.
In particular A^+: p(x)->p(x+1) and A^-:p(x)->p(x-1) are linear maps, as well as the identity id:p(x)->p(x).
Writing the equation in terms of the linear operations, we need some vector p(x)=(a_0,...,a_n)^T, such that
(1,0,...,0)^T+id*p(x)=(A^+*p(x)+A^-*p(x))/2.
Rearranging and using linearity:
(A^+ + A^- -2*id)/2p(x)=(1,0,...,0)^T.
As soon as we know how the corresponding basis matrices for A^+ and A^- look like, we just need to solve a linear system of equations.
It turns out (by applying the operations to the basis vectors) that for k>l a_{kl}^{+/-}=0 and for k>=l a_{k,l}^{+/-)=(+/-1)^{l-k}*binom(l,k). In particular,
M=(A^+ +A^- -2*id)/2 has only zero entries below and on the diagonal starting at entry 1,2, while the diagonal starting at 1,3 has only nonzero entries with the entry 1,3 being 1.
(stop here and explicitly write down A^+, A^- and M to see what I mean)
This means that the matrix M is already in echelon form, so we can solve the equation
Mp(x)=(1,0,...,0)^T by simple substitution starting at the bottom. But this means that a_n=0 => a_n-1=0 => ... => a_{3}=0 and a_{2}=1, while a_1 and a_0 are free parameters. Hence, independent of n, we get the 2-dimensional affine space of solutions +(0,0,1,0,...)^T, which translates back to x^2+ax+b.
The same proof shows that for any fixed polynomial q, the set of all polynomials p satisfying
q(x)+p(x)=(p(x+1)-p(x-1))/2 is a 2-dimensional affine vectorspace. (For the proof, n must be chosen bigger than deg(q)+2))
For the first question take y=kx, then divide both the equations. Simplify. You will get a cubic equations in k. Use synthetic division. You will get k= -2/3. Now solve the quadratic equation to get the other two roots. Now you have y in terms of x. Proceed to solve.
Yeah, that’s what I did exactly too
At 19:20 you have a(n-2) + 1 = a(0) + 1 = a(2) + a(0)
This gives a(2) = 1, but you wrote down a(2) = 0, although you do go on to using 1 as the coefficient of x^2 (which is correct).
Since the order is the same for both E1 and E2 on the left hand side, what I did was just calculate E2/E1, let k=y/x, then it’s very easy to rewrite it as a function of k, now we have 1 variable and 1 equation so we can solve for k. I did some trial and error to find -2/3 is a root, so we can factorize it to find the other 2 roots. The rest is the same.
The first problem has an obvious symmetry between multipliers and exponents. That's why simplification 2x = u and 3y = t works well. Then assigning equations equal gives 3u^2t + 3ut^2 = u^3 + t^3, which is much easier to recognize as a cube of something. Knowing that (u + t)^3 = u^3 + 3u^2t + 3ut^2 + t^3 leads to an apparent try (u^2 - 2ut + t^2)(u + t) and when it fails a little bit to the correct (u + t)(u^2 - 4ut + t^2).
Here one very easy approach
Substitute y=tx in bothe the eqn
Divide both the eqn
U will notice a perfect cube in terms of t forming (3t+2)^3=0
t=-2/3
Subst t value in any eqn u will get the correct answer
20:18
Yo, what’s up today? Here’s the homework...
Your friend has chosen at random a card from a standard deck of 52 cards but keeps this card concealed. You have to guess what card it is. Before doing so, you can ask your friend either the question whether the chosen card is red or the question whether the card is the ace of spades. Your friend will answer truthfully. What question would you ask?
It doesnt matter which question we ask. The probability of geeting the right answer in each case is 1/26.
Explanation:
Case1
we ask if the card is red.
the prob. of YES and NO is 1/2 each.
after knowing the color, we have a 1/26 prob. of guessing the correct card in each case( Yes or No)
so total prob= (1/2)*(1/26) + (1/2)*(1/26) = 1/26
Case2
we ask of the card is ace of spade.
prob of YES = 1/52
prob of NO = 51/52
if yes:
prob of guess =1(as we now know the exact card)
if no:
prob of guess= 1/51(1 card out of remaining 51 cards)
total prob = (1/52)*(1) + (51/52)*(1/51) = 1/26
If anyone finds any mistake in my solution pls point it out.
@@divyanshaggarwal6243 Why do I think you messed the second set of calculations up
I think they’re the same. If you ask the first question the probability that you’re right is 1/26. For the second question, there’s a 1/52 chance you’ll have the answer right there, and if he says no (51/52) you’ll have a 1/51 chance of being right, meaning the second probability is (1/52)+(51/52)(1/51)=1/26.
@@hamiltonianpathondodecahed5236 I dont see any issue. Can you please elaborate?
The second one is easier if you set Q(x) = P(x+1) - P(x)
Q(x+1)-Q(x)=2 -> Q(x)=2x+a -> P(x)=x^2+ax+b
I try to give quick sketch of proof.
1.Rewriting initial equation we get 2=-[P(x)-P(x-1)]+[P(x+1)-P(x)], it looks like in some way as a sum of two "consecutive" polynomials. Substituting W(x)=P(x+1)-P(x), we get: 2=-W(x-1)+W(x).
*W(x) is polynomial as it is a linear combination of two polynomials.
2.Analyzing expression W(x)=W(x-1)+2 we can see, that in some way it is "shapely periodic function". I.e. if we consider W(x) in interval (x-1,x) and then W(x) in interval (x,x+1) then their looks, considering the shapes (considering the derivatives), are the same. The consecutive shapes are moved by vector [1,2]. But because W(x) is a polynomial, we can suspect whether W(x) is not linear. So we will try to prove this. We must show, that deg W(x)=1
3.In proving this fact we will use easy to prove theorem, that if A(x)=B(x) for x in R, then corresponding polynomial coefficients of A and B are the same.
4.Let n=deg W(x). Lets consider an(x-1)^2 in W(x-1). Expending this (using binomial coefficients) we can see, that the terms standing next to x^(n-1) in W(x) and W(x-1) are different. So the term standing next to x^(n-1) in W(x)-W(x-1) is non-zero. But because of W(x)-W(x-1)=2=2x^0 and point 3 we can see, that n-1=0, and thus def W(x)=1. So we have W(x)=a1x+a0. After substituting this to W(x)-W(x-1)=2, we can see, that W(x)=2x+a0, (a0 free).
5.Substituting above back to 2x+a0=W(x)=P(x+1)-P(x) we can see, that we play the same game as in the point 4, and we get, that P(x) is a quadratic function. And substituting back quadratic function P(x) to 2x+a0=P(x+1)-P(x) we can see, that: P(x)=x^2+alpha*x+beta, alpha, beta free.
Yes but even better with f(x)= P(x+1)-P(x) - 2x. Then f(x+1)= f(x) so since f is a polynomial f is constant. That is f’(x)=0. Then P’’(x+1) = P’’(x) so P’’(x) is constant which means P(x)= ax^2 + bx + c. By insertion a=1. Qed.
Where the heck does he get those three cases at 15:46??
I was able to solve the first one but truly, I didn't believe I had done the calculation correct until I saw the solution
I could of course check my solutions by putting the values but I was lazy enough not to do that
ok so I can't be the only one who expanded (2x - 3y i)³ and did some algebra for P1
Same here 🙋♂️
Same.
Me too
Explain pls
@@Aramil4 If you expand (2x-3iy)^3 you will see that (2x-3iy)^3=8+8i then, since x and y are real, you can solve the problem by matching real and imaginary parts.
Divide the two eqn and let y/x =t
Q2 is solvable in minutes if one knows the techniques of finite differences. I suspect this subject area is not taught anywhere these days. In days of old, whole books were written about it but, with the advent of computers, the knowledge has been “lost”.
0:54
When you do a lot of Calculus...
lmaaooooooo
Yes but homogenous equations also mean when everything has the same power, which he does have there. Pls don’t whoosh me, this is a math channel and I want to help.
For Q4, we can see that P(x+1)-P(x)-P(1)+P(0)=2x. So P(x+1)-P(x)=2x+P(1)-P(0). Add +1-1 on the RHS and rearranging, we get P(x+1)-P(x)=(2x+1)+(P(1)-P(0)-1). Using discrete calculus, and taking the "antiderivative", we get P(x)=x^2+(P(1)-P(0)-1)x+b. Substituting x=0, we get b=P(0). So P(x)=x^2+(P(1)-P(0)-1)x+P(0)
Q2.We can see the solution x=(-3/2)*y from
3*((2x)^2)*(3y)-(3y)^3=2^3
3*((3y)^2)*(2x)-(2x)^3=-2^3
Q4: solution
1+p(x) = (p(x+1) + p(x-1))/2
or p(x+1) - p(x) = p(x) - p(x-1) + 2
or f(x+1) = f(x) + 2 [say f(x) = p(x) - p(x-1)]
=> f(x) = 2*x + c [ obviously] [this can also be seen as f'(x+1) = f'(x) => slope is constant => linear function]
or p(x) - p(x-1) = 2*x + c
=> p'(x) - p'(x-1) = 2
=> p'(x) = 2*x + c [same as above]
=> p(x) = x^2 + cx + a
Thanks for olimpiad problems. I love them.
What ? Olimpiad !
Ohh 😄
if you plot 4x^3 - 27x^2 = 4, then you get a figure which looks like a hyperbola on the right, but has two branches that are sort of like hyperbolas on the left. So maybe it could be called a "triperbola."
The first thing when I saw this problem is that I didn't know what Moldova is.
Man I was so close to getting the first problem, but I assumed adding the equations would give something of the form (Ax+By)^3=0 and that A and B would be + or - 2 and 3, but no combination would actually work
turns out if that -18 at 5:55 were a +18, then (2x + 3y)^3 would = 0 so anything on the line y=(-3/2)x are solutions. I guess that'd be too easy lol
I solved it by differentiating both sides and comparing the coefficients of x^(n-3), obviously after assuming that n is larger than 2, and thus showing that a_n must be zero (contradiction).
Try the P4 of the OIM 2020
Yes but even better with f(x)= P(x+1)-P(x) - 2x. Then f(x+1)= f(x) so since f is a polynomial f is constant. That is f’(x)=0. Then P’’(x+1) = P’’(x) so P’’(x) is constant which means P(x)= ax^2 + bx + c. By insertion a=1. Qed.
I’d like some insight into those binomial coefficients. That was maybe a little unclear
To this particular solution, one must add the homogeneous solution. It grows asymptotically as exp(x)..
9:00 why do you use only α+ ? As α+ is not equal to -α-, (α+)² is not equal to (α-)². Do I make a mistake ?
No, you didn't, since (α+)² =/= (α-)²
Sir please bring some olympiad geometry problems also .
^^ this. I always struggle with geometry and I think it would be helpful for me to see how to work through Olympiad geometry
@@captainsnake8515 I’ve always found Olympiad geo to be largely just a long, LONG list of obscure theorems to memorize. Dunno if that would be best for Mr. Penn’s teaching style.
Q2 is easier if you replace x2 and y2 with a and b in the beginning
the second one was a very tricky one
i dont understand why u compared an-2
Because the first thing to try is a_n, which is the same on both sides. Then a_{n-1}, which again is the same on both sides. So you’re left with trying a_{n-2}.
Sir a2=1 sir
You could also work with a tyalorexpansion around x, yada yada yada , 2nd derivative....
f''(x) = lim_{h->0) ((f(x+h)-2*f(x)+f(x-h))/h)
(x+1)^2 + (x-1)^2 = 2x^2 + 2; yeah it's interesting how basically the ONLY thing that the x^2 terms do is to generate this random +1 on the left-hand side, but other than that the other terms bear no significance whatsoever.
I know right.
Why do you cancel 27 by 3 down to a 3 shouldn't it be to a 9?
in case n=2 must be a2=1
Elegant 👍👨🏻🏫
In the last, if you say a2 = 0, then P(x) does not exist, because the first term is an.
??
2022 ayt 2. soru ve 13. sorular bu videoda
Nice nice nice nice......... up to infinity
This is an homogenous equation
Just put
y=mx
And solve it done
Yay!
Totally false.
@@robertgerbicz why please explain
@David Schmitz Yes, 2*E2-E1 is homogenous, but not E1 nor E2.
@@siddharthabhattacharya3787 pj sir style y=mx(homogeneous eqñ)....👌👌
@@siulibasak3804
General Theory hai shayad yeh...
these are the easiest exercises from the mathematical olympiad from Moldova
Next video: IMO 2020 problem 6
Hard problems need to wait a bit until the semester is over and it is easier to think more!
Please also try to give explanation of IIT(Indian Institute of Technology) Advance paper questions' solution. It will really help
I have watched many of your videos. I finally got fucked when i had no idea what "N choose 2" was. It must be some short hand notation for multiplication of powers in a polynomial. I'll have to sleep and think if I need to look that up.
geeeez that first problem was pretty miserable. very ugly and no clever insight in the solution except maybe the factoring part. cant help but feel like the problem designer(s) had a different method in mind that might have been nicer
So for the first one
Start with
36ix^2y-27iy^3=8i
and 8x^3-54xy^2=8
then adding you get
8x^3+36ix^2y-54xy^2-27iy^3=8+8i
(2x+3iy)^3=8e^(ipi/4)
2x+3iy=2*2^(1/6)*(cos(pi/12)+isin(pi/12)),
2x+3iy=2*2^(1/6)*(-cos(pi/4)+isin(pi/4)),
2x+3iy=2*2^(1/6)*(-cos(5pi/12)-isin(5pi/12))
x=2^(1/6)cos(pi/12), y=2*2^(1/6)sin(pi/12)/3,
x=-1/cubrt(2),y=2/3cubrt(2),
x=-2^(1/6)cos(5pi/12),y=-2*2^(1/6)sin(5pi/12)
Can you use computer screen rather than the blackboard?
Blackboard is the charming part.
Why you need technology
I loved the way of solving questions on blackboard especially by Michael 😇