Hello Kanav, for the first question, is there a solution using coefficient matching? Assume the coefficients are real, then (a_n)^2 = a_n, from the leading coeffecient, and so a_n = 1. Similarly a_0 = 0 or 1. Also, notice P(x^2) is composed completely of even powers, so we can set all odd-power coefficients to 0. Thus, the sum of roots of P(x^2) or a_2n-1 = 0. But notice, a_2n-1 = 2(a_n-1 + n). So the sum of roots of P(x) and P(x+2) is both n. Is this a feasible method?
Let P(x) = a_nx^n + a_n-1x^n_1 . . . . . . . + a0. P(x+2) = a_n(x+2)^n + a_n-1(x+2)^n-1. . . . .a0. Notice, from the Binomial expansion, the coefficient of x^n is still a_n. Similarly for the RHS, the leading coefficient is also a_n. Thus a_n^2 = a_n.
As a better explanation P(x) and P(x+2) have the same leading coefficient, and P(x^2) = P(x)*P(x+2). We may equate the leading coefficients of both sides. Notice the leading coefficient of P(x^2) to be a_n. Why? As a_n(x^2)^n is the first term. Then equation the leading terms: a_n(x^2n) = a_n(x^2)*a_n(x^2). Thus a_n^2 = a_n and a_n = 1.
is my reasoning correct in the first question that in second case the polynomial would have been a zero polynomial too if its roots have not cycled back so by a equation k^2^m=k^2^n we get absolute value of k is 1 or 0 hence by algebraicly proceeding towards the solution
Kanav Bhai Orz!!!
Hello Kanav, for the first question, is there a solution using coefficient matching? Assume the coefficients are real, then (a_n)^2 = a_n, from the leading coeffecient, and so a_n = 1. Similarly a_0 = 0 or 1. Also, notice P(x^2) is composed completely of even powers, so we can set all odd-power coefficients to 0. Thus, the sum of roots of P(x^2) or a_2n-1 = 0. But notice, a_2n-1 = 2(a_n-1 + n). So the sum of roots of P(x) and P(x+2) is both n. Is this a feasible method?
why (an)^2=an you cant get this until some other statements are given
for example let n = 3x+4y and n=5m+9n so can you say without sufficient conditions that 3x=5m or 4y=9n Hope it helps
Let P(x) = a_nx^n + a_n-1x^n_1 . . . . . . . + a0. P(x+2) = a_n(x+2)^n + a_n-1(x+2)^n-1. . . . .a0. Notice, from the Binomial expansion, the coefficient of x^n is still a_n. Similarly for the RHS, the leading coefficient is also a_n. Thus a_n^2 = a_n.
As a better explanation P(x) and P(x+2) have the same leading coefficient, and P(x^2) = P(x)*P(x+2). We may equate the leading coefficients of both sides. Notice the leading coefficient of P(x^2) to be a_n. Why? As a_n(x^2)^n is the first term. Then equation the leading terms: a_n(x^2n) = a_n(x^2)*a_n(x^2). Thus a_n^2 = a_n and a_n = 1.
you cannot basicly compare any equations if you want i can prove
Loved this session Kanav bhayya , Ur a great teacher :D
is my reasoning correct in the first question that in second case the polynomial would have been a zero polynomial too if its roots have not cycled back so by a equation k^2^m=k^2^n we get absolute value of k is 1 or 0 hence by algebraicly proceeding towards the solution
I have only one doubt in the first question how absolute value of every root of p(x) is 1
god himself
Hello
First
orz