relationship between zeros and coefficient

relationship between zeros and coefficient : The population correlation coefficient X,Y between two ... of a decreasing linear relationship, and some value between -1 ... but their correlation is zero ; they are uncorrelated. .....   More from Wikipedia

  it always has the trivial solution, of course, a1, a2 equals zero. Now, we don't want that trivial solution because if a1 and a2 are zero, then so are x and y zero. Now that is a solution. Unfortunately, it is of no interest. If the solution were x, y zero, it corresponds to the fact that this is an ice bath. The yoke is at zero, the white is at zero and it stays that way for all time until the ice melts. So that is the solution we don't want. We don't want the trivial solution. Well, when ...

  the bell curve. If we start at u equals zero, we are going to integrate as far as we need to out here. And the integral from zero to infinity of this thing is root pi over two. And we want this to go from zero to one, so we will put the factor two over root pi out in front. That way, when we integrate this from zero to infinity, erf of infinity becomes one. Erf runs from zero to one. And that is the area under this curve. That is why some people call this the Gaussian error function ...

Question : In general, what is the relationship between the zeros of two polynomials of degree n with their coefficients reversed in order, with the leading coefficient and constant coeffiecients both not zero? (Assume the coefficients are all real numbers) Im pretty sure that they are reciprocals of each other but I dont know how to show that. My teacher said that we should use compositions of functions to show this and that one good function to use would be 1/x. But I don't know what he means or how t..

Answer : Consider the polynomial y = ax^n + bx^(n-1) + ... + cx + d ...(1) Now consider y/x^n = a + b/x + ... + c/x^(n-1) + d/x^n Let z = 1/x y/x^n = a + bz + ... + cz^(n-1) + dz^n ...(2) The LHS of both equations will be zero at the same time. So if x = t makes the LHS of eqn 1 zero, then z = 1/t makes the LHS of eqn 2 zero. So yes the roots are reciprocals. Verify with: x^2 + 3x + 2 and 2x^2 + 3x + 1 The only problem would be if x=0 is a root, however since the constant term is non zero, that is not hte case...   More from Yahoo Answers