Let V be the (real) vector space of all polynomial functions from R into R of de

Question

Let V be the (real) vector space of all polynomial functions from R into R of degree 2 or less, i.e., the space of all functions f of the form f(x) = c_0 + c_1*x + c_2*x^2. Let t be a fixed real number and define g_1(x) = 1, g_2(x) = x + t, g_3(x) = (x+t)^2. Prove that B = {g_1, g_2, g_3} is a basis for V. If f(x) = c_0 + c_1*x + c_2*x^2.
what are the coordinates of f in this ordered basis B?
Let V be the (real) vector space of all polynomial functions from R into R of degree 2 or less, i.e., the space of all functions f of the form f(x) = c_0 + c_1*x + c_2*x^2. Let t be a fixed real number and define g_1(x) = 1, g_2(x) = x + t, g_3(x) = (x+t)^2. Prove that B = {g_1, g_2, g_3} is a basis for V. If f(x) = c_0 + c_1*x + c_2*x^2.
what are the coordinates of f in this ordered basis B?

Explanation / Answer

let fx= a *g_1(x) + b*g_2(x) + c *g_3(x) ---(1) then if for every fx there is existence a,b,c then these span V so try n find out that is there sol. of a,b,c (1) can be written as c_0 + c_1*x + c_2*x^2=a*1 +b*( x + t)+ c* (x+t)^2. c_0 + c_1*x + c_2*x^2=a*1 +bt + ct^2 +x(b+ 2ct) +c*x^2 this will be true for every x if c_0 =a*1 +bt + ct^2 c_1=b+ 2ct c_2=c using three u will get solution of a,b,c which is unique c=c_2 b=c_1- 2c_2t a=c_0 - c_1t + c_2t^2 so it is basis coordinates= (a,b,c)

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