Try to determine the values of constants a, B and C such that e ^ x * (1 + BX + CX ^ 2) = 1 + ax + ο (x ^ 3), where ο (x ^ 3) is the infinitesimal of higher order than x ^ 3 when x → 0 There is an answer (e^x)*(1+Bx+Cx^2)=(1+x+x^2/2!+x^3/3!+o(x^3))*(1+Bx+Cx^2)=1+(1+B)x+(1/2+B+C)x^2+(1/6+B/2+C)x^3+o(x^3)=1+Ax+ο(x^3) So, there are 1 + B = a, 1 / 2 + B + C = 0, 1 / 6 + B / 2 + C = 0 The solution is: a = 1 / 3, B = - 2 / 3, C = 1 / 6 But I can't understand the second equal sign

Try to determine the values of constants a, B and C such that e ^ x * (1 + BX + CX ^ 2) = 1 + ax + ο (x ^ 3), where ο (x ^ 3) is the infinitesimal of higher order than x ^ 3 when x → 0 There is an answer (e^x)*(1+Bx+Cx^2)=(1+x+x^2/2!+x^3/3!+o(x^3))*(1+Bx+Cx^2)=1+(1+B)x+(1/2+B+C)x^2+(1/6+B/2+C)x^3+o(x^3)=1+Ax+ο(x^3) So, there are 1 + B = a, 1 / 2 + B + C = 0, 1 / 6 + B / 2 + C = 0 The solution is: a = 1 / 3, B = - 2 / 3, C = 1 / 6 But I can't understand the second equal sign

The second equal sign is (1 + X + x ^ 2 / 2! + x ^ 3 / 3! + O (x ^ 3)) * (1 + BX + CX ^ 2) = 1 + (1 + b) x + (1 / 2 + B + C) x ^ 2 + (1 / 6 + B / 2 + C) x ^ 3 + O (x ^ 3). It is calculated by expanding the brackets. Without looking at the part o (x ^ 3) on the left side of the equal sign, the part on the right side of the equal sign other than o (x ^ 3) will be obtained. Then, O (x ^ 3) is multiplied by (1 + BX + CX ^ 2) to get o (x ^ 3)
o(x^3)*1=o(x^3),
o(x^3)*Bx=o(x^4),
o(x^3)*Cx^2=o(x^5),
o(x^3)+o(x^4)+o(x^5)=o(x^3).
In other words, if there are small o signs in F and G, then f = g and g = f are two different things. Moreover, O (x ^ 3) cannot be regarded as a specific number or variable, so it cannot be eliminated

It is known that when x tends to zero, (e ^ (x ^ 2) - (AX ^ 2 + BX + C)) is infinitesimal of higher order than x ^ 2. Try to determine the constants a, B, C

Let X - > 0, ax & # 178; + BX + c-cosx be infinitesimal of higher order than X & # 178;, find the value of constant a, B, C? Detailed process

The Taylor expansion of cosx at zero is cosx = 1-x ^ 2 / 2! + x ^ 4 / 4! + When x approaches 0, ax & # 178; + BX + c-cosx = ax & # 178; + BX + C - (1-x & # 178; / 2! + x ^ 4 / 4! + )=(a+1/2)x²+bx+(c-1)-x^4/4!+…… Because ax + BX + c-cosx is infinitesimal of higher order than x