binomial expansion at a level?
(1-2x)^-1/2 - (1+2x)^1/2 = ax^2 + bx^3
where a and b are constants and x is sufficiently small that terms in x^4 and higher powers of x can be ignored, find the values of a and b.
(1-2x)^-1/2 - (1+2x)^1/2 = ax^2 + bx^3
where a and b are constants and x is sufficiently small that terms in x^4 and higher powers of x can be ignored, find the values of a and b.
Josh K
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1/sqrt[1-2x] - sqrt[1+2x] = ax^2 + bx^3
(1-sqrt[1-4x^2]) / sqrt[1-2x] = ax^2 + bx^3 {square both sides}
[1-2sqrt[1-4x^2] + 1-4x^2] / 1-2x = a^2x^4 + 2abx^5 + b^2x^6
1-2sqrt[1-4x^2] + 1-4x^2 = a^2x^4 + 2abx^5 + b^2x^6 - [2a^2x^5 + 4abx^6 + 2b^2x^7]
1-2sqrt[1-4x^2] + 1-4x^2 = a^2x^4 + (2ab-2a^2)x^5 + (b^2-4ab)x^6 - 2b^2x^7 {now isolate the radical}
-2sqrt[1-4x^2] = 2 - 4x^2 + a^2x^4 + (2ab-2a^2)x^5 + (b^2-4ab)x^6 - 2b^2x^7
Now you need to square both sides, which for obvious reasons I'm not going to do.
My guess is they want you to ignore all powers of x greater than 3 at each stage, but that seems pretty hokey to me.
Making this
-2sqrt[1-4x^2] = 2 - 4x^2 {then square}
4*(1-4x^2) = 2-16x^2 + 16x^4 {once again ignore the x^4}
4-16x^2 = 2-16x^2
4 = 2 which is a contradiction, so if the method of ignoring powers of x greater than 3 is true, there is no solution that satisfies this problem.