Find the fourth term in the expansion x^2 -2y^1/2
(x2−2y)12=[x2(1−2yx2)]12=x(1−2yx2)12Using binomial expansion, we expand the expression above=x[1−12.2yx2+12(12−1).(2yx2)22!+12.(12−1).(12−2).(2yx2)23!]=x[1−yx2−y22x4−y33x6]=x−yx−y22x3−y33x5+...(x^2-2y)^{\frac{1}{2}}=[x^2(1-\frac{2y}{x^2})]^{\frac{1}{2}} \\=x(1-\frac{2y}{x^2})^{\frac{1}{2}} \\\text{Using binomial expansion, we expand the expression above} \\=x[1-\frac{1}{2}.\frac{2y}{x^2} + \frac{\frac{1}{2}(\frac{1}{2}-1).(\frac{2y}{x^2})^2}{2!}+ \frac{\frac{1}{2}.(\frac{1}{2}-1).(\frac{1}{2}-2).(\frac{2y}{x^2})^2}{3!}] \\=x[1-\frac{y}{x^2}-\frac{y^2}{2x^4}-\frac{y^3}{3x^6}] \\= x-\frac{y}{x}-\frac{y^2}{2x^3}-\frac{y^3}{3x^5}+...(x2−2y)21=[x2(1−x22y)]21=x(1−x22y)21Using binomial expansion, we expand the expression above=x[1−21.x22y+2!21(21−1).(x22y)2+3!21.(21−1).(21−2).(x22y)2]=x[1−x2y−2x4y2−3x6y3]=x−xy−2x3y2−3x5y3+...
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment