Page:Elementary algebra (1896).djvu/411

 Equating the coefficients of like powers of $$x$$, we have

$$ \begin{align} A = 1; 2AB=1, B^2+2A&& C=0,&& 2AD+2BC=0,&& \\ \therefore B=\frac{1}{2};&& \therefore C=-\frac{1}{8},&& \therefore D= \frac{1}{16},&& \\ C^2 + 2BD+2AE = 0; \therefore E=-\frac{5}{128}+\dots. \end{align} $$

thus \sqrt{1+x} = 1+\frac{x}{2}} +\frac{x^2}{8} ++\frac{x^3}{16} - \frac{5x^4}{128} + \cdots.

Note, The expansion can be readily effected by the use of the Binomial Theorem [Art. 421].

Expand the following expressions to four terms :

1. $$\sqrt{1-x}$$. 2. $$\sqrt{a-x}$$. 3. $$\sqrt{a^2-x^2}$$. 4. $$\sqrt[3]{2+x}$$. 5. $$(1 + x)^{\frac{3}{2}}$$. 6. $$(1 + x + x^2)^{\frac{1}{2}}$$.

REVERSION OF SERIES.

490. To revert a series $$y = ax + bx^2 + cx^3 + \cdots$$ is to express $$x$$ in a series of ascending powers of $$y$$.

Revert the series

Assume x = Ay+ By^2 + Cy^3 + Dy^4 +.

Substituting in this equation the value of y as given in (1), we have

$$ \begin{align} x&=A(x-2x^2+3x3-4x^4+\cdots)&=A(x-2x^2+3x^3+4x4+\cdots&)&& \\ &+B(x-2x^2+3x^3-4x^4+\cdots)^2&=B(x^2+4x^4-4x^3+6x^4+\cdots&)&& \\ &+C(x-2x^2+3x^3-4x^4+\cdots)^3&=C(x^3-6x^4-\cdots&)&& \\ \end{align} $$

Equating the coefficients of like powers of x,

$$ \begin{align} & A = 1; & B-2A = 0, && C-4B + 3A = 0,&& \\ & & \therefore B = 2;&& \therefore C = 5;&& \\ & & D-6C-10B-4A = 0, \\ & & \therefore D = 14. \end{align} $$

Hence $$x = y + 2 y^2 + 5 y^3 + 14y^4 + \cdots$$.