Page:Elementary algebra (1896).djvu/411

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

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

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].

EXAMPLES XLII. c.

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

$y=x-2x^{2}-3x^{3}-4x^{4}+\dots$

(1)

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

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

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

Equating the coefficients of like powers of x,

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

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