This is in regards to another question on here, Maclaurin expansion of $\arcsin x$
In the solution accepted by the OP, the expression
$\sqrt {1-x}$
is rewritten "using substitution" as
$\frac{1}{\sqrt{1-x^2}}$
I don't see how this is done?
Asked
Active
Viewed 44 times
0
-
Pray, provide some details!! What don't you see about it? Do you know what substitution is? Can you see that $x=t^2 $ is a valid substitution? If not, why do you think it's wrong? If yes, that's exactly what's done there, only with a slightly different notation, $x$ substituted to $x^2$. – user0 Sep 19 '18 at 15:20
2 Answers
1
Maybe the OP is simply referring to binomial expansion
$$(1 + x)^\alpha = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \cdots$$
with $\alpha =-\frac12$ and $x=y^2$ which leads to
$$ \frac{1}{\sqrt{1-y^2}} = 1 + \frac{y^2}{2} + \frac{3y^4}{8} + \frac{5y^6}{16} + \frac{35y^8}{128} + \dotsc $$
user
- 154,566
0
The OP uses: $$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + \dotsc$$ to get: $$\frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \frac{35x^8}{128} + \dotsc$$ $===================================$
First, substitute $x=-x^2$ to get: $$\sqrt{1-x^2} = 1 - \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^6}{16} - \frac{5x^8}{128} - \dotsc$$ Second, consider its reciprocal: $$\frac{1}{\sqrt{1-x^2}}=\frac{1}{1 - \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^6}{16} - \frac{5x^8}{128} - \dotsc}=a_0+a_1x+a_2x^2+a_3x^3+\cdots \Rightarrow \\ (1 - \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^6}{16} - \frac{5x^8}{128} - \dotsc)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)=1 \Rightarrow \\ \begin{align}a_0&=1;\\ a_1&=0;\\ a_2-\frac{a_0}{2}=0 \Rightarrow a_2&=\frac12;\\ a_3-\frac{a_1}{2}=0 \Rightarrow a_3&=0;\\ a_4-\frac{a_2}{2}-\frac{a_0}{8}=0 \Rightarrow a_4&=\frac38;\\ a_5-\frac{a_3}{2}-\frac{a_1}{8}=0 \Rightarrow a_5&=0;\\ a_6-\frac{a_4}{2}-\frac{a_2}{8}-\frac{a_0}{16}=0 \Rightarrow a_6&=\frac{5}{16};\\ \vdots \end{align}$$
farruhota
- 31,482