Approximate the positive branch of y²(1-x²)=x² by its

Approximate the positive branch of $y^2(1-x^2)=x^2$ by its degree-3 Taylor (Maclaurin) polynomial about $x=0$, and estimate the error for $|x|\lt 10^{-1}$.

Step 1: Explicit form

Solving for $y>0$, $|x|\lt 1$: \[y = f(x) = \frac{x}{\sqrt{1-x^2}}. \tag{1}\]

Step 2: Derivatives at $x=0$

Differentiating: $f'(x)=(1-x^2)^{-3/2}\Rightarrow f'(0)=1$; $f''(x)=3x(1-x^2)^{-5/2}\Rightarrow f''(0)=0$; $f'''(x)=(3+12x^2)(1-x^2)^{-7/2}\Rightarrow f'''(0)=3$.

Step 3: Degree-3 Maclaurin polynomial

\[P_3(x) = x+\frac{x^3}2\]

This is confirmed directly: $(1-x^2)^{-1/2}=1+\tfrac{x^2}2+\tfrac{3x^4}8+\cdots$, so $f(x)=x\bigl(1+\tfrac{x^2}2+\cdots\bigr)=x+\tfrac{x^3}2+\cdots$, matching $P_3$ term for term.

Step 4: Lagrange remainder

\[ R_3(x) = \frac{f^{(4)}(\xi)}{4!}x^4 = \frac{\xi(45+60\xi^2)}{24(1-\xi^2)^{9/2}}x^4, \qquad \xi\text{ between }0\text{ and }x. \tag{2} \]

Step 5: Bound for $|x|\lt 10^{-1}$

For $|\xi|\lt \tfrac1{10}$: the numerator $|\xi|(45+60\xi^2)\lt 4.56$, and $(1-\xi^2)^{9/2}\gt (0.99)^{9/2}\approx0.956$, so $|f^{(4)}(\xi)|\lt 4.77\lt 5$, giving

\[ |R_3(x)| \lt \frac5{24}\left(\frac1{10}\right)^4 \approx 2.1\times10^{-4}. \tag{3} \]

Step 6: Verification

\[x=0.1\text{: }P_3(0.1)=0.1005,\ f(0.1)=0.1/\sqrt{0.99}\approx0.100504,\ |{\rm error}|\approx4\times10^{-6}\lt 2.1\times10^{-4}\ \checkmark\]