Given y=x²sqrt(1-x): 1° write the degree- n Maclaurin

Given $y=x^2\sqrt{1-x}$: 1° write the degree-$n$ Maclaurin polynomial and the remainder; 2° for $n=4$, find the error bound for $|x|\lt \tfrac1{10}$; 3° verify by multiplying the series for $f(x)=x^2$ and $\varphi(x)=\sqrt{1-x}$.

Step 1: Part 1°: general Maclaurin polynomial

Writing $c_k$ for the coefficient of $x^k$ in $\sqrt{1-x}$ ($c_0=1$, $c_1=-\tfrac12$, $c_k=-\tfrac{1\cdot1\cdot3\cdots(2k-3)}{2^kk!}$ for $k\ge2$): \[\sqrt{1-x} = 1-\frac x2-\frac{x^2}8-\frac{x^3}{16}-\frac{5x^4}{128}-\cdots \tag{1}\]

Multiplying by $x^2$ shifts every exponent up by two, so the coefficient of $x^k$ in $x^2\sqrt{1-x}$ is $c_{k-2}$ ($k\ge2$): \[P_n(x) = \sum_{k=2}^n c_{k-2}\,x^k = x^2-\frac{x^3}2-\frac{x^4}8-\frac{x^5}{16}-\cdots-c_{n-2}x^n, \qquad |R_n(x)| \le |c_{n-1}|\,|x|^{n+1}\]

Step 2: Part 2°: error bound for $n=4$, $|x|\lt \tfrac1{10}$

Here $P_4(x)=x^2-\tfrac{x^3}2-\tfrac{x^4}8$; the first neglected term is $-\tfrac{x^5}{16}$, so $|c_3|=\tfrac1{16}$ and

\[ |R_4(x)| \le \frac{|x|^5}{16} \lt \frac{(1/10)^5}{16} \approx 6.25\times10^{-7}. \tag{2} \]

Step 3: Part 3°: verification by series multiplication

Writing $y=f(x)\varphi(x)$ with $f(x)=x^2$ (already its own series) and $\varphi(x)=\sqrt{1-x}$ as in (1), the term-by-term product reproduces (1) shifted by $x^2$ exactly, confirming every coefficient above.

Step 4: Verification

\[x=0.1\text{: }P_4(0.1)=0.00949375,\ (0.01)\sqrt{0.9}\approx0.00948683,\ |{\rm error}|\approx6.9\times10^{-6}\text{, order }x^5\text{ as expected}\ \checkmark\]