Approximate f(x)=(1+x²) sqrt(1+x) by its degree-4

Approximate $f(x)=(1+x^2)\sqrt{1+x}$ by its degree-4 Maclaurin polynomial and estimate the error for $x\in[0,\tfrac12]$.

Step 1: Series for $\sqrt{1+x}$

\[ \sqrt{1+x} = 1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+\cdots \tag{1} \]

Step 2: Multiply by $(1+x^2)$

Adding the shift-by-two contribution from $x^2\cdot\sqrt{1+x}$ and collecting by degree: \[x^0{:}\,1;\ \ x^1{:}\,\tfrac12;\ \ x^2{:}\,{-}\tfrac18+1=\tfrac78;\ \ x^3{:}\,\tfrac1{16}+\tfrac12=\tfrac9{16};\ \ x^4{:}\,{-}\tfrac5{128}-\tfrac18=-\tfrac{21}{128}. \tag{2}\]

Step 3: Degree-4 polynomial

\[P_4(x) = 1+\frac12x+\frac78x^2+\frac9{16}x^3-\frac{21}{128}x^4\]

Step 4: Fifth derivative via the Leibniz rule

Writing $f=uv$ with $u=1+x^2$ ($u'=2x$, $u''=2$, higher derivatives zero) and $v=(1+x)^{1/2}$: \[f^{(5)} = \binom50 u\,v^{(5)} + \binom51 u'v^{(4)} + \binom52 u''v^{(3)} = \frac{105(1+x^2)}{32}(1+x)^{-9/2} - \frac{150x}{16}(1+x)^{-7/2} + \frac{60}{8}(1+x)^{-5/2}. \tag{3}\]

Step 5: Bound on $[0,\tfrac12]$

For $\xi\in[0,\tfrac12]$, all factors $(1+\xi)^{-k}\le1$; with $1+\xi^2\le\tfrac54$ and $\xi\le\tfrac12$: \[|f^{(5)}(\xi)| \le \frac{525}{128}+\frac{75}{16}+\frac{60}8 = \frac{2085}{128} \approx 16.3. \tag{4}\]

With $x^5\le(\tfrac12)^5=\tfrac1{32}$: \[|R_4(x)| \le \frac{16.3}{120}\cdot\frac1{32} \approx 4.25\times10^{-3} \lt 5\times10^{-3}. \tag{5}\]

Step 6: Verification

\[x=0.5\text{: }P_4(0.5)\approx1.528809,\ (1.25)\sqrt{1.5}\approx1.530931,\ |{\rm error}|\approx2.1\times10^{-3}\lt 5\times10^{-3}\ \checkmark\]