glasskerfu

2021-09-03

Find the Taylor polynomials of orders 0, 1, 2, and 3 generated by ƒ at a.
$f\left(x\right)=\sqrt{x},a=4$

2k1enyvp

Expert

Consider the provided function,
$f\left(x\right)=\sqrt{x},a=4$
Find the Taylor polynomials of orders 0, 1, 2, and 3 generated by ƒ at a.
The Taylor polynomial is shown below,
${P}_{n}\left(x\right)=f\left(a\right)+\frac{f{}^{″}\left(a\right)}{2!}{\left(x-a\right)}^{2}+\dots +\frac{{f}^{k}\left(a\right)}{k!}{\left(x-a\right)}^{k}+\dots$
First find the derivative of the function,
$f\left(x\right)=\sqrt{x}$
${f}^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$
$f{}^{″}\left(x\right)=-\frac{1}{4{x}^{\frac{3}{2}}}$
$f{}^{‴}\left(x\right)=\frac{3}{8{x}^{\frac{5}{2}}}$
Substitute a=4 in the above functions we get,
$f\left(4\right)=\sqrt{4}⇒2$
${f}^{\prime }\left(4\right)=\frac{1}{2\sqrt{4}}⇒\frac{1}{4}$
$f{}^{″}\left(4\right)=\frac{1}{4{\left(4\right)}^{\frac{3}{2}}}⇒\frac{1}{32}$
$f{}^{‴}\left(4\right)=\frac{3}{8{\left(4\right)}^{\frac{5}{2}}}⇒\frac{3}{256}$
So, the Taylor polynomials of orders 0, 1, 2, and 3 is shown below.${P}_{0}=2$
${P}_{1}=2+\frac{1}{4}\left(x-4\right)$
${P}_{2}=2+\frac{1}{4}\left(x-4\right)+\frac{1}{64}{\left(x-4\right)}^{2}$
${P}_{3}=2+\frac{1}{4}\left(x-4\right)+\frac{1}{64}{\left(x-4\right)}^{2}+\frac{3}{1536}{\left(x-4\right)}^{3}$
Hence.

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