Answer
(a) ${p_n}\left( x \right) = {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + \cdot\cdot\cdot + {c_n}{x^n}$
(b) ${p_n}\left( x \right) = {c_0} + {c_1}\left( {x - 1} \right) + {c_2}{\left( {x - 1} \right)^2} + {c_3}{\left( {x - 1} \right)^3} + \cdot\cdot\cdot + {c_n}{\left( {x - 1} \right)^n}$
Work Step by Step
(a) We take the derivatives of
$f\left( x \right) = {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + \cdot\cdot\cdot + {c_n}{x^n}$:
$f'\left( x \right) = {c_1} + 2{c_2}x + 3{c_3}{x^2} + \cdot\cdot\cdot + n{c_n}{x^{n - 1}}$, ${\ \ \ }$ $f'\left( 0 \right) = {c_1}$
$f{\rm{''}}\left( x \right) = 2{c_2} + 2\cdot3{c_3}x + \cdot\cdot\cdot + n\left( {n - 1} \right){c_n}{x^{n - 2}}$, ${\ \ \ }$ $f{\rm{''}}\left( 0 \right) = 2{c_2}$
$f{\rm{'''}}\left( x \right) = 2\cdot3{c_3} + \cdot\cdot\cdot + n\left( {n - 1} \right)\left( {n - 2} \right){c_n}{x^{n - 3}}$, ${\ \ \ }$ $f{\rm{'''}}\left( 0 \right) = 2\cdot 3{c_3}$
$\cdot\cdot\cdot$ ${\ \ \ }$ $\cdot\cdot\cdot$
${f^{\left( n \right)}}\left( x \right) = n\left( {n - 1} \right)\left( {n - 2} \right){c_n}$, ${\ \ \ }$ ${f^{\left( n \right)}}\left( 0 \right) = n!{c_n}$
By Definition 9.7.2, the $n$th Maclaurin polynomial for $f$ is
${p_n}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$
${p_n}\left( x \right) = {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + \cdot\cdot\cdot + {c_n}{x^n}$
Thus, ${p_n}\left( x \right) = f\left( x \right)$.
(b) We take the derivatives of
$f\left( x \right) = {c_0} + {c_1}\left( {x - 1} \right) + {c_2}{\left( {x - 1} \right)^2} + {c_3}{\left( {x - 1} \right)^3} + \cdot\cdot\cdot + {c_n}{\left( {x - 1} \right)^n}$:
$f'\left( x \right) = {c_1} + 2{c_2}\left( {x - 1} \right) + 3{c_3}{\left( {x - 1} \right)^2} + \cdot\cdot\cdot + n{c_n}{\left( {x - 1} \right)^{n - 1}}$, ${\ \ \ }$ $f'\left( 1 \right) = {c_1}$
$f{\rm{''}}\left( x \right) = 2{c_2} + 2\cdot3{c_3}\left( {x - 1} \right) + \cdot\cdot\cdot + n\left( {n - 1} \right){c_n}{\left( {x - 1} \right)^{n - 2}}$, ${\ \ \ }$ $f{\rm{''}}\left( 1 \right) = 2{c_2}$
$f{\rm{'''}}\left( x \right) = 2\cdot 3{c_3} + \cdot\cdot\cdot + n\left( {n - 1} \right)\left( {n - 2} \right){c_n}{\left( {x - 1} \right)^{n - 3}}$, ${\ \ \ }$ $f{\rm{'''}}\left( 1 \right) = 2\cdot 3{c_3}$
$\cdot\cdot\cdot$ ${\ \ \ }$ $\cdot\cdot\cdot$
${f^{\left( n \right)}}\left( x \right) = n\left( {n - 1} \right)\left( {n - 2} \right){c_n}$, ${\ \ \ }$ ${f^{\left( n \right)}}\left( 1 \right) = n!{c_n}$
By Definition 9.7.3, the $n$th Taylor polynomial about $x=1$ for $f$ is
${p_n}\left( x \right) = f\left( 1 \right) + f'\left( 1 \right)\left( {x - 1} \right) + \dfrac{{f{\rm{''}}\left( 1 \right)}}{{2!}}{\left( {x - 1} \right)^2} + \dfrac{{f{\rm{'''}}\left( 1 \right)}}{{3!}}{\left( {x - 1} \right)^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( n \right)}}\left( 1 \right)}}{{n!}}{\left( {x - 1} \right)^n}$
${p_n}\left( x \right) = {c_0} + {c_1}\left( {x - 1} \right) + {c_2}{\left( {x - 1} \right)^2} + {c_3}{\left( {x - 1} \right)^3} + \cdot\cdot\cdot + {c_n}{\left( {x - 1} \right)^n}$
Thus, ${p_n}\left( x \right) = f\left( x \right)$.
