Chapter 9.2, Problem 45E

$\text{(a)}$

To determine

To prove: The statement: $f^{‴}(x)=m(m-1)(m-2){(1+x)}^{m-3}$

Explanation of Solution

Given information:

  $f(x)={(1+x)}^m$ for some non zero constant $m$ .

Calculation:

The function's first three derivatives are determined by the Power Rule and are:

  $\begin{array}{}$ f^{\prime }(x)=m{(1+x)}^{m-1}$$ f^{″}(x)=m(m-1){(1+x)}^{m-2}$$ f^{‴}(x)=m(m-1)(m-2){(1+x)}^{m-3}$\end{array}$

Therefore, the required statement is proved.

$\text{(b)}$

To determine

To prove: The statement $f^{(k)}(0)=m(m-1)(m-2)\cdots (m-k+1)$

Explanation of Solution

Given information:

  $f(x)={(1+x)}^m$ for some non zero constant $m$ .

Calculation:

The $k$ th derivative of the function in part $(a)$ can be determined by using the Power Rule $k$ times, thus,

  $f^{(k)}(x)=m(m-1)(m-2)\dots (m-k+1){(1+x)}^{m-k}$

By substitution $x=0$ ,

obtain

  ${(1+0)}^{m-k}=1$

Since,

  $f^{(k)}(0)=m(m-1)(m-2)\dots (m-k+1)$

By differentiating $k$ times and substitution $x=0$ , obtain the statement.

$(c)$

To determine

To find: The coefficient of $x^k$ .

Answer to Problem 45E

The coefficient of $x^k$ is $\left(\begin{array}{c}m\\ k\end{array}\right)$ .

Explanation of Solution

Given information:

  $f(x)={(1+x)}^m$ for some non zero constant $m$ .

Calculation:

Given that the part $\text{(b)}$ the $kth$ derivative of the function at $x=0$

Is:

  $f^{(k)}(0)=m(m-1)(m-2)\dots (m-k+1)$

Since the coefficient of $x^k$ in the Maclaurin series is $\frac{f^{(k)}(0)}{k!}$ ,

To have,

  $\begin{array}{c}\frac{f^{(k)}(0)}{k!}=\frac{m(m-1)(m-2)\dots (m-k+1)}{k!}\\ =\left(\begin{array}{l}m\hfill \\ k\hfill \end{array}\right)\end{array}$

Therefore, the required coefficient of $x^k$ is $\left(\begin{array}{c}m\\ k\end{array}\right)$ .

$(d)$

To determine

To prove: The statement: The Maclaurin series is generated by $f(x)={(1+x)}^m$ is .

Explanation of Solution

Given information:

  $f(x)={(1+x)}^m$ for some non zero constant $m$ .

Calculation:

Based on part $\text{(b)}$ and $(c)$ , the coefficient of $x^k$ in the Maclaurin series is:

  $\begin{array}{c}\frac{f^{(k)}(0)}{k!}=\frac{m(m-1)(m-2)\dots (m-k+1)}{k!}\\ =\left(\begin{array}{c}m\\ k\end{array}\right)\end{array}$

For each $k\in \mathbb{N}$ ,

Thus the series is:

  

It can be shown to use the results in part $\text{(b)}$ and $(c)$ .