Question

Find the Maclaurin series for the following functions $\mathit{c}\mathit{o}\mathit{s}\mathbf{\left[}\mathit{l}\mathit{n}\mathbf{\right(}\mathbf{\text{1 + x}}\mathbf{\left)}\mathbf{\right]}$:

Short answer

Maclaurin series of $\cos \left[\ln \right(1+x\left)\right]$is $\left(\frac{x^2}{2}+\frac{x^3}{2}-\frac{5x^4}{12}+\dots \dots \right)$.

Step-by-step solution

Concept and formula used to find the Maclaurin series of the given function 

The Maclaurin series of$\mathit{c}\mathit{o}\mathit{s}\mathit{x}$is expressed as follows:

$\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{=}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}$

The Maclaurin series of $\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}$is expressed as follows:

$\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}$

$\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}$

Calculation to find the Maclaurin series of the function cos [1n(1 + x)]:

Substitute$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\dots \dots .$ for$x$ in expansion of $\cos x$as follows:

role="math" localid="1664264756169" $$\begin{gathered} \cos \left[\ln \right(1+x\left)\right]=1-\frac{{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\dots \dots \right)}^2}{2}+\frac{{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\dots \dots \right)}^4}{24}+\dots \dots \\ \cos \left[\ln \right(1+x\left)\right]=1-\frac{\left(-576x^2+576x^3-480x^4\right)}{1152}+\dots \dots . \\ \cos \left[\ln \right(1+x\left)\right]=\frac{x^2}{2}+\frac{x^3}{2}-\frac{5x^4}{12}+\dots \dots . \end{gathered}$$

Thus, the Maclaurin series of role="math" localid="1664264808184" $\cos \left[\ln \right(1+x\left)\right]$ is $\left(\frac{x^2}{2}+\frac{x^3}{2}-\frac{5x^4}{12}+\dots \dots \right)$.