Chapter 8 Mean Value Theorems-II

Formulas for the expansion

Taylor’s theorem in finite form with Lagrange’s form of remainder

f(a+h)=f(a)+hf′(a)+h22!f″(a)+…+hn−1(n−1)!fn−1(a)+hnn!fn(a+θh)$$\begin{array}{cc}f(a+h)& =f\left(a\right)+h{f}^{\prime }\left(a\right)+\frac{h^2}{2!}{f}^{\prime \prime }\left(a\right)+\dots +\frac{h^{n-1}}{(n-1)!}{f}^{n-1}\left(a\right)+\\ & \frac{h^n}{n!}{f}^n(a+\theta h)\end{array}$$

where 0<θ<1$0<\theta <1$

The term hnn!fn(a+θh)$\frac{h^n}{n!}{f}^n(a+\theta h)$ is the remainder Rn$R_n$.

Taylor’s theorem in finite form with Cauchy’s form of remainder

f(a+h)=f(a)+hf′(a)+h22!f″(a)+…+hn−1(n−1)!fn−1(a)+hn(n−1)!(1−θ)n−1fn(a+θh)$$\begin{array}{cc}f(a+h)& =f\left(a\right)+h{f}^{\prime }\left(a\right)+\frac{h^2}{2!}{f}^{\prime \prime }\left(a\right)+\dots +\frac{h^{n-1}}{(n-1)!}{f}^{n-1}\left(a\right)+\\ & \frac{h^n}{(n-1)!}(1-\theta {)}^{n-1}{f}^n(a+\theta h)\end{array}$$

where 0<θ<1$0<\theta <1$

The term hn(n−1)!(1−θ)n−1fn(a+θh)$\frac{h^n}{(n-1)!}(1-\theta {)}^{n-1}{f}^n(a+\theta h)$ is the remainder Rn$R_n$.

Maclaurin’s series in finite form with Lagrange’s form of remainder

f(x)=f(0)+x1!f′(0)+x22!f″(0)+…+xn−1(n−1)!fn−1(0)+xnn!fn(θx)$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+\frac{x}{1!}{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\dots +\frac{x^{n-1}}{(n-1)!}{f}^{n-1}\left(0\right)+\\ & \frac{x^n}{n!}{f}^n\left(\theta x\right)\end{array}$$

where 0<θ<1$0<\theta <1$

The term xnn!fn(θx)$\frac{x^n}{n!}{f}^n\left(\theta x\right)$ is the remainder Rn$R_n$.

Maclaurin’s series in finite form with Cauchy’s form of remainder

f(x)=f(0)+x1!f′(0)+x22!f″(0)+…+xn−1(n−1)!fn−1(0)+xn(n−1)!(1−θ)n−1fn(θx)$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+\frac{x}{1!}{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\dots +\frac{x^{n-1}}{(n-1)!}{f}^{n-1}\left(0\right)+\\ & \frac{x^n}{(n-1)!}(1-\theta {)}^{n-1}{f}^n(\theta x)\end{array}$$

where 0<θ<1$0<\theta <1$

The term xn(n−1)!(1−θ)n−1fn(θx)$\frac{x^n}{(n-1)!}(1-\theta {)}^{n-1}{f}^n(\theta x)$ is the remainder Rn$R_n$.

Taylor’s series in infinite series

f(x+h)=f(x)+hf′(x)+h22!f″(x)+h33!f‴(x)+…$$\begin{array}{cc}f(x+h)& =f\left(x\right)+h{f}^{\prime }\left(x\right)+\frac{h^2}{2!}{f}^{\prime \prime }\left(x\right)+\frac{h^3}{3!}{f}^{\prime \prime \prime }\left(x\right)+\dots \end{array}$$

Maclaurin’s series in infinite series

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \end{array}$$

Maclaurin Series Expansion is a Taylor Series Expansion centered at 0.

8.0.1 Few expansions to remember

ex=1+x+x22!+x33!+x44!+…sinx=x−x33!+x55!−x77!+…cosx=1−x22!+x44!−x66!+…tanx=x+x33+2x515+17315x7+…log(1+x)=x−x22+x33−x44+x55−…$$\begin{array}{cc}{e}^x& =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \\ \sin x& =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots \\ \cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \\ \tan x& =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17}{315}{x}^7+\dots \\ \log (1+x)& =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots \end{array}$$

8.0.2 Question 1

For each of the following functions prove that Rn→0$R_n\to 0$ as n→∞$n\to \infty$ and write down the power series expansion of each.

1. cosx$\cos x$

Let f(x)=cosx$f\left(x\right)=\cos x$.

We know the Maclaurin’s series with Lagrange’s form of remainder,

f(x)=f(0)+x1!f′(0)+x22!f″(0)+…+xn−1(n−1)!fn−1(0)+xnn!fn(θx)$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+\frac{x}{1!}{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\dots +\frac{x^{n-1}}{(n-1)!}{f}^{n-1}\left(0\right)+\\ \frac{x^n}{n!}{f}^n\left(\theta x\right)& \end{array}$$

Then,

fn(x)=cos(x+nπ/2)fn(θx)=cos(θx+nπ/2)Rn=xnn!fn(θx)=xnn!cos(θx+nπ/2)$$\begin{array}{cc}{f}^n\left(x\right)& =\cos (x+n\pi /2)\\ f^n\left(\theta x\right)& =\cos (\theta x+n\pi /2)\\ R_n& =\frac{x^n}{n!}{f}^n\left(\theta x\right)\\ & =\frac{x^n}{n!}\cos (\theta x+n\pi /2)\end{array}$$

As n→∞$n\to \infty$, Rn→0$R_n\to 0$ for all values of x$x$. Thus conditions for Maclaurin’s expansion for infinite series is valid.

Also,

f(0)=1f′(0)=cos(0+π/2)=0f″(0)=cos(0+2π/2)=−1f‴(0)=cos(0+3π/2)=0f4(0)=cos(0+4π/2)=1$$\begin{array}{cc}f\left(0\right)& =1\\ f^{\prime }\left(0\right)& =\cos (0+\pi /2)=0\\ f^{\prime \prime }\left(0\right)& =\cos (0+2\pi /2)=-1\\ f^{\prime \prime \prime }\left(0\right)& =\cos (0+3\pi /2)=0\\ f^4\left(0\right)& =\cos (0+4\pi /2)=1\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…=1+0−x22!+0+x44!+0−x66!+…cosx=1−x22!+x44!−x66!+…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ & =1+0-\frac{x^2}{2!}+0+\frac{x^4}{4!}+0-\frac{x^6}{6!}+\dots \\ \cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \end{array}$$

2. sin2x${\sin }^{2}x$

Let f(x)=sin2x=1−cos2x2$f\left(x\right)={\sin }^{2}x=\frac{1-\cos 2x}{2}$.

fn(x)=0−12×2ncos(2x+nπ/2)=−2n−1cos(2x+nπ/2)fn(θx)=−2n−1cos(2θx+nπ/2)Rn=xnn!fn(θx)=−xnn!2n−1cos(2θx+nπ/2)$$\begin{array}{cc}{f}^n\left(x\right)& =0-\frac{1}{2}\times 2^n\cos (2x+n\pi /2)=-2^{n-1}\cos (2x+n\pi /2)\\ f^n\left(\theta x\right)& =-2^{n-1}\cos (2\theta x+n\pi /2)\\ R_n& =\frac{x^n}{n!}{f}^n\left(\theta x\right)\\ & =-\frac{x^n}{n!}{2}^{n-1}\cos (2\theta x+n\pi /2)\end{array}$$

As n→∞$n\to \infty$, Rn→0$R_n\to 0$ for all values of x$x$. Thus conditions for Maclaurin’s expansion for infinite series is valid.

Also,

f(0)=0f′(0)=−20cos(0+π/2)=0f″(0)=−22−1cos(0+2π/2)=2f‴(0)=−22cos(0+3π/2)=0f4(0)=−23cos(0+4π/2)=−23f5(0)=−24cos(0+5π/2)=0f6(0)=−25cos(0+6π/2)=25$$\begin{array}{cc}f\left(0\right)& =0\\ f^{\prime }\left(0\right)& =-2^0\cos (0+\pi /2)=0\\ f^{\prime \prime }\left(0\right)& =-2^{2-1}\cos (0+2\pi /2)=2\\ f^{\prime \prime \prime }\left(0\right)& =-2^2\cos (0+3\pi /2)=0\\ f^4\left(0\right)& =-2^3\cos (0+4\pi /2)=-2^3\\ f^5\left(0\right)& =-2^4\cos (0+5\pi /2)=0\\ f^6\left(0\right)& =-2^5\cos (0+6\pi /2)=2^5\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…sin2x=0+0+2x22!+0−23x44!+0+25x66!−…=12{(2x)22!+0−(2x)44!+0+(2x)66!−…}=12{(2x)22!−(2x)44!+(2x)66!−…}$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ {\sin }^{2}x& =0+0+\frac{2x^2}{2!}+0-\frac{2^3{x}^4}{4!}+0+\frac{2^5{x}^6}{6!}-\dots \\ & =\frac{1}{2}\{\frac{(2x{)}^2}{2!}+0-\frac{(2x{)}^4}{4!}+0+\frac{(2x{)}^6}{6!}-\dots \}\\ & =\frac{1}{2}\{\frac{(2x{)}^2}{2!}-\frac{(2x{)}^4}{4!}+\frac{(2x{)}^6}{6!}-\dots \}\end{array}$$

sin2x=12{(2x)22!−(2x)44!+(2x)66!−…}$$\begin{array}{}\begin{array}{cc}{\sin }^{2}x& =\frac{1}{2}\{\frac{(2x{)}^2}{2!}-\frac{(2x{)}^4}{4!}+\frac{(2x{)}^6}{6!}-\dots \}\end{array}& \text{(8.1)}\end{array}$$

3. cos2x${\cos }^{2}x$

We know,

cos2x=1−sin2x$$\begin{array}{cc}{\cos }^{2}x& =1-{\sin }^{2}x\end{array}$$

From the expansion of sin2x${\sin }^{2}x$ in (8.1), we have,

cos2x=1−12{(2x)22!−(2x)44!+(2x)66!−…}$$\begin{array}{cc}{\cos }^{2}x& =1-\frac{1}{2}\{\frac{(2x{)}^2}{2!}-\frac{(2x{)}^4}{4!}+\frac{(2x{)}^6}{6!}-\dots \}\end{array}$$

4. log(1−x)$\log (1-x)$

Let y=f(x)=log(1−x)$y=f\left(x\right)=\log (1-x)$.

General equation for higher derivatives of the given function is,

fn(x)=(n−1)!(−1)2n−1(1−x)n$$\begin{array}{cc}{f}^n\left(x\right)& =\frac{(n-1)!(-1{)}^{2n-1}}{(1-x{)}^n}\end{array}$$

So,

f(0)=0f′(0)=−1f″(0)=−1f‴(0)=−2!f4(0)=−3!$$\begin{array}{cc}f\left(0\right)& =0\\ f^{\prime }\left(0\right)& =-1\\ f^{\prime \prime }\left(0\right)& =-1\\ f^{\prime \prime \prime }\left(0\right)& =-2!\\ f^4\left(0\right)& =-3!\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…log(1−x)=0−x1!−x22!−2!x33!−3!x44!−…log(1−x)=−x−x22−x33−x44−…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ \log (1-x)& =0-\frac{x}{1!}-\frac{x^2}{2!}-\frac{2!x^3}{3!}-\frac{3!x^4}{4!}-\dots \\ \log (1-x)& =-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\dots \end{array}$$

5. 1(1+x)$\frac{1}{(1+x)}$

Let y=f(x)=11+x$y=f\left(x\right)=\frac{1}{1+x}$.

f′(x)=(−1)(1+x)−2f″(x)=(−1)22!(1+x)−3f‴(x)=(−1)33!(1+x)−4$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =(-1)(1+x{)}^{-2}\\ f^{\prime \prime }\left(x\right)& =(-1{)}^{2}2!(1+x{)}^{-3}\\ f^{\prime \prime \prime }\left(x\right)& =(-1{)}^{3}3!(1+x{)}^{-4}\end{array}$$

So general form for higher derivatives is,

fn(x)=(−1)nn!(1+x)n+1$$\begin{array}{cc}{f}^n\left(x\right)& =\frac{(-1{)}^{n}n!}{(1+x{)}^{n+1}}\end{array}$$

Calculating higher derivatives for x=0$x=0$,

f(0)=1f′(0)=−1f″(0)=2!f‴(0)=−3!$$\begin{array}{cc}f\left(0\right)& =1\\ f^{\prime }\left(0\right)& =-1\\ f^{\prime \prime }\left(0\right)& =2!\\ f^{\prime \prime \prime }\left(0\right)& =-3!\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…11+x=1−x+2!x22!−3!x33!+…=1−x+x2−x3+x4−…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ \frac{1}{1+x}& =1-x+\frac{2!x^2}{2!}-\frac{3!x^3}{3!}+\dots \\ & =1-x+x^2-x^3+x^4-\dots \end{array}$$

8.0.3 Question 2

Expand in ascending powers of x

1. (1+x)m$(1+x{)}^m$

Let y=f(x)=(1+x)m$y=f\left(x\right)=(1+x{)}^m$.

Differentiating successively,

f′(x)=m(1+x)m−1f″(x)=m(m−1)(1+x)m−2f‴(x)=m(m−1)(m−2)(1+x)m−3fn(x)=m!(m−3)!(1+x)m−n$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =m(1+x{)}^{m-1}\\ f^{\prime \prime }\left(x\right)& =m(m-1)(1+x{)}^{m-2}\\ f^{\prime \prime \prime }\left(x\right)& =m(m-1)(m-2)(1+x{)}^{m-3}\\ f^n\left(x\right)& =\frac{m!}{(m-3)!}(1+x{)}^{m-n}\end{array}$$

Calculating higher derivatives at x=0$x=0$,

f(0)=1f′(0)=mf″(0)=m(m−1)f‴(0)=m(m−1)(m−2)$$\begin{array}{cc}f\left(0\right)& =1\\ f^{\prime }\left(0\right)& =m\\ f^{\prime \prime }\left(0\right)& =m(m-1)\\ f^{\prime \prime \prime }\left(0\right)& =m(m-1)(m-2)\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…(1+x)m=1+mx+m(m−1)x22!+m(m−1)(m−2)x33!+…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ (1+x{)}^m& =1+mx+\frac{m(m-1)x^2}{2!}+\frac{m(m-1)(m-2)x^3}{3!}+\dots \end{array}$$

2. tanhx$\tan \text{h}x$

Let tanhx=sinhxcoshx=f(x)g(x)$\tan \text{h}x=\frac{\sin \text{h}x}{\cos \text{h}x}=\frac{f\left(x\right)}{g\left(x\right)}$. We have to find individual expansion of f(x)=sinhx$f\left(x\right)=\sin \text{h}x$ and g(x)=coshx$g\left(x\right)=\cos \text{h}x$ and divide f(x)$f\left(x\right)$ by g(x)$g\left(x\right)$ to find the required expansion.

We have,

f(x)=sinhx⟹f(0)=0f′(x)=coshx⟹f′(0)=1f″(x)=sinhx⟹f″(0)=0f‴(x)=coshx⟹f‴(0)=1f4(x)=sinhx⟹f4(0)=0$$\begin{array}{ccc}f\left(x\right)& =\sin \text{h}x& ⟹f\left(0\right)=0\\ f^{\prime }\left(x\right)& =\cos \text{h}x& ⟹f^{\prime }\left(0\right)=1\\ f^{\prime \prime }\left(x\right)& =\sin \text{h}x& ⟹f^{\prime \prime }\left(0\right)=0\\ f^{\prime \prime \prime }\left(x\right)& =\cos \text{h}x& ⟹f^{\prime \prime \prime }\left(0\right)=1\\ f^4\left(x\right)& =\sin \text{h}x& ⟹f^4\left(0\right)=0\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…sinhx=x+x33!+x55!+…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ \sin \text{h}x& =x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots \end{array}$$

Similarly,

g(x)=coshx⟹g(0)=1g′(x)=sinhx⟹g′(0)=0g″(x)=coshx⟹g″(0)=1g‴(x)=sinhx⟹g‴(0)=0g4(x)=coshx⟹g4(0)=1$$\begin{array}{ccc}g\left(x\right)& =\cos \text{h}x& ⟹g\left(0\right)=1\\ g^{\prime }\left(x\right)& =\sin \text{h}x& ⟹g^{\prime }\left(0\right)=0\\ g^{\prime \prime }\left(x\right)& =\cos \text{h}x& ⟹g^{\prime \prime }\left(0\right)=1\\ g^{\prime \prime \prime }\left(x\right)& =\sin \text{h}x& ⟹g^{\prime \prime \prime }\left(0\right)=0\\ g^4\left(x\right)& =\cos \text{h}x& ⟹g^4\left(0\right)=1\end{array}$$

Using Maclaurin’s expansion for infinite series,

g(x)=g(0)+xg′(0)+x22!g″(0)+x33!g‴(0)+x44!g4(0)+x55!g5(0)+x66!g6(0)+…coshx=1+x22!+x44!+x66!+…$$\begin{array}{cc}g\left(x\right)& =g\left(0\right)+x{g}^{\prime }\left(0\right)+\frac{x^2}{2!}{g}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{g}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{g}^4\left(0\right)+\frac{x^5}{5!}{g}^5\left(0\right)+\frac{x^6}{6!}{g}^6\left(0\right)+\dots \\ \cos \text{h}x& =1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots \end{array}$$

Now,

tanhx=sinhxcoshx=f(x)g(x)=x+x33!+x55!+…1+x22!+x44!+x66!+…=x−23!x3+165!x5−…$$\begin{array}{cc}\tan \text{h}x& =\frac{\sin \text{h}x}{\cos \text{h}x}=\frac{f\left(x\right)}{g\left(x\right)}\\ & =\frac{x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots }{1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots }\\ & =x-\frac{2}{3!}{x}^3+\frac{16}{5!}{x}^5-\dots \end{array}$$

How this division is carried out is shown here,

Figure 8.1: Division of two expansions

For inspiration of such division, see this video.

8.0.4 Question 3

1. Expand tanx$\tan x$ in the ascending integral powers of x$x$ using Maclaurin’s series and hence get the expansion of sec2x${\sec }^{2}x$.

We know tanx=sinxcosx$\tan x=\frac{\sin x}{\cos x}$. We also know the expansions as follow:

sinx=x−x33!+x55!−…cosx=1−x22!+x44!−x66!+…$$\begin{array}{cc}\sin x& =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \\ \cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \end{array}$$

Thus,

tanx=x−x33!+x55!−…1−x22!+x44!−x66!+…$$\begin{array}{cc}\tan x& =\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots }{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots }\end{array}$$

By division as shown in figure below, we have

tanx=x+13x3+215x5+17315x7+…$$\begin{array}{cc}\tan x& =x+\frac{1}{3}{x}^3+\frac{2}{15}{x}^5+\frac{17}{315}{x}^7+\dots \end{array}$$

Figure 8.2: Division of sinx$\sin x$ by cosx$\cos x$

Differentiating both sides w.r.t x$x$,

sec2x=1+3x23+2×5x415+17×7x6315+…=1+x2+23x4+1745x6+…$$\begin{array}{cc}{\sec }^{2}x& =1+\frac{3x^2}{3}+\frac{2\times 5x^4}{15}+\frac{17\times 7x^6}{315}+\dots \\ & =1+x^2+\frac{2}{3}{x}^4+\frac{17}{45}{x}^6+\dots \end{array}$$

2. Obtain the series expansion of esinx$e^{\sin x}$ by Maclaurin’s theorem as far as the term x4$x^4$.

Let f(x)=esinx$f\left(x\right)=e^{\sin x}$.

f′(x)=esinxcosx=f(x)cosxf″(x)=f′(x)cosx−f(x)sinxf‴(x)=f″(x)cosx−f′(x)sinx−(f(x)cosx+f′(x)sinx)=f″(x)cosx−2f′(x)sinx−f′(x)f4(x)=f‴(x)cosx−f″(x)sinx−2(f′(x)cosx+f″(x)sinx)−f″(x)=f‴(x)cosx−3f″(x)sinx−2f′(x)cosx−f″(x)$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =e^{\sin x}\cos x=f\left(x\right)\cos x\\ f^{\prime \prime }\left(x\right)& =f^{\prime }\left(x\right)\cos x-f\left(x\right)\sin x\\ f^{\prime \prime \prime }\left(x\right)& =f^{\prime \prime }\left(x\right)\cos x-f^{\prime }\left(x\right)\sin x-\left(f\right(x)\cos x+f^{\prime }(x\left)\sin x\right)\\ & =f^{\prime \prime }\left(x\right)\cos x-2f^{\prime }\left(x\right)\sin x-f^{\prime }\left(x\right)\\ f^4\left(x\right)& =f^{\prime \prime \prime }\left(x\right)\cos x-f^{\prime \prime }\left(x\right)\sin x-2\left(f^{\prime }\right(x)\cos x+f^{\prime \prime }(x\left)\sin x\right)-f^{\prime \prime }\left(x\right)\\ & =f^{\prime \prime \prime }\left(x\right)\cos x-3f^{\prime \prime }\left(x\right)\sin x-2f^{\prime }\left(x\right)\cos x-f^{\prime \prime }\left(x\right)\end{array}$$

Calculating higher order derivatives at x=0$x=0$,

f(0)=1f′(0)=1×1=1f″(0)=1−0=1f‴(0)=1−0−1=0f4(0)=0−0−2−1=−3f5(0)=f4(0)cos0−3f″(0)cos0−2f″(0)cos0−f‴(0)=−3−3−2−0=−8$$\begin{array}{cc}f\left(0\right)& =1\\ f^{\prime }\left(0\right)& =1\times 1=1\\ f^{\prime \prime }\left(0\right)& =1-0=1\\ f^{\prime \prime \prime }\left(0\right)& =1-0-1=0\\ f^4\left(0\right)& =0-0-2-1=-3\\ f^5\left(0\right)& =f^4\left(0\right)\cos 0-3f^{\prime \prime }\left(0\right)\cos 0-2f^{\prime \prime }\left(0\right)\cos 0-f^{\prime \prime \prime }\left(0\right)\\ & =-3-3-2-0=-8\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…esinx=1+x+x22!×1+0+x44!×(−3)+x55!×(−8)…=1+x+x22−x48−x515…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ e^{\sin x}& =1+x+\frac{x^2}{2!}\times 1+0+\frac{x^4}{4!}\times (-3)+\frac{x^5}{5!}\times (-8)\dots \\ & =1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}\dots \end{array}$$

I have answered this question in Math StackExchange too.

3. Find the power series expansion of exsinx$e^x\sin x$ by Maclaurin’s theorem upto x4$x^4$.

Let y=f(x)=exsinx$y=f\left(x\right)=e^x\sin x$.

f′(x)=excosx+exsinxf″(x)=excosx−exsinx+excosx+exsinx=2excosxf‴(x)=2(excosx−exsinx)f4(x)=2(excosx−exsinx−excosx−exsinx)=−4exsinxf5(x)=−4(excosx+exsinx)$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =e^x\cos x+e^x\sin x\\ f^{\prime \prime }\left(x\right)& =e^x\cos x-e^x\sin x+e^x\cos x+e^x\sin x\\ & =2e^x\cos x\\ f^{\prime \prime \prime }\left(x\right)& =2(e^x\cos x-e^x\sin x)\\ f^4\left(x\right)& =2(e^x\cos x-e^x\sin x-e^x\cos x-e^x\sin x)\\ & =-4e^x\sin x\\ f^5\left(x\right)& =-4(e^x\cos x+e^x\sin x)\end{array}$$

Other way of finding successive derivatives is the formula in the chapter on higher derivatives,

f(x)=exsinxfn(x)=(12+12)n/2exsin(x+ntan−11)=2n/2exsin(x+nπ/4)fn(0)=2n/2sin(nπ4)$$\begin{array}{cc}f\left(x\right)& =e^x\sin x\\ f^n\left(x\right)& =(1^2+1^2{)}^{n/2}{e}^x\sin (x+n{\tan }^{-1}1)\\ & =2^{n/2}{e}^x\sin (x+n\pi /4)\\ f^n\left(0\right)& =2^{n/2}\sin \left(\frac{n\pi }{4}\right)\end{array}$$

Calculating higher derivatives at x=0$x=0$,

f(0)=0f′(0)=1f″(0)=2f‴(0)=2f4(0)=0f5(0)=−4$$\begin{array}{cc}f\left(0\right)& =0\\ f^{\prime }\left(0\right)& =1\\ f^{\prime \prime }\left(0\right)& =2\\ f^{\prime \prime \prime }\left(0\right)& =2\\ f^4\left(0\right)& =0\\ f^5\left(0\right)& =-4\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…exsinx=0+x+x22!×2+x33!×2+0+x55!×(−4)…exsinx=x+x2+x33−x530…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ e^x\sin x& =0+x+\frac{x^2}{2!}\times 2+\frac{x^3}{3!}\times 2+0+\frac{x^5}{5!}\times (-4)\dots \\ e^x\sin x& =x+x^2+\frac{x^3}{3}-\frac{x^5}{30}\dots \end{array}$$

4. Expand log(1+sinx)$\log (1+\sin x)$ with the help of Maclaurin’s theorem as far as the term involving x4$x^4$.

Let f(x)=log(1+sinx)$f\left(x\right)=\log (1+\sin x)$.

f′(x)=cosx1+sinxf″(x)=−sinx(1+sinx)−cos2x(1+sinx)2=−sinx−sin2x−cos2x(1+sinx)2=−11+sinxf‴(x)=cosx(1+sinx)2f4(x)=−sinx(1+sinx)2−cosx(2(1+sinx)cosx)(1+sinx)4=(1+sinx)(−sinx−sin2x−2cos2x)(1+sinx)4=−1+sinx+cos2x(1+sinx)3$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =\frac{\cos x}{1+\sin x}\\ f^{\prime \prime }\left(x\right)& =\frac{-\sin x(1+\sin x)-{\cos }^{2}x}{(1+\sin x{)}^2}\\ & =\frac{-\sin x-{\sin }^{2}x-{\cos }^{2}x}{(1+\sin x{)}^2}\\ & =-\frac{1}{1+\sin x}\\ f^{\prime \prime \prime }\left(x\right)& =\frac{\cos x}{(1+\sin x{)}^2}\\ f^4\left(x\right)& =\frac{-\sin x(1+\sin x{)}^2-\cos x(2(1+\sin x)\cos x)}{(1+\sin x{)}^4}\\ & =\frac{(1+\sin x)(-\sin x-{\sin }^{2}x-2{\cos }^{2}x)}{(1+\sin x{)}^4}\\ & =-\frac{1+\sin x+{\cos }^{2}x}{(1+\sin x{)}^3}\end{array}$$

Calculating higher derivatives at x=0$x=0$,

f(0)=0f′(0)=1f″(0)=−1f‴(0)=1f4(0)=−2$$\begin{array}{cc}f\left(0\right)& =0\\ f^{\prime }\left(0\right)& =1\\ f^{\prime \prime }\left(0\right)& =-1\\ f^{\prime \prime \prime }\left(0\right)& =1\\ f^4\left(0\right)& =-2\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…log(1+sinx)=0+x+x22!×(−1)+x33!×(1)+x44!×(−2)+…log(1+sinx)=x−x22+x36−x412+…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ \log (1+\sin x)& =0+x+\frac{x^2}{2!}\times (-1)+\frac{x^3}{3!}\times \left(1\right)+\frac{x^4}{4!}\times (-2)+\dots \\ \log (1+\sin x)& =x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{12}+\dots \end{array}$$

Instead of going through cumbersome successive derivatives, we can approach the problem in other way too,

ddxlog(1+sinx)=cosx1+sinx=1−x22!+x44!−x66!+…1+x−x33!+x55!−x77!+…$$\begin{array}{cc}\frac{d}{dx}\log (1+\sin x)& =\frac{\cos x}{1+\sin x}\\ & =\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots }{1+x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots }\end{array}$$

Integrating both sides after the result of division gives the desired result.

22. Apply Maclaurin’s series to find the expansion of ex1+ex$\frac{e^x}{1+e^x}$ as far as the term in x3$x^3$ and hence the expansion of log(1+ex)$\log (1+e^x)$.

The expansion of ex$e^x$ is,

ex=1+x+x22!+x33!+x44!+…$$\begin{array}{cc}{e}^x& =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \end{array}$$

So,

1+ex=1+1+x+x22!+x33!+…=2+x+x22!+x33!+…$$\begin{array}{cc}1+e^x& =1+1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ & =2+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \end{array}$$

Thus,

ex1+ex=1+x+x22!+x33!+…2+x+x22!+x33!+…=12+x4−x348+x496…$$\begin{array}{cc}\frac{e^x}{1+e^x}& =\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots }{2+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots }\\ & =\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+\frac{x^4}{96}\dots \end{array}$$

Figure 8.3: Figure showing divisions of ex$e^x$ by 1+ex$1+e^x$ expansion

For second part, let f(x)=log(1+ex)$f\left(x\right)=\log (1+e^x)$.

f(x)=log(1+ex)f(0)=log2f′(x)=ex1+ex=12+x4−x348+x496…f′(0)=12f″(x)=14−3x248+4x396…f″(0)=14f‴(x)=−6x48+12x296…f‴(0)=0f4(x)=−648+24x96…f4(0)=−648$$\begin{array}{cc}f\left(x\right)& =\log (1+e^x)\\ f\left(0\right)& =\log 2\\ f^{\prime }\left(x\right)& =\frac{e^x}{1+e^x}=\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+\frac{x^4}{96}\dots \\ f^{\prime }\left(0\right)& =\frac{1}{2}\\ f^{\prime \prime }\left(x\right)& =\frac{1}{4}-\frac{3x^2}{48}+\frac{4x^3}{96}\dots \\ f^{\prime \prime }\left(0\right)& =\frac{1}{4}\\ f^{\prime \prime \prime }\left(x\right)& =-\frac{6x}{48}+\frac{12x^2}{96}\dots \\ f^{\prime \prime \prime }\left(0\right)& =0\\ f^4\left(x\right)& =-\frac{6}{48}+\frac{24x}{96}\dots \\ f^4\left(0\right)& =-\frac{6}{48}\end{array}$$

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…log(1+ex)=log2+12x+x22!×14+0+x44!×(−648)+…=log2+x2+x28−x4192+…$$\begin{array}{cc}f\left(x\right)& =f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{\prime \prime }\left(0\right)+\frac{x^3}{3!}{f}^{\prime \prime \prime }\left(0\right)+\\ & \frac{x^4}{4!}{f}^4\left(0\right)+\frac{x^5}{5!}{f}^5\left(0\right)+\frac{x^6}{6!}{f}^6\left(0\right)+\dots \\ \log (1+e^x)& =\log 2+\frac{1}{2}x+\frac{x^2}{2!}\times \frac{1}{4}+0+\frac{x^4}{4!}\times (-\frac{6}{48})+\dots \\ & =\log 2+\frac{x}{2}+\frac{x^2}{8}-\frac{x^4}{192}+\dots \end{array}$$

8.0.5 Question 4

Assuming the validity of expansion, prove the following series using Maclaurin’s theorem.

1. secx=1+12x2+524x4+…$\sec x=1+\frac{1}{2}{x}^2+\frac{5}{24}{x}^4+\dots$

We know,

cosx=1−x22!+x44!−x66!+…$$\begin{array}{cc}\cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \end{array}$$

Let,

secx=a0+a1x+a2x2+a3x3+a4x4+…$$\begin{array}{}\begin{array}{cc}\sec x& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\dots \end{array}& \text{(8.2)}\end{array}$$

1=(a0+a1x+a2x2+a3x3+a4x4+…)cosx1=(a0+a1x+a2x2+a3x3+a4x4+…)(1−x22!+x44!−x66!+…)

Equating the constant term and the coefficients of x,x2,x3,x4, we have,

a0=1a1=0−a02!+a2=0,a2=12−a12!+a3=0,a3=0a04!−a22!+a4=0,a4=524

Replacing values of a0,a1,a2,a3,a4 in (8.2), we have,

secx=1+0+12x2+0+524x4+…=1+12x2+524x4+…

2. log(1+tanx)=x−x22!+43!x3−…

Approaching this problem via successive derivatives is cumbersome. So we use algebraic method for expansion utilising what we know already.

What we know,

tanx=x+x33+2x515+17315x7+…log(1+x)=x−x22+x33−x44+x55−x66+x77−…

So,

log(1+tanx)=tanx−(tanx)22+(tanx)33−(tanx)44+(tanx)55−(tanx)66+(tanx)77−…

Lets find values of numerators in the fractions,

(tanx)2=(x+x33+2x515+17315x7+…)2=x2+2x(x33+2x515+17315x7+…)+(x33+2x515+17315x7+…)2=x2+2x43+4x615+…+x69+…=x2+2x43+17x645+…(tanx)3=(x+x33+2x515+17315x7+…)3=x3+3x2(x33+2x515+…)+3x(x33+2x515+…)2+(x33+2x515+…)3=x3+3x53+6x715+…+3x79+12x945+…+x927+…=x3+3x53+11x715+…(tanx)4=(x+x33+2x515+17315x7+…)4=x4+4x3(x33+2x515+…)+…=x4+4x63+…(tanx)5=(x+x33+2x515+17315x7+…)5=x5+5x4(x33+2x515+…)+…=x5+5x73+…(tanx)6=(x+x33+2x515+17315x7+…)6=x6+…(tanx)7=(x+x33+2x515+17315x7+…)7=x7+…

Lets put these values in equation (8.3),

log(1+tanx)=tanx−(tanx)22+(tanx)33−(tanx)44+(tanx)55−(tanx)66+(tanx)77−…=(x+x33+2x515+17315x7+…)−12(x2+2x43+17x645+…)+13(x3+3x53+11x715+…)−14(x4+4x63+…)+15(x5+5x73+…)−16(x6+…)+17(x7+…)

Now collecting coefficients of x,x2,x3,x4,x5,x6,x7, we have,

x	x2	x3	x4	x5
1	−12	13+13=23	−13−14=−712	215+13+15=23

x6	x7
−1790−13−16=−3145	17315+1145+13+17=244315

So the expansion is,

log(1+tanx)=x−12x2+23x3−712x4+23x5−3145x6+244315x7−…

For a much easier approach, derivative of the function log(1+tanx) is given by

ddxln(1+tan(x))=sec2(x)1+tan(x)=1+tan2(x)1+tan(x)=1+(x+13x3+215x5)21+x+13x3+215x5+O(x7)=1+x2+23x4+1745x61+x+13x3+215x5+O(x7)=1−x+2x2−73x3+103x4−6215x5+24445x6+O(x7)

Integrating both sides,

log(1+tanx)=x−12x2+23x3−712x4+23x5−3145x6+244315x7−…

3. exsecx=1+x+x2+23x3+…

exsecx=(1+x+x22!+x33!+x44!+…)(1+12x2+524x4+…)=1+x+(12+12!)x2+(12+13!)x3+(524+12!2+14!)x4+…=1+x+x2+46x3+1124x4+…=1+x+x2+23x3+1124x4+…

This video on YouTube is good to learn the concept of multiplication of two expansions.

4. eaxcosbx=1+ax+(a2−b2)x22!+a(a2−3b2)x33!+…

y=eaxcosbxy1=aeaxcosbx−beaxsinbx=ay−beaxsinbxy2=ay1−b(beaxcosbx+aeaxsinbx)=ay1−b2y−abeaxsinbxy3=ay2−b2y1−ab(beaxcosbx+aeaxsinbx)=ay2−b2y1−ab2y−a2beaxsinbx

(y)0=1(y1)0=a(y2)0=a2−b2(y3)0=a(a2−b2)−ab2−ab2−0=a(a2−3b2)

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…eaxcosbx=1+ax+(a2−b2)x22!+a(a2−3b2)x33!+…

22. log(1+x+x2)=x+x22−23x3+x44+…

We know the expansion for log(1+x) is,

log(1+x)=x−x22+x33−x44+x55−…

So,

log(1+x+x2)=log{1+(x+x2)}=(x+x2)−(x+x2)22+(x+x2)33−(x+x2)44+(x+x2)55−…

Lets expand the expressions in the numerator using binomial expansion,

Figure 8.4: Binomial expansion

log(1+x+x2)=(x+x2)−(x+x2)22+(x+x2)33−(x+x2)44+(x+x2)55−…=x+x2−x2+2x3+x42+x3+3x5+3x4+x63−x4+4x5+6x6+4x7+x84+x5+5x6+10x7+10x8+5x9+x105−…

Bringing together the coefficients of x,x2,x3,x4,x5…,

log(1+x+x2)=1x+(1−12)x2+(−1+13)x3+(−12+1−14)x4+(1−1+15)x5…=x+12x2−23x3+14x4+15x5…

8.0.6 Question 5

Prove that sin−1x=x+x33!+9x55!+…

Let y=sin−1x.

y1=1√1−x2(1−x2)(y1)2=1

Differentiating w.r.t x,

(1−x2)2y1y2−2x(y1)2=0(1−x2)y2−xy1=0

Differentiating n times using Leibnitz’s theorem,

yn+2(1−x2)+nyn+1(−2x)+n(n−1)2yn(−2)+0−(yn+1x+nyn)=0(1−x2)yn+2−2nxyn+1−(n2−n)yn−xyn+1−nyn=0(1−x2)yn+2−(2n+1)xyn+1−n2yn=0

Putting x=0,

(yn+2)0=n2(yn)0(y)0=0(y1)0=1(y2)0=0(y3)0=(y1)0=1(y4)0=4(y2)0=0(y5)0=9(y3)0=9

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…sin−1x=0+x+0+x33!×1+0+x55!×9+…=x+x33!+9x55!+…

and hence show that cos−1x=π2−x−x33!−…

From the formula on inverse sum identities, we know that,

sin−1x+cos−1x=π2, orcos−1x=π2−sin−1x

We have already found the expansion for sin−1x, so,

cos−1x=π2−x−x33!−9x55!−…

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8.0.7 Question 6i

Show that xex−1=1−12x+112x2−1720x4+…

We know,

ex=1+x+x22!+x33!+x44!+x55!+…ex−1=x+x22!+x33!+x44!+x55!+…

So,

xex−1=xx+x22!+x33!+x44!+x55!+…

Let

xex−1=a0+a1x+a2x2+a3x3+a4x4+a5x5+…

Equating (8.4) and (8.5),

xx+x22!+x33!+x44!+x55!+…=a0+a1x+a2x2+a3x3+a4x4+a5x5+…x=(a0+a1x+a2x2+a3x3+a4x4+a5x5+…)(x+x22!+x33!+x44!+x55!+…)

Equating the coefficients of x,x2,x3,x4,x5,

a0=1a02!+a1=0⟹a1=−12a03!+a12!+a2=0⟹a2=112a04!+a13!+a22!+a3=0⟹a3=0a05!+a14!+a23!+a32!+a4=0⟹a4=−1720

Substituting these values in (8.5), we get

xex−1=a0+a1x+a2x2+a3x3+a4x4+a5x5+…=1−12x+112x2+0−1720x4+…=1−12x+112x2−1720x4+…

8.0.8 Question 6ii

Show that esin−1x=1+x+x22!+2x33!+54!x4+…

Let y=esin−1x.

Differentiating w.r.t x,

y1=esin−1x1√1−x2(1−x2)(y1)2=y2

Differentiating again w.r.t x,

2y1y2(1−x2)−2x(y1)2=2yy1(1−x2)y2−xy1=y

Differentiating n times using Leibnitz’s theorem,

yn+2(1−x2)−nC12xyn+1−nC22yn−xyn+1−nC1yn=yn(1−x2)yn+2−(2n+1)xyn+1−(n2+1)yn=0

Putting x=0,

(yn+2)0=(n2+1)(yn)0(y)0=e0=1(y1)0=e0=1(y2)0=(y)0=1(y3)0=(12+1)(y1)0=2(y4)0=(22+1)(y2)0=5

Using Maclaurin’s expansion for infinite series,

f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!f4(0)+x55!f5(0)+x66!f6(0)+…esin−1x=1+x+12!x2+23!x3+54!x4+…

8.0.9 Question 6iii

If y=emtan−1x=a0+a1x+a2x2+a3x3+… show that

• (1+x2)y1=my

Given equation is y=emtan−1x. Differentiating w.r.t x,

y1=emtan−1x×m×11+x2y1=y×m×11+x2

(1+x2)y1=my

• (n+1)an+1+(n−1)an−1=man

Differentiating equation (8.6) n times using Leibnitz’s theorem, we have,

(1+x2)yn+1+nyn×2x+n(n−1)2yn−1×2+0=myn

(1+x2)yn+1+2nxyn+n(n−1)yn−1=myn

(1+x2)yn+1+(2nx−m)yn+n(n−1)yn−1=0

Differentiating equation (8.6) again w.r.t x,

(1+x2)y2+2xy1=my1(1+x2)y2+(2x−m)y1=0

Also,

y=a0+a1x+a2x2+a3x3+…⟹(y)0=a0y1=a1+2a2x+3a3x2+…⟹(y1)0=a1y2=2a2+6a3x+…⟹(y2)0=2a2y3=6a3+…⟹(y3)0=6a3

Thus a pattern emerges, (yn)0=n!an.

Putting x=0 in (8.7),

(yn+1)0−m(yn)0+n(n−1)(yn−1)0=0(n+1)!an+1−mn!an+n(n−1)(n−1)!an−1=0n!(n+1)an+1−n!man+(n−1)(n−1)!nan−1=0n!(n+1)an+1−n!man+n!(n−1)an−1=0(n+1)an+1+(n−1)an−1=man

Also obtain the expansion of emtan−1x.

From the derivatives, we have

(y)0=1=a0(y1)0=m(y)0=m=a1(y2)0=m(y1)0=m2=2a2

From nth derivative (8.7), putting n=2,

(y3)0=m(y2)0−2(y1)0=m×m2−2m=m(m2−2)=6a3

So substituting values of a0,a1,a2,a3,… in y=a0+a1x+a2x2+a3x3+…, we have:

emtan−1x=1+mx+m22!x2+m(m2−2)3!x3+…

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8.0.10 Question 7

Show that the following functions cannot be expanded in Maclaurin’s infinite series.

• f(x)=√x

f(x)=√x⟹f(0)=0f′(x)=12√x⟹f′(0) does not exist

Since f′(0) does not exist, so √x cannot be expanded in Maclaurin’s infinite series.

• f(x)=x5/2

f(x)=x5/2⟹f(0)=0f′(x)=52x3/2⟹f′(0)=0f″(x)=52×32x1/2⟹f″(0)=0f‴(x)=52×32×12√x⟹f‴(0) does not exist

Since f‴(0) does not exist, so x5/2 cannot be expanded in Maclaurin’s infinite series.