91Ó°ÊÓ

(a).

According to the figure a strong chain and a convenient tree, Fasten the chain to the car and to the tree. Pull with a force $F$ at the center of the chain. So it has,$F=2T\sin \mathrm{θ}$

, and$T=\frac{F}{2\sin \mathrm{θ}}$ where, $\text{T}$ is the tension in the chain.

Taylor series for the$\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{y}}$,

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{y}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{3}}\mathbf{+}\mathbf{…}\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{y}}^{\mathbf{n}}$for all $\mathit{n}$

From mechanics,

$\mathit{F}\mathbf{=}\mathbf{2}\mathit{T}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{,}\mathbf{\text{and}}\mathit{T}\mathbf{=}\frac{\mathbf{F}}{\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}$

Using the figure,

$$\begin{gathered} \sin \mathrm{θ}=\frac{x}{10}n \\ T=\frac{F}{2(x/10)}n \\ T=\frac{10F}{2x}n \\ T=\frac{5F}{x}n \\ T=5F{x}^{-1} \end{gathered}$$

It knows the Taylor series for $\frac{1}{1-y}â–¹$.

Take $1-y=t$,

$$\begin{gathered} y=-(t-1)n \\ T=5F\left[1-(t-1)+{(t-1)}^2-{(t-1)}^3+…\right] \\ T=\underset{n=0}{\overset{∞}{∑}}{(-1)}^n(t-1{)}^n,\text{for all}n \end{gathered}$$

Here $t=x$.

Hence, $T=5F\left[1-(x-1)+{(x-1)}^2-{(x-1)}^3+…\right]=\underset{n=0}{\overset{∞}{∑}}{(-1)}^n(x-1{)}^n,\text{for all}n$

(b).

As given: Using the figure in part (a), it can draw right angle triangle.

Using part (a),

$T=5F\left[1-(x-1)+{(x-1)}^2-{(x-1)}^3+…\right]=\underset{n=0}{\overset{∞}{∑}}{(-1)}^n(x-1{)}^n,\text{for all}n$

It knows,$\sin \mathrm{θ}=\frac{x}{10}$ .

For small angle$(\mathrm{θ}<<<),\sin \mathrm{θ}≈\mathrm{θ}$ .

Then, $\mathrm{θ}≈\sin \mathrm{θ}=\frac{x}{10}$.

$$\begin{gathered} x=10.\mathrm{θ}n \\ T=\frac{F}{2(x/10)} \\ =\frac{F}{2\mathrm{θ}} \\ =\frac{F}{2}{\mathrm{θ}}^{-1}n \\ T=\frac{F}{2}\left[1-(\mathrm{θ}-1)+{(\mathrm{θ}-1)}^2-{(\mathrm{θ}-1)}^3+…\right] \\ =\underset{n=0}{\overset{∞}{∑}}{(-1)}^n(\mathrm{θ}-1{)}^n,\text{for all}n \end{gathered}$$

Hence, $T=\frac{F}{2}\left[1-(\mathrm{θ}-1)+{(\mathrm{θ}-1)}^2-{(\mathrm{θ}-1)}^3+…\right]=\underset{n=0}{\overset{∞}{∑}}{(-1)}^n(\mathrm{θ}-1{)}^n,\text{for all}n$.