Question

Find $\mathit{T}$as${\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{=}\mathbf{1}\mathbf{/}\mathit{x}\mathbf{)}$ times a series of powers of$\mathit{x}$ .

(b). Find$\mathit{T}$ as${\mathit{\theta }}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{=}\mathbf{1}\mathbf{/}\mathit{\theta }\mathbf{)}$ times a series of powers of$\mathit{\theta }$ .

Short answer

(a) .$T=5F\left[1-(x-1)+{(x-1)}^2-{(x-1)}^3+\dots \right]=\sum _{n=0}^{\infty }{(-1)}^n(x-1{)}^n,\text{for all}n$

(b) .$T=\frac{F}{2}\left[1-(\theta -1)+{(\theta -1)}^2-{(\theta -1)}^3+\dots \right]=\sum _{n=0}^{\infty }{(-1)}^n(\theta -1{)}^n,\text{for all}n$

Step-by-step solution

xFind the T  as x-1(=1/x)  times a series of powers of  .

(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,$F=2T\sin \theta$

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

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

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{y}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{y}}^{\mathbf{n}}$for all $\mathit{n}$

From mechanics,

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

Using the figure,

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

Determine the  T as x-1(=1/x)  times a series of powers of  xby using Taylor series

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

Take $1-y=t$,

$$\begin{gathered} y=-(t-1)n \\ T=5F\left[1-(t-1)+{(t-1)}^2-{(t-1)}^3+\dots \right] \\ T=\sum _{n=0}^{\infty }{(-1)}^n(t-1{)}^n,\text{for all}n \end{gathered}$$

Here $t=x$.

Hence, $T=5F\left[1-(x-1)+{(x-1)}^2-{(x-1)}^3+\dots \right]=\sum _{n=0}^{\infty }{(-1)}^n(x-1{)}^n,\text{for all}n$

Determine  T as  θ-1(=1/θ) times a series of powers of θ . 

(b).

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

Using part (a),

$T=5F\left[1-(x-1)+{(x-1)}^2-{(x-1)}^3+\dots \right]=\sum _{n=0}^{\infty }{(-1)}^n(x-1{)}^n,\text{for all}n$

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

For small angle$(\theta <<<),\sin \theta \approx \theta$ .

Then, $\theta \approx \sin \theta =\frac{x}{10}$.

$$\begin{gathered} x=10.\theta n \\ T=\frac{F}{2(x/10)} \\ =\frac{F}{2\theta } \\ =\frac{F}{2}{\theta }^{-1}n \\ T=\frac{F}{2}\left[1-(\theta -1)+{(\theta -1)}^2-{(\theta -1)}^3+\dots \right] \\ =\sum _{n=0}^{\infty }{(-1)}^n(\theta -1{)}^n,\text{for all}n \end{gathered}$$

Hence, $T=\frac{F}{2}\left[1-(\theta -1)+{(\theta -1)}^2-{(\theta -1)}^3+\dots \right]=\sum _{n=0}^{\infty }{(-1)}^n(\theta -1{)}^n,\text{for all}n$.