4.1 HW 1

4.1.1 Problems listing

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4.1.2 Problem 1.6.1

Expand the function $f(x)=\frac{\sin (x)}{\cosh (x)+2}$ in Taylor series around the origin going up to $x^3$. Calculate $f\left(0.1\right)$ from this series and compare to the exact answer obtained by using a calculator

Solution

The Taylor series of function $f(x)$ around origin is given by (1.3.16) ($\approx$ is used throughout this HW to mean that the left side is the Taylor series approximation of $f(x)$).$$f(x)\approx \sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}{f}^{\left(n\right)}(0)$$ Where $f^{(n)}(0)$ is the $n^{th}$ derivative of $f(x)$ evaluated at $x=0$.

For $n=0$, $f^{(0)}(x)=f\left(x\right)=\frac{\sin (x)}{\cosh (x)+2}$, therefore $f(0)=0$.

For $n=1$ $$\begin{array}{rl}{f}^{(1)}(x)& =\frac{d}{dx}\left(\frac{\sin (x)}{\cosh (x)+2}\right)\\ & =\frac{\cos (x)(\cosh (x)+2)-\sin (x)\sinh (x)}{{(\cosh (x)+2)}^2}\\ & =\frac{\cos (x)(\cosh (x)+2)}{{(\cosh (x)+2)}^2}-\frac{\sin (x)\sinh (x)}{{(\cosh (x)+2)}^2}\\ & =\frac{\cos (x)}{\cosh (x)+2}-\frac{\sin \left(x\right)\sinh (x)}{{(\cosh (x)+2)}^2}\end{array}$$

The above evaluated at $x=0$ becomes $$\begin{array}{rl}{f}^{(1)}(0)& =\frac{1}{1+2}-\frac{0}{{(1+2)}^2}\\ & =\frac{1}{3}\end{array}$$

For $n=2$ $$\begin{array}{rl}{f}^{(2)}(x)& =\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{\sin (x)}{\cosh (x)+2}\right)\right)\\ & =\frac{d}{dx}(\frac{\cos (x)}{\cosh (x)+2}-\frac{\sin (x)\sinh (x)}{{(\cosh \left(x\right)+2)}^2})\\ & =\frac{-\sin (x)(\cosh (x)+2)-\cos (x)\sinh (x)}{{(\cosh (x)+2)}^2}\\ & -\frac{(\cos (x)\sinh (x)+\sin \left(x\right)\cosh (x)){(\cosh (x)+2)}^2-\sin (x)\sinh (x)(2(\cosh (x)+2)\sinh (x))}{{(\cosh (x)+2)}^4}\\ & =\frac{-\sin (x)(\cosh (x)+2)}{{(\cosh (x)+2)}^2}-\frac{\cos (x)\sinh (x)}{{(\cosh (x)+2)}^2}-\frac{\cos (x)\sinh (x){(\cosh \left(x\right)+2)}^2}{{(\cosh (x)+2)}^4}\\ & -\frac{\sin (x)\cosh (x){(\cosh \left(x\right)+2)}^2}{{(\cosh (x)+2)}^4}+\frac{\sin (x)\sinh (x)(2(\cosh (x)+2)\sinh (x))}{{(\cosh (x)+2)}^4}\\ & =\frac{-\sin (x)}{\cosh (x)+2}-\frac{\cos (x)\sinh (x)}{{(\cosh (x)+2)}^2}-\frac{\cos (x)\sinh (x)}{{(\cosh (x)+2)}^2}-\frac{\sin (x)\cosh (x)}{{(\cosh (x)+2)}^2}+\frac{2\sin (x)\sinh (x)\sinh (x)}{{(\cosh (x)+2)}^3}\\ & =\frac{-\sin (x)}{\cosh (x)+2}-2\frac{\cos (x)\sinh (x)}{{(\cosh (x)+2)}^2}-\frac{\sin (x)\cosh (x)}{{(\cosh (x)+2)}^2}+\frac{2\sin (x){\sinh }^2(x)}{{(\cosh (x)+2)}^3}\end{array}$$

The above evaluated at $x=0$ becomes$$\begin{array}{rl}{f}^{(2)}(0)& =\frac{-0}{1+2}-2\frac{0}{{(1+2)}^2}-\frac{0}{{(1+2)}^2}+\frac{0}{{(1+2)}^3}\\ & =0\end{array}$$

For $n=3$ $$\begin{array}{rl}{f}^{(3)}(x)& =\frac{d}{dx}\left(\frac{d^2}{d{x}^2}\left(\frac{\sin (x)}{\cosh (x)+2}\right)\right)\\ & =\frac{d}{dx}(\frac{-\sin (x)}{\cosh (x)+2}-2\frac{\cos (x)\sinh (x)}{{(\cosh \left(x\right)+2)}^2}-\frac{\sin (x)\cosh (x)}{{(\cosh (x)+2)}^2}+\frac{2\sin (x){\sinh }^2(x)}{{(\cosh (x)+2)}^3})\\ & =\frac{-\cos (x)(\cosh (x)+2)+\sin (x)\sinh (x)}{{(\cosh (x)+2)}^2}\\ & -2\frac{(-\sin (x)\sinh (x)+\cos \left(x\right)\cosh (x)){(\cosh (x)+2)}^2-\cos (x)\sinh (x)(2(\cosh (x)+2)\sinh (x))}{{(\cosh (x)+2)}^4}\\ & -\frac{(\cos (x)\cosh (x)+\sin \left(x\right)\sinh (x)){(\cosh (x)+2)}^2-\sin (x)\cosh (x)(2(\cosh (x)+2)\sinh (x))}{{(\cosh (x)+2)}^4}\\ & +2\frac{(\cos (x){\sinh }^2(x)+2\sin (x)\cosh (x)){(\cosh \left(x\right)+2)}^3-(\sin (x){\sinh }^2\left(x\right))(3{(\cosh (x)+2)}^2\sinh (x))}{{(\cosh (x)+2)}^6}\end{array}$$

The above evaluated at $x=0$ gives$$\begin{array}{rl}{f}^{(3)}(0)& =\frac{-1(1+2)+0}{{(1+2)}^2}-2\frac{(-0+1){(1+2)}^2-0}{{(1+2)}^4}-\frac{(1+0){(1+2)}^2-0}{{(1+2)}^4}+2\frac{(0+0){(1+2)}^3-(0)\left(3{(1+2)}^{2}0\right)}{{(1+2)}^6}\\ & =\frac{-(3)}{(3{)}^2}-2\frac{\left(1\right)(3{)}^2}{(3{)}^4}-\frac{\left(1\right)(3{)}^2}{(3{)}^4}+2\frac{0}{{\left(3\right)}^6}\\ & =\frac{-1}{3}-2\frac{1}{3^2}-\frac{1}{3^2}\\ & =-\frac{2}{3}\end{array}$$

The process stops here, because the problem is asking for $n=3$.  Substituting all the derivatives $f^{(n)}(0)$ values above into$$f(x)\approx \sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}{f}^{\left(n\right)}(0)$$ For up to $n=3$ gives the following$$\begin{array}{rl}f(x)& \approx f(0)+x{f}^{(1)}(0)+\frac{x^2}{2}{f}^{(2)}(0)+\frac{x^3}{3!}{f}^{(3)}(0)+\cdots \\ & \approx 0+x\frac{1}{3}+\frac{x^2}{2}(0)+\frac{x^3}{3!}(-\frac{2}{3})\\ & \approx x\frac{1}{3}-\frac{2}{3}\frac{x^3}{6}\\ & \approx \frac{x}{3}-\frac{x^3}{9}\end{array}$$

When $x=\frac{1}{10}$ the above becomes$$\begin{array}{rl}{f}_{n=3}\left(\frac{1}{10}\right)& \approx \frac{1}{30}-\frac{1}{\left(1000\right)9}\\ & \approx \frac{1}{30}-\frac{1}{9000}\\ & \approx \frac{300-1}{9000}\\ & \approx \frac{299}{9000}\end{array}$$

From the calculator$$\frac{299}{9000}\approx 0.0332222$$ And from the exact expression $$\begin{array}{rl}\frac{\sin (x)}{\cosh (x)+2}& =\frac{\sin \left(0.1\right)}{\cosh \left(0.1\right)+2}\\ & =0.0332224\end{array}$$

The error is about $1.67\times {10}^{-7}$.

4.1.3 Problem 2

Consider $f(x)={(1+x)}^p$ for (a) $p=\frac{1}{3}$ and (b) $p=-2$, respectively.  (1) Find the Taylor series of $f\left(x\right)$ around $x=0$. (2) From the form of the general term, find the interval of convergence of the series. (3) How many terms in the series do you need to estimate $f\left(0.1\right)$ to within 1% ? Check that the difference between your estimate and the actual result has approximately the same magnitude as the next term in the series.

Solution

4.1.3.1 Case $p=\frac{1}{3}$

$$f(x)={(1+x)}^{\frac{1}{3}}$$ Part (1) The Taylor series is given by $$\begin{array}{}\text{(1)}& f(x)\approx f(0)+x{f}^{\prime }(0)+\frac{x^2}{2!}{f}^{\prime \prime }(x)+\frac{x^3}{3!}{f}^{\prime \prime \prime }(x)+\cdots \end{array}$$ Where $f(0)=1$ and $f^{\prime }(x)=\frac{1}{3}{(1+x)}^{-\frac{2}{3}}$. Hence $f^{\prime }(0)=\frac{1}{3}$ and $f^{\prime \prime }(x)=\frac{1}{3}(-\frac{2}{3}){(1+x)}^{-\frac{5}{3}}$. Hence $f^{\prime \prime }(0)=-\frac{(2)}{3^2}$, and $f^{\prime \prime \prime }(x)=\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3}){(1+x)}^{-\frac{8}{3}}$, hence $f^{\prime \prime \prime }(0)=\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})=\frac{(2)\left(5\right)}{3^3},$ and $f^{(4)}(x)=\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})(-\frac{8}{3}){(1+x)}^{-\frac{11}{3}}$, hence $f^{\left(4\right)}(0)=\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})(-\frac{8}{3})=-\frac{1}{3^4}((2)(5)(8))$ and on. The series in (1) becomes$$\begin{array}{rl}f(x)& \approx 1+\frac{1}{3}x-\frac{(2)}{3^2}\frac{x^2}{2!}+\frac{(2)(5)}{3^3}\frac{x^3}{3!}-\frac{(2)(5)(8)}{3^4}\frac{x^4}{4!}+\frac{(2)(5)\left(8\right)\left(11\right)}{3^5}\frac{x^5}{5!}-\frac{(2)(5)(8)\left(11\right)\left(14\right)}{3^6}\frac{x^6}{6!}-\cdots \\ \text{(2)}& & \approx 1+\frac{1}{3}x-\frac{1}{3^2}{x}^2+\frac{5}{3^4}{x}^3-\frac{10}{3^5}{x}^4+\frac{22}{3^6}{x}^5-\frac{154}{3^8}{x}^6+\cdots \end{array}$$

The general term is found by comparing the above to the general term obtained from binomial expansion. Since $$\begin{array}{}\text{(3)}& {(1+x)}^p=\left(\begin{array}{c}p\\ 0\end{array}\right)x^0+\left(\begin{array}{c}p\\ 1\end{array}\right)x+\left(\begin{array}{c}p\\ 2\end{array}\right)x^2+\cdots \end{array}$$ Comparing (2,3) shows that the general term is the binomial coefficient $\left(\begin{array}{c}\frac{1}{3}\\ n\end{array}\right)$. Therefore the Taylor series for ${(1+x)}^{\frac{1}{3}}$ can be written as$$f(x)\approx \sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}p\\ n\end{array}\right)x^n$$ For $p=\frac{1}{3}$ the above becomes$$f(x)\approx \sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}\frac{1}{3}\\ n\end{array}\right)x^n$$ Part(2) $$\begin{array}{rl}R& =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_n}{a_{n+1}}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\left(\begin{array}{c}p\\ n\end{array}\right)}{\left(\begin{array}{c}p\\ n+1\end{array}\right)}\right|\end{array}$$

The Binomial coefficient $\left(\begin{array}{c}p\\ n\end{array}\right)=\frac{p!}{n!(p-n)!}$, for when $p$ is integer. This is not the case here. For non-integer $p$ The Binomial coefficient becomes $\left(\begin{array}{c}p\\ n\end{array}\right)=\frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(p-n+1)}$ where $\mathrm{\Gamma }(p)$ is the Gamma function. The above ratio now becomes$$\begin{array}{rl}R& =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(p-n+1)}}{\frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(n+2)\mathrm{\Gamma }(p-n)}}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\mathrm{\Gamma }(n+2)\mathrm{\Gamma }(p-n)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(p-n+1)}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{n\mathrm{\Gamma }(p-n)}{\mathrm{\Gamma }(p-n+1)}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{n}{p-n}\right|\end{array}$$

But $p=\frac{1}{3}$, hence the above becomes$$\begin{array}{rl}R& =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{n}{\frac{1}{3}-n}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{n}{n-\frac{1}{3}}\right|\\ & =1\end{array}$$

Therefore the radius of convergence is $1$. This means the Taylor series found above converges to $f(x)$ for $\left|x\right|<1$.

Part 3

$$f(x)={(1+x)}^{\frac{1}{3}}$$ When $x=0.1$ $$\begin{array}{rl}f\left(0.1\right)& ={\left(1.1\right)}^{\frac{1}{3}}\\ & =1.032280115\end{array}$$

one percent of the above is $$\frac{1}{100}\left(1.032280115\right)=0.01032280115$$ The value $n$ is now found such that$$|R_n(x)|\le M\frac{{\left(0.1\right)}^{n+1}}{(n+1)!}\le 0.01032280115$$ Where $R_n(x)$ is the Taylor series remainder using $n$ terms.  $M$ is the upper bound for the $n+1$ derivative of $f(x)$ any where between $[0,0.1]$. Instead of trying to find $M$, few calculations are used to find how many terms are needed.

For $n=0$, $\tilde{f}\left(0.1\right)=1$ and the error is $1.032280115-1=032280115$.

For $n=1,$ $\tilde{f}\left(0.1\right)=1+\left(\begin{array}{c}\frac{1}{3}\\ 1\end{array}\right)0.1=1.0333333$, and the error is $|1.032280115-1.0333333|=0.001053218$. Because this is smaller than $R_n(x)$ then only two terms are needed in the Taylor series to obtained the required accuracy. Therefore$$f(x)\approx 1+\frac{1}{3}x$$

4.1.3.2 Case $p=-2$

$$f(x)={(1+x)}^{-2}$$ Part (1) The Taylor series is $$f(x)\approx f(0)+x{f}^{\prime }(0)+\frac{x^2}{2!}{f}^{\prime \prime }(x)+\frac{x^3}{3!}{f}^{\prime \prime \prime }(x)+\cdots$$ But $f(0)=1$ and $f^{\prime }(x)=(-2){(1+x)}^{-3}$. Hence $f^{\prime }(0)=-2$ and $f^{\prime \prime }(x)=(-2)(-3){(1+x)}^{-4}$. Hence $f^{\prime \prime }(0)=(-2)(-3)$, and $f^{\prime \prime \prime }(x)=-2(-3)(-4){(1+x)}^{-5}$, hence $f^{\prime \prime \prime }(0)=(-2)(-3)(-4)(-5)$ and so on. The above becomes$$\begin{array}{rl}f(x)& \approx 1+(-2)x-(-2)(-3)\frac{x^2}{2!}+(-2)(-3)(-4)\frac{x^3}{3!}+\cdots \\ & \approx 1-2x+(2)(3)\frac{x^2}{2!}-\left(2\right)(3)(4)\frac{x^3}{3!}+\cdots \\ & \approx 1-2x+3x^2-4x^3+\cdots \end{array}$$

The general term is therefore$$f(x)\approx \sum _{n=0}^{\mathrm{\infty }}{(-1)}^n(n+1)x^n$$ Part(2) $$\begin{array}{rl}R& =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_n}{a_{n+1}}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{(n+1)}{(n+2)}\right|\\ & =1\end{array}$$

Hence the series converges to $f(x)$ for $\left|x\right|<1$.

Part 3

$$f(x)={(1+x)}^{-2}$$ For $x=0.1$ $$\begin{array}{rl}f\left(0.1\right)& ={\left(1.1\right)}^{-2}\\ & =\frac{1}{{1.1}^2}\\ & =0.82644628\end{array}$$

One percent of the above is $$\frac{1}{100}\left(0.8264462810\right)=0.0082644628$$ The value $n$ is now found such that$$|R_n(x)|\le M\frac{{\left(0.1\right)}^{n+1}}{(n+1)!}\le 0.0082644628$$ Where $R_n(x)$ is the Taylor series remainder using $n$ terms.  $M$ is the upper bound for the $n+1$ derivative of $f\left(x\right)$ any where between $[0,0.1]$. Doing few calculations gives

For $n=0$, $\tilde{f}\left(0.1\right)=1,$ the error is $|0.82644628-1|=0.1735537190$.

For $n=1,$ $\tilde{f}\left(0.1\right)=0.8,$ the error is $|0.82644628-0.8|=0.02644628$.

For $n=2$, $\tilde{f}\left(0.1\right)=0.83,$ the error is $|0.82644628-0.83|=0.0035537190$. Because this is within $1$% then only three terms are needed. Therefore$$f(x)\approx 1-2x+3x^2$$

4.1.4 Problem 3

Expand $f(x)=\tan \left(x^2\right)$ to order $x^6$ using (a) direct Taylor expansion. (b) The Taylor series for $\sin (x)$ and $\cos x$ with appropriate substitution.

Solution

4.1.4.1 Part a

Using Taylor series$$f(x)\approx \sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}{f}^{\left(n\right)}(0)$$ Where $f(x)=\tan \left(x^2\right)$ and the expansion is around $x=0$. The Taylor series for $f(u)=\tan (u)$ is found instead of $\tan \left(x^2\right)$, and then at the end $u$ is replaced by $x^2$. This is called the substitution method. This simplifies the derivations. Therefore $f(0)=0$. The first derivative is$$\begin{array}{rl}{f}^{\prime }(u)& =\frac{d}{du}\tan (u)\\ & =\frac{d}{du}\left(\frac{\sin u}{\cos u}\right)\\ & =\frac{{\cos }^{2}u+{\cos }^{2}u}{{\cos }^{2}u}\\ & =\frac{1}{{\cos }^{2}u}\end{array}$$

At $u=0$ this gives $f^{\prime }(0)=1$.

The next derivative using the above result gives$$\begin{array}{rl}{f}^{\prime \prime }(u)& =\frac{d}{du}\left(\frac{1}{{\cos }^{2}u}\right)\\ & =\frac{2\cos u\sin u}{{\cos }^{4}u}\\ & =\frac{2\sin u}{{\cos }^{3}u}\end{array}$$

At $u=0$ this gives $f^{(2)}(0)=0$. The next derivative gives$$\begin{array}{rl}{f}^{(3)}(u)& =2\frac{d}{du}\left(\frac{\sin u}{{\cos }^{3}u}\right)\\ & =2\frac{\cos u{\cos }^{3}u-\sin u(3{\cos }^{2}u(-\sin u))}{{\cos }^{6}u}\\ & =2\frac{{\cos }^{4}u+3{\sin }^{2}u{\cos }^{2}u}{{\cos }^{6}u}\\ & =\frac{2{\cos }^{4}u}{{\cos }^{6}u}+\frac{6{\sin }^{2}u{\cos }^{2}u}{{\cos }^{6}u}\\ & =\frac{2}{{\cos }^{2}u}+\frac{6{\sin }^{2}u}{{\cos }^{4}u}\\ & =\frac{2}{{\cos }^{2}u}+\frac{6(1-{\cos }^{2}u)}{{\cos }^{4}u}\\ & =\frac{2}{{\cos }^{2}u}+\frac{6}{{\cos }^{4}u}-\frac{6}{{\cos }^{2}u}\\ & =-\frac{4}{{\cos }^{2}u}+\frac{6}{{\cos }^{4}u}\end{array}$$

At $u=0$ this gives $f^{(3)}(0)=-\frac{4}{1}+\frac{6}{1}=2$. Since the problem is asking for order $x^6$ the process stops here, as this is the same as order $u^3$ when $u$ is replaced by $x^2$.

Therefore the Taylor series for $\tan (u)$ is (for up to $n=3$)$$\begin{array}{rl}f(u)& \approx f(0)+u{f}^{{}^{\prime }}\left(0\right)+\frac{u^2}{2!}{f}^{(2)}(0)+\frac{u^3}{3!}{f}^{(3)}(0)+\cdots \\ & \approx 0+u+0+2\frac{u^3}{3!}\\ & \approx u+\frac{1}{3}{u}^3\end{array}$$

Replacing $u=x^2$, gives the Taylor series for $\tan \left(x^2\right)$ for up to $x^6$ term as $$\tan \left(x^2\right)\approx x^2+\frac{1}{3}{x}^6$$

4.1.4.2 Part b

To obtain the above result using the Taylor series for $\sin \left(x^2\right),\cos \left(x^2\right)$, the Taylor series for $\sin \left(x^2\right)$ and $\cos \left(x^2\right)$ is found, and long division is applied using the definition of $\tan \left(x^2\right)=\frac{\sin \left(x^2\right)}{\cos \left(x^2\right)}$. Terms with order larger than $x^6$ are ignored. The Taylor series for $\sin (x)$ is$$\sin (x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$ Using the substitution method, the Taylor series for $\sin \left(x^2\right)$ becomes$$\begin{array}{rl}\sin \left(x^2\right)& \approx x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\cdots \\ \text{(1)}& & \approx x^2-\frac{x^6}{6}+\frac{x^{10}}{120}-\cdots \end{array}$$

The Taylor series for $\cos (x)$ is$$\cos (x)\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$ Using the substitution method, the Taylor series for $\cos \left(x^2\right)$ becomes $$\begin{array}{rl}\cos \left(x^2\right)& \approx 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots \\ \text{(2)}& & \approx 1-\frac{x^4}{2}+\frac{x^8}{24}-\cdots \end{array}$$

Since $\tan \left(x^2\right)=\frac{\sin \left(x^2\right)}{\cos \left(x^2\right)}$ then the Taylor series for $\tan \left(x^2\right)$ is$$\tan \left(x^2\right)\approx \frac{x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\cdots }{1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots }$$ Performing long division and stopping when the remainder has powers larger than $x^6$ gives$$\tan \left(x^2\right)\approx x^2+\frac{1}{3}{x}^6+\cdots$$ Which is same result as part(a).

Figure 4.1:Polynomals long division

4.1.5 Problem 4

A particle of mass $m$ moves along the $+x$ axis (i.e. $x>0$) with potential energy$$V(x)=\frac{a}{2x^2}-\frac{b}{x}$$ Where $a$ and $b$ are positive parameters. (a) Find the equilibrium position $x_0$. (b) Show that the particle executes harmonic oscillations near $x=x_0$. (c) Find the angular frequency of oscillations.

Solution

4.1.5.1 Part a

Equilibrium position is where the slope of the potential energy is zero. This position $x_0$ is found by solving for $x$ from$$\frac{dV}{dx}=0$$ But $$\begin{array}{rl}\frac{dV}{dx}& =\frac{a}{2}(-2x^{-3})-b(-x^{-2})\\ & =\frac{-a}{x^3}+\frac{b}{x^2}\\ & =\frac{-a+bx}{x^3}\end{array}$$

Hence$$\begin{array}{rl}\frac{-a+bx}{x^3}& =0\\ bx& =a\end{array}$$

Therefore $$x_0=\frac{a}{b}$$

4.1.5.2 Part b

Approximating $V(x)$ around $x_0$ using Taylor series gives$$V(x)\approx V\left(x_0\right)+(x-x_0)V^{\prime }\left(x_0\right)+\frac{{(x-x_0)}^2}{2!}{V}^{\prime \prime }\left(x_0\right)+\cdots$$ But $\frac{dV}{dx}$ evaluated at $x_0$ is zero, since this the equilibrium point. The above simplifies to$$\begin{array}{}\text{(A)}& V(x)\approx V\left(x_0\right)+\frac{{(x-x_0)}^2}{2!}{V}^{\prime \prime }\left(x_0\right)+\cdots \end{array}$$ Higher terms are ignored, because $(x-x_0)$ is assumed small and mass remain close to $x_0$. But$$V\left(x_0\right)=\frac{a}{2x_0^2}-\frac{b}{x_0}$$ And since $x_0=\frac{a}{b}$ from part (a), the above simplifies to$$\begin{array}{rl}V\left(x_0\right)& =\frac{a}{2{\left(\frac{a}{b}\right)}^2}-\frac{b}{\left(\frac{a}{b}\right)}\\ & =\frac{a{b}^2}{2a^2}-\frac{b^2}{a}\\ & =\frac{b^2}{2a}-\frac{b^2}{a}\\ \text{(A1)}& & =-\frac{1}{2}\frac{b^2}{a}\end{array}$$

And $$\begin{array}{rl}\frac{d^{2}V}{d{x}^2}& =\frac{d}{dx}(\frac{-a}{x^3}+\frac{b}{x^2})\\ & =\frac{3a}{x^4}-\frac{b}{x^3}\end{array}$$

At $x=x_0$ the above becomes$$\begin{array}{rl}{V}^{\prime \prime }\left(x_0\right)& =\frac{3a}{{\left(\frac{a}{b}\right)}^4}-\frac{b}{{\left(\frac{a}{b}\right)}^3}\\ \text{(A2)}& & =\frac{b^4}{a^3}\end{array}$$

Using (A1,A2) into A gives$$\begin{array}{rl}V(x)& \approx -\frac{1}{2}\frac{b^2}{a}+\frac{{(x-x_0)}^2}{2!}\frac{b^4}{a^3}+\cdots \\ & \approx -\frac{1}{2}\frac{b^2}{a}+\frac{{(x-\frac{a}{b})}^2}{2}\frac{b^4}{a^3}+\cdots \\ & \approx -\frac{1}{2}\frac{b^2}{a}+\frac{1}{2}(x^2+\frac{a^2}{b^2}-2x\frac{a}{b})\frac{b^4}{a^3}+\cdots \\ & \approx -\frac{1}{2}\frac{b^2}{a}+\frac{1}{2a}{b}^2+\frac{1}{2a^3}{b}^4{x}^2-\frac{1}{a^2}{b}^{3}x+\cdots \\ & \approx \frac{b^4}{2a^3}{x}^2-\frac{b^3}{a^2}x+\cdots \end{array}$$

Therefore near $x_0$ the potential energy is approximated as$$\begin{array}{}\text{(1)}& V(x)\approx \frac{b^4}{2a^3}{x}^2-\frac{b^3}{a^2}x\end{array}$$ The force on the mass is given by $$F=-\frac{dV}{dx}$$ Using $V(x)$ in (1) the force becomes$$F=-\frac{b^4}{a^3}x-\frac{b^3}{a^2}$$ But $F=m\frac{d^{2}x}{d{t}^2}$. Hence we obtain the equation of motion as$$\begin{array}{rl}m\frac{d^{2}x}{d{t}^2}& =F\\ & =-\frac{b^4}{a^3}x-\frac{b^3}{a^2}\end{array}$$

Therefore$$\begin{array}{rl}m\frac{d^{2}x(t)}{d{t}^2}+\frac{b^4}{a^3}x(t)& =-\frac{b^3}{a^2}\\ \text{(B)}& \frac{d^{2}x(t)}{d{t}^2}+\left(\frac{b^4}{m{a}^3}\right)x(t)& =-\frac{b^3}{m{a}^2}\end{array}$$

Let $$\frac{b^4}{m{a}^3}={\omega }^2$$ The equation of motion (B) becomes$$\frac{d^{2}x(t)}{d{t}^2}+{\omega }^{2}x(t)=-\frac{b^3}{m{a}^2}$$ But this is standard second order ode whose solution is $$x(t)=A\cos \left(\omega t\right)+B\sin \left(\omega t\right)+x_p(t)$$ Where $x_p(t)$ is the particular solution due to the forcing function $-\frac{b^3}{m{a}^2}$ and $A,B$ are constants of integrations found from initial conditions. Since the forcing function is just constant, and not function function of time, the above becomes$$\begin{array}{rl}x(t)& =A\cos \left(\omega t\right)+B\sin \left(\omega t\right)+F_p\\ & =A\cos (\omega t+\varphi )+F_p\end{array}$$

Therefore the motion is simple harmonic motion since $\cos (\omega t+\varphi )$ is harmonic. The forcing function $F_p$ has no effect on the nature of the harmonic motion, other than adding an extra constant displacement shift to $x(t)$ for all time. Since there is no damping, the particle will continue this motion forever.

The following is a plot of the solution for $10$ seconds using arbitrary values for $a,b,m$ and with initial conditions $x(0)=1,x^{\prime }(0)=0$. The solution shows the motion is harmonic as expected.

Figure 4.2:Plot of solution

4.1.5.3 Part c

The angular frequency of oscillation is$$\omega =\sqrt{\frac{b^4}{m{a}^3}}$$ In radians per second. The quantity $\frac{b^4}{a^3}$ can be called the stiffness $k$ (Newton per meter). Hence $\omega =\sqrt{\frac{k}{m}}$.

4.1.5.4 Appendix

An easier way to do part b, is to keep $(x-x_0)$ intact and replace this with $y$ at the end. Like this

Using (A1,A2) into A gives$$V(x)\approx -\frac{1}{2}\frac{b^2}{a}+\frac{{(x-x_0)}^2}{2!}\frac{b^4}{a^3}+\cdots$$ The force on the mass is given by $$\begin{array}{rl}F& =-\frac{dV}{dx}\\ & =-(x-x_0)\frac{b^4}{a^3}\end{array}$$

But $F=m\frac{d^{2}x}{d{t}^2}$. Hence we obtain the equation of motion as$$\begin{array}{rl}m\frac{d^{2}x}{d{t}^2}& =F\\ & =-(x-x_0)\frac{b^4}{a^3}\end{array}$$

Now let $y=x-x_0$. the above becomes$$\begin{array}{rl}m\frac{d^{2}y}{d{t}^2}& =-y\frac{b^4}{a^3}\\ m\frac{d^{2}y}{d{t}^2}+y\frac{b^4}{a^3}& =0\\ \frac{d^{2}y}{d{t}^2}+y\frac{b^4}{m{a}^3}& =0\end{array}$$

Which is SHM. Using this method, it is faster to show.

4.1.6 key solution for HW 1

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