Expand the function \(f\relax (x) =\frac {\sin \relax (x) }{\cosh \relax (x) +2}\) 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 \(f\relax (x) \) 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 \(f\relax (x) \)).\[ f\relax (x) \approx \sum _{n=0}^{\infty }\frac {x^{n}}{n!}f^{\left ( n\right ) }\relax (0) \] Where \(f^{\relax (n) }\relax (0) \) is the \(n^{th}\) derivative of \(f\relax (x) \) evaluated at \(x=0\).
For \(n=0\), \(f^{\relax (0) }\relax (x) =f\left ( x\right ) =\frac {\sin \relax (x) }{\cosh \relax (x) +2}\), therefore \(f\relax (0) =0\).
For \(n=1\) \begin {align*} f^{\relax (1) }\relax (x) & =\frac {d}{dx}\left (\frac {\sin \relax (x) }{\cosh \relax (x) +2}\right ) \\ & =\frac {\cos \relax (x) \left (\cosh \relax (x) +2\right ) -\sin \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}\\ & =\frac {\cos \relax (x) \left (\cosh \relax (x) +2\right ) }{\left (\cosh \relax (x) +2\right ) ^{2}}-\frac {\sin \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}\\ & =\frac {\cos \relax (x) }{\cosh \relax (x) +2}-\frac {\sin \left ( x\right ) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}} \end {align*}
The above evaluated at \(x=0\) becomes \begin {align*} f^{\relax (1) }\relax (0) & =\frac {1}{1+2}-\frac {0}{\left ( 1+2\right ) ^{2}}\\ & =\frac {1}{3} \end {align*}
For \(n=2\) \begin {align*} f^{\relax (2) }\relax (x) & =\frac {d}{dx}\left (\frac {d}{dx}\left (\frac {\sin \relax (x) }{\cosh \relax (x) +2}\right ) \right ) \\ & =\frac {d}{dx}\left (\frac {\cos \relax (x) }{\cosh \relax (x) +2}-\frac {\sin \relax (x) \sinh \relax (x) }{\left (\cosh \left ( x\right ) +2\right ) ^{2}}\right ) \\ & =\frac {-\sin \relax (x) \left (\cosh \relax (x) +2\right ) -\cos \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}\\ & -\frac {\left (\cos \relax (x) \sinh \relax (x) +\sin \left ( x\right ) \cosh \relax (x) \right ) \left (\cosh \relax (x) +2\right ) ^{2}-\sin \relax (x) \sinh \relax (x) \left (2\left ( \cosh \relax (x) +2\right ) \sinh \relax (x) \right ) }{\left ( \cosh \relax (x) +2\right ) ^{4}}\\ & =\frac {-\sin \relax (x) \left (\cosh \relax (x) +2\right ) }{\left (\cosh \relax (x) +2\right ) ^{2}}-\frac {\cos \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}-\frac {\cos \relax (x) \sinh \relax (x) \left (\cosh \left ( x\right ) +2\right ) ^{2}}{\left (\cosh \relax (x) +2\right ) ^{4}}\\ & -\frac {\sin \relax (x) \cosh \relax (x) \left (\cosh \left ( x\right ) +2\right ) ^{2}}{\left (\cosh \relax (x) +2\right ) ^{4}}+\frac {\sin \relax (x) \sinh \relax (x) \left (2\left ( \cosh \relax (x) +2\right ) \sinh \relax (x) \right ) }{\left ( \cosh \relax (x) +2\right ) ^{4}}\\ & =\frac {-\sin \relax (x) }{\cosh \relax (x) +2}-\frac {\cos \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}-\frac {\cos \relax (x) \sinh \relax (x) }{\left ( \cosh \relax (x) +2\right ) ^{2}}-\frac {\sin \relax (x) \cosh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}+\frac {2\sin \relax (x) \sinh \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{3}}\\ & =\frac {-\sin \relax (x) }{\cosh \relax (x) +2}-2\frac {\cos \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}-\frac {\sin \relax (x) \cosh \relax (x) }{\left ( \cosh \relax (x) +2\right ) ^{2}}+\frac {2\sin \relax (x) \sinh ^{2}\relax (x) }{\left (\cosh \relax (x) +2\right ) ^{3}} \end {align*}
The above evaluated at \(x=0\) becomes\begin {align*} f^{\relax (2) }\relax (0) & =\frac {-0}{1+2}-2\frac {0}{\left ( 1+2\right ) ^{2}}-\frac {0}{\left (1+2\right ) ^{2}}+\frac {0}{\left ( 1+2\right ) ^{3}}\\ & =0 \end {align*}
For \(n=3\) \begin {align*} f^{\relax (3) }\relax (x) & =\frac {d}{dx}\left (\frac {d^{2}}{dx^{2}}\left (\frac {\sin \relax (x) }{\cosh \relax (x) +2}\right ) \right ) \\ & =\frac {d}{dx}\left (\frac {-\sin \relax (x) }{\cosh \relax (x) +2}-2\frac {\cos \relax (x) \sinh \relax (x) }{\left (\cosh \left ( x\right ) +2\right ) ^{2}}-\frac {\sin \relax (x) \cosh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}+\frac {2\sin \relax (x) \sinh ^{2}\relax (x) }{\left (\cosh \relax (x) +2\right ) ^{3}}\right ) \\ & =\frac {-\cos \relax (x) \left (\cosh \relax (x) +2\right ) +\sin \relax (x) \sinh \relax (x) }{\left (\cosh \relax (x) +2\right ) ^{2}}\\ & -2\frac {\left (-\sin \relax (x) \sinh \relax (x) +\cos \left ( x\right ) \cosh \relax (x) \right ) \left (\cosh \relax (x) +2\right ) ^{2}-\cos \relax (x) \sinh \relax (x) \left (2\left ( \cosh \relax (x) +2\right ) \sinh \relax (x) \right ) }{\left ( \cosh \relax (x) +2\right ) ^{4}}\\ & -\frac {\left (\cos \relax (x) \cosh \relax (x) +\sin \left ( x\right ) \sinh \relax (x) \right ) \left (\cosh \relax (x) +2\right ) ^{2}-\sin \relax (x) \cosh \relax (x) \left (2\left ( \cosh \relax (x) +2\right ) \sinh \relax (x) \right ) }{\left ( \cosh \relax (x) +2\right ) ^{4}}\\ & +2\frac {\left (\cos \relax (x) \sinh ^{2}\relax (x) +2\sin \relax (x) \cosh \relax (x) \right ) \left (\cosh \left ( x\right ) +2\right ) ^{3}-\left (\sin \relax (x) \sinh ^{2}\left ( x\right ) \right ) \left (3\left (\cosh \relax (x) +2\right ) ^{2}\sinh \relax (x) \right ) }{\left (\cosh \relax (x) +2\right ) ^{6}} \end {align*}
The above evaluated at \(x=0\) gives\begin {align*} f^{\relax (3) }\relax (0) & =\frac {-1\left (1+2\right ) +0}{\left (1+2\right ) ^{2}}-2\frac {\left (-0+1\right ) \left (1+2\right ) ^{2}-0}{\left (1+2\right ) ^{4}}-\frac {\left (1+0\right ) \left ( 1+2\right ) ^{2}-0}{\left (1+2\right ) ^{4}}+2\frac {\left (0+0\right ) \left (1+2\right ) ^{3}-\relax (0) \left (3\left (1+2\right ) ^{2}0\right ) }{\left (1+2\right ) ^{6}}\\ & =\frac {-\relax (3) }{\relax (3) ^{2}}-2\frac {\left ( 1\right ) \relax (3) ^{2}}{\relax (3) ^{4}}-\frac {\left ( 1\right ) \relax (3) ^{2}}{\relax (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 {align*}
The process stops here, because the problem is asking for \(n=3\). Substituting all the derivatives \(f^{\relax (n) }\relax (0) \) values above into\[ f\relax (x) \approx \sum _{n=0}^{\infty }\frac {x^{n}}{n!}f^{\left ( n\right ) }\relax (0) \] For up to \(n=3\) gives the following\begin {align*} f\relax (x) & \approx f\relax (0) +xf^{\relax (1) }\relax (0) +\frac {x^{2}}{2}f^{\relax (2) }\relax (0) +\frac {x^{3}}{3!}f^{\relax (3) }\relax (0) +\cdots \\ & \approx 0+x\frac {1}{3}+\frac {x^{2}}{2}\relax (0) +\frac {x^{3}}{3!}\left (-\frac {2}{3}\right ) \\ & \approx x\frac {1}{3}-\frac {2}{3}\frac {x^{3}}{6}\\ & \approx \frac {x}{3}-\frac {x^{3}}{9} \end {align*}
When \(x=\frac {1}{10}\) the above becomes\begin {align*} 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 {align*}
From the calculator\[ \frac {299}{9000}\approx 0.0332222 \] And from the exact expression \begin {align*} \frac {\sin \relax (x) }{\cosh \relax (x) +2} & =\frac {\sin \left (0.1\right ) }{\cosh \left (0.1\right ) +2}\\ & =0.0332224 \end {align*}
The error is about \(1.67\times 10^{-7}\).
Consider \(f\relax (x) =\left (1+x\right ) ^{p}\) 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
\[ f\relax (x) =\left (1+x\right ) ^{\frac {1}{3}}\] Part (1) The Taylor series is given by \begin {equation} f\relax (x) \approx f\relax (0) +xf^{\prime }\relax (0) +\frac {x^{2}}{2!}f^{\prime \prime }\relax (x) +\frac {x^{3}}{3!}f^{\prime \prime \prime }\relax (x) +\cdots \tag {1} \end {equation} Where \(f\relax (0) =1\) and \(f^{\prime }\relax (x) =\frac {1}{3}\left (1+x\right ) ^{-\frac {2}{3}}\). Hence \(f^{\prime }\relax (0) =\frac {1}{3}\) and \(f^{\prime \prime }\relax (x) =\frac {1}{3}\left ( -\frac {2}{3}\right ) \left (1+x\right ) ^{-\frac {5}{3}}\). Hence \(f^{\prime \prime }\relax (0) =-\frac {\relax (2) }{3^{2}}\), and \(f^{\prime \prime \prime }\relax (x) =\frac {1}{3}\left (-\frac {2}{3}\right ) \left (-\frac {5}{3}\right ) \left (1+x\right ) ^{-\frac {8}{3}}\), hence \(f^{\prime \prime \prime }\relax (0) =\frac {1}{3}\left (-\frac {2}{3}\right ) \left (-\frac {5}{3}\right ) =\frac {\relax (2) \left ( 5\right ) }{3^{3}},\) and \(f^{\relax (4) }\relax (x) =\frac {1}{3}\left (-\frac {2}{3}\right ) \left (-\frac {5}{3}\right ) \left (-\frac {8}{3}\right ) \left (1+x\right ) ^{-\frac {11}{3}}\), hence \(f^{\left ( 4\right ) }\relax (0) =\frac {1}{3}\left (-\frac {2}{3}\right ) \left ( -\frac {5}{3}\right ) \left (-\frac {8}{3}\right ) =-\frac {1}{3^{4}}\left ( \relax (2) \relax (5) \relax (8) \right ) \) and on. The series in (1) becomes\begin {align} f\relax (x) & \approx 1+\frac {1}{3}x-\frac {\relax (2) }{3^{2}}\frac {x^{2}}{2!}+\frac {\relax (2) \relax (5) }{3^{3}}\frac {x^{3}}{3!}-\frac {\relax (2) \relax (5) \relax (8) }{3^{4}}\frac {x^{4}}{4!}+\frac {\relax (2) \relax (5) \left ( 8\right ) \left (11\right ) }{3^{5}}\frac {x^{5}}{5!}-\frac {\relax (2) \relax (5) \relax (8) \left (11\right ) \left (14\right ) }{3^{6}}\frac {x^{6}}{6!}-\cdots \nonumber \\ & \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 \tag {2} \end {align}
The general term is found by comparing the above to the general term obtained from binomial expansion. Since \begin {equation} \left (1+x\right ) ^{p}=\begin {pmatrix} p\\ 0 \end {pmatrix} x^{0}+\begin {pmatrix} p\\ 1 \end {pmatrix} x+\begin {pmatrix} p\\ 2 \end {pmatrix} x^{2}+\cdots \tag {3} \end {equation} Comparing (2,3) shows that the general term is the binomial coefficient \(\begin {pmatrix} \frac {1}{3}\\ n \end {pmatrix} \). Therefore the Taylor series for \(\left (1+x\right ) ^{\frac {1}{3}}\) can be written as\[ f\relax (x) \approx \sum _{n=0}^{\infty }\begin {pmatrix} p\\ n \end {pmatrix} x^{n}\] For \(p=\frac {1}{3}\) the above becomes\[ f\relax (x) \approx \sum _{n=0}^{\infty }\begin {pmatrix} \frac {1}{3}\\ n \end {pmatrix} x^{n}\] Part(2) \begin {align*} R & =\lim _{n\rightarrow \infty }\left \vert \frac {a_{n}}{a_{n+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac {\begin {pmatrix} p\\ n \end {pmatrix} }{\begin {pmatrix} p\\ n+1 \end {pmatrix} }\right \vert \end {align*}
The Binomial coefficient \(\begin {pmatrix} p\\ n \end {pmatrix} =\frac {p!}{n!\left (p-n\right ) !}\), for when \(p\) is integer. This is not the case here. For non-integer \(p\) The Binomial coefficient becomes \(\begin {pmatrix} p\\ n \end {pmatrix} =\frac {\Gamma \left (p+1\right ) }{\Gamma \left (n+1\right ) \Gamma \left ( p-n+1\right ) }\) where \(\Gamma \relax (p) \) is the Gamma function. The above ratio now becomes\begin {align*} R & =\lim _{n\rightarrow \infty }\left \vert \frac {\frac {\Gamma \left ( p+1\right ) }{\Gamma \left (n+1\right ) \Gamma \left (p-n+1\right ) }}{\frac {\Gamma \left (p+1\right ) }{\Gamma \left (n+2\right ) \Gamma \left ( p-n\right ) }}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac {\Gamma \left (n+2\right ) \Gamma \left (p-n\right ) }{\Gamma \left (n+1\right ) \Gamma \left ( p-n+1\right ) }\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac {n\Gamma \left (p-n\right ) }{\Gamma \left (p-n+1\right ) }\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac {n}{p-n}\right \vert \end {align*}
But \(p=\frac {1}{3}\), hence the above becomes\begin {align*} R & =\lim _{n\rightarrow \infty }\left \vert \frac {n}{\frac {1}{3}-n}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac {n}{n-\frac {1}{3}}\right \vert \\ & =1 \end {align*}
Therefore the radius of convergence is \(1\). This means the Taylor series found above converges to \(f\relax (x) \) for \(\left \vert x\right \vert <1\).
Part 3
\[ f\relax (x) =\left (1+x\right ) ^{\frac {1}{3}}\] When \(x=0.1\) \begin {align*} f\left (0.1\right ) & =\left (1.1\right ) ^{\frac {1}{3}}\\ & =1.032280115 \end {align*}
one percent of the above is \[ \frac {1}{100}\left (1.032280115\right ) =0.01032280115 \] The value \(n\) is now found such that\[ \left \vert R_{n}\relax (x) \right \vert \leq M\frac {\left (0.1\right ) ^{n+1}}{\left (n+1\right ) !}\leq 0.01032280115 \] Where \(R_{n}\relax (x) \) is the Taylor series remainder using \(n\) terms. \(M\) is the upper bound for the \(n+1\) derivative of \(f\relax (x) \) any where between \(\left [ 0,0.1\right ] \). 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 \(n=1,\) \(\tilde {f}\left (0.1\right ) =1+\begin {pmatrix} \frac {1}{3}\\ 1 \end {pmatrix} 0.1=1.0333333\), and the error is \(\left \vert 1.032280115-1.0333333\right \vert =0.001053218\). Because this is smaller than \(R_{n}\relax (x) \) then only two terms are needed in the Taylor series to obtained the required accuracy. Therefore\[ f\relax (x) \approx 1+\frac {1}{3}x \]
\[ f\relax (x) =\left (1+x\right ) ^{-2}\] Part (1) The Taylor series is \[ f\relax (x) \approx f\relax (0) +xf^{\prime }\relax (0) +\frac {x^{2}}{2!}f^{\prime \prime }\relax (x) +\frac {x^{3}}{3!}f^{\prime \prime \prime }\relax (x) +\cdots \] But \(f\relax (0) =1\) and \(f^{\prime }\relax (x) =\left ( -2\right ) \left (1+x\right ) ^{-3}\). Hence \(f^{\prime }\relax (0) =-2\) and \(f^{\prime \prime }\relax (x) =\left (-2\right ) \left (-3\right ) \left (1+x\right ) ^{-4}\). Hence \(f^{\prime \prime }\relax (0) =\left ( -2\right ) \left (-3\right ) \), and \(f^{\prime \prime \prime }\relax (x) =-2\left (-3\right ) \left (-4\right ) \left (1+x\right ) ^{-5}\), hence \(f^{\prime \prime \prime }\relax (0) =\left (-2\right ) \left ( -3\right ) \left (-4\right ) \left (-5\right ) \) and so on. The above becomes\begin {align*} f\relax (x) & \approx 1+\left (-2\right ) x-\left (-2\right ) \left (-3\right ) \frac {x^{2}}{2!}+\left (-2\right ) \left (-3\right ) \left (-4\right ) \frac {x^{3}}{3!}+\cdots \\ & \approx 1-2x+\relax (2) \relax (3) \frac {x^{2}}{2!}-\left ( 2\right ) \relax (3) \relax (4) \frac {x^{3}}{3!}+\cdots \\ & \approx 1-2x+3x^{2}-4x^{3}+\cdots \end {align*}
The general term is therefore\[ f\relax (x) \approx \sum _{n=0}^{\infty }\left (-1\right ) ^{n}\left ( n+1\right ) x^{n}\] Part(2) \begin {align*} R & =\lim _{n\rightarrow \infty }\left \vert \frac {a_{n}}{a_{n+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac {\left (n+1\right ) }{\left ( n+2\right ) }\right \vert \\ & =1 \end {align*}
Hence the series converges to \(f\relax (x) \) for \(\left \vert x\right \vert <1\).
Part 3
\[ f\relax (x) =\left (1+x\right ) ^{-2}\] For \(x=0.1\) \begin {align*} f\left (0.1\right ) & =\left (1.1\right ) ^{-2}\\ & =\frac {1}{1.1^{2}}\\ & =0.82644628 \end {align*}
One percent of the above is \[ \frac {1}{100}\left (0.8264462810\right ) =0.0082644628 \] The value \(n\) is now found such that\[ \left \vert R_{n}\relax (x) \right \vert \leq M\frac {\left (0.1\right ) ^{n+1}}{\left (n+1\right ) !}\leq 0.0082644628 \] Where \(R_{n}\relax (x) \) 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 \(\left [ 0,0.1\right ] \). Doing few calculations gives
For \(n=0\), \(\tilde {f}\left (0.1\right ) =1,\) the error is \(\left \vert 0.82644628-1\right \vert =0.1735537190\).
For \(n=1,\) \(\tilde {f}\left (0.1\right ) =0.8,\) the error is \(\left \vert 0.82644628-0.8\right \vert =0.02644628\).
For \(n=2\), \(\tilde {f}\left (0.1\right ) =0.83,\) the error is \(\left \vert 0.82644628-0.83\right \vert =0.0035537190\). Because this is within \(1\)% then only three terms are needed. Therefore\[ f\relax (x) \approx 1-2x+3x^{2}\]
Expand \(f\relax (x) =\tan \left (x^{2}\right ) \) to order \(x^{6}\) using (a) direct Taylor expansion. (b) The Taylor series for \(\sin \relax (x) \) and \(\cos x\) with appropriate substitution.
Solution
Using Taylor series\[ f\relax (x) \approx \sum _{n=0}^{\infty }\frac {x^{n}}{n!}f^{\left ( n\right ) }\relax (0) \] Where \(f\relax (x) =\tan \left (x^{2}\right ) \) and the expansion is around \(x=0\). The Taylor series for \(f\relax (u) =\tan \relax (u) \) 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 \(f\relax (0) =0\). The first derivative is\begin {align*} f^{\prime }\relax (u) & =\frac {d}{du}\tan \relax (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 {align*}
At \(u=0\) this gives \(f^{\prime }\relax (0) =1\).
The next derivative using the above result gives\begin {align*} f^{\prime \prime }\relax (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 {align*}
At \(u=0\) this gives \(f^{\relax (2) }\relax (0) =0\). The next derivative gives\begin {align*} f^{\relax (3) }\relax (u) & =2\frac {d}{du}\left (\frac {\sin u}{\cos ^{3}u}\right ) \\ & =2\frac {\cos u\cos ^{3}u-\sin u\left (3\cos ^{2}u\left (-\sin u\right ) \right ) }{\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\left (1-\cos ^{2}u\right ) }{\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 {align*}
At \(u=0\) this gives \(f^{\relax (3) }\relax (0) =-\frac {4}{1}+\frac {6}{1}=2\). 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 \(\tan \relax (u) \) is (for up to \(n=3\))\begin {align*} f\relax (u) & \approx f\relax (0) +uf^{^{\prime }}\left ( 0\right ) +\frac {u^{2}}{2!}f^{\relax (2) }\relax (0) +\frac {u^{3}}{3!}f^{\relax (3) }\relax (0) +\cdots \\ & \approx 0+u+0+2\frac {u^{3}}{3!}\\ & \approx u+\frac {1}{3}u^{3} \end {align*}
Replacing \(u=x^{2}\), gives the Taylor series for \(\tan \left (x^{2}\right ) \) for up to \(x^{6}\) term as \[ \tan \left (x^{2}\right ) \approx x^{2}+\frac {1}{3}x^{6}\]
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 \(\sin \relax (x) \) is\[ \sin \relax (x) \approx x-\frac {x^{3}}{3!}+\frac {x^{5}}{5!}-\cdots \] Using the substitution method, the Taylor series for \(\sin \left ( x^{2}\right ) \) becomes\begin {align} \sin \left (x^{2}\right ) & \approx x^{2}-\frac {x^{6}}{3!}+\frac {x^{10}}{5!}-\cdots \nonumber \\ & \approx x^{2}-\frac {x^{6}}{6}+\frac {x^{10}}{120}-\cdots \tag {1} \end {align}
The Taylor series for \(\cos \relax (x) \) is\[ \cos \relax (x) \approx 1-\frac {x^{2}}{2!}+\frac {x^{4}}{4!}-\cdots \] Using the substitution method, the Taylor series for \(\cos \left ( x^{2}\right ) \) becomes \begin {align} \cos \left (x^{2}\right ) & \approx 1-\frac {x^{4}}{2!}+\frac {x^{8}}{4!}-\cdots \nonumber \\ & \approx 1-\frac {x^{4}}{2}+\frac {x^{8}}{24}-\cdots \tag {2} \end {align}
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).
A particle of mass \(m\) moves along the \(+x\) axis (i.e. \(x>0\)) with potential energy\[ V\relax (x) =\frac {a}{2x^{2}}-\frac {b}{x}\] 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
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 {align*} \frac {dV}{dx} & =\frac {a}{2}\left (-2x^{-3}\right ) -b\left ( -x^{-2}\right ) \\ & =\frac {-a}{x^{3}}+\frac {b}{x^{2}}\\ & =\frac {-a+bx}{x^{3}} \end {align*}
Hence\begin {align*} \frac {-a+bx}{x^{3}} & =0\\ bx & =a \end {align*}
Therefore \[ x_{0}=\frac {a}{b}\]
Approximating \(V\relax (x) \) around \(x_{0}\) using Taylor series gives\[ V\relax (x) \approx V\left (x_{0}\right ) +\left (x-x_{0}\right ) V^{\prime }\left (x_{0}\right ) +\frac {\left (x-x_{0}\right ) ^{2}}{2!}V^{\prime \prime }\left (x_{0}\right ) +\cdots \] But \(\frac {dV}{dx}\) evaluated at \(x_{0}\) is zero, since this the equilibrium point. The above simplifies to\begin {equation} V\relax (x) \approx V\left (x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2!}V^{\prime \prime }\left (x_{0}\right ) +\cdots \tag {A} \end {equation} Higher terms are ignored, because \(\left (x-x_{0}\right ) \) 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 {align} V\left (x_{0}\right ) & =\frac {a}{2\left (\frac {a}{b}\right ) ^{2}}-\frac {b}{\left (\frac {a}{b}\right ) }\nonumber \\ & =\frac {ab^{2}}{2a^{2}}-\frac {b^{2}}{a}\nonumber \\ & =\frac {b^{2}}{2a}-\frac {b^{2}}{a}\nonumber \\ & =-\frac {1}{2}\frac {b^{2}}{a} \tag {A1} \end {align}
And \begin {align*} \frac {d^{2}V}{dx^{2}} & =\frac {d}{dx}\left (\frac {-a}{x^{3}}+\frac {b}{x^{2}}\right ) \\ & =\frac {3a}{x^{4}}-\frac {b}{x^{3}} \end {align*}
At \(x=x_{0}\) the above becomes\begin {align} V^{\prime \prime }\left (x_{0}\right ) & =\frac {3a}{\left (\frac {a}{b}\right ) ^{4}}-\frac {b}{\left (\frac {a}{b}\right ) ^{3}}\nonumber \\ & =\frac {b^{4}}{a^{3}} \tag {A2} \end {align}
Using (A1,A2) into A gives\begin {align*} V\relax (x) & \approx -\frac {1}{2}\frac {b^{2}}{a}+\frac {\left ( x-x_{0}\right ) ^{2}}{2!}\frac {b^{4}}{a^{3}}+\cdots \\ & \approx -\frac {1}{2}\frac {b^{2}}{a}+\frac {\left (x-\frac {a}{b}\right ) ^{2}}{2}\frac {b^{4}}{a^{3}}+\cdots \\ & \approx -\frac {1}{2}\frac {b^{2}}{a}+\frac {1}{2}\left (x^{2}+\frac {a^{2}}{b^{2}}-2x\frac {a}{b}\right ) \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 {align*}
Therefore near \(x_{0}\) the potential energy is approximated as\begin {equation} V\relax (x) \approx \frac {b^{4}}{2a^{3}}x^{2}-\frac {b^{3}}{a^{2}}x \tag {1} \end {equation} The force on the mass is given by \[ F=-\frac {dV}{dx}\] Using \(V\relax (x) \) in (1) the force becomes\[ F=-\frac {b^{4}}{a^{3}}x-\frac {b^{3}}{a^{2}}\] But \(F=m\frac {d^{2}x}{dt^{2}}\). Hence we obtain the equation of motion as\begin {align*} m\frac {d^{2}x}{dt^{2}} & =F\\ & =-\frac {b^{4}}{a^{3}}x-\frac {b^{3}}{a^{2}} \end {align*}
Therefore\begin {align} m\frac {d^{2}x\relax (t) }{dt^{2}}+\frac {b^{4}}{a^{3}}x\relax (t) & =-\frac {b^{3}}{a^{2}}\nonumber \\ \frac {d^{2}x\relax (t) }{dt^{2}}+\left (\frac {b^{4}}{ma^{3}}\right ) x\relax (t) & =-\frac {b^{3}}{ma^{2}} \tag {B} \end {align}
Let \[ \frac {b^{4}}{ma^{3}}=\omega ^{2}\] The equation of motion (B) becomes\[ \frac {d^{2}x\relax (t) }{dt^{2}}+\omega ^{2}x\relax (t) =-\frac {b^{3}}{ma^{2}}\] But this is standard second order ode whose solution is \[ x\relax (t) =A\cos \left (\omega t\right ) +B\sin \left (\omega t\right ) +x_{p}\relax (t) \] Where \(x_{p}\relax (t) \) is the particular solution due to the forcing function \(-\frac {b^{3}}{ma^{2}}\) 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 {align*} x\relax (t) & =A\cos \left (\omega t\right ) +B\sin \left (\omega t\right ) +F_{p}\\ & =A\cos \left (\omega t+\phi \right ) +F_{p} \end {align*}
Therefore the motion is simple harmonic motion since \(\cos \left ( \omega t+\phi \right ) \) 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 \(x\relax (t) \) 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 \(x\relax (0) =1,x^{\prime }\relax (0) =0\). The solution shows the motion is harmonic as expected.
The angular frequency of oscillation is\[ \omega =\sqrt {\frac {b^{4}}{ma^{3}}}\] In radians per second. The quantity \(\frac {b^{4}}{a^{3}}\) can be called the stiffness \(k\) (Newton per meter). Hence \(\omega =\sqrt {\frac {k}{m}}\).
An easier way to do part b, is to keep \(\left (x-x_{0}\right ) \) intact and replace this with \(y\) at the end. Like this
Using (A1,A2) into A gives\[ V\relax (x) \approx -\frac {1}{2}\frac {b^{2}}{a}+\frac {\left ( x-x_{0}\right ) ^{2}}{2!}\frac {b^{4}}{a^{3}}+\cdots \] The force on the mass is given by \begin {align*} F & =-\frac {dV}{dx}\\ & =-\left (x-x_{0}\right ) \frac {b^{4}}{a^{3}} \end {align*}
But \(F=m\frac {d^{2}x}{dt^{2}}\). Hence we obtain the equation of motion as\begin {align*} m\frac {d^{2}x}{dt^{2}} & =F\\ & =-\left (x-x_{0}\right ) \frac {b^{4}}{a^{3}} \end {align*}
Now let \(y=x-x_{0}\). the above becomes\begin {align*} m\frac {d^{2}y}{dt^{2}} & =-y\frac {b^{4}}{a^{3}}\\ m\frac {d^{2}y}{dt^{2}}+y\frac {b^{4}}{a^{3}} & =0\\ \frac {d^{2}y}{dt^{2}}+y\frac {b^{4}}{ma^{3}} & =0 \end {align*}
Which is SHM. Using this method, it is faster to show.