学前教育

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学前教育
2.5 Find the equation of the tangent to the curve $y = x^2$ at (1,1).
Solution. $y'=2x$, so the gradient of tangent to the curve at (1,1) is 2, so the equation of tangent is $y-1=2(x-1)$, which is $y=2x-1$.
4.2 Classify the stationary point of $y = x−2\ln x$, where $x > 0$.
Solution. The root of $y'=1-\frac2x$ is $x=2$, so $(2,2-2\ln2)$ is a stationary point. When $x>0$, $\frac2x$ is decreasing, so $y'$ is increasing, so $(2,2-2\ln2)$ is a minimum point.
6.7 Show that the mapping $w = z +\frac cz$, where $z = x + iy, w = u + iv$ and $c$ is a real number, maps the circle $|z| = 1$ in the $z$ plane into an ellipse in the $w$ plane and find its equation.
Solution. Since $z\overline z=1$, $z +\frac cz=z+c\bar z=x+iy+c(x-iy)$, we get $\begin{aligned}u&=(1+c)x\\v&=(1-c)y\end{aligned}$ From $|z|=1$, we get $x^2+y^2=1$, so $\frac{u^2}{(1+c)^2}+\frac{v^2}{(1-c)^2}=1$, which is the equation of an ellipse.
8.7 Find the shortest distance from the origin to the line given in vector parametric form by $\boldsymbol r = \boldsymbol a+t\boldsymbol b$, where $\boldsymbol a = (1, 2, 3)$ and $\boldsymbol b = (1, 1, 1)$, and $t$ is the parameter (Hint: use a calculus method, with $t$ as the independent variable.)
Solution. From $\boldsymbol r = \boldsymbol a+t\boldsymbol b$, we get $\boldsymbol r=(1+t,2+t,3+t)$, so $|\boldsymbol r|^2=(1+t)^2+(2+t)^2+(3+t)^2=3t^2+12t+14=3(t+2)^2+2\ge2\Rightarrow |\boldsymbol r|\ge\sqrt2$.
10.1 For the vectors $\boldsymbol a$ and $\boldsymbol b$, show $|\boldsymbol{a}+\boldsymbol{b}|^{2}+|\boldsymbol{a}-\boldsymbol{b}|^{2}=2\left(|\boldsymbol{a}|^{2}+|\boldsymbol{b}|^{2}\right)$ and $\boldsymbol{a}\cdot\boldsymbol{b}=\frac{1}{4}\left(|\boldsymbol{a}+\boldsymbol{b}|^{2}-\left|\boldsymbol{a}-\boldsymbol{b}\right|^{2}\right)$.
Proof. From $|\boldsymbol{a}+\boldsymbol{b}|^{2}=(\boldsymbol{a}+\boldsymbol{b})\cdot(\boldsymbol{a}+\boldsymbol{b})=|\boldsymbol{a}|^2+2\boldsymbol{a}\cdot\boldsymbol{b}+|\boldsymbol{b}|^2$ and $|\boldsymbol{a}-\boldsymbol{b}|^{2}=|\boldsymbol{a}|^2-2\boldsymbol{a}\cdot\boldsymbol{b}+|\boldsymbol{b}|^2$, we get $\begin{aligned}|\boldsymbol{a}+\boldsymbol{b}|^{2}+|\boldsymbol{a}-\boldsymbol{b}|^{2}&=2\left(|\boldsymbol{a}|^{2}+|\boldsymbol{b}|^{2}\right)\\|\boldsymbol{a}+\boldsymbol{b}|^{2}-\left|\boldsymbol{a}-\boldsymbol{b}\right|^{2}&=4\boldsymbol{a}\cdot\boldsymbol{b}\end{aligned}$
10.3 What is the geometrical significance of $\boldsymbol a\times\boldsymbol b=\boldsymbol 0$?
Solution. $|\boldsymbol a|\cdot|\boldsymbol b|\sin\theta=0$, therefore $|\boldsymbol a|=0$ or $|\boldsymbol b|=0$ or $\theta=k\pi$, and in the last case $\boldsymbol{a,b}$ are collinear.
10.5 Let $\boldsymbol{a, b, c}$ be three non-coplanar vectors, and $\boldsymbol{v}$ be any vector. Show that $\boldsymbol{v}$ can be expressed as $\boldsymbol v=X\boldsymbol a+Y\boldsymbol b+Z\boldsymbol c$. where $X, Y, Z$, are constants given by $X=\boldsymbol{v} \cdot(\boldsymbol{b} \times \boldsymbol{c}) / \mathrm{D}$,$Y=\boldsymbol{v} \cdot(\boldsymbol{c} \times \boldsymbol{a}) / D$, $Z=\boldsymbol{v} \cdot(\boldsymbol{a} \times \boldsymbol{b}) / D$, where $D=\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol c)$.
Proof. Doing dot product of $(\boldsymbol{a} \times \boldsymbol{b})$ with both sides of $\boldsymbol v=X\boldsymbol a+Y\boldsymbol b+Z\boldsymbol c$ and using $\boldsymbol{a}\cdot(\boldsymbol{a} \times \boldsymbol{b})=\boldsymbol{b}\cdot(\boldsymbol{a} \times \boldsymbol{b})=0$, we get $\boldsymbol v\cdot(\boldsymbol{a} \times \boldsymbol{b})=X\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol c)$, so $X=\boldsymbol{v} \cdot(\boldsymbol{b} \times \boldsymbol{c}) / \mathrm{D}$.
Calculus 1: Q3. The curve $C$ in the $xy$-plane has equation $x^2 + xy + y^2 = 1.$
By solving $\frac{\mathrm dy}{\mathrm dx} = 0$, show that the maximum and minimum values taken by $y$ are $±\frac2{\sqrt3}$.
By changing to polar coordinates, ($x = r \cos \theta, y = r \sin \theta$),sketch the curve $C$.
What is the greatest distance of a point on $C$ from the origin?
Solution. From $x^2 + xy + y^2 = 1$, we have $2x + y+x\frac{\mathrm dy}{\mathrm dx}+ 2y\frac{\mathrm dy}{\mathrm dx} = 0$, we let $\frac{\mathrm dy}{\mathrm dx}=-\frac{2x+y}{x+2y}=0\Rightarrow x=-\frac12y$, substituting into equation, $\frac{y^2}4-\frac{y^2}2+ y^2 = 1\Rightarrow y=\pm\frac2{\sqrt3}$.
So the maximum and minimum values taken by $y$ are $±\frac2{\sqrt3}$.
From $x = r \cos \theta, y = r \sin \theta$, we have $xy=\frac12r^2\sin2\theta$ and $x^2+y^2=r^2$. So $x^2 + xy + y^2 = 1$⇔$r^2+\frac12r^2\sin2\theta=1$⇔$r=\left(1+\frac12\sin2\theta\right)^{-1/2}$.
Since $\sin2\theta\le1$, we have $r\le\sqrt{\frac23}$. Equality occurs when $\theta=\frac\pi4+2k\pi,k\in\mathbb Z$.
Calculus 3: Q4. Let $t = \tan \frac12\theta$. Show that $\sin \theta=\frac{2 t}{1+t^{2}}, \quad \cos \theta=\frac{1-t^{2}}{1+t^{2}}, \quad \tan \theta=\frac{2 t}{1-t^{2}}$ and that $\mathrm{d} \theta=\frac{2 \mathrm{~d} t}{1+t^{2}}$
Use the substitution $t=\tan \frac{1}{2} \theta$ to evaluate $\displaystyle\int_{0}^{\pi / 2} \frac{\mathrm{d} \theta}{(1+\sin \theta)^{2}}$
Solution. From the double angle formula,
$\sin \theta
= 2\sin \frac{\theta}{2}\cos \frac{\theta}{2} =\frac{2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}+\sin^2\frac{\theta}{2}}
= \frac{2\sin \frac{\theta}{2}\cos \frac{\theta}{2} \div \cos^2 \frac{\theta}{2}}{(\cos^2 \frac{\theta}{2}+\sin^2\frac{\theta}{2})\div \cos^2 \frac{\theta}{2}}  
=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\frac{2t}{1+t^2}$
$\cos\theta=\cos^2\frac{\theta}2-\sin^2\frac{\theta}2=\frac{\cos^2\frac{\theta}2-\sin^2\frac{\theta}2}{\sin^2\frac\theta2+\cos^2\frac\theta2}=\frac{\left(\cos^2\frac{\theta}2-\sin^2\frac{\theta}2\right)\div \cos^2\frac{\theta}2}{\sin^2\frac\theta2+\cos^2\frac\theta2\div \cos^2\frac{\theta}2}=\frac{1-\tan^2\frac{\theta}2}{1+\tan^2\frac{\theta}2}=\frac{1-t^2}{1+t^2}$
$\tan\theta=\frac{2\tan\frac12\theta}{1-\tan^2\frac12\theta}=\frac{2t}{1-t^2}$ (we can derive the first two formulae from this one, as $\sin\theta=\frac{\tan\theta}{1+\tan^2\theta}$ and $\cos\theta=\frac1{1+\tan^2\theta}$)
$\frac{\mathrm{d} \theta}{\mathrm dt}=\frac{\mathrm{d}(2\arctan t)}{\mathrm dt}=\frac{2}{1+t^{2}}$
$\displaystyle\int_{0}^{\pi / 2} \frac{\mathrm{d} \theta}{(1+\sin \theta)^{2}}=\int_{0}^{1} \frac{2}{\left(t^2+1\right) \left(\frac{2 t}{t^2+1}+1\right)^2}~\mathrm d t=\int_{0}^{1} \frac{2 \left(t^2+1\right)}{(t+1)^4}~\mathrm d t=\int_{0}^{1} \left(\frac{4}{(t+1)^4}-\frac{4}{(t+1)^3}+\frac{2}{(t+1)^2}\right)~\mathrm d t=\left[-\frac{4}{3 (t+1)^3}+\frac{2}{(t+1)^2}-\frac{2}{t+1}\right]_{t=0}^1=\frac23$
Dynamics 1: Q2. Consider a particle of mass $m$ moving in one vertical dimension with height $y(t)$. It moves under gravity, so that its acceleration always satisfies $\frac{\mathrm d^2y}{\mathrm dt^2} = −g$. Initially the particle is projected from ground-level with speed $v$. That is, at $t = 0$, we have $y = 0$ and $\frac{\mathrm dy}{\mathrm dt} = v$.
(i)   Determine $y(t)$.
(ii)  What is the greatest height achieved by the particle?
(iii) Find the time taken to return to ground-level.
(iv) Show that the quantity$$E=\frac{1}{2} m\left(\frac{\mathrm{d} y}{\mathrm{~d} t}\right)^{2}+m g y$$is constant throughout the motion.
Solution. Integrating $\frac{\mathrm d^2y}{\mathrm dt^2} = −g$, we get $\frac{\mathrm dy}{\mathrm dt}=-gt+v$(since initial speed=$v$), integrating, $y=-\frac12gt^2+vt$(since initial height=0).
The root of $\frac{\mathrm dy}{\mathrm dt}=-gt+v=0$ is $t=\frac{v}g$, so $y_{\max}=-\frac12g\left(\frac{v}g\right)^2+v\cdot\frac{v}g=\frac{v^2}{2g}$.
The root of $y=-\frac12gt^2+vt=0$ is $t=0$ and $t=\frac{2v}g$, therefore the time taken to return to ground-level is $\frac{2v}g$.
$\frac{\mathrm dE}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\left(\frac{1}{2} m\left(-gt+v\right)^{2}+m g y\right)=-g m (-g t + v) + m g (-g t + v)=0$, so $E$ is constant throughout the motion.
Dynamics 1: Q4. A particle of mass $m$ moves along the $x$-axis under a force $F(x)$, when at position $x$, given below$$F(x)=\left\{\begin{array}{cc}-k x^{3} & -a<x<a \\ 0 & |x| \geqslant a .\end{array}\right.$$Initially, at $t = 0$, we have $x = 0$ and $\frac{\mathrm dx}{\mathrm dt }= u\ge0$.
(i) Let $v =\frac{\mathrm dx}{\mathrm dt}$. Show that $\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=v \frac{\mathrm{d} v}{\mathrm{~d} x}$
(ii) From Newton’s Second Law, show that $\frac{1}{2} m v^{2}+\frac{1}{4} k x^{4}=E$ is constant throughout the motion.
(iii) Find the minimum value $U$ of $u$ such that the particle moves outside of the interval $−a \lt x \lt a$. If $u \lt U$ what is the maximum value of $x$?
(iv) Show that if $u = U$ then the time $T$ taken for the particle to reach $x = a$ equals $\displaystyle T=\sqrt{\frac{2 m}{k}} \int_{0}^{a} \frac{\mathrm{d} x}{\sqrt{a^{4}-x^{4}}}$
Solution. From $v=\frac{\mathrm dx}{\mathrm dt}$, we have$\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm dx}{\mathrm dt}\frac{\mathrm dv}{\mathrm dx}=v \frac{\mathrm{d} v}{\mathrm{~d} x}$
From (i) and Newton’s Second Law, $\frac{\mathrm dv}{\mathrm dx}=\left\{\begin{array}{cc}-\frac {kx^3}{vm} & -a<x<a \\ 0 & |x| \geqslant a .\end{array}\right.$
When $-a<x<a$, $\frac{\mathrm dE}{\mathrm dx}=mv\frac{\mathrm dv}{\mathrm dx}+kx^3=0\Rightarrow E$ is constant.
$\displaystyle\frac{\mathrm dv}{\mathrm dx}=-\frac {kx^3}{vm}\Rightarrow \int v~\mathrm dv=-\int \frac{kx^3}{m}~\mathrm dx\Rightarrow v^2=-\frac{kx^4}{2m}+C$
From (ii), $E=\frac12mu^2=\frac12mv^2+\frac14kx^4\Rightarrow \frac12mu^2\ge\frac14ka^4\Rightarrow u\ge a^2\sqrt{\frac k{2m}}$
Let $v=0$, we get $x=\sqrt{u\sqrt{\frac{2m}k}}$, which is the maximum value of $x$, if $u\lt U$.
If $u = U$, from $v^2=-\frac{kx^4}{2m}+C$, we get $\displaystyle v=\sqrt{\left(a^2\sqrt{\frac k{2m}}\right)^2-\frac{kx^4}{2m}}=\sqrt{\frac{k}{2m}}\sqrt{a^4-x^4}\Rightarrow\frac{\mathrm dt}{\mathrm dx}=\frac{\sqrt{\frac{2m}{k}}}{\sqrt{a^4-x^4}}\Rightarrow T=\sqrt{\frac{2 m}{k}} \int_{0}^{a} \frac{\mathrm{d} x}{\sqrt{a^{4}-x^{4}}}$
Dynamics 2: Q4.(Simple gravity pendulum) Consider a mass $m$ at the end of a light inextensible rod of length $l$ making small swings under gravity; let $θ$ denote the angle the rod makes with the vertical.
(i) Note that $\mathbf{r}=(l \sin \theta,-l \cos \theta)$. What do $\mathrm d\mathbf r/\mathrm dt$ and $\mathrm d^2\mathbf r/\mathrm dt^2$ equal?
(ii) Use Newton’s Second Law to show that $l\frac{\mathrm{d}^2\theta}{\mathrm{d} t^{2}}=-g \sin\theta$ and find an expression for the tension in the rod.
(iii) Show throughout the motion that $\frac12ml^2\left(\frac{\mathrm d\theta}{\mathrm dt}\right)^2-mgl\cos\theta=E$ is constant.
(iv) Say that the pendulum’s oscillations are small enough that the approximation $\sin\theta\approx\theta$ applies. Show that the pendulum’s swings have period $2 \pi \sqrt{l / g}$.
(v) More generally if the particle starts off with $\theta=\alpha,\mathrm dθ/\mathrm dt=0$, show that the oscillations have exact period$$4 \sqrt{\frac{l}{2g}}\int_0^{\alpha}\frac{\mathrm{d} \theta}{\sqrt{\cos \theta-\cos \alpha}}$$Solution.
(i) $\frac{\mathrm d\mathbf r}{\mathrm dt}=\left(l\cos\theta\frac{\mathrm d\theta}{\mathrm dt},l\sin\theta\frac{\mathrm d\theta}{\mathrm dt}\right)$,$\frac{\mathrm d^2\mathbf r}{\mathrm dt^2}=\left(l\cos\theta\frac{\mathrm d^2\theta}{\mathrm dt^2}-l\sin\theta\left(\frac{\mathrm d\theta}{\mathrm dt}\right)^2,l\sin\theta\frac{\mathrm d^2\theta}{\mathrm dt^2}+l\cos\theta\left(\frac{\mathrm d\theta}{\mathrm dt}\right)^2\right)$
(ii) Use Newton’s Second Law $m\ddot{\mathbf r}=-mg\mathbf k+\mathbf T$, where $\mathbf T$ is the tension, $\mathbf k$ is unit vector of $z$-axis.
Dot product of $\left(\cos\theta,\sin\theta\right)$ with both sides will eliminate $\mathbf T$ and allow us to show $\ddot\theta=-\frac gl\sin\theta$
(iii) Because velocity $\dot{\mathbf r}$ is perpendicular to $\mathbf T$, $m\dot{\mathbf r}\cdot\ddot{\mathbf r}=-mg\mathbf k\cdot\dot{\mathbf r}\Rightarrow\frac{\mathrm d}{\mathrm dt}\left(\frac12m\dot{\mathbf r}\cdot\dot{\mathbf r}\right)=\frac{\mathrm d}{\mathrm dt}\left(-mgy\right)\Rightarrow\frac12m|\dot{\mathbf r}|^2+mgy$ is constant.
(iv)Substituting for $\sin θ$ using the small-angle approximation $\sin \theta \approx \theta$ yields the equation for a harmonic oscillator,$${\frac {\mathrm d^2\theta }{\mathrm dt^2}}+{\frac {g}{\ell }}\theta =0.$$The error due to the approximation is of order $θ^3$ (from the Taylor expansion for $\sin θ$). Let the starting angle be $θ_0$. If it is assumed that the pendulum is released with zero angular velocity, the solution becomes$$\theta (t)=\theta _{0}\cos \left({\sqrt {\frac {g}{\ell }}}\,t\right)\quad \quad \quad \quad \theta _{0}\ll 1.$$The motion is simple harmonic motion where $θ_0$ is the amplitude of the oscillation (that is, the maximum angle between the rod of the pendulum and the vertical). The corresponding approximate period of the motion is then$$\displaystyle T_{0}=2\pi {\sqrt {\frac {\ell }{g}}}\quad \quad \quad \quad \quad \theta _{0}\ll 1$$which is known as Christiaan Huygens's law for the period. Note that under the small-angle approximation, the period is independent of the amplitude $θ_0$; this is the property of isochronism that Galileo discovered.

(v)For amplitudes beyond the small angle approximation, one can compute the exact period by first inverting the equation (iii) for the angular velocity obtained from the energy method, ${\frac{\mathrm dt}{\mathrm d\theta }}={\sqrt {\frac {\ell }{2g}}}{\frac {1}{\sqrt {\cos \theta -\cos \theta _{0}}}}$ and then integrating over one complete cycle, $T=t(\theta _{0}\rightarrow 0\rightarrow -\theta _{0}\rightarrow 0\rightarrow \theta _{0})$, or twice the half-cycle $T=2t(\theta _{0}\rightarrow 0\rightarrow -\theta _{0})$, or four times the quarter-cycle $T=4t(\theta _{0}\rightarrow 0)$, which leads to $T=4\sqrt{\frac{\ell}{2g}}$. Note that this integral diverges as $θ_0$ approaches the vertical $\displaystyle\lim_{\theta _{0}\rightarrow {\frac {\pi }{2}}}T=\infty$, so that a pendulum with just the right energy to go vertical will never actually get there. (Conversely, a pendulum close to its maximum can take an arbitrarily long time to fall down.)
This integral can be rewritten in terms of elliptic integrals as $T=4{\sqrt{\frac{\ell}{g}}}F\left({\frac{\pi}{2}},\sin{\frac{\theta_{0}}{2}}\right)$
where $F$ is the incomplete elliptic integral of the first kind defined by $F(\varphi,k)=\int _{0}^{\varphi }{\frac {1}{\sqrt {1-k^{2}\sin ^{2}u}}}\,\mathrm du$.
Or more concisely by the substitution $\sin{u}=\frac{\sin{\frac{\theta}2}}{\sin{\frac{\theta _0}2}}$ expressing $θ$ in terms of $u$, $T={\frac {2T_{0}}{\pi }}K(k),\qquad \mathrm {where} \quad k=\sin {\frac {\theta _{0}}{2}}.$
Here $K$ is the complete elliptic integral of the first kind defined by $K(k)=F\left({\frac {\pi }{2}},k\right)=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {1-k^{2}\sin ^{2}u}}}\,\mathrm du$
For comparison of the approximation to the full solution, consider the period of a pendulum of length 1 m on Earth (g = 9.80665 m/s²) at initial angle 10 degrees is
$4{\sqrt {\frac {1{\text{ m}}}{g}}}\ K\left(\sin {\frac {10^{\circ }}{2}}\right)\approx 2.0102{\text{ s}}$
The linear approximation gives $2\pi {\sqrt {\frac {1{\text{ m}}}{g}}}\approx 2.0064{\text{ s}}.$
The difference between the two values, less than 0.2%, is much less than that caused by the variation of $g$ with geographical location.
From here there are many ways to proceed to calculate the elliptic integral.
Legendre polynomial solution for the elliptic integral
Given Eq. 3 and the Legendre polynomial solution for the elliptic integral:
$K(k)={\frac {\pi }{2}}\sum _{n=0}^{\infty }\left({\frac {(2n-1)!!}{(2n)!!}}k^{n}\right)^2$
where $n!!$ denotes the double factorial, an exact solution to the period of a pendulum is:\begin{alignedat}{2}T&=2\pi {\sqrt {\frac {\ell }{g}}}\left(1+\left({\frac {1}{2}}\right)^{2}\sin ^{2}{\frac {\theta _{0}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{2}\sin ^{4}{\frac {\theta _{0}}{2}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right)^{2}\sin ^{6}{\frac {\theta _{0}}{2}}+\cdots \right)\\&=2\pi {\sqrt {\frac {\ell }{g}}}\cdot \sum _{n=0}^{\infty }\left(\left({\frac {(2n)!}{(2^{n}\cdot n!)^{2}}}\right)^{2}\cdot \sin ^{2n}{\frac {\theta _{0}}{2}}\right).\end{alignedat}Figure 4 shows the relative errors using the power series. $T_0$ is the linear approximation, and $T_2$ to $T_{10}$ include respectively the terms up to the 2nd to the 10th powers.

Power series solution for the elliptic integral
Another formulation of the above solution can be found if the following Maclaurin series:
$\sin {\frac {\theta _{0}}{2}}={\frac {1}{2}}\theta _{0}-{\frac {1}{48}}\theta _{0}^{3}+{\frac {1}{3\,840}}\theta _{0}^{5}-{\frac {1}{645\,120}}\theta _{0}^{7}+\cdots$.
is used in the Legendre polynomial solution above. The resulting power series is:[5]\begin{alignedat}{2}T&=2\pi {\sqrt {\frac {\ell }{g}}}\left(1+{\frac {1}{16}}\theta _{0}^{2}+{\frac {11}{3\,072}}\theta _0^4+{\frac{173}{737\,280}}\theta _{0}^{6}+{\frac{22\,931}{1\,321\,205\,760}}\theta _{0}^{8}+{\frac {1\,319\,183}{951\,268\,147\,200}}\theta _{0}^{10}+{\frac {233\,526\,463}{2\,009\,078\,326\,886\,400}}\theta _{0}^{12}+\cdots \right)\end{alignedat}more fractions available in OEIS: A223067|A223068.
Arithmetic-geometric mean solution for elliptic integral
Given Eq. 3 and the arithmetic–geometric mean solution of the elliptic integral: $K(k)={\frac {\frac {\pi }{2}}{M(1-k,1+k)}},$ where $M(x,y)$ is the arithmetic-geometric mean of $x$ and $y$.
This yields an alternative and faster-converging formula for the period:[6][7][8]
$T={\frac {2\pi }{M\left(1,\cos {\frac {\theta _{0}}{2}}\right)}}{\sqrt {\frac {\ell }{g}}}$.
The first iteration of this algorithm gives$$T_1=\frac{2T_0}{1+\cos {\frac {\theta _{0}}2}}$$This approximation has the relative error of less than 1% for angles up to 96.11 degrees.[6]
Since $\frac{1+\cos(\theta _0/2)}2=\cos^2{\frac{\theta _0}4}$, the expression can be written more concisely as
$T_{1}=T_{0}\sec ^{2}{\frac {\theta _{0}}{4}}$.
The second order expansion of $\sec^2(\theta _{0}/4)$ reduces to $T\approx T_{0}\left(1+{\frac {\theta _{0}^{2}}{16}}\right)$.
A second iteration of this algorithm gives $T_{2}={\frac {4T_{0}}{1+\cos {\frac {\theta _{0}}{2}}+2{\sqrt {\cos {\frac {\theta _{0}}{2}}}}}}={\frac {4T_{0}}{\left(1+{\sqrt {\cos {\frac {\theta _{0}}{2}}}}\right)^{2}}}$.
This second approximation has a relative error of less than 1% for angles up to 163.10 degrees.
Reference:Wikipedia.