Lab Sheet 9

from header import *

Exercise 1

Here we simplify $x_0^2+x_1^2+x_2^2$, where $x_0=\sin(\theta)\sin(\phi)$ and $x_1=\sin(\theta)\cos(\phi)$ and $x_2=\cos(\theta)$. (This is relevant for spherical polar coordinates.)

theta, phi = sp.symbols('theta phi', real=True)

x = [
 sp.sin(theta) * sp.sin(phi),
 sp.sin(theta) * sp.cos(phi),
 sp.cos(theta)
]

sp.simplify(x[0]**2 + x[1]**2 + x[2]**2)

$\displaystyle 1$

Exercise 2

x, y, t = sp.symbols('x y t', real=True)

u = 8-(x-y)**2*(x+y)**2*(16-4*x**2-y**4)
u

$\displaystyle - \left(x - y\right)^{2} \left(x + y\right)^{2} \left(- 4 x^{2} - y^{4} + 16\right) + 8$

v = u.subs({x : 2 * sp.cos(t), y : 2 * sp.sin(t)})
display(v)
display(sp.trigsimp(v))
display(sp.trigsimp(v - 8*sp.cos(8*t)))

$\displaystyle - \left(- 2 \sin{\left(t \right)} + 2 \cos{\left(t \right)}\right)^{2} \left(2 \sin{\left(t \right)} + 2 \cos{\left(t \right)}\right)^{2} \left(- 16 \sin^{4}{\left(t \right)} - 16 \cos^{2}{\left(t \right)} + 16\right) + 8$

$\displaystyle 8 - 16 \sin^{2}{\left(4 t \right)}$

$\displaystyle 0$

del x, y, t, u, v

Exercise 3

x, y, z, a, b, c = sp.symbols('x y z a b c')
eqs = [
  sp.Eq(a*x + b*y + c*z, 1),  
  sp.Eq(b*x + c*y + a*z, 1),  
  sp.Eq(c*x + a*y + b*z, 1)  
]
display(Latex("Equations:"))
for eq in eqs:
    display(eq)
sol = sp.solve(eqs,[x, y, z])
display(Latex("Solution:"))
display(sol)

Equations:

$\displaystyle a x + b y + c z = 1$

$\displaystyle a z + b x + c y = 1$

$\displaystyle a y + b z + c x = 1$

Solution:

{x: 1/(a + b + c), y: 1/(a + b + c), z: 1/(a + b + c)}

del x, y, z, a, b, c, eqs, sol

Exercise 4

Find an approximate root for $x^4+\sin(x)=10^4$ close to $x=10$.

sc.optimize.root_scalar(lambda x: x**4+np.sin(x)-10**4, x0=10).root

np.float64(10.000136031035714)

Exercise 5

Find numerical approximations for $f(5)$, and then for $f(0),\dotsc,f(50)$, where $f(x)=(e^x+x^7)/(e^x-x^7)$

f = lambda x: (np.exp(x) + x**7)/(np.exp(x) - x**7)
print(f(5))
print([f(n) for n in range(51)])

-1.003806608242687
[np.float64(1.0), np.float64(2.163953413738653), np.float64(-1.1225271249844706), np.float64(-1.0185383754323816), np.float64(-1.0066870973639803), np.float64(-1.003806608242687), np.float64(-1.0028864528505674), np.float64(-1.0026667590803273), np.float64(-1.0028469097578356), np.float64(-1.0033940571609772), np.float64(-1.0044150178829998), np.float64(-1.0061639189591798), np.float64(-1.0091258262711527), np.float64(-1.0142012857935216), np.float64(-1.0230802015421432), np.float64(-1.0390119982885597), np.float64(-1.0684733803053261), np.float64(-1.1250956551879607), np.float64(-1.2402662528741273), np.float64(-1.4989794130955643), np.float64(-2.220795044726118), np.float64(-6.469168971016049), np.float64(5.574474435059179), np.float64(2.074081155298447), np.float64(1.4188051469748693), np.float64(1.1852317897189424), np.float64(1.0855823585356026), np.float64(1.040109647695872), np.float64(1.018834821691931), np.float64(1.0088142586895557), np.float64(1.0041014178880834), np.float64(1.0018960266415378), np.float64(1.000870653683825), np.float64(1.0003971878104394), np.float64(1.0001800566340657), np.float64(1.0000811366472568), np.float64(1.000036354154339), np.float64(1.0000162012958154), np.float64(1.0000071833399449), np.float64(1.00000316955388), np.float64(1.0000013921016913), np.float64(1.0000006087561573), np.float64(1.0000002650982411), np.float64(1.0000001149864026), np.float64(1.0000000496868584), np.float64(1.0000000213927154), np.float64(1.0000000091788581), np.float64(1.0000000039253136), np.float64(1.0000000016733375), np.float64(1.0000000007111693), np.float64(1.0000000003013672)]

Exercise 6

Plot the curve $x^4+y^4=4$, together with the curve given by $x=(2+\sin(8t))\cos(t)$ and $y=(2+\sin(8t))\sin(t)$ (for $0\leq t\leq 2\pi$).

xs0, ys0 = np.meshgrid(np.linspace(-3,3,200),np.linspace(-3,3,200))
zs0 = xs0**4 + ys0**4
ts = np.linspace(0,2*np.pi,200)
xs1 = (2 + np.sin(8*ts)) * np.cos(ts)
ys1 = (2 + np.sin(8*ts)) * np.sin(ts)
fig = go.Figure()
fig.update_layout(width=600, height=600, showlegend=False)
fig.add_trace(go.Contour(x=xs0[0], y=ys0[:,0], z=zs0, contours=dict(type='constraint', value=4), line_color='red'))
fig.add_trace(go.Scatter(x=xs1, y=ys1, mode='lines', line=dict(color='blue')))

Exercise 7

Plot $y=\tan(\pi x)$, $y=\cot(\pi x)$, $y=-1$ and $y=1$ on the same graph, for $x$ from $-4$ to $4$.

xs = np.linspace(-4,4,1000)
ys0 = np.tan(np.pi * xs)
ys1 = 1/np.tan(np.pi * xs)
ys0[np.abs(ys0) > 10] = np.nan
ys1[np.abs(ys1) > 10] = np.nan
fig = go.Figure()
fig.update_yaxes(range=[-2,2])
fig.add_hline( 1, line_color='green')
fig.add_hline(-1, line_color='magenta')
fig.add_trace(go.Scatter(x=xs, y=ys0, mode='lines', line=dict(color='red'), name=r'$\tan(\pi x)$'))
fig.add_trace(go.Scatter(x=xs, y=ys1, mode='lines', line=dict(color='blue'), name=r'$\cot(\pi x)$'))

Exercise 8

Simplify some composites of Möbius functions

t = sp.symbols('t')

f = lambda t: ((2+sp.sqrt(3))*t-1)/(2+sp.sqrt(3)+t)
g = sp.lambdify(t, sp.simplify(f(f(f(t)))))
h = sp.lambdify(t, sp.simplify(g(g(t))))
k = sp.lambdify(t, sp.simplify(h(h(t))))
display(Latex("$f(t)=" + sp.latex(f(t)) + "$"))
display(Latex("$g(t)=f(f(f(t)))=" + sp.latex(g(t)) + "$"))
display(Latex("$h(t)=g(g(t))=" + sp.latex(h(t)) + "$"))

$f(t)=\frac{t \left(\sqrt{3} + 2\right) - 1}{t + \sqrt{3} + 2}$

$g(t)=f(f(f(t)))=\frac{t - 1}{t + 1}$

$h(t)=g(g(t))=- \frac{1}{t}$

del t, f, g, h, k

Exercise 9

Find and simplify $\displaystyle\left(\frac{1}{y}\frac{d^8y}{dx^8}\right)^{1/8}$, where $y=\sin(10x)$.

x = sp.symbols('x')
y = sp.sin(10*x)
r = (sp.diff(y,x,8)/y)**sp.Rational(1,8)
sp.simplify(r)

$\displaystyle 10$

del x, y, r

Exercise 10

Plot the curves $y=n^{3/2}x^n(1-x)^2$ for $n=1,\dotsc,10$ on the same graph, with $x$ running from $0$ to $1$.

xs = np.linspace(0,1,200)
ns = np.arange(1,11).reshape(10,1)
ys = ns ** (3/2) * xs ** ns * (1 - xs)**2
fig = go.Figure()
fig.update_layout(width=600, height=600, showlegend=False)
for i in range(10):
    fig.add_trace(go.Scatter(x=xs, y=ys[i], mode='lines'))
fig.show()

Exercise 11

Here we investigate the function $\displaystyle y=\frac{x^4(1-x)^4}{1+x^2}$. This is interesting because it is clear that $y\geq 0$ for $x\in[0,1]$ and the calculation below gives $\frac{22}{7}-\pi=\int_0^1 y\,dx\geq 0$ which proves that $\pi\leq\frac{22}{7}$.

x = sp.symbols('x')
y = x**4*(1 - x)**4/(1+x**2)
display(Latex("$y=" + sp.latex(y) + "$"))
Y = sp.integrate(y, x)
display(Latex("$\\int y\\,dx=" + sp.latex(Y) + "$"))
Y0 = sp.integrate(y, (x,0,1))
display(Latex("$\\int_0^1 y\\,dx=" + sp.latex(Y0) + "$"))
dy = sp.factor(sp.diff(y, x))
display(Latex("$\\frac{dy}{dx}=" + sp.latex(dy) + "$"))
xs = np.linspace(0,1,100)
ys = [float(y.subs(x,xi)) for xi in xs]
fig = go.Figure()
fig.update_layout(width=600, height=600, showlegend=False)
fig.add_trace(go.Scatter(x=xs, y=ys, mode='lines'))

$y=\frac{x^{4} \left(1 - x\right)^{4}}{x^{2} + 1}$

$\int y\,dx=\frac{x^{7}}{7} - \frac{2 x^{6}}{3} + x^{5} - \frac{4 x^{3}}{3} + 4 x - 4 \operatorname{atan}{\left(x \right)}$

$\int_0^1 y\,dx=\frac{22}{7} - \pi$

$\frac{dy}{dx}=\frac{2 x^{3} \left(x - 1\right)^{3} \left(3 x^{3} - x^{2} + 4 x - 2\right)}{\left(x^{2} + 1\right)^{2}}$

dy_fun = sp.lambdify(x, dy, 'numpy')
x0 = sc.optimize.root_scalar(dy_fun, x0=0.5).root
y0 = y.subs(x,x0)
display(Latex(f"The hump is at $(x,y)=(" + str(x0) + "," + str(y0) + ")$"))

The hump is at $(x,y)=(0.47580841099465915,0.00315543153223710)$

Exercise 12

The plot below illustrates the fact that the sequence $a_n=(1+\frac{1}{n})^n/e$ tends to $1$ as $n\to\infty$.

nn = np.arange(1,101)
aa = (1 + 1/nn) ** nn / np.exp(1)
fig = go.Figure()
fig.update_layout(width=600, height=600, showlegend=False)
fig.add_trace(go.Scatter(x=nn, y=aa, mode='markers'))

We now find the first $n$ for which $a_n>0.99$.

n0 = nn[np.argmax(aa > 0.99)]
print(n0)
nn1 = nn[((n0 - 5) < nn) & (nn < (n0 + 5))]
aa1 = aa[((n0 - 5) < nn) & (nn < (n0 + 5))]
fig = go.Figure()
fig.update_layout(width=600, height=600, showlegend=False)
fig.add_trace(go.Scatter(x=nn1, y=aa1, mode='markers', marker=dict(color='red')))
fig.add_hline(0.99, line_color='green', line_dash='dash')
fig.add_vline(n0, line_color='blue', line_dash='dash')

50