sample questions for test - 3

We should use these packages:

using Plots
using Roots
Base.ctranspose(f::Function) = D(f)

ctranspose (generic function with 38 methods)

Newton's method

• Find the value of the first step of Newton's method when $f(x) = x^2 - x - 1$ and $x_0=2$.

f(x) = x^2 - x - 1
x = 2
x = x - f(x) / f'(x)

1.6666666666666667

• Find the value after 5 steps of Newton's method when $f(x) = x^3 - 2x - 5$ and $x_0=2$.

f(x) = x^3 - 2x - 5
x = 2
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x, f(x)

(2.0945514815423265,-8.881784197001252e-16)

• Run Newton's method until convergence starting at $x=2$ for the function $f(x) = x^6 - x^5 - x^4 - x^3 - x^2 - x - 1$.

f(x) = x^6 - x^5 - x^4 - x^3 - x^2 - x - 1
x = 2
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x = x - f(x) / f'(x)
x, f(x)

(1.9835828434243263,6.661338147750939e-16)

(It only takes 4 steps. If you do another, x's value isn't changed.)

• Newton's method is guaranteed to converge when $f''(x) > 0$ for all $x$ but it may take some time. How many steps are needed for convergence for the function $f(x) = x^{20}-1$ starting at $x=2$? (Use newton(f, x0, verbose=true).)

f(x) = x^20 - 1
x = 2
newton(f, x, verbose=true)  # 19 steps

1.0

• Use Newton's method to find the largest intersection point of $f(x) = e^x$ and $g(x) = x^6$.

We need to plot to find a good starting point:

f(x) = exp(x)
g(x) = x^6
h(x) = f(x) - g(x)
plot(h, 10, 20) # can't tell but looks like h(16) is negative
plot(h, 16, 20) # answer is around 17 -- use that
newton(h, 17)

16.99888735229605

extrema

• On $[1,5]$ find the absolute extrema of $f(x) = 3x^4 - 26 x^3 + 60x^2 -11$ Do they happen at endpoints? Critical points? (From Lamar)

f(x) = 3x^4 - 26x^3 + 60x^2 - 11
plot(f, 1, 5)   # two critical points
cps = fzeros(f', 1, 5)
f(1), f(2.5), f(4.0), f(5)

(26,74.9375,53.0,114)

Absolute maximum is at the endpoint $x=5$, absolute minimum is also at a critical point $x=1$

• Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen? (From UCDavis )

The constraint is $2w + h = 500$, the area is $wh$:

h(w)= 500 - 2w
A(w) = w * h(w) # 0 <= w <= 250
plot(A, 0, 250)  # max is at the lone critical point, near x = 125
a = fzero(A', 125)
a, h(a)   # dimensions

(125.0,250.0)

• A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume? (Ditto)

We have square cut from corner of size $x$. When assembled box will have height $x$ and $(3-2x)$ by $(4-2x)$. So:

V(x) = x * (3 - 2x) * (4 - 2x)  # 0 <= x <= 3/2
plot(V, 0, 3/2)  # max volume is at a critical point near x=0.5
a = fzero(V', 0.5)
a, (3-2a), (4 - 2a) # dimensions

(0.5657414540893351,1.8685170918213299,2.86851709182133)

• Construct a window in the shape of a semi-circle over a rectangle. If the distance around the outside of the window is 12 feet, what dimensions will result in the rectangle having largest possible area? (Again from UCDavis)

We have perimeter is 12. If we label the rectangle with y for height, x for width, then $P = x + 2y + (1/2)*2pi(x/2)$ (why?) and the area is $A = x\cdot y + 1/2 \pi (x/2)^2$. (Again, as r of the semicircle is $x/2$.

r(x) = x/2
y(x) = 12 - (x + (1/2)*2 * pi * r(x))
A(x) = x * y(x) + 1/2 * pi * r(x)^2  # 0 <= x <= 12/pi
plot(A, 0, 12/pi)  # max at a critical point near 2.75
a = fzero(A', 2.75)
a, y(a)   # dimensions

(2.754697942669256,4.918232647556417)

• Find the point $(x, y)$ on the graph of $ y=\sqrt{x} $ nearest the point $(4, 0)$. (Again from UCDavis)

f(x) = sqrt(x)
x0, y0 = 4, 0
d(x) = sqrt( (x - x0)^2 + (f(x) - y0)^2 )
plot(d, 1, 8)   # smallest value at critical point near 3.5
a = fzero(d', 3.5)
a, f(a)

(3.5,1.8708286933869707)

Integration

• For the integral $\int_0^1 x^2 dx$ compare the right riemann sum with $n=3$ to the left riemann sum with $n=3$. How big is the difference? Do this "by hand". For example, the left riemann sum is found with:

f(x) = x^2
delta = (1-0)/3
f(0)*delta + f(1/3)*delta + f(2/3)*delta

Left riemann sum:

f(x) = x^2
delta = (1-0)/3
lr = f(0)*delta + f(1/3)*delta + f(2/3)*delta

0.18518518518518517

Right-Riemann sum

f(x) = x^2
delta = (1-0)/3
rr = f(1/3)*delta + f(2/3)*delta + f(3/3)*delta

0.5185185185185185

rr - lr

0.3333333333333333

• For the integral $\int_0^1 x^2 dx$ compare the value of the "trapezoid" rule to that of "Simpsons" rule using $n=3$. How big is the difference? (You should use the riemann function from the project.)

Need to copy-and-paste this:

function riemann(f::Function, a::Real, b::Real, n::Int; method="right")
  if method == "right"
     meth(f,l,r) = f(r) * (r-l)
  elseif method == "left"
     meth(f,l,r) = f(l) * (r-l)
  elseif method == "trapezoid"
     meth(f,l,r) = (1/2) * (f(l) + f(r)) * (r-l)
  elseif method == "simpsons"
     meth(f,l,r) = (1/6) * (f(l) + 4*(f((l+r)/2)) + f(r)) * (r-l)
  end

  xs = a + (0:n) * (b-a)/n
  as = [meth(f, l, r) for (l,r) in zip(xs[1:end-1], xs[2:end])]
  sum(as)
end

riemann (generic function with 1 method)

tr = riemann(f, 0, 1, 3, method="trapezoid")
sr = riemann(f, 0, 1, 3, method="simpsons")
tr, sr

(0.35185185185185186,0.33333333333333326)

• Which of the last four answers (right, left, trapezoid, simpsons) is closest to $1/3$?

lr - 1/3, rr - 1/3, tr - 1/3, sr - 1/3  # Simpson's wins

(-0.14814814814814814,0.18518518518518517,0.018518518518518545,-5.551115123125783e-17)

(This isn't really fair. Simpsons is exact for quadratic polynomials.)

• For the function f(x) = airy(x) we wish to estimate the area under its curve between 0 and 2. Use a right-Riemann sum with $n=10$ to do so.

riemann(airy, 0, 2, 10, method="right")

0.28120907426713126

• For the function f(x) = airy(x) we wish to estimate the area under its curve between 0 and 2. Use a Julia's quadgk function do so.

ans, err = quadgk(airy, 0, 2)

(0.31253275578067924,3.414490912234669e-13)

• Use quadgk to approximate the integral $\int_0^1 \pi \cdot (2 - x^2)^2 dx$.

f(x) = pi * (2 - x^2)^2
quadgk(f, 0, 1) # 9.005898940290741

(9.005898940290741,1.7763568394002505e-15)

• Use quadgk to approximate the integral $\int_0^1 \pi \cdot (2 - \sin(x)^2)^2 dx$. Compare the error to the last answer. Which is greater?

f(x) = pi * (2 - sin(x))^2
quadgk(f, 0, 1)

(7.646274817115874,1.7763568394002505e-15)

The quadgk function is exact for polynomials, but not so for non-polys. So we would be really surprised if the last one were smaller. In this case they are the same.

• The volume of a glass whose radius is described in terms of its height by $r(h)$ is $V=\int_a^b \pi \cdot r(h)^2 dh$. (picture ) A fluted glass has radius given by $(1 + h)^{3/2}$. Find the volume if the height of the glass is $10$?

r(h) = (1 + h)^(3/2)
dv(h) = pi * r(h)^2
quadgk(dv, 0, 10)

(11498.229112138642,0.0)

• A wavy sided glass has radius given by $r(h) = 1 - 1/2 \cdot e^{h/20} \cdot \sin(3\pi/20 \cdot h)$. Find the volume if the height of the glass is $10$?

r(h) = 1 - (1/2) * exp(h/20) * sin(3pi/20 * h)
dv(h) = pi * r(h)^2
quadgk(dv, 0, 10)

(32.88695672273968,8.27523827240384e-10)