Sample questions for test 4

using MTH229

Newton

• From the graph, estimate the value of the first step of Newton's method starting at $1$

f(x) = x^x - e^x
plot(f, 2, 3.2)
plot!(zero, 2, 3.2)
scatter!([3], [f(3)])

ANS: It is about 2.8. Note you are guessing where the tangent line would hit the graph – not the actual intersection point which is about 2.71.

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

ANS: Five steps means you need to do x=x-f(x)/f'(x) 5 times:

f(x) = 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 = x - f(x)/f'(x)
x, f(x)

(1.618033988749895,0.0)

The last line displays the value for the approximation, x, and the value of the function at this, f(x). That f(x) is 0.0 says the algorithm has converged and 1.618033988749895 is the approximate zero found.

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

ANS: This is the same pattern:

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)

Again, the small value for f(x) indicates this algorithm has converged.

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

ANS: We can do this with newton:

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

• 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).)

ANS: It takes 19 iterations. This is because the second derivative is large until x gets close to the answer of 1.

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

ANS: Let $h(x) = e^x - x^6$. A graph shows the answer is around 16:

h(x) = e^x - x^6
newton(h, 16)

(This problem is kinda fussy. If you pick a bad starting point, say 15, you will not get this answer, but a different one.)

• Consider the graph. Does the point look like a good initial guess to the zero?

f(x) = x^5 - x - 1
x = 0
plot(f, -1, 1.5)
plot!(zero, -1, 1.5)
scatter!([0], [f(0)])

ANS: No, and really, no. Why? Because the initial point is on the other side of a minimum than the zero. The algorithm may get trapped and it does, try it.

Optimization

• What inscribed rectangle in a half circle of radius 1 has maximum area?

ANS: A point $(x,y)$ on the circle determines a rectangle with height $y$ and width=$2x$, so we have $A = 2x \cdot y$ with $x,y$ related by being on a circle: $x^2 + y^2 = 1^2$. So...

A(x) = 2x * sqrt(1 - x^2)

A (generic function with 1 method)

Graphing will show a peak around $0.7$, which we find by looking at the value where the derivative is $0$:

x0 = fzero(A', 0.7)
2*x0, sqrt(1 - x0^2), A(x0)  # width, height, area

(1.414213562373095,0.7071067811865476,1.0)

• 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)

ANS: Well, we can get the critical points by finding when the derivative is 0:

f(x) = 3x^4 - 26x^3 + 60x^2 -11
cps = fzeros(f', 1, 5)

2-element Array{Float64,1}:
 2.5
 4.0

So we need to look at the function values at the endpoints 1 and 5 and the critical points 2.5 and 4:

f(1), f(5), f(2.5), f(4)

(26,114,74.9375,53)

The largest and smallest happen at endpoints – not critical points!

• 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 )

ANS: Okay, this problem is not my favorite, but it shouldn't be hard: we have the main function $A=x\cdot y$ and the constraint $P=500 = x + 2y$, allowing us to solve for $A$ as a function of $y$ alone by substituting or $x$:

A(y) = (500 - 2y)*y

A (generic function with 1 method)

Oh boy, a parabola with zeros at 0 and 250 – the max is in the middle, so occurs when $y=125$. This makes $x=250$. Don't believe that? Well, maybe this will convince you:

fzero(A', 0, 250)

• 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)

ANS: We cut of $d$ by $d$ squares in each corner and will get the resulting volume:

V(d) = (3 - 2d) * (4 - 2d) * d

V (generic function with 1 method)

As all dimensions must be non-negative, we have $0 \leq d \leq 3/2$, so we can look:

cps = fzeros(V', 0, 1.5)

1-element Array{Float64,1}:
 0.565741

The max must occur at this lone critical point, as the values of $V$ at the endpoints are 0:

x0 = cps[1]
V(0), V(1.5), V(x0)

(0,0.0,3.0323024659641433)

• 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)

ANS: Let the rectangle have dimensions $x$ by $y$ then the circle will have radius $r=x/2$. The perimeter will be $x + 2y + (1/2)\cdot 2\pi r = x + 2y + \pi\cdot x/2$. As this is 12, we see that $y = (12 - x - \pi\cdot x/2)$ giving:

A(x) = x * (12 - pi * x/2)

A (generic function with 1 method)

Again, this is a parabola with zeros at $0$ and $24/\pi$, so the midpoint is the peak and will be at $12/\pi$. So $x$ is $12/\pi$ and:

x = 12/pi
y = (12 - pi * x/2)

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

ANS: The distance formula means:

x0, y0 = 4, 0
f(x) = sqrt(x)
d(x) = sqrt((x - x0)^2 + (f(x) - y0)^2)

d (generic function with 1 method)

We start with a guess of $x=4$, and see:

cp = fzero(d', 4)

The point is then

cp, f(cp)

(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

ANS: by hand means we have:

x0, x1, x2, x3 = 0, 1/3, 2/3, 1
f(x) = x^2
delta = (1-0)/3
lft = f(x0) * delta + f(x1)*delta + f(x2) * delta
rht = f(x1)*delta + f(x2) * delta + f(x3) * delta
lft, rht, rht - lft

(0.18518518518518517,0.5185185185185185,0.3333333333333333)

The difference can be seen or you could directly compute

f(x3) * delta - f(x0) * delta  # why?

• 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.)

ANS: Find the values:

f(x) = x^2
trap = riemann(f, 0, 1, 3, method="trapezoid")
simp = riemann(f, 0, 1, 3, method="simpsons")
trap - simp # their difference

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

ANS: We just see:

lft - 1/3, rht - 1/3, trap - 1/3, simp - 1/3

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

(Simpson's will be exact for quadratic and cubic polynomials, the trapezoid exact for linear functions.)

• 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.

ANS: We can use riemann for this:

riemann(airy, 0, 2, 10)

• 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: quadgk will give a better estimate:

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)

(9.005898940290741,1.7763568394002505e-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)^2
quadgk(f, 0, 1)

(9.529465228074825,2.673417043297377e-12)

The one with the sin function. A fact about polynomial functions is that quadgk should be fully accurate.

• Use quadgk and the formula $\int_a^b \sqrt{1 + f'(x)^2} dx$ to approximate the length of the graph of $e^x$ from 0 to 4.

ANS: we first define $f$:

f(x) = exp(x)

f (generic function with 1 method)

We do not do this integral quadgk(f, 0, 4), as that would give the area. Rather, we integrate the related function:

l(x) = sqrt(1 + f'(x)^2)
quadgk(l, 0, 4)

(54.05615249432967,3.033418991549297e-7)

That's it!

• Use quadgk and the formula $\int_a^b \sqrt{1 + f'(x)^2} dx$ to approximate the length of the graph of $\sqrt{1 + x^4}$ from -1 to 1.

ANS: Similarly:

f(x) = sqrt(1 + x^4)
l(x) = sqrt(1 + f'(x)^2)
quadgk(l, -1, 1)

(2.2941933656397393,3.921931668315892e-10)

• Use quadgk and the formula $\int_a^b \sqrt{1 + f'(x)^2} dx$ to approximate the length of the graph of $\sqrt{1 + x^2}$ from -1 to 1.

ANS: This follows the same pattern:

f(x) = sqrt(1 + x^2)
l(x) = sqrt(1 + f'(x)^2)
quadgk(l, -1, 1)

(2.199374827478408,9.352554286579107e-10)