Day 20 in Mat229

• I'll be emailing you your labs separately. I wanted to start with the in-class version.
  • The Exam
  • Key
  • Mean: 73.27
  • Median: 77.5
  • Some comments:
    1. Problem 1: limits
       • Is \(0 \cdot 0\) is indeterminate?
       • Need quotients to use L'Hopital's rule.
    2. Problem 2: derivatives
       • Just checking to see if you know your chain rule
       • Also properties of logs make that second one easy....
    3. Problem 3: Integrals
       1. Straight integration by parts
       2. Two possible trig substitutions: it's nice to have choices
       3. That third one is 0.
    4. Problem 4: Tangent lines
       1. Use points slope - you've got the points, and the slopes are multiplicative inverses of each other (i.e. 2 and 1/2).
       2. Graphs are reflections in the mirror.
    5. Problem 5: hyperbolic tangent
       1. Graph is a reflection in the mirror.
       2. You have to use the log in the right place. Reverse domains and ranges of the hyperbolic tangent for its inverse.
       3. You can compute the derivative in either of two ways.

• Per the lab exams:
  • First of all, if it helps it's 1/3 of the total exam grade; if it hurts, it's 1/10 or the exam.
  • "Divide Cell" is a command some of us need, so that you can execute just a little snippet of code. That way, if there's a fatal error, it doesn't stop further progress. We'll talk about it Friday.
  • I'm sure that copy and paste is your friend in this situation; being familiar with the commands in your previous labs might have handled most situations. Why not start a file called "how to do various tasks"?

  • I had to grade a little creatively, because it seemed unfair to punish too much for failure to plot something which wasn't defined....
  • "Local linearization" was not well-understood:

    Local because it's the tangent line function (i.e. linear), and represents the function very well at a point and in a (generally small) neighborhood around that point.

  • For extrema and inflection points, you can see from the plot that there are local max and min, and three inflection points.

  • Some of you struggled with the antiderivative. The "easy" solution is to use Mathematica's Integrate command, and then take the output and jam it into a new function.

  • The details (Wyatt wrote the key, nicer than mine!):
    • Key (.nb) (pdf)
    • Mean: 66.3
    • Median: 59.4

• Left-endpoint Rectangle Rule.
• Right-endpoint Rectangle Rule.
• Midpoint Rule.
• Trapezoidal Rule.
• Simpson's Rule.

but as the image to the right (above) and the graphical insight below show, we can think of the Midpoint rule as being a "Tangent rule", another trapezoidal-type rule:

Midpoint Rule: \[ \int_a^b f(x) dx \approx M_n = \Delta x [f(\overline{x_1})+f(\overline{x_2})+\cdots+f(\overline{x_n})] \] where \[ \Delta x = \frac{b-a}{n} \] and $\overline{x_i}=\frac{1}{2}(x_{i-1}+x_i)=$ midpoint of $[x_{i-1},x_i]$.

This gives rise to an important observation, which I encourage you to remember:

(their arithmetic average, in this case). Notice that the two methods (LRR and RRR) make errors in the opposite sense in this case (and generally): one method tends to overestimate (the left, in this case), while the other tends to underestimate (RRR here); the average cancels their errors, to produce a more accurate estimate (in general).

Trapezoidal Rule: \[ \int_a^b f(x) dx \approx T_n = \frac{\Delta x}{2} [f(a)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n-1})+f(b)] \] where \[ \Delta x = \frac{b-a}{n} \] and $x_i=a + i \Delta x$. This is just the average of the RRR and LRR: \[ LRR = \Delta x [f(a)+f(x_1)+f(x_2)+\cdots+f(x_{n-1})] \] \[ RRR = \Delta x [f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(b)] \] (and note that each of those methods ignores one endpoint).

Be careful however not to confuse the midpoint and trapezoidal rule. Each is effectively a trapezoid method, but the midpoint method uses "tangent trapezoids", while the trapezoid method uses "endpoint trapezoids" (and it turns out that the tangent trapezoids give better results, generally).

Notice that the number of subintervals in Simpson's rule must be even.

Now we can go further (with Long's rule!:):

(their weighted arithmetic averages).

• Let's approximate the integral \[ \int_0^1 e^x d x \] using a step-size of 1.
• This is called a single "panel" -- we'll just use a single rectangle. Let's compare the results using our five methods.

It turns out that the midpoint method is twice as good as the trapezoidal method, and the errors tend to be of opposite sign. This analysis is based off of an approximating parabola:

Notice that we're adding two copies of the Midpoint rule to one copy of the Trapezoidal rule -- so Simpson's rule has three estimates of the integral in the numerator (so we must divide that by 3 to produce a single estimate, Simpson's rule). Because the error of the Trapezoidal rule tends to be about twice the error of the Midpoint rule, and of opposite sign, the errors cancel in a beautiful way to create a cool new rule!

• Trapezoidal and Midpoint rules get integrals of linear functions exactly right.
• Simpson's rule gets integrals of cubics exactly right! We leap over quadratics, straight to cubics. This seems almost magical....

• Let's check the errors against the three methods' error bounds.
• We'll need some derivatives -- which ones?
• How do the actual errors compare to their bounds?

1. Numerical Integration (pdf)
2. Numerical Integration (nb)

• Mathematica code
• Check midpoint calculation
• Let's do the calculations by the formulas, for
  • Trapezoid
  • Simpson's
• Check that we get the same things using the averages
• Calculate the true value of the integral, and compare to the approximations. What's surprising?