Day 18 in MAT227

Last time: Section 7.2/7.3: Sehnert Lecture

Next time: Section 7.8: Inverse Trig Functions

Today:

Announcements:
- You have a homework assignment due Today (7.3).
- You have a homework assignment due Monday (7.4 -- applications).
- Homework Section 7.2 returned: Graded #3, 6, 28
  - For 3, show a graph! Close your interval....
  - For 6: no need to do this twice -- f(x)=g(x) -- this function is self-invertible! (Can you think of a cryptographic function that is self-invertible?)
  - You'll need to use the theorem (since they asked you to), and you'll need to solve for ${f^{-1}(x)}$ .
- How did you like the Sehnert lecture?
Inverses:
- Navah made an interesting comment: the functions for incryption should be easy to apply, but difficult to invert. I don't think that she meant that -- what she meant was this: it's difficult to find the inverse (but not to use it).
We'll start today, with Lab 16: Logs, part II:
- Today's lab includes examples of using logs to model some interesting phenomena.
- Included is this
- To do the limit in the lab, we need to use a result which you may recall from your earlier calculus days:
  The limit of a composition of continuous functions is the composition of the limits.
  How does that play out in this case?
Section 7.7: L'Hopital's Rule (Sometimes limits are kind of tricky, however!)
- In financial mathematics (e.g. compound interest problems, subject of section 7.5, which we're not covering!), we tend to use an unusual base for an exponential:
  
  ${P(t)=P_0\left(1+\frac{r}{n}\right)^{nt}}$
- You may have encountered a few laws such as The rule of 72 (70, 71, 69.3,...) to calculate doubling time: forget them! Just do the math: the question is
  
  For what value of t is ${P(t)=P_0\left(1+\frac{r}{n}\right)^{nt}=2P_0}$ ?
  
  Solving t is ${P_0\left(1+\frac{r}{n}\right)^{nt}=2P_0}$ for t, we get
  ${t=\frac{\ln 2}{n\ln(1+\frac{r}{n})}}$
  
  where n is the number of compoundings per year, and r is given as a decimal (e.g. 9% is represented by .09). This is the doubling time.
  When compounding is continuous (i.e. ${n \to \infty}$ ), this reduces to the very lovely rule
  ${\frac{\ln 2}{r}}$ which is found in your book on p. 369.
- Now, how do we know that ${\lim_{n \to \infty}\frac{\ln 2}{n\ln(1+\frac{r}{n})}=\frac{\ln 2}{r}}$ ? That is, how do we know that
  
  ${\lim_{n \to \infty} \ {n\ln(1+\frac{r}{n})}={r}}$ ?
  The answer, of course, is L'Hopital's Rule, which is useful in solving certain indeterminate limits:
  
  Let's rewrite it a little: we want to show that
  
  ${\lim_{n \to \infty}\frac{\ln(1+\frac{r}{n})}{\frac{1}{n}}={r}}$
- Here's the general situation:
- And if the limit of the quotient of the derivatives is indeterminate, then iterate -- that is, do it again! Then work with the ratio of the second (or higher) derivatives.
- Examples:
  - Demonstration that ${\lim_{h \to 0} \frac{e^h-1}{h} = 1}$ (remember that this limit figured into the definition of $f'(x),$ where $f(x)=e^x$ )
  - #2, p. 390
  - #6, p. 390
  - #20, p. 390
  - #42, p. 390
  - A better proof of the fact that
    
    (take logs! Better than induction....)
  - Proof of L'Hopital's Rule (requires the limit definition of the derivative)

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