#### Annual Interest Limit
1. As the number of dividend payments per year increases, the effective annual interest approaches infinity.
Answered False
Class Discussion and Explanation:
The growth factor with n compounding periods per year is:
$(1 + \frac{1}{n})^n$
As $n \rightarrow \infty$
$$
\lim_{n \to \infty}(1 + \frac{1}{n})^n = e ≈ 2.71828
$$
So, if we start with $100, the maximum possible ending balance after one year at 100% annual interest compounded continously is:
$100 \cdot e ≈ 271.83$
Even though the number of compounding periods increases without bound, the total annual growth does not blow up to infinity. Instead it levels off at e times the principal

#### Infinitely Compounding Growth
2. Recall that $f(r)=er$ takes in a growth rate, r and returns the amount of infinitely compounding growth. It is UNLIKELY that this function of growth rate $(f(r)=er)$ has an inverse. That is, there is no function $g(x) = r$ that returns the growth rate needed to attain a certain amount of growth, x
Answer False
Class Discussion and Explanation
Given: $f(r) = e^r$
This function is strictly increasing and continous for all real r
The range of this function is $(0,\infty)$
This function does not have an inverse because this is one-to-one function.
The inverses of this function will be a natuaral logarithm
$g(x) = ln(x)$
so that:
$g(f(r)) = g(e^r) = ln(e^r) = r$
and $f(g(x)) = f(ln(x)) = e^(ln(x)) = x$
for all $x > 0$
Example:
Suppose we want to know the growth rtate that yields a doubling of the investment(x=2). Then:
$r = g(2) = ln(2) ≈ 0.693$
Therefore, a ~69.3% continously compounding rate produces doubling in one year.

#### Divergence
3. This description of e is an example of divergence. So, we can safely say that "compounding growth within a time period diverges as n approaches infinity".
Answer False
Not Discussed in Class and Explanation:
The description of e in the compounding-interest framework is actually an example of convergence. Because:
- The growth factor of n compounding periods is:
$(1 + \frac{1}{n})^n$
- As $n \rightarrow \infty$, this expression does not approaches infinity (Grow without bound/limit). Instead, it approaches a finite constant:
$$
\lim_{n \to \infty}(1 + \frac{1}{n})^n = e ≈ 2.71828
$$
- So, no matter how many times we subdivide the compounding (daily, hourly, per second etc.), the growth stabilizes at a maximum multiplier of e for a 100% annual rate.
- For example:
- n = 1: factor = 2.0
- n = 2: factor = 2.25
- n = 4: factor ≈ 2.441
- n = 100: factor ≈ 2.705
- n $\rightarrow \infty$: factor $\rightarrow$ 2.718 (not infinity)

#### Function Inverses
4. The two interactive plots "Subdivided Interest ..." and "Total Growth ..." illustrate two functions that are inverses of one another. So, if one is f then the other is $f −1$
Answered False
Class Discussion and Explanation:
Left Plot is Subdivided Interest Plot:
$y_{left}(k) = (1 + \frac{1}{n})^k, k = 0,1,...,n$
In this plot n is fixed and we vary k(the subdivision within the year)
It shows how growth accumulates within a single year as compounding periods pass
The domain of this plot is subdivisions within a fixed n.
The right plot is Total Growth Plot:
$y_{right}(x) = (1 + \frac{1}{x})^x
, x = 1, 2,..., n$
In this plot we vary x(the number of subdivision per year)
It shows how the final annual growth factor depends on the number of compounding periods.
The domain of right plot is different choices of n.
Therefore, these plots are not inverses of each other. One is not undoing the other - they're describing different relationships:
- one tracks growth inside a fixed year(progression over time)
- The other tracks growth as n changes(effect of compoudning frequency)

#### Asymptotic Behavior
5. The functions $n \text{ and } n+log2(n)n +log2(n)$ both share the same asymptotic end behavior. I.e., the impact of $log⁡2(n)$ is negligible when n is large enough.
Answer True
Class Discussion and Explanation:
We are comparing:
$f(n) = n \text{ and } g(n) = n + log_2(n)$
For large n, the linear term n dominates $log_2(n)$
Formally,
$$\lim_{n \to \infty}\frac{g(n)}{f(n)} = \lim_{n \to \infty}\frac{n + log_2(n)}{n} = \lim_{n \to \infty}(1 + \frac{log_2(n)}{n}) = 1$$
Since the ration goes to 1, the "extra" log_2(n) is negligible in the asymptotic sense.

Example:
- At $n = 1000: log_2(1000) ≈ 10$, thereforem $n + log_2(n) = 1010$, which is only a 1% increase over n.
- At $n = 1,000,000: log_2(10^6)≈20$, therefore, $1,000,000 + 20$ is just 0.002% larger.
The gap grows in absolute terms(10, 20,...), but the relative distance shrinks to 0.
Therefore, $n \text{ and } n + log_2(n)$ share the same asymptotic behavior. The logarithmic term is negligible compare with n as $n \to \infty$

#### Asymptotic Growth
6. The following functions of n are listed in (descending) order of their "asymptotic growth", as $n→∞$.
* n
* log 2 (n)
* 10
i.e., "at infinity", (1) will always be larger than (2), and (2) will always be larger than (3) ...
Answer True
Not Discussed in Class and Explanation:
We compare growth by ratios as $n \to \infty$
- $n$ vs. $log_2(n)$
$$\lim_{n \to \infty} \frac{log_2(n)}{n} = 0 \Rightarrow \text{n dominates } log_2(n).$$
("The long term is negligible next to a root/linear/etc." - same idea we used when contrasting $\sqrt{n} + log(n) \text{ with } \sqrt{n} + 10$)
- $log_2(n) \text{ vs. constant } 10:$
$$ \lim_{n \to \infty} \frac{10}{log_2(n)} = 0 \Rightarrow log_2(n) \text{ dominates any constant}
$$
Thus the descending asymptotic order is:
$n \succ log_2(n) \succ 10$

In situation 2 we emphasized that lower-order terms(constants, logs) become negligible relative to higher-order terms(roots, linear, etc.)
Here, the linear term n outgrows $log_2(n)$ without bound (their ratio $\frac{n}{log_2(n)} \to \infty)$, and $\log_2(n)$ outgrows any constant.
So at infinity, $(1) > (2) > (3)$ holds.

#### Trignometric Limits
7. Suppose we let $t_0=2π$.Then, the $\lim_{ω→∞}sin(ωt_0)$ exists.
Answered False
Class Discussion and Explanation:
We set:
$t_0 = \frac{\pi}{2}$
We want:
$$\lim_{n \to \infty} sin(\omega t_0) = \lim_{\omega \to \infty} sin(\frac{\pi}{2}\omega)$$
- As $\omega$ increases, the argument $\frac{\pi}{2} \omega$ grows without bound.
- Multiples of $\frac{\pi}{2}$ cycle through sine values:
$sin(\frac{\pi}{2} \cdot 1) = 1, sin(\frac{\pi}{2} \cdot 2) = 2,
sin(\frac{\pi}{2} \cdot 3) = -1,
sin(\frac{\pi}{2} \cdot 4) = 0, ...$
So, as $\omega \to \infty$, the sequence of values oscillates between -1, 0, +1 infintely often.
Therefore, the limit does not exist, because $sin(\omega t_0)$ does not settle to a single value as $\omega \to \infty$.

#### Trignometric Domain
8. The function $sin⁡(ωx)$ is only valid for integer (e.g., 1, 7, -3) values of ω.
Answer False
Class Discussion and Explanation
The sine function is defined for all real numbers as input - not just integers.
$sin: \mathbb{R} \to [-1, 1]$
So for
$f(x) = sin(\omega x)$
the variable $\omega$ (the angular frequency) can be any real number. In physics and signal processing, $\omega$ is typically measured in radians per second, and can take on non-integer values like $2.5, \pi, or 0.001$.
Example:
- if $\omega = 2: f(x) = sin(2x)$
- if $\omega = \pi: f(x) = sin(\pi x)$
- if $\omega = 0.5: f(x) = sin(0.5 x)$
These all are perfectly valid
Therefore, the function $sin(\omega x)$ is valid for all real values of $\omega$, not just integers.