Subsections

• Exponential growth
  • Continuous Compounded Interest
  • Compound Interest with Regular Payments
  • Maple commands
  
  
• Inverse Functions
• Exercises

The exponential and logarithmic functions

Exponential growth

The simple model for growth is exponential growth, where it is assumed that $y'(t)$ is proportional to $y$. That is,

$$\frac{dy}{dt} = ky\quad\hbox{where $k$\ is a positive constant.}$$

Separating the variables and integrating, we have

$$\int\frac{dy}{y} = \int k\,dt$$

so that

$$\ln \mid y \mid = kt + C$$

In the case of exponential growth, we can drop the absolute value signs around $y$, because $y$ will always be a positive quantity. Solving for $y$, we obtain

$$\mid y \mid = e^{kt + C}$$

which we may write in the form $y = Ae^{kt}$, where $A$ is an arbitrary positive constant. The same formula is used for exponential decay, except the decay constant $k$ is negative.

Continuous Compounded Interest

Suppose you invest a principal amount $P$ at an interest rate $r$ and want to compute the future value $A$ after $t$ years. Assuming that the interest is compounded once a year, the future value of the investment would be given by the formula $A(t)=\displaystyle P(1+r)^t$. When the interest is compounded more frequently, say $n$ times per year, then formula for the future value of the investment would be

$$A(t)=P(1+\frac{r}{n})^{nt}$$

If the case of continuous compound interest or $n \rightarrow \infty$, then the future value of the investment would be

$$A(t) = \lim_{n \rightarrow \infty} P (1+\frac{r}{n})^{nt} = P e^{rt}$$

Compound Interest with Regular Payments

Now suppose you start with your initial investment $P$ and add to that by making a regular investment of $x$ dollars per compound period. Then the future value $A(t)$ after $t$ years with interest rate $r$ compounded $n$ times per year is given by the formula:

$$A(t) = \frac{x\left[(1+\frac{r}{n})^t-1\right]}{r/n}$$

Suppose you want to figure out how long it would take to reach a desired future value, let's call it $FV$. Then solving the above equation for $t$ would give how many years $t$ it would take to reach this amount making regular investments of $x$ dollars per compound period with interest rate $r$ compounded $n$ times per year.

$$t=\frac{\ln(FV r +nx)-\ln(nx)}{\ln(n+r)-\ln(n)}$$

Maple commands

The main functions you need are the natural exponential and natural logarithm. The Maple commands for these functions are exp and ln. Here are a few examples.

> A:=t->P*(1+r/n)^(n*t);
> subs({P=50,r=0.06,n=12},A(10));
> A:=t->P*exp(r*t);
> subs({P=50,r=0.06},A(10));
> A:=t->x*((1+r/n)^t-1)/(r/n);
> subs({x=50,r=0.06,n=1},A(10));
> t_yrs:=(ln(FV*r+n*x)-ln(n*x))/(ln(n+r)-ln(n));
> subs({FV=1000,r=0.06,x=50,n=1},t_yrs);

Inverse Functions

Consider the functions $f,g$ defined by

$$f(x)=e^x+e^{-x},g(x)=e^x-e^{-x}$$

To be able to get an inverse the function must be one-to-one. You can plot the functions to get a hint as to whether they are invertible or not.

> f:=x->exp(x)+exp(-x);
> plot(f(x),x=-5..5);
> g:=x->exp(x)-exp(-x);
> plot(g(x),x=-5..5);

Both satisfy the vertical-line test but $f(x)$ is not invertible since it does not satisfy the horizontal-line test. Indeed $f$ is not one-to-one, for instance $f(0.5)=f(-0.5)$. From the plot it seems that the function $g$ is one-to-one. In order to determine its inverse we solve for x.

> solve(g(x)=y,x);

We observe that one of the solutions is not defined since the arguement of the logarithm can only be positive. Thus:

$$g^{-1}(y)=ln(\frac{y}{2}+\frac{\sqrt(y^2+4}{2})$$

> ginv:=y->ln(y/2+sqrt(y^2+4)/2);

Let's look at the plot along with the line $y=x$ to see if our functions seem to make sense.

> plot({x,g(x),ginv(x)},x=-20..20,y=-20..20,scaling=constrained);

Let's check that we have computed the right inverse. By definition the composot ion of the functions should be the line $y=x$ since an inverse is the reflectio n about this line.

$$(g\circ g^{-1})(y)=y=(g^{-1}\circ g)(x)=x$$

> g(ginv(y));
> simplify(%);
> ginv(g(x));
> simplify(%);

We are having difficulty getting $x$ for the last composition. Think about what issue the computer has in dealing with this simplification as you will come across this in the exercises.

Exercises

1. Caclulate the future value of an ivestment of $1,500. earning 4.75% interest over a 9 year period using the formula for compound interest that compounds the interest for each of the following: yearly, semi-anually, quarterly, monthly, daily, hourly, and continuously. Explain what you observe.

2. 1. Use Maple's solve command to verify the formula in the background that calculates the time $t$ in years that it would take to reach a desired future value $FV$ making regular payments of $x$ dollars per compound period, compounded $n$ times per year with interest rate $r$.
   2. Use this formula to approximate the number of years it would take to save $50,000 if you invested $1200 per year with an average rate of return at 3.5% compounded yearly.

3. 
   
   $$f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$
   
   
   
   
   $$g(x)=\frac{ln(x^2-4)}{x}$$
   
   

   A)

   Plot the function $f$ over the interval $-5 \leq x \leq 5$ and plot the function $g$ over the interval $0 \leq x \leq 50$. Which function is not invertible and why?

   B)

   Find the inverse of the invertible function.

   C)

   Plot the function and its inverse along with the line $y=x$ on the domain of $- 3 \leq x \leq 3$.

   D)

   Show that you have the correct inverse by using the composite definition. (When you come across a simplifying problem and have figured out why the computer won't simplify ask your lab instructor how to bypass this problem.)