Compound interest, e, and i/Chapter 11

I must say that I don't know everything, and make a lot of mistakes, in spite of what my students might believe. So I wasn't able to give Ian a quick answer, but he and I worked very hard to solve his problem. We started this way, and came to a place where most people can do this- patterns are so important and make things simple. Just last evening I asked Will what compound interest was. He is in an advanced algebra class doing this. He gave me a formula, but didn't really understand what it meant. I went through the following discussion with him (he was trying to fight it most of the way, but finally saw the light!)

1. Simple interest. Find the amount you have in the bank after 2 years if you put in $1, at a 6% annual rate of interest. (It will always be an annual rate of interest with all the problems below.)
The Interest = Principal  *  rate  *  time
The Interest (for 2 yr) = $1 * .06 * 2 = $.12 and the amount (A) you would have after 2 years
A = P + I = $1 + $.12 = $1.12 Notice that the interest is not added each year to get a new principal.

2. Compound interest- leading to a very important infinite sequence. Here the interest is added after each compounding period.

The Interest = Principal  *  rate  *  time = P*r*t and the Amount = P + I

Find the amount you have in the bank after 1 year, with a principal of $1, at 6% annual rate of interest, compounded semi-annually or 2 times per year.

The interest earned during the first (1/2) year = I(first 1/2 year) = $1*.06*(1/2) = .06/2 ($.03)
Now find the Amount you have in the bank after the first (1/2) year
Amount = P + I = 1 + .06/2 = $1.03 which will also be the new principal for the 2nd 1/2 year. [If you don't do the arithmetic, and leave out the dollar sign, you will find the pattern quickly].
Interest (2nd 1/2 year) = (1 + .06/2)* .06/2
Amount (after 2-(1/2)years)= P + I = (1 + .06/2) + (1 + .06/2)*.06/2
Factoring out (1 + .06/2) from each term above, we get (1 + .06/2)(1 + .06/2),  

The algebra looks a little tricky, but is just the distributive property. a is the common factor.

We have a + a*b = a*(1+b) because the right side is a*1+ a*b = a + a*b. In the case above

a= (1 + .06/2)  and b= .06/2. So if we factor 

(1 + .06/2)*1 + (1 + .06/2)*.06/2 =  (1 + .06/2)(1 + .06/2)

So the Amount (after 2 -(1/2)years, or 1 year) = (1 + .06/2)(1 + .06/2) = (1 + .06/2)² = 1.124 (more than the simple interest after 2 years).

What would the amount you have in the bank after 1 year, putting in $1, at 6%, compounded quarterly (4 times per year)? Can you predict what this will be? Let's look at it.

The interest earned during the first (1/4) year = Interest(first 1/4 year) = P*r*t =1*.06*(1/4) = .06/4  (Notice: the r*t = .06*(1/4) = .06/4)

Now find the Amount you have in the bank after the first (1/4) year
Amount (after the 1st (1/4))= P + I = 1 + .06/4,  which will be the new principal for the 2nd 1/4 year.

Interest(2nd 1/4 year) = (1 + .06/4)* .06/4

Amount(after 2-(1/4)years)= P + I = (1 + .06/4) + (1 + .06/4)*.06/4 Factoring out the term 

(1 + .06/4), we get (1 + .06/4)(1 + .06/4), so the 
Amount (after 2 -(1/4)years, or 1/2 year) = (1 + .06/4)²

Interest(3rd 1/4 year) = (1 + .06/4)² *.06/4

Amount(after 3-(1/4)years)= P + I = (1 + .06/4)² +  (1 + .06/4)² *.06/4 Factoring out the term (1 + .06/4)² , we get (1 + .06/4)²(1 + .06/4), so the 
Amount (after 3 -(1/4)years) = (1 + .06/4)³ See the pattern?

What would the amount you have in the bank after 1 year, putting in $1, at 6%, compounded monthly (12 times per year)? (1 + .06/12)¹² = 1.061677812
What would the amount you have in the bank after 1 year, putting in $1, at 6%, compounded daily (365 times per year)? (1 + .06/365)³⁶⁵ = 1.061831311
What would the amount you have in the bank after 1 year, putting in $1, at 6%, compounded 10,0000 times per year? (1 + .06/10,000)^10,000 = 1.061836355
What would the amount you have in the bank after 1 year, putting in $1, at 6%, compounded continuously, (an infinite number of times per year)?
We end up with an infinite sequence
1.124, 1.061363551, 1.061677812, 1.061831311, 1.061836355, ... which converges to

limit_n->inf(1 + .06/n)ⁿ = 1.0618363547... and equals the very important number e^.06 !!

So e¹ = e = limit_n->inf(1 + 1/n)ⁿ

What would be the amount you have in the bank after 2 years, putting in $1, at 6%, compounded monthly (12 times per year)? (1 + .06/12)^12*2 = $1.127159776

What would be the amount you have in the bank after 2 years, putting in $500, at 6%, compounded monthly (12 times per year)? 500(1 + .06/12)^12*2 = $563.5798881

What would be the amount (A) you have in the bank after t years, putting in P dollars, at an annual rate of interest r, compounded n times per year)?

A = P*(1+ r/n)^n*t

What would be the amount (A) you have in the bank after t years, putting in P dollars, at an annual rate of interest r, compounded continuously? We need  

limit_n->inf(1 + 1/n)ⁿ  to get e.

We'll do a little algebra, starting with A = P*(1+ r/n)^n*t  , using the fact that r/n = 1/(n/r)

then A = P*(1+ r/n)^n*t  = P*(1+ 1/(n/r))^n*t. 

Now let m =n/r, and substitute m in place of n/r, we get 

A = P*(1+ r/n)^n*t  = P*(1+ 1/(n/r))^n*t  = A = P*[(1+ 1/m)]^n*t .

Since n=m*r, we can substitute (m*r) for n in the exponent A = P*[(1+ 1/m)]^n*t obtaining

A=P*[(1+ 1/m)]^m*r*t . Now using the identity for raising a power to a power, this can be written as A =P*[(1+ 1/m)^m]^r*t , then compounded continuously, 

A =P*[limit_m->inf(1+ 1/m)^m]^r*t

 The amount (A) you have in the bank after t years, putting in P dollars, at an annual rate of interest r, compounded continuously, is

A=P*e^r*t

^{See Katie's work on compound interest #3}

^{Kirsten works on (1+.07/x)x}

On Sept. 10, 1996 Kirsten, at age 15, (she had started with Don at age 5) was working on this problem (using .07 instead of .06) for her calculus class. We graphed (1 + .07/x)^x in Derive, but couldn't import it into a paint program and then to a gif file. So we graphed it in Mathematica and got this:

  The graph of (1 + 1/n)ⁿ  by Geoffrey in Mathematica' . ( also see chapter 11)

Naming e and e^x as an infinite series

Using the binomial expansion from chapter 9, the first 6 terms of (A + B)ⁿ are:

Substituting 1-> A and 1/n-> B

We'll simplify this: in the terms with 1 to any power, that's 1, and the powers of n are exchanged with the factorials since they are multiplied, to get:

We've gone out 6 terms of the binomial expansion, now we'll look at what happens when we let n -> infinity. All the fractions with n's will go to 1 (subtracting 1 and 2, etc. in the top as n gets very large, won't affect that), and we get:

The above gave us e¹.  

Can you find the infinte series for e^x using the binomial expansion? Just substitute the same1->A, and x/n ->B  and see what you get. We'll come back to this in a bit.

Patterns with i

In the history of mathematics, someone came along with the equation 3 + x = 1, and people said "Oh, you can't do that". Someone else invented the integers and was able to solve the equation putting ^-2 in for x to make the sentence true. Sometime later, someone came up with the equation x² = x^.x = ^-1 and remember rule for substitution, in one open sentence you have to put the same number in for each x. If you try 1x1= 1, and ^-1 x ^-1= 1, neither of these work. Some kids try ^-1/2 x ^-1/2, but that is 1/4, not ^-1. Darshan, a 4th grader, said the answer was unreal, and invented the symbol below- looks like tilted "i glasses"- to solve the problem and proceeded to find powers of i. Don encourages students to make up other symbols, because the symbols are not the important thing, but what it stands for.

Darshan found a pattern in the powers of i, just above. i⁰=1, i¹ =i, i² =^-1, i³ =^-i, then this pattern continues in groups of 4. Don left Darshan with the problem of finding i¹²⁷. He found it to be ^-i, which he drew below with a lot of flair!! Notice also that 4 = magic #, which he found as the pattern above.

Fine work, Darshan!

Shaleen, a 6th grader, graphs iⁿ, finds i^1/2, and finds the graph is not a square!

Don asked Shaleen to graph iⁿ for n = whole numbers. [Note: the vertical axis is the imaginary axis with the scale 4 units = 1i, and the horizontal axis is the real axis, with the scale 4 units = 1]. He knew that i⁰=1, i¹ =i, i² =^-1, i³ =^-i and i⁴ =1, so he plotted those points below. Most youngsters believe right away that these points are the corners of a square, which he drew. When Don asked Shaleen where i^1/2 

would be, he put a point half-way between i⁰=1 and i¹ =i on the square. This point is represented by the complex number 1/2 + (1/2)i and he thought would be i^1/2 . Don said that if i^1/2 is squared, that would = (i^1/2)² which = i because raising a power to a power, you multiply the exponents (which he knew). So Don had Shaleen square 1/2 + (1/2)i below.

Don uses the area model for (A+B)² = (A+B)(A+B), and Shaleen substituted 1/2->A and 1/2 i ->B. He found that [1/2 + (1/2)i]²  = 1/2 i, NOT i. So 1/2 + (1/2)i is not = i^1/2 ! At this point Don suggested a way Shaleen could find i^1/2 . Since the point is half-way between i⁰=1 and i¹ =i, we can use the point (A + Ai) to keep it on the y=x line, square it and set that number equal to i to find A. Shaleen proceeded to do this below.

Note, if 2A²i =  something times i, the something must = 1, so 2A² = 1. He got A= the square root of (1/2) = .707... So i^1/2 = .707..+ .707...i, which Shaleen was able to plot on the first graph, repeated below. Don had Shaleen use the calulator to find the sqrt(2) then divide by 2 and he got .707... THAT NUMBER AGAIN! They talked about if he knows that point, what other points can he plot? He plotted points at i^1.5, i^2.5 = ^-.707..- .707..i , and i^3.5 = .707..- .707..i . 

At one point Shaleen put in the calculator ^-.707..- .707..i, hit ENTER, and got ^-1- i which he knew was wrong. What happened? It turned out that the calculator was set for 0 decimal places, and had rounded off the numbers to 1. What is (.707..+ .707..i)² ? Shaleen got out the compasses and constructed a circle of radius 1 that went through these points. Don drew a right triangle in the first quadrant, mentiond that the length of the hypotenuse was 1, the radius of the circle, then asked Shaleen the measure of the other 2 angles, besides the right angle. They turned out to be 45' each because the sides are both equal to .707..Then he asked him to find the length of the 2 sides that form the right angle. At first he said 1, because 1² + 1² = 1² but quickly realized that couldn't be and said x² + x² = 1² and 2x² = 1, so x = sqrt (1/2) = .707... Don had Shaleen look up on his calculator the cosine of 45' which is .707.. We had to stop, and will continue the trigonometry we had started earlier.

 Fine work Shaleen!

                                   Graph of iterating iⁿ, starting with n = i
                                                     

Before Mathematica I never would have seen this graph! Notice, another spiral! (Actually there are 4 spirals here if you look at the endpoints of the segments). And iⁱ is a real number! (Can you prove that?) Amazing stuff here. One of the finest math sites on the internet, which has interactive applets, is IES in Japan. One applet they made was inspired by the graph above, is under complex number i^(i^i..). You can drag the point at 0+1 i  around the complex plane to get some amazing things!! (You need to download Java (free) to see these).

Back to Ian's problem,,,what would his Dad's monthly payment on a $10,000 house loan at 10% for a 30 year mortgage be? We did it with a lot of work- M is the monthly payment, n is the number of payments (12x30=360), R is the monthly rate of interest= .10/12, P= $10,000

^{Generalizing for n payments, we get  M = and our problem is solved,} 

M came out to be about $87+/month.

See Chapter 11 Questions and answers.