Euler’s Initial

In my last post, the irrational number e was used. Let’s define and explain e a bit more in this post.

If you deposit $1 in the bank which pays 100% interest once each year (I need to find this bank!), then at the end of 1 year, you will have the original$1 plus 100% of that which is another $1, for a total of$2. Now bear with me, but an equivalent expression that will give me this same answer is

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{1}}\right)}^{1}\hspace{0.33em}{=}\hspace{0.33em}{(}{2}{)}^{1}\hspace{0.33em}{=}\hspace{0.33em}{2}$

What happens if this generous bank computes part of the interest during the year and that interest is added to your original $1 to be included in subsequent interest calculations? When banks do this, this is called compounding interest. That is the interest made compounds, or is added to, the original investment to get even more interest. Now suppose the bank computes the interest twice per year. In 6 months, the bank will compute your interest but since only half the year has gone by, only half the interest, or 50%, is used. So in 6 months you have$1.50. At the end of the year, 50% interest is again computed but on $1.50 now. This gives a total of$2.25 which is better than the $2 you would get if the bank didn’t compound the interest semi-annually. Now the fractional equivalent of 50% is 1/2, so again bear with me, but an algebraic equivalent expression that computes the amount you will have at the end of 1 year when the bank compounds semi-annually is ${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{2}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{(}{1}{.}{5}{)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{25}$ Suppose the bank compounds interest four times a year (quarterly)? Again, the interest used each quarter will only be 1/4 of the annual 100% interest but the interest will compound each quarter. At the end of the first quarter, you will have$1.25. This is now the amount to be used at the end of the second quarter, and so on. The algebraic equivalent expression to compute what you have after the entire year is

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{4}}\right)}^{4}\hspace{0.33em}{=}\hspace{0.33em}{(}{1}{.}{25}{)}^{4}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{44}$

This is better still! But do you notice the pattern in the algebraic expressions? The denominator of the fraction in the brackets and the exponent (power) are the same as the number of times interest is compounded during the year. So in general, if interest is compounded n times per year, the amount you will have at the end of the year is

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{n}}\right)}^{n}$

You may have also noticed that the larger n is, the more money you have. Well if you’re not greedy enough, let’s find a bank that compounds daily. If your dollar is compounded daily, at the end of 1 year you will have

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{365}}\right)}^{365}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{71}$

That’s great but you may have thought that would be a larger amount. The problem is that though the power over the bracket stuff is increasing which has the effect of increasing the amount, the fraction part in the brackets is getting smaller, making the stuff in the brackets closer to 1. Raising 1 to a power is just 1. So there are two competing forces here, one that increases the value of the expression and one that decreases it.

Now we can compound more frequently than daily. We can compound half-daily, etc. What happens to the expression as n increases to infinity, ∞?

Well maths does have a process for that, it’s called limits. Let me just show that and then explain it:

$\mathop{\lim}\limits_{{n}\rightarrow\infty}{\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{n}}\right)}^{n}$

This is read: “What is the limit of this expression as n goes to infinity?”. Now you can get an approximate answer to this by putting larger and larger numbers in for n on your calculator. It turns out, that there is no exact answer that can be written in decimal numbers because the answer to the above is an irrational number like 𝜋. This was proven to be the case in 1737 by Leonhard Euler. Because of his work with this number, it is given the symbol e in his honour.

It turns out that to 50 decimal places e = 2.71828182845904523536028747135266249775724709369995…

So you see that the most you can make with your dollar is \$2.72.

This number was first calculated by Jacob Bernoulli in 1683 to solve the very problem about interest we just went through. But Euler did a lot more work with it.

e is a very important number in calculus, probability, finance, and the interesting world of complex numbers.

Posted on Categories Algebra, Pre-VCE