Slide 0

A good day to you all, and welcome to a new episode in which we talk about mathematics. We dedicate this episode to looking at functions or – in concrete terms – growth processes and exponential functions.

Slide 1

We’ll focus our attention on what exponential growth means. Recently, the term has become more familiar among the general public due to the Covid-19 pandemic.

Slide 2

Let’s begin and take a look at what exponential growth actually means.

Imagine that you save very consistently. You put one cent in your piggy bank on the first day, two cents on the second day, four cents on the third day, eight cents on the fourth day, and so on. Every day you stow double the amount of the previous day in your piggy bank.

Can you become a millionaire in this way?

Just in case, how long would this take? Take a guess and write it down.

Slide 3

But first let’s talk about Ms. Wood and her dog, Lotte. Lotte is a Weimaraner and needs a lot of exercise. Because Ms. Wood has very little time, she is looking for someone who will take Lotte for a walk every day for an hour. Mara, a dog walker, does this professionally and charges 8 euros per hour for her services.

Slide 4

Ann, the girl next door, is also applying for the job. She would like 10 euro cents for the first hour, 20 euro cents for the second hour, 40 euro cents for the third hour, and 80 euro cents for the fourth hour. And so, it would continue; each hour she would simply like to earn double the amount she earned the previous hour.

Slide 5

Ms. Wood calculates the cost: She needs the support only five days a week.

With Mara, the dog walker, the first week will cost 5 • 8 € = 40 €. “Not exactly cheap,” Ms. Wood thinks to herself.

With Ann, the girl next door, the first week will cost

10 euro cents + 20 euro cents + 40 euro cents + 80 euro cents + 160 euro cents = 3,10 €. That’s significantly cheaper. Does that mean that Ann is offering a bargain?

Slide 6

However, Ms. Wood looks ahead and calculates further:

With Mara, the dog walker, the first two weeks will cost 10 • 8 € = 80 €.

With Ann, the girl next door, they will cost 10 euro cents + 20 euro cents + 40 euro cents + 80 euro cents + 160 euro cents + 320 euro cents + 640 euro cents + 1,280 euro cents + 2,560 euro cents + 5,120 euro cents= 10,230 euro cents = 102,30 €.

Slide 7

Oh, no. Ms. Wood is astonished. That no longer looks like a bargain.

Slide 8

However, there are more nice people in the neighborhood, and so Ms. Wood talks with Mike. Each hour, Mike would also simply like to earn double the amount he earned the previous hour. But he starts more cautiously and asks for 1 euro cent for the first hour, 2 euro cents for the second hour, 4 euro cents for the third hour, and thus 8 euro cents for the fourth hour. And so on.

Slide 9

Ms. Wood examines the offer and calculates further with hope:

With Mara, the dog walker, the first two weeks will still cost 10 • 8 € = 80 €.

In contrast, with Mike they will cost

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1023 euro cents = 10,23 €.

This time, Ms. Wood assumes she has a true bargain.

Slide 10

However, she’s not that completely trusting, and so she calculates further:

With Mara, the dog walker, the first three weeks will cost 3 • 5 • 8 € and that’s a total of 120 €.

With Mike, for the same performance she would pay

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1,024 + 2,048 + 4,096 + 9,192 + 18,384, thus 36,767 euro cents = 367,67 € .

That can’t be true, right? It starts with a single euro cent, and after only three weeks, Ms. Wood would have to dig very deep in her wallet.

Slide 11

What is the reason behind this? With Mara, the dog walker, the billed amount is proportional to performance. For each hour that Mara takes Lotte for a walk, the bill rises by the same (fixed) amount. Mathematically, this looks as follows: f(n) = 8 • n for each number of hours n. And 8 • n is exactly 8 + 8 + 8 + etc. Therefore, we add 8 n-times in a row. Each new sum differs from the previous sum by this fixed addend 8.

We call this linear growth. Students become familiar with functions of this type early on at school: 2 kg of apples cost twice as much as 1 kg, 3 bars of chocolate cost three times as much as one bar.

Slide 12

With Ann and Mike, the billed amount is not proportional to performance. For each hour that Ann or Mike takes Lotte for a walk, the bill rises by double the amount most recently charged for an hour. We call this exponential growth.

After ten hours with Mike, this looks as follows:

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1,023.

We can also write this using powers:

2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ = 2¹⁰ – 1.

In this case, the amount is multiplied by the same number in each step. If this number is greater than 1, then the result grows significantly faster than for simple addition of a positive number.

Slide 13

In general, Mike charges for any number n of hours

2⁰ + 2¹ + 2² + … + 2^n-1 = 2ⁿ – 1 euro cents.

The sum is low for small numbers n, but it grows very very quickly for larger numbers n.

Slide 14

Do we have to limit ourselves to doubling? No, of course not.

If triple the amount is charged each time, then the earnings grow faster. After ten hours, it would look as follows:

1 + 3 + 9 + 27 + 81 + 243 + 729 + 2,187 + 6,561 + 19,683 = 29,524.

We could also write this, using powers:

3⁰ + 3¹ + 3² + … + 3⁹.

But no matter how we write it, the amount is a remarkable 29,524 euro cents or 295,24 € for ten hours of work.

Want to know what three weeks would cost? More than 70.000 Euros and this is definitely far from a bargain.

Slide 15

Oh yeah, let’s go back to the piggy bank. Did you make a guess? Then let’s take a look at what the calculation says.

Clearly, it’s all about when the sum 2⁰ + 2¹ + 2² + 2³ + 2³ + … etc.

is greater than 100 million (we’re calculating in euro cents, after all). There are tables for this, and we find that 2²⁷ = 134,217,728. Thus, on day 28 you would already have a million.

The only problem: Even a fat piggy bank would have burst long before that.

Slide 15a

[Coins fall to the ground]

Slide 16

That’s all for today. It’s nice that you were here. Thank you for your interest and remain cautious when growth is too fast.

But wait a moment. If you would rather not finish with the sad sight of the piggy bank, then I would like to invite you to a special session. Everyone who is interested in a little more math is welcome.

Slide 17

Do you remember? During the episode, we used the fact that the sum of the power of two from 2⁰ to 2⁹ is exactly 2¹⁰ - 1. We can prove this using induction.

We would like to generally show that the sum of the first n powers of 2 – starting with 0 and ending with n-1 – is exactly 2ⁿ - 1. That is the statement.

The sum of all 2ⁱ from i = 0 to i = n-1 is equal to 2ⁿ – 1.

Slide 18

For the base case, we set n = 1 and check whether the statement is true. In this case, n-1 = 1-1 = 0. Now, 2⁰ = 1 = 2 -1 = 2¹ – 1. Thus is clearly true.

Let’s try it, even if that is actually unnecessary, with n = 2, thus n-1 = 1.

2⁰ + 2¹ = 1 + 2 = 3 = 4 -1 = 2² - 1. It therefore works in this case as well.

So that we don’t have to try out all natural numbers, we can use a general method called induction.

Slide 19

We now assume that the statement is true for a natural number.

Thus, our assumption is true: 2⁰ + 2¹ + 2² + … + 2^m-1 = the sum of 2ⁱ for all i from 0 to m-1 ist equals 2^m - 1.

We can do this because we have already shown this for a particular m – and here that was m = 1. If we can now extrapolate from the validity of the statement for any (but fixed) m to the validity for m+1, then we have shown everything. From the already shown validity for m = 1 then follows the validity for m = 1+1 = 2, from this the validity for m = 2+1 = 3 etc.

Slide 20

We call the necessary step here the induction step or inductive conclusion of m to m+1. This is the core of the argumentation.

In this case, this step is not difficult and goes as follows:

The sum of all 2ⁱ for i from 0 to m-1 is exactly 2^m – 1, and this follows from the inductive assumption. Thus, the sum of all 2ⁱ for i from 0 to m is 2^m – 1 + 2^m. Now, 2^m – 1 + 2^m is exactly 2 • 2^m – 1, which is equal to 2^m+1 – 1.

This is precisely what we wanted to prove.

And now we are finally finished for today. See you soon!