Natural Exponentials and Infinite Series

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In the previous part, we constructed the sequence of numbers:

The first few terms of the sequence are: We claimed that there is a number called Euler's number: e, which is approximately 2.718281828459045 with the property that

In order to show that, we will show two things:

1)

2) For all n,

The second statement shows that the sequence is bounded above (by 3) and so it has a least upper bound from a basic property of real numbers. The first statement guarantees that the sequence gets and remains as close as we like to that least upper bound.

Exploration: The Euler Sequence

In order to motivate the theorems that follow, let us experiment a little with the Euler sequence. You will find a control panel where you may define sequences with indices varying from -99 to 1000.

The Euler sequence (1+1/k)^k is already defined in the Define the sequence a(k) := field. We claim that this sequence converges to the Euler number e. We can calculate the first 10 terms of the sequence, starting at k=1 by pressing the Iterate! button. Do it. You see

The graph on the left indicates that the sequence is indeed increasing from 2 to roughly 2.5937. That graph always adjusts so that it can fit the output.

The MathEdit on the right lists the first 10 values. From that, you can see that it is increasing also. Experiment with more points. To that by entering the start and stop indices in the fields:

You should press Reset before each new trial to clear the windows. Does the sequence seem always to be increasing? Can you get values larger than 3 ?

Also, if you decide to try a new sequence, remember that the system uses the index k. Just write your expression for a(k) in terms of k in the indicated field,

select the start and stop indices, (these can vary from -99 to 1000) and press Iterate! For example, try (1+1/k)^k. What do you think it converges to? We will come back to this one later. If you would like to change the precision (up to 19 places), enter the value you want and press Enter in the field to register it.

Before we step through the proof, observe that

where . This obvious fact will be the basis for our proofs of (1) and (2).

Theorem 1:

Proof:

To show (1) we will show that for all the proposition is true where

and argue inductively. Certainly is true:

Now can be written

So it will be enough to show that

and this means

Now the binomial theorem says that

where

So it will now be enough to show that

That is

Factoring from both sides, we have to show

Now

So it will be enough to show that

or

And since the result will follow if

and as we observed above, , so

End of Proof

This shows that Next, we show that all of the terms are smaller than 3.

Theorem 2: For all n,

Proof:

We use the binomial theorem again.

Obviously, it will be enough to show that

Again, for

and so since , we conclude that

End of Proof

The actual upper limit is called Euler's number: e, which is approximately 2.718281828459045 It is a "transcendental number," an infinite decimal expression that does not repeat itself in any simple way.

Exploration: Calculating powers of e

We will shortly have a method for calculating e with great precision. We saw above that to 15 places, the value is 2.718281828459045. If you converge the sequence (1+1/k)^k with indices from 1 to 1000, you will see 2.7169239322355936. This is almost accurate to 1 part in 1000. Not too bad, but not great either. The Euler sequence converges very slowly.

Still, it provides, for now, a nice way to estimate powers of e. For example, change the sequence to (1+2/k)^k and calculate the 1000th term. You will see (with 15-place precision set) 7.3743123903546159. If you go to the Calculator

and type in the yellow command field

calc 2.7169^2; (and Press Enter)

you will see 7.381545609999999. We will do much better shortly.

Natural Exponential Function

Now this calculation seemed to indicate that perhaps

We would like to show that for any number x, the sequence of numbers:

approaches a limit as m increases without bound. We will show that the limit is in fact . This finally will justify the terminology on the title of the page. We will call the natural exponential function.

Our discussion of compound interest on the previous page might prepare us to understand the nature of this limit. If r is an annual rate of interest, and if A dollars is invested for 1 year in an account, then

is the amount received after 1 year if the interest is compounded r times in that year. This amount cannot be larger that some amount (that we will call dollars).

Now our approach to showing that the limit of exists could retrace the steps of the previous two theorems, but we prefer to do it differently here in order to introduce the idea of convergent series. Given a sequence of numbers we may form a new sequence: (called the sequence of partial sums)

The new sequence is called the series based on and the series is said to be convergent if the sequence has a limit. The series is then said to converge to that limit. We write

in that case.

Now suppose that is a fixed nonnegative number. Construct the following sequence of numbers

First we will show that the series

converges to a limit. Next, we will show that this limit is the limit of the sequence

And after that, we will give a proof that these common limits are the number using continuity of exponentiation. Finally, we extend these observations to negative x.

There is a general test that is often useful to show that a series converges. Let us illustrate it with a famous series, called a "geometric series." Suppose that a is a fixed number, and r is a number satisfying . Then from the sequence form the series

The partial sums have the form since

It is fairly obvious that since , these partial sums approach the limit

Question 1: If this last assertion is not obvious, prove it.

End of Question

The r term in the geometric series above is called the constant ratio. When that ratio has absolute value less than 1, the series converges, tout entiere.

Now the sequence we will consider is not geometric. We have instead a varying ratio, that we can consider to be a sequence itself.

Semi-geometric series

Definition 1: Call a series semi-geometric if there is a number a and a sequence such that together, they define a new recursive sequence (Recall the discussion in Art of Approximation, Section 2 of recursive, as opposed to iterative, sequences.)

and we form the series .

In the case of the series we recognize that , and

Exploration: Geometric and semi-geometric series.

Let us sum a geometric series. To sum a geometric or semi-geometric series, enter the start value a in the field:

In this case, we use 1. This will be the 0th term of the series: . Now for series, the start index should be 0. Suppose we sum the geometric series with ratio 1/3 to the 10th index

Enter the (constant) ratio term in the series definition field

and press Sum the Series. You see

The series is converging to 3/2. Well, geometric series are easy. Let us look at some semi-geometric series. For example, we will be studying the series that defines the Euler number shortly. In that case, the ratio term r(k) varies. It is 1/k. So enter that

and use initial value 1 as before, but sum to the 20th term:

You will see:

This is very fast convergence! The 16th term is accurate to 15 places. It is much faster than the Euler sequence. But as we will see, the Euler sequence has the advantage that, with it, it is easy to prove the basic properties of the natural exponential function. We will show that the Euler sequence defines the function .

We can use this utility to sum many series, including Taylor-Maclaurin series. For example, change the ratio to -k/(k+1)

and sum to the 100th term, starting at 1

You see

This famous series is called the alternating harmonic series. It sumsthe sequence: 1, -1/2, 1/3, -1/4, 1/5, ... Can you guess what iit is converging to ? Hint: Go to the calculator and type

calc ln(2);

We will understand what this means in the next part: Natural logarithms and Area Integrals.

Theorem 3: Let be a semi-geometric series with Then if there is a positive number and a positive integer N with the property that

then converges to a limit.

Proof: This condition is sufficient, but not necessary for convergence. For example the alternating harmonic series

is a semi-geometric series that converges, with ratio , that does not satisfy the condition.

Cauchy convergence criterion

Our proof requires the Cauchy convergence criterion for sequences in the real numbers (We have already used it several times, for example, to prove that Riemann sums converge to limits for continuous approximating functions on closed intervals.) The criterion is easy to state, and is reasonably intuitive.

Cauchy Convergence Criterion: A sequence of numbers converges to a limit if and only if for each there is a positive integer N such that whenever

Given this criterion, it will be enough to show that for each there is a positive integer N such that

whenever since expressions like are the absolute value of differences of partial sums.

Now suppose given . And suppose first that is an integer such that

Then the semi-geometric series is just a tail or final part of the original series in the sense that there is a constant such that

Clearly, it is enough to show that the series converges. Now choose any i and j such that . Then

and

Since is constant, we see that we should choose so large that it satisfies the condition:

This condition can be easily satisfied since it simply requires finding a with

and since it follows that

Then for any i and j such that . Then

End of Proof

Thus consider the semi-geometric series with , and

It is clear that whatever we choose, will be smaller than, say, as soon as

This guarantees that converges for all real x.

Theorem 4: For non-negative x, the limit of the series is equal to the limit of the sequence

Proof:

Recall from the binomial theorem,

We may write this

The first thing to observe is that, just as with the sequence

the sequence is increasing for positive x. We can ignore the case where x = 0.

We will show that for all the proposition is true where

and argue inductively. Certainly is true:

Now can be written

So it will be enough to show that

and this means

Now the binomial theorem says that

So it will now be enough to show that

That is

Factoring from both sides, we have to show

Now

So we will show that

or

or

And since the result will follow if

and as we observed before, , so

Next, since

we wish to show that for any pre-assigned

when m is large enough. First, using the Cauchy Convergence Criterion, choose so large that

Next, let be chosen arbitrarily. Then

can be written

Now the second term

and our job is to choose the that

Clearly, however we choose m,

since for each ,

Now, and are constant, and so we can arrange for

to be as close as we like to 1 by choosing m large enough. Therefore, choose m so large that

End of Proof

We now know that for non-negative x, the limits

exist. Suppose we call the limit

for .

We have not yet justified any "exponential" interpretation. We will do that momentarily.

Exploration: Graphing exponential functions.

Go over to the calculator and define and graph .

You see the graph:

Now define the Euler sequence approximation (say the 100th term)

The graphs are essentially identical!

First, let us now extend the result to negative x. Consider the sequence

It is easy to see that

If we can show that

we will be able to conclude that

Question 2: Prove this last statement.

End of Question

From the binomial theorem,

We may write this

Now for any i,

So

Since

we have the result: and so

So far, we know that

Now suppose k is a positive integer. Then is the limit (that, as we know, exists)

This can be rewritten

if we believe that

whenever exists. This is easily seen to be true because the function is continuous.

Thus, for positive integers k,

An identical argument shows that for positive integers k,

That is

So for arbitrary integers k, .

Now, if then we know from Theorem 3, that

exists. So

Therefore,

And we argue as before that (for positive p first)

Question 3: Extend this argument to negative p.

End of Question

We have the result for all rational , is defined, and

This is as far as we will go in this section. The extension of exp to irrational values is straightforward but technical, and we will have a much prettier way to do it in the next part, Natural Logarithms and Area Integrals. For now, we assume that it has been done, and that for all , the specification

defines the function . In the next part, we will take up the question: "What is the derivative of the function ?".

The point of this discussion is to show that even for rational arguments can be computed effectively using the infinite sequence (or the infinite semi-geometric series ). This was Euler's amazing accomplishment (among many others). His number e is the natural one to use as the base for exponentiation.