Natural Logarithms and Area Integrals

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In the previous section, we showed that for any real number x, the sequence of numbers:

approaches a limit as m increases without bound. When x = 1, we call the limit: Euler's number e.

And we showed that for r and s rational numbers (that is, ) we have the following facts

From the first fact, it follows that for

so that for rational numbers at least, has the exponential property: equal ratios for equal changes in r. In fact, by properties (2) and (4) above, we may identify

Further, we know how to approximate for any real x using semi-geometric series or the Euler sequence. Suppose we call : . Then the following statements can be proved true using somewhat technical arguments, for all real numbers:

We will not follow this technical course, however. In this section, we will approach the problem of defining the exponential function for all real numbers and exhibiting its properties in a less technical and surprisingly beautiful way. The approach we will use will recruit what we know about Riemann integrals and change of variables and will also exhibit a fundamental interpretation of integrals as a method of estimating area. We will also learn, of course, how to differentiate this function.

The basic properties of that we will prove are:

We will then be justified to write : for all . Such a function "transforms" addition to multiplication, as you can see from (2) and (3). In itself, this is not so useful. Addition is easier than multiplication. But an inverse for this function, call it log, (once we discover how to find it) would satisfy

Logarithms

That is, if so that , then so that

This would be very useful for practical calculation. To multiply two numbers x and y, find the values of and add those. Then find the number whose log is the result! This led to the creation of tables of logarithms that greatly simplified the calculation of products and powers (and so, as you saw, the calculation of compound interest, amortization schedules, and so on). It also led to the invention of the slide rule, which led to another "art of approximation." But for all of these practical applications of the log function, the theoretical contributions (for example, to analyze the Logarithmic Spiral in the next part) far outweigh them.

Thus, let's get to work. We seek a function with properties (1) -- (4). To see that such a function exists, let us ask what properties the inverse of an "exponential function" should have. Suppose we have a function

that is the inverse of an exponential function (a function that gives equal ratios for equal changes). Then we would expect that L would have the inverse property:

for any

This means that if two pairs of positive numbers have equal ratios, then the differences in their L-values must be the same. It is clear that if L is such a function, then if K > 0 is arbitrary, the function will also have this property, and so, since we also want we replace if the condition is not satisfied.

So now we want a function satisfying the two conditions

and

It is easy to see that if we have such a function, then

so

Such a function would provide a table of logarithms by the above formula.

If the function had an inverse then

and if and we could define

This is the approach that we will take. It happens that Riemann integrals, when used to interpret areas under the graphs of continuous functions, give us just the function we need.

Area and Volume Integrals

Riemann integration is usually first introduced in Calculus as a method for calculating areas and volumes. This approach has a simple geometric appeal and we will discuss it now before we apply it to solve our problem.

Area of an ellipse

Suppose we wish to calculate the area of an ellipse. We will use the earlier picture of an ellipse with semi-major axis a and semi-minor axis b. We will want to show that the area is

This example is chosen to exercise our change of variable formula for integration.

Let us consider this ellipse then to be the curve (discussed in Conic Sections)

where we choose . Now in this representation it is natural to choose parameter interval [-a,a] and to imagine that the parameter variable is x as it varies from -a to a.

For each value x of x, where -a £ x £ a, associate the area between the lines x = -a and x = x
of the ellipse. Call this quantity of area Q(x). Then what we want to do is calculate Q(a). We know of course that Q(-a)=0. This is the first step, determining the parameter interval (the domain of integration).

Next, we need an approximating function

What should we choose? Consider for each x inŽ [-a,a], the length of the intersection of the line x=x with the ellipse. This is a "tranverse slice" of the ellipse over the parameter value x. We'll see why in a moment.

If we make this choice, then a little calculating tells us that

Now if this is to be an approximating function, then

• It must be continuous, and
• For pairs of parameter values sufficiently close, say, (if we say then we are requiring that be small) the area between the lines which we call should be equal to for some . That is, there is at least one such that

Is this true for our approximating function f ?

Consider the following picture (remember, we are allowed to insist that Dx be small).

It is clear that the area (the shaded part of the ellipse between points egjd) contains the rectangle efhg and is likewise contained within the rectangle cdji. The widths of these latter two rectangles are both Dx. But the height of efhg is the length of eg which is , for our approximating function f. Also, the height of rectangle cdji is the length of dj, and this is apparently .

Now this implies that

Since f is easily seen to be continuous, we may conclude from the Intermediate Value Theorem that there is a point , such that

And that's all there is to it. For this reason, it is apparent that f is a good approximating function for this area measurement. We know from what was said earlier that if we construct a partition

and we choose any points whatsoever, then the sum (for those points)

approximates the area of the ellipse.

As we let the mesh of the partition approach 0 these Riemann sums converge to the number

Which is the area of the ellipse.

Now, let us be more explicit. The approximating function

So we may write this area as

And if we could find an anti-derivative P(x) for

we could simply evaluate to get the answer. For that we can use our change of variable formula. Consider the function

According to our formula:

and we can do the calculation if we can anti-differentiate . But we see that the latter expression is

since is non-negative on

We saw that

and so an anti-derivative for is

and the area is

This fact will be important later for Newton's deduction. For now it tells us that for area and volume calculations, it is sometimes very easy to find a (simple) approximating function, once we choose the variable of integration and identify the "transverse slice."

Volume of a sphere

Let us derive next the formula for the volume of a sphere of radius R. We begin again by recalling that the surface area of a sphere of radius R is . Now the volume of the sphere is "swept out" by the surface as we expand the radius uniformly.

Thus the quantity Q(x) is the volume of the sphere of radius x with center at the origin, just as in the previous case. We have determined that our parameter is the radius x and it varies in the interval . What is our approximating function?

We let that be the surface area of this sphere of radius x. If a < b are two parameter values, we want to approximate . Now it is fairly easy to see (at least intuitively) that

when a is close to b. This indicates that the continuous function of x,

is a transverse approximating function for this problem. That is because (by the Intermediate Value Theorem) we are guaranteed that for some . That is, there is at least one such that

That being the case, our job again is to calculate the limit of Riemann Sums

This limit is what we have called

And we know that the function Q(x) is an antiderivative of and so, we observe that another anti-derivative of f(x) is

Thus, by the Fundamental Theorem of Calculus,

is the volume of the sphere of radius R.

Natural logarithms as area integrals

Returning to logarithms, we claimed that a certain area integral would give us a function that has the "inverse" property of exponentials. It gives equal changes for equal ratios:

for any

This means that if two pairs of positive numbers have equal ratios, then the differences in their L-values must be the same. We will show that the function L will satisfy conditions (1) -- (4) below.

To this end, consider the function

This function is continuous and its graph looks like:

Now consider for each

(1)   if , the area of the region from the line to the line and between the graph and the t-axis

(2)   if , the negative of area of the region from the line to the line and between the graph and the t-axis

We will call this function and according to our observations just above, if our variable of integration for measuring if is also if then if gives a transverse slice to the area, so that, we may write

(By convention, if the integral is the negative of the area, that is ).

Obviously, . We will show next that if c > 0 then

This means that for

These two facts will guarantee that this is a "logarithm" function satisfying conditions (2) and (3) above.

Before we give the proof (which follows easily from a change of variable) , let us explore this surprising fact.

Exploration: Areas and logarithms

The control panel on this page shows two graph2D windows and a few buttons:

You may approximate the integral for on the interval by entering the numbers x and y in the fields A = and B =. For example, let us approximate the integral on . Enter the numbers in the fields:

Then press the button below that field:

You will see the window dimensions change to accommodate the interval, and then an approximating Riemann sum will be drawn and summed up.

It reports that this approximation is 1.3859044586. It is drawn with 30 rectangles. You may control the number of rectangles drawn (up to 100) by changing the value on the slider before pressing Sum for [ A, B].

On the other hand, if you would like to do the integration directly using the MathScript Integrate() function, press Integral for [ A, B]

and you will see: 1.3862950642. The "exact" result is only slightly larger.

Now the point of the exercise is to compare this Riemann sum or integral with the one obtained with endpoints on the interval for c some positive number.

The points will have the same ratio as do, and so our claim is that the integrals will be the same. Now to do the experiment, choose a factor c. Here, we will choose 2. Enter that factor in the C = field under the right hand graph2D.

When you press the button:

You will see the Riemann sum on [1, 4] in yellow on the right.

Both the sums and the integrals themselves are equal, but they are different from each other. This illustrates the "equal ratios give equal changes" for our area function L.

Now let us prove the result.

Theorem 1: If c > 0 then

. This means that for

Proof: Consider the change of variable by . The change of variable formula says, for approximation formula , that

End of Proof

The proof was so simple that this hardly deserves to be called a theorem, except for its remarkable consequences.

First of all, properties (2) and (3) of L are now obvious as we showed earlier,

What about the statement:

Since , we see that it is an anti-derivative for the approximating function . Therefore L is differentiable, and

Since this derivative is positive, the Mean value theorem guarantees that L is 1-1, that is, if then

The graph of the function L looks like:

As we will soon see, it is the reflection about the line of the graph of the function . In any case, we can see that it is rising as x increases, and as x approaches 0, it falls sharply. But how do we know that the function is onto ?

That is an interesting question. Suppose that M is a positive integer. It is not too difficult to see from the definition of L as an area function that

Here it is illustrated for M = 10.

Now the series of numbers is called the "harmonic series". It has the property that it "goes to infinity." In particular, for any positive integer K, there is a term in the sequence such that

The proof is easy. Notice that

and so on. It follows that for any positive integer K when M is sufficiently large. Since L is differentiable, it is continuous, and so whenever y > 0 is a positive number, one can solve the equation

Now it is not difficult to see from the fact that the ratio of is the same as the ratio of 1 to x, that

So since it follows (for positive x) that

This shows that for all real y, one can solve the equation . So .

We have one thing left to show. And that is property: . This latter fact relates the natural logarithm function that we have just defined, with the natural exponential function whose exponential properties we have so far only proved for rational r.

Notice that when the exploration opens, we have set A = 1 and B = e If you set it back to those values and press Integral for [A, B], you will see

The "exact" integral is equal to 1 to 1 place in 1,000,000. Experiment with powers of e , and so on.

Now of course, this exploration is not a proof. For that, we must take a new kind of limit.

Theorem 2:

Proof: Recall that Euler's number e is defined as a limit.

We will calculate for each positive integer m,

Since the function L is continuous, and the Euler sequence does converge, we know that

Thus, let us evaluate . From the property it is obvious that

Now the area equal to is between the numbers as a sketch of the graph will show. Therefore

We see that as

Therefore, .

End of Proof

We are almost finished. We now define to be the inverse of . We now know that for all real numbers

And we may write colloquially,

We have one final observation to make. Another amazing one! What is the derivative

We know by the (deep) inverse function theorem that this derivative exists. And the chain rule tells us what it is. Recall that

Therefore

So it follows that since

so

The natural exponential function is equal to its derivative!