A log that is natural

“Algebra is but written geometry and geometry is but figured algebra.” - Sophie German

From an infinite sound to an infinite staircase

We just reviewed how the integers can be viewed as music, where 1 is the frequency of the fundamental, 2 is the frequency of the first overtone, 3 is the frequency of the second overtone, etc. Then we can view the reciprocal of these numbers and think about it in terms of the wavelength. Thus, 11\frac{1}{1}represents the wavelength of the fundamental, 12\frac{1}{2} represents the wavelength of the first overtone, 13\frac{1}{3} represents the wavelength of the second overtone, etc. This is consistent with how Pythagoras first connected the integers to pleasant sound produced by an urn that was full, 12\frac{1}{2} full, 13\frac{1}{3} full, etc.

Rather than hearing these sounds separately, we considered adding the sound of the fundamental with all the overtones to create this expression:


Since positive integers continue forever, we can imagine the sum of these numbers to continue forever. We are in the Expanse portion of Lazarus Math where we expand our horizons to the ultimate limit. So here is an opportunity to think really big. As mentioned, we give the sum of these numbers a special name: the harmonic series. Even though our human ear is only capable of hearing a few of the overtones, this sum shows the complete expansion for all the overtones. Math people get excited about this infinite sum because it is directly connected to the positive integers. It’s like how body builders get excited about protein and artists get excited about color. It’s part of our math DNA.

A natural question whispers to math people: What is the answer to this sum? Often, there are two possible answers to these types of problems. Either the series will “converge” to a number or it won’t. For example, will this sum ever exceed 100? If we consider the first 1000 terms, the sum is less than 7.49. If we jump to a million terms, the sum is less than 14.4. If we jump to 10 million terms, the sum is less than 16.7, and if we include 100 million terms, the sum is less than 19. It seems like the sum will never reach 50, let alone 100. However, surprising as it may be, this series never converges to a number. That means whatever positive number you choose, eventually this sum will exceed that number if you add enough terms. It is a very slow process, but it will eventually get there. In short, this sum goes to infinity. We will get a better grasp on why this sum approaches infinity later in Lazarus Math.

Before we move closer to that idea, let’s pause and appreciate our progress. Before Lazarus Math, you may have viewed this sum as just a sum of fractions. But we have connected this sum to music and can imagine that it is also the sound of a fundamental with all of the overtones playing music in harmony. We might even think of this sum as a series of waves. As a reminder, Figure 1 is the graph of the waves for the fundamental and first two overtones.

Figure 1. Fundamental and first two overtones

Isn’t it neat that we can now view an infinite sum as a chorus of notes playing in harmony? What a nice change in perspective. And this is only the beginning of how we will change our perspective on an infinite collection of numbers.

Another perspective on the harmonic series

Let’s view this sum in a different way. Would you agree that the sum of the harmonic series does not change if we multiply each number in the sum by 1?

=111+112+113+114+...= 1\cdot\frac{1}{1}+1\cdot\frac{1}{2}+1\cdot\frac{1}{3}+1\cdot\frac{1}{4}+...

I’m sure you’re wondering why we would pursue such nonsense and clutter up a perfect musical harmonic series. By introducing the factor 1, each term is now 1 multiplied by the original fraction of the harmonic series. As a result, each term contains two numbers multiplied together: 1 and the fraction. It appears that 1 is just clutter and wasted space. But let’s resist the urge to remove the clutter and think about a different way to consider two numbers multiplied together.

Applying geometry, we can think of the product of two numbers as representing the area of a rectangle, where one number is the height of the rectangle and the other is the width. In other words, we can think of each term in the series as a measurement of a two-dimensional space: an area. So rather than summing an infinite series of numbers, we are summing an infinite series of areas. Actually, this is returning to the Greek mindset—thinking of two numbers multiplied together as an area of a rectangle. However, Pythagoras and the Greeks would not have taken this sum to infinity but would have kept it at some finite number of terms.

Notice we now have three perspectives for this infinite sum. The first is just thinking of it as adding numbers, perhaps an algebraic perspective. The second is to think of it from a musical perspective, as a collection of sounds. Now, our third perspective is to think of it as the area for an infinite number of rectangles, each one having a side of length 1.

The first rectangle is 1 by 1, actually a square, and its area is 1. The second rectangle is 1 by 12\frac{1}{2}, and its area is 12\frac{1}{2}. Then let your imagination run and perform this for each term. Let’s be sure we understand the consequence of this change in perspective.

When we view the harmonic series as a sum of infinite areas rather than the sum of infinite numbers, we are trusting that the math for areas will yield the same result as the math for numbers. This is not a small change if you think about it. When we think of area, we often think of geometry. Thus, we could state that we are implying algebra and geometry are not two different math concepts. Rather, we’re trusting that there is only one math but two different perspectives. This is a mighty leap of faith—one that required thousands of years and many great math minds to wrestle with it and finally arrive at by the 16th century.

Once we are comfortable that changing the problem from a sum of numbers to a sum of areas produces the same result, we are ready to proceed.

Area math

If our goal is to better understand the harmonic series, did we make any progress by converting it to a sum of the area of an infinite number of rectangles? Adding an infinite number of concrete things, like the area of rectangles, may seem worse than adding an infinite number of abstract things, like numbers. But at least we have a different perspective. We may start thinking about how to gather an infinite number of rectangles. We could randomly create rectangles and present them in some haphazard way. But why not do it in an orderly manner?

Let’s start with a few rectangles to create a process. Begin the process by lining the rectangles in a row, like kids at school lining in order of height, tallest to shortest. And while we are ordering from tallest to smallest, let’s throw them on the xyxy-coordinate system just because it is a playground we are comfortable in. You can see some of the rectangles organized this way in Figure 2.

Figure 2. Converting an infinite sum to an infinite area

We chose different scales for the vertical axis and the horizontal axis so it would be easier to view the areas of the smaller rectangles. But the first rectangle is 1x1, so it is actually a square. Regardless of the scale, the graph, if we want to call it that, looks like an infinite staircase going down from left to right. The important thing is the width of each rectangle is 1 and the height is the original fraction. This displays nicely on the xyxy-coordinate system because the width of each rectangle is one unit. Thus, the area for each step, or rectangle, represents the number for each term in the harmonic series. Because the area of each “step” represents a number in our harmonic series, the total area under this graph as we extend to infinity is the same as the sum of the infinite series.

Even though we haven’t made much progress in understanding this sum, wouldn’t you agree that this is at least a creative way of thinking about the same problem? In other words, not only did we recast the original series as an area problem, but it is an area problem that displays well on this familiar coordinate system. That means we can think about all the tools we have at our disposal within this coordinate system. We might be blazing new trails, but we’re using our trusted old compass! Also, we converted a completely abstract formula into a picture we can visualize.

Now that we have a different visual perspective, let’s remember that we stated that the sum of the infinite series goes to infinity. Since our intuition is that algebra and geometry play well together and yield the same solution, we also believe the sum of the area of these rectangles must also go to infinity as the number of rectangles along the xx-axis approaches infinity. This is our intuition because we know for any finite positive integer nn, the sum of the first nn terms of the harmonic series will also equal the area of the first nn rectangles. Since this is true for any nn, no matter how large, and it is also true for the next (n+1)th(n+1)^{th} term, we can be confident that the infinite area will equal the infinite sum. Since the infinite sum does not converge, then the infinite area does not converge.

I try to minimize notation in Lazarus Math because I want to focus on the concepts, and sometimes notation can distract us from the concepts. But notation serves a useful purpose in math, and it can be fascinating once you adjust to it. Here, I am going to introduce notation because it will help us better understand future ideas. We can rewrite the area of the first 5 rectangles in this concise way:


I’m using the letter xx to represent the integers rather than nn. More importantly, the Greek letter sigma is a fancy way to write a sum of terms. This notation is just a concise way of writing the sum of the area of the first 5 rectangles. In other words, we want all terms starting with x=1x=1 up to and including the term for x=5x=5. If we want the sum to go to infinity, it’s easy to change 5 to infinity and write:


Why did we introduce this notation? Isn’t it wonderful how concise this expression is? We have built up to this concept with graphs and area to get a different perspective. But we also have the capacity to be extremely brief, clear, and to the point. Also, isn’t it neat we can just define notation as a way to do the impossible? We certainly cannot write every term in this sum because we are finite and we live in finite time and space. But using our imagination, we are able to do something we can’t do in the physical world. Of course, we can attempt to compute this sum. It is actually quite easy to compute a relatively large number of finite terms. For example, I was able to calculate the first 100 million terms instantly as approximately 18.9978964139 using a software package. The problem does become difficult as we think of the sum that exceeds our software limits, especially when we think about infinity. But as long as we are within the limits of our computer, this sum is easy to calculate.

From rectangles to a function

Now that we have new notation, we are strategically positioned to change course and consider something new. Even though this is something new in the context of this problem, it should feel like a natural progression because this is how we have been playing with math. Our new notation has prepared us for this new idea, but we need to make one more change.

Even though we have recast our harmonic series as an area of a series of rectangles, we can think of this area in a different way. Consider the top horizontal line for each rectangle. Notice this line is at y=1y=1 in the region where xx is between 0 and 1. Then for the next rectangle, this line is at y=12y=\frac{1}{2} where xx is between 1 and 2. Then for the next rectangle, this line is at y=13y=\frac{1}{3} where xx is between 2 and 3. So another way we can think of the area of the rectangles is to think of this top line as a function, and then we can calculate the area between the function and the xx-axis.

Specifically, the top of the rectangle is the yy-value for the function y=1xy=\frac{1}{x} where xx is a positive integer. The xx-axis is the bottom of the rectangle, which is also the line y=0y=0. But we want xx to be a real number since it is in the xyxy-coordinate system. Again, math has a solution for this problem of inputting a real number and outputting an integer. The math solution is to use the function “floor” to round down to the next integer. So the real number 1.75 becomes 1 and 2.25 becomes 2. Now, we can rewrite our previous equation using the real numbers as input to produce the same result as when we used integers as input:

y=11+floor(x) {x>0}y=\frac{1}{1+\operatorname{floor}\left(x\right)}\ \left\{x>0\right\}

This yy-value will produce the top line for the rectangles in Figure 2. Then we calculate the area between this function and the xx-axis everywhere the function is defined, which is all the real numbers greater than 0. Remember that when we calculate the area between this function and the xx-axis, we create the same harmonic series we started with. In other words, we are still calculating the same area depicted in Figure 2. But now, it is in the world of a function with positive real numbers as input. So we can still view this as an area problem, but we use a function in the xyxy-coordinate system rather than rectangles. That means we have again left the Greek mindset and returned to modern-day algebra, or more precisely, analytic geometry.

Beyond integers

Hopefully, it has been interesting to recast the same problem using different perspectives as we think about the math people of the past. But let’s gear up for a change in the problem itself and not just a change in our perspective. Our current formula now looks rather messy because we are calculating an area connected to integers by using real numbers. It’s nice that we can still perform that math using real numbers as input rather than integers. But by using the floor function, we are kind of fitting a square peg in a round hole. It is not exactly the clean and pristine look we seek. We made this rather awkward transition in order to pursue a new idea, and we created a new problem that is no longer whispering at us but nearly shouting at us.

Rather than making this awkward transition of calculating an area connected to integers, wouldn’t it be interesting if we just calculated that area based on real numbers? Yes, this problem started with Pythagoras and integers. It is certainly an interesting problem using integers, but are you curious to find out what the problem becomes using real numbers?

Now that we have presented this integer problem in a format that contains real numbers, it will be natural to convert this to be just about the real numbers. So our goal is to create a new problem that is based on real numbers as a natural progression from the problem based on integers.

If it helps, you can think of it this way. We can think of the graph in Figure 2 as representing a walk downstairs, assuming our perspective is from left to right. Walking downstairs is essentially a way to arrive at a lower elevation. There are two common ways to arrive at a lower elevation. We could either walk downstairs or walk down a hill. The harmonic series certainly represents the stair approach. But what if we think about walking down a nice smooth hill? Can we convert this function from something bumpy to something smooth? Integers are designed for bumpy rides, but real numbers are engineered for smooth rides. Let me propose writing a new function based on real numbers, which is consistent with, but not equal to, the function based on integers:

y=1x {x>0}y=\frac{1}{x}\ \left\{x>0\right\}

Note that the graph of this new function and the graph of the old function intersect at integer values for xx. For example, when x=1x=1, the result is 11=1\frac{1}{1}=1, and when x=2x=2, the result is 12=0.5\frac{1}{2}=0.5. That means when xx is an integer, the output is the same, but when xx is not an integer, the output is different. Certainly, this has nice “curb appeal” compared to our previous function which had the floor function cluttering things up. This function is amazingly simple, and we might naively think it is boring. But we now consider this simple function as an extension of the harmonic series, which began with Pythagoras, his love of integers, and the pleasant music produced by integers. However, with this simple and hopefully interesting function, we are extending the work of Pythagoras, expanding it to real numbers, and getting the area between this function and the xx-axis.

Figure 3. Area under the graph of y=1xy=\dfrac{1}{x}

We graphed this function in Figure 3 and shaded the area that we will be considering. We accomplished our goal of creating a nice smooth curve. I should mention again that the graph is not drawn to scale as the vertical axis is much more stretched out than the horizontal axis. This is because I want to emphasize the shape of the curve as it goes beyond x>10x>10. If we drew the two axes using the same scale, the slope to the right of x=1x=1 up to around x=5x=5 would represent a more moderate hill that may resemble a downhill in your neighborhood. The curve will appear nearly flat to the right of x=5x=5. Feel free to edit this graph and change the scale to different scaling perspectives.

Before we get into the details of calculating the area, take a moment and appreciate the journey we are on. Imagine being the first person to consider this graph and wondering what the area of the shaded region is. What would you be thinking? When I was thinking about this, I had great anticipation and excitement. After all, we started this journey by following in Pythagoras’s shoes and seeing some of the things he saw. But as we go from integers to real numbers, we are going further than Pythagoras. It should give you a bit of a pioneer spirit. Of course, pioneers experience a tremendous amount of uncertainty. Because we are just experimenting, this may not lead to anything important and take us to a dead end. But it feels like a “natural” experiment to go further on this path.

We may wonder why Pythagoras stopped this journey with positive integers. Is there something he knew that we don’t? Can we think of something interesting that Pythagoras didn’t? Perhaps if we were in a conversation with Pythagoras and he was sitting with his urn, we could tell him, “Rather than banging on your urn when it is full, emptying ½, then banging again, and so on, let’s put a slow leak in the urn to let the water out so that at any time xx, the amount of water in the urn is exactly 1x\frac{1}{x}, where xx is a positive real number.” What would Pythagoras have thought? The music may not have been pleasant to hear, but there would have been a nice smooth transition in sound.

OK, history aside, let’s dig in and try to solve this problem. We have rather bravely presented a really difficult problem, so now we must identify how to calculate this area. The graph is certainly curved, and we know calculating the area under curved things is difficult. It took some of the greatest minds centuries to calculate the area for a circle. What hope do we have of calculating the area under this curve in Figure 4?

Figure 4. What is the area from x=0 to infinity?

Calculating the area of a curved region

You may recall in the foundations of Lazarus Math we considered the graph of y=exy=e^x. Its graph was curved, and the area under it appeared difficult to calculate. But, surprisingly, we learned that the area under this graph, from the far left (negative infinity) to any point xx, had the unusual and quite amazing property of being equal to exe^x. In fact, y=exy=e^x is the only function with that property—the area under its graph equals its value. So we can’t expect to get anything that clever here. Realistically, we should just hope to be able to somehow calculate the area, even if the result is not clean and pristine.

Calculating the area likely appears difficult, if not impossible. We hope that the area approaches some number, even if we don’t know what that number is. What do you think the area will be? Could it somehow equal an integer? Will π\pi show up? There certainly could be a circle hidden in here somewhere since we have a nice round curve. Or maybe ee will somehow show up. Recall ee originated when we doubled an investment. We’re certainly not doubling an investment, but one never knows with ee.

In order to calculate this area, like it often is with difficult problems, we must decide how to begin. Remember when we calculated the area for a circle, we started with something we knew the area of, which was a chessboard. Is there anything we can compare this curve to that we know something about? Well, don’t overthink this too much. An idea is staring us in the face. Recall we started this journey by considering the graph that represented the harmonic series. So let’s consider putting these two graphs together in one graph and compare the two.

Figure 5. Rethinking the area under the graph y=1xy=\dfrac{1}{x}

Figure 5 shows the graph of y=1xy=\frac{1}{x} as well as the rectangles that represent the harmonic series. Superimposing Figure 2 and Figure 3 is a small step, but it may be enough to help you guess the area under the graph of the continuous function y=1xy=\frac{1}{x}.

Comparing areas

Remember we know the sum of the harmonic series is infinite, and we know the area of the graph of rectangles matches the sum of the infinite series. So we know the area for the series of rectangles is infinite. Notice that for each integer value, the curve for y=1xy=\frac{1}{x} intersects the rectangles. We identified that principle before from an algebraic perspective. Now we can see that intersection visually. But for all values in between the integer value, the curve is higher and thus has greater area than the rectangles. That leads us to think the area under the graph of y=1xy=\frac{1}{x} must be greater than the area of the sum of the rectangles. But we know the area for the sum of the rectangles is infinity. Since the area under the graph of y=1xy=\frac{1}{x} must be greater than the area of the sum of the rectangles, that means the area under the graph must also be infinity.

Surprisingly, what appeared to be a difficult problem is actually quite simple. The sum of the area under the graph of y=1xy=\frac{1}{x} does not have a bound and is therefore infinite. Again, that may be surprising when we see the right tail of the graph quickly approaches, though never touches, the xx-axis. Perhaps it is disappointing the area sum does not converge to a number since it would have been neat if it equaled a cool number like π\pi.

This may be a letdown as we had great anticipation for something interesting. It was interesting that the solution was easy to determine. It seems like our story is over just as we were getting started. But we are far from being done with this story. First, let’s get a feel for why the area is infinite.

Perhaps you think that the area under the graph of y=1xy=\frac{1}{x} is infinite because the area increases from right to left without bound between x=0x=0 and x=1x=1. But we can measure the area starting at x=1x=1 and still arrive at the same conclusion. The reason is we are only removing the first rectangle, which has an area of 1. Then the harmonic series becomes 12+13+14+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ...

Notice when we subtract 1 from the harmonic series, we are looking for infinity minus 1 since the sum of the harmonic series is infinity. Subtracting 1 from infinity does not rescue the sum from entering the black hole of infinity. Thus, the sum of the series 12+13+14+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ... is still infinity. This means that even if we start at ½, when we sum the area of all the rectangles, we will still get an infinite area.

Since the area under the graph of y=1xy=\frac{1}{x} starting at xx = 1 is greater than the sum of the area of the rectangles, we conclude that the area under y=1xy=\frac{1}{x} must also be infinity. This yields the surprising conclusion that for both the harmonic series and the graph of y=1xy=\frac{1}{x}, the infinite area results from the right tail, even though the right tail approaches 0 quite quickly. I did not present the graphs to scale so we can appreciate how quickly the right tail approaches the xx-axis and the surprising result that the area does not converge.

Let’s review the strategy we used to calculate the area under our continuous function. We performed a mystery move by converting the harmonic series from algebra to geometry, which changed the problem to calculating an area. Then, we modified this area by considering real numbers rather than integers. Most importantly, we were able to compare the areas for the two graphs. This was a handy trick because we knew the area when the graph was connected to integers. Then, when we converted to real numbers, we were able to leverage knowledge from the integers into an answer based on real numbers.

This is a handy trick math people often do. Rather than calculating something directly, they compare it to something they've already calculated and use that information to arrive at the answer. I suppose you can view it as a form of laziness; I think of it as being clever.

So we’ve investigated the continuous curve defined by 1x\frac{1}{x} and found it interesting because it expands the harmonic series to the real numbers. We also identified the area under the curve is infinite. But is this curve used for anything important?

Actually, this is where the delightful part of the story occurs. In my opinion, if there was a math hall of fame, I am confident that the function yy = 1x\frac{1}{x} would be enshrined into that hall of fame. So what makes this function so important? The reason this function is so important is revealed when we consider only real numbers in a specific range.

If we considered all the positive real numbers as input, the area is infinite. But this function becomes hall-of-fame material when we consider only a subset of real numbers. You may think talking about hall-of-fame math as silly talk. But one criterion for being important enough to be enshrined in the math hall of fame is to be important enough to create a calculator key just for the occasion. And almost all calculators have reserved precious real estate for a button connected to this function. I’m not referring to the button 1x\frac{1}{x}, which simply converts a number to the denominator of a fraction. Which button on the calculator is connected to the area under the function y=1xy=\frac{1}{x}? It is a button that often gets used but is not widely understood. That function is the log function. In other words, the log function is intimately connected to the function y=1xy=\frac{1}{x}. Depending on your relationship with math, you may or may not be well acquainted with the log function.

Reintroducing the log

I’ll assume you, like most people, are a bit rusty on the log function. So allow me the privilege of reintroducing you to the log function. First, log is a nickname; it's short for logarithm. But what is the log function? From one perspective, the log function’s main job in math is to undo things. When we use the log function button on the calculator, we’re usually thinking of undoing something.

If the function is y=10×xy = 10 \times x and we want to solve for xx, then we must “undo” multiplication. We know division can “undo” or reverse the result of multiplication, so we divide both sides by 10 to get x=y/10x = y/10.

What if the operation uses exponents rather than multiplication? How do we “undo” the exponent? If the equation is y=10xy=10^x and you were given the value for yy, how would you solve for xx? In this example, we define 10 as a base and xx is an exponent. When xx is an exponent, then we need something that will “undo” the exponent. That is what a log function does.

Next, we can take the log of both sides of the equation and write log(y)log(y) = log(10x)log(10^x). We can do this because if a=ba=b, then log(a)=log(b)log(a)=log(b). One neat property of logs is we can rewrite log(ab)log(a^b) as b×log(a)b\times log(a). In other words, if we take the log of a number raised to an exponent, we can move the exponent out of the log operation, make it a multiplicative factor, and get the same result. That means we can write log(y)log(y) = log(10x)=x×log(10)log(10^x)=x\times log(10).

Logs are also quite versatile because we have a log for every base. By definition, log(10)=1log(10)=1 if the log is to base 10. Then, the right side simplifies to just xx. In other words, the log will “undo” the exponentiation. That means x=log(y)x=log(y) where the log(y)log(y) is for base 10. Notice that logs just performed the one main goal of algebra, which is to solve for xx, specifically when xx is resting as an exponent.

Because we have a log for every base, we are not limited to base 10 when we use logs to “undo” an exponent. If the equation was y=9xy=9^x , we would follow the same recipe but conclude that x=log(y)x=log(y) using the log for base 9 rather than base 10.

The log’s hidden past

We all have things in our past that are hidden. But it may be helpful for some of these hidden things to be brought to light. The log has a past, and I think the past is hidden from most of us. There are actually books written on its past, and these books are delightful to read. I will only uncover the highlight of the past but feel free to research this further. (One book I recommend is “e: The story of a number” by Eli Maor. Amazon link). This hidden past is directly connected to the area below the graph of y=1xy=\frac{1}{x}. What is this hidden past and how is it connected to undoing an exponent?

Let’s say we wanted to know the value of log(4)log(4). Believe it or not, the answer is actually equal to the area under the graph in Figure 6. This is the graph of y=1xy=\frac{1}{x} for xx between 1 and 4, and the shaded area is precisely log(4)log(4). This is only true for a specific base for the log, and you may notice that I have not stated that base yet.

Figure 6. Area between x=1 and x=4 is log(4).

So what is this unknown base which connects the log function to the area below y=1xy=\frac{1}{x}? What base could possibly be somehow connected to “undoing” an exponent? Which base receives this honor?

A hidden friend reveals itself

Remember when we studied ee, we recognized ee was special in that the area under the curve always equaled its value at that point. In other words, the area under the curve of exe^x from negative infinity to xx is exe^x. We also learned that the slope of the tangent line to this curve at any point is the value of the function at that point. There are many smooth curves, but when xx is the exponent for base ee, it is the perfect smooth curve.

That is all background information we have already covered. But now, the interesting conclusion to our area problem is the base we are looking for is none other than base ee. It may seem normal, or even natural, to think of an integer as the base. Often that is true. But, when it comes to math, and not only the imaginary world of math but also the real world, the most natural base to use is base ee. In fact, it is so natural that we refer to the undoing process, the log with base ee, as the natural log. We even give it the special notation lnln rather than just loglog. Thus, if we write ln(4)ln(4), we know we are taking the natural log of 4, which is the log of 4 using base ee. You will notice most math calculators have an “ln” key reserved and engineered to perform such calculations. So if you enter 4 and then the “ln” key, you will get an approximate value for the natural log of 4. But we usually don’t think about it being the area under the graph of y=1xy=\frac{1}{x} between x=1x=1 and x=4x=4. Isn’t that a surprising result?

Of course, this is not a one-off for the number 4. This works for any positive number greater than or equal to 1. In other words, it is true not just for positive integers but for any real number greater than or equal to 1. That means we can make the wonderful statement that the definition of the area under the function y=1xy=\frac{1}{x} from 1 to cc is log(c)log(c) with base ee, or ln(c)ln(c), where cc is a real number greater than or equal to 1.

For example, if we decided to change 4 to 4.1 and want to calculate ln(4.1)ln(4.1), then we just extend the curve out a little farther and calculate the area between 1 and 4.1. Since we are calculating the area under a continuous curve, we use the magic of calculus to calculate the area. Thus, we are not limited to integer values of xx.

At first, it may appear odd to have ee as a base, so the name "natural" feels anything but natural. However, the more you dive into math, the more "natural" it becomes. As an actuary, I have needed to “undo an exponent” frequently, so the log function was a regular tool. And actuaries rarely use a base other than ee.

Even though it appears that the two stories we’ve discussed regarding exe^x and log(x)log (x) to base ee are separate stories, they are actually connected. If you think about it, we identified exe^x as the special exponential function because of how the slope and area are related. Then, since the purpose of the log function is to “undo” the exponentiation process, it shouldn’t be surprising that there is one base for the log that emerges as special.

In short, ee is special when we exponentiate and when we undo the exponentiation. You can think of ee as the “secret sauce” of the exponentiating process.

By the way, the graph of the function y=1xy=\frac{1}{x} is a hyperbola. You may have studied hyperbolas in high school trigonometry. That means if we take a cross section of a cone, we not only find a hyperbola but also expose the function y=1xy=\frac{1}{x} and the wonderful story about the natural log and ee. We won’t discuss hyperbolas in Lazarus Math, but it is worth noting that a hyperbola has two components, and we only focused on the component where xx is a positive real number. There is a second component to the hyperbola in the region where xx is a negative real number.

An approximate area

If calculating the area under 1x\frac{1}{x} and labeling it as the natural log is new to you, it likely appears quite abstract. To help make it more concrete, let’s estimate the natural log of 4, or ln(4)ln(4), from our graph. Again, we are at the starting point and need to decide how to start. Like before, our starting point isn’t starting with something new but something we have already discussed. Let’s think about how we can estimate ln(4)n(4) by using a graph as a visual help.

We can return to where we started in this section and consider using rectangles. If we made rectangles at every integer value, we have two extreme choices. We could create 3 rectangles that underestimate the area as shown in Figure 7.

Figure 7. Estimating the area from x=1 to x=4

We are underestimating because the curve is never lower than the rectangles at any xx value. The area using this method is 12+13+14\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.

Alternatively, we can overshoot the area by making the height of the rectangle match the highest point of the function as in Figure 8.

Figure 8. Overstating the area from x=1 tp x=4.

The area using this method is 1+12+13.1+\frac{1}{2}+\frac{1}{3}. Clearly, the first method undershot the target and the second method overshot the target. Let's use these two extreme values to improve our estimate.

One simple approximation is to average these two extremes, which is

12(12+13+14)+12(11+12+13)\frac{1}{2} \cdot (\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{2} \cdot (\frac{1}{1}+\frac{1}{2}+\frac{1}{3}). We can easily do the math here. But we gain more insight by recognizing that we just take 12\frac{1}{2} of the extreme points 1 and 14\frac{1}{4}, and then the full value of the in-between values. Thus, the average is 12(11+14)+(12+13)=58+56=1524+2024=35241.46\frac{1}{2} \cdot (\frac{1}{1}+\frac{1}{4})+(\frac{1}{2}+\frac{1}{3}) = \frac{5}{8}+\frac{5}{6}=\frac{15}{24}+\frac{20}{24} = \frac{35}{24} \cong 1.46.

Since ln(4)ln(4) to two decimal places is 1.39, we got a pretty good approximation given how simple the process was. If we wanted to get a better approximation, we could follow a similar plan but use more rectangles with narrower widths. For example, if we change the width from 1 to 0.5, then we would double the rectangles from 3 to 6 and improve our approximation. In fact, we could get extremely close to the exact number by making the width of the rectangles quite small. However, we would still overestimate the exact result.

In the next section, we will identify why this approximate solution is larger than the correct value. But even before we dig into this mystery, there is a clue from Figure 8 that may lead you to the answer. Can you identify from the picture in Figure 8 as to why the approximate method we used overstated the actual value? If not, we will answer that question in the next section.

Understanding the function better

Did approximating the result help make the area calculation, which is an abstract concept, easier to understand? Thinking about why our approximation overstated the actual result will also help you understand this abstract concept. Another way to clarify is to identify boundaries for our area.

We can identify two boundaries for the lnln function. Since we’re measuring the area starting at x=1x=1, we may be inclined to ask, “What is the value of ln(1)ln(1)?” Because the width of the area when x=1x=1 is 11=01-1=0 , there is no area, so ln(1)=0ln(1)=0. This should make it easier to remember that ln(1)=0ln(1)=0.

The second boundary is at the other end where xx approaches infinity. What is ln(x)ln(x) as xx approaches infinity? We can see from the rectangles in Figure 7 that the area for these rectangles is the same as the sum of the harmonic series starting with the second term (ignoring 1). We already identified this sum is infinite. Notice the graph of y=1xy= \frac{1}{x} is always greater than or equal to every point of the rectangles for any xx value in the range. Thus, the area under the graph of y=1xy= \frac{1}{x} is greater than the area for the rectangles. Since the area for the rectangles is infinite, then the area under the graph of y=1xy= \frac{1}{x} starting at x=1x=1 is likewise infinite.

Also, there’s one value in between the two extremes that we can easily deduce. If we take the log of a number with the base of that number, the result is always 1. For example, the log(10)log(10) with base 10 is 1. Thus, we know the ln(e)=1.ln(e)=1. How can we use this fact to calculate the area under the graph shown in Figure 9? This tells us the area under the graph of 1x\frac{1}{x} from 1 to ee is 1.

Figure 9. Area from x=1x=1 to x=ex=e of y=1xy=\frac{1}{x}.

If you want to collect images for what the number 1 represents, this is a creative image to add to your collection. Of course, if we want to use this tactic, we could also easily identify the area under the graph between 1 and e2e^2 is 2, or even the area under the graph between 1 and e1.5e^{1.5} is 1.5, since there isn’t anything special about the exponent 1. This example highlights what is actually occurring with our area function.

To better appreciate the concepts at play, rewrite the value on the xx-axis, our input, in terms of base ee raised to a number. So, convert x=1x=1 to the equivalent e0e^0. Likewise, write the output of the function y=1xy= \frac{1}{x} in terms of base ee raised to a number. So if the input is x=1x=1, then write the output as the equivalent e0e^{-0}. We negated the exponent even though -0 and +0 are the same number. The reason will be clear in a moment. Then the point on the graph is (e0,e0)(e^{0},e^{-0}), and the area under the graph from x=1x=1 to x=1x=1 is 0, which is the positive value of our exponent. This idea works for any value xx greater than or equal to 1.

For example, consider the input value e2e^{2}. If we use this number as input, then the output is e2e^{-2} and the area under the graph from x=1x=1 to x=e2x=e^2 is 2, which is the positive value of our exponent. To picture this, the value of e2e^{2} is about 7.3890560989. Let’s say we wanted to know the area under the graph between x=1x=1 and some number that is not an integer exponent to base ee. For example, how would we calculate the area under the graph between x=1x=1 and x=10x=10? This is where we “reverse engineer” the process. We want to know what exponent to base ee produces the number 10? This is the job of the log function, which is to undo the exponential. That means the exponent would be ln(10)ln(10), or about 2.30258509299. That means the area under the graph from x=1x=1 to x=10x=10 is ln(10)ln(10), which again is the positive value of our exponent. This is because eln(10)=10e^{ln(10)}=10.

Do you see beauty in all this?

The log

The log function, especially the natural log function, is rich in meaning and complexity. The more you can understand this function from different perspectives, the easier it will be to use it in the many ways it appears in math. We will advance this concept in the remaining parts of Lazarus Math beginning in the next subsection where we will again dance with infinity but return to earth with something surprisingly concrete. I hope you are enjoying our voyage into the vast world of infinity. The best minds in math have thought much about these things.

In case you are wondering, this topic is covered in calculus class when studying the definite integral (or the antiderivative). The neat thing about calculus is that it specializes in calculating the area under a curve and it is done by using the definite integral. Usually the textbook just states that the definite integral of 1x\frac{1}{x} is ln(x)ln(x). This is another fact that is often memorized, just like the definite integral of exe^x is exe^x. As we stated in the foundation section of Lazarus Math, so much is lost when we reduce this interesting story to a mere fact or formula.

One last thing. Perhaps now it is clear why I recast the harmonic series, which is about numbers, to something that is about area. In other words, we started with the work of Pythagoras who studied integers and created this infinite harmonic series. We then converted this same infinite sum to an infinite area in the xyxy-coordinate system using a stairstep pattern.

This may not have seemed like a step forward, but it was a change in perspective. The benefit of this change was to shift our thinking from adding numbers to adding areas. Even though we were adding the same numbers and getting the same result, converting this to an area problem opened the door to solving a more challenging problem.

This more challenging problem was to consider the area under a smooth function based on real numbers rather than the step function for integers. A key concept in all this is that we returned to the ancient Greek mindset and saw the area of our rectangles was a fancy way to think about addition. Since the area of a rectangle is two numbers multiplied together, the net result was adding a series of numbers multiplied together. Then we extended this concept to a smooth function involving the real numbers. We can’t solve this using our usual methods. We cannot use addition to “count” how much rain we have received, but we can use a rain gauge to measure it. Similarly, we need the right tool for our problem, so we use the magic of calculus, which specializes in calculating the area under a curve.

In other words, we can calculate the infinite sum of 1x\frac{1}{x} when xx is limited to positive integers. But when we expanded xx to positive real numbers, we could not create this sum since we cannot order real numbers like we can integers. This is because we know the lowest positive integer is 1, the next positive integer is 2, and so on. But we don’t know the lowest positive real number, so we can’t write it as an infinite sum. That is why we recast the problem as an area under a graph of a continuous function.

Doesn’t math produce some brilliant ways to think?

We have more to explore with the log function later in Lazarus Math. But we’ll need to cover a couple more topics first in order to appreciate additional wonders and beauty.

Lazarus Math Part 2

  1. 01
    Harmonic Series
    40 min read
  2. 02
    A log that is natural
    60 min read
  3. 03
    Euler–Mascheroni constant
    50 min read
  4. 04
    Gamma function
    50 min read
  5. 05
    A closer look at the area function
    25 min read
  6. 06
    Dancing with numbers
    60 min read
  7. 07
    From 1/4 to 1/3
    40 min read
  8. 08
    Geometric Sum
    50 min read
  9. 09
    Madhava–Leibniz series
    60 min read
  10. 010
    Connecting to something of interest
    45 min read