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Euler’s Number (the mathematical constant ‘e’)

Last week, in this blogpost here, I tried to explain why the scale height of the atmosphere $H$ is defined as the altitude one has to go up for the pressure to drop by a factor of $1/e \;(\approx 37\%)$ of its value at sea level. After posting that blog I decided I would write a blog to say a little bit more about the mathematical constant e, a very important number in mathematics.

Probably the best known mathematical constant is $\pi$, which is defined as the ratio of the circumference of a circle to its diameter. Pretty much every school child comes across $\pi$; but it is only people who study maths at a more advanced level who come across e, so let me try in this blogpost to explain what e is and why it is so important.

Euler’s number

e is also know as Euler’s number, named after the Swiss mathematician Leonhard Euler who lived from 1707 to 1783. Euler was one of the great mathematicians, but it was not he who discovered e. The constant was discovered by Jacob Bernoulli (from the same family as the name attached to “the Bernoulli effect” which causes lift in the wings of an aeroplane), but it was Euler who started to use the letter ‘e’ to represent the constant, in 1727 or 1728.

The number itself is defined as the solution to the following sum

$\displaystyle e \equiv \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + .... = 2.718281828...$

where $2! \text{ (called "two factorial") } = 2 \times 1$, $3! = 3 \times 2 \times 1, \; 4! = 4 \times 3 \times 2 \times 1$ etc. $0! \equiv 1$ by definition. This is an example of a converging series. It is summed to infinity, but each term is smaller than the one before; but it never ends. So, as you go to more and more terms in the series you only affect numbers which are maybe 100 or even 1,000 places to the right of the decimal point. Calculators usually display numbers to 7 or 8 decimal places, so you would not need to go very far in this series to get the number displayed by your calculator (try it to find out how many!)

The mathematical constant e is the sum of 1/n! where n goes from zero to infinity ($\displaystyle \sum_{n=0}^{\infty} \frac{1}{n!}$). It is equal to 2.718281828…….

Just like with $\pi$, e is a transcendental number (I will explain what that is in more detail in a future blogpost). Briefly, this means that it carries on forever and does not repeat, but it is slightly more complicated than that. Unlike with $\pi$, where people have competitions to remember it to thousands of decimal places (the current world record is 70,000 decimal places achieved by Rajveer Meena on 21 March 2015!!), no one seems that concerned in remembering e.

Why is e important in mathematics?

Mathematicians often love numbers and formulae for their own sake, sometimes just for their beauty. So, for example, the solution to something like

$\displaystyle a = \sum_{n=0}^{\infty} \left( 1 + \frac{1}{n^{3}} \right)$

(which I just made up) may not have any importance mathematically, but still mathematicians may enjoy playing and exploring such a series. But, in the case of

$\displaystyle e \equiv \sum_{n=0}^{\infty} \frac{1}{n!}$

the number which comes form this, 2.718281828……., is important mathematically (and physically), and here I will try to explain why.

Compound interest

I mentioned above that e was discovered by Jacob Bernoulli. His discovery was made in 1683 when he was investigating a question concerning compound interest. Remember, compound interest is when you get a certain percentage interest on the amount you have in e.g. a bank, but the interest is calculated not on the initial amount but on the amount after the previous period’s interest has been added.

Suppose we invest £10 in a bank account which pays an interest of 5% per year, and we leave it there for 3 years. Assuming the interest is added just once a year, after the first year our £10 will have earned £0.50 interest, so we will have £10.50. At the end of the second year the interest earned will be 5% of £10.50 which is £0.53 (rounding up to the nearest penny), so we will now have £11.03 at the start of year 3. The interest at the end of year 3 is going to be 5% of £11.03 which is £0.55, so the total at the end of the 3  years is £11.58.

Doing this for just 3 years manually is quite easy, but if we wanted to do it for e.g. 25 years, there would be a lot of tedious calculation. Also, often the interest is added more than once a year. So, for example, you may have an annual interest rate of 5% but it is added each quarter. You can quickly see that even doing this manually over a 3-year period would be a lot of calculating.

Thankfully, there is a simple formula for calculating the total accumulated value, which is

$P(1 + i/n)^{nt}$

where $P$ is the initial amount invested (the ‘principal’), $i$ is the rate of interest, $n$ is how often each year the interest is added (called the ‘compounding frequency’) and $t$ is the time for which we are making the calculation, expressed in years.

If we go back to our example above, and stick to $n=1$, we have $P=10, \; i=0.05, \; t=3$ and so

$P(1 + i/n)^{nt} = 10(1+0.05)^{3} = 10(1.05)^{3} = 10(1.157625) = 11.58$

exactly as we calculated manually.

But, you may be wondering, what is the similarity between the formula

$P(1 + i/n)^{nt}$

and the formula for $e$

$\displaystyle e \equiv \sum_{n=0}^{\infty} \frac{1}{n!}?$

What Bernoulli noticed was that, if you make $n$ larger and larger (do the compounding daily instead of quarterly or once a year), the sequence approaches a limit. So, for example, in the above example, with $n=1$ we found the amount at the end was £11.58. If we made $n=4$ (compounding quarterly) we would get £11.61 and if we compounded the interest every day ($n=365$) we would get £11.62. If we compounded every hour (!!!) ($n=8760$) we would get £11.62, the same answer as if we compound every day, so we have reached the limit.

What Bernoulli did was consider the formula with $P=1, \; i=100\%$ and $t=1$. If we do this for different values of $n$ we get the following curves. The first one is just showing n from 0 to 20, and it is clear that the values are flattening out. The second plot goes from n=0 to 500, and I have just shown on the y-axis values from 2.5 to 2.72. This shows even more clearly that, as n gets larger, the value of $1(1+1/n)^{n}$ tends towards a particular value, and that value turns out to be 2.71828… (which is e).

$y = 1(1+1/n)^{n}$ for n from 0 to 20

$y=1(1+1/n)^{n}$ for n from 0 to 500, but note the y-axis is only displayed between 2.5 and 2.72. The curve is clearly tending towards a value, and that value is $e=2.71828...$

e in calculus

Even if you don’t know how to do calculus, you have probably heard of it. It was co-invented by Isaac Newton and Gottfried von Leibniz (see my blogpost “Who Invented Calculus” for the fascinating story of the 30-year feud between Newton and Leibniz). Without going into too much detail about all the various things one can do with calculus, one thing is that it gives is the gradient (slope) of a curve at a given point (something which is sometimes called the derivative).

There are an infinite number of different mathematical functions, for example $y=x^{2}$, or $y=x^{3} - 2x + 3$, and we can use differentiation to determine the gradient of these functions for any particular value of $x$. But, the function $y=e^{x}$ is unique; it is the only mathematical function whose derivative is the same as the function. To put this another way, the slope of the curve $y=e^{x}$ is $e^{x}$ for any value of $x$, and there is no other mathematical function whose derivative is the same as the function, only $f(x)=e^{x}$.

A plot of $e^{x}$ as a function of $x$. At $x=0, \; e^{x}=1$. As $x \rightarrow -\infty, \; x \rightarrow 0$.

In addition, when we integrate something like $dx/x$, we get a logarithm; but the base of that logarithm is e, not base 10 (our usual base of counting). In fact, we call logarithms in base e natural logarithms. Because the derivation of the variation of pressure with altitude involves integrating $dP/P$, we find that the vertical distribution of pressure is logarithmic, but in base e, $P=P_{0}e^{-z/H}$, where $H$ is the scale height and $P_{0}$ is the pressure at sea level. It is because of the pressure’s exponential dependence on altitude that $H$ is usually expressed as the value for the pressure to drop by a factor of $1/e$.

The normal (or gaussian) distribution

As one last example of where e pops up in mathematics, it arises in the equation which describes the normal or gaussian distribution. I blogged about that distribution in this blogpost here “What does a 1-sigma, 3-sigma or 5-sigma detection mean?”. The function which describes the normal distribution has the form

$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^{2}/2}$

where e is our friend, Euler’s number.

The normal, or Gaussian, distribution $y = \frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}$

Who invented calculus?

One of the physicists in our book Ten Physicists Who Transformed Our Understanding of Reality (follow this link for more information on the book) is, not surprisingly, Isaac Newton. In fact, he is number 1 in the list. One could argue that he practically invented the subject of physics. We decided to call him the ‘father of physics’, with Galileo (whose life preceded Newton’s) being given the title of ‘grandfather’.

Newton was, clearly, a man of genius. But he was also a nasty, vindictive bastard (not to mince my words!). He didn’t really have any close friends in his life; there were plenty of people who admired him and respected him, and of course he had colleagues. But, apart from a niece whom he seemed to dote on in later life, and two men with whom he probably had love affairs, he was not a man who sought company. He was probably autistic, but lived at a time before such conditions were diagnosed or talked about.

Isaac Newton (1643-1727), the ‘father of physics’. He relished in feuding with other scientists

One sort of interaction that he did seem to enjoy with other people though was feuds. In fact, he seemed to thrive on feuding with other scientists. He loved to argue with others, which is not uncommon amongst academics. He had strong opinions which he liked to defend; this is normal. But, Newton took these disputes to an extreme; if he fell out with someone he would do everything he could to destroy that person.

Although I am sure that he had many ‘minor’ arguments, he had three main feuds with fellow scientists. These three men were

• Robert Hooke – curator of experiments at the Royal Society
• Gottfried (von) Leibniz – the German mathematician
• John Flamsteed – the first Astronomer Royal

In each case, he did his level best to destroy the other man. Each of these feuds is discussed in more detail in our book, but in this blogpost I will give a brief summary of his feud with Leibniz.

The feud came about because Newton refused to believe that Leibniz had independently come up with the mathematical idea of calculus. It was a recurring theme throughout Newton’s life that he sincerely believed that he was special. He had deep religious views (some would say extreme religious views). As part of these views, he believed that he had been specially chosen by God to understand things that others would never be able to understand.

Thus, when he heard that Leibniz had developed a mathematics similar to his own ‘theory of fluxions’ (as Newton called it), he naturally assumed that the German had stolen it from him. There then ensued a 30-year dispute between the two men, with Newton very much the aggressor.

Gottfried (von) Leibniz (1646-1716), German mathematician and co-inventor of calculus

It escalated from a dispute to a feud, and culminated in the Royal Society commissioning an ‘official investigation’ to establish propriety for the invention of calculus. When the report came out in 1713 it came out in Newton’s favour. But, by this time Newton was not only President of the Royal Society, but he had secretly authored the entire report. It was anything but impartial. Leibniz died the following year, a broken man from Newton’s relentless attacks.

One should, of course, be able to to admire a person for their work but not admire them in the least for the person that they were. Newton, in my mind, falls very firmly into this category. His contribution to physics is unparalleled, but I don’t think he was the kind of person one would want to know or even come across if one could help it!

Ten Physicists Who Transformed Our Understanding of Reality is available now. Follow this link to order

Derivation of the volume of a sphere – method 2

In this blog here, I derived the expression for the volume of a sphere, $V= \frac{4}{3} \pi r^{3}$ using spherical polar coordinates. There is, however, another way to do it which some people may find easier, and that is to use something called the volume of rotation. Remember, a circle with radius $r$ which is centred on the origin has the equation $x^{2} + y^{2} = r^{2}$, but as I pointed out in my blog on deriving the area of a circle, because we can’t integrate $\sqrt{ \left( r^{2} - x^{2} \right) } \; dx$ we are stuck in trying to use conventional Cartesian coordinates, which is why we had to use polar coordinates instead.

However, if you are not comfortable with spherical polar coordinates to derive the volume of a sphere, there is a trick to get around using them. That is to use the volume of revolution. The idea is quite simple, imagine the strip which has area $dA = y \; dx$ as shown below.

Now, imagine rotating this trip about the x-axis, to produce a disk as shown in the figure below. The surface area of this disk is just the area of a circle with radius $y$, and so its area $dA = \pi y^{2}$. The volume element of the disk is then just $dV = dA \; \cdot dx = \pi y^{2} \; dx$, but as we can write $y^{2} = r^{2} - x^{2}$ this becomes $dV = \pi (r^{2} - x^{2}) \; dx$, which is easy to integrate.

This leads to the total volume $V$ being

$V = \int_{-r}^{+r} (r^{2} - x^{2}) dx = \left[ r^{2}x - \frac{x^{3}}{3} \right]_{-r}^{+r} = (r^{3} - \frac{r^{3}}{3}) - ( -r^{3} + \frac{r^{3}}{3})$

$V = 2r^{3} - \frac{2r^{3}}{3} = \; \boxed{ \; \frac{4}{3} \pi r^{3} }$

just as we had before. QED!

Newton’s equations of motion – revisited

Last week, I showed how one could derive 3 of Newton’s equations of motion. As a colleague of mine pointed out to me on FaceBook, the 3rd equation I showed can, in fact, be derived using algebra from the 1st and 2nd. Also, if you go to the Wikipedia page on the equations of motion, you will find 5 listed, not just the 3 I showed. It turns out that the only two “fundamental” ones are the 1st and 2nd that I showed, the other three (including my 3rd, shown as the 4th in the image below) can be derived from the 1st and 2nd ones.

This is often the case in mathematics, there is more than one way to do something. So, although the method I showed to derive the 3rd equation is perfectly correct, and it also shows how we can rewrite $dv/dt \text{ as } (dv/ds \cdot ds/dt)$, which is a useful technique, I will today show how it can also be derived by algebraically combining equations (1) and (2).

Remember, equations (1) and (2) were

$v = u + at \text{ (Equ. 1)}$

$s = ut + \frac{1}{2} at^{2} \text{ (Equ. 2)}$

The first step is to square equation (1), which gives us

$v^{2} = (u + at)^{2} = u^{2} + 2uat + a^{2}t^{2} \text{ (Equ. 3a)}$

Next, we multiply equation (2) by 2 to get rid of the fraction

$2s = 2ut + at^{2} \text{ (Equ. 3b)}$

We next multiply each term in equation (3b) by $a$ to give

$2as = 2uat + a^{2}t^{2} \text{ (Equ. 3c)}$

Comparing equations (3a) and (3c) we can see that we can substitute the 2nd and 3rd terms of equation (3a) by $2as$ and so we have

$v^{2} = u^{2} + 2as \text{ (Equ. 3)}$

which is our equation (3) from the previous blog. As they say in mathematics, Quod erat domonstradum (QED) 🙂

Derivation of the volume of a sphere

In this blog, I derived the expression for the surface area of a sphere, $A = 4 \pi r^{2}$. In today’s blog, I will derive the expression for the volume of a sphere. Actually, once one has understood how to derive the surface area of a sphere using spherical polar coordinates, deriving the volume is pretty straight forward. It only involves one extra step, and that is to create a volume element $dV$ with the same surface area $dA$ that we had before, but with a thickness $dr$, and to integrate over $r$ in addition to integrating over $\theta \text{ and } \phi$.

As we can see from the figure below, the volume element $dV$ is given by $dr \cdot dA$ where $dA$ is the same surface element we derived before, namely $dA = r^{2} \; \cos \theta d \theta \; d \phi$. So, the expression for our volume element is

$\boxed{ dV = dr \cdot dA = r^{2} \; \cos \theta d \theta \; d \phi \; dr = r^{2} dr \; \cos \theta d \theta \; d \phi }$

The volume element $dV$ is given by $dA \cdot dr = r^{2}dr \; \cos \theta d \theta \; d \phi$

We need to integrate this volume element over all three variables $r, \theta \text{ and } \phi$ so we have

$\text{total volume of a sphere} = \int d V = \int_{0}^{r} r^{2} dr \; \int_{-\pi/2}^{+ \pi/2} \cos \theta d \theta \; \int_{0}^{2 \pi} d \phi$

$\text{total volume of a sphere} = \left[ \frac{ r^{3} }{ 3 } \right]_{0}^{r} \left[ \sin \theta \right]_{-\pi / 2 }^{ + \pi / 2 } \left[ \phi \right]_{0}^{2 \pi} = \frac{ r^{3} }{ 3 } \cdot ( 1 + 1 ) \cdot 2 \pi = \boxed{ \frac{ 4 }{ 3} \pi r^{3} }$

as required.

Derivation of the surface area of a sphere

In this blog, I used polar coordinates to derive the well-known expression for the area of a circle, $A =\pi r^{2}$. In today’s blog, I will go from 2 to 3-dimensions to derive the expression for the surface area of a sphere, which is $A = 4 \pi r^{2}$. To do this, we need to use the 3-dimensional equivalent of polar coordinates, which are called spherical polar coordinates.

Let us imagine drawing a line in 3-D space of length $r$ into the positive part of the $x,y,z$ coordinate system. We will draw this line at an angle $\theta$ above the x-y (horizontal) plane, and at an angle $\phi$ to the y-z (vertical) plane (see the figure below). When we drop a vertical line from our point $(x,y,z)$ onto the x-y plane it has a length $r \cos \theta$, as shown in the figure below.

We then increase the angle $\theta$ by a small amount $d\theta$, and increase the angle $\phi$ by a small amount $d \phi$. As the figure shows, the small surface element $dA$ which is thus created is just $r d\theta$ multiplied by $r \; \cos \theta \; d\phi$, so $dA = r^{2} \; \cos \theta \; d\theta d\phi$.

Using spherical polar coordinates, the area element $dA$ on the surface of a sphere is given by $r^{2} \cos \theta d \theta d\phi$.

.

To find the surface area of the sphere, we need to integrate this area element over the entire surface of the sphere. Therefore, we keep $r \text{ constant}$ and we vary $\theta \text{ and } \phi$. We can go from a $\theta \text{ of } -\pi/2 \text{ to } +\pi/2$ (the negative $z-direction$ to the positive $z-direction$), and from a $\phi \text{ of } 0 \text{ to } 2\pi$ (one complete rotation about the z-axis on the x-y plane), so we have

$\text{ total surface area } = \int dA = r^{2} \int_{-\pi/2}^{+\pi/2} \cos \theta d \theta \int_{0}^{2\pi} d\phi$

$\text{ total surface area } = r^{2} \left[ \sin \theta \right]_{-\pi/2}^{+\pi/2} \cdot \left[ \phi \right]_{0}^{2\pi} = r^{2} \cdot (1-(-1)) \cdot (2\pi) = r^{2} \cdot (2) \cdot (2\pi)$

so, the total surface area of a sphere is

$\boxed{ \text{surface area } = 4\pi r^{2} }$

as required. In a future blog I will use spherical polar coordinates to derive the volume of a sphere, where $r$ will no longer be constant as it is here.

Derivation of the area of a circle

As most people reading this blog probably know, the area of a circle is given by

$\boxed{ \text{ area of a circle} = \pi r^{2} }$

where $\pi$ is the ratio of the diameter of a circle to its circumference, and $r$ is the radius of the circle. But, how would we go about proving this well known formula? To do so, we can use something called integration, which is a branch of calculus.

The area under a curve

The area under any curve between the x-axis, the curve, and the lines $x_{1} \text{ and } x_{2}$ (see the figure below) is given by

$\int_{ x_{1} }^{ x_{2} } f(x) \; dx$

where $f(x)$ is the function (the curve), and $dx$ is an infinitesimally small width. Effectively, what we are doing is summing a series of rectangles of area $ydx$ from the lower limit $x_{1}$ to the upper limit $x_{2}$.

The area under any curve $y=f(x)$ can be found by summing infinitesimally small rectangles, each of height $y$ and of width $dx$ between the lower x-limit $x_{1}$ and the upper x-limit $x_{2}$

The equation of a circle

To do this for a circle, we need to write the equation for a circle. To make things easier we will centre the circle at the origin. In this case, if the circle has a radius $r$ the equation which describes this circle is just

$x^{2} + y^{2} = r^{2}$

Therefore, to find the area of the circle, all we need to do is integrate between $x=0 \text{ and } x=r$, and then multiply the answer by 4 (as we have only found the area of quarter of the circle).

To find the area of a circle, in principle all we need to do is sum the rectangles $ydx$ from $x=0$ to $x=r$ between the x-axis and the circle, then multiply the answer by 4.

The integration to do this is

$\int_{0}^{r} y \; dx \; = \; \int_{0}^{r} \sqrt{ r^{2} - x^{2} } \; dx$

This is not an integral which we can do, so we appear to be stuck!

Changing variables to use “polar coordinates”

Luckily for us, there is a way around this problem. Rather than using x-y co-ordinates (more correctly known as Cartesian co-ordinates), we can change the co-ordinate system to something called polar co-ordinates, and when we do this we get an integral that we can do.

To see how to go from Cartesian to polar co-ordinates, consider the figure below.

The point $(x,y)$ on the circle can be written in terms of the radius $r$ and the angle $\theta$. We can write that $x = r \cos \theta$ and $y = r \sin \theta$

We write the $x \text{ and } y$ coordinates in terms of two new variables $r \text{ and } \theta$, where $r$ is the radius of the circle and $\theta$ is the angle between the line from the centre of the circle to the point and the x-axis. When we do this, we can write that $x = r \cos \theta$ and $y = r \sin \theta$.

Then, to determine the area of the quadrant were $x \text{ and } y$ are positive, we can instead integrate using $r \text{ and } \theta$. To see how we do this, consider the figure below.

To find the area of the circle, we add a series of slices (the shaded region), each with an infinitesimally small angle $d\theta$ and radius $r$ between an angle of $\theta = 0$ and $\theta = \pi/2$, then multiply the answer by 4

One of the reasons the method of using polar coordinates is easier is that we now have only one variable, $\theta$, as the radius $r$ is a constant. When we were using $x \text{ and } y$ as our variables, moving along the circle involved both variables changing, but with polar coordinates only one variable changes, $\theta$.

Instead of finding the area of a rectangle and summing those, we instead consider a slice of the circle, where the angle in the slice is $d\theta$, an infinitesimally small angle, and the radius of each slice is $r$. This is shown by the shaded region in the figure above. We then sum these slices between an angle of $\theta =0$ and $\theta = 90^{\circ}$. But, when using calculus, we do not use degrees, but rather we have to use radians, which as I explained in this blog, are a more natural unit for measuring an angle.

As $d\theta$ becomes infinitesimally small, the slice becomes a triangle, and the area of a triangle is given by $\text{ half the base } \times \text{ the height}$. The height is, of course, just the radius $r$, but what about the base? The base is the length of the arc, which you will recall from the definition of a radian is just $\text{ base } = r \; d\theta$.

The integral we wish to do is therefore

$\int_{0}^{\pi/2} \frac{1}{2} r \times r \; d\theta \; = \; \int_{0}^{\pi/2} \frac{1}{2} r^{2} \; d\theta = \frac{ r^{2} }{ 2 } \int_{0}^{\pi/2} d\theta = \frac{ r^{2} }{ 2 } [ \frac{ \pi }{2} - 0 ] = \frac{ \pi r^{2} }{ 4 }$

But, remember, this is the area of just the positive quadrant, so the area of the whole circle is going to be 4 times this, or

$\boxed { 4 \times \frac{ \pi r^{2} }{ 4 } = \pi r^{2} }$

just as the famous formula states!

When I derived the acceleration of an object moving in a circle (the centripetal acceleration)

$\vec{a} = \frac{ v^{2} }{ |\,\vec{r}\,| } \hat{r}$

one of the comments stated that I could have derived the same formula using polar coordinates. Now that I have introduced polar coordinates, I will in a future blog re-derive the formula using them.