Welcome to another Mathologer video. Have a look at this. First squish this square down by a factor
of 2 and then stretch the resulting rectangle horizontally, also by a factor of 2. Is this rectangle larger or smaller than the
square we started with? Easy, right? The answer is “neither”. They have the same area. First, squishing by a factor of 2 gives a
shape of half the original area. Then stretching by the same factor doubles
that “one half” and gets us back to where we started from. Of course, the same is true for any shape
and any factor. For example, take this cute cat shape and
choose the factor to be three. Then squishing by a factor of 3 followed by
stretching by the same factor gives this very different looking cat monster that has exactly
the same area as the cute cat we started with. Here’s an interesting question for you:
Do anti-shapeshifters exist? Is there a shape that is immune to these squish-and-stretch
transformations? Immune in the sense that the shape is exactly the same before and after
applying the transformation. Is there such a shape? And I don’t mean trivial stuff like a point
or a line. Obviously those are immune. No, a shape that actually has a non-zero area. Believe it or not, anti-shapeshifters do exist. AND these anti-shapeshifters form the natural,
visual heart of logarithms, hyperbolic functions, the theory of relativity and much more. And now that I’ve got you hooked, I am sure
that you can’t wait for me to dazzle you with all the amazing anti-shapeshifter maths
that nobody ever mentions in school and university :) Happy to oblige. Okay, so prepare for the grand entrance of
the prototype anti-shapeshifter! For this I need a Rubik’s cube and a drill
:) Looks familiar? Yes, that’s our good old friend 1/x. Nice grand entrance, hmm. I bet you didn’t know that 1/x was hiding
in a spinning cube. I also bet you didn’t know that 1/x leads
a second life in anti-shapeshifter land :) To see 1/x in action in anti-shapeshifter land
let’s focus on its branch in the first quadrant. I claim that this orange
shape is an anti-shapeshifter, that this shape is immune
to any squish-and-stretch transformation. Let’s do a quick spot check. Let’s squish and then stretch by a factor
of 2. But first let’s checkerboard the shape to
also see what’s going on within the shape as it transforms. Okay, squish down.. and stretch to the right. Very satisfying, isn’t it. That our 1/x shape stays
unchanged under squish-and-stretch transformations, no matter what factor we use is easy to check algebraically. Really just a one-liner. Maybe one of you year 10 students watching
can do this in the comments as homework. So both the overall shape and the area of this figure stay unchanged
whenever we squish-and-stretch. But what IS the area of this shape. Well there is a really nifty way to figure
this out. Let me show you. Focus on this spike here. There the blue spike spiking off to infinity
to the right. Okay squish-and-stretch by a factor of 2. Squish ... Stretch. As always, the new blue spike has the same
area as the original spike. But, at the same time, the new spike is contained
within the original spike. There. Curious :) Can you already see where this
is going? Well, we can express the fact that one spike
is contained in the other like this: original blue area equals the second blue area plus
the the excess green area. green + blue equals blue. Cancel the blue on both sides and you get. Hmm, we conclude that the green shape has
zero area. Of course this is nonsense. How can we avoid this nonsense? Well, let’s have another look at our equation. Now, IF the blue is an ordinary number, then
we can definitely cancel, and the result is nonsense. So the blue must be an extraordinary number,
right? Extraordinary number? Hmm, what comes to mind there? Yes, INFINITY. Right? The only way the nonsense can be avoided is
if the blue spike has infinite area. Of course this implies that the area of the
entire shapeshifter is infinite, too. And now one of the most
iconic truths of mathematics is lurking just around the corner. And so let’s quickly chase down that iconic
truth before we continue with the main arc of today’s story. Okay so our spike has infinite area. Zoom in. Now, have a look at this infinite staircase. What is ITS area? Obvious, right? Infinite, too, since the staircase contains
the infinite spike. On the other hand, the area of the staircase
is also the sum of the areas of its infinitely many rectangular steps. And what are those areas? Also easy. Base times height. All the steps are one unit wide and so their
areas are equal to their heights. And what are those heights? Well, our staircase is constructed to align
with 1/x. And so the heights are 1, 1/2, 1/3, and so
on. Now, as we already figured out, the areas
of the steps are the same as their heights. And so this shows that the super-famous infinite
sum 1 plus 1/2 plus 1/3 and so on sums to infinity. As I said, one of the most iconic truths in
mathematics. Super neat proof don’t you think? Those of you who know a bit of calculus may
have seen this proof before. But I hope you all appreciate that what’s
extra nifty about the way I’ve presented this proof is that it skips all the calculus
:) Okay, let’s continue with our main story. Let’s focus on the area under the curve
between 1 and some number x. Let’s call this area A(x). Guess what the A stands for? Anti-shapeshifter? Of course not :) “A” stands for area :) Oookay. So this is A(3) and this is A(2). Now, prepare yourself for something super
pretty. I am going to calculate A(2) plus A(3) a la
shapeshifter. What does that mean? Okay, here we go. First A(2). At x=2 our 1/x is equal to 1/2. Move in the green shape. We adjust the height of the green shape by
squishing by a factor of 2. There. And we compensate areawise by also stretching
by 2. Now shift the green like this. A perfect fit because of the anti-shapeshifter
which shows that A(2)+A(3) equals A(6). Let’s do a couple more experiments with
this visual adding a la shapeshifter. First off, let’s switch the order, so we
begin with A(3). So at 3 our 1/x is equal to 1/3. Squish-and-stretch by 3. Squish and stretch. And shift to align. And, as expected, we get A(6) again. Although we know why this works it still looks
very magical, doesn’t it? Next, A(2) plus A(2). Let’s do this on autopilot. A(2)+A(2) is A(4). One more A(3) plus A(3). A(3) plus A(3) is A(9). Great. So, what’s the general rule? Can you guess? Pretty obvious, right? A of x + A of y = A of x times y. Sums somehow turn into products. I leave it to the keen among you to check
that this really always works. Again, very easy and again, to make sure that
you really know what is going on, please write up your proof in the comments. Anyway, sums turning into products, that sounds
familiar doesn’t it. Where have you encountered something like
that before? Yes, logarithms. And of course it’s exactly this so-called product formula for logarithms
that made logarithms so so important in the old days, since, reading
from right to left, this formula allowed us to translate the painful task of multiplying
two numbers into the much simpler task of adding the logarithms of the two numbers. Anyway looks like our mystery area function
may be a logarithm in disguise and what we’ve done so far amounts to a very beautiful visual
way to introduce logarithms. And, so, is our area function a logarithm? Well, let’s see, apart from this product
formula, there is also the quotient formula for logs, that one here. And it’s also easy to visually see that
the corresponding formula works for our area function as well. Looking good. What about the power formula for logs? That one there. Well that also works for our area function. Let me just show you quickly WHY this works
in the simple case of N=2. There 2 times A(x) equals A of x squared. Here is the simple proof. 2 times A(x) that’s just A(x) + A(x). But then our product formula tells us that
this is equal to A of x times x. And since X times X equal X squared we are
done. We can now iterate this argument to show that
the power formula is true for all positive integers. 3, 4, 5, and so on. But now challenge for the keen among you:
How do you show that the power formula works for ALL integers not just the positive ones? And then what about all the rational numbers? And all the real numbers? Leave your thoughts in the comments. Anyway, with enough work we can convince ourselves
that our area function works like a logarithm and quacks like a logarithm. So, presumably it is a logarithm. But then we have a puzzle. What puzzle? Well, there is not just one logarithm, there
are infinitely many, each of them pinned down by their base. So if our area function is a logarithm what
is its base? Let me show you a super nice way of answering
this question. Okay, so apart from the product, quotient and
and power formulas what other famous properties do logarithms have? Well there are some special values. Okay let’s see how rusty your logs are :) Log
of 1 is equal to what? Well, any log at 1 is equal to 0 no matter
what the base is. How about A(1)? Well, that’s also 0. Pretty obvious. No area and so A(1)=0. Great. Now what’s that other special value? Remember? Almost? Ok, I’ll help you. The log to the base of b at b is always 1. Confused? Take my word for it, you once knew this :) Anyway,
IF our area function is really a logarithm that should give us a way to figure out what
its base is. Right, we simply have to figure out for which
x our area function spits out a value of 1. Okay, let’s first convince ourselves that
there actually is such a special value. That’s easy. We know that A(x) is 0 when x = 1. Now as we push x to infinity there is more
and more area and so A(x) gets larger and larger. In fact because the blue spike has infinite
area we know that A(x) explodes to infinity. But then, on the way from 0 to infinity our
area function must come across a value of x where A(x) = 1. Cool. Again, it’s that special value that we are
hunting. Okay we want area 1. Well here then, for easy reference, is a unit
square, which has area 1. Now let’s have a look at A(x) for different
values of x. Let’s start with x=2. Is 2 our special number? Well, obviously not since A(2) is smaller
than 1. There. The orange A(2) fits into the unit square
like this and so the orange area is smaller than 1. This means that 2 is smaller than the special
number we are looking for. Okay, so what about A(3)? There. Well let’s see. Aha, we get an overlap and so A(3) is too
big. So what that means is that our special value
is somewhere between 2 and 3. Nice visual argument, don’t you think? And now comes an even prettier argument that
even the jaded know-it-all calculus demons among you will like, guaranteed. Okay, let’s step back and put x halfways
between 1 and 2. There x is equal to 1+1/2. Shift the orange into the unit square. Here comes the first niftiness. Fit a second copy of the orange into the square. Like this. There, now the total orange area is 2 times
A(1+1/2). But now, remember the power rule? 2 times A of stuff equals A of stuff squared. And what’s stuff squared in this case? Well 1+1/2 squared that’s 2.25.What does
that mean? Well together the orange bits don’t fill
the unit square and so 2.25 is still less than the special value we are looking for. But we’ve definitely made some progress,
right? We now know that the special value is between
2.25 and 3. What’s next? Can you guess? Well, instead of 1+1/2 we are now going to
go with 1+1/3. Then we can snugly fit three copies of the
orange into the unit square. And in the formula all 2s turns into 3s. Evaluate (1+1/3)^3. That’s 2.37 and a bit. Progress again :) Neat. What comes after 3? Well 4 of course, then 5, 6, 7, etc. And hopefully we’ll get closer and closer
to our special value. Let me just show you what things look like
for 9. There, As you can see most of the unit square
is filled at this point and it is clear that as we go 10, 11, 12 to infinity we’ll completely
fill the square. And so what’s that special number that we
are going for here? Well, after calculating 1+1/N to the power
N for some really large N, we find that the special number is approximately … 2.71828. And a lot of you will recognise this number,
it’s the super famous number e, the base of the natural logarithm. Of course, for those of you who know some
calculus, bells will have been ringing for quite a while now. Right? Up there, that’s how e is commonly defined
in terms of that crazy limit. And on closer inspection our area function
really turns out to be the natural logarithm, the logarithm to the base e. And all you calculus demons will know that
the relationship between 1/x and the area function is expressed in terms of this basic
integral. One natural question :) pun intended-natural
:( Apart from the natural logarithm, what about all the other logarithms? Where are they hiding? The answer to this natural :) question highlights another must-know property
of our anti-shapeshifter. So, let’s also quickly chase down those
other logarithms and then finish off with some really crazy stuff like hyperbolic rotations
and hyperbolic functions that naturally fit in with what we’ve been talking about today. Okay, so where are all those other logarithms
hiding? Well, remember, it all started with the 1/x
anti-shapeshifter over there. What was the first thing we did? We first checkerboarded and then squished
by a factor of 2. Let’s do that again but this time without
the checkerboarding. Here we go, squiiiiish. Hmm, that somehow didn’t look right, did
it? But what exactly seemed off here? Let’s watch this squishing down action one
more time. Squiiish. Although I definitely squished down, that’s
not what it looked like, right? It looked like I was contracting the whole
shape uniformly towards the origin, the bottom left corner of the shape. There, again. And that’s it, that’s that other remarkable
property of our anti-shapeshifter. As far as the overall shape is concerned,
squishing down by a factor of 2 is the same as overall scaling down by a factor of, well
what? As you can easily check the factor is root
2. And then stretching horizontally by a factor
of 2 is the same as overall scaling up, again by the same factor of root 2. Aha, so what this means for our shapeshifter
is that stretching is the reverse of squishing :) Nice. Again easy to check with just one line of
algebra. What this also means is that the intermediate
shapes that we get at the end of squishing are just scaled versions of
our original anti-shapeshifter. But since squishing and stretching just moves
us between these scaled shapes, it is also clear that these scaled shapes are themselves
anti-shapeshifters. Interesting :) That means that we could have
started with ANY of these anti-shapeshifters and derived things in exactly the same way
as we just did. And that works all the way up to the point
of determining the base. For example, let’s have another look at
the anti-shapeshifter that arises when we squish by a factor of 2. Factor 2 squishing means that our new area
function is just 1/2 times the natural logarithm. And now, since this function is a logarithm
itself, we can find its base as before. The base is the number x for which this function
takes on the value 1. And so solving for x, we get this. And so our anti-shapeshifter corresponds to
the logarithm to the base e squared. And, in general the different scaled shapes
correspond to the different logarithms. Great stuff, don’t you think? And, in fact, following the path that we tracked
today it’s possible to quickly DEFINE and create all the incredibly important number
e and natural logarithm related mathematics from scratch. Right, just to really drive home this very
important point, let me do a quick recap. Let’s pretend we don’t know any e-related
maths. Start with our 1/x anti-shapeshifter, show
that its area is infinite, show that the area function satisfies the product, quotient and
power rule. Now, remember, we are pretending that we don’t
know logarithms yet and we are trying to define all things logarithm and e from scratch. And so, at this point, we give the area function
a name. We call it the natural logarithm which is
usually written ln(x) which is short for the Latin logarithmus naturalis. Now we find the special number that makes
the natural logarithm 1 and call it e. We define e^x to be the inverse function of
the natural logarithm and maybe also introduce all the other logarithms and exponential functions
at this point. Now, from the way we defined the natural logarithm
it is clear that we have this identity. From this it follows, give or take the fundamental
theorem of calculus, that the derivative of the natural logarithm is 1/x. From this it follows, give or take the fundamental
theorem of calculus, that the derivative of the natural logarithm is 1/x. And then also that the derivative of e^x is
e^x. At which point we’ve got pretty much all
the basic super important e related mathematics under control. I hope you all agree with me that this visual
approach is super slick :) Also, there is a lot of other beautiful e-related stuff lurking
just around the corner. In regular calculus courses this extra material
is often covered badly because it’s hard to motivate within the standard framework
of introducing e-related mathematics. And so, let me finish off by also by chasing
down some more of these value-adding e-related beauties. Have a look at this. I’ve arranged for all those wedges over
there to have the same area. Okay. Now, any squish-and-stretch combo will transform
all of these wedges into other wedges of the exactly the same area. Right? If we do this continuously we get this. So at all times all these wedges have the
same area. Does this remind you of something? No? How about a circle turning? There all those circle wedges also have the
same area at all times (obviously). Very similar, right? Now the observation that the the graph of
1/x exhibits circle-like behaviour inspired the 18th-century mathematicians Vincenzo Riccati
and Johann Heinrich Lambert to investigate hyperbolic counterparts of
the trigonometric functions. And so let me finish off by showing you a
pretty natural anti-shapeshifter powered way to go from the familiar trigonometric functions
sine and cosine to their hyperbolic counterparts, hyperbolic sine and hyperbolic cosine. Okay, for the trigonometric functions, also
called the circular functions, we start out with the unit circle and an angle alpha. There. Then the sine and cosine of this angle are
these lengths. In calculus the angle is usually captured
by the length of this arc. Of course that length is just the angle in
radians. Now expressing an angle by the length of the
sector’s arc that’s an interesting idea in itself. But you are all familiar with that, right? BUT what you may not be familiar with is the
idea of expressing the angle in terms of the sector’s AREA. When you think about that for a moment you’ll
see that this is just as natural as going via the arc length. Right? Different angles correspond to different areas
and so we can also express angles in terms of areas. What’s also nice is that there is a simple
visual connection between that arc angle and the area angle. And here it is. Scale the circle and everything attached to
it by the square root of 2. There, everything got “timesed” by root
2 and the circle now touches the hyperbola at a point. Now you can easily check that the value of
the area of the scaled sector is equal to the original arc length. Again, the green area of this expanded sector
is equal to the angle in radians, the length of the original arc. Now, with the circle and the hyperbola touching
we are almost where we want to be. Okay, the point where the two curves touch,
that’s the super special point of the 1/x hyperbola. And so what we want to do now is to align
everything in sight with this point. And we do this by rotating everything through
45 degrees. Now extend the circular sector to a hyperbolic
sector and define the two hyperbolic trigonometric functions as lengths just like the trig functions
were defined in the circle. The names of these functions are obvious,
just the trig names with an added h. But how should we SAY these hyperbolic functions. Well spelling it out, how about cosh and sin-huh. Hmm. cosh is pretty good, but sin-huh doesn’t
sound that great does it? And so the hyperbolic sin function is often
called SHINE: we cheat by putting the h with the s. Shine, I like it! :) Anyway, names aside, all pretty natural,
right? Again, here the angle alpha is the area of
the green hyperbolic sector. And now pretty much any property of sine or
cosine has a counterpart for these hyperbolic trigonometric functions. For example, take Euler’s formula. The hyperbolic counterpart of Euler’s formula
is super pretty. Just get rid of the i s. Whoa! The exponential function is just the sum of
cosh and shine. And so the exponential function itself also
has a nice geometrical interpretation in term of the sum of those two lengths in our diagram. I’ll give an animated proof for this fact
at the end of this video. Actually, for all this to make sense we have
to declare that when the angle flips to the other side both the angle and shine are to
be interpreted as negative numbers. But that doesn’t come as a surprise, right? After all, we do the same sort of thing with
sine and cosine. Anyway with this convention in place, let’s
replace alpha by minus alpha. On the right side things simplify. Since cosh(-alpha) equals cosh(alpha). And shine(-alpha) is equal to minus shine(alpha)
we get this second pretty identity up there. Now if we add the two identities the plus
and minus shines on the right cancel and we get the famous way of expressing cosh in terms
of the exponential function. On the other hand, when we subtract the two
equations we can solve for shine and get this. A minus instead of a plus on the left. Very nice. And also, with what I’ve shown you it’s
now straightforward to prove the hyperbolic counterparts of all the other basic trigonometric
identities, like, for example, the counterpart of the famous sine squared plus cosine squared
= 1 identity. This one here. Great stuff, right? And why is all this important? Well, hyperbolic functions pop up all over
the place in physics and engineering when it comes to describing natural phenomena. For example, the graph of cosh describes the
shape of hanging chains and cables. And those squish and stretch hyperbolic rotations
lead a second life as Lorentz transformations in the theory of special relativity. To see this in action check out the super
nice Minute Physics video that features that mechanical Lorentz transformation simulator
over there. A couple more hours worth of amazing stuff
to talk about. But let’s stop here for today. I hope you enjoyed this Mathologer video. Please let me know what worked and what did
not work for you and ask all your questions in the comments. Until next time :)
