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Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #30 on: December 18, 2010, 02:30:05 PM »
The next article in this series is on calculus. Since a good deal of this stuff is, shall we say, fairly horrific, I'll try to cut out most of the working and equations - but unfortunately some of that can't be avoided.



As is usual with almost every subject in mathematics, it all starts with the Ancient Greeks. This time it was one of those three problems I posted earlier:

i)   How do you find a cube twice the size of another cube?
ii)   How do you trisect an angle?
iii)   How do you construct a square the same area as a given circle?

The third problem is the only one that leads anywhere seriously interesting. Why? Because it's about finding areas contained by curved lines rather than straight lines.

The Greeks quickly found that all three of these problems were impossibly, at least when using their favourite tools of a straight edge, pencil and compasses. So impossible, in fact, that in the 3rd or 4th Century AD, in one Greek play one character insults another by calling him "circle-squarer!", or, in modern speak, "time waster".

Flash forward 2,000 years and we know, of course, that the area of a circle is (pi)*r2, or roughly 3.14 times the radius squared. But, as we also know, the circumference of a circle is given by 2*(pi)*r. These results were known to the ancient Greeks too. Two questions come from this:

  • Why, to move from the formula for the area to the formula for the circumference, have we only needed to move the 2 to the left and down a bit?
  • And, for that matter, why is pi in the area formula at all? After all, it comes specifically from the circumference - pi is how many diameters there are in the circumference - so you might think it a bit odd that it appears again.

These quirks don't end there. For a sphere, the surface area is 4*(pi)*r2, while the volume is (4/3)*(pi)*r3. These too are uncannily similar. It's as if you can get one directly from the other and back again - that, in a way, they are the same problem. Why should this be?

The clue lies in the last article I wrote. We saw there that a right-angled triangle with a fixed hypotenuse and a fixed point, if the other two lengths are allowed to change, traces out a circle:



Now, we remember Pythagoras' Theory, that "the square on the hypotenuse is equal to the sum of the squares on the other two sides in a right-angled triangle" and write this using algebra as:

x2 + y2 = r2 (1)

where r is our hypotenuse or radius, x is the base of the triangle and y is the height.

Why have I done this? The point is that Pythagoras' Theorem in fact describes a circle, and by writing the circle in this way we can deal with the problem a lot more easily, basically because now we can mess around with the formula labelled (1) and get:

y = sqrt(r2 - x2) (2)

Where "sqrt" is the square root. We'll come back to this a bit later.



The fact that the area of a circle is given by (pi)*r2 suggests that it might be handy to look at the graph of y=x2. This can be thought of as "if we choose a number x, then the number y is the square of x". So, for x=2, y=4, and for x=4, y=16, and so on. If we draw this on a graph we get:



Now it would be nice to know a bit about this graph, and in particular we want to know two things: how steep it is at any given point (for curved lines this is obviously changing all the time), and what areas we get if we draw three straight lines like so:



The reason we might want to know these things is that, for example, when a car is travelling along a motorway, then you could track its journey in terms of how fast it's going at any given moment, and draw this on a graph. In this example, the steepness (or, from now on, gradient) of the graph tells you how much the car is accelerating, and the area under the graph tells you how much distance has been travelled. So these are useful problems to solve.

The area problem links to the area of a circle, and will be looked at later. The gradient problem is the one we'll solve now.

You may have looked at "speed-time" graphs in school, and these would probably have straight lines in them. The gradient of the line is found by working out how much the speed increased in, say, one minute. We can write this as:

(3)

For a curved graph, we can get a reasonable value for the gradient at any point by using straight lines like so:



To get a better value for the gradient, we just shrink the size of the line:



until, eventually, our line is so small that it has no length. We can't work out how steep that is, obviously, but if you take several measurements of gradients and get results of, say, 2.005, 2.0033, 2.002, 2.000001, and so on, then it makes sense to say that the gradient at that point is almost certainly 2. How do we prove this result?

Think about what we were doing. We took two points on the graph and drew a line between them. If the two points lie on the graph, then we can work out the coordinates of those points. For the graph y=x2, we can say that the gradient of that line is:



The numbers in subscript are just a way of  saying, for example, “the second value of x”.

This isn't too hepful yet, but we can make things simpler by, instead of using x1 and x2, we could just say that x2 is "x1 plus a little bit more". Call that little bit h, say, and now (the clever bit) think of y2 as (x+h)2. Now our gradient formula becomes:

[(x+h)2 – x2 ]/ h  (4)

Expand the brackets, cancel a few things out, and you get:

Gradient (of the line) = 2x + h


And now all we do is let h get smaller and smaller, and so the line we’re using gets smaller and smaller too – and when h is 0, we are left with:

Gradient (of the curve y=x2) = 2x

Success! not only have we found the gradient of a curved graph, but we've also answered the problem about why the area of a circle and circumference are so similar. It comes from this result.

We can use this method to find the gradient of all sorts of curved lines, from the simple to the ridiculously complicated, so long as we know the equation that matches the line.



In the first part we left off with the formula y = sqrt(r2 - x2). This traces out a circle:



What is the area of this circle? we'll answer that by taking the top semicircle, finding the area of that, then doubling it.

Again, from high school, you might be given a curved shape and told to find the area under it. You can't do this exactly but you can get a good estimate by drawing the shape on squared paper and counting up the squares inside the shape. We'll do a similar sort of thing, only using rectangles, and we'll make sure these rectangles have the same base length:



Obviously the smaller those rectangles are the better the estimate we'll get:



Until, as with the gradient problem, we get rectangles that are so small they have virtually no area.

How to represent this in algebra? Since the area of a rectangle is the base times the height, call the base h, and the height can be found by the value of y at the top-left corner of the rectangle. We then sum up the areas of these rectangles. The next steps are fairly ugly (or at any rate messy) so I will skip them, but the end result is that the area depends, a little surprisingly, only on the shape the curve takes and the two points at either end:



Here the "dx" is the length of the base of the rectangles, the curly line stands for "integral", which just means a sum, and the values a and b are the values of x at either end of the "curve segment" (part of the curve) we're interested in. During the working we would find that this problem is effectively the same as the gradient problem, only backwards! As a result we can start from the other end, which is usually much easier. That's what the third part of the expression means, where a capital Y is a sort of "anti-gradient" - usually we can find this by looking it up in a handy table somewhere. If you can find that then you can find the area.

For our circle, we replace a and b by –r and r (because we want the whole semicircle and that’s the value of x where the circle meets the x-axis)  the result is rather strange-looking (you can work this out but I'd rather not bother):



You may not be able to make much of this, but you should notice that there's a "sin" appeared. This is the same "sin" as in the last article, only again backwards. "sin-1" means that we know the value of the sine ratio and want to find the angle to match this.

For our problem, we're somewhat helped by the far that we've used x=r and x=-r. Squaring these gives r2 both times, so that horrible-looking first bit disappears, thanks goodness. The second bit just becomes:



so now we're left wondering what sin-11 might be. Look again at this diagram:



So for the sine of some angle to be one, we need the opposite side to be equal to the hypotenuse. The only way we can get this is for the right angle itself! Now, we write this angle in radians, and if there are pi radians in 180 degrees (see last article), then 90 degrees is (1/2)*pi radians. So we get sin-11 = (1/2)*pi. Meanwhile,  sin-1-1 is just -pi, and so the area of the semicircle turns out to be (1/2)*pi*r2. Double this to get the area of the circle...

So there you have it. The area of a circle looks similar to the circumference because one gives you the other - they are effectively the same problem.



What does this have to do with ei*(pi)+1=0 ?

Suppose, instead of squaring x, x2, you swapped the 2 and the x round, 2x? What would that look like, and what is its gradient?

Here's what it looks like:



And using the gradient formula (4) above, with all letters meaning the same things as earlier but swapping a few things around, we find that:

Gradient = [(2h – 1)/h] * 2x

Using a basic power law. If we now let h get smaller and smaller, this eventually becomes:

Gradient = 0.693… * 2x

So the gradient of 2x is almost the same as what we started with. What’s more, if you do the same thing with 3x then you get:

Gradient = 1.098…*3x

From this it makes sense to think that there’s a number between 2 and 3 where we’d just have got the result that the gradient is the same as what we started with. And there is, this number is called e.

What’s e for? Well, without going into details, e appears in finance, probability, calculus, radioactivity, and indeed anything involving exponential changes. So it turns out to be a very useful number indeed.



We’ll see in the final article how e even turns up in trigonometry, despite its only popping up as a sort of fluke.

« Last Edit: January 19, 2011, 02:27:58 PM by BFM_three60 »
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Offline Lucky

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Re: three60's Short introduction to... corner
« Reply #31 on: December 18, 2010, 04:48:10 PM »
 :Way Cool:


Offline BFM_Kiwi

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Re: three60's Short introduction to... corner
« Reply #32 on: December 18, 2010, 05:45:15 PM »

*applause*

Well done three60!



(thanks TrkKing)


Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #33 on: December 19, 2010, 03:42:08 AM »
Forgot to mensh - there are two more articles on their way in this "ei*pi+1 = 0" series, the first of these about "i", and the second about the whole equation.
Check out my Short introduction... corner and my "Historical figures who should perhaps be better-known" thread!!

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Offline BFM_Kiwi

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Re: three60's Short introduction to... corner
« Reply #34 on: December 19, 2010, 10:40:42 AM »

*applause dies out abruptly*




(thanks TrkKing)


Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #35 on: December 19, 2010, 02:56:41 PM »
Hmm... I think the next article will be more likely easier to follow though... fewer equations for a start.
Check out my Short introduction... corner and my "Historical figures who should perhaps be better-known" thread!!

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Offline BFM_Edison

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Re: three60's Short introduction to... corner
« Reply #36 on: December 19, 2010, 03:03:53 PM »
Methinks the forum should support LaTeX.
52.87   60.07   46.40   72.73   68.23   55.10   98.27   84.73

Offline BFM_Mil

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Re: three60's Short introduction to... corner
« Reply #37 on: December 19, 2010, 03:51:37 PM »
Why going to school when you have three60 here? ;D

Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #38 on: December 29, 2010, 08:34:32 AM »
I'm currently busy grinding through the diagrams for the next article - on the origins of i and the mysterious nature of numbers themselves...

I think I may have it ready this week. Eyes peeled, please!
« Last Edit: January 19, 2011, 02:28:28 PM by BFM_three60 »
Check out my Short introduction... corner and my "Historical figures who should perhaps be better-known" thread!!

Exciting videos: 1.1 / 1.2 / 2 / 3 / 4 / 5 / 6



               

Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #39 on: December 30, 2010, 02:50:50 PM »
Last time I explored the origins of calculus, and also cleared up the meaning of “e” in the equation ei*pi+1 = 0. Now it’s the turn of  “i” to be explained.

The number i is usually introduced in terms of quadratic equations, though I’m going to be more historical about it. Having said that, let’s just explain what a quadratic equation is, what it is used for, and how it is solved. I’ll also look at something rather curious that you might not have realised was even worth noticing.

“Sqrt” means “square root”, throughout.



Let’s say that you have a rectangle-shaped field with an area of 600 square metres, and you are told that one side is 10 metres longer than the other. How long is the shorter side?

The usual first step is to draw a diagram:



When you look at the problem like this, hopefully it becomes a lot clearer. We’ll call the shorter side “x” and the longer side “x+10”, and because the area is found be multiplying the lengths of the two sides together:

x(x+10) = 600.

This is a quadratic equation, which just means that there’s an x2 term in the equation, but no higher power of x than that (such as x3). At this point you have to expand the brackets (parentheses for my US readers), take everything to one side, and then factorise the equation, but I’ll skip straight to the answer since you can learn all about the rest of that in school. We find that x is 20 metres, and if you like you can try this.

For reasons that will be clear shortly, I’ll also introduce the so-called “quadratic formula”. Earlier I went from a real problem straight to how it would be written in algebra. So let’s take the most general quadratic equation possible:

ax2 + bx + c = 0

This means that we are trying to find x, while the new letters a, b and c are given to us in the question. So you might see something like “Solve for x if 3x2 + 10x + 3 = 0”. a has been replaced by 3, b by 10 and c by 3.

The quadratic formula says that if you see this sort of equation, you can find x by putting in the values of a, b and c into the following equation:



Don’t worry about the  thing for now, though it just means “plus or minus”. Also the thick line is a division, and the squiggly line is a square root, just to clarify. The main thing to notice is that, with a little bit of work (and a calculator), you can solve any quadratic equation you can think of.

Except for these ones:



For all of those equations, the quadratic formula goes wrong. Let’s take the simplest of those, x2+1=0, and we’ll bang this into the quadratic equation:

x = (1/2)*sqrt(-4) = sqrt(-1)

Which doesn’t exist. Alternatively, if we look at the graph of y=x[sup2][/sup] from last time:



The curve doesn’t go below the x-axis, which means that there’s no value of x that, when squared, will give a negative number.*

So that’s quadratic equations for you. They solve problems mainly to do with areas, but it turns out that they crop up in other places.

If we return to the problem we started with, and go through the steps, the last step before the solution turns out to be:

(x+30)(x-20)=0

Now this is true when at least one of the brackets works out to be 0, and the “x=20” solution comes from the second of these. But if you look at the first bracket, you’ll see that there is another possible value of x, and that is x= -30. No trouble there then… except that the problem we were thinking about was that of a field, and there’s no way you can have a field with a side that is –30 metres long. So… what is that solution all about then? What does a negative number even mean?

Think about that, dear reader, while I move quickly into the world of cubic equations…

*For those of you who already know what i is, and are wondering how I can possibly get there when I just said that it doesn’t exist, read on.



So much for problems with areas, but what about volumes?

Let’s take a box that has a volume of 168 cubic metres. The longest edge is 10 metres longer than the shortest edge, and 5 metres longer than the third edge. What is the length of the shortest edge?

Again, this becomes clearer with a diagram:



And, since the volume is given by multiplying the lengths of the three edges together, we find that:

x(x+5)(x+10) = 168

With a bit of guesswork you’ll find that x is 2 metres. If we expand the brackets (multiply everything out) we find that:

x3+15x2+50x = 168

This is a cubic equation because now the highest power of x is the x3 term. Of course, to solve problems involving volumes we might want to study these equations too, but they turn out to be more complicated than the quadratic equations were.

Why else might we need cubic equations? Let’s return once again to those three “unsolvable” problems that the Greeks were worried about:


i)   How do you find a cube twice the size of another cube?
ii)   How do you trisect an angle?
iii)   How do you construct a square the same area as a given circle?

To be clear about this, these problems can only be attempted when using a pencil, a straight edge and compasses. Last time we looked at the problem of the circle, but what about the other two?

The problem of the cube is obviously about volumes, but doesn’t go very far: say that we had a cube with all sides a metre long. Then to double the volume of the cube we’re looking for a cube with sides of some length x where:

x3 = 2

Then we just take the cube root of both sides and find that x is the cube root of 2. The only reason the Greeks couldn’t solve this problem is that there’s no way, using just a ruler and compasses, that you can draw this length. So let’s not worry about that one.

Instead, let’s look at the problem of trisecting an angle. This problem is about dividing any angle into three equal parts:



Let’s call the angle we’re given 3x, and the angle we’re looking to construct (draw using compasses, ruler etc., but NOT a protractor) is x. Note that we don’t even know what size the angle 3x is.

How is this helpful? Let’s draw a few more lines on the diagram…



Within this mess you should see a few things, most importantly the fact that triangles have appeared. This means that the problem of trisecting an angle is also a problem involving triangles, and those things we know about. Because of this link we can re-write the problem using trigonometry, and specifically by talking about the cosine (cos ratio from my first article) of the angle. Why? Well, with a few rules, such as “the addition formula for the cosine ratio" (whatever that is) we’ll eventually find that:

cos(3x) = 4*(cos(x))3 – 3*cos(x)

This ugly-looking thing can be made a little simpler if, we write, say, q for cos(3x) and t for cos(x):

4t3 –3t = q

By this point I should say that we’ve left behind the world of compasses – which is why the Greeks couldn’t solve this problem either. In the meantime, if we even wanted to solve this equation then q would have been given to us, so we might see “find t if 4t3 –3t – 1 = 0”. We aren’t going to solve this, but you can see that understanding cubic equations might help us understand this problem too.

In the same way that the quadratic equation has a “quadratic formula”, the type of cubic equation that looks like:

x3 + px + q = 0

(in other words the sort of equation that we just saw pop out of trisecting an angle) can be solved with its own formula. Again, p and q are given and you’re trying to find x. This formula, though, is a little more complicated…



Ouch. However, again, the main thing to see is that you can solve this if you wanted to.



Let’s think about a simple question: “Are there any numbers who are their own cubes?” This question can also be written:

Find x if x3 = x

We can solve this first by just looking at it, and you should see that x=1 works, and so do x=0 and x=-1. But we could also use the formula for solving cubic equations above…



But here’s a problem, what’s that square root of –1 doing there? I just said that there’s no such thing, after all.

To make sure that we haven’t messed up it might be worth while checking that this hideous thing fits in with the equation we started with. We were looking at x3 = x, so if we cube this we should get what we started with (you’ll have to trust my working here, I’m afraid…):



That looks fairly promising since the same ugly expression that we started with is back on the right. There’s just the small matter of that sqrt(-1)+1/sqrt(-1) bit. But wait! –



So that first bit just disappears and (when you cancel out the 3's) we’re left with what we started with! This means that this non-existent number is actually a solution to x3 = x, and what is more it must be one of 1, 0 or –1, because there are never any more than 3 solutions to a cubic equation. So how can we have got this result, a number that doesn’t exist working out to be 1 or 0 or –1, numbers that we’re happy to work with?

Let’s backtrack for a minute, though, and think about the poser earlier: what does a negative number mean? After all, it’s no use when talking about lengths, or areas, or the like. For that matter, even when they do appear usefully such as when dealing with money, what does -$5 look like?

For an even more confusing thought, think about the following joke:

“Three people, a biologist, a physicist and a mathematician, are made to watch a room with only one entrance. They are told that the room is empty. Two people walk in, and three people walk out. How did they explain this?

The biologist says, “Maybe they reproduced?”
The physicist says, “Perhaps I counted wrongly.”
The mathematician says, “Send one more person into the room and it will be empty again.”

:LOL: :LOL: :LOL: :LOL: :siderofl: :siderofl: :siderofl: :siderofl:

When you have finished laughing your head off, think about the mathematician’s answer. If you write what happened as a sum:

0 (the empty room) + 2 people – 3 people = -1 person.

So his answer is to add one more person to make the room empty again, because at the moment there are –1 people in the room. Which is a silly thing to say, really.

The ultimate point of this is to make you realise something: Negative numbers, that many people are happy to work with daily without a second thought, don’t really exist. They’re just useful to help us explain what happens when shops (or banks) make a loss, or how you slow down, or any time you’re reducing something. They’re just a tool, in other words, to help us make sense of the world.

It’s not even that silly to doubt the existence or use of negative numbers. In fact, for many years mathematicians had arguments about whether they were useful or not, whether they existed or not. You might see people talking about (to go back that that field problem earlier) the “real solution” (20 metres) and the “imaginary solution” (-30 metres). Then, later, when that cubic formula turned up, the person who first wrote it up still didn’t even accept negative numbers as being worth thinking about. You can imagine how hard it was to contemplate square roots of them!

Eventually, out of sheer curiosity, mathematicians started looking into this quirk that was the square root of –1. Where this took them we’ll see in the next article.

Does the square root of –1 exist? Yes, it exists in the same way that negative numbers do. That means that it exists only if you want it to and if it’s useful. Which in turn means that any number exists if you can find a use for it.

This all goes back to another earlier topic: axioms. An axiom is something that is assumed to be true without bothering to prove it. In High School it’s assumed, for example, that there’s no such thing as the square root of –1. This is true. In more advanced maths, it’s assumed that there is such a number. This allows you to do a lot more, as we shall see.



So the number i is just the letter we use to represent the square root of –1. I stands for “imaginary”, by the way.

We’ll be thinking about what we might be able to do with this new “number” in the next article.
« Last Edit: January 19, 2011, 02:36:49 PM by BFM_three60 »
Check out my Short introduction... corner and my "Historical figures who should perhaps be better-known" thread!!

Exciting videos: 1.1 / 1.2 / 2 / 3 / 4 / 5 / 6



               

Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #40 on: January 10, 2011, 09:28:44 AM »
The next article is still not ready yet, I expect it to be finished within a fortnight though.

Any thoughts on the last one?
Check out my Short introduction... corner and my "Historical figures who should perhaps be better-known" thread!!

Exciting videos: 1.1 / 1.2 / 2 / 3 / 4 / 5 / 6



               

Offline BFM_Ben1

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Re: three60's Short introduction to... corner
« Reply #41 on: March 07, 2011, 02:58:21 PM »
Something sciency would. Be interesting, maybe how electrons shufflefrom shell to shell during fireworks (or something like that...)

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Offline BFM_three60

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Re: three60's Short introduction to... corner
« Reply #42 on: March 07, 2011, 03:53:58 PM »
They do?

Check out my Short introduction... corner and my "Historical figures who should perhaps be better-known" thread!!

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Offline MrT©

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Re: three60's Short introduction to... corner
« Reply #43 on: April 06, 2011, 01:00:13 PM »
Ah huh, i dont really understand.
But i get this bit:

"Does the square root of –1 exist? Yes, it exists in the same way that negative numbers do."

:D

Offline Trael

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Re: three60's Short introduction to... corner
« Reply #44 on: April 08, 2011, 01:58:46 PM »
Maybe the next could be something about how an audio signal travels from the input ex. the computer to the output device ex. the speaker?

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