OK. Lets have a go at imaginary numbers. These are NOT straightforwards. The name is also very annoying because, they are very real indeed, and I will try to satisfy you that that is the case in due course.
First of all what the hell are they? We have already seen four different types of numbers. We have seen whole numbers such as one, two, three, ten and so on. These are obvious in our every day world. How many pens are on the desk? Three. How many cows are in that field? Twenty. They are used to count separate and distinct objects. Looked at formally, the \(x\) in the following equation is a whole number:
\[x+5=7\]
We can see without too much effort that \(x\) is two. Two is obviously a whole number so we are quite happy with that as a value of \(x\) that makes the equation work.
When we looked at the opposite of addition, we found a different type of number, a negative number. We agreed that saying you had negative five cows in a field did not make sense. The concept was so odd that for a long time mathematicians refused to accept that these numbers really existed. Nowadays we are perfectly happy with the concept (unless your bank account is very, very, negative, in which case you will be perfectly unhappy with the concept).
So what about a slightly different equation which works with a negative value of x?
\[x+5=2\]
To work, we need to set \(x\) equal to negative three. \(x\) cannot be a positive number, because there are no positive numbers that are five smaller than two. Again, we are quite happy with this, but weirdly even just a few hundred years ago the best mathematicians in the world would have said that there was NO ANSWER to that problem.
OK, next we considered fractions. We agreed that while you could have half a cow, it would not be very pleasant to look at. More fundamentally we realised that fractions are ratios between two whole numbers. So if I have five apples, and my friend has ten apples then I have half as many apples as my friend. That is a statement about the ratio of our apple collections to each other. What would this look like stated as an algebra question?
\[3\cdot x=2\]
This is just a little bit trickier because it is saying three what's are two? Or, what is two divided by three? We are quite happy with the answer two thirds. And in general people have always been comfortable with this idea. After all, you are just comparing two whole numbers.
OK, moving on we then learned about irrational numbers. These are numbers that cannot be written as one whole number divided by another whole number. We satisfied ourselves that the number which you multiply by itself to get two is one of these numbers. The equation which has one of these as an answer looks like this:
\[x^2-2=0\]
For the hard of thinking, you add two to both sides and then take the square root of both sides getting \(x\) equal to the square root of two. The ancient greeks really did not like this. They felt that all numbers should be rational, and they were really disturbed to find out that was not the case. it's a bit more abstract today, but we are generally not bothered by the idea that there are some numbers which would just go on for ever if you tried to write them out.
Now then. What is the answer to the following puzzle. What number, if you multiply it by itself, and then add one, gives zero? Or algebraically:
\[x^2+1=0\]
It is very similar to the square root of two equation just above isn't it? So what do we do? We subtract one from each side, getting:
\[x^2=-1\]
And we then take the square root of each side:
\[x=\sqrt{-1}\]
So the answer is the number, that if you multiply it by itself, makes negative one. OK, so what would that be then? One multiplied by itself is one, so is negative one multiplied by itself negative one? We need to think about multiplying negative things to get an answer to that.
We said that multiplying is just a special type of addition. So that three multiplied by four is:
\[3+3+3+3=12\]
Notice that there are four threes there. So what would negative three multiplied by four look like? Well, we would just add together four negative threes. That would look like this:
\[(-3)+(-3)+(-3)+(-3)=-12\]
Remember that adding a negative is the same as subtracting a positive. And also remember that you do stuff in brackets first. So while you have plus signs in between each set of brackets, the fact that there the numbers INSIDE the brackets are negative means that you end up subtracting. Didn't we say though that it doesn't matter which way round you multiply things? So what does four multiplied by negative three look like as addition? Well, sticking with our definitions, it is four added together negative three times. How do you add something a negative amount of times? Remember that a negative something is the same as the something subtracted from zero. So, for positive multiplication you add up a group of things, but for negative multiplication you subtract your number from zero the same amount of times you are supposed to negatively multiply it by. So it looks like this:
\[(0)-(4)-(4)-(4)=-12\]
(The zero looks a bit odd there, and I suppose it could be implied in the same way that positive one is implied to be zero plus one. So we could have written the four threes above as zero plus the four threes.)
As we would expect that is also negative twelve. So we have now looked at a positive multiplied by a positive (where both the number and the sign between the numbers are positive). That gives a positive result. We have looked at a negative multiplied by a positive (the numbers are negative but the sign between them is positive). And we just looked at a positive multiplied by a negative (the numbers are positive but the sign in between them changes to a subtraction instead of addition sign). What about the last option, multiplying a negative number, a negative amount of times? What does that look like as an addition?
Well, you will be multiplying a negative number, so the numbers IN the brackets are going to be negative. And we are going to be doing it a negative amount of times so the numbers BETWEEN the brackets will also be negative. The sum looks like this:
\[(0)-(-3)-(-3)-(-3)-(-3)=12\]
Because we are duplicating a negative number a negative amount of times we end up with two negative signs. Adding a negative is the same as subtracting the number, so subtracting a negative is the same as adding the number. Sound weird? Not really, if I lend you £10, then you owe me £10 (lets call that negative £10). If I then subtract, or cancel, the debt I have effectively gifted you £10. So I turn a negative obligation (you have to give me £10) into a positive benefit (I have given you £10).
So what you actually get when you subtract all those negative numbers is a positive number. In my minds eye, I see the two minus signs combine into a plus sign, with one of them rotating through ninety degrees. So for every two minus signs you create a plus.
So what about our hypothesis that if you multiply negative one by itself negative one times you get negative one? That would mean that you subtract (-) negative (-) one (1) from zero once (1). That would look like this:
\[(0)-(-1)\]
The two minus signs combine to form a plus, and you get positive one. So the square root of negative one cannot be negative one. It cannot be positive one either, because two positive numbers multiplied together always give a positive answer. So if the answer cannot be negative and cannot be positive, then what the hell is it? All the numbers that we know about - all of the numbers above, are either less than zero or greater than zero (or zero itself, I grant you). So how can a number be neither positive or negative?
Oh dear. It appears that we are stumped. And indeed for a long time people treated equations which produced answers that were the square roots of negative numbers in the same way as they treated equations which gave negative numbers themselves as answers. In other words they ignored them.
We do not do that any more though. Instead we say that there IS a number which is the square root of negative one, or a number which if you multiply it by itself gives negative one. We have a symbol for that number, and the symbol is \(i\). What we say, and just work with me on this, is that each number has a 'real' part which is a multiple of one, and an 'imaginary' part which is a multiple of \(i\). The number that is actually the square root of negative one is a number with no real part (or technically a real part multiplied by zero), and an imaginary part which is \(i\) multiplied by one. We say that \(i\) is not positive or negative in the sense of being more or less than zero. Instead we say that it has a whole positive and negative spectrum all to itself. So the following equation makes perfect sense:
\[i-2i=-i\]
This all sounds a bit abstract. What would such a number look like, and how does it relate to the other numbers that we are familiar with? I'll try and make it more visual next time.
Monday, 18 July 2011
Monday, 11 July 2011
Are Infinite Sums Infinite?
So, infinite sums. We covered these. These are an infinitely long list of numbers which you add up. If we want to add up all of the positive whole numbers we would write out:
\[\sum_{x=1}^\infty x\]
If we wanted to add up all the even numbers we would write out:
\[\sum_{x=1}^\infty 2\cdot x\]
Of course, both of these additions is a pointless exercise, because the answers are themselves infinite. There are an infinite amount of whole numbers, and if you add them all up you get infinity. Hell, even if you just added one to itself an infinite amount of times you would also get infinity:
\[\sum_{x=1}^\infty \tfrac{x}{x}\]
(Any number divided by itself is automatically one).
Does this always hold true? Do infinite sums ALWAYS add up to infinity? What about this sum:
\[\sum_{x=0}^\infty \frac{1}{2^x}\]
Looks a bit more complicated doesn't it? First of all notice that I am going to start adding from the zero'th position in the series. So first of all I plug in 0 for \(x\). I get one divided by two to the power of zero. Remember that anything to the power of zero is one. So the first term in this series is one divided by one, or one.
The next term is one divided by two to the power of one. Anything to the power of one is a just one copy of itself. So this is just two. So the term is one divided by two or a half. So far our sum is one plus a half.
The next term is one divided by two squared. Two squared is four, so this is one quarter. The next term is going to be one over two cubed, or an eighth and so on. Basically the series is one plus a whole long list of the inverses of the powers of two. Looks a bit like this:
\[1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\tfrac{1}{32}+\tfrac{1}{64}+\tfrac{1}{128}+\tfrac{1}{256}+\tfrac{1}{512}+\ldots\]
Does that add up to infinity as well? Hmm. Maybe not - look at each term, they all get smaller very very quickly. If they get smaller quickly enough, then adding them all up may not reach infinity.
Lets try some mathematical wizardry. Lets create a variable, which we will call \(x\). Actually, no, lets call it \(S\) for series instead. now let us give make the variable equal to the series that we have created from our sum. To keep things simple we'll cut down the number of terms on display:
\[S=1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots\]
OK, now we will divide each side by two:
\[\frac{S}{2}=\frac{1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots}{2}\]
If we divide a long series of additions by two, that is the same as dividing each individual number in the series by two. Starting with the one at the beginning, that will become a half, and then the half becomes a quarter and the quarter an eighth and so on. Can you see that we are effectively just throwing away the one, and moving every other entry in the series one to the left. Because the series is infinitely long, it is still infinitely long because infinity minus one is still infinity. So once we have done all our divisions by two we get:
\[\frac{S}{2}=\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots\]
OK. What can we do with that then? Look up two equations to the one where we first defined what \(S\) was going to be. You can see that it is one plus a string of fractions. Now look at the equation above. We have established that one half of \(S\) is a string of fractions. Now consider the string of fractions itself. It looks similar. In fact it looks identical, and it will remain identical no matter how many terms you write down for either one. So far, so good. We can now say that \(S\) is one plus that string of fractions, but we know that the string of fractions is actually \(\tfrac{S}{2}\) so we can actually write:
\[S=1+\frac{S}{2}\]
And if we subtract one half of \(S\) from both sides we get:
\[S-\frac{S}{2}=1\]
\(S\) minus half of \(S\) is obviously just the other half of \(S\), so:
\[\frac{S}{2}=1\]
\[S=1\cdot2\]
\[S=2\]
So we have proved that the whole series up there adds up to two, even though there are an infinite amount of terms in it.
\[\sum_{x=1}^\infty x\]
If we wanted to add up all the even numbers we would write out:
\[\sum_{x=1}^\infty 2\cdot x\]
Of course, both of these additions is a pointless exercise, because the answers are themselves infinite. There are an infinite amount of whole numbers, and if you add them all up you get infinity. Hell, even if you just added one to itself an infinite amount of times you would also get infinity:
\[\sum_{x=1}^\infty \tfrac{x}{x}\]
(Any number divided by itself is automatically one).
Does this always hold true? Do infinite sums ALWAYS add up to infinity? What about this sum:
\[\sum_{x=0}^\infty \frac{1}{2^x}\]
Looks a bit more complicated doesn't it? First of all notice that I am going to start adding from the zero'th position in the series. So first of all I plug in 0 for \(x\). I get one divided by two to the power of zero. Remember that anything to the power of zero is one. So the first term in this series is one divided by one, or one.
The next term is one divided by two to the power of one. Anything to the power of one is a just one copy of itself. So this is just two. So the term is one divided by two or a half. So far our sum is one plus a half.
The next term is one divided by two squared. Two squared is four, so this is one quarter. The next term is going to be one over two cubed, or an eighth and so on. Basically the series is one plus a whole long list of the inverses of the powers of two. Looks a bit like this:
\[1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\tfrac{1}{32}+\tfrac{1}{64}+\tfrac{1}{128}+\tfrac{1}{256}+\tfrac{1}{512}+\ldots\]
Does that add up to infinity as well? Hmm. Maybe not - look at each term, they all get smaller very very quickly. If they get smaller quickly enough, then adding them all up may not reach infinity.
Lets try some mathematical wizardry. Lets create a variable, which we will call \(x\). Actually, no, lets call it \(S\) for series instead. now let us give make the variable equal to the series that we have created from our sum. To keep things simple we'll cut down the number of terms on display:
\[S=1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots\]
OK, now we will divide each side by two:
\[\frac{S}{2}=\frac{1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots}{2}\]
If we divide a long series of additions by two, that is the same as dividing each individual number in the series by two. Starting with the one at the beginning, that will become a half, and then the half becomes a quarter and the quarter an eighth and so on. Can you see that we are effectively just throwing away the one, and moving every other entry in the series one to the left. Because the series is infinitely long, it is still infinitely long because infinity minus one is still infinity. So once we have done all our divisions by two we get:
\[\frac{S}{2}=\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots\]
OK. What can we do with that then? Look up two equations to the one where we first defined what \(S\) was going to be. You can see that it is one plus a string of fractions. Now look at the equation above. We have established that one half of \(S\) is a string of fractions. Now consider the string of fractions itself. It looks similar. In fact it looks identical, and it will remain identical no matter how many terms you write down for either one. So far, so good. We can now say that \(S\) is one plus that string of fractions, but we know that the string of fractions is actually \(\tfrac{S}{2}\) so we can actually write:
\[S=1+\frac{S}{2}\]
And if we subtract one half of \(S\) from both sides we get:
\[S-\frac{S}{2}=1\]
\(S\) minus half of \(S\) is obviously just the other half of \(S\), so:
\[\frac{S}{2}=1\]
\[S=1\cdot2\]
\[S=2\]
So we have proved that the whole series up there adds up to two, even though there are an infinite amount of terms in it.
Monday, 4 July 2011
Inifinite Sums
We are moving on to the last symbol shortly, \(i\). Before we do so I just want to lay a little ground work with infinite sums.
You do not need to actually know about infinite sums to understand about imaginary numbers. So why are we bothering to look at them now? Two reasons. Firstly because the notation for these sums looks frightening, but isn't, and it crops up from time to time in maths texts. So it helps to know what we are looking at. Secondly, and more importantly, imaginary numbers sound a bit, well, imaginary, and actually they are very, very real and can be demonstrated using infinite sums - which do not actually involve any imaginary numbers at all.
So what the hell are we talking about? We all know what sums are - right? Colloquially they are arithmetical exercises, and also perhaps algebraic exercises. You "do your sums" if you are carrying out these types of exercises. In a more formal sense "sum" can be used as a synonym of "add". So if I "sum" two numbers I add them together. So, infinite sums are not homework that never, never, ends, but instead an addition operation that never, never ends. By way of example this is a sum:
\[1+2+3+4+5+6\]
and this is an infinite sum:
\[1+2+3+4+5+6+\ldots\]
The dots at the end just mean, and so on. Incidentally, these dots are called an ellipsis. The pattern is obvious - the numbers being added increase by one each time. So instead of writing out all the natural numbers (which would take literally for ever), you just stick the dots on the end. The result of the first sum is twenty one, the result of the second sum is infinity.
Even if we are just writing out the first sum above, it does take up quite a bit of space. We can use a notation to represent this sum in much less space, albeit it looks quite scary to start with. So, what we want to do is firstly work out how we describe each individual number to be added. We intuitively know that the numbers above are a pattern. There is an obvious logic behind the numbers. Each one is one larger than the last one. A list of numbers like this, with a logic behind how you get the next one, is called a series of numbers. So what we want to do is work out how to get any particular number in the series if we are just told which position it has in the series. How do we show that? We use functions! Remember a function just takes a variable (like the position in the series) and then does things to it to produce a result.
Once we have done that, we need to show that we are adding lots of things together. We need to have some sort of symbol to represent lots of addition. We also need to show which position in the series we are going to start counting from, and when to stop. That should do it.
In fact the symbol used in maths is this:
\[\sum\]
All that says is that what follows it is going to be added up. We put our formula for working out the numbers to the right of the symbol like this:
\[\sum f(x)\]
We then show what position we start in the series below the symbol:
\[\sum_{x=1} f(x)\]
Which for us in our example above is the first position, so we start with \(x\) equal to one. Remember, the one just means the first position in the series which goes into the function, not the outcome of the function. Lastly we need to show at which position in the series we want to stop adding entries. We do this by putting that number at the top of the symbol:
\[\sum_{x=1}^6 f(x)\]
So that complicated mess of stuff just means take a series where you generate each entry by putting the position of the entry in the series into the function, put the numbers one to six into the function in turn, and add up all the outcomes. To generate the series above the function \(f(x)\) is just \(x\) because the output of the function is the same as the input.
To symbolise adding up the infinite series illustrated above, you would put infinity at the top of the summation symbol to show that you just kept adding and adding forever:
\[\sum_{x=1}^\infty f(x)\]
What if you wanted to add up all the even numbers? Remember in our proof of the irrationality of \(\sqrt{2}\) we said that any even number was a number that could be divided by two to give another whole number. So \(f(x)\) would be \(2\cdot x\). That would make the first entry in the series two multiplied by one, the second two multiplied by two and the third two multiplied by three, or 2, 4, 6 etc etc ellipsis.
It is traditional that instead of \(x\) the variable we use to denote the position in the sequence is \(n\). Typically we would also dispense with the \(f(n)\) stuff and we would write \(a_n\) instead. That's a bit bloody weird though, using two variables for one number. All that means is that \(a\) is the \(n^{th}\) number in the series. I prefer to stick with \(f(x)\) or at least set out the function, because that way you can be sure what is generating the series of numbers.
You do not need to actually know about infinite sums to understand about imaginary numbers. So why are we bothering to look at them now? Two reasons. Firstly because the notation for these sums looks frightening, but isn't, and it crops up from time to time in maths texts. So it helps to know what we are looking at. Secondly, and more importantly, imaginary numbers sound a bit, well, imaginary, and actually they are very, very real and can be demonstrated using infinite sums - which do not actually involve any imaginary numbers at all.
So what the hell are we talking about? We all know what sums are - right? Colloquially they are arithmetical exercises, and also perhaps algebraic exercises. You "do your sums" if you are carrying out these types of exercises. In a more formal sense "sum" can be used as a synonym of "add". So if I "sum" two numbers I add them together. So, infinite sums are not homework that never, never, ends, but instead an addition operation that never, never ends. By way of example this is a sum:
\[1+2+3+4+5+6\]
and this is an infinite sum:
\[1+2+3+4+5+6+\ldots\]
The dots at the end just mean, and so on. Incidentally, these dots are called an ellipsis. The pattern is obvious - the numbers being added increase by one each time. So instead of writing out all the natural numbers (which would take literally for ever), you just stick the dots on the end. The result of the first sum is twenty one, the result of the second sum is infinity.
Even if we are just writing out the first sum above, it does take up quite a bit of space. We can use a notation to represent this sum in much less space, albeit it looks quite scary to start with. So, what we want to do is firstly work out how we describe each individual number to be added. We intuitively know that the numbers above are a pattern. There is an obvious logic behind the numbers. Each one is one larger than the last one. A list of numbers like this, with a logic behind how you get the next one, is called a series of numbers. So what we want to do is work out how to get any particular number in the series if we are just told which position it has in the series. How do we show that? We use functions! Remember a function just takes a variable (like the position in the series) and then does things to it to produce a result.
Once we have done that, we need to show that we are adding lots of things together. We need to have some sort of symbol to represent lots of addition. We also need to show which position in the series we are going to start counting from, and when to stop. That should do it.
In fact the symbol used in maths is this:
\[\sum\]
All that says is that what follows it is going to be added up. We put our formula for working out the numbers to the right of the symbol like this:
\[\sum f(x)\]
We then show what position we start in the series below the symbol:
\[\sum_{x=1} f(x)\]
Which for us in our example above is the first position, so we start with \(x\) equal to one. Remember, the one just means the first position in the series which goes into the function, not the outcome of the function. Lastly we need to show at which position in the series we want to stop adding entries. We do this by putting that number at the top of the symbol:
\[\sum_{x=1}^6 f(x)\]
So that complicated mess of stuff just means take a series where you generate each entry by putting the position of the entry in the series into the function, put the numbers one to six into the function in turn, and add up all the outcomes. To generate the series above the function \(f(x)\) is just \(x\) because the output of the function is the same as the input.
To symbolise adding up the infinite series illustrated above, you would put infinity at the top of the summation symbol to show that you just kept adding and adding forever:
\[\sum_{x=1}^\infty f(x)\]
What if you wanted to add up all the even numbers? Remember in our proof of the irrationality of \(\sqrt{2}\) we said that any even number was a number that could be divided by two to give another whole number. So \(f(x)\) would be \(2\cdot x\). That would make the first entry in the series two multiplied by one, the second two multiplied by two and the third two multiplied by three, or 2, 4, 6 etc etc ellipsis.
It is traditional that instead of \(x\) the variable we use to denote the position in the sequence is \(n\). Typically we would also dispense with the \(f(n)\) stuff and we would write \(a_n\) instead. That's a bit bloody weird though, using two variables for one number. All that means is that \(a\) is the \(n^{th}\) number in the series. I prefer to stick with \(f(x)\) or at least set out the function, because that way you can be sure what is generating the series of numbers.
Monday, 27 June 2011
e (by gum)
Now that we have got that fairly abstract stuff about functions and limits out of the way we can finally talk about \(e\). This is another number like \(\pi\) in that it is a constant. \(e\) is always the same number like 2 is two or 5 is five. It does not change like the \(x\)'s we have been looking at.
So what number is it? Well, lets talk about the interest on your bank account. We want to arrive at an exact figure for \(e\), not a multiple of something else, so we will start with a unit value in our bank account. If you remember a unit is always one. So lets say you have £1 in your bank account. It could be Euros, Dollars, Yen, or Sestertii. What is important is that we have one of them.
Your bank pays you interest on the money that it holds for you. it does this by giving your money to other people to set up businesses or buy houses, and charging them for the service. Because the money was yours it passes on some of the payment for this service to you. In these financially desperate times, of course, what the bank does is charge a large amount to lend out money, gives you a tiny amount back and keeps the vast amount of the charge for itself. Anyway, what we want to know is how much you will have in your account at the end of a period of investment. Traditionally the period of investment used to compare different methods of saving is one year, so let's use that.
Now, we also need to know what interest rate is going to be applied to your account. Lets say that you have pictures of your bank manager in flagrante delicto with a local farm animal, and you have been given an interest rate of 100%. (In reality, by contrast, an interest rate of one half of one percent would be more likely.)
So how much do you have in your account at the end of the year? The answer seems to be obvious - if you get 100% and you start with £1 you will have your original £1 and 100% of that again (£1) in interest, giving a total of £2 at the end of the year.
Things, however, are not that simple. We have made an assumption that your interest will be calculated and paid at the end of the year. That may not necessarily be the case. Instead, let's consider what happens if you negotiate with the bank to be paid your interest in two lump sums, one halfway through the year and then at the end of the year.
After six months of the year have passed, you will get one half of 100% of the interest on £1. That would be 50% of the total of £1 or 50p. Just what we would have expected. BUT WAIT a moment, because when you come to get your interest for the second six months, you are not earning interest on £1 any more, you now have £1.50 in your account. So you get half of 100% of the interest on £1.50. 100% of that interest would be £1.50 so you get 75p. So at the end of the year you have £1.50 plus 75p, or £2.25.
Hey, there is an extra 25p in interest over and above what you get if the interest is worked out once at the end of the year. Where did that come from? It comes from getting paid interest on the interest that you have already earned. 25p is half of 100% of the interest that you could have got on the 50p, had you started the year with that in your account. That sounds excellent, more money from seemingly nothing. Can you try that trick again?
Let's go back to the bank manager and show him the CCTV footage you have of him attending a very private party with the coach of the local college's ladies volleyball team. And the whole team. He agrees to calculate your interest payments every month. So at the end of the first month you get one twelfth of 100% of the interest on £1 or about 8p. At the end of the second month you get one twelfth of 100% of the interest on £1.08, or about 9p. At the end of the third month you get one twelfth of 100% of the interest on £1.17 (£1.08 plus 9p), which is nearly 10p. Note that each month the interest you get is a little bit more than what you got the previous month, because each month you get a little bit of interest paid on the interest you earned in the previous month.
So what do you get at the end of the year? If you do all of the adding up, you get a little bit more than £2.61. So that's pretty good. An extra 36p over the 25p we earned by calculating it twice per year. But notice - we are into the territory of diminishing returns. We haven't earned twelve lots of 25p extra. What is happening is that we are squeezing smaller and smaller amounts of interest out each time we calculate. So the increase over calculating it one per year is not that much more.
Can we work out what is happening here mathematically? Let's go back to payment once a year. How to we mathematically represent payment of 100% of the interest at the end of the year? We need to show that we still have our original £1, and we have been given an extra £1. That would be:
\[1+1=2\]
We could also write that down as:
\[1\cdot 2=2\]
because we end up with twice as many ones as we started with. Adding 100% to something means multiplying that thing by two, because you have your original thing, and you have a 100% copy of it, so you have two of your original thing.
Right. What happens with payment twice a year, after six months and then after twelve? Well, after the first six months you get half of 100% of the annual interest. How do we show that part? Well, we are not doubling our original thing, but we are adding half of it to the original. If you want to add half of something you would multiply it by one and a half. The one shows you are keeping your original, and the half gives you the half. So after six months we get:
\[1\cdot 1\tfrac {1}{2} = 1\tfrac{1}{2}\]
That's right, because we had £1.50 after the first six months. So what happens at the end of the year? Well, we want to take what we had at the halfway point and add half of 100% of the annual interest rate on that amount. Again that means we need to multiply the amount we had after six months by one (to show that that amount stays in our account) and a half (to show we are getting half of 100% of the annual rate). This looks like this:
\[1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
And that works out again, because you had £2.25 at the end of the year where you got interest after each six months. Let's try to work out what is happening. We wrote out the maths for the second interest payment above as the amount you had after six months multiplied by one and a half. But we already know that the amount you had after six months was one and a half times one. Remember that it does not make a difference in what order you multiply things. So we could have written the whole year out as:
\[1\cdot 1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
That works out mathematically because:
\[1\tfrac{1}{2} = \tfrac{3}{2}\]
So (remembering that to multiply two fractions together you just multiply the bottoms together and the tops together):
\[1\cdot 1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
\[1\cdot \tfrac{3}{2}\cdot \tfrac{3}{2}=2\tfrac{1}{4}\]
\[1\cdot \tfrac{9}{4}=2\tfrac{1}{4}\]
\[1\cdot (\tfrac{8}{4}+\tfrac{1}{4})=2\tfrac{1}{4}\]
\[1\cdot (2+\tfrac{1}{4})=2\tfrac{1}{4}\]
\[(2+\tfrac{1}{4})=2\tfrac{1}{4}\]
Anyway, back to the result. We also know how to write one number multiplied by itself don't we? That's just the square of the number, or the number raised to power two. That would make it:
\[1\cdot (1\tfrac{1}{2})^2=2\tfrac{1}{4}\]
Notice that we now have two twos on the left hand side of the equals sign. One is the denominator of the fraction, and the other is the power to which we raise the number in the brackets. We also get our interest paid twice a year. Coincidence? No! The fraction shows how much of the total annual interest we get applied each time, and the power number shows how many times we get paid the interest. So they are always going to be the same. If we get paid interest twice we get half each time. If we get paid four times, we get a quarter each time. Any if we get paid monthly we get a twelfth each time. If we write out the sum for monthly interest payments, it looks like this:
\[\begin{multline}
1\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot\\
1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}=\tfrac{23298085122481}{8916100448256}
\end{multline}\]
The fraction at the end looks insane because:
\[1\tfrac{1}{12} = \tfrac{13}{12}\]
and when you multiply \(\tfrac{13}{12}\) by itself twelve times you get \(\frac{13^{12}}{12^{12}}\) which is the fraction at the end of that line. The maths works though, because if you divide the number on the top of the fraction by the number on the bottom of the fraction you get about 2.61, and you got £2.61 at the end of the year with monthly payments. That all looks like a hell of a mess though, so lets tidy it up a bit.
\[1\cdot (1\tfrac{1}{12})^{12}=\tfrac{23298085122481}{8916100448256}\]
Ladies and gentleman we have ourselves a pattern. If we calculate and pay interest \(n\) times a year, then at the end of the year we will have \(1\cdot(1\tfrac{1}{n})^n\) pounds, dollars, euros or whatever. We can describe those instructions as function, and give it the name \(f\), and the variable in question is \(n\), so:
\[f(n)=1\cdot(1\tfrac{1}{n})^n\]
Which we can tidy up a bit. The 1 at the beginning is just redundant. Look at the examples above - it does nothing. It just hangs about telling you you have one of whatever comes. Remember when we decided to use a unit value - this is why. We can just get rid of the one. If we started with £2 in our account we would need to bring the two in at this point in place of the one, but we are just working in units of one now. So removing the one we get:
\[f(n)=(1\tfrac{1}{n})^n\]
Also as we saw above, one and a half is the same as one plus a half. So to complete the tidying:
\[f(n)=(1+\tfrac{1}{n})^n\]
So we can calculate what we would get if we got paid every day, by sticking in 365 for \(n\):
\[f(365)=(1+\tfrac{1}{365})^{365}\]
\[f(365)\approx 2.71456748\]
The \(\approx\) just means approximately. As we saw, just going to \(n=12\) generates stupidly big fractions which have very long decimal expansions, so I am just rounding off the result. We could also calculate the amount we get if interest was paid every second of every minute of every hour of every day of the year:
\[60\cdot60\cdot24\cdot365=31536000\]
\[f(31536000)=(1+\tfrac{1}{31536000})^{31536000}\]
\[f(31536000)\approx 2.7182817\]
I should point out at this stage that there is nothing special about the choice of a year as the period we are using. It could just as well have been a week or an hour or a decade. The important thing is how many times during that period we pay interest.
As you should hopefully be able to see, the bigger the number that we plug into our function the closer we get to a fixed number. We have gone from \(2 \to 2.25 \to 2.61 \to 2.71456748 \to 2.7182817\). We are getting closer and closer to a number just a little bit bigger than 2.7. If we were paid interest infinitely often during the year, in other words if interest constantly accrued into our account every moment, then the amount we would have at the end of the year would be the number that we are approaching. This number, then, is the limit of the function that we have worked out, which is approached as the size of the number that we plug in gets larger and larger. Remember that we write this as:
\[\lim_{n \to \infty} (1+\tfrac{1}{n})^n\approx2.7182817\]
I still have to use the \(\approx\) symbol because the number just keeps going and going. Just like \(\pi\). Just like \(\pi\) we use a letter instead to represent the number exactly. That letter is \(e\). So:
\[e=\lim_{n \to \infty} (1+\tfrac{1}{n})^n\]
What does that actually mean? It means that if you have an interest rate of 100% and you get paid interest infinitely often during any period, then you will end up with \(e\) times your original balance. It is not a ratio like \(\pi\) is, so we have to reach it by a more circuitous route, but there it is an actual number represented by an actual symbol.
So what number is it? Well, lets talk about the interest on your bank account. We want to arrive at an exact figure for \(e\), not a multiple of something else, so we will start with a unit value in our bank account. If you remember a unit is always one. So lets say you have £1 in your bank account. It could be Euros, Dollars, Yen, or Sestertii. What is important is that we have one of them.
Your bank pays you interest on the money that it holds for you. it does this by giving your money to other people to set up businesses or buy houses, and charging them for the service. Because the money was yours it passes on some of the payment for this service to you. In these financially desperate times, of course, what the bank does is charge a large amount to lend out money, gives you a tiny amount back and keeps the vast amount of the charge for itself. Anyway, what we want to know is how much you will have in your account at the end of a period of investment. Traditionally the period of investment used to compare different methods of saving is one year, so let's use that.
Now, we also need to know what interest rate is going to be applied to your account. Lets say that you have pictures of your bank manager in flagrante delicto with a local farm animal, and you have been given an interest rate of 100%. (In reality, by contrast, an interest rate of one half of one percent would be more likely.)
So how much do you have in your account at the end of the year? The answer seems to be obvious - if you get 100% and you start with £1 you will have your original £1 and 100% of that again (£1) in interest, giving a total of £2 at the end of the year.
Things, however, are not that simple. We have made an assumption that your interest will be calculated and paid at the end of the year. That may not necessarily be the case. Instead, let's consider what happens if you negotiate with the bank to be paid your interest in two lump sums, one halfway through the year and then at the end of the year.
After six months of the year have passed, you will get one half of 100% of the interest on £1. That would be 50% of the total of £1 or 50p. Just what we would have expected. BUT WAIT a moment, because when you come to get your interest for the second six months, you are not earning interest on £1 any more, you now have £1.50 in your account. So you get half of 100% of the interest on £1.50. 100% of that interest would be £1.50 so you get 75p. So at the end of the year you have £1.50 plus 75p, or £2.25.
Hey, there is an extra 25p in interest over and above what you get if the interest is worked out once at the end of the year. Where did that come from? It comes from getting paid interest on the interest that you have already earned. 25p is half of 100% of the interest that you could have got on the 50p, had you started the year with that in your account. That sounds excellent, more money from seemingly nothing. Can you try that trick again?
Let's go back to the bank manager and show him the CCTV footage you have of him attending a very private party with the coach of the local college's ladies volleyball team. And the whole team. He agrees to calculate your interest payments every month. So at the end of the first month you get one twelfth of 100% of the interest on £1 or about 8p. At the end of the second month you get one twelfth of 100% of the interest on £1.08, or about 9p. At the end of the third month you get one twelfth of 100% of the interest on £1.17 (£1.08 plus 9p), which is nearly 10p. Note that each month the interest you get is a little bit more than what you got the previous month, because each month you get a little bit of interest paid on the interest you earned in the previous month.
So what do you get at the end of the year? If you do all of the adding up, you get a little bit more than £2.61. So that's pretty good. An extra 36p over the 25p we earned by calculating it twice per year. But notice - we are into the territory of diminishing returns. We haven't earned twelve lots of 25p extra. What is happening is that we are squeezing smaller and smaller amounts of interest out each time we calculate. So the increase over calculating it one per year is not that much more.
Can we work out what is happening here mathematically? Let's go back to payment once a year. How to we mathematically represent payment of 100% of the interest at the end of the year? We need to show that we still have our original £1, and we have been given an extra £1. That would be:
\[1+1=2\]
We could also write that down as:
\[1\cdot 2=2\]
because we end up with twice as many ones as we started with. Adding 100% to something means multiplying that thing by two, because you have your original thing, and you have a 100% copy of it, so you have two of your original thing.
Right. What happens with payment twice a year, after six months and then after twelve? Well, after the first six months you get half of 100% of the annual interest. How do we show that part? Well, we are not doubling our original thing, but we are adding half of it to the original. If you want to add half of something you would multiply it by one and a half. The one shows you are keeping your original, and the half gives you the half. So after six months we get:
\[1\cdot 1\tfrac {1}{2} = 1\tfrac{1}{2}\]
That's right, because we had £1.50 after the first six months. So what happens at the end of the year? Well, we want to take what we had at the halfway point and add half of 100% of the annual interest rate on that amount. Again that means we need to multiply the amount we had after six months by one (to show that that amount stays in our account) and a half (to show we are getting half of 100% of the annual rate). This looks like this:
\[1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
And that works out again, because you had £2.25 at the end of the year where you got interest after each six months. Let's try to work out what is happening. We wrote out the maths for the second interest payment above as the amount you had after six months multiplied by one and a half. But we already know that the amount you had after six months was one and a half times one. Remember that it does not make a difference in what order you multiply things. So we could have written the whole year out as:
\[1\cdot 1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
That works out mathematically because:
\[1\tfrac{1}{2} = \tfrac{3}{2}\]
So (remembering that to multiply two fractions together you just multiply the bottoms together and the tops together):
\[1\cdot 1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
\[1\cdot \tfrac{3}{2}\cdot \tfrac{3}{2}=2\tfrac{1}{4}\]
\[1\cdot \tfrac{9}{4}=2\tfrac{1}{4}\]
\[1\cdot (\tfrac{8}{4}+\tfrac{1}{4})=2\tfrac{1}{4}\]
\[1\cdot (2+\tfrac{1}{4})=2\tfrac{1}{4}\]
\[(2+\tfrac{1}{4})=2\tfrac{1}{4}\]
Anyway, back to the result. We also know how to write one number multiplied by itself don't we? That's just the square of the number, or the number raised to power two. That would make it:
\[1\cdot (1\tfrac{1}{2})^2=2\tfrac{1}{4}\]
Notice that we now have two twos on the left hand side of the equals sign. One is the denominator of the fraction, and the other is the power to which we raise the number in the brackets. We also get our interest paid twice a year. Coincidence? No! The fraction shows how much of the total annual interest we get applied each time, and the power number shows how many times we get paid the interest. So they are always going to be the same. If we get paid interest twice we get half each time. If we get paid four times, we get a quarter each time. Any if we get paid monthly we get a twelfth each time. If we write out the sum for monthly interest payments, it looks like this:
\[\begin{multline}
1\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot\\
1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}=\tfrac{23298085122481}{8916100448256}
\end{multline}\]
The fraction at the end looks insane because:
\[1\tfrac{1}{12} = \tfrac{13}{12}\]
and when you multiply \(\tfrac{13}{12}\) by itself twelve times you get \(\frac{13^{12}}{12^{12}}\) which is the fraction at the end of that line. The maths works though, because if you divide the number on the top of the fraction by the number on the bottom of the fraction you get about 2.61, and you got £2.61 at the end of the year with monthly payments. That all looks like a hell of a mess though, so lets tidy it up a bit.
\[1\cdot (1\tfrac{1}{12})^{12}=\tfrac{23298085122481}{8916100448256}\]
Ladies and gentleman we have ourselves a pattern. If we calculate and pay interest \(n\) times a year, then at the end of the year we will have \(1\cdot(1\tfrac{1}{n})^n\) pounds, dollars, euros or whatever. We can describe those instructions as function, and give it the name \(f\), and the variable in question is \(n\), so:
\[f(n)=1\cdot(1\tfrac{1}{n})^n\]
Which we can tidy up a bit. The 1 at the beginning is just redundant. Look at the examples above - it does nothing. It just hangs about telling you you have one of whatever comes. Remember when we decided to use a unit value - this is why. We can just get rid of the one. If we started with £2 in our account we would need to bring the two in at this point in place of the one, but we are just working in units of one now. So removing the one we get:
\[f(n)=(1\tfrac{1}{n})^n\]
Also as we saw above, one and a half is the same as one plus a half. So to complete the tidying:
\[f(n)=(1+\tfrac{1}{n})^n\]
So we can calculate what we would get if we got paid every day, by sticking in 365 for \(n\):
\[f(365)=(1+\tfrac{1}{365})^{365}\]
\[f(365)\approx 2.71456748\]
The \(\approx\) just means approximately. As we saw, just going to \(n=12\) generates stupidly big fractions which have very long decimal expansions, so I am just rounding off the result. We could also calculate the amount we get if interest was paid every second of every minute of every hour of every day of the year:
\[60\cdot60\cdot24\cdot365=31536000\]
\[f(31536000)=(1+\tfrac{1}{31536000})^{31536000}\]
\[f(31536000)\approx 2.7182817\]
I should point out at this stage that there is nothing special about the choice of a year as the period we are using. It could just as well have been a week or an hour or a decade. The important thing is how many times during that period we pay interest.
As you should hopefully be able to see, the bigger the number that we plug into our function the closer we get to a fixed number. We have gone from \(2 \to 2.25 \to 2.61 \to 2.71456748 \to 2.7182817\). We are getting closer and closer to a number just a little bit bigger than 2.7. If we were paid interest infinitely often during the year, in other words if interest constantly accrued into our account every moment, then the amount we would have at the end of the year would be the number that we are approaching. This number, then, is the limit of the function that we have worked out, which is approached as the size of the number that we plug in gets larger and larger. Remember that we write this as:
\[\lim_{n \to \infty} (1+\tfrac{1}{n})^n\approx2.7182817\]
I still have to use the \(\approx\) symbol because the number just keeps going and going. Just like \(\pi\). Just like \(\pi\) we use a letter instead to represent the number exactly. That letter is \(e\). So:
\[e=\lim_{n \to \infty} (1+\tfrac{1}{n})^n\]
What does that actually mean? It means that if you have an interest rate of 100% and you get paid interest infinitely often during any period, then you will end up with \(e\) times your original balance. It is not a ratio like \(\pi\) is, so we have to reach it by a more circuitous route, but there it is an actual number represented by an actual symbol.
Monday, 20 June 2011
What is a Limit, Part Two?
There is one further thing we need to say about limits before we move on. Instead of making your input number get closer and closer to the number at which everything falls apart, you can also make your input number just get bigger and bigger. This can also help us answer the unanswerable, because you cannot really make your input number infinitely large (you would need an infinitely powerful calculator to work out the answer), but you can say what would happen if you did.
Lets look again at our original function:
\[x+1\]
Lets switch the sign again:
\[x-1\]
Lets add an \(x\) to it:
\[2x-1\]
And lets divide it by \(x\):
\[\frac{2x-1}{x}\]
This time, lets call the function \(g(x)\), so we distinguish it from the last one. . What is \(g(1)\)? Well that is two times one (two) less one (one) divided by one. One divided by one is one. Algebraically:
\[\frac{2\cdot 1 -1}{1}\]
\[\frac{2-1}{1}\]
\[\frac{1}{1}\]
\[1\]
What about \(g(10)\)? That is two times ten (twenty) less one (nineteen) divided by ten. Nineteen divided by ten is one and nine tenths. So:
\[\frac{2\cdot 10 -1}{10}\]
\[\frac{20-1}{10}\]
\[\frac{19}{10}\]
\[1.9\]
Multiplying the input number by 10 has not made a massive difference to the output - it just added on 0.9. The reason why is that while the input number gets larger it appears on both the top AND bottom of the function so it multiplies and divides. Each operation (multiplication and division) just about cancels the other out. So what if we go really big? Lets try \(g(10000)\):
\[\frac{2\cdot 10000 -1}{10000}\]
\[\frac{20000-1}{10000}\]
\[\frac{19999}{10000}\]
\[1.9999\]
So no matter how big the number you stick in, you double it and divide it by itself which nearly cancels one of the copies completely, and would cancel it BUT FOR the one that you take away on the top line. This also works for whatever number you stick in - it does not have to be 1, 10, 100, 1000 or so on, it is just that the results from other numbers look a bit messy. What we are seeing here is that the bigger and bigger the starting number you put in, the closer and closer the output is to two. In mathematical terms we describe this as:
\[\lim_{x \to \infty} \frac{2x-1}{x}=2\]
Which means that as \(x\) gets closer to infinity (\(\infty\)) the output of the function gets closer to two.
Lets look again at our original function:
\[x+1\]
Lets switch the sign again:
\[x-1\]
Lets add an \(x\) to it:
\[2x-1\]
And lets divide it by \(x\):
\[\frac{2x-1}{x}\]
This time, lets call the function \(g(x)\), so we distinguish it from the last one. . What is \(g(1)\)? Well that is two times one (two) less one (one) divided by one. One divided by one is one. Algebraically:
\[\frac{2\cdot 1 -1}{1}\]
\[\frac{2-1}{1}\]
\[\frac{1}{1}\]
\[1\]
What about \(g(10)\)? That is two times ten (twenty) less one (nineteen) divided by ten. Nineteen divided by ten is one and nine tenths. So:
\[\frac{2\cdot 10 -1}{10}\]
\[\frac{20-1}{10}\]
\[\frac{19}{10}\]
\[1.9\]
Multiplying the input number by 10 has not made a massive difference to the output - it just added on 0.9. The reason why is that while the input number gets larger it appears on both the top AND bottom of the function so it multiplies and divides. Each operation (multiplication and division) just about cancels the other out. So what if we go really big? Lets try \(g(10000)\):
\[\frac{2\cdot 10000 -1}{10000}\]
\[\frac{20000-1}{10000}\]
\[\frac{19999}{10000}\]
\[1.9999\]
So no matter how big the number you stick in, you double it and divide it by itself which nearly cancels one of the copies completely, and would cancel it BUT FOR the one that you take away on the top line. This also works for whatever number you stick in - it does not have to be 1, 10, 100, 1000 or so on, it is just that the results from other numbers look a bit messy. What we are seeing here is that the bigger and bigger the starting number you put in, the closer and closer the output is to two. In mathematical terms we describe this as:
\[\lim_{x \to \infty} \frac{2x-1}{x}=2\]
Which means that as \(x\) gets closer to infinity (\(\infty\)) the output of the function gets closer to two.
Monday, 13 June 2011
What is a Limit, Part One?
The limit of something, is the maximum amount of something you can have. It can be natural or artificial. There is usually a drink drive limit, which is an artificial level of alcohol set by the state, that you are allowed to have in your system and legally drive. So if you have more alcohol than that you are known as being over the limit. Then there are natural limits, such as the limit of human hearing - dogs can hear sounds beyond that limit.
In maths limits are important because they help us get the answers to questions that we cannot ask. If that sounds odd, it is supposed to. There are some questions that we want to ask, but if we do we get nonsensical answers. As an example, lets build a function. Lets start with the function we used when talking about functions the first time:
\[x+1\]
Now, lets change the sign from addition to subtraction:
\[x-1\]
Lets divide it by itself:
\[\frac{x-1}{x-1}\]
And finally lets multiply the \(x\) on top by itself once.
\[\frac{x^2-1}{x-1}\]
Right. Lets call this \(f(x)\). What is \(f(2)\)? Well it is two times itself (four) less one (three) divided by two less one (one), so three divided by one. Or in algebra:
\[\frac{2^2-1}{2-1}\]
\[\frac{4-1}{2-1}\]
\[\frac{3}{1}\]
\[3\]
What \(f(5)\)? Well...
\[\frac{5^2-1}{5-1}\]
\[\frac{25-1}{5-1}\]
\[\frac{24}{4}\]
\[6\]
Looking at this there seems to be an obvious connection between \(x\) and \(f(x)\). The function just seems to add one to \(x\). 3=2+1, and 6=5+1. We started with \(x+1\), and it appears we have got back there! So \(f(1)\) should be two. Lets see if it is:
\[\frac{1^2-1}{1-1}\]
\[\frac{1-1}{1-1}\]
\[\frac{0}{0}\]
Hmm. Zero divided by zero. That's just a mess. Is that infinity or zero? It is meaningless. It does not work because as soon as \(x\) is one, you get zeros everywhere because you are subtracting one from one on the top and bottom. So \(f(1)\) is special because of the way the function is built.
How does the concept of a limit help us to get an answer to \(f(1)\)? Well, we could ask what \(f(0.5)\) and \(f(1.5)\) are. The answers are 1.5 and 2.5 respectively. That matches what we thought - the function ends up adding one to your starting number. It also supports our guess for \(f(1)\) before we tried it, which was two. Lets get a bit closer to one, and try again this time with \(f(0.9)\) and \(f(1.1)\). The results are 1.9 and 2.1 respectively. Again, just what we would expect. The closer we get to \(f(1)\) the closer the result gets to two. If we try with something silly like \(f(0.99999)\) we get 1.99999. That pattern is going to repeat for as many nines as we add onto the end of the number we plug in.
The way we answer this question is to say that as the number we plug into the function gets closer and closer to one, the result of the function gets closer and closer to two, but because it breaks when we put in exactly one, we never get to exactly two. That's what a limit is. In mathematical script it looks like this:
\[\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}=2\]
Spoken out loud that says the limit of the function (blah blah describe the function blah blah) as ecks approaches one is two.
In maths limits are important because they help us get the answers to questions that we cannot ask. If that sounds odd, it is supposed to. There are some questions that we want to ask, but if we do we get nonsensical answers. As an example, lets build a function. Lets start with the function we used when talking about functions the first time:
\[x+1\]
Now, lets change the sign from addition to subtraction:
\[x-1\]
Lets divide it by itself:
\[\frac{x-1}{x-1}\]
And finally lets multiply the \(x\) on top by itself once.
\[\frac{x^2-1}{x-1}\]
Right. Lets call this \(f(x)\). What is \(f(2)\)? Well it is two times itself (four) less one (three) divided by two less one (one), so three divided by one. Or in algebra:
\[\frac{2^2-1}{2-1}\]
\[\frac{4-1}{2-1}\]
\[\frac{3}{1}\]
\[3\]
What \(f(5)\)? Well...
\[\frac{5^2-1}{5-1}\]
\[\frac{25-1}{5-1}\]
\[\frac{24}{4}\]
\[6\]
Looking at this there seems to be an obvious connection between \(x\) and \(f(x)\). The function just seems to add one to \(x\). 3=2+1, and 6=5+1. We started with \(x+1\), and it appears we have got back there! So \(f(1)\) should be two. Lets see if it is:
\[\frac{1^2-1}{1-1}\]
\[\frac{1-1}{1-1}\]
\[\frac{0}{0}\]
Hmm. Zero divided by zero. That's just a mess. Is that infinity or zero? It is meaningless. It does not work because as soon as \(x\) is one, you get zeros everywhere because you are subtracting one from one on the top and bottom. So \(f(1)\) is special because of the way the function is built.
How does the concept of a limit help us to get an answer to \(f(1)\)? Well, we could ask what \(f(0.5)\) and \(f(1.5)\) are. The answers are 1.5 and 2.5 respectively. That matches what we thought - the function ends up adding one to your starting number. It also supports our guess for \(f(1)\) before we tried it, which was two. Lets get a bit closer to one, and try again this time with \(f(0.9)\) and \(f(1.1)\). The results are 1.9 and 2.1 respectively. Again, just what we would expect. The closer we get to \(f(1)\) the closer the result gets to two. If we try with something silly like \(f(0.99999)\) we get 1.99999. That pattern is going to repeat for as many nines as we add onto the end of the number we plug in.
The way we answer this question is to say that as the number we plug into the function gets closer and closer to one, the result of the function gets closer and closer to two, but because it breaks when we put in exactly one, we never get to exactly two. That's what a limit is. In mathematical script it looks like this:
\[\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}=2\]
Spoken out loud that says the limit of the function (blah blah describe the function blah blah) as ecks approaches one is two.
Monday, 6 June 2011
Irrationality of the Square Root of Four?
OK, last time we proved that the square root of two is not one whole number divided by another whole number, because we could continue dividing for ever, and we know that we cannot do that because we would eventually run out of numbers. It does all sound a bit fishy though. Is this not just some trick of the maths? To double check, lets try the same process, but with the square root of four instead. The square root of four is easy - it is just two, because two multiplied by two is four.
Anyway, we are assuming that the square root of four can be written as one number over another number. So lets use the same letters as last time.
\[\frac{a}{b}=\sqrt{4}\]
Excellent, again. Some fraction equals the square root of four. That's the assumption that we want to make. Nothing wrong so far. Let's now do a mathematical operation on this equation.
What we want to do is get the right hand side to just be four. This means that we need to multiply it by itself to get four. Obviously the square root of four multiplied by the square root of four is four - because that's where it came from. Let's skip the not equals sign stuff this time around, you get the idea.
\[\frac{a^2}{b^2}=4\]
And again, we know that we can multiply both sides by \(b^2\) to get:
\[a^2=4\cdot b^2\]
Now at this point can we say that \(a\) is an even number? Yes, we can run through all of our odds and evens arguments and still say that, because four is just two multiplied by two, so the left hand side is still some other number on the right hand side multiplied by two. So that gives us:
\[(2\cdot c)^2=4\cdot b^2\]
And we can square out contents of the brackets on the left hand side:
\[2\cdot 2\cdot c\cdot c=4\cdot b^2\]
\[2\cdot 2\cdot c^2=4\cdot b^2\]
\[4\cdot c^2=4\cdot b^2\]
OK. Last time when hunting for the square root of two we had this:
\[4\cdot c^2=2\cdot b^2\]
What is the difference here? The right hand side is multiplied by the SAME NUMBER as the left hand side - four. What can we gather from this? Well we can divide both sides by four to get this:
\[c^2 = b^2\]
And we can square root each side to get this:
\[c = b\]
So we now know that \(c\) is the same as \(b\). We also know that \(a\) is the same as \(2\cdot c\). So we can now convert both \(a\) and \(b\) into their equivalents of \(c\) and stick those back in the very first equation:
\[\frac{2c}{c}=\sqrt{4}\]
And from our arguments last time we also know that is the same as:
\[2\cdot \frac{c}{c}=\sqrt{4}\]
\[2\cdot 1=\sqrt{4}\]
\[2=\sqrt{4}\]
So we have established that two is the square root of four. Which it is. So our assumption this time, that the square root of four can be written as one number divided by another number IS correct. In fact we know that it is always correct as long as the top number is the bottom number multiplied by two.
Anyway, we are assuming that the square root of four can be written as one number over another number. So lets use the same letters as last time.
\[\frac{a}{b}=\sqrt{4}\]
Excellent, again. Some fraction equals the square root of four. That's the assumption that we want to make. Nothing wrong so far. Let's now do a mathematical operation on this equation.
What we want to do is get the right hand side to just be four. This means that we need to multiply it by itself to get four. Obviously the square root of four multiplied by the square root of four is four - because that's where it came from. Let's skip the not equals sign stuff this time around, you get the idea.
\[\frac{a^2}{b^2}=4\]
And again, we know that we can multiply both sides by \(b^2\) to get:
\[a^2=4\cdot b^2\]
Now at this point can we say that \(a\) is an even number? Yes, we can run through all of our odds and evens arguments and still say that, because four is just two multiplied by two, so the left hand side is still some other number on the right hand side multiplied by two. So that gives us:
\[(2\cdot c)^2=4\cdot b^2\]
And we can square out contents of the brackets on the left hand side:
\[2\cdot 2\cdot c\cdot c=4\cdot b^2\]
\[2\cdot 2\cdot c^2=4\cdot b^2\]
\[4\cdot c^2=4\cdot b^2\]
OK. Last time when hunting for the square root of two we had this:
\[4\cdot c^2=2\cdot b^2\]
What is the difference here? The right hand side is multiplied by the SAME NUMBER as the left hand side - four. What can we gather from this? Well we can divide both sides by four to get this:
\[c^2 = b^2\]
And we can square root each side to get this:
\[c = b\]
So we now know that \(c\) is the same as \(b\). We also know that \(a\) is the same as \(2\cdot c\). So we can now convert both \(a\) and \(b\) into their equivalents of \(c\) and stick those back in the very first equation:
\[\frac{2c}{c}=\sqrt{4}\]
And from our arguments last time we also know that is the same as:
\[2\cdot \frac{c}{c}=\sqrt{4}\]
\[2\cdot 1=\sqrt{4}\]
\[2=\sqrt{4}\]
So we have established that two is the square root of four. Which it is. So our assumption this time, that the square root of four can be written as one number divided by another number IS correct. In fact we know that it is always correct as long as the top number is the bottom number multiplied by two.
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