Real or Imaginary Part 4

Last time we managed to graph the function \(f(x)=\frac{1}{1-x^2}\). First we calculated some values for the function around zero in intervals of a quarter. Then we drew a line and marked off these intervals to scale and then drew circles above each interval at a point that was to scale with the value of the function when we used that particular interval as the input to the function.

Now we are going to try that with the other function that we created. This is the other one that is exactly equal to the result of an infinite sum, but lets give it a different name to avoid confusion: \(g(x)=\frac{1}{1+x^2}\). The first job is to work out the values of the function at the same intervals as before, to get a rough idea about how this function behaves.

\[g(0)=1\]
\[g(\tfrac{1}{4})=\frac{16}{17}\]
\[g(\tfrac{1}{2})=\frac{4}{5}\]
\[g(\tfrac{3}{4})=\frac{16}{25}\]
\[g(1)=\frac{1}{2}\]

Hey look! This time the function is perfectly calculable at one. No breaking. What about a quarter more than one?

\[g(\tfrac{5}{4})=\frac{16}{41}\]
OK, so it is still going then. Let's do one more just to check.
\[g(\tfrac{3}{2})=\frac{4}{13}\]
Yep, still going strong, and it seems to be getting smaller and smaller. What about negative numbers as inputs to the function? Well again the only variable that we plug into this function is immediately squared, so it matters not a jot whether or not the input is positive or negative, the square is always going to be positive. So the function should be symmetrical around zero, because there is no difference in the output between a positive or negative input.

So let's set up the same graph as we did last time. This time we have calculated the values of the function at up to one and a half. We'll stop there for now, and make our line stretch from minus one and a half to plus one and a half. That makes our line look like this:

Now we have our line, lets mark off, using circles, the values of the function that we worked out above - and their negative mirrors which we know have the same value.

Those plotted points look very different to the set of points for the other function, don't they? Instead of a smiley face it is a moustache. There is a hump in the middle at \(x\) equal to zero and then the ends just seem to tail off. Let's fill in the gaps in-between our plotted values by getting the computer to calculate and draw the hundreds of points needed to form what looks like a line:

As we thought the line that appears bulges in the middle and then tails off at either end towards zero. Notice that it does not go nuts at plus and minus one by heading off to infinity, it just smoothly sails through those values.

As with the other function, we can construct estimates of our function from the first so many terms of the infinite sum. These should get closer and closer to our function just like when we first tested to see if we could get to \(\tfrac{4}{3})\) by adding up terms of the sum. Remember we saw that the more terms we added onto the infinite sum, the closer the total did indeed get to the answer we got from our function. We can do that again, this time starting by graphing the function \(g(x)=1-x^2\). That looks like this:

OK, that looks a bit different to our last one as well. Why? Well look what we do with our \(x^2\). We subtract it instead of adding it. That is why the line in this one disappears off below our number line instead of heading off to the top. You can see, as it was last time, that around the zero value the function we just drew is a pretty good fit but it all goes to hell shortly afterwards. Ok, let's add \(x^4\) onto our function and see where that takes us (in cyan):

Bloody hell, what's happened here then? Well the plus \(x^4\) drowns out the other parts of the function, so this time it stays positive. Again it looks like it is trying to match the real function around the zero point but then it heads off at high speed up the way. Lets now subtract the next term which is \(x^6\). We'll draw resulting line in green:

That is starting to look distinctly odd. The green one seems to flatten out along the real line and then it spears off down the way. Like adding the previous term, the subtraction of the sixth power of \(x\) is now dominating, so the function is only very briefly positive. Also, can you see that the lines at either end head down the way at a much more pronounced angle than the blue line? OK, lets add \(x^8\) to the mix to see what happens. We'll do this one in red:

It looks much like we have come to expect. Very similar to the \(x^4\) addition, but with the ends tilted in much more - like the previous one where we subtracted \(x^3\). What is going on with these line ends? They are acting much like the estimate lines in the previous function when they were getting closer and closer to the real line that went off to infinity. But HERE the real line doesn't go off to infinity at all. Just to make what seems to be happening clearer, I'll graph the functions that would result if we worked out all the terms to our sum up to subtracting \(x^{18}\) in green, and up to adding \(x^{20}\) in red. We'll take the other functions we've drawn out, apart from the real one, so we can see what is going on:

That's pretty clear. The more and more powers of \(x\) that you add on to the sum, the faster and more dramatically they head off to positive and negative infinity. And look where they are doing it! Right at plus and minus one. But why is this? The real function that short cuts to the answer is nice and smooth at plus and minus one. In fact it is nice and smooth all along the graph. Why on earth does the shortcut to the answer stay normal at plus and minus one, but the actual totals of the infinite sum go daft at plus and minus one?

The answer is that the number \(x\) that we plug into the function to give us our graph is, like every number, made up of a real and an imaginary part. What we are seeing at plus and minus one on that graph above, is the effect of the imaginary part of \(x\) sending the function off to infinity. I thought in school that imaginary numbers were just tacked onto real numbers to answer questions like what is the square root of minus one. In other words, that they were an artificial invention of mathematicians. But here you have no square roots of negative numbers at all. You just have plus and minus various powers of \(x\). Nevertheless though, the imaginary part of \(x\) is making itself known by sending the totals of the sums closer and closer to infinity at plus and minus one. So imaginary numbers are not imaginary at all - they exist and have an effect even when we don't invoke them, or expect them, at all.

Why does this happen? Well, very basically, remember that each number has a real and imaginary part. With the function \(f(x)=\frac{1}{1-x^2}\) when the REAL part of \(x\) is plus or minus the square root of one (i.e. either positive or negative one), you square it, getting one, which you then subtract from one getting zero on the bottom of the equation which breaks it. With \(g(x)=\frac{1}{1+x^2}\), when the IMAGINARY part of \(x\) is plus or minus the square root of minus one (i.e. plus or minus \(i\)) you then square the square root of minus one getting ... minus one, you cretin. When you add minus one to one and you get ... zero on the bottom of the equation which breaks it just as badly as when the REAL part of \(x\) is zero.

But hang on I hear you cry, the function goes nuts around the REAL numbers plus and minus one. As I was at pains to say this number line is for real numbers only. It does not show imaginary numbers. If the imaginary part of a number is responsible for sending the function off to infinity, then the whole number must square to negative one. If it does so then its real part must be zero. In other words why doesn't the function go daft at the real value zero?

The answer involves the absolute value of numbers and something called the circle of convergence. Which sounds like it should be drawn on a dusty basement floor in pigs blood and surrounded by hooded figures illuminated by large dribbly white candles. Both of these concepts will require pictures. More of which next time.

Real or Imaginary Part 3

Last time we took two infinitely long sums and we worked out general equations which allow us to jump straight to the answer without having to add up an infintely large amount of individual terms. We have expressed our series as a list of \(x\) variables to various powers, and unsurprisingly our formulas are also in terms of the \(x\) variable. Because we are working with formulas we can now call them functions which will be \(f(x)\) and \(g(x)\). This is because there is only one variable in the formula - the \(x\) bit.

Remember that an infinite sum can either add up to some specific number, or it can go nuts and head off to infinity. Now that we have these to functions, if you want to know what will happen with a specific infinite sum is to pick a number, any number, stick it in the function in place of the \(x\). The output of the function will tell you whether or not the infinite sum for that number adds up to any specific number, or disappears off to infinity. When I say "pick a number" this would be the number that would appear in place of the half in the brackets in this description of the sum:

\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\]

What we are interested in, for the purpose of thinking about imaginary numbers, is at what point the infinite sums we are looking at go from a finite to an infinite answer. When I say "what point" I really mean which numbers when plugged in in place of \(x\) give infinite answers and which give finite answers. What we could do to find this out is write down a list of the numbers we are interested in and, next to them, the output from the function we calculated last time. We could then consider the answers. We've already worked out \(f(x)=\frac{1}{1-x^2}\) is four thirds, so let's try numbers around a half to start with.

\[f(0)=1\]
\[f(\tfrac{1}{4})=\tfrac{16}{15}\]
\[f(\tfrac{1}{2})=\tfrac{4}{3}\]
\[f(\tfrac{3}{4})=\tfrac{16}{7}\]
\[f(1)=\]
Oh dear, it is broken. We cannot divide by zero, which is what one minus one squared is. We could consider this to be an infinite answer. And because we square the \(x\) component, we know that whether the \(x\) starts as positive or negative it will square to the same number. In other words we know that the result for \(f(-\tfrac{1}{4})\) is still going to be \(\tfrac{16}{15}\)

That all looks a bit boring though, and it hardly gives us a feeling for what is actually happening with the output of the function as the numbers that we put into it change. Lets make things a bit more dramatic by making them graphic! How do we do that then? Lets draw a line just like when we were making our number line:

Now lets mark the centre of the line as zero:

Now let's mark off the other values for which we have just calculated the function along the line which passes through zero, using the left of zero as negative numbers and the right of zero as positive numbers. Lets make these all proportionate, so the distance between a half and one is the same as negative three quarters and negative one quarter. In other words lets mark these off to scale:

Note that this is just a one dimensional line. It does not have an imaginary vertical axis like our number plain. Any dots that appear above the number line are NOT imaginary numbers. Instead we will mark off the \(f(x)\) values for each of the values on the number line. We will do so by just placing a dot the same distance above the line that the value turned out to be. So for instance, if you stuck a half into the function the output was four thirds. So our dot will be four thirds of a unit above the line. A "unit" here is just the distance from zero to one on the line. This is just going to be a simple graph to give us a visual idea of how the output of the function behaves depending on changes to the input. Let's start with \(f(\tfrac{1}{2})\):

See the dot? That represents our value for \(f(\tfrac{1}{2})\). It is four thirds of a unit directly above the marker for a half along the line. Lets fill in dots for all the other values that we know about:

Hey! We got a smiley face! Can we see what is happening now though? The dots start off high (technically infinite at negative one) and get lower as they get closer to zero. Then they start getting higher again. Of course we have only marked off seven values. Those seven values fall on an invisible line that would appear if we calculated ALL the infinitely many \(f(x)\)'s between negative one and one. We can still draw that line though, by telling the computer to calculate hundreds of different \(f(x)\)'s. We are still drawing dots, but now we are drawing so many that it will just look like a line. That line looks like this:

The black lines at the sides are just the computers way of trying to draw infinity. It just draws that as a line going directly up and down. But, do you see the rest of the curved line going through all the dots that we plotted? That line is called the plot of the function \(f(x)=\frac{1}{1-x^2}\). This gives you a much better idea of what is going on. As we calculate values closer and closer to plus and minus one, the function just takes off towards infinity. To give an example of this, lets calculate two more values of \(x\) - between the two extremes on the outside (plus and minus three quarters) and plus and minus one. If we make them mid point between those, we will end up with plus and minus seven eighths. Which is fine.

What we should see is plus and minus one divided by one minus seven eighths squared. Seven eighths squared is seven times seven (forty nine) divided by eight times eight (sixty four). Take that away from one and you get fifteen sixty fourths. One divided by fifteen sixty fourths is the same as sixty four fifteenths. So the answer at plus and minus seven eighths is sixty four fifteenths. That looks like:

The image is now much taller, to fit in the dots getting higher and higher. See how the gap between three quarters and seven eighths is much bigger than all the other gaps. I also had to make the text at the bottom a bit smaller so it would all fit in.

Just in case we are getting too far away from where we started, lets remember that the right most dot in that picture represents the total that you get if you add up:

\[1+(\tfrac{7}{8})^2+(\tfrac{7}{8})^4+(\tfrac{7}{8})^6+(\tfrac{7}{8})^8+\ldots\]

What we can also do is draw a plot on the graph that represents the first so many terms of the infinite sum. Think back to when we were first checking that we had successfully turned an infinite sum into a function. Remember the bit where we added up the first few terms of the infinite sum by working out what they were:

\[1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}=\frac{64+16+4+1}{64}=\tfrac{85}{64}\]

And then we asked if they were getting close to our prediction of \(\tfrac{4}{3}\). So we asked what \(\tfrac{4}{3}\) was in sixty-fourths. We chose sixty-fourths because they were the smallest fraction that we had reached so far in the infinite sum. We got the answer: \(\frac{85\tfrac{1}{3}}{64}\), so we confirmed that we were moving towards \(\tfrac{4}{3}\) as we added more terms to the infinite sum.

We can now do that using the graph, by plotting functions that gradually have more and more terms of the infinite sum in them, to see if we get close to the prediction. Last time we were doing it with individual numbers, this time it is whole functions. To start, we plot the function \(f(x)=1+x^2\). That looks like this:

I have drawn it in blue so it stands out. As you can see it is really close to the proper curve down by zero, but it gets less accurate quickly, the further you move out. What about adding another term in a different colour? Lets do \(f(x)=1+x^2+x^4\) in Cyan:

As we may have expected the line gets close again to the true total of the infinite sum. Lets add two more powers of \(x\): \(f(x)=1+x^2+x^4+x^6\) and \(f(x)=1+x^2+x^4+x^6+x^8\) in green and red respectively:

Again, much as we expected, the lines are getting progressively closer to the "true" line.

So what does this tell us? Well, the functions we are generating by adding more powers of \(x\) are getting closer and closer to infinity at plus and minus one. We know why. It is because the function for working out the actual answer to the infinite sum goes off to infinity at that very point. This is because you end up dividing by zero at plus and minus one. Because the "true" function goes to infinity, then all our estimates of that function get closer and closer to infinity at that point. All we are doing each time we add a new power of \(x\) to the estimate term is getting one step closer to the real answer, which is demonstrated by the real line.

So what does this have to do with imaginary numbers? To find out we will have to look at what happens when we try to graph our OTHER function: \(g(x)=\frac{1}{1+x^2}\)

Real or Imaginary Part 2

OK, last time we looked at these two formulas:
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\]
and
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\cdot-1^n\]
We worked out that the first one adds up to \(\tfrac{4}{3}\) and the second one \(\tfrac{4}{5}\).

Now lets replace the \(\tfrac{1}{2}\) in each one with \(x\). This makes both sums:
\[\sum_{n=0}^\infty x^{2n}\]
and
\[\sum_{n=0}^\infty x^{2n}\cdot-1^n\]
When we do the expansion they are going to look like this:
\[1+x^2+x^4+x^6+\ldots\]
and
\[1-x^2+x^4-x^6+\ldots\]
We can then set both sides equal to a variable. Best give them slightly different names so we do not get too confused doing this simultaneously. I think we can also dispense with the "ands".
\[S_a=1+x^2+x^4+x^6+\ldots\]
\[S_b=1-x^2+x^4-x^6+\ldots\]
OK. Now multiply both sides of the equation by \(x^2\). Hang on though - didn't we divide by four last time? Well, yes, but dividing by four is the same as multiplying by \(\tfrac{1}{4}\) and \(\tfrac{1}{2}^2=\tfrac{1}{4}\), so this step is actually the same as we did last time. It gives us:
\[x^2\cdot S_a=x^2\cdot (1+x^2+x^4+x^6+\ldots)\]
\[x^2\cdot S_b=x^2\cdot (1-x^2+x^4-x^6+\ldots)\]
If we then multiply out the right hand side we get:
\[x^2\cdot S_a=x^2+x^4+x^6+x^8+\ldots\]
and
\[x^2\cdot S_b=x^2-x^4+x^6-x^8+\ldots\]
Now we have defined our long series of fractions we can substitute into \(S_a\) and \(S_b\) above.
\[S_a=1+x^2\cdot S_a\]
\[S_b=1-x^2\cdot S_b\]
We then subtract, or as the case may be add the \(x^2\) multiple of \(S_{a/b}\) to both sides:
\[S_a-x^2\cdot S_a=1\]
\[S_b+x^2\cdot S_b=1\]
We then look at the left hand side of the equation and we spot that the \(S\) term appears twice. We can actually reform the left hand side by working out what we need to multiply by the \(S\) term. Here it looks like this:
\[S_a(1-x^2)=1\]
\[S_b(1+x^2)=1\]
We then divide both sides by the expression in the brackets. (Look carefully and if you multiply the stuff in brackets by the \(S\) term, you get the same left hand side as we had in the last step.
\[S_a=\frac{1}{1-x^2}\]
\[S_b=\frac{1}{1+x^2}\]
What we have now achieved are two functions of \(x\). You can see this because only the right hand side has an \(x\) in it. We have named these things already, but now they are functions, lets give them proper names:
\[f(x)=\frac{1}{1-x^2}\]
\[g(x)=\frac{1}{1+x^2}\]
OK. Lets check that we have done this right by sticking in \(\tfrac{1}{2}\) for \(x^2\) in each equation and seeing if we get the same results as last time:
\[f(\tfrac{1}{2})=\frac{1}{1-\tfrac{1}{2}^2}\]
\[g(\tfrac{1}{2})=\frac{1}{1+\tfrac{1}{2}^2}\]
Then:
\[f(\tfrac{1}{2})=\frac{1}{1-\tfrac{1}{4}}\]
\[g(\tfrac{1}{2})=\frac{1}{1+\tfrac{1}{4}}\]
Then:
\[f(\tfrac{1}{2})=\frac{1}{\tfrac{3}{4}}\]
\[g(\tfrac{1}{2})=\frac{1}{\tfrac{5}{4}}\]
And remember that dividing by a number is the same as multiplying by the reciprocal:
\[f(\tfrac{1}{2})=1\cdot \tfrac{4}{3}\]
\[g(\tfrac{1}{2})=1\cdot \tfrac{4}{5}\]
And finally:
\[f(\tfrac{1}{2})=\tfrac{4}{3}\]
\[g(\tfrac{1}{2})=\tfrac{4}{5}\]
Which is exactly what we worked out last time. So our formulas work just fine. Now that we have a formula for working out the solution to the infinite sum based on one variable, we can make a graph of that function and it will show us at what point our infinite sum goes from a sensible proper answer to an infinite answer. That fun will have to wait for next time. For now, lets sum up (haha) what we have established so far:
\[\sum_{n=0}^\infty x^{2n}=\frac{1}{1-x^2}=f(x)\]
and
\[\sum_{n=0}^\infty x^{2n}\cdot-1^n=\frac{1}{1+x^2}=g(x)\]

Real or Imaginary Part 1

OK, so we just made up a number \(i\) which if you multiplied it by itself you get negative one. That sounds like the laziest way of solving the problem of the square root of negative one - just make shit up. Does it actually exist (if any number can be said to exist)? let's see if I can convince you it does.

First of all remember our infinite sums. We have decided that some of them add up to infinity, and others add up to a finite number, like two. I want to consider two more infinite sums. Lets look at these:
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\]
and
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\cdot-1^n\]
Right. Those look a bit odd. First of all they do not have \(x\) in them, it has been replaced with \(n\). Why? Well, we'll come back to that. I want to do a general version of these two in due course, where I will replace \(\tfrac{1}{2}\) with \(x\), so I don't want to confuse things by having \(x\) appear now.

So what do these look like expanded out? Lets work through the first one. For \(n\) equal to zero, we get one, because the exponent is two times zero, which is zero and everything to the power of zero is one. So we have:
\[1+\ldots\]
Good. Now lets go to \(n\) as one. Now we have two times one as the exponent, which is just two. So we need to square a half. Half of a half is a quarter. So we are now up to:
\[1+\tfrac{1}{4}+\ldots\]
Great. OK, with \(n\) as two, we are going to have the exponent equal to two times two, or four. This means that we need to do a half of a half of a half of a half, or a sixteenth.
\[1+\tfrac{1}{4}+\tfrac{1}{16}+\ldots\]
Fantastic. Lets do one more with \(n\) equal to three. We get an exponent equal to six, and I will just tell you that a half to the power of six is a sixty-fourth.
\[1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots\]
This looks quite like our previous infinite sum, in that the amounts added are getting smaller very quickly. In fact each term is a quarter of the last term. We can work out what number this is going to approach. We do so in the same way as before. Lets pick a name for the series \(S\):
\[S=1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots\]
Now let us divide both sides of the equation by four:
\[\frac{S}{4}=\frac{1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots}{4}\]
Again, we can just divide each term on top of the large fraction by four. Just like we saw before what will happen is that the one drops off the end and all the rest of the infinite terms march one place to the left. But an infinitely long list of numbers less one is still an infinitely long list of numbers.
\[\frac{S}{4}=\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\tfrac{1}{256}+\ldots\]
Again we can spot that this term \(\frac{S}{4}\) is what we add to one to get \(S\) above. I mean, you look up to the first place where we say what \(S\) is. \(S\) is one plus a whole long list of fractions. If you compare that long list of fractions to the list that we now know is equal to \(\frac{S}{4}\), you should see that they are identical. So we can just replace the long list of fractions with \(\frac{S}{4}\), which tidies things up dramatically.
\[S=1+\frac{S}{4}\]
If that is the case then to get rid of the fraction we can just multiply both sides by four:
\[4\cdot S=4\cdot(1+\frac{S}{4})\]
\[4S=4+S\]
We can then subtract an \(S\) from both sides.
\[3S=4\]
Finally we can divide both sides by three to find out what \(S\) actually is:
\[S=\tfrac{4}{3}\]
Does this make sense? Lets look at the results of adding up the terms that we worked out above:
\[1=1\]
\[1+\tfrac{1}{4}=\frac{4+1}{4}=\tfrac{5}{4}\]
\[1+\tfrac{1}{4}+\tfrac{1}{16}=\frac{16+4+1}{16}=\tfrac{21}{16}\]
\[1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}=\frac{64+16+4+1}{64}=\tfrac{85}{64}\]
What number are those results moving towards? Well let me put it this way. What is \(\tfrac{4}{3}\) in sixty-fourths?
\[\frac{85\tfrac{1}{3}}{64}\]
So, yes we are moving towards \(\tfrac{4}{3}\) as we add more terms. OK, so let's look at the second infinite sum I put up there.
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\cdot-1^n\]
The only difference is that it multiplies each term by \(-1^n\). What effect does that have on proceedings? Well, remember that the zeroth term (where \(n=0\)) will be multiplied by \(-1^0\) and ANYTHING raised to the power of zero is one. So the term is multiplied by one, so remains the same. But what happens when \(n=1\)? Well negative one to the power one is just negative one. So whatever was our next term will be multiplied by negative one. That means we are going to have to subtract it instead of adding it. The next term will be multiplied by negative one squared, which is one (as we saw before), so it stays the same. The next term will be multiplied by negative one cubed. What will that be? Well, one way to think about it is negative one squared multiplied by negative one. We know what negative one squared is - it is just one. One multiplied by negative one is negative one. So we will end up subtracting this term instead of adding it.

Can you see what is going to happen here? This \(-1^n\) multiple that I have introduced is going to alternate between one and negative one all the way to infinity. The effect is that it is going to flip every other addition sign between the terms in our equation to a subtraction. But that is all. So instead of the term:
\[S=1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots\]
We are going to get:
\[S=1-\tfrac{1}{4}+\tfrac{1}{16}-\tfrac{1}{64}+\ldots\]
We can divide both sides by four again to get this:
\[\frac{S}{4}=\frac{1-\tfrac{1}{4}+\tfrac{1}{16}-\tfrac{1}{64}+\ldots}{4}\]
BUT look what happens when we divide through all the terms. Each term moves to the left BUT IT LEAVES ITS SIGN BEHIND!
\[\frac{S}{4}=\tfrac{1}{4}-\tfrac{1}{16}+\tfrac{1}{64}-\tfrac{1}{256}+\ldots\]
So this time when we look up, we cannot spot a duplicate of our long list of fractions because the signs have switched. What to do?

Well, lets look at what happens when you take a series of additions and subtractions and deduct them all from zero. Lets use letters to stand in for any numbers at all. I am going to use \(a,b,c,d\) to save confusion. Lets start with:
\[a-b+c-d=E\]
Now, lets take away both sides from zero:
\[0-(a-b+c-d)=0-E\]
Now get rid of the brackets:
\[(0-a)+(0--b)+(0-c)+(0--d)=0-E\]
\[-a+b-c+d=-E\]
We know what \(E\) is, so we can write:
\[-a+b-c+d=-(a-b+c-d)\]
So can you see what happens when you negate a whole string of numbers? Each number in the string flips its sign from positive to negative. Just like we saw with our long list of fractions in this case. So we can actually say that:
\[S=1-\frac{S}{4}\]
Remember that the FIRST term in \(\frac{S}{4}\) is \(+\tfrac{1}{4}\) so the NEGATIVE sign between the one and the \(\frac{S}{4}\) turns the \(+\tfrac{1}{4}\) into \(-\tfrac{1}{4}\), and vice versa for all the other entries in that string of fractions. Guess what? All the signs have now switched back again - hooray! So what next? Well lets multiply through by four again:
\[4\cdot S=4\cdot (1-\frac{S}{4})\]
\[4S=4-S\]
This time we have to ADD \(S\) to each side:
\[5S=4\]
And then divide both sides by five:
\[S=\frac{4}{5}\]
Again lets check that makes sense:
\[1=1\]
\[1-\tfrac{1}{4}=\frac{4-1}{4}=\tfrac{3}{4}\]
\[1-\tfrac{1}{4}+\tfrac{1}{16}=\frac{16-4+1}{16}=\tfrac{13}{16}\]
\[1-\tfrac{1}{4}+\tfrac{1}{16}-\tfrac{1}{64}=\frac{64-16+4-1}{64}=\tfrac{51}{64}\]
What number are those results moving towards? Well let me put it this way. What is \(\tfrac{4}{5}\) in sixty-fourths?
\[\frac{51\tfrac{2}{5}}{64}\]
Yay!
OK, next time we'll reintroduce the \(x\) variable in place of the \(\tfrac{1}{2}\) that we used this time.