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: \(g(x)=\frac{1}{1+x^2}\)
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