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.
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