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.
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