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
\[\sum_{n=0}^\infty x^{2n}\cdot-1^n=\frac{1}{1+x^2}=g(x)\]