Does $x-x^2+x^2-x^2+x^2 - \dots=\frac14 x^2$?

Question

The following is just some context; you don't need to read it.

I had to find the locus of P(x,y) when x=2t and y=2t-t². I went about solving it, but I stumbled upon:

x=2t
y=2t-t²

y=x-t²
y=x-x²+3t²
y=x-x²+x²-t²
y=x-x²+x²-x²+3t²
...

y=x-x²+x²-x²+x²...

I knew this was wrong because none of the previous answers looked like this, so I tried it again.

x=2t
y=2t-t²

y=x-t²
y=x-x²/4

Instead of adding and subtracting, I divided for this attempt. I thought, were both my attempts correct in some sense? My steps seemed correct, and I don't think I divided by zero in any step. Does this mean x-x²+x²-x²+x²=x-x²/4?

Answer 1

The first approach is correct (though not that useful); but you make a mistake by extrapolating from your finite series to an infinite series. You correctly derived the following family of equations: $$ \begin{eqnarray} y &=& x-t^2 \\ y &=& x-x^2+3t^2 \\ y &=& x-x^2+x^2-t^2 \\ y &=& x-x^2+x^2-x^2+3t^2 \\ y &=& x-x^2+x^2-x^2+x^2-t^2 \\ &\cdots& \end{eqnarray} $$ But you can't go from these to $y=x-x^2+x^2-x^2+x^2-\ldots$. The final term, since it's not becoming smaller with each equation, still matters! In fact, since $-x^2+x^2$ can be cancelled when it appears, you really only have two distinct equations, the first two. Those can both be correct only if $-t^2=-x^2+3t^2$, or $t^2=x^2/4$; and then each reduces to the same answer as your second approach, $y=x-x^2/4$.