Find $x_{50}$ quadratic equation

Question

I don't really know how to title this question

You are provided the quadratic equation $ax^2 + x -1 = 0$ with a root $x$, with formula $x= \frac{-1+\sqrt{1+4a}}{2a}$ for each $a = 2^{-p}$,Where $(p=1,2,...,n)$ Let's define the root $x_{p}$ by the formula where $$\lim_{p\to\infty} x_{p}=1$$ (i) Assume you are using a computer that uses IEEE double precision arithmetic with rounding, find $x_{50}$

I'm not sure where to start with this problem, I've begun with placing $a=2^{-n}$ inside our formula for x.

So, $x= \frac{-1+\sqrt{1+4*2^{-p}}}{2*2^{-p}}$

I don't know where the limit comes into this but if I wanted to find $x_{50}$ would i plug in $p=50$

Any guidance on this problem would be greatly appreciated..

Answer 1

Where the limit comes in? That quadratic has two roots. One of them is close to $1$. The other is close to $-2^p$. Saying that the limit is $1$ is saying that we take the former, with the positive square root.

And then, yes, you plug in $p=50$. I'm pretty sure this exercise is about showing the impact of a bad subtraction - your answer will be something that isn't close to $1$.

I tested this in an Excel spreadsheet I happened to have open. Result: Exactly $1$. Still $1$ up to $p=53$, then becomes $0$ at $p=54$. Don't take this as gospel, though - I wouldn't trust Excel to implement the IEEE standard exactly.

Answer 2

Your formula, $\lim_{p\to\infty} x_p= 1$ does NOT define "$x_p$"! The "previous formula" was $x_p= \frac{-1+ \sqrt{1+ 4*2^{-p}}}{2*2^{-p}}$ although you never say that. So $x_{50}= \frac{-1+ \sqrt{1+ 4*2^{-50}}}{2*2^{-50}}= 2^{48}\left(-1+ \sqrt{1+ 2^{-48}}\right)= -2^{48}+ \sqrt{2^{96}+ 2^{48}}$.

Answer 3

I would plug in the 50 as you did, which can do no harm but get that term in the sequence, which comes out to 1 exactly (rounded beyond the number of digits my computer holds). One can confirm the limit of $\underset{p\rightarrow \infty}{lim} x_p = 1$ via L'Hopital's rule to yield $\underset{p\rightarrow \infty}{lim}(1+4*2^{-p})^{-1/2}$. It is pretty crazy to find this out numerically as it gives NaN for p = 100000. Perhaps the problem is showing the value of analysis over numerics. If you want to see the computations after applying L'Hopital's rule I can show them but I think you wanted to know the point of this and I am guessing as to such...

Answer 4

The point of the exercise seems to me that the equation

$$ \frac{-1+\sqrt{1+4a}}{2a} = \frac{2}{1+\sqrt{1+4a}} \tag{1}$$

holds true and if $\,|a|<<1\,$ then the numerator of the left side suffers from catastrophic cancellation, while the denominator of the right side is well behaved.

By the way, the power series expansion of the expression is the generating function of the alternating signed Catalan numbers which is OEIS sequence A168491.