Finite-precision direct factorizations and least squares — Examples

The exact first sum is 2^-1022-(2^-1022+2^-1074)=-2^-1074, a nonzero subnormal; [F1] therefore returns range-failure.

The exact sum N_max+N_max exceeds the largest finite normal magnitude, so the root triggers the same status under [F2].

The value 2^-53 is half an ulp at 1, so 1+2^-53 rounds to 1. Ordinary summation then returns 0.

Neumaier step 2 selects |s_1|>=|x_2| and gives t_2=1,q_2=0,a_2=2^-53,c_2=2^-53; step 3 gives t_3=0,q_3=1,a_3=0,c_3=2^-53.

The final exact addition 0+2^-53 is 2^-53, which is the exact sum; [F2] records this correction without claiming uniform relative accuracy.

The displayed quadratic form proves positive definiteness. The first pivot is $1$, ${\widehat{l}}_{21}=(1\mathrm{/}2)\mathrm{/}1=1\mathrm{/}2$, and the second pivot is $1-(1\mathrm{/}2{)}^2=3\mathrm{/}4$ exactly, so the procedure succeeds through the square-root step.

Every binary64 value is rational, whereas sqrt(3/4)=sqrt(3)/2 is irrational; hence l-hat_22^2!=3/4. Multiplication of the displayed factor gives exactly the stated nonzero E, to which [F2] applies.

The spacing above $1$ is $2^{-52}$, so $1+2^{-53}$ is the midpoint between $1$ and $1+2^{-52}$. The even-significand endpoint is $1$; hence $s=1$. The next exact operands give $z=1-1=0$, $v=1-0=1$, $w=1-1=0$, and $t=2^{-53}-0=2^{-53}$. Each is representable, so each subtraction is exact.

The final exact intermediate is $w+t=2^{-53}$, hence $e=2^{-53}$. The first exact sum $1+2^{-53}$ and every later nonzero exact intermediate ($1$ or $2^{-53}$) are normal and within range; the zero intermediates are allowed by [F3]. Thus the call succeeds, its format satisfies [F2], and $s+e=1+2^{-53}=a+b$ exactly, also directly verified by these outputs.

The determinant is 2^-27, while direct multiplication gives (A^T A)_22=1+2^-54.

Since 2^-54<2^-53, the exact entry is strictly below the RNE midpoint above 1 and rounds to 1. The rounded Gram matrix has equal rows and is singular.

The exact result [F1] acts on the full-rank A without a Gram formation. The computed conclusion [F2] applies only if its separate successful-step hypotheses hold.

Direct multiplication by A^-1=diag(1,2^40) gives x=A^-1b=(1,0)^T and x-tilde=A^-1b-tilde=(1,1)^T; uniqueness follows from [F1].

The input difference and x norm are both measured against unit vectors: ||b-tilde-b||_2/||b||_2=2^-40 and ||x-tilde-x||_2/||x||_2=1.

For every $z$, $\mathrm{\parallel }Az{\mathrm{\parallel }}_2^2=z_1^2+2^{-80}{z}_2^2\le \mathrm{\parallel }z{\mathrm{\parallel }}_2^2$, with equality at $e_1$, so [F4] gives $\mathrm{\parallel }A{\mathrm{\parallel }}_2=1$. Likewise $\mathrm{\parallel }{A}^{-1}z{\mathrm{\parallel }}_2^2=z_1^2+2^{80}{z}_2^2\le 2^{80}\mathrm{\parallel }z{\mathrm{\parallel }}_2^2$, with equality at $e_2$, hence $\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2=2^{40}$. Thus [F3] gives ${\kappa }_2(A)=2^{40}$. The exact data calculation proves order-one forward change under a relative $2^{-40}$ right-hand-side perturbation; backward stability [F2] alone is not a forward-accuracy guarantee.