Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29
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A backward-stable computed solution of an ill-conditioned system can still have large forward error

Example

Fix an arithmetic model with unit roundoff 0<u<1 and set ε=u. Take A=diag(1,ε) and b=(1,ε), whose exact solution is x=(1,1). On the singleton input class consisting of this system, let an algorithm return x^=(1,0). Its normwise backward error is at most u/2, so this algorithm is backward stable on the declared class, yet its relative forward error is 1/2: ill conditioning is not instability, and backward stability does not remove ill conditioning.

Facts & Assumptions

Given: An arithmetic model with unit roundoff u(0,1), the parameter ε=u, the matrix A=diag(1,ε), the vector b=(1,ε), the exact solution x=(1,1), and an algorithm on the singleton input class I={(A,b)} that returns x^=(1,0).

[L1]

The normwise backward error is η2(x^)=r2/(A2x^2+b2) for r=bAx^ (Residual formulas for normwise and componentwise backward error).

[L2]

Along admissible perturbations tending to zero, relative forward error is at most (κrel+o(1)) times relative backward error (Local conditioning times backward error controls forward error to first order).

[A1]

For the fixed linear map f(b)=A1b, one has f(b+h)f(b)=A1h by linearity.

Verification

technique · direct
1.1

The residual is r=bAx^=(0,ε) and r2=ε; also A2=1, x^2=1 and b2=1+ε2.

algebra
1.2

The forward error is xx^2=(0,1)2=1 and the relative forward error 1/2, order one even as ε0+.

algebra
1.3

The conditioning conversion. For the problem map f(b)=A1b the absolute condition number is κabs=A12=1/ε, attained in the second coordinate, and the computed answer is exact for the perturbed input b~:=Ax^=(1,0)=b+h with h=(0,ε). By [A1], f(b+h)f(b)=A1h=(0,1), and direct substitution gives 1/2=κrelh2/b2.

A1algebra
2.1

By [L1], the normwise backward error is η2(x^)=ε/(1+1+ε2)ε/2=u/2. Thus the one constant C=1/2 works for every input in the declared singleton class I, so the algorithm is normwise backward stable there in the specialized sense of Forward and backward stability for a problem family under an arithmetic model.

givenL1step 1.1algebra
3.1

Steps 2.1, 1.2 and 1.3 show that this backward-stable algorithm on an ill-conditioned problem class has order-one forward error even though its backward error is order u; [L2] records the general first-order conditioning principle illustrated exactly here. Backward stability controls the data perturbation, and only conditioning controls the answer.

L2step 2.1step 1.2step 1.3

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