Learning-rate stability

Problem

Let f(u,v)=12(u2+9v2)f(u,v)=\frac{1}{2}(u^2+9v^2). Derive the fixed learning-rate interval that converges from every start. From (1,1)(1,1), compute one update for η=0.2\eta=0.2 and explain what happens for η=0.25\eta=0.25.

Reveal answer or reference solution

The updates are

unext=(1−η)u,vnext=(1−9η)v.u_{\text{next}}=(1-\eta)u,\qquad v_{\text{next}}=(1-9\eta)v.

Both multipliers must have magnitude less than 11, giving

0<η<29.0<\eta<\frac{2}{9}.

At η=0.2\eta=0.2, the next point is (0.8,−0.8)(0.8,-0.8). At η=0.25\eta=0.25, the vv multiplier is −1.25-1.25, so nonzero vv oscillates with growing magnitude and diverges.

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