Truncated RM Algorithm and TS Method

In Section 1.2 we considered the root-seeking problem where the sought-for root may be any point in If the region belongs

to is known, then we may use the truncated algorithm and the growth rate restriction on can be removed.

Let us assume that and is known. In lieu of (1.2.2) we now consider the following truncated RM algorithm:

where the observation is given by (1.2.1), is a given point,

and

The constant used in (1.4.1) will be specified later on.

The algorithm (1.4.1) means that it coincides with the RM algorithm when it evolves in the sphere but if exits the sphere then the algorithm is pulled back to the fixed point

We will use the following set of conditions:

A1.4.1 The step size satisfies the following conditions

A1.4.2 There exists a continuously differentiable Lyapunov function

(not necessarily being nonnegative) such that

and for (which is used in (1.4.1)) there is such that

A1.4.3 For any convergent subsequence of

where is given by (1.3.2);

A1.4.4 is measurable and locally bounded.

We first compare these conditions with A1.3.1–A1.3.4. We note that A1.4.1 is the same as A1.3.1, while A1.4.2 is weaker than A1.2.2.

The difference between A1.3.3 and A1.4.3 consists in that Condition

(1.4.2) is required to be verified only along convergent subsequences, while (1.3.3) in A1.3.3 has to be verified along the whole sequence

It will be seen that A1.4.3 in many problems can be verified while A1.3.3

is difficult to verify.

Comparing A1.4.4 with A1.3.4 we find that the conditions on have now been weakened. The growth rate restriction used in Theorem 1.2.1 and the boundedness assumption on imposed in Theorem 1.3.1 have been removed in the following theorem.

Theorem 1.4.1 Assume Conditions A1.4.1, A1.4.2, and A1.4.4 hold

and the constant in A1.4.2 is available. Set for (1.4.1). If for some sample path A1.4.3 holds, then given by (1.4.1) converges to for this

Proof. We say that crosses an interval

if and

We first prove that the number of truncations in (1.4.1) may happen at most for a finite number of steps. Assume the converse: there are infinitely many truncations occurring in (1.4.1). Since

by A1.4.2, there is an interval such that

and there are infinitely many that cross

Since is bounded, we may extract a convergent subsequence from Let us denote the extracted convergent subsequence still

by It is clear that

Since the limit of is located in the open sphere there is an such that

for all sufficiently large

Since is bounded by Al.4.4 and the boundedness of using (1.4.2) we have

if is small enough and is large enough. This incorporating with (1.4.5) implies that

Therefore, the norm of

cannot reach the truncation bound In other words, the algorithm (1.4.1) turns to be an untruncated RM algorithm (1.4.7) for

for small and large

By the mean theorem there exists a vector with components located in-between the corresponding components of and such that

Notice that by (1.4.2) the left-hand side of (1.4.6) is of for all sufficiently large since is bounded. From this it follows that i) for small enough and large enough

and hence and ii) the last term in (1.4.8) is of

since as From (1.4.7) and (1.4.8) it then

follows that

Since the interval does not contain the origin. Noticing

that we find

for sufficientlysmall and all large enough Then by A1.4.2 there is such that

for all large and small enough As mentioned above from (1.4.9) we have

for sufficiently large and small enough where denotes a mag- nitude tending to zero as

Taking (1.4.4) into account, from (1.4.10) we find that

for large However, we have shown that

The obtained contradiction shows that the number of truncations in (1.4.1) can only be finite.

We have proved that starting from some large the algorithm (1.4.1) develops as an RM algorithm

and is bounded.

We are now in a position to show that converges. Assume it were not true. Then we would have

Then there would exist an interval not containing the origin

and would cross for infinitely many

Again, without loss of generality, assuming by the same argument as that used above, we will arrive at (1.4.9) and (1.4.10) for large and obtain a contradiction. Thus, tends to a finite limit

as

It remains to show that

Assume the converse that there is a subsequence

Then there is a such that for all sufficiently large We still have (1.4.8), (1.4.9), and (1.4.10) for some

Tending in (1.4.10), by convergence of we arrive at a contradictory inequality:

This means

In this section we have demonstrated an analysis method which is different from those used in Sections 1.2 and 1.3. This method is based on analyzing the sample-path behavior, and conclusions on the whole sequence are deducted from the local behaviors of estimates that are obtained immediately after which denotes a convergent subsequence of We call this method as Trajectory-Subsequence (TS) Method. The TS method is the main tool to be used in subsequent chapters for analyzing more general cases. It will be seen that the TS method is powerful in dealing with complicated errors including both random noise and structural inaccuracy of the function.

The obvious weakness of Theorem 1.4.1 is the assumption on the avail- ability of the upper bound for This limitation will be removed later on.