Ritz Method implies Not Worse Approximation with Increased Number of Functions

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Theorem

Consider the Ritz method.

Let $\eta_n = \boldsymbol \alpha \boldsymbol \phi$.

Let $J \sqbrk {\eta_n} = \mu_n$.


Then:

$\mu_n \ge \mu_{n + 1}$


Proof

Denote $\eta_{n + 1} = \eta_n + \alpha_{n + 1} \phi_{n + 1}$.

For $\alpha_{n + 1} = 0$ we have:

$\eta_n = \eta_{n + 1}$

Suppose that $J \sqbrk {\eta_n}$ has been minimised with respect to $\boldsymbol \alpha$.

If:

$\exists \boldsymbol \alpha \in \R^n: \nexists \alpha_{n + 1} \ne 0: J \sqbrk {\eta_n} > J \sqbrk {\eta_{n + 1} }$

then $J \sqbrk {\eta_{n + 1} }$ is minimised for previously determined $\boldsymbol \alpha$ and $\alpha_{n + 1} = 0$:

$J \sqbrk {\eta_{n + 1} } = J \sqbrk {\eta_n}$

where $\eta_{n + 1} = \eta_n$.


Suppose:

$\exists \boldsymbol \alpha \in \R^n: \exists \alpha_{n + 1} \ne 0: J \sqbrk {\eta_n} > J \sqbrk {\eta_{n + 1} }$

Then $\eta_{n + 1} \ne \eta_n$, because their respective $\boldsymbol \alpha$ differ by at least one value: $\alpha_{n + 1}$.

Hence, for this supposition:

$J \sqbrk {\eta_{n + 1} } > J \sqbrk {\eta_n}$

Finally, both cases together imply that:

$J \sqbrk {\eta_{n + 1} } \ge J \sqbrk {\eta_n}$

or equivalently:

$\mu_n \ge \mu_{n + 1}$

$\blacksquare$


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