User:IssaRice/Chain rule proofs: Difference between revisions

Revision as of 01:24, 28 November 2018

Using Newton's approximation

Since $g$ is differentiable at $y_{0}$ , we know $g^{'} (y_{0})$ is a real number, and we can write

$g (y) = g (y_{0}) + g^{'} (y_{0}) (y - y_{0}) + [g (y) - (g (y_{0}) + g^{'} (y_{0}) (y - y_{0}))]$

If we define $E_{g} (Δ y) : = g (y) - (g (y_{0}) + g^{'} (y_{0}) (y - y_{0}))$ we can write

$g (y) = g (y_{0}) + g^{'} (f (x_{0})) (y - y_{0}) + E_{g} (Δ y)$

Newton's approximation says that $| E_{g} (Δ y) | \leq ϵ | y - y_{0} |$ as long as $| y - y_{0} | \leq δ$ .

Since $f$ is differentiable at $x_{0}$ , we know that it must be continuous at $x_{0}$ . This means we can keep $| f (x) - y_{0} | \leq δ$ as long as we keep $| x - x_{0} | \leq δ^{'}$ .

Revision as of 01:22, 28 November 2018 (view source) IssaRice (talk \| contribs) (Created page with "==Using Newton's approximation== Since <math>g</math> is differentiable at <math>y_0</math>, we know <math>g'(y_0)</math> is a real number, and we can write <math>g(y) = g(y...")		Revision as of 01:24, 28 November 2018 (view source) IssaRice (talk \| contribs) No edit summary Newer edit →
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	Newton's approximation says that <math>\|E_g(\Delta y)\| \leq \epsilon\|y-y_0\|</math> as long as <math>\|y-y_0\|\leq \delta</math>.		Newton's approximation says that <math>\|E_g(\Delta y)\| \leq \epsilon\|y-y_0\|</math> as long as <math>\|y-y_0\|\leq \delta</math>.

			Since <math>f</math> is differentiable at <math>x_0</math>, we know that it must be continuous at <math>x_0</math>. This means we can keep <math>\|f(x)-y_0\|\leq \delta</math> as long as we keep <math>\|x-x_0\|\leq \delta'</math>.