User:EGM6341.S11.team5.cavalcanti/Mtg15

Mtg 15: Wed, 2 Feb 11

[[media:Nm1.s11.mtg15.djvu | Page 15-1]]

[[media:Nm1.s11.mtg15.djvu | Page 15-2]] Proof LIET cont'd

(4) p.14-3 → $$\left(n+1\right)$$zeros of G(.) $$G\left({x}_{i}\right)=\mathrm{0,}i=\mathrm{0,}\mathrm{...},n$$

(1) p.14-3 $$g\left(t\right)=e\left(t\right)-\frac{{q}_{n+1}\left(t\right)}{{q}_{n+1}\left(t\right)}e\left(t\right)=0 \Leftarrow$$ one more zero! → G(.) has n + 2 zeros

Remember G(.) vs qn+1(.)


 * qn+1(.) polynomial of degree n + 1


 * qn+1(xi)=0 i = 0, …, n → (n + 1) zeros


 * qn+1(t) not real zero

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positive polynomial of degree n + 1 (due to qn+1(x), with $$\begin{array}{c}\frac{1}{{q}_{n+1}\left(t\right)}{e}^{L}+n\left(f;t\right)=\mathit{constant}\\ t=\mathit{constant}\end{array}$$(fixed)) "End Note"
 * G(.) = not poly in general ($$\color{red}{e}_{n}^{L}\left(f;x\right)$$)
 * G(.) = 0 at {t, x0, …, xn} → n + 2 zeros

Rolle's theorem → Derivative of G(.), ie, G(1)(.) has at least (n + 1) zeros.

Ex. f(.) has 3 zeros. Rolle's thm → f'(.) has at least 3 – 1 = 2 zeros. Here, f'(.) has 4 zeros.