Tema 12

Esercizio 9.44

Data la funzione $f:\mathbb {R} \to \mathbb {R}$$f:\mathbb {R} \to \mathbb {R}$

$$f(x)=(\sin x)^{2}$$

$$f(x)=(\sin x)^{2}$$

1. interpolarla con un polinomio $p(x)$$p(x)$ di grado 2 nei punti di ascissa $\pi /4$$\pi /4$, $\pi /2$$\pi /2$, $3\pi /4$$3\pi /4$ usando il metodo di Lagrange.
2. calcolare l'errore nel punto $x=\pi /3$$x=\pi /3$.
3. fornire una maggiorazione dell'errore di interpolazione sull'intervallo $[\pi /4,3\pi /4]$$[\pi /4,3\pi /4]$

 

Polinomio interpolante: I nodi sono $x_{1}=\pi /4,y_{1}=1/2$$x_{1}=\pi /4,y_{1}=1/2$; $x_{2}=\pi /2$$x_{2}=\pi /2$, $y_{2}=1$$y_{2}=1$; $x_{3}=3\pi /4$$x_{3}=3\pi /4$, $y_{3}=1/2$$y_{3}=1/2$.

$$p(x)=1/2*{\frac {(x-3\pi /4)(x-\pi /2)}{(\pi /4-3\pi /4)(\pi /4-\pi /2)}}+1*{\frac {(x-\pi /4)(x-3\pi /4)}{(\pi /2-\pi /4)(\pi /2-3\pi /4)}}+1/2*{\frac {(x-\pi /4)(x-\pi /2)}{(3\pi /4-\pi /4)(3\pi /4-\pi /2)}}$$

$$p(x)=1/2*{\frac {(x-3\pi /4)(x-\pi /2)}{(\pi /4-3\pi /4)(\pi /4-\pi /2)}}+1*{\frac {(x-\pi /4)(x-3\pi /4)}{(\pi /2-\pi /4)(\pi /2-3\pi /4)}}+1/2*{\frac {(x-\pi /4)(x-\pi /2)}{(3\pi /4-\pi /4)(3\pi /4-\pi /2)}}$$

$$p(x)=8/2*{\frac {(x-3\pi /4)(x-\pi /2)}{\pi ^{2}}}-16{\frac {(x-\pi /4)(x-3\pi /4)}{\pi ^{2}}}+8/2*{\frac {(x-\pi /4)(x-\pi /2)}{\pi ^{2}}}$$

$$p(x)=8/2*{\frac {(x-3\pi /4)(x-\pi /2)}{\pi ^{2}}}-16{\frac {(x-\pi /4)(x-3\pi /4)}{\pi ^{2}}}+8/2*{\frac {(x-\pi /4)(x-\pi /2)}{\pi ^{2}}}$$

$$p(x)={\frac {4(x^{2}+3/8\pi ^{2}-5\pi x/4-16(x^{2}+3\pi ^{2}/16-\pi x)+4(x^{2}+\pi ^{2}/8-3\pi x/4)}{\pi ^{2}}}$$

$$p(x)={\frac {4(x^{2}+3/8\pi ^{2}-5\pi x/4-16(x^{2}+3\pi ^{2}/16-\pi x)+4(x^{2}+\pi ^{2}/8-3\pi x/4)}{\pi ^{2}}}$$

$$p(x)={\frac {4x^{2}+3/2\pi ^{2}-5\pi x-16x^{2}-3\pi ^{2}+16\pi x+4x^{2}+\pi ^{2}/2-3\pi x}{\pi ^{2}}}$$

$$p(x)={\frac {4x^{2}+3/2\pi ^{2}-5\pi x-16x^{2}-3\pi ^{2}+16\pi x+4x^{2}+\pi ^{2}/2-3\pi x}{\pi ^{2}}}$$

$$p(x)={\frac {-8x^{2}+8\pi x-\pi ^{2}}{\pi ^{2}}}$$

$$p(x)={\frac {-8x^{2}+8\pi x-\pi ^{2}}{\pi ^{2}}}$$

Errore di interpolazione nel punto $x=\pi /3$$x=\pi /3$:

$$|f(\pi /3)-p(\pi /3)|=|(\sin \pi /3)^{2}-{\frac {-8*(\pi /3)^{2}+8\pi *\pi /3-\pi ^{2}}{\pi ^{2}}}|$$

$$|f(\pi /3)-p(\pi /3)|=|(\sin \pi /3)^{2}-{\frac {-8*(\pi /3)^{2}+8\pi *\pi /3-\pi ^{2}}{\pi ^{2}}}|$$

$$=|3/4-{\frac {-8/9*\pi ^{2}+8/3\pi ^{2}-\pi ^{2}}{\pi ^{2}}}|$$

$$=|3/4-{\frac {-8/9*\pi ^{2}+8/3\pi ^{2}-\pi ^{2}}{\pi ^{2}}}|$$

$$=|3/4-7/9|=1/36$$

$$=|3/4-7/9|=1/36$$

Stima dell'errore complessivo sull'intervallo:

$$E_{r}={\frac {f^{(3)}(\xi )}{3!}}*(x-\pi /4)(x-\pi /2)(x-3\pi /4)$$

$$E_{r}={\frac {f^{(3)}(\xi )}{3!}}*(x-\pi /4)(x-\pi /2)(x-3\pi /4)$$

$$f'(x)=2*\sin x*\cos x$$

$$f'(x)=2*\sin x*\cos x$$

$$f''(x)=2*(\cos ^{2}x-\sin ^{2}x)=2*\cos(2x)$$

$$f''(x)=2*(\cos ^{2}x-\sin ^{2}x)=2*\cos(2x)$$

$$f^{(3)}(x)=-4*\sin(2x)$$

$$f^{(3)}(x)=-4*\sin(2x)$$

$$|-4\cos(2x)|<4\forall x$$

$$|-4\cos(2x)|<4\forall x$$

$$E_{r}\leq 4/6*(x^{2}+\pi ^{2}/8-3/4\pi x)(x-3\pi /4)$$

$$E_{r}\leq 4/6*(x^{2}+\pi ^{2}/8-3/4\pi x)(x-3\pi /4)$$

$$E_{r}\leq 4/6*(x^{3}-3\pi /4x^{2}+\pi ^{2}/8x-3\pi ^{3}/32-3/4\pi x^{2}+9/16\pi x)$$

$$E_{r}\leq 4/6*(x^{3}-3\pi /4x^{2}+\pi ^{2}/8x-3\pi ^{3}/32-3/4\pi x^{2}+9/16\pi x)$$

$$E_{r}\leq 4/6*(x^{3}-3\pi /2x^{2}+(\pi ^{2}/8+16\pi )x-3\pi ^{3}/32)$$

$$E_{r}\leq 4/6*(x^{3}-3\pi /2x^{2}+(\pi ^{2}/8+16\pi )x-3\pi ^{3}/32)$$

$$E_{r}\leq 2/3*x^{3}-\pi x^{2}+2/3(\pi ^{2}/8+16\pi )x-\pi ^{3}/16$$

$$E_{r}\leq 2/3*x^{3}-\pi x^{2}+2/3(\pi ^{2}/8+16\pi )x-\pi ^{3}/16$$

Esercizio 9.45

Data la formula di quadratura,

$$I=af(0)+bf(1/2)+cf(1)+df'(0)$$

$$I=af(0)+bf(1/2)+cf(1)+df'(0)$$

per il calcolo approssimato di

$$I=\int _{0}^{1}f(x)\,dx$$

$$I=\int _{0}^{1}f(x)\,dx$$

con $f\in C^{1}([0,1])$$f\in C^{1}([0,1])$. Determinare i coefficienti $a,b,c,d$$a,b,c,d$ in modo che $I$$I$ abbia grado di precisione massimo. Di che formula si tratta? Utilizzare la formula di quadratura ottenuta per approssimare l'integrale definito

$$I(f)=\int _{0}^{1}\sin(\pi x)\,dx$$

$$I(f)=\int _{0}^{1}\sin(\pi x)\,dx$$

infine si valuti il numero minimo di intervalli necessari per calcolare $I(f)$$I(f)$ con un errore assoluto $\leq 10^{-3}$$\leq 10^{-3}$ utilizzando la formula dei trapezi composita.

 

$$C_{0}=\int _{0}^{1}1\,dx=1=a+b+c$$

$$C_{0}=\int _{0}^{1}1\,dx=1=a+b+c$$

$$C_{1}=\int _{0}^{1}x\,dx=1/2=b/2+c+d$$

$$C_{1}=\int _{0}^{1}x\,dx=1/2=b/2+c+d$$

$$C_{2}=\int _{0}^{1}x^{2}\,dx=1/3=b/4+c$$

$$C_{2}=\int _{0}^{1}x^{2}\,dx=1/3=b/4+c$$

$$C_{3}=\int _{0}^{1}x^{3}\,dx=1/4=b/8+c$$

$$C_{3}=\int _{0}^{1}x^{3}\,dx=1/4=b/8+c$$

$${\begin{cases}a+b+c=1\\b/2+c+d=1/2\\1/3=b/4+c\\1/4=b/8+c\end{cases}}$$

$${\begin{cases}a+b+c=1\\b/2+c+d=1/2\\1/3=b/4+c\\1/4=b/8+c\end{cases}}$$

Eguagliando le ultime due equazioni ricavo:

$$1/3-b/4=1/4-b/8$$

$$1/3-b/4=1/4-b/8$$

$$1/3-1/4=b/8$$

$$1/3-1/4=b/8$$

$$b/8=1/12,\,b=2/3$$

$$b/8=1/12,\,b=2/3$$

$$c=1/3-b/4=1/3-2/12=1/6$$

$$c=1/3-b/4=1/3-2/12=1/6$$

$$a=1-b-c,\,a=1-2/3-1/6={\frac {6-4-1}{6}}=1/6$$

$$a=1-b-c,\,a=1-2/3-1/6={\frac {6-4-1}{6}}=1/6$$

$$d=1/2-c-b/2=1/2-1/6-1/3=0$$

$$d=1/2-c-b/2=1/2-1/6-1/3=0$$

quindi ottengo la formula di Simpson

$$I=1/6f(0)+2/3f(1/2)+1/6f(1)$$

$$I=1/6f(0)+2/3f(1/2)+1/6f(1)$$

$$\int _{0}^{1}\sin(\pi x)\,dx=1/\pi *[-\cos(\pi x)]_{0}^{1}=1/\pi *[-\cos \pi +\cos 0]=2/\pi$$

$$\int _{0}^{1}\sin(\pi x)\,dx=1/\pi *[-\cos(\pi x)]_{0}^{1}=1/\pi *[-\cos \pi +\cos 0]=2/\pi$$

Con la formula di quadratura:

$$I\sim 1/6\sin 0+2/3\sin \pi /2+1/6\sin \pi =2/3$$

$$I\sim 1/6\sin 0+2/3\sin \pi /2+1/6\sin \pi =2/3$$

Con la formula dei trapezi composita:

$$E_{k,m}(f)=-h/12*H^{2}*f^{(2)}(\psi )$$

$$E_{k,m}(f)=-h/12*H^{2}*f^{(2)}(\psi )$$

$$h=1,\,H=1/m$$

$$h=1,\,H=1/m$$

$$f'(x)=\pi *\cos(\pi x)$$

$$f'(x)=\pi *\cos(\pi x)$$

$$f^{(2)}(x)=|-\pi ^{2}*\sin(\pi x)|\leq \pi ^{2}$$

$$f^{(2)}(x)=|-\pi ^{2}*\sin(\pi x)|\leq \pi ^{2}$$

$$-h/12*H^{2}*f^{(2)}(\psi )\leq 10^{-3}$$

$$-h/12*H^{2}*f^{(2)}(\psi )\leq 10^{-3}$$

$$1/12*\pi ^{2}*1/m^{2}\leq 10^{-3}$$

$$1/12*\pi ^{2}*1/m^{2}\leq 10^{-3}$$

$$1/m^{2}\leq {\frac {12}{\pi ^{2}*10^{3}}}$$

$$1/m^{2}\leq {\frac {12}{\pi ^{2}*10^{3}}}$$

$$m^{2}\geq {\frac {\pi ^{2}10^{3}}{12}}$$

$$m^{2}\geq {\frac {\pi ^{2}10^{3}}{12}}$$

$$m\geq \pi *{\sqrt {10^{3}/12}}=28,7$$

$$m\geq \pi *{\sqrt {10^{3}/12}}=28,7$$

quindi considero $29$$29$ intervalli.