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Culegere de probleme de Analiz˘a numeric˘a

Culegere de probleme de Analiz˘a numeric˘a

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68 Calculul cu diferent¸e<br />

Problema 5.0.7 Aplicat¸ie. Vom stabili o formulă explicită pentru calculul sumei<br />

Sm,r = 1 r +2 r +3 r +···+m r<br />

cu ajutorul diferent¸elor lui 0.<br />

r<br />

<br />

m+1<br />

Sm,r = ∆<br />

i+1<br />

i=1<br />

i 0 r<br />

p<br />

<br />

p<br />

f(ap) = ∆<br />

k<br />

j=0<br />

k hf(a) ∆m m<br />

h f(a) = (−1)<br />

i=0<br />

m−i<br />

<br />

m<br />

f(a+ih)<br />

i<br />

f(x) = xr pr p<br />

<br />

p<br />

= f(p) = ∆<br />

k<br />

k=0<br />

k 0 r , p = 1,2,...,m<br />

1r <br />

1<br />

= ∆<br />

0<br />

0 0 r <br />

1<br />

+ ∆<br />

1<br />

1 0 r<br />

2r <br />

2<br />

= ∆<br />

0<br />

0 0 r <br />

2<br />

+ ∆<br />

1<br />

1 0 r <br />

2<br />

+ ∆<br />

2<br />

2 0 r<br />

...<br />

mr <br />

m<br />

= ∆<br />

0<br />

0 0 r <br />

m<br />

+ ∆<br />

1<br />

1 0 r <br />

m<br />

+···+ ∆<br />

m<br />

m 0 r<br />

m<br />

<br />

j j +1 m<br />

Sm,r = + +···+ ∆<br />

j j j<br />

j 0 r =<br />

j=1<br />

j=1<br />

Deoarece dacă m > r, ∆o0r = 0 pentru j = r +1,m iar pentru m < r,<br />

r<br />

<br />

m+1<br />

∆<br />

j +1<br />

j 0 r<br />

<br />

m+1<br />

= 0, pentru j = m+1,m+2,...,r.<br />

j +1<br />

Cazuri particulare <br />

m+1 m+1<br />

Sm,1 = ∆0 = =<br />

2 2<br />

m(m+1)<br />

, ∆0 = 1<br />

<br />

2<br />

2 m+1 ∆0<br />

Sm,2 =<br />

2 1 +<br />

2 2 m+1 ∆ 0 m(m+1)(2m+1)<br />

=<br />

3 2 6<br />

3 m+1 ∆0<br />

Sm,3 =<br />

2 1 +<br />

2 3 m+1 ∆ 0<br />

3 6 +<br />

3 3 m+1 ∆ 0<br />

4 6 =<br />

<br />

m(m+1)<br />

2<br />

Problema 5.0.8 Să se <strong>de</strong>monstreze formula<br />

(prin induct¸ie).<br />

∆ m h<br />

1<br />

x =<br />

(−1) m m!h m<br />

x(x+h)...(x+mh)<br />

2

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