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354 Answers to Exercises and Solutions to Problems

Answers to Exercises

Chapter 7

1.5.

p.(t)= (t -ai)...(tG).+*(t- a,)

I

(Ui -Ul)***(Qi-0,) *

1.6.

q(t) = (7/15)(t - l)(t - 3) - (1/3)(t + 2)(t - 3) + (l/lo)(t + 2)(t - 1)

= (7/30)t2 - (43/30)t + (1615).

1.7. 45, 55, 66, 78,....

1.8. nth terms from 6th column back to 1st column: 0, 3,3n+B, (3/2)n2 +
(13/2)n + 7, (1/2)n3 + (5/2)n2 + 4n + 2, I(n) = (l/8)n4 + (7/12)n3 +
(7/8)n2 + (5/12)n.

1.10. (e)

k> (4

Akf(t) = &(-I)‘-’ ( : ) f(t + r).

r=O

f(n) = 2 + 9(n - 1) + (15/2)(n - l)(n - 2) + (ll/S)(n - l)(n - 2)(n - 3)

+ (l/S)(n - l)(n - 2)(n - 3)(n - 4)

= 2 + 9(n - 1) + (15/2)(n2 - 3n + 2) + (ll/6)(n3 - 6n2 + lln - 6)

+ (1/8)(n4 - 10n3 + 3571~ - 50n + 24)

which is the same as the answer to Exercise 1.8.

1.11. 6 + 44(n - 1) + (131/2)(n - 1)c2) + 34(n - 1)c3) + (22/3)(n - 1)c4) +
(2/3)(n - 1)c5).

1.12. (a) -0.00107, -0.00003, -0.00031.
(b) 1.32827.
(c) 1.32832.

1.13. log 1.25 = 0.22314; logo.75 = -0.28768; log2.1 = 0.74194;
log2.71828 = 1.00000.

1.15. Yes. Both polynomials have degree 6 and are each the uniquely de-
termined polynomial taking the assigned values.

1.16. (b) t4 = d4) + 6d3) + 7d2) + t, t5 = tc5) + 10tc4) + 25d3) + 15tc2) + t.
(c) The proof is by induction. Suppose all powers of t up to t” can be
written as a linear combination of factorial powers. Since tk+’ - dk+‘)