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Answers to Exercises; Chapter (^7 357)
Remark.
BUM) = f&(i) ( ; ) C-V-” ( :r,” )tr
= gg(-l)p-kf (:) ( ; ) ( ; ) t’
= g (e ) [ipr-k ( ; ) f (:)I it’
= Arf(0)tr = (1 + tA)“f(O)
where Af(t) = f(t + l/n) - f(t) in this situation. This gives a handy
way of computing the Bernstein polynomials in ascending powers of t. It
can be readily seen that deg B(f,n; t) 5 deg f(t) when f(t) is itself a
polynomial. In particular, when f(t) = t2, then f(0) = 0, Af(0) = I/n2
and A2f(0) = 2/n2 so that
B(t2, n;t) = n(l/n2)t +
Suppose B(f, n; t) = kf(t) for a polynomial f(t) = at’ + bt’-l +. .. of
degree r. Differentiating both sides of the equation r times leads to
Now, f(t) = at’+... can be written in terms of factorial powers as at(‘)+...
where, here, t(‘) = t(t - l/n)(t - 2/n).. -(t - (r - 1)/n) (to take account
of the changed differencing interval). Then t(‘) = (r/n)t(‘-‘), so that
A’&‘) = r(r - l)(r - 2)... (r - s + l)n-“t(‘-“).
Hence A”f(0) = ur!n-’ and k = n(n- 1)... (n-r+l)n-‘. Every eigenvalue
must have this form.
3.1. (a) (x2 - y2)2 + (22 - w2)2 + 2(xy - %W)2.
(b) If a, b, c, d > 0, then we can find real x, y, z, w such that a = x2,
etc. From (a) (a + b + c + d)/4 > xyzw = (ubcd)li4 as required.
3.3. Consider the case n = 3k. We have that
(yf” + YZk + y,““) + (y,“” +.. -) + ... 2 ~[(YIYzY~)~ +(Y4Y5Ys)" + ...I
1 3kylym....
3.4. The inequality holds for n = 2 and n = 3. For n = 2’3$, we can use
Exercise 2 repeatedly.