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By J. Todd

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**Example text**

N. It is clear that F(x) = 0 and F(xc) = 0, i = 0, 1, 2, ... , n. It follows from RoLLE's Theorem that F'(z) vanishes for (n + 1) distinct values of z, none of which coincicle with any xc. It is obvious that F'(xc) = 0, i = 0, 1, 2, ... , n. x,, i = 0, 1, 2, ... , n. Repeated application of RoLLE's Theorem shows that p< 2 n+ 2l(z) must vanish at some point Cin this interval. We note that H(f, z) is a polynomial of degree (2 n + 1) at most and so H< 211 + 2> = 0. Since II (z - xc)8 is a polynomial of degree (2 n + 2) with leading coefficient unity, it follows that its (2 n + 2)th derivative is (2 n + 2) !

067621 . ] If where a0 , a 1 , ••• , are arbitrary complex numbers, show that If(z) I ;;:: Iao I + Ian I and there is equality if and only if f (z) = a + an zn. max l•I ~ 1 0 36 CHAPTER 4 The Theorems of the Markoffs Let Pn(x) be a polynomial of degree n, such that j Pn(x) j ~ 1 , - 1 ~ x ~ 1. Does this imply any restriction on the bounds of the derivatives of Pn(x) for - 1 ~ x ~ 1 ? This question was raised by the chemist MENDELIEFF for the case of p~(x), and was answered by A. A. MARKOFF in 1890.

We shall now discuss a little further the concept of completeness introduced earlier. 8) If the range [a, b] is finite the corresponding sequence {:n;n(x)} is complete (for continuous functions). Proof. Suppose f(x) is continuous in [a, b]. Take any e > 0. 6) b J[J(x) - fn(x)] 8 w(x) dx ~ o b J[f(x) - p(x)] 8 w(x) dx a and so J[f(x) b b fn(x)]• w(x) dx a =/ fl f•(x) w(x) dx - " J; a¥< µ r-0 Hence b 0 :::;;: / f•(x) w(x) dx - .. 00 ,~ aP < µ 0 e• . f f b 00 8 (x) w(x) dx = ,~ a: . C. of any continuous f(x) vanish, then so does f.