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Unformatted text preview: Math 17B
Kouba Taylor Polynomials QUESTION : Consider the function y 2: f (:12) and assume that mathematical circum—
stances would require that this function be replaced with an nthdegree polynomial whose
values closely approximate the values of y = f (:13). How would we determine the coefﬁcients
of this unknown polynomial ? ANSWER : Assume that y = f (:c) is a given function and constant “a” is known. Deter
mine a list of real numbers (10, a1, a2, a3,    , an so that the polynomial is Pn(m) = a0 + a1(a: — a) + (12(11,‘ —‘ a)2 + a3(x — (1)3 +    + an(:c — a)" ,
which satisfies the condition
f($) W RICE),
i.e., (T) f(m) z a0 +a1(a: — a) +a2(:v—a)2 +a3($— (1)3 ++an(m — a)". If we assume that “ z ” is “ =” and substitute a: = a in equation (T), we get
f(0) = 00 + (11(0) + a2(0)2 + ay3(0)3 + . . . + an(0)n = a0,
i.e., a0 = f(a) . Now differentiate equation (T) term by term getting f’(x) = a1 + 2a2(a: — a) + 3a3(a: — (1)2 + 4a4(:1: — a)3 +    + nan(m — a)"'1.
If we substitute :5 = a in this equation, we get f’(a) = a1 + 2a2(0) + 3ag,(0‘)2 + 4a4(0)3 +    + nan(0)"'1 = a1 ,
i.e., a1 = f’(a) .
Now differentiate again term by term getting
f”(a:) = 2a2 + 3 2a3(:t: — a) + 4  3a4(a: — (1)2 + 5  4a5(a: —— (1)3 +   + n (n  1)anx"’2. If we substitute :1: = a in this equation, we get f”(a) = 2612 +3  2a3(0) +43a4(0)2 +54a4(0)3 +   .+ (n — 1) nan(0)”'2 = 202, = [7(0)
2! ' Now differentiate again term by term getting i.e., 02 f”’(a:)=32a3+432a4(x—a)+5~43a5(:1:—a)2+~~+n(n—1)(n—2)ana:""3. 1 If we substitute a: = a in this equation, we. get f’”(a) = 32a3+432a4(0) +543a5(0)2 +~ n(n—1)(n2)an(0)"‘3 = 3~2a3, i.e., Continuing this term by term differentiation and substitution process results in the fact
that (k) a ) forki=0,1’2,3’..,n' DEFINITION : The Taylor Polynomial of degree n centered at x = a for function
y = f (:13) is given by (P) Pn(m) = a0 + a1(a: ~— a) + (12(30 — (1)2 + a3(w  (1)3 +    + an(:r —— a)“
and (k)
(S) ak=fk'(a) forlc=0,1,2,3,~,n. QUESTION : Consider an ordinary function y = f and Pn(:v), its Taylor Polynomial
of degree n centered at a: = a. We assume that f z Pn Let the error be given by
Rn+1(:r;a) = f —— Pn(:c). It can be shown that the absolute error (Taylor Error) for their diﬂerence is Rn+1($; 0). = (1’3 ‘ a)n+1 3 where c is some number between :1: and a. (Note that the number c appears as a result of
using the Intermediate Value Theorem to derive this error formula.) ...
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 Winter '07
 Kouba

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