(* CALCULUS 265 Honors Fall Semester 2000 8:00 to 8:50 MTTF Carver 0002 Instructor: Hentzel Office: 432 Carver Office Hours: 9:00-10:00 MTTF E-mail hentzel@iastate.edu phone 294-8141 Website http://www.math.iastate.edu/hentzel/honors.265 Tuesday, November 28: Section 8.7 assignment: Find P(x) the taylor polynomial of degree 10 which approximates f[x]. The error formula is (11) error(z) = f (z) --------- 11! To study the Taylor error formula, (a) graph f[x] and P[x] on the same graph for x in the range given. (b) Mark on the graph the interval of convergence. (c) What is the maximal error on the interval (-0.1,0.1) as given by the error formula. Do this for each of the following functions. (a) Sqrt[1+x] Plot range (-2,2). (b) Log[1+x] Plot range (-2,2). (c) Tan[x] Plot range (-Pi,Pi). *) f[x_] := Sqrt[1+x]; P[x_] = Normal[Series[f[x],{x,0,10}]]; temp = f[x]; Do[ temp = D[temp,x], {i,1,11}]; (* CORRECTION *) error[x_] = (0.1)^11 temp/11!; (* ADD FACTOR OF x^(n+1) since error is *) (* f^(n+1)(z)/(n+1)! x^(n+1) *) a1 = Plot[{f[x],P[x]},{x,-2,2}]; a2 = Plot[error[x],{x,-0.1,0.1}]; (* Main idea: You can tell how good the approximation is. Key words: Taylor error formula. Goals: Learn Taylor error formula. It depends on the size of the interval, the accuracy desired, and the number of terms used. Our goal is to study "good enough" mathematics. We take a horrendous function and replace it with a simple polynomial which is a "good enough" representation of the original function. The rest of this course will be to show that everything behaves as it should, and also whatever could be expected to go wrong probably does at some stage or other. Previous Assignment: (a) Given that SUM ai c^i converges, show that SUM ai x^i converges for all |x| < |c|. Proof: Suppose SUM ai c^i converges. We will show That if |x|<|c| then SUM ai x^i converges absolutely. SUM | ai x^i | = SUM |ai||c|^i |x/c|^i Now since SUM ai c^i converges, then Limit ai c^i = 0. Eventually |ai c^i| < 1. Therefore: SUM | ai x^i | = SUM |ai||c|^i |x/c|^i <= SUM |x/c|^i . (eventually) SUM |x/c|^i is a geometric series, r = |x/c| < 1 by assumption. It converges. The original series converges by the direct comparison test. (b) Show that this proves that if SUM ai d^i diverges, then SUM ai x^i diverges if |x| > |d|. Proof: We will proof the theorem by contradiction. Assume that SUM ai d^i diverges, but that for some x with |x| > |d| the series SUM ai x^i converges. By the above proof, since the series converges at "x" it must converge absolutely at all values less than |x|. This contradicts that the series diverges at d. (c) Show that if SUM ai c^i converges then SUM i ai x^(i-1) also converges if |x| < |c|. Proof: Suppose that SUM ai c^i converges and that |x| < |c|. The derived series converges if x = 0 so let us further assume that x =/= 0. Sum | i ai x^(i-1) | = SUM i |ai c^i| |x/c|^i |c/x| and since SUM ai c^i converges, limit ai c^i = 0. Therefore eventually |ai ci| < 1. Thus SUM | i ai x^(i-1) | = SUM i |ai c^i| |x/c|^i |c/x| < SUM i |x/c|^i |c/x| (eventually) = |c/x| SUM i |x/c|^i Using the ratio test: (i+1) |x/c|^(i+1) i+1 Limit ---------------- = Limit --- |x/c| = |x/c| < 1 i->inf i |x/c|^i i->inf i Therefore SUM i |x/c|^i converges by the ratio test and so |c/x| SUM i |x/c| = SUM i |x/c|^i |c/x| converges. Therefore SUM |i ai x^(i-1) | converges by the direct comparison test. By the same argument as in parts (a) and (b) we know that the original series and the derived series have the same radius of convergence. Taylor's error formula for a Maclaurin series is: The error for the degree n polynomial approximation at x=c c is: (n+1) f (z) error = ---------- where z is between 0 and c. (n+1)! There are three ways to use this formula. The series for e^x is 1 + x + x^2/2 + x^3/3! + ... + x^n/n! + ... (a) Bound the error for the approximation e^x = 1 + x + x^2/2 + x^3/3! on the interval [0,1]. (b) How many terms should I use so that the error of the approximation on [0,1] is less than 0.00001? (c) Express the value of e to 6 decimal places. *)