(* 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 Friday, December 1: Section 8.8 Three applications of the Taylor error 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]. Error = f^(4)[z]/4! x^4 <= e^1 /4! 1^4 <= 3/4! = 0.125 which is not very good. If we use more terms but keep the length of the interval the same, the accuracy increases. n error estimate 1 3. 2 1.5 3 0.5 4 0.125 5 0.025 6 0.00416667 7 0.000595238 8 0.0000744048 -6 9 8.2672 10 -7 10 8.2672 10 -8 11 7.51563 10 (b) How many terms should I use so that the error of the approximation on [0,1] is less than 0.00001? 0.00001 <----bigger Use 8 terms. When n+1 = 0 the error is less than 0.0000082672 (c) Express the value of e to 6 decimal places. 0.0000001 error occurs beyond the 6th place. n+1 = 11 7.51563 10^(-8) = 0.0000000751563 So degree 10 should give six place accuracy at least. In[4]:= Sum[x^n/n!,{n,0,10}] 2 3 4 5 6 7 8 9 10 x x x x x x x x x Out[4]= 1 + x + -- + -- + -- + --- + --- + ---- + ----- + ------ + ------- 2 6 24 120 720 5040 40320 362880 3628800 9864101 e = -------- = 2.718281801 3628800 /|\ |error in 8 th decimal place | Show that a power series can be differentiated term by term. | f(t)-f(c) | Essentially we wish to show | --------- - f'(c) | -->0 as t->c. | t-c | To find the derivative at c, we work with a t close to c and certainly within the radius of convergence. --------(----------------0-------------------)------------- -R c t R i i-1 We use f[x] = SUM a x and f'[x] = SUM i a x . i i We do not know that f'[x] is actually the derivative of f. To check that f'[x] is the derivative of f[x] we have to show that | f(t)-f(c) | | --------- - f'(c) | ---> 0 as |t-c| --->0. | t-c | i i | SUM a t - SUM a c | | i i i-1 | | ----------------------- - SUM i a c | | t-c i | | | i i | a t - a c | | i i i-1 | = | SUM ------------------- - SUM i a c | | t-c i | i i | / a t - a c \ | | i i i-1 | | = |SUM | ------------------- - i a c | | | \ t-c i / | i i | a t - a c | | i i i-1 | <= SUM | --------------- -i a c | | t-c i | i i | (t - c ) i-1 | = SUM |a -------------- -i a c | | i t-c i | | | i i | (t - c ) i-1 | = SUM |a | | --------------- -i c | i | t-c | Now we use the mean value theorem to argue that i for the function x we know that i i t - c i-1 ---------- = i t t-c i for some t between c and t. i t i --------(----------------0-------------------)------------- -R c t R | i-1 i-1 | SUM |a | | i t - i c | i | i | We factor out the i: | i-1 i-1 | SUM |a | i | t - c | i | i | t i --------(----------------0-------- ----------)------------- -R c| t R | u i i-1 Now we look at the function x . Again we apply the mean value theorem saying that there is some value u between the i previous c and t such that i i-1 i-1 i-2 t - c = (i-1) u (t -c) i i i | i-2 | SUM i |a | | (i-1) u (t -c) | i | i i | Now |t -c| is smaller than |t-c| since t was chosen between i i c and t. Furthermore, the values u are i all between c and t and we can pick T to be the maximum of |c| and |t|. We substitute these values into the sum making the value of the sum larger. | i-2 | <= SUM i |a ||(i-1) T |t-c| | i | | __ __ | | | i-2 | <= |SUM i(i-1)|a | T | |t-c| | i | |__ __| Now the first part converges since it is the second derivative and T is inside the Interval of convergence. Thus | f(c) - f(t) | | ____________ - f'(c)| <= M |t-c| where M is the | c-t | i-2 SUM | i(i-1)a T |. i The limit of M |t-c| is zero as t-> c. *)