(*
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

Thursday, November 30:    Section  8.8

Previous Assignment:  

Find P(x) the taylor polynomial of degree 10  which approximates f[x].  

The error formula says that the error can be captured as 

               (11)
              f    (z)   n+1
  error =    ---------  x 
                11! 

The value of z is unknown, but we do know that it lies between 0 and x.

This means that if we graph the function over the interval of interest

               (11)
              f    (z)   
             ---------   
                11! 

we can look where it is largest.  The maximum usually occurs at one

end or in the middle.  But it could occur in strange places and that

is why we graph it.  We do not really need to know the maximum.  We

only need to know a bound M.  

       In our example, the interval of interest is (-0.1,0.1) and

on the interval the maximum value of x is 0.1.  Thus the error

of the series approximation will be less than 
   
            M*(0.1)^11 

over the whole interval.

    


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}];     
error[x_] = (0.1)^11 temp/11!;        

a1 = Plot[{f[x],P[x]},{x,-2,2}]; 
a2 = Plot[error[x],{x,-0.1,0.1}];


(*
Main idea:  You can bound the error for the Taylor polynomial! 

Key words: Taylor error formula. 

Goals: Develop some interesting in seeing the proof of the

Taylor error formula.


Taylor's error formula for a Maclaurin series is: 


The error for the degree n polynomial approximation at x=c is: 

              (n+1)
             f     (z)
error =     ----------    where z is  between 0 and c. 
              (n+1)!



Proof:   We wish to establish that the error for the

degree n approximation of the series can be captured in the

form
        
       f^(n+1) (z)/(n+1)! c^(n+1)  where z is between 0 and c.


The error at c is of course

         f[c] - P[c]  = error at c.  

We choose to write this error in the form

        f[c] - P[c] = R  for some constant R.

We pull the following expression out of thin air.  Do not worry

where this expression came from.  It seems to work well. 






                                                  (n)
                            f"[t]     2          f  [t]      n 
f[c] - f[t] - f'[t] (c-t) - ---- (c-t)  - ... - ------  (c-t)   
                             2                    n!
                   (n+1) 
              (c-t)     
           -R -----
                 (n+1)
                c

Notice that when t = 0 the expression is zero.  (By choice of R)

Notice when t=c the expression is zero.  (By all of the (c-t) terms.)


Now simply use the mean value theorem to say that the derivative is 

zero somewhere between 0 and c.  Check it out.


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. 

*)