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Unformatted text preview: Homework #1
Section 1.1
√
2.
Computing first 6 derivatives
and evaluating them at we get the answer from the
general formula for Taylor series 5. .
√ 9. There are 2 possible ways of solving this problem : one
can approximate
and use
or
√ and use . For the first choice the remainder would be
√ , then
√ , . √ . Therefore one would get
() () 10a. () We have
√ √ √ 11a. . ( ) ( and √ )
, , so
( ) . , . Such remainder could be bounded for all 11b. and
. , . Remainder cannot be bounded for all
because of the division by zero. 12c. and
( Remainder , )( .
), so that . To get desired accuracy one needs to
consider large amount of terms which is very impractical.
However, change of the interval would allow to get
accuracy with much less number of terms.
() 15. . Define
as the approximation generated by
using an m term Gregory series to approximate
( ) and n term Gregory series for
Then we have () .
( ) where is the remainder in the Gregory series. Therefore
. If we
require both expansions to be equally accurate, then we
have that and
Solving these unequalities for appropriate accuracy we get the number of terms in Gregory series. Section 1.2
1. From Taylor’s Theorem for
√

example  ,
 . Then for all sufficiently small x. For
 √ for 3. Expanding cosine in Taylor series one gets
therefore

Letting   we get that 10. We are given the following statement: 
. . which means that there exist such a
constant :
. We multiply both sides of
the unequality with
    This means that .
from definition of Section 1.3
1b.
Abs error rel error 2b.
Exactly: ( ) Chopping: ( ) Rounding: ( ) 3b. √ Function is suspectible to subtractive cancellation which
will be amplified by division by a small number. In order
to avoid this problem one can use Taylor expansion to make the subtraction and division both explicit
operations. Expanding √ around one gets
. 7. I was using Matlab for this problem (using default
settings ‘format short’). The following numbers were
considered:
a=1; b=0.00002; c=0.00003;
In short format Matlab uses 4 digits after the decimal
point (with rounding). Therefore
a+b=1.0000; (a+b)+c=1.0000+0.00003=1.0000
However b+c=0.00005 which Matlab rounds to 0.0001
and the final result a+(b+c)=1.0001 (a+b)+c.
Problem 4
In order to improve the code one can notice that
expanding the function about instead of required less terms in Taylor expansion. Looking at the
general formula for the remainder one can see that it has
a factor of
factorial,
th derivative of the , where in principle
If we consider
would be
maximum value of , and the factor of
then the maximum value of
However, for
would be the
and therefore remainder would faster reach the desired accuracy. ...
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This note was uploaded on 01/15/2012 for the course MAE 107 taught by Professor Rottman during the Spring '08 term at UCSD.
 Spring '08
 Rottman

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