hw11_solutions

hw11_solutions - Homework #1 Section 1.1 √ 2. Computing...

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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.

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