Diary outline for Math 152,

Diary outline for Math 152, - Math 152 diary, spring 2009...

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Unformatted text preview: Math 152 diary, spring 2009 Later material Much later material In reverse order: the most recent material is first. Wednesday, February 18 (Lecture #9) The old QotD, given by the vicious lecturer Well, I suppose that is better than the viscous lecturer. The old QotD was to compute [1/(3x 2 +2x)]dx. In the first lecture, two nice people ( Mr. Cicero and Mr. Schultz ) wrote a solution for the old QotD which about 50% of people did wrong. Sigh. They did it right. This was duplicated by what Ms. Du and Mr. Haberman did in the later lecture. Factoring the bottom is easy. Breaking up the function into a nice sum and then finding the unknowns is easy. The nasty thing is that [1/(3x+2)]dx is (1/3)ln(3x+2)+C and if you don't believe this, differentiate! What is an ALGORITHM ? Since there may be some confusion, please read this . Another example or two The example I did last time ended with 11x+7 divided by x 2 2x3. The bottom (denominator?) here can be factored easily as (x3)(x+1). Let me change the example by changing the bottom because otherwise the full horror (!) of partial fractions will not be revealed. (11x+7)/[(x-3)(x+1)(5x+2)]dx? Here is the sort of sum we need to investigate: 11x+7 A B C---------------- = --- + --- + ------ (x-3)(x+1)(5x+2) x-3 x+1 5x+2 Let's combine the fractions on the right using the least common multiple. The right side combines like this: A(x+1)(5x+2)+B(x-3)(5x+2)+C(x-3)(x+1)------------------------------------ (x-3)(x+1)(5x+2) This is supposed to be equal to the fraction we had above (on the left-hand side of the equation). Now the bottoms of the fractions are the same, so the tops must agree. Therefore we know 11x+7=A(x+1)(5x+2)+B(x3)(5x+2)+C(x3)(x+1) Now x=1 tells me that 11(1)+7=0+B(13)(5+2)+0 so we can deduce the value of B, which is 3. And x=3 tells me that 11(3)+7=A(3+1)(5(3)+2)+0+0 so that A=40/(417). Finally, x=2/5 will let me deduce the value of C. Therefore we get a partial fraction decomposition of (11x+7)/[(x3)(x+1)(5x+2)]. Notice that each piece can be easily antidifferentiated: [A/(x3)]+[B/(x+1)]+[C/(5x+2)]dx=Aln(x3)+Bln(x+1)+(C/5)ln(5x+2). Whew. The only trick is the /5 because of the Chain Rule again. (11x+7)/[(x-3)(x+1) 2 ]dx? Now one of the factors has a higher power than 1. Some people say that 1 is a root of the bottom polynomial which has multiplicity 2. In this case, the symbolic sum has to be modified. There is one term for each power going up from 1 to the multiplicity. So here: 11x+7 A B C---------- = --- + --- + ------ Diary outline for Math 152, spring 2009 http://www.math.rutgers.edu/~greenfie/vnx/math152/diary.html 1 of 47 7/26/2009 8:15 AM (x-3)(x+1) 2 x-3 x+1 (x+1) 2 This is step 2, the symbolic sum. Step 3 requires that specific values be found for A and B and C. What I do here is good for small examples. So I would combine the fractions on the right-hand side, using the least common denominator. You need to be careful about this. I have made mistakes lots of times.denominator....
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This note was uploaded on 07/26/2009 for the course CALC 2 taught by Professor Zeilberg during the Spring '08 term at Rutgers.

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Diary outline for Math 152, - Math 152 diary, spring 2009...

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