MAC Calculus II Summer All you need to know on partial fractions and more

Transcription

1 MC -75-Calulus II Summer 00 ll you need to know on partial frations and more What are partial frations? following forms:.... where, α are onstants. Partial frations are frations of one of the + α, ( + α n, where, α are onstants and n is a natural number (positive integer. If n =, we are bak at ase. + B + β + γ, where, B, β, γ are onstants and + β + γ is an irreduible quadrati polynomial. + B ( + β + γ n, where, B, β, γ are onstants, + β + γ is an irreduible quadrati polynomial and n is a natural number (positive integer. If n =, we are bak at ase. n irreduible quadrati polynomial is a polynomial + β + γ of degree whih does not fator into linear fators. s one an see, this is equivalent to the equation + β + γ = 0 having NO real roots. If one ompletes the square, an irreduible quadrati polynomial an always be rewritten in the form + β + γ = ( + b +. For eample: + + is irreduible; if we omplete the square we get + + = ( + +, so b = /, =. On the other hand, + + is not irreduible; ompleting the square leads to + + = ( + and annot be epressed in the form (with real. For eample, all of the following are partial frations: 5 7, + 9, 8 ( +, e π ( + 8, + = +, (5 + + = 5 ( The interest, importane, of partial frations for alulus is due to two fats:

2 Fat. The fat that every quotient of polynomials in whih the degree of the numerator is < than the degree of the numerator an be deomposed (written as a sum of partial frations, in an essentially unique way. In other words, if f( = a m m + a m m + + a + a 0 b n n + b n n + + b + b 0 where m < n, then f( an be written as a sum of frations of type,,,. Don t worry about the essential uniqueness part; it simply means that two people doing the deomposition independently of eah other, will ome up with the same result, eept for some silly differenes, suh as the order of the terms. Fat. One knows (or an know how to integrate the partial frations. Integrating the partial frations. Memorizing the formulas preeded by an asterisk (* in parentheses may not be a terribly bad idea. Or, you should be able to get them fast from general priniples. I. Case From (* one gets d = ln + α + C + α d = ln + α + C + α II. Case ssume now n, the ase n = having been done. Sine (* ( + α n d = n ( + α n + C if n, it follows that ( + α n d = (n ( + α n + C for n =,,.... III. Case The following two formulas play a role here: (* and (* + α d = ln( + a + C + α d = ( α tan + C. α The first one follows from u du = ln u + C by the substitution u = + α, the seond one follows from (u + du = tan u + C by the substitution = αu. Using these formulas, here is how we integrate a partial fration + B + β + γ

3 We omplete the square in the denominator: + β + γ = ( + β + ( (γ β = + β + with = γ (β /. Don t memorize these formulas!, just omplete squares in eah individual ase. If by some hane γ is not stritly larger than b / then your quadrati polynomial wasn t irreduible. If β = 0, then the square is already ompleted and there is nothing to do. Our fration now looks like + B ( + b + with b = β/. This epression an be integrated quite easily by the substitution u = + b, so = u b, thanks to the formulas we mentioned above: + B (u b + B ( + b + d = u + d u + ( b + B = u + d u = u d + ( b + B + u + d = ln(u + + b + B ( tan u + C = ln(( + b + + b + B tan ( + b + C. Comments In the last epression, one does not need an absolute value inside the logarithm. The epression ( + b + only assumes positive values. Moreover, what is inside the logarithm is the original denominator; one ould re-epand and write ln( + β + γ instead of ln(( + b +. One an avoid (and perhaps should avoid the substitution if one an memorize a slightly more general version of the last two formulas that have been given an asterisk, namely (* + b ( + b + α d = ln[( + b + a ] + C and (* ( + b + α d = α tan ( + b α + C. We do two eamples, using these formulas in the first eample. Eample. Evaluate d. One feature of this eample is that the leading oeffiient in the denominator is not. There are several ways to deal with this; I would reommend just fatoring it out. We do this in the first step (first equality below, labelled (. In the seond step, step (, we omplete the square in the denominator. In step ( we adjust the numerator so we an use our formulas. If there is an +b squared in the denominator, and the numerator is of the form +B, just write = + b b; then + B = ( + b + ( b + B. In ( we separate

4 the integral into two to whih the formulas apply, and (5 has the final result. The numbers don t look pretty beause ontrary to what tetbooks do, we did not arrange things so everything works out very niely and looks beautiful. lot of things in real life are not pretty; one has to deal with them anyway. (5 + d = 5 + ( d = 5 + ( ( d 5( + ( ( (5 = = 5 ( d + ( d 8 = 5 ln ( ( tan ( d ( C. Eample. Evaluate d. This is a muh simpler eample, beause the square in the denominator is already omplete. In fat, the first two starred formulas of ase apply d = 7 + d = + d 7 + d = ln( + 7 ( tan + C. IV. Case, n. We won t enounter this ase too many times in this ourse, at least not for n. However, it should be more or less lear that to deal with it we need to know how to evaluate the following two types of integrals: ( + a d, and n ( + a n d when n =,,.... The first one is easy; by the substitution u = + a one sees it works out to ( + a n d = (n ( + a + C. n The seond one an be transformed into a trigonometri integral whih an then be solved by the methods of Setion 7. of our tetbook, via the substitution = a tan θ. We only do one eample. Eample. Evaluate ( + d. We substitute = tan θ; then + = (tan θ + = se θ, d = se θ dθ. J. Stewart, Calulus, Early Transendentals, th. Ed., 999

5 We get ( + d = se θ dθ = os θ dθ. We an integrate os θ using the redution formula, or by os θ = ( + os θ dθ = θ + sin θ + C. Thus, so far, ( + d = (θ + os θ sin θ + C where we used that sin θ = sin θ os θ. We draw a triangle with an angle θ suh that tan θ = /. + θ From the triangle we see that os θ = / +, sin θ = / + so that ( + d = ( tan + ( + + C. This onludes the setion on how to integrate the individual partial frations. We turn to the general ase, how to integrate a funtion f( = P ( Q( where P, Q are polynomials and the degree of P is stritly less than the degree of Q. (If the degree of P is larger than or equal to the degree of Q, one first has to divide out P by Q, obtaining an epression G( + [R(/Q(], where G is a polynomial hene easy to integrate and the degree of R is now < the degree of Q. Step. Fator Q ompletely into distint powers of linear fators and/or of irreduible quadrati fators. This, of ourse, is easier said than done and in all the eerises you ll have to do Q will either be already fatored, or partially fatored, or fatorization will be quite easy. Step. Deompose into partial frations by writing out sums of terms (eah one a partial fration as follows. For eah power of a linear fator ( + a m appearing in Q, add the following terms to the sum: + a + ( + a + + m ( + a m. You don t have to use the symbols,..., m for the onstants, but the symbols you use for the onstants in every partial fration you write down should be different from those used in any other partial fration of the whole list. For 5

6 eample, if Q( has the fator ( + a (power only, you just add a single term + a. If a previous term already used the onstant, you have to use a different symbol. If Q( has the fator ( + a, you add + a + B ( + a + C ( + a. For every power ( + β + γ m of an irreduible quadrati fator, add the terms + B + β + γ + + B ( + β + γ + + m + B m ( + β + γ m. Step. In the sum of partial frations you now have a whole bunh of unknown onstants, eatly as many as the degree of Q. Take ommon denominator. The ommon denominator should be Q(, the numerator will be a (possibly horrendous polynomial involving all these onstants we alled,, B,...; of degree one less than the degree of Q. Collet equal powers of in the numerator and equate the oeffiient of eah power of with the orresponding oeffiient of P (. This will give you a system of linear equations, as many as there are unknown onstants, for these unknown onstants. Step. If instrutions were followed eatly, the theory states that the system of equations you found in step has one and only one solution. Find it by solving the system; obtaining the values of the unknown onstants,,.... This again, is easier said than done. Systems of up to four equations in four unknowns are relatively easy to solve. The time it takes to solve them inreases fast; systems of ten equations in ten unknowns should probably not be done without the aid of a omputer. It should also be mentioned that there is a slew of shortuts available to ompute these oeffiients, so that one basially never has to atually solve diretly the system of linear equations they satisfy. Final step 5. Now that you know all the onstants in the partial fration deomposition of f = P/Q, just integrate f by integrating term by term eah partial fration. We do one eample. Determine the partial fration deomposition of P ( Q( = ( + + ( ( ( + + ( +. The denominator is partially fatored. + + is not irreduible, in fat it fators into ( +, hene ( + + fators as ( +. Neither is irreduible, it fators as ( ( +. To make a longish story short, we an keep on fatoring Q( getting Q( = ( + ( + ( ( ( + ( + + ( + = ( + ( + ( ( ( + ( + ( + + ( +. The polynomials + and + + are irreduible, so those fators stay as they are. But we are not done yet, we still have to ollet equal fators so that the omplete fatorization of Q is Q( = ( + ( ( + ( + ( + +.

7 We would now write, for the partial fration deomposition, ( + + ( ( ( + + = ( + = + + B ( + + C ( + + D ( + + E ( F ( + + G + H ( + J + + K + L + + M + N P + Q ( R + S ( + +. You might notie there are 7 unknown onstants, B, C, D, E, F, G, H, J, K, L, M, NP, Q, R, S. Not oinidentally, the degree of Q is also 7. We ompleted steps and above. To keep on going would be a bit razy, so we stop here. However, just in ase some of you happened to be urious on how the atual deomposition into partial frations looked, we used Maple to get it. nd here it is: ( + + ( ( ( + + ( + = ( + 8 ( ( + 5 ( ( ( ( ( ( ( + + ( + + 7