Expanding definite integrals in a power series about a parameter

*To*: mathgroup at smc.vnet.net*Subject*: [mg35444] Expanding definite integrals in a power series about a parameter*From*: "Carl K. Woll" <carlw at u.washington.edu>*Date*: Fri, 12 Jul 2002 04:29:17 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Hi newsgroup, I have developed a package called SeriesSolve that can do two neat things (I think). There are a few bugs in the package, so that it is not quite ready for MathSource. The package works sufficiently well for me as it is, so I am checking with you guys to see if the functionality provided by the package would interest you. First, given a polynomial with coefficients which are themselve power series in another variable, the function SeriesSolve will find the roots of the polynomial in a series in that variable. As a very simple example, suppose you have the polynomial (a+a^2)+x+a x^2+(1+a^2)x^3 and are interested in the roots of this polynomial. Using SeriesSolve, this is simple (you will need to view all of the following formulas in a fixed font to understand what's going on): rt=SeriesSolve[(a+a^2)+x+a x^2+(1+a^2)x^3,x,{a,0,5}]//Chop 2 4 5 6 {-1. a - 1. a + 1. a + 3. a + O[a] , 2 3 1. I + (0.5 - 0.5 I) a + (0.5 + 0.5 I) a - 4 5 6 (0.5 - 1.25 I) a - 2. a + O[a] , 2 3 -1. I + (0.5 + 0.5 I) a + (0.5 - 0.5 I) a - 4 5 6 (0.5 + 1.25 I) a - 2. a + O[a] } Let's check that the putative solutions are indeed correct In[4]:= (1+a^2)Times@@(x-rt)//ExpandAll//Chop Collect[%,x] Out[4]= 3 2 3 2 6 (1. x + x ) + (1. + 1. x ) a + (1. + x ) a + O[a] Out[5]= 2 2 2 3 1. a + 1. a + 1. x + 1. a x + (1 + a ) x Admittedly, it is not difficult to produce the above output for low order power series in a. However, for high order power series, for power series with multiple roots (at lowest order) or for power series where the coefficient of the highest order term in the polynomial vanishes at leading order in the power series, it does become difficult to produce the roots of the polynomials. My package can handle all of these situations. For a slightly more difficult example, consider In[11]:= rt=SeriesSolve[(1+a)x^3+a x^2+a^3 x+a^4,x,{a,0,5}]//Chop Out[11]= 3/2 3 7/2 4 {1. I a + 0.5 a - 0.5 I a - 0.5 a - 9/2 5 11/2 6 0.125 I a - 1. a + 2.125 I a + O[a] , 3/2 3 7/2 4 -1. I a + 0.5 a + 0.5 I a - 0.5 a + 9/2 5 11/2 6 0.125 I a - 1. a - 2.125 I a + O[a] , 2 3 4 5 6 -1. a + 1. a - 2. a + 2. a + 1. a + O[a] } and checking the solution: In[12]:= (1+a)Times@@(x-rt)//ExpandAll//Chop Out[12]= 3 2 3 3 4 6 x + (1. x + x ) a + 1. x a + 1. a + O[a] I hope that gives you a sufficient idea about how SeriesSolve works. In particular, when the coefficient of the highest order term vanishes at leading order in the expansion, it becomes quite difficult to find the series form of the roots. You might wonder why I developed this package. Consider evaluating an integral of the form Integrate[ q^m / poly^n, {q,0,Infinity}] where poly stands for some polynomial in q with coefficients which are themselves polynomials in another variable, say a. Then, using the calculus of residues it is possible to evaluate these integrals, and in particular one can find the series expansion in a of the integral. This is exactly what the function SeriesRationalIntegrate does. For example, if the polynomial is a+a q+q^2+q^3, then let In[13]:= int=SeriesRationalIntegrate[a+a q+q^2+q^3,q,{a,0,5}]; where I put a semicolon at the end as the output of SeriesRationalIntegrate is rather unappetizing. Now, int[n,m] will return the series expansion of the integral Integrate[q^m/(a+a q+q^2+q^3)^n,{q,0,Infinity}] Note that when a=0, the above integral is divergent. Hence, you can't simply expand the integrand in a series in a and then do term by term integration. At any rate, we find for example, In[14]:= int[1,1/2] Out[14]= 2.22144 1/4 3/4 ------- - 3.14159 + 2.22144 a - 2.22144 a + 1/4 a 5/4 7/4 2 3.14159 a - 2.22144 a + 2.22144 a - 3.14159 a + 9/4 11/4 3 2.22144 a - 2.22144 a + 3.14159 a - 13/4 15/4 4 2.22144 a + 2.22144 a - 3.14159 a + 17/4 19/4 5 21/4 2.22144 a - 2.22144 a + 3.14159 a + O[a] Deriving this series expansion is not at all trivial. Of course, there may not be much interest in doing a series on a root object (SeriesSolve) or doing a series on a particular kind of definite integral (SeriesRationalIntegrate). If there is, I will polish up the package and post it sometime. Carl Woll Physics Dept U of Washington

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