# How to replace a parameter with a function in a Series expansion?

Posted 10 years ago
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 I have expanded the product of two power series: Series[1/(1+a)^3,{a,0,5}]*Series[1/Sqrt[1+b],{b,0,5}]. the "a" and "b" parameters are actually function. I wish to substitute for "a" and "b" in the product. When I replace a->(z2-z1), Mathematica does not recognize (z2-z1) as a legitimate variable. "b" is a more complex expression. I have tried defining or setting a and b and setting rules and performing the substitutions, but nothing seems to work. z1 and z2 are subsequent integration variables.My goal is to integrate the product of the series term by term over the variables within a and b: there are six variables. The direct integrals of the original functions have no analytical or numerical solutions using Mathematica, hence the series expansion. The total integrals remain finite in each term, so with enough terms, I can find the convergence point of the integrated series.Any help would be appreciated.Regards,
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Posted 10 years ago
 Hi Luther,how about:(Series[1/(1 + a)^3, {a, 0, 5}]*Series[1/Sqrt[1 + b], {b, 0, 5}]) /. a -> HoldForm@(z2 - z1) Henrik
Posted 10 years ago
 That would work, but as I set up the model, the issue is a bit more complex. I have to increase the index or power on the second series for each term in the first series. In other words, the first term in the first series has a power for the second series. The next term in the first series requires the second series power to be indexed by two...and so forth. I can do it manually, and Craigs suggestions and yours are what I needed to move forward...slowly but surely.It appears that if I try to bundle too many operations into one statement, Mathematic runs forever and I don't know if it will eventually be able to do what I am asking. However, I notice two issues. First, I do a double angular integral and sometimes I get a numberical answer and someties I get powers of Pi. I would like to always get powers of Pi and not the decimal value multiplied time the rest of the numerical values that are calculated. Turns out the integrals are not useful using DIntegrate but stay finite and regular with Integrate.Second issue is that when I try to same a large amount of material to a notebook page, after a dozen lines the material is squashed over into a single column on the right side of the notebook page. How do I save fifty or more  In calculations withoiut this distortion. It is not useable when it is squashed together and run on.Thanks for any suggestions.Luther
Posted 10 years ago
 Your suggestion should work, but using the original series directly and replacing a with z2-z1 does not work. Mathematica tells me it is not a proper variable.I will experiment with your suggestion and see if it does what I wish. It looks like it should. I am replacing both a and b, so the syntax may require experimentation. I am not familiar with what many things in Mathematica actually do, so I am not sure what I need to be looking for.Thanks and regards.
Posted 10 years ago
 Hello Luther,I am not sure what you are after here.  Does this not do what you wish it to?Clear[a, b];originalSeries = Series[1/(1 + a)^3, {a, 0, 5}]*Series[1/Sqrt[1 + b], {b, 0, 5}]originalSeries /. {a -> (z1 - z2)}I imagine that you might wish to have this instead:originalSeriesNormal = Normal[originalSeries]originalSeriesNormal /. {a -> (z1 - z2)}But, this may be even better for your purposes:expansion[ax_, bx_] := Normal[Series[1/(1 + ax)^3, {ax, 0, 5}]* Series[1/Sqrt[1 + bx], {bx, 0, 5}]]expansion[z1 - z2, b]Integrate[expansion[z1 - z2, b], z1]I hope this is helpful to you.Craig