Revisions to Simplifying expressions involving Sum

deleted 1017 characters in body

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edited Aug 27, 2013 at 19:55

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(*  (x - x0)^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)

(x - x0)^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n\)]\ a[n]\)\) + (x - 
    x0)^(-1 + α) α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)
*)

enter image description here

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

enter image description here

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + α \)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + α \)]\ α \ a[n]\)\)
*)

enter image description here

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
    n + α \)]\ \((n + α )\)\ a[n]\)\)
*)

enter image description here

(*  (x - x0)^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)

(x - x0)^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n\)]\ a[n]\)\) + (x - 
    x0)^(-1 + α) α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + α \)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + α \)]\ α \ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
    n + α \)]\ \((n + α )\)\ a[n]\)\)
*)

enter image description here

deleted 101 characters in body

Source Link

edited Aug 27, 2013 at 15:11

Artes

57.9k
13
159
247

 expr1 = (x - x0)^\[Alpha]^α  Sum[a[n] (x - x0)^n, {n, 0, Infinity}]
expr2 = D[expr1, x]

(*  (x - x0)^\[Alpha]^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)

(x - x0)^\[Alpha]^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n\)]\ a[n]\)\) + (x - 
    x0)^(-1 + \[Alpha]α) \[Alpha]α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)
*)

expr3 = expr2 /. u_*\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]v_\) -> \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\((u* v)\)\)

*UnderoverscriptBox[([Sum]), (n = 0), ([Infinity])]v_) -> !( *UnderoverscriptBox[([Sum]), (n = 0), ([Infinity])]((u* v))) thisthis is the result:

    (*
    \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
    \*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
        n]\)\) + \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
    \*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
         n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\α \)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + \[Alpha]\α \)]\ \[Alpha]\α \ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
    n + \[Alpha]\α \)]\ \((n + \[Alpha]α )\)\ a[n]\)\)
*)

 expr1 = (x - x0)^\[Alpha] Sum[a[n] (x - x0)^n, {n, 0, Infinity}]
expr2 = D[expr1, x]

(*  (x - x0)^\[Alpha] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)

(x - x0)^\[Alpha] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n\)]\ a[n]\)\) + (x - 
    x0)^(-1 + \[Alpha]) \[Alpha] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)
*)

expr3 = expr2 /. u_*\!\(

*UnderoverscriptBox[([Sum]), (n = 0), ([Infinity])]v_) -> !( *UnderoverscriptBox[([Sum]), (n = 0), ([Infinity])]((u* v))) this is the result:

    (*
    \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
    \*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
        n]\)\) + \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
    \*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
         n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
    n + \[Alpha]\)]\ \((n + \[Alpha])\)\ a[n]\)\)
*)

 expr1 = (x - x0)^α  Sum[a[n] (x - x0)^n, {n, 0, Infinity}]
expr2 = D[expr1, x]

(*  (x - x0)^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)

(x - x0)^α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n\)]\ a[n]\)\) + (x - 
    x0)^(-1 + α) α  \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)
*)

expr3 = expr2 /. u_*\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]v_\) -> \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\((u* v)\)\)

this is the result:

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + α \)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + α \)]\ α \ a[n]\)\)
*)

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
    n + α \)]\ \((n + α )\)\ a[n]\)\)
*)

Source Link

created Aug 26, 2013 at 13:45

Alexei Boulbitch

40k
2
48
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One way to do it is application of series of rules to the expression, just to "teach" Mathematica of how to transform the expression. I will break the chain of transformations into steps to make it better visible. These are your functions:

 expr1 = (x - x0)^\[Alpha] Sum[a[n] (x - x0)^n, {n, 0, Infinity}]
expr2 = D[expr1, x]

and here are the results:

(*  (x - x0)^\[Alpha] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)

(x - x0)^\[Alpha] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n\)]\ a[n]\)\) + (x - 
    x0)^(-1 + \[Alpha]) \[Alpha] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(n\)]\ a[n]\)\)
*)

Let us do the first transformation:

expr3 = expr2 /. u_*\!\(

*UnderoverscriptBox[([Sum]), (n = 0), ([Infinity])]v_) -> !( *UnderoverscriptBox[([Sum]), (n = 0), ([Infinity])]((u* v))) this is the result:

    (*
    \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
    \*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
        n]\)\) + \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
    \*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
         n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

The second transformation:

 expr4 = expr3 /. (x - x0)^m_ \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(s_*\ 
\*SuperscriptBox[\((x - x0)\), \(q_\)]\)\) -> \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(s*\ 
\*SuperscriptBox[\((x - x0)\), \(q + m\)]\)\)

yielding

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(n\ 
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + n + \[Alpha]\)]\ a[
    n]\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
     n + \[Alpha]\)]\ \[Alpha]\ a[n]\)\)
*)

and the last one:

 expr5 = expr4 /. \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(u1_*
      v_\)\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(u2_*
      v_\)\) -> \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 
      0\), \(\[Infinity]\)]\(\((u1 + u2)\)*v\)\)

gives the result:

(*
\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((x - x0)\), \(\(-1\) + 
    n + \[Alpha]\)]\ \((n + \[Alpha])\)\ a[n]\)\)
*)

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