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Definite summation (particular solutions): Zeilberger, Gosper, Ramanujan, etal)  P/  -Indefinite summation parallels to integrationintegration of polynomials integration of rational functions difference operator D parallels the derivative S and s are similar But not similar enough for some purposes! HQ9,Some simple examples (Macsyma, in this case) A more elaborate example  Start simply.if we need g(n) = i=an f(i) we approach by finding the indefinite summation h(x) = i=0x-1 f(i) then g(x) = h(n+1)  h(a). Sidestepping any issues of singularities. Note also that D h(x) defined by h(x+1)-h(x) is f(x) ,  >  _&,AAlso D-1 f(x) = h(x) = i=0x-1 f(i) Note parallel: we can obtain an expression for the summation by anti-differencing; compare to integration by anti-differentiation. hSimple Properties of D$sUnique up to addition of functions whose first difference is zero Constants functions with period 1, e.g. sin (p x)<B2B-#We also need the shift operation, E/E f(x) = f(x+1) hence D f(x) = Ef(x)-f(x) H!  More properties of D$  Occasionally useful propertyIThe chain rule D f(g(x)) = D f(g(x)) where D f(x,y) = f(x+h,y)-f(x,y)bJ( d(  4The simplest non-trivial form to sum is a polynomial'A(x) = ai xi The analogy to differential calculus is to integrate, term by term: easy since Dxn = nxn-1. Differences of powers are not so concise: D(xn) = (x+1)n-xn = (binomial(n,i),i=0..n-1) and so neither is summation. INSTEAD consider factorial functions. [x]n = x(x-1)(x-2)....(x-n+1).T*F  + _> T6/What is the difference of a factorial function? D [x]n = E [x]n  [x]n =n[x]n-1. Proof: E[x]n = (x+1)x(x-1)(x-2).... (x-n+2) [x]n = x(x-1)(x-2)....(x-n+2)(x-n+1). All the terms in red are the same, and one can factor them out. they are [x]n-1. The remaining factor is simply (x+1)-(x-n+1) = n. )  952To sum a polynomial, convert it to factorial form:<one way is to expand a bunch of factorial functions [x]1+x, [x]2= x2-x, etc, solve for powers of x, e.g. x2= [x]2-[x]1, and we are done. Another is to use Newton s divided difference interpolation formula in a table: f(x)=sum([x]i/i! Di f(0)) where we use higher differences this way: 7&o33+Divided difference table for f= 3*x^3-2*x+1 +Divided difference table for f= 3*x^3-2*x+1 XConverting to conventional polynomial form is done by expanding [x]i and combining terms$YC[x]1 =x [x]2 = - x+ x2 3[x]3 = 6x  9 x2 +3x3 [x]4 = -9/2x  21/4 x2-9/2x3+ x4 total is 2x-55/4*x2-3/2x3+3/4x4w   Sums of rational functions Define factorial operators on functions... [f(x)]k = f(x) f(x-1) ... f(x-k+1) for k >0 extend the operator by noticing [f(x)]k = [f(x)]r [f(x-r)]k-r Define [f(x)]0 to be 1 and use the previous line as an identity. Then for k=0 we get [f(x)]-r = 1/[f(x+r)]r 1:   N Differences of factorials (What does this mean for summation (D-1)?.)#{If we can get rational expressions so they look like the RHS of that equation, we can find their summation, namely [f(x)]-l*|y 6We need to use Shift Free Decomposition to go further.Given a product of functions, we can decompose it into a product of factorial functions. Let S=a b c where a,b,c are mutually relatively prime and Ea=b. Then shift S: ES = (Ea)(Eb)(Ec) = b Eb Ec GCD(S,ES) = b So we can divide out b and a from S and express S=[b]2 c . aN G> LFIf we apply this observation repeatedly, we can get S to be shift freeS=[s1]1 [s2]2 ... [sk]k where the individual sk are shift-free. Analogous to partial fraction decomposition in the differential calculus Hermite integration process, we can form a shift-free partial fraction for some rational function we wish to sum. That is, A(x)/S(x) = (Ai/[si]i), i=1..k and a  complete decomposition A(x)/S(x) = (Aij/[si]j), i=1..k,j=1..i jvZ5bZ7 IShift-1 independence is not enough. We need to show S(x) is k-shift-freeCompute resultant of S(x) and S(x+k) with respect to k. If there is an integer k>1 shift, then fill in the terms for numerator and denominator. e.g. if S = x*(x+3), change it to [x+3]4 and multiply numerator by (x+1)(x+2).0-'Summation by parts Similar to Hermite integration using D-1(u D v) = u v - D (Ev D u) can be used to reduce denominators of the form [xi]j to [xi]1 EVENTUALLY... one gets a rational function plus an indefinite summation of terms with shift-free denominators of factorial degree 1.%d&5> -8The transcendental partSDefine ym(x)= Dm(logG(x+1)), m>0 where n!=G(n+1) is the well-known gamma function dT(I4The sum of a negative power of x+1 finishes the task D-1(x+1)-m = (-1)m-1/(m-1)! ym(x) The ym functions are known as polygamma functions and serve a role similar to logs in Hermite integration. Rational summation is pretty much solved, though people still look for fast ways of doing some of the steps (shift-free decomposition). P /EThis is not the end of the story: what about more elaborate summands?Gosper s algorithm looks at ai = D-1ai by seeking a  telescoping function f(n). Let an = D g(n) = g(n+1)-g(n) then suppose g(n)=f(n)*an. We have to solve the functional equation C(n)=an+1/an = (f(n)+1)/f(n+1) Only the ratio of 2 terms is used (easily computed). If C(n) is rational in n, then this is called hypergeometric summation. U   0 , 1   6 Restrictions/ ExtensionsNote that the terms an can be far more general than rational; just an+1/an is rational Gosper s work is the basis for a decision procedure, widely used in computer algebra systems. More recent work by Zeilberger (see refs) pushes this further. H.,Wk "PL  ` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dd@z?" dd@  " @ ` n?" dd@   @@``PR    @ ` `p>>  &(     6dbB  "  ` B  T Click to edit Master title style! !$  0dB  "  B  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     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S  T ee P    x*  A!!AAbb  Tx1ee     z*  A!!AAbbH  0޽h9 ? ̙3380___PPT10.e"Y< `8(    Nċ ee P#    `*  A!!AAbb  N_1ee  #   b*  A!!AAbb  TT4p ee P    `*  A!!AAbb  THp ee   p  b*  A!!AAbbH  0޽h9 ? ̙3380___PPT10.e"\= 0$(  r  S @ p   r  S    `     H  0޽h ? ̙33  P$(  r  S F    `    r  S TG      H  0޽h ? ̙33  `0$(  0r 0 S L    `    r 0 S L      H 0 0޽h ? ̙33  B:p4(  4r 4 S O    `      4  xA?texpointfig"* jb  SOURCE\documentclass{slides}\pagestyle{empty} \begin{document} % 'SUM(I,I,1,N) = N*(N+1)/2 $$ \sum_{i=1}^{n}{i}={{n\,\left(n+1\right)}\over{2}} $$ $$ \sum_{i=1}^{n}{{{i}\over{2^{i}}}}=-{{n}\over{2^{n}}}-{{2}\over{2^{% n}}}+2 $$ \end{document} 6EXTERNALNAMEEdittex$ BLEND False0TRANSPARENT False6BITMAPFORMATbmpmono8 DEBUGINTERACTIVETrue(ORIGWIDTH223H 4 0޽h ? ̙33  8:(  8r 8 S     `  >  8  xA?texpointfig" @ r SOURCEV\documentclass{slides}\pagestyle{empty} \begin{document}% 'SUM(N^4*4^N/BINOMIAL(2*N,N),N,0,N) = 2*(N+1)*(63*N^4+112*N^3+18*N^2-22*N+3)*4^N/(693*BINOMIAL(2*N,N))-2/231 $$ \sum_{n=0}^{n}{{{n^{4}\,4^{n}}\over{{2\,n\choose n}}}}={{2\,\left(% n+1\right)\,\left(63\,n^{4}+112\,n^{3}+18\,n^{2}-22\,n+3\right)\,4^{% n}}\over{693\,{2\,n\choose n}}}-\mathchoice {{2}\over{231}}{{2% }\over{231}}{2/231}{2/231} $$ \end{document} 6EXTERNALNAMEEdittex$ BLEND False0TRANSPARENT False6BITMAPFORMATbmpmono8 DEBUGINTERACTIVETrue(ORIGWIDTH667H 8 0޽h ? ̙33  <$(  <r < S a    `    r < S b      H < 0޽h ? ̙33  @$(  @r @ S o    `    r @ S lp      H @ 0޽h ? ̙33  D$(  Dr D S  u    `    r D S u      H D 0޽h ? ̙33  H$(  Hr H S |    `    r H S }      H H 0޽h ? ̙33"  Tb(  Tr T S (    `     T  xA?texpointfig"P  SOURCE~\documentclass{slides}\pagestyle{empty} \begin{document}$$\Delta k\,f(x)=k\,\Delta f (x)~~~~k \in ~F$$ $$\Delta(g(x)+f(x)% )=\Delta g(x)+\Delta f(x )$$ $$\Delta(f(x)\cdot g(x)% )=E g(x)\cdot \Delta f(x)+f(x)\cdot \Delta g(x)$$ $$\Delta\left({{1}\over{g(x)}}\right)=-{{\Delta% g(x)}\over{g(x)\,E g(x )}}$$ $$\Delta\left({{f(x)}\over{g(x)% }}\right)=-{{f(x)\,\Delta g(x)+g% (x)\,\Delta f(x)}\over{g(x% )\,Eg(x)}}$$ \end{document} 6EXTERNALNAMEEdittex$ BLEND False0TRANSPARENT False6BITMAPFORMATbmpmono8 DEBUGINTERACTIVETrue(ORIGWIDTH457H T 0޽h ? ̙33  JB X(  Xr X S ,    `    r X S        X H ?"` p0Z  7x 2 X Ht ?"0  *  :g(x) 2 X H ?" p0  71 2  X H ?"P h $  HD g(x)& 2  X H ?"   7x 2  X Hĩ ?"  ;h 2H X 0޽h ? ̙33  \$(  \r \ S     `    r \ S ز      H \ 0޽h ? ̙33  `$(  `r ` S     `    r ` S ܒ      H ` 0޽h ? ̙33  d$(  dr d S D    `    r d S       H d 0޽h ? ̙33!    ,fiN (  hr h S 8V   `   t    fi #"&   h NY ?"   > @` h N`j ?"   > @` h NXr ?"  > @` h Nz ?"  H76 @` h Nh ?"   G3 @` h N ?"   > @` h N ?"   > @` h N ?"  H55 @` h N ?"  H21 @` h N ?"   G2 @` h N ?"   > @` h N8 ?"   H36 @` h N3 ?"   H19 @` h N ?"   G2 @` h N0 ?"   G1 @` h N ?"   H18 @`  h N ?"   H18 @`  h N ?"   G1 @`  h N ?"   G1 @`  h N ?"   G0 @`  h N ?"   `D3f(x)( @` h N8 ?"   `D2f(x)( @` h N ?"   VD f(x) @` h N ?"   Jf(x) @` h N` ?"   Gx @`B h H1 ?"  B #h Ho?"  B $h Ho?"  B )h Ho?"  B 7i Ho?"  B ?i Ho?"  B Ai Ho?"  B Ii Ho?"  B Ki Ho?"  B Si Ho?"  B Ui Ho?"  B ]i Ho?"  B _i Ho?"  B ai Ho?"  B ci Ho?"  B ei Ho?"  H h 0޽h ? ̙33$  ##0./lD#(  l -l N?"pPx l c $h   `   z    l# #"&   l N ?"   > @` l NT ?"   > @` l NTs ?"  > @` l N ?"  H76 @` l N ?"   G3 @`  l N ?"   > @`  l N ?"   > @`  l NQ ?"  H55 @`  l NY ?"  H21 @`  l Na ?"   G2 @` l N k ?"   > @` l Nhc ?"   H36 @` l N8c ?"   H19 @` l NXc ?"   G2 @` l Ndac ?"   G1 @` l Nc ?"   H18 @` l N1c ?"   H18 @` l Nc ?"   G1 @` l Nc ?"   G1 @` l Nc ?"   G0 @` l Nc ?"   `D3f(x)( @` l Nlc ?"   `D2f(x)( @` l Nc ?"   VD f(x) @` l Nc ?"   Jf(x) @` l N@c ?"   Gx @`B l H1 ?"  B l Ho?"  B l Ho?"  B  l Ho?"  B !l Ho?"  B "l Ho?"  B #l Ho?"  B $l Ho?"  B %l Ho?"  B &l Ho?"  B 'l Ho?"  B (l Ho?"  B )l Ho?"  B *l Ho?"  B +l Ho?"  B ,l Ho?"  V /l HpHc ?" @D FD-1 f =sum([x]i/i! Di f(0))=1*[x]1+ 1/2*[x]2 +18/3!*[x]3 +18/4!*[x]4 .LG 2    33 33 33 33 2H l 0޽h ? ̙33  @p$(  pr p S c    `  c  r p S c    c  H p 0޽h ? ̙33  Pt$(  tr t S xc    `  c  r t S yc    c  H t 0޽h ? ̙33  pxN(  xr x S ~c    `  c  X x S 0d4Ahpcap0200 < x H r  S DB    B  H  0޽h ? ̙33rPZKLU*gy biklryv{b}N0:dޒʔvm Oh+'0  hp    (4<4Introduction to Programming Languages and Compilers Alex AikennlexRichard Fateman152Microsoft PowerPointamm@P/A@Qɵ-_@f"Gg  <  -- @ !--'@Times New Roman-. <2 a#Richard Fateman CS 282 Lecture 19            ."System-@Times New Roman-.  2 1_ .-@BComic Sans MS-. 02 FClosed Forms for Summations  .-@BComic Sans MS-. 2  Lecture 19 .-՜.+,0    DOn-screen ShowDigital Integrity, Inc{ !Times New RomanComic Sans MSSymbolcmsy10icfp99Closed Forms for Summations'Two categories (maybe 3) of Summations.Indefinite summation parallels to integration-Some simple examples (Macsyma, in this case)A more elaborate exampleStart simply.AlsoSimple Properties of D$We also need the shift operation, EMore properties of DOccasionally useful property5The simplest non-trivial form to sum is a polynomial0What is the difference of a factorial function?3To sum a polynomial, convert it to factorial form:,Divided difference table for f= 3*x^3-2*x+1,Divided difference table for f= 3*x^3-2*x+1YConverting to conventional polynomial form is done by expanding [x]i and combining termsSums of rational functionsDifferences of factorials)What does this mean for summation (D-1)?7We need to use Shift Free Decomposition to go further.GIf we apply this observation repeatedly, we can get S to be shift freeJShift-1 independence is not enough. We need to show S(x) is k-shift-freeSummation by partsThe transcendental part5The sum of a negative power of x+1 finishes the taskFThis is not the end of the story: what about more elaborate summands?Restrictions/ Extensions  Fonts UsedDesign Template Slide Titles'_rp Richard FatemanRichard Fateman  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrtuvwxyz{|}~ Root EntrydO)PicturesCCurrent UserSummaryInformation(PowerPoint Document(sDocumentSummaryInformation8