[MUG] getting hold of exponents in ifactor

[MUG] Re: getting hold of exponents in ifactor
Author: Maple User Group    Posted: Fri, 9 Aug 2002 17:55:12 -0400 (
>> From: Maple User Group "maple_gr" -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 7 Aug 2002 18:00:32 -0400 (EDT) From: Carl Devore "devore" To: "maple-list" "moree" Subject: getting hold of exponents in ifactor On Tue, 6 Aug 2002 "moree" wrote: > I wonder how to get hold of the primes and exponents that one gets on > applying IFACTOR. Use ifactors. > To give an example, suppose I want to compute \sum_{p|n}(e_p(n))^2*p, > where e_p(n) denotes the exponent of the prime p in the prime > factorisation of n, how do I most efficiently program this in Maple ? With ifactors this is trivial: sq_e:= n-> add(f[2]^2*f[1], f= ifactors(n)[2]); > The number n is any non-zero rational number. Don't you mean integer? -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 7 Aug 2002 16:47:27 -0700 (PDT) From: Robert Israel "israel" To: "maple-list" Subject: getting hold of exponents in ifactor It is slightly easier using ifactors rather than ifactor. The result of ifactors is a list. The first element is the sign of n, which doesn't concern us; the second is a list of lists [p,e] where p is prime and p^e occurs in the factorization. So what you want can be done as add(op(2,t)^2*op(1,t), t=op(2,ifactors(n))); Robert Israel "israel" Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 7 Aug 2002 22:11:29 -0400 (EDT) From: Edwin Clark "eclark" To: "maple-list" Subject: getting hold of exponents in ifactor I guess you mean ifactor? Do you mean n is an integer greater than 1? If I understand your question the following procedure should work: f:=n->add(x[2]^2*x[1],x = ifactors(n)[2]); This uses ifactors(n)[2] which gives the prime factorization as a list of lists [p,e_p(n)] Compare ifactor with ifactors: > n:=3^2*5^4; n := 5625 > ifactor(n); 2 4 (3) (5) > ifactors(n); [1, [[3, 2], [5, 4]]] > ifactors(n)[2]; [[3, 2], [5, 4]] ------------------------------------------------------------ W. Edwin Clark, Math Dept, University of South Florida, http://www.math.usf.edu/~eclark/ ------------------------------------------------------------ -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Stanley J Houghton "S.J.Houghton" Date: Fri, 9 Aug 2002 10:27:11 +0100 To: "maple-list" Subject: getting hold of exponents in ifactor Pieter I expect others will produce more efficient code but if I understand you correctly you could use the approach below. The trick in this extraction is to recognise that the terms produced by ifactor are of the form ``(prime)^exponent or ``(prime) if the exponent is unity, where `` is in invisible function (the identity function when expanded by `expand`) used to retain the brackets in the result. You use `op` to extract the components from the terms. The code below turns the result of ifactor into a list of terms that are converted (map) to the form [prime,exponent]. These may be accessed by index and manipulated as you require. I define p and e to make the expression for the sum more readable (note the quotes to avoid premature evaluation, i.e. before the sum). > n:=450; > # elaborate the list of factors as [prime,exponent] > # in two steps, to make it more obvious, but > # expressions may be combined for brevity > f:=[op(ifactor(n))]; #list of factor terms > # now convert to list of lists of form [p,e] > f:=map(x-> > if type(x,`^`) then # for exponent terms > [op(op(1,x)),op(2,x)] > else # for simple prime terms > [op(op(x)),1] fi > ,f); > # now f[i][1] is the prime p > p:='f'['i'][1]; > # and f[i][2] the exponent e > e:='f'['i'][2]; > # thus the sum comes from for example > sum(e^(2*p),i=1..nops(f)); Hope it helps Regards Stan -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Fri, 09 Aug 2002 10:30:39 -0400 From: "Douglas B. Meade" "meade" To: "maple-list" "moree" Subject: getting hold of exponents in ifactor Pieter and MUG, It is easier to use ifactors, not ifactor. The first example shows how to use the output from ifactors. I also provide a procedure that makes all of this automatic and makes slightly more efficient use of the ifactors output. > q := ifactors(1025); q := [1, [[5, 2], [41, 1]]] > P := map2( op, 1, q[2] ); P := [5, 41] > E := map2( op, 2, q[2] ); E := [2, 1] > add( E[i]^(2*P[i]), i=1..nops(P) ); 1025 > S := ( n::posint ) -> add( t[2]^(2*t[1]), t=ifactors( n )[2] ): > S( 1025 ); 1025 I hope this is of use to you, Doug ----------------------------------------------------------------------- Prof. Douglas B. Meade Phone: (803) 777-6183 FAX: (803) 777-6527 Department of Mathematics URL: http://www.math.sc.edu/~meade/ USC, Columbia, SC 29208 E-mail: "mailto:meade"

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