Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to (x+x^2)*x^(1/2). I am willing to assume(x>0). Maple does recognize these as being equal: > p:=x^(3/2)+x^(5/2): > q:=(x+x^2)*sqrt(x): > simplify(p-q); 0 Here's a clumsy way to do it: > p:=x^(3/2)+x^(5/2): > assume(t>0): > subs(x=t^2,p); > simplify(%,{t^2=x}); > collect(%,t); Is there something simpler? (This has undoubtedly come up before, but I cannot find it anywhere.) --Edwin

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12/29/2003 10:02:34 PM

If you do not insist upon your incomplete factorization: > x^(3/2) + x^(5/2); (3/2) (5/2) x + x > factor(%); (3/2) x (1 + x) All the best Jurgen Edwin Clark schrieb: > Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to > (x+x^2)*x^(1/2). > I am willing to assume(x>0). > > Maple does recognize these as being equal: > > >>p:=x^(3/2)+x^(5/2): >>q:=(x+x^2)*sqrt(x): >>simplify(p-q); > > > 0 > > Here's a clumsy way to do it: > > >>p:=x^(3/2)+x^(5/2): >>assume(t>0): >>subs(x=t^2,p); >>simplify(%,{t^2=x}); >>collect(%,t); > > > Is there something simpler? > > (This has undoubtedly come up before, but I cannot find it anywhere.) > > --Edwin > > > >

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12/30/2003 12:51:42 AM

I wrote: > Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to > (x+x^2)*x^(1/2). > I am willing to assume(x>0). ------ > > Here's a clumsy way to do it: > > > p:=x^(3/2)+x^(5/2): > > assume(t>0): > > subs(x=t^2,p); > > simplify(%,{t^2=x}); > > collect(%,t); When I copied this bit of code in for some reason the last line was not copied. It should include: > subs(t=sqrt(x),%); > > Is there something simpler? > > (This has undoubtedly come up before, but I cannot find it anywhere.) > > --Edwin

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12/30/2003 1:10:10 AM

Edwin Clark wrote: >| Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to >| (x+x^2)*x^(1/2). >| I am willing to assume(x>0). Well, you may do the following: > restart; > p:=x^(3/2)+x^(5/2); (3/2) (5/2) p := x + x > series(p/sqrt(x),x)*sqrt(x); 2 1/2 (x + x ) x > whattype(%); * Helmut

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12/30/2003 4:04:09 AM

"Edwin Clark" <eclark@math.usf.edu> wrote in message news:<_v1Ib.153184$b01.3400427@twister.tampabay.rr.com>... > Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to > (x+x^2)*x^(1/2). > I am willing to assume(x>0). > Is there something simpler? > --Edwin Why not x^(3/2)+x^(5/2)=x^(3/2)*(1+x)? > x^(3/2)+x^(5/2): %=factor(%,sqrt(x)) assuming x>0; x^(3/2)+x^(5/2)=x^(3/2)*(1+x) Ken

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12/30/2003 4:10:45 AM

Helmut Kahovec wrote: >| Edwin Clark wrote: >| >| >| Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to >| >| (x+x^2)*x^(1/2). >| >| I am willing to assume(x>0). >| >| Well, you may do the following: >| >| > restart; >| > p:=x^(3/2)+x^(5/2); >| >| (3/2) (5/2) >| p := x + x >| >| > series(p/sqrt(x),x)*sqrt(x); >| >| 2 1/2 >| (x + x ) x >| >| > whattype(%); >| >| * Sorry! I forgot converting the series object back to a polynomial: > restart; > p:=x^(3/2)+x^(5/2); (3/2) (5/2) p := x + x > series(p/sqrt(x),x)*sqrt(x); 2 1/2 (x + x ) x > map(convert,%,polynom); 2 1/2 (x + x ) x Helmut

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12/30/2003 4:31:19 AM

"Helmut Kahovec" wrote > Edwin Clark wrote: > > >| Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to > >| (x+x^2)*x^(1/2). > >| I am willing to assume(x>0). > > > Well, you may do the following: > > > restart; > > p:=x^(3/2)+x^(5/2); > > (3/2) (5/2) > p := x + x > > > series(p/sqrt(x),x)*sqrt(x); > > 2 1/2 > (x + x ) x > Thanks Helmut, I notice that you can also get the result by expand(p/sqrt(x))*sqrt(x). But both of these run into trouble if p has a constant term say p=x^(3/2)+x^(5/2) + 1; I was also thinking of things like a + b*x^(3/2) + c*y^(5/2) + d*x^(7/2)*y^(5/2). In general, how does one convert all terms of the form z^((2*n+1)/2) to z^n*sqrt(z) and maintain this form while collecting coefficients of sqrt(z) . Maple does this automatically if x, y , z are positive integers. For example: > 2^(5/2) + 3^(7/2)*5^(5/2); 1/2 1/2 1/2 4 2 + 675 3 5 --Edwin

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12/30/2003 4:57:14 AM

"Ken Lin" wrote > "Edwin Clark" wrote > > Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to > > (x+x^2)*x^(1/2). > > I am willing to assume(x>0). > > Why not x^(3/2)+x^(5/2)=x^(3/2)*(1+x)? Because I want to express everything in terms of sqrt(x). As I mentioned in my reply to Helmut I really would like all subexpressions like x^(2*n+1) to be converted to x^n*sqrt(x) where there could be several such in an expression and several variables not just x. The following works except for subexpressions like y^(7/2)*x^(5/2): > p:= 1+x^(3/2) + x^(7/2) + 2*y^(7/2)*x^(5/2) + y^(9/2)+3*y^(5/2): > assume(t>0,r>0): > subs({x=t^2,y = r^2},p): > simplify(%,{t^2=x,r^2=y}): > collect(%,[t,r], distributed): > subs({t=sqrt(x),r=sqrt(y)},%); (7/2) (5/2) 3 1/2 4 2 1/2 1 + 2 y x + (x + x) x + (y + 3 y ) y --Edwin

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12/30/2003 5:48:47 AM

Edwin Clark wrote: >| I notice that you can also get the result by expand(p/sqrt(x))*sqrt(x). >| >| But both of these run into trouble if p has a constant term say >| p=x^(3/2)+x^(5/2) + 1; >| >| I was also thinking of things like a + b*x^(3/2) + c*y^(5/2) + >| d*x^(7/2)*y^(5/2). In general, how does one convert all terms of the form >| z^((2*n+1)/2) to z^n*sqrt(z) and maintain this form while collecting >| coefficients of sqrt(z) . Maple does this automatically if x, y , z are >| positive integers. For example: >| >| > 2^(5/2) + 3^(7/2)*5^(5/2); >| >| 1/2 1/2 1/2 >| 4 2 + 675 3 5 Well, I do not think that this will be possible, since Maple automatically rewrites expressions like x*sqrt(x) as x^(3/2). It is analogous to automatically expanding 2*(x+y) to 2*x+2*y. However, you may try the following: > restart; > p:=1+x^(3/2)+x^(7/2)+2*y^(7/2)*x^(5/2)+y^(9/2)+3*y^(5/2); (3/2) (7/2) (7/2) (5/2) (9/2) (5/2) p := 1 + x + x + 2 y x + y + 3 y You may assign the order term of the series objects any sufficiently large number: > N:=1000: Remember that expressions within parentheses are of type series: > subsindets( p, `^`, u->series(u/op(1,u)^(1/2),op(1,u),N)*op(1,u)^(1/2) ); 1/2 3 1/2 3 1/2 2 1/2 4 1/2 1 + (x) x + (x ) x + 2 (y ) y (x ) x + (y ) y 2 1/2 + 3 (y ) y Transforming back to polynomials triggers Maple's automatic simplifier: > subsindets( %, series, u->convert(u,polynom) ); (3/2) (7/2) (7/2) (5/2) (9/2) (5/2) 1 + x + x + 2 y x + y + 3 y Helmut

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12/31/2003 3:09:15 AM

Edwin Clark wrote: >| The following works except for subexpressions like y^(7/2)*x^(5/2): >| >| > p:= 1+x^(3/2) + x^(7/2) + 2*y^(7/2)*x^(5/2) + y^(9/2)+3*y^(5/2): >| > assume(t>0,r>0): >| > subs({x=t^2,y = r^2},p): >| > simplify(%,{t^2=x,r^2=y}): >| > collect(%,[t,r], distributed): >| > subs({t=sqrt(x),r=sqrt(y)},%); >| >| (7/2) (5/2) 3 1/2 4 2 1/2 >| 1 + 2 y x + (x + x) x + (y + 3 y ) y Well, as I indicated in my previous message, you may alternatively do the following: > restart; > p:=1+x^(3/2)+x^(7/2)+2*y^(7/2)*x^(5/2)+y^(9/2)+3*y^(5/2); (3/2) (7/2) (7/2) (5/2) (9/2) (5/2) p := 1 + x + x + 2 y x + y + 3 y > subsindets( p, `^`, u->series(u/op(1,u)^(1/2),op(1,u),1000)*op(1,u)^(1/2) ); 1/2 3 1/2 3 1/2 2 1/2 4 1/2 1 + (x) x + (x ) x + 2 (y ) y (x ) x + (y ) y 2 1/2 + 3 (y ) y > collect(%,[sqrt(x),sqrt(y)],distributed); 3 1/2 4 2 1/2 1 + ((x) + (x )) x + ((y ) + 3 (y )) y 3 1/2 2 1/2 + 2 (y ) y (x ) x > subsindets( %, series, u->convert(u,polynom) ); 3 1/2 4 2 1/2 (7/2) (5/2) 1 + (x + x ) x + (y + 3 y ) y + 2 y x Helmut

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12/31/2003 3:24:27 AM

hi normal(1/(1-2*x) + x/ (1-x-x^2),expanded); 1/(1-2*x) + x/(1-x-x^2) => (1-3*x^2)/(1-3*x+x^2+2*x^3) how to do the reverse? dillogimp@gmail.com writes: > normal(1/(1-2*x) + x/ (1-x-x^2),expanded); > 1/(1-2*x) + x/(1-x-x^2) => (1-3*x^2)/(1-3*x+x^2+2*x^3) > how to do the reverse? convert(%,parfrac,x); -- Joe Riel ...

The CAS "simplifies" x^2-x-x^3 to -(x^3-x^2+x) (flag -114 clear). Is there a way to set it to simplify this to -x^3+x^2-x ? John Pressman wrote: > The CAS "simplifies" x^2-x-x^3 to -(x^3-x^2+x) (flag -114 clear). Is > there a way to set it to simplify this to -x^3+x^2-x ? FDISTRIB Regards, -- Beto Reply: Erase between the dot (inclusive) and the @. Responder: Borra la frase obvia y el punto previo. ...

Thanks! In article <1140494713.980133.168190@f14g2000cwb.googlegroups.com>, loric <dr.huiliu@gmail.com> wrote: >Thanks! It isn't clear exactly what your question is, but: > `union`({2*x+y=3},{3*x+2*y=5}); {2 x + y = 3, 3 x + 2 y = 5} > Union := proc() { seq( `if`(s::'set',op(s),s), s=args ) } end proc: > Union(2*x+y=3,3*x+2*y=5); {2 x + y = 3, 3 x + 2 y = 5} > Union(2*x+y=3,{3*x+2*y=5}); {2 x + y = 3, 3 x + 2 y = 5} ...

The CAS "simplifies" x^2-x-x^3 to -(x^3-x^2+x) (114 clear). Is there a way to set it to simplify this to -x^3+x^2-x ? In article <2cdd7b99.0312120050.4f899916@posting.google.com>, voldermort@hotmail.com (John Pressman) wrote: > The CAS "simplifies" x^2-x-x^3 to -(x^3-x^2+x) (114 clear). Is there a > way to set it to simplify this to -x^3+x^2-x ? On the hp49/HP49+ use FDISTRIB. ...

HoldForm[] is loosing the 1* when it apparently should not: In[1]:= HoldForm[1*2*3] Out[1]= 2 x 3 In[2]:= HoldForm[1*1*1] Out[2]= 1 x 1 x 1 In[3]:= HoldForm[3*2*1] Out[3]= 3 x 2 In[4]:= HoldForm[2*2*2] Out[4]= 2 x 2 x 2 In[5]:= HoldForm[2*1*3] Out[5]= 2 x 3 In[6]:= HoldForm[1*2] Out[6]= 1 x 2 In[7]:= HoldForm[1*2*1] Out[7]= 1 x 2 x 1 Q.E.D. Indeed, the same happens with Hold and HoldComplete. I'd say this is a bug. Cheers -- Sjoerd On Feb 25, 11:07 am, "Q.E.D." <a...@netzero.net> wrote: > HoldForm[] is loosing the 1* when it...

Hi Since the power is more relevant to analyze than the amplitude I have come = up with the idea of using signed square ss(x) =3D x^2 for positive x and -x= ^2 for negative x or ss(x) =3D (x^2)^(1/2)*x =3D sgn(x)*x^2 Instead of analyzing the power as the square of the amplitude the signed po= wer could be used instead. This would preserve more information of the orig= inal signal. The original amplitude signal can be recreated from the signed= square power signal with signed square root. I can imagine signed square to be applicable to codec evaluation and signal= compression s...

Hi, any suggestion to make the integral of: Exp[-(x-m)^2/(2 s^2)] x (1+x^2)^-1 Exp[-(x-m)^2/(2 s^2)] x^2 (1+x^2)^-1 Exp[-(x-m)^2/(2 s^2)] x^3 (1+x^2)^-1 between -inf and +inf (or indefinite)? Look like it is not possible, but it is too long time I do not make integrals with more advanced techinques (as going to the complex plane)... so if you have suggestions (wonderful a solution :) ).... THANKS Ale ...

makes this any difference ? x**2 or x**2.0 or x*x x is real Cheers -Ralf "Ralf Schaa" <schaa@geo.uni-koeln.de> wrote in message news:1133832989.230157.34130@g49g2000cwa.googlegroups.com... > makes this any difference ? > > x**2 or x**2.0 or x*x > x is real > > Cheers > -Ralf > It might be. Some would implement the first as x*x. Some would also implement the second as X*x as a special case of the more general expansion of exp(ln(x)*2)) (providing my math is correct anyway).. Jim Ralf Schaa wrote: > makes this any difference ? > > x...

Question How can i do that? I've been trying, solve, explot and subs. I can get the graph, but i cannot get the (x,y) pairs that I want. When I use function solve to solve the equation in term of x or y and then substitute the numeric value for one of them to find the other, I got the problem about complex numer. This might be because that numeric value that I substitute is not the number that satisfy the equation, so the program give me the result in complex number. Background. I need to use the pairs to find the optimal value of my objective function. I tried to use the fminco...

x = x - (x+1) + x + 2: Numbers: 101 (TM) Code. -100=-101+1==-99-1===-100 -099=-101+2==-98-1===-99 -098=-101+3==-97-1===-98 -097=-101+4==-96-1===-97 -096=-101+5==-95-1===-96 -095=-101+6==-94-1===-95 -094=-101+7==-93-1===-94 -093=-101+8==-92-1===-93 -092=-101+9==-91-1===-92 -091=-101+10==-90-1===-91 -090=-101+11==-89-1===-90 -089=-101+12==-88-1===-89 -088=-101+13==-87-1===-88 -087=-101+14==-86-1===-87 -086=-101+15==-85-1===-86 -085=-101+16==-84-1===-85 -084=-101+17==-83-1===-84 -083=-101+18==-82-1===-83 -082=-101+19==-81-1===-82 -081=-101+20==-80-1===-81 -080=-101+21==-79-1===-80 -079=-101+22==-...

Hi all, My problem is very easy. But I don't know quite MatLab to know if there is a function to solve that: 0 = a(1) + a(2)*x + a(3)*x^2 + a(4)*x^3 +...? "a" is vector known. Thank you for your help Fabien. Look at the "roots" function. "Fabien Blarel" <eezfb@gwmail.nottingham.ac.uk> wrote in message news:eec6364.-1@WebX.raydaftYaTP... > Hi all, > > > My problem is very easy. But I don't know quite MatLab to know if > there is a function to solve that: > > > 0 = a(1) + a(2)*x + a(3)*x^2 + a(4)*x^3...

Please check Table[{x, 2*x^2 - x^3, (2*x^2 - x^3)^ (1/3), Re[(2*x^2 - x^3)^(1/3)]}, {x, -4, 4}] Best regards, MATTHIAS BODE S 17.35775=B0, W 066.14577=B0 > Date: Sat, 28 Aug 2010 07:01:34 -0400 > From: bernard.vuilleumier@edu.ge.ch > Subject: Plot of (2 x^2 - x^3)^(1/3) > To: mathgroup@smc.vnet.net > > I see in Calcul Diff==E9rentiel et int==E9gral, N. Piskounov, Editions > MIR, Moscou 1970, on p. 210 the graph of (2 x^2 - x^3)^(1/3) with > negative values for x > 2, something like : > > Plot[Piecewise[{{(2 x^2 - x^3)^(1/3...

Risch(sqrt(x+sqrt(x^2+a^2))/x, x) seems to hang on my 49G+. Does anyone know what it's trying to do, since (AFAIK) the algebraic case of the Risch algorithm is not implemented? Thanks, Bhuvanesh. The Risch algorithm is only a partial implementation. It does handle sqrt very easily AFAIK. I think it is documented in the latest Erable 3.024 distribution for the 48. Regards, Terry "Bhuvanesh" <lalu_bhatt@yahoo.com> wrote in message news:662e00ed.0408171050.4fabb8db@posting.google.com... > Risch(sqrt(x+sqrt(x^2+a^2))/x, x) seems to hang on my 49G+. Does > anyone know...

Hi group, The integral Integrate[ x^2 * Erfc[a x] *(BesselJ[1, b x])^2 , {x, 0, Inf}] has a solution in Mathematica (in terms of HypergeometricPFQ) My question: how did Mathematica know? I tried the usual suspects: i) for example Gradshteyn and Ryzhik (1965), p.769, 6.784 is very close, but not close enough ii) Abramowitz & Stegun hasn't got it on the menu either Then how about expressing Erfc[a*x] in terms of... Erfc(z)=-1/(sqrt(pi)) * IncompleteGamma[1/2,x^2]+1 (Grads.+Ryzh. p. 942) ....and hoping to get lucky? ... no alas! sorry, this is a cross-post (als...

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