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 ...

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 If you do not insist upon your incomplete factoriz...

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. ...

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. ...

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...

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, 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 ...

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 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...

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...

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} ...

where g(x), k0 and k1 are known and * denotes the convolution operation. From the chirp point view, exp(j k x^2) is a linear FM chirp. Is there a method that we can recover f0(x) from the composite? ...

where g(x), k0 and k1 are known and * denotes the convolution operation. From the chirp point view, exp(j k x^2) is a linear FM chirp. Is there a method that we can recover f0(x) from the composite? Tristan Imp wrote: > Can we recover f0(x) or f1(x), given g(x)=f0(x)*exp(j k0 > x^2)+f1(x)*exp(j k1 x^2)? where g(x), k0 and k1 are known and * > denotes the convolution> operation. > > From the chirp point view, exp(j k x^2) is a linear FM chirp. > > Is there a method thatwe can recover f0(x) from the composite? As far as I can tell, you have one equation and two unk...

What languages might use a scheme similar to this? Could I build my own? 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...

So I've been looking at execution time for various algorithms, and I found this interesting result: bigarr = fltarr(1000,1000) t1 = systime(/seconds) t = bigarr^2.0 t2 = systime(/seconds) t = bigarr*bigarr t3 = systime(/seconds) print,t2-t1 print,t3-t2 IDL prints: 0.024163008 0.010262012 Apparently multiplying an array by itself is twice as fast as using the carat operator! Anyone know why this is? Is it a memory issue or something? On Jul 9, 12:32 pm, Conor <cmanc...@gmail.com> wrote: > So I've been looking at execution time for various algorithms, and I &...

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...

On 4/16/10 at 5:53 AM, klaus.engel@tiscali.it (Klaus Engel) wrote: >I tried to simplify an awkward looking integral with "Mathematica 7" >using its "(Full)Simplify[...]" function. Unfortunately it failed to >do so, even though I know that this would be possible. I boiled down >the problem to the following very simple example ("f" is just a >generic, undefined function): The input >Integrate[2 f[x], {x, 0, 1}]/2 // FullSimplify >returns just the input >Integrate[2 f[x], {x, 0, 1}]/2 The problem is f is undefined. Consequen...

Hi! Why Maple developers decided to evaluate int(1/x, x) as ln(x) and not ln(abs(x))? Thanks, Humberto. Humberto Bortolossi wrote: > Hi! > > Why Maple developers decided to evaluate > > int(1/x, x) > > as > > ln(x) > > and not > > ln(abs(x))? > > Thanks, Humberto. Because ln(abs(x)) is wrong if x is complex. If x is not dimensionless then both ln(x) and ln(abs(x)) are both wrong, whether x is complex or not. Instead int(1/x,x) is ln(x/c) where c is an arbitrary constant with the same dimensions...

Hello my name is Don Robison. I am having a problem with the way matlab evaluates expressions. It seems if I put a polynomial on the command prompt and I have declared the variables to be matrices the evaluation will be fine, but if I declare the polynomial to be a symbol then try to evaluate the symbolic expression instead of a literal they are not the same thing. Is there something other than eval I should be using to plug the matrices into the symbolic expression? eval seems to not be evaluating the exponent of the matrix. Don Robison <joe_blow_281@hotmail.com> wrote in messag...

Hi, ArcCos[x/Sqrt[x^2+y^2]] and Pi/2-ArcTan[x/Abs[y]] are the same. But I can not get the first expression be simplified to the second one. And the following command can not be simplified to zero in mathematica as well. FullSimplify[ArcCos[x/Sqrt[x^2 + y^2]] - ( Pi/2 - ArcTan[x/Abs[y]]), Element[x, Reals] && Element[y, Reals]] I'm wondering if there is any walkaround to do this simplification. Thanks, Peng Peng Yu schrieb: > Hi, > > ArcCos[x/Sqrt[x^2+y^2]] > > and > > Pi/2-ArcTan[x/Abs[y]] > > are the same. &...

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...

Hello, All! There are some sources of pseudo network interface driver develkoped for 2.2.x kernel. I have to port it onto 2.4.x kernel. I have explored the structure 'net_device' from linux/net/netdevice.h, there are few major changes. The main problem for now is - how to check that interface is ready to transmit data (in 2.2.x 'tbusy' field was used fot this and function 'test_and_set_busy' checking the 'tbusy' state). But thus field was removed from 2.4.x kernel, and now you should use the function 'netif_start_queue' or 'netif_stop_queu...

z=sqrt(x^2+y^2) c=(z<15) imshow(c) gives a square cut diagonally half with lower right side being white and other being black z=sqrt(x.^2+y.^2) c=(z<15) imshow(c) gives a square with circle in centre What is the difference between (x^2+y^2) and (x.^2+y.^2) in matlab? "Sagar Chand" <sagarchand_9@rediffmail.com> wrote in message <n8efgg$100$1@newscl01ah.mathworks.com>... > z=sqrt(x^2+y^2) > c=(z<15) > imshow(c) > gives a square cut diagonally half with lower right side being white and other being black > > z=sqrt(x.^2...

Hi, I thought that x +=3D ... was the same than x =3D x + ..., but today I have= realized it is not true when operating with mutable objects. In Python 3.3 or 2.7 IDLE (Windows) compare: >>> a =3D [3] >>> b =3D a >>> a =3D a + [1] >>> b [3] and >>> a =3D [3] >>> b =3D a >>> a +=3D [1] >>> b [3, 1] Is this behaviour explained in the Python documentation?=20 Thanking you in advance, Bartolom=E9 Sintes bartolome.sintes@gmail.com writes: > Hi, > > I thought that x += ... was t...

In computing , signed number representations are required to encode negative numbers in binary number systems. In mathematics , negative numbers ...

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