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?

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1/29/2016 1:33:04 AM

"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+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? The ^ operator (without the period) is matrix power. The .^ operator (with the period) is element-by-element power. Hence you will get different results if x or y are not scalar. James Tursa

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1/29/2016 2:12:03 AM

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} ...

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...

Hi Fellas: What happens is that I need to put the value for Indep:X or Indep:Y ? I have tried both of them and I am receiving an error message. I can plot it just using X or Y as Indep value, also the graph is not pointing at the righ side. This is a parabola but both sides of the parabola are pointing to -y, It should be to -x. I think Indep values must be xand y at the same time but the calculator does not allow me to put x and y together, why? Help please... Thanks everybody for your time. Jp In plot setup, chose tye: CONIC This should solve your problem Arnaud Arnaud wrote: > In...

hi, is there a way to plot this funtion?: f(x,y)=x^2+x*y+y^2 (just like a polynomial for instance) In article <fp55a4$6lo$1@fred.mathworks.com>, Rafa <l_black_flowers_l@hotmail.com> wrote: >is there a way to plot this funtion?: f(x,y)=x^2+x*y+y^2 >(just like a polynomial for instance) xr = -5:.1:5; yr = -3:.2:8; [x, y] = ndgrid(xr, yr); f = x.^2 + x.*y + y.^2; surf(xr, yr, f); -- "Let me live in my house by the side of the road -- It's here the race of men go by. They are good, they are bad, they are weak, they are strong Wise, foolish -- so am I;&q...

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. &...

I have a 2D video stack of images (8 bit greyscale) and would like to take out vertical and horizontal trends. I can easily take the average of the stack to get one image, but is there an easy way of finding this 2D array's best fit into an equation of the form g(x,y)=A+B*x+C*y+D*x^2+E*y^2+F*x*y ? I know there is a program called polyfit that is on the right track for what I want to do, but it is not exactly what I want for this data analysis. Is there something better out there for what I am wanting to do? Thanks, Stephen Stephen wrote: > I have a 2D video stack of images (8 bit...

Does anybody knows how to plot (x^2+y^2-1)^3 - x^2*y^3 =0 in gunplot? Thanks in advance On 27.10.2012 18:51, Shin wrote: > Does anybody knows how to plot (x^2+y^2-1)^3 - x^2*y^3 =0 in gunplot? gnuplot generally only does explicit or parametric curves, not implicit ones like the above. But you can fake it to some extent by asking for z=0 contour of the surface z = f(x,y)=(x^2+y^2-1)^3 - x^2*y^3: gnuplot> f(x,y) = (x**2+y**2-1)**3 - x**2*y**3 gnuplot> set contour base gnuplot> set cntrparam lev discrete 0 gnuplot> set isosamp 50 gnuplot> set view map; uns...

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...

Hallo, I have solved an equation and got solution N(e,r)or number of particles(N) as a function of energy(e) and position(r). I want to transfer this solution to N(e,x,y,z) where r=sqrt(x^2+y^2+z^2). How do I do it? On 29 Mrz., 09:45, "Sunipa Som" <sunipa_...@yahoo.com> wrote: > Hallo, I have solved an equation and got solution N(e,r)or number of particles(N) as a function of energy(e) and position(r). I want to transfer this solution to N(e,x,y,z) where r=sqrt(x^2+y^2+z^2). How do I do it? It's not clear to me what you are trying to do. N(e,x,y,z) is your solution, ev...

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. ...

Assume X; Y and Z are jointly Gaussian mean-zero random variables with the (joint) variance given by K := Var(X; Y;Z) =[2 1 0;1 2 1;0 1 2 ] (a) Determine the distribution of U = X + Y + Z. (b)Let W = X + Y and V = X + Y . Compute the joint distribution of W and V . after assiging the values to k matrix how can i find u "pramod kumar" <pramod.kilu@gmail.com> wrote in message news:jpbdev$hqh$1@newscl01ah.mathworks.com... > Assume X; Y and Z are jointly Gaussian mean-zero random variables with the > (joint) variance given > by > K := Var(X; Y;Z) =[2 1 0;1...

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...

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...

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