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2017-05-26 20:26:21 -0600 | asked a question | Unable to Use Array without Issue(s) bar turing red In my Sage/CoCalc article (Sage has recently merged with CoCalc), I am unable to create an Array without a red issue bar popping up. Morover, I am unable to cross out element in my array. The issue(s) is The whole Document is below. I have never had this problem with SageMath before it merged with CoCalc. Please help. |
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2015-12-24 12:34:02 -0600 | commented answer | Maximize the integral of an implicit relation with two parameters @Slelievere How did you figure this out by using sage software. Can you show me? |
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2015-01-08 18:38:12 -0600 | commented question | Maximize the integral of an implicit relation with two parameters I doubt my question became clearer. It's been months. If u and v has values that make the function discontinuous isn't part of the answer I'm looking for. |
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2014-12-21 15:48:10 -0600 | commented question | Maximize the integral of an implicit relation with two parameters Is there anyway I can solve this problem? |
2014-12-17 11:52:53 -0600 | commented question | Maximize the integral of an implicit relation with two parameters Here I made some edits. Hopefully, this could help. |
2014-12-14 13:33:04 -0600 | asked a question | Maximize the integral of an implicit relation with two parameters Suppose we have an equation: $${(r\cos{(t)}+u})^{2}+{(-(r\sin{(t)}+v)^3+1)}^{\frac{2}{3}}=1$$ Where $x=r\cos{t}$ and $y=r\sin{t}$ The graph of this implicit relation has two regions, where one part is above the x-axis, and the other part is below the x-axis. Suppose we're are trying to find the area of the relation above the x-axis, between the x-values ${0}\le{x}\le{2\pi}$. How can one solve for the values of u and v that can give the highest area for this equation that's is above the x-axis, so that u and v as point (u,v) is still inside the implicit relation $$(1-x^2)^3=(1-y^3)$$. Note: The blue stripes should stop at $2\pi$, and if the equation has regions that are "UNDEFINED", try to ignore it. I MADE EDITS! |
2014-11-24 21:00:32 -0600 | asked a question | How does one graph this? I am trying to parametrize $x^2+y^2+sin(4x)+sin(4y)=4$. I need to find a way of taking the intersections between $x^2+y^2+\sin(4x)+\sin(4y)=4$, and $\tan(nx)$. As n increases from $0\le{n}\le{2\pi}$, I can take the following in coordinate-form.... $$(n,\text{The x-intersection value})$$ $$(n,\text{The y-intersection value})$$ Finally I need to take the following to graph its parametric derivative. Which is... $$\frac{({\text{The x-intersection value}})^2+4\cos(4(\text{The x-intersection value}))}{-(\text{The y-intersection value})^2-4\cos(4(\text{The y-intersection}))}$$ I have little knowledge with how to use sage. If someone can help I'll be thankful. |
2014-09-14 17:04:59 -0600 | commented question | Need help with an error? I want to plot a 3-d parameter. Try and think of it this way. I'm trying to find the area under a specific curve. That is A. Then if you see u, and v in inside find_root(find_root(((y*cos(x))^2+u)/4+((y*sin(x))^2+v)/9-1,0,4)), I need highest value of A, at u, and v...... x=u y=v z=A Just look above, and it will make sense. If not I'll figure it out myself. This is as much as I can clarify it. You'll only understand if you were literally with me. This is clearly not the case! |
2014-09-14 14:47:53 -0600 | asked a question | Need help with an error? In sage math I put in the following software However, I was given the following error... Is there a way I can graph this? If so is there also a way to find the maximum values of u and v, from this parametric function? |
2014-08-12 19:27:21 -0600 | commented answer | Finding the maximum of a parameter Let's say I used the variables u, v, A, from the first code-block in my question, could I take those and make them a 3-d parameter. If there's something wrong with what is on the top how can I get the 3-d parameter I need? |
2014-08-12 08:57:30 -0600 | commented question | Finding the maximum of a parameter I want too be able to know to fix the error message and make a function, out of u,v, and A. Then use the program for solving a maximum and minimum of that function. |
2014-08-11 15:30:17 -0600 | commented question | Finding the maximum of a parameter Is it possible to answer my question now? |
2014-08-09 17:22:29 -0600 | commented question | Finding the maximum of a parameter If there's something that doesn't make sense, please say so. |
2014-08-08 14:23:25 -0600 | asked a question | Finding the maximum of a parameter I was trying to use integrals, in order to find the average radius. What I did was I took a function, like $$f'(x,y)$$, and then I replaced f'(rcos(theta),rsin(theta)). Then I replaced r with y, and theta with x. This was divided into the positives above the x-axis, and the negatives below. Using computer programming I got used it all inorder to find the area under the positive curve. Where a is the start of the integral, b is the end of it, h is the intervals for the integral, s is the root to identify the region for integration, the float is the summation,A is the area, and u and v are parameters.. Note: This is an 2-D implicit function, you can't use a direct integral, but you can use the definition. I'm trying to use the co-ordinates to find the maximum of u, and v in terms of A, or the area. I would like to find the co-ordinates of the maximum that comes out of this function. You can think of it as $$x=u$$ $$y=v$$ $$z=A$$. I tried this (at the very top of this post) on sage, but it gave... |
2014-07-28 09:54:45 -0600 | commented question | Need help finding maximum values over 3-d parameters? If my question is not clear please be free to say it. |
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2014-07-27 15:24:10 -0600 | asked a question | Need help finding maximum values over 3-d parameters? If you look into my work so far I was trying to solve under a specific section of a function using the left-endpoint rule, since it can't be computed explicitly. In this case e is the change of the function by x, and f is the change by y. And z is equal to the area under an equation from $a=0$, to $b=2\pi$, where the area is positive. You can see here: https://www.desmos.com/calculator/kv4... I tried to make a 3-d parameter by making $m(x)=e$, $m(y)=f$, and $m(z)=q$, and tried to find the maximum values of e, and f. I've tried using sage's programming, but there is something wrong with what I did as seen here: https://cloud.sagemath.com/projects/1... Is there a way of finding the maximum value of e, and f values? If it is done correctly both of them should be calculated as $e=0$, and $f=0$, since this should have the maximum value of $q$. |