## How solve nonlinear equations by 14 unknowns?

This is my code, i want to solve system by 14 equations and 14 unknowns(rr list).But the code doesn’t run…

Clear["Global*"] T[0, t_] = 1; T[1, t_] = t; T[n_, t_] := 2 *t*T[n - 1, t] - T[n - 2, t]; For[n = 0, n <= 7, n++, Print["T[", n, ",t]= ", T[n, t], "\n"]]; tableoft = tt /.NSolve[T[7, tt], tt]; tableoft[]; Subscript[z, 1][t_] = Sum[Simplify[Subscript[a, j]*T[j, t]], {j, 0, 6}]; Subscript[z, 2][t_] = Sum[Simplify[Subscript[b, l]*T[l, t]], {l, 0, 6}]; f[t_] = (t^4/6 - t^3/3 + t) /. t -> 1/2*(\[Tau] + 1); Subscript[k, 1][t_, s_] = s^3 /. s -> 1/4*(\[Tau] + 1)*(r + 1); Subscript[k, 1][t, s]*Subscript[z, 1][r]; p[\[Tau]_] =   Integrate[Subscript[k, 1][t, s]*Subscript[z, 1][r], {r, -1, 1}] Subscript[k, 2][t_, s_] = -2 (t - s) /. t -> 1/2*(\[Tau] + 1) /.    s -> 1/4*(\[Tau] + 1)*(r + 1); Subscript[k, 2][t, s]*Subscript[z, 2][r];     pp[\[Tau]_] =   Integrate[Subscript[k, 2][t, s]*Subscript[z, 2][r], {r, -1, 1}]; Subscript[g, 2][\[Tau]_] =   Expand[(f[t] + p[\[Tau]] + pp[\[Tau]])^2] // N; Subscript[\[Delta],    1][\[Tau]_] = ((Subscript[z,      1][\[Tau]])*((f[t] + p[\[Tau]] + pp[\[Tau]])) - 1) // N; Subscript[\[Delta],    2][\[Tau]_] = (Subscript[z, 2][\[Tau]] -Subscript[g, 2][\[Tau]]) // N;  Table[N[Subscript[\[Delta], 1][tableoft[[i]]]], {i, 1, 7}];  Table[N[Subscript[\[Delta], 2][tableoft[[i]]]], {i, 1, 7}];  r = Flatten[Table[N[{Subscript[\[Delta], 1][tableoft[[i]]],   Subscript[\[Delta], 2][tableoft[[i]]]}], {i, 1, 7}]];  rr =Table[Simplify[r[[i]]] == 0, {i, 1, 14}]  list =Flatten[Table[{Subscript[a, i], Subscript[b, i]}, {i, 0, 6}]];  Solve[rr, list] 

How to get solution fast?

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## What is an efficient way of planning solo-trips with so many unknowns?

I love traveling, and have always traveled since I was a kid. Now that I have more freedom of movement, I’d like to take the chance to see the rest of the world. While living in Europe, it’s easy and accessible to travel within the continent over long weekends or for short breaks, however I find the process of planning the trip very tedious and exhausting or even impossible(!) to the point that I end up going nowhere.

For instance I would like to go skiing, but I haven’t skied much in the Alps. Last time (which was my first time skiing in the Alps), I was fairly lucky to have a friend’s friend’s friend to help me get to the mountains, etc. but that’s not always the case and anyhow it takes so much time to research, read online, post questions, and eventually choose from the so many options. This time I tried narrowing down my options by asking a friend who’s a ski instructor but I only got a couple of one-word answers and and in the end also I noticed the transport is quite impossible (well, possible at the cost of 600-1000 euros from the airport to the resort!). My usual process which I don’t think is particularly efficient or useful is:

1. Think and think and think about that upcoming long holiday that will happen some time in the future
2. Getting closer to the date start stressing that I have to plan or it’ll be too late
3. Start looking at the options and panic about all those unfamiliar names and hundreds of options that I can choose
4. Try and ask someone with some local knowledge where they’d also make a list of names (still many to choose from) without much info or say: “they’re all good” or something like that
5. Eventually while the panic monster is running after me, I force myself to sit down and choose and book a flight while taking the options of accommodation into consideration
6. Once the flights are booked, including making the decisions on how many days to stays, I start planning the things I would like to do during the trip and the accommodation
7. In this last instance (which I still would like to happen this coming weekend!), I was about to purchase the tickets when I realized that I need to figure out my transport from the airport to the resort as well, and then it became clear that I don’t have many choices to use public transport, and I also don’t have any choice over the flight times as they happen only on certain hours to certain destination
8. Call it a failure at making the plans and start over from step 1!

I’ve got to the point that I believe my assessment/understanding of how long things can take or how I can/should plan things do not comply with the realtime world. To get away for 3-4 days, it seems like I have to spend a similar amount of time to research and plan. Of course this would have been a lot easier if I were traveling in groups, but that’s not much of an option at the moment, so I have to plan and do everything on my own.

So, what is the average/usual thought process of someone who wants to plan a trip? What is the average time one would/should spend on planning a trip and any other tips to make this process more efficient that you would suggest? I would also like to know how you initially pick a destination and a period of time in the future to plan things far ahead? (there’s probably an element of fear of uncertainties and commitment that probably becomes bolder by age!)

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## Solving a system of 3 linear inequalities with 3 unknowns

find lowest possible x, y and z whole number variables where:  x  =< 2y+2z  6y =< x+z    3z =< x+y 

I am trying to solve this system of 3 linear equations with 3 unknowns and get the ratio that x, y and z have to be in for the inequalities to work. So far I have been able to get a working ratio

Q1 : but I am trying to get a ratio that satisfies the equations below.

x  = 2y+2z  6y = x+z    3z = x+y 

Q1, a : And if no such ratio exists how can I mathematically determine that?

The formular I used to get a working ratio for the inequality is by solving the equation. I made the equations all equal to each other as follows:

0.5x + x  =  x + y + z    6y + y  =  x + y + z      3z + z  =  x + y + z 

This way I was able to get the equation below.

1.5x = 7y = 4z   `

The equation above satisfies the first inequality but not the corresponding equation.

PS: I am not very conversant with the tags if I have left any out please add. Thank you.

## Is the solution inconsistent or there is a way to find the unknowns?

The system of homogenous linear equations is 4x+4y+2z=0; 4x+4y+2z=0; 2x+2y+z=0

The row echelon form for this is |1 1 1/2| |0 0 0 | |0 0 0 |

Is the solution possible or not, and how?

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## Solving underdetermined nonlinear system of 2 equation with 3 unknowns.

I’ve gotten into a problem I haven’t really worked with before in my numerics classes.

I have an underdetermined nonlinear system of equations with 3 parameters.

Newtons method, Boydens method etc. all include the inverse of the jacobian, but if the system is underdetermined this is not defined as far as I understand as: \begin{align} \begin{cases} A=\cos(\alpha)e^{i\phi}\ B=\sin(\alpha)e^{i\chi} \end{cases} \end{align} where $$A$$ and $$B$$ are known parameters.

Is there any straightforward way or trick to solve this kind of problems?

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## Six coupled linear PDE for three unknowns

Let $$x \in (0,L)$$, $$t \in (0,T)$$, and let $$u_0 = u_0(x) \in \mathbb{R}^3$$, $$g=g(t) \in \mathbb{R}^3$$, $$P = P(x,t) \in \mathbb{R}^3$$ and $$Q = Q(x,t) \in \mathbb{R}^3$$ be continuously differentiable functions.

Denote by $$\hat{P}, \hat{Q}$$ the matrices \begin{align*} \hat{P} = \begin{bmatrix} 0 & -P_3 & P_2 \ Q_3 & 0 & -P_1 \ -P_2 & P_1 & 0 \end{bmatrix}, \qquad \hat{Q} = \begin{bmatrix} 0 & -Q_3 & Q_2 \ Q_3 & 0 & -Q_1 \ -Q_2 & Q_1 & 0 \end{bmatrix}. \end{align*}

My question is:

Assuming that the following four conditions hold:

Condition 1: \begin{align*} \partial_x P – \partial_t Q = \hat{P}Q \qquad \text{in }(0,L) \times (0,T), \end{align*} Condition 2: \begin{align*} u_0(0) = g(0), \end{align*} Condition 3: \begin{align*} u_0′ = – \hat{Q}u_0 \qquad \text{for } x \in (0,L), \end{align*} Condition 4: \begin{align*} g’ = – \hat{P}g \qquad \text{for } t \in (0,T), \end{align*} is there a solution $$u = u(x,t) \in \mathbb{R}^3$$ to \begin{align*} \begin{cases} \partial_t u = -\hat{P}u &\text{in }(0,L)\times(0,T) \ \partial_x u = – \hat{Q}u &\text{in }(0,L)\times(0,T) \ u(x,0) = u_0(x) & \text{for } x \in (0,L)\ u(0,t) = g(t) & \text{for }t \in (0,T)? \end{cases} \end{align*}

In the above, $$f’$$, $$\partial_t f$$a and $$\partial_x f$$ denote the derivative, partial time derivative and partial space derivative respectively; and $$P_k$$, $$Q_k$$, $$u_k$$, $$u_{0k}$$ and $$g_k$$ denote the components of $$P$$, $$Q$$, $$u$$, $$u_0$$ and $$g$$ respectively.

My question with all terms developed rewrites as follows. Assuming that the following four conditions hold:

Condition 1: \begin{align*} \begin{cases} \partial_x P_1 – \partial_t Q_1 = P_2 Q_3 – P_3 Q_2\ \partial_x P_2 – \partial_t Q_2 = P_3 Q_1 – P_1 Q_3\ \partial_x P_3 – \partial_t Q_3 = P_1 Q_2 – P_2 Q_1 \end{cases} \qquad \text{in }(0,L) \times (0,T), \end{align*} Condition 2: \begin{align*} u_0(0) = g(0), \end{align*} Condition 3: \begin{align*} \begin{cases} u_{01}’ = \ \ Q_3(\cdot,0) u_{02} – Q_2(\cdot,0) u_{03} \ u_{02}’ = -Q_3(\cdot,0) u_{01} + Q_1(\cdot,0) u_{03} \ u_{03}’ = \ \ Q_2(\cdot,0) u_{01} – Q_1(\cdot,0) u_{02} \end{cases} \qquad \text{for } x \in (0,L), \end{align*} Condition 4: \begin{align*} \begin{cases} g_1′ &= \ \ P_3(0,\cdot) g_2 – P_2(0,\cdot) g_3 \ g_2′ &= -P_3(0,\cdot) g_1 + P_1(0,\cdot) g_3 \ g_3′ &= \ \ P_2(0,\cdot) g_1 – P_1(0,\cdot) g_2 \end{cases} \qquad \text{for } t \in (0,T), \end{align*} is there a solution $$u = u(x,t) \in \mathbb{R}^3$$ to the problem: \begin{align*} \begin{cases} \partial_t u_1 = \ \ P_3 u_2 – P_2 u_3 &\text{in }(0,L)\times(0,T) \ \partial_t u_2 = -P_3 u_1 + P_1 u_3 &\text{in }(0,L)\times(0,T) \ \partial_t u_3 = \ \ P_2 u_1 – P_1 u_2 & \text{in }(0,L)\times(0,T) \ \partial_x u_1 = \ \ Q_3 u_2 – Q_2 u_3 & \text{in }(0,L)\times(0,T) \ \partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &\text{in }(0,L)\times(0,T) \ \partial_x u_3 = \ \ Q_2 u_1 – Q_1 u_2 & \text{in }(0,L)\times(0,T) \ u(x,0) = u_0(x) & \text{for } x \in (0,L)\ u(0,t) = g(t) & \text{for }t \in (0,T). \end{cases} \end{align*}

What I started. I started reasoning the following way. For $$x \in (0,L)$$ fixed, a function of the form \begin{align*} u_1(x,t) &= u_{01}(x) + \int_0^t P_3(x,s)u_2(x,s) – P_2(x,s)u_3(x,s)ds\ u_2(x,t) &= u_{02}(x) + \int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\ u_3(x,t) &= u_{03}(x) + \int_0^t P_2(x,s)u_1(x,s) – P_1(x,s)u_2(x,s)ds, \end{align*} satisfies the initial value problem involving time derivatives, with $$u(\cdot, 0) = u_0$$, while for $$t \in (0,T)$$ fixed, a solution of the form \begin{align*} \begin{aligned} u_1(x,t) &= g_1(t) + \int_0^x Q_3(\xi,t)u_2(\xi,t) – Q_2(\xi, t)u_3(\xi, t)d\xi \ u_2(x,t) &= g_2(t) + \int_0^x -Q_3(\xi, t)u_1(\xi, t) + Q_1(\xi, t) u_3(\xi, t)d\xi\ u_3(x,t) &= g_3(t) + \int_0^x Q_2(\xi, t)u_1(\xi, t) – Q_1(\xi, t)u_2(\xi, t)d\xi. \end{aligned} \end{align*} satisfies the initial value problem involving space derivatives, with $$u(0, \cdot) = g$$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as \begin{align*} u_1(x,t) &= u_{01}(0) + \int_0^x Q_3(\xi, 0)u_{02}(\xi) – Q_2(\xi, 0)u_{03}(\xi)d\xi \ &\quad + \int_0^t P_3(0,s)g_2(s) – P_2(0,s)g_3(s)ds + \int_0^t \int_0^x \partial_x (P_3 u_2 – P_2 u_3 )(\xi, s)d\xi ds\ u_2(x,t) &= u_{02}(0) + \int_0^x -Q_3(\xi, 0)u_{01} + Q_1(\xi, 0)u_{03}(\xi)d\xi \ &\quad + \int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + \int_0^t \int_0^x \partial_x(-P_3 u_1 + P_1 u_3)(\xi, s)ds\ u_3(x,t) &= u_{03}(0) + \int_0^x Q_2(\xi, 0)u_{01}(\xi) – Q_1(\xi, 0)u_{02}(\xi)\ &\quad + \int_0^t P_2(0,s)g_1(s) – P_1(0,s)g_2(s)ds + \int_0^t \int_0^x \partial_x (P_2 u_1 – P_1 u_2)(\xi,s)d\xi ds, \end{align*} and \begin{align*} u_1(x,t) &= g_1(0) + \int_0^t P_3(0,s)g_2(s) – P_2(0,s)g_3(s) ds \ &\quad + \int_0^x Q_3(\xi,0)u_{02}(\xi) – Q_2(\xi, 0)u_{03}(\xi)d\xi + \int_0^x \int_0^t \partial_t ( Q_3 u_2 – Q_2 u_3)(\xi, s)dsd\xi\ u_2(x,t) &= g_2(0) + \int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\ &\quad + \int_0^x -Q_3(\xi, 0)u_{01}(\xi) + Q_1(\xi, 0) u_{03}(\xi)d\xi + \int_0^x \int_0^t \partial_t (-Q_3 u_1 + Q_1 u_3)(\xi,s)dsd\xi\ u_3(x,t) &= g_3(0) + \int_0^t P_2(0,s)g_1(s) – P_1(0,s)g_2(s) ds \ &\quad + \int_0^x Q_2(\xi, 0)u_{01}(\xi) – Q_1(\xi, 0)u_{02}(\xi)d\xi + \int_0^x \int_0^t \partial_t ( Q_2 u_1 – Q_1 u_2)(\xi, s)ds d\xi \end{align*} respectively. It seems that both expressions would coincide if the following equalities hold: \begin{align*} \partial_x (P_3 u_2 – P_2 u_3 ) &= \partial_t ( Q_3 u_2 – Q_2 u_3)\ \partial_x(-P_3 u_1 + P_1 u_3) &= \partial_t (-Q_3 u_1 + Q_1 u_3)\ \partial_x (P_2 u_1 – P_1 u_2) &= \partial_t ( Q_2 u_1 – Q_1 u_2). \end{align*} To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.

Any suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.

(This question is also on Mathematics Stack Exchange: https://math.stackexchange.com/questions/3160600/six-coupled-linear-pde-for-3-unknowns)

## System of three non linear equations with three unknowns with random coefficients

Ax + By + Cz = D

Exy + Fxz + Gyz = H

Ixyz = J

A,B,C,D,E,F,G,H,I,J can have random integer values from 1 to 9.

x,y,z are the three variables that have to be functions of the letters above, and the system has to meet equality in all cases.