how can i get properly expressed answer(only expressed with constants) in ti-nspire cas(system of linear equations) - ti-nspire

I'm struggling with ti linear equation solving. I want to solve system of equations of two variables(i,v).
following is two equations.
ai=v+iq
v=(i-1-bv)p
When I use ti-nspire cas function 'solve' answer i=,v= still contains i and v. But I want to express them in unknown constants(a,b,p,q) only. How can I do that??
I tried to express them in other ways, But it doesn't work.
And some times answer of system of equation contains expression like'x=c5'. I wonder What c5 is meaning.
thank you.

Remember to use explicit mulitplication:
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Related

How to solve this system of differential equations in matlab?

My task is to model a certain physical problem and use matlab to solve it's differential equations. I made the model but it seems far more complex than what I've learned so far so I have no idea how to solve this.
The black color means it's a constant
I assume that by "solve" you seek a closed form solution of the form x(t) = ..., z(t) = ... Unforunately, it's very likely you cannot solve this system of differential equations. Only very specific canonical systems actually have a closed-form solution, and they are the most simple (few terms and dependent variables). See Wikipedia's entry for Ordinary Differential Equations, in particular the section Summary of exact solutions.
Nevertheless, the procedure for attempting to solve with Matlab's Symbolic Math Toolbox is described here.
If instead you were asking for numerical integration, then I will give you some pointers, but you must carry out the math:
Convert the second order system to a first order system by using a substitution w(t) = dx/dt, allowing you to replace the d2x/dt2 term by dw/dt. Example here.
Read the documentation for ode15i and implement your transformed model as an implicit differential equation system.
N.B. You must supply numerical values for your constants.

Matlab cannot compute an infinite integral?

I am studying Stochastic calculus, and occasionally we need to compute an integral (from -infinity to +infinity) for some complex distribution. In this case, it was
with the answer on the right. This is the code I put into Matlab (and I have the symbolic math toolbox), which Matlab simply cannot process:
>> syms x t
>> f = exp(1+2*x)*(1/((2*pi*t)^0.5))*exp(-(x^2)/(2*t))
>> int(f,-inf,inf)
ans =
-((2^(1/2)*pi^(1/2)*exp(2*t + 1)*limit(erf((2^(1/2)*((x*1i)/t - 2i))/(2*(-1/t)^(1/2))), x, -Inf)*1i)/(2*(-1/t)^(1/2)) - (2^(1/2)*pi^(1/2)*exp(2*t + 1)*limit(erf((2^(1/2)*((x*1i)/t - 2i))/(2*(-1/t)^(1/2))), x, Inf)*1i)/(2*(-1/t)^(1/2)))/(2*pi*t)^(1/2)
This answer at the end looks like nonsense, while Wolfram (via their free tool), gives me the answer that the picture above has. Am I missing something fundamental about doing such integrations in Matlab that the basic Mathworks pages don't cover? Where am I proceeding incorrectly?
In order to explain what is happening, we need some theory:
Symbolic systems such as Matlab or Mathematica calculate integrals symbolically by the Risch algorithm (yes, there is a method to mechanically calculate integrals, just like derivatives).
However, the Risch algorithms works differently than using derivation rules. Strictly spoken, it is not an algorithm but a semi-algorithm. This is, it is not deterministic one (as algorithms are).
This (semi) algorithm makes a series of transformations on the input expression (the one to be integrated), and in a specific point, it requires to ask if the transformed expression is equals to zero, because if it were zero, it cannot continue (the input is not integrable using a finite set of terms).
The problem (and the reason of the "semi-algoritmicity") is that, the (apparently simple) equation:
E = 0
Is indecidable (it is also called the constant problem). It means that there cannot exist a formal method to solve the constant problem, for any expression E. Of course, we know to solve the constant problem for specific forms of the expression E (i.e. polynomials), but it is impossible to solve the problem for the general case.
It also means that the Risch algorithm cannot be perfect (being able to solve any integral -integrable in finite terms-). In other words, the Risch algorithm will be as powerful as our ability to solve the constant problem for as many forms of the expression E as we can, but losing any hope of solving for the general case.
Different symbolic systems have similar, but different methods to try to solve equations (and therefore the constant problem), it explains why some of them can "solve" different sets of integrals than others.
And generalizing, because no symbolic system will never be able to solve the constant problem for the general case, it will neither be able to solve any integral (integrable in finite terms).
The second parameter of int() needs to be the variable you're integrating over (which looks like t in this case):
syms x t
f = exp(1+2*x)*(1/((2*pi*t)^0.5))*exp(-(x^2)/(2*t))
int(f,'t',-inf,inf) % <- integrate over t

Explanation of two integral equations and implementation

I have a problem with these two equations showing in the pictures.
I have two vectors represented the C(m) and S(m) in the two equations. I am trying to implement these equations in Matlab. Instead of doing a continuous integral operation, I think I should do the summation. For example, the first equation
A1 = sqrt(sum(C.^2));
Am I right? Also, I am not sure how to implement equation two that contains a ||dM||. Please help.
What are the mathematical meaning of these two equations? I think the first one may related to the 'sum of squares', if C(m) is a vector then this equation will measure the total variance of the random variable in vector C or some kind of average of vector C? What about the second one?
Thanks very much for your help!
A.
In MATLAB there are typically two different ways to do an integration.
For people who have access to the symbolic toolbox, algebraic integration is an option. If this is the case for you, I would look into help int and see which inputs you need.
For the rest, numerical integration is available, this basically means that you just calculate a function on a lot of points and then take the mean of the function values in these points.
For the mathematical meaning some more context would be helpful, and you may want to ask this question at math.stackexchange.com or on the site of whatever field you are in. (stats, physics?)

Find complex roots from a nonlinear equation

I need to find the roots from the equations as follows (Mathematica):
Sqrt[3]/2*x-(I-x*Sqrt[3]/2*c^2)*I/Sqrt[2*Pi]/d^3*Integrate[t*Exp[-t^2/2/d^2]/(Sqrt[3]/2*x+I*(t+b0)),{t,-Inf,Inf}]=0
i.e. as the picture shows:
where c, d, and b0 is given parameters, x is a complex root needs to find.
I have tried several methods, including scanning the real and imagine part of x and the iteration approach, but non of them could resolve all the cases.
Are there any general approaches that could solve such kind of equation efficiently, or by MATLAB/Mathematica?
did you try Matlab's mupad? it is a powerful symbolic tool, very similar to Maple wich gives really good results in non-numerical mathematics. Try there. declare the equation, give assumptions to the software ,i.e assume c real positive (don't copy this, I dont remember the proper syntax) and then solve! It will very likely find a solution if it exits (sometimes in some mathematical cases that you even don't know!)

How can I numerically solve equations containing Bessel functions in MATLAB?

I have confronted an equation containing Bessel functions of the first type on one side and modified Bessel functions of the second type on the other. I want to know its exact solutions (values of u). The equation is as follows:
u*besselj(s-1,u)/besselj(s,u)=-w*besselK(s-1,w)/besselk(s,w)
where s is an arbitrary integer number, for example 2.
w can be written as a function of u:
w=sqrt(1-u^2);
and so this equation has only one variable: u
I'm new to MATLAB. I have no idea about how I should approach this. Could anyone please help me?
A quick thing to try may be the FZERO function, a generic nonlinear zero finder. To learn how to use it, you can implement the examples given in the documentation. Then, rewrite your function so it can be input to fzero and see what you get..
(Note: I haven't tried this, but I just noticed there were no replies yet so maybe it's better than nothing.)