Let's assume a is a constant and x is my variable with respect to time, so basically x(t).
Then in Maxima , what is the best way to replace 'diff(a*x,t) with a*'diff(x,t) automatically without use subst command.
The reason I don't to use subst is that I have many variables and higher order derivatives. It is not efficient to use subst to replace all the occurrences.
Thanks.
UPDATE
I have tried with depends(x,t) command, but it only works with the simple case. Here is an minimal example of my situation.
depends([x,y],t);
eq1:diff(x,t)-b=c;
eq2:subst([x=a*y],eq1);
sol_dy=solve(eq2,diff(y,t))
Of course here a,b,c are constants and x, y are variables on t.
Maxima can not solve diff(y,t) directly. How do deal with it?
I see that 'diff(...) (i.e. derivative noun expression) isn't linear (doesn't distribute over + and doesn't factor out constants) but diff(...) (verb expression) is linear. That's a misfeature, at least.
I was going to suggest declare(nounify(diff), linear) but that makes derivatives come out as 0 in your example ... this is probably a bug, I'll have to think more about it.
Try ev(eq2, nouns); to re-evaluate the derivatives as verbs -- I think that should cause the constant to factor out.
Related
I'm wondering if it's possible to define a natural variable n in TI-Nspire CAS. For example I'd like to write:
You can't define your own natural variables. However, Nspire has the following special variables you can use:
#n0...#n255: Restricted to natural numbers
#c0...#c255: Restricted to real numbers
You can replace the original variables with them by hand or for convience just put |x=#n0 and y=#n1 at the end of line.
Example: You are calculating fourier coefficients and know that variable k will only get real numbers from Σ operation. Replacing k with #n1 will simpilfy the function.
Picture
(Calculator needs to be in RAD mode if you want to try)
The answer is no. Variables in NSpire store a value. A variable has no type. Solve might return #n1 in a result to indicate an arbitrary natural number, but you can tell solve to look for integer solutions only.
I have a lengthy symbolic expression that involves rational polynomials (basic arithmetic and integer powers). I'd like to simplify it into a single (simple) rational polynomial.
numden does it, but it seems to use some expensive optimization, which probably addresses a more general case. When tried on my example below, it crashed after a few hours--out of memory (32GB).
I believe something more efficient is possible even if I don't have a cpp access to matlab functionality (e.g. children).
Motivation: I have an objective function that involves polynomials. I manually derived it, and I'd like to verify and compare the derivatives: I subtract the two expressions, and the result should vanish.
Currently, my interest in this is academic since practically, I simply substitute some random expression, get zero, and it's enough for me.
I'll try to find the time to play with this as some point, and I'll update here about it, but I posted in case someone finds it interesting and would like to give it a try before that.
To run my function:
x = sym('x', [1 32], 'real')
e = func(x)
The function (and believe it or not, this is just the Jacobian, and I also have the Hessian) can't be pasted here since the text limit is 30K:
https://drive.google.com/open?id=1imOAa4VS87WDkOwAK0NoFCJPTK_2QIRj
I am trying to compare two simple expressions using Matlab symbolic toolbox. For some reason, the code returns 0. Any idea ?
syms a b c
A = (a/b)^c
B = a^c/b^c
isequal(A,B)
It seems like MATLAB has a hard time telling that two expressions are the same when (potentially) fractional exponents are involved.
So, one solution, as suggested by Mikhail is to restrict the values of c to be only integers although, as discussed in the Math.SE question jodag posted, there is nothing wrong with fractional exponents in this case.
Hence, since this restriction to integers is not necessary for the statement to be true, another solution is to use simplify function on the expression for B but allowing it to run more simplification steps in order to get the most simplified expression.
syms a b c
A = (a/b)^c
B = a^c/b^c
isequal(A,simplify(B,'step',4))
Four steps is actually the smallest number that worked for me, but that could vary across versions of MATLAB I'm assuming. To be sure, I would include more, but for really large expressions, this could become computationally intensive, so some judgment is necessary. Note that, you could also use the 'Seconds' option to limit the amount of time allowed for simplification.
In general what you wrote isn't true, under the right "assumptions" it becomes true: for example, assuming c is an integer you can trick MATLAB into expanding A
clc; clear all;
syms a
syms b
syms c integer
A = (a/b)^c;
B = simplify((a^c)/(b^c));
disp(isequal(A,B));
disp(A);
disp(B);
1
a^c/b^c
a^c/b^c
I want to edit this to get numberOfCircuits on its own on the left. Is there a possible way to do this in MATLAB?
e1=power(offeredTraffic,numberOfCircuits)/factorial(numberOfCircuits)/sum
The math for this problem is given in https://math.stackexchange.com/questions/61755/is-there-a-way-to-solve-for-an-unknown-in-a-factorial, but it's unclear how to do this with Matlab's functionality.
I'm guessing the easy part is rearranging:
fact_to_invert = power(offeredTraffic,numberOfCircuits)/sum/e1;
Inverting can be done, for instance, by using fzero. First define a continuous factorial based on the gamma function:
fact = #(n) gamma(n+1);
Then use fzero to invert it numerically:
numberOfCircuits_from_inverse = fzero(#(x) fact(x)-fact_to_invert,1);
Of course you should round the result for safe measure, and if it's not an integer then something's wrong.
Note: it's very bad practice (and brings 7 years bad luck) to name a variable with a name which is also a built-in, such as sum in your example.
There is one thing I do not like on Matlab: It tries sometimes to be too smart. For instance, if I have a negative square root like
a = -1; sqrt(a)
Matlab does not throw an error but switches silently to complex numbers. The same happens for negative logarithms. This can lead to hard to find errors in a more complicated algorithm.
A similar problem is that Matlab "solves" silently non quadratic linear systems like in the following example:
A=eye(3,2); b=ones(3,1); x = A \ b
Obviously x does not satisfy A*x==b (It solves a least square problem instead).
Is there any possibility to turn that "features" off, or at least let Matlab print a warning message in this cases? That would really helps a lot in many situations.
I don't think there is anything like "being smart" in your examples. The square root of a negative number is complex. Similarly, the left-division operator is defined in Matlab as calculating the pseudoinverse for non-square inputs.
If you have an application that should not return complex numbers (beware of floating point errors!), then you can use isreal to test for that. If you do not want the left division operator to calculate the pseudoinverse, test for whether A is square.
Alternatively, if for some reason you are really unable to do input validation, you can overload both sqrt and \ to only work on positive numbers, and to not calculate the pseudoinverse.
You need to understand all of the implications of what you're writing and make sure that you use the right functions if you're going to guarantee good code. For example:
For the first case, use realsqrt instead
For the second case, use inv(A) * b instead
Or alternatively, include the appropriate checks before/after you call the built-in functions. If you need to do this every time, then you can always write your own functions.