I'm an occasional Mathematica user and I am trying to transform an expression from spherical to Cartesian coordinates.
My function is defined as:
g[theta_, phi_] := Cos[phi](Sin[theta])^2 Sin[phi]
I'm hoping to transform that function using the following rules:
Sin[theta]Sin[phi] -> x
Cos[theta]-> y
Sin[theta]Cos[phi]-> z
in order to get the result:
zx
Here is the code I'm using to do that:
g[theta, phi] //. {Sin[theta]Sin[phi] -> x, Cos[theta] -> y, Sin[theta] Cos[phi] -> z}
And the result I get is:
Cos[phi] Sin[phi] Sin[theta]^2
So no transformation occurred.
Is there a function or an option I could add to help Mathematica figure out that the transformation is possible?
Thanks!
Perhaps this will be sufficient
Assuming[Sin[theta]Sin[phi]==x&&Cos[theta]==y&&Sin[theta]Cos[phi]==z,
Simplify[Cos[phi]Sin[theta]^2 Sin[phi]]]
which instantly returns
x z
That doesn't show you the steps or rules it used to arrive at that result, but because it considered x z to be "simpler" than your trig expression the evaluation process went in that direction.
There is a slightly more compact way of doing the same thing, if that matters. Simplify can accept a second argument which are the things which are assumed to be true during the process of simplification. Thus
Simplify[Cos[phi]Sin[theta]^2 Sin[phi],
Sin[theta]Sin[phi]==x&&Cos[theta]==y&&Sin[theta]Cos[phi]==z]
will give you exactly the same result
Related
I HAVE EQUATION
y - 7(e^x/x)dydx=0
How to find analytic solution in Mathematica?
My work :
I simplify the equations become y'=yx/7e^x
I run in Mathematica,
DSolve[y'[x] == (y[x] x)/(7 e^x), y[x], x]
I get result:
{{y[x] -> E^(1/7 E^-x (-1 - x)) C[1]}}.
Questions:
Are my simplifications correct?
Do I type correct code in step 2?
How to find exact value of C in the result, because I want to use to find y' for given x value
Thank you for the answer
I have a source file like (without loss of generality (only to image a possible syntax)):
function a()
return g // global variable without any internal structure exactly
end
function b(x, y)
local z = x * y
return z + 1
end
function c(z, t)
return b(z * z, a())
end
// ...etc
I want to defferentiate any function WRT to some variable.
All the formal parametres we can treat as a functions with unknown at derive time internal structure.
If I stand correct further, then the following is truth (for depending symbols ' is part of symbol, for global variables is operator during substitute time stage (def: g{g} is one, but g{y} is zero)):
function a'()
return g';
end
function b'(x, y, x', y')
local z' = x' * y + x * y'
return z' + 0
end
But what to do with last function? Namely, with actual parameters in substitution of function b?
Is there any ready to use implementations of general algorithm to work with the above? What to do with higher order derivatives (especially interesting, how to handle the formal parameters)? Are there any other possible unclear cases?
I would suggest having your parameters be symbolic expressions that know how to respond to derivatives, and having all operations take functions and return functions. Then you will get a final expression that knows how to be represented as a derivative. Furthermore you can do things like partial derivatives at a later point because you have the symbolic expression.
For a real example of what I mean, see http://www.elem.com/~btilly/kelly-criterion/js/advanced-math.js for a library that I wrote to solve a calculus problem in JavaScript, and search for "Optimize if requested" in the source for http://www.elem.com/~btilly/kelly-criterion/betting-returns2.html to see how I used it. See http://www.elem.com/~btilly/kelly-criterion/ for an explanation of why I was writing that code.
In that example I, of course, was not working from infix notation. But that is a standard parsing problem that I think you know how to solve.
I have a differential equation A*dx/dt + B(y-y0) = 0
Where x is a very complicated function of y.
How can I use Mathematica to rearrange y to get a function x in order to solve this?
Thanks
There are two or three different problems here that you might be asking:
Option 1: The subject line
First, if you really do have a function f[x] defined and you want to rearrange it, you would be doing something like this:
f[x_]=2+x+x^2;
Solve[y==f[x],x]
However, even here you should notice that inverse functions are not necessarily unique. There are two functions given, and the domain of each is only for y>=7/4.
Option 2: Solving a DE
Now, the equation you give is a differential equation. That is not the same as "rearranging a function y=f[x] into x=g[y]" because there are derivatives involved.
Mathematica has a built-in differential-equation solver:
DSolve[a y'[t] + b (y[t] - y0) == 0, y[t], t]
That will give you a function (in terms of constants $a,b,y_0$) that is the answer, and it will include the unspecified constant of integration.
Your system seems to refer to two functions, x(t) and y(t). You cannot solve one equation for two variables, so it is impossible to solve this (Mathematica or otherwise) without more information.
Option 3: Rearranging an expression
As a third alternative, if you are trying to rearrange this equation without solving the differential equation, you can do that:
Solve[a x'[t] + b(y[t]-y0)==0,x'[t]]
This will give you $x'(t)$ in terms of the other constants and the function $y(t)$, but in order to integrate this (i.e. to solve the differential equation) you will need to know more about y[t].
I just started to use Mathematica and wanted to try out if I got a result correct from a residue integral i made. I have two poles in the UHP, actually looking more complicated but I just wanted to see if Mathematica could to this. This is my code:
x1 = Iw;
x2 = 2 Iw;
Integrate [e^(Ixt)/((x - x1) (x - x2)), {x, -infinity, infinity}, Assumptions -> t > 0]
Is this integral possible to do in Mathematica?
Mathematica sees Iw as a two character variable name, but I w as the product of two things.
Mathematica is fanatic about correct capitalization and spelling.
x1=I w; x2=2 I w;
Integrate[E^(I x t)/((x-x1)(x-x2)), {x,-Infinity,Infinity}, Assumptions-> t>0]
returns a large complicated result depending on the signs and whether w might be zero. If you can supply additional domain information the result might be simplified.
I am trying to gain a deeper understanding of how Mathematica expressions are represented internally, and am puzzled by the logic of the Level command in Mathematica. If we have the following input:
In[1]:= a = z*Sin[x + y] + z1*Cos[x1 + y1]
Out[1]= z1 Cos[x1 + y1] + z Sin[x + y]
In[2]:= FullForm[a]
Out[2]= Plus[Times[z1,Cos[Plus[x1,y1]]],Times[z,Sin[Plus[x,y]]]]
In[3]:= TreeForm[a]
We get the following tree:
If we ask Mathematica to return Level 4 only, we get:
In[4]:= Level[a,{4}]
Out[4]= {x1,y1,x,y}
I understand that we are 4 levels down from the "stem" (the Plus operator at Level 0). In fact, I think I understand that positive indexes are always in relation to the stem position of the tree. (I hope I'm correct about that??)
In contrast, when you ask for a negative level, there is no common reference point (like the stem above), because different branches of the tree are of varying lengths. So, if you ask Mathematica to provide only Level -1, we get:
In[6]:= Level[a,{-1}]
Out[6]= {z1,x1,y1,z,x,y}
I was surprised by this output, when I had guessed that I should get back {x1, y1, x, y} (without z1 & z). But ok, if I try to understand this, I take -1 to mean "the end of each branch". If this is so, then I would expect Level[a,{-2}] to return:
{z1*Cos[x1+y1],z*Sin[x+y],x1+y1,x+y}
But, this is not what I get back, Mathematica yields:
In[8]:= Level[a,{-2}]
Out[8]= {x1+y1,x+y}
So, now I am confused, and don't see a consistent way of understanding the output of negative levels.
Is there a consistent, easier way of understanding this topic? Is there a certain "correct" way I should be reading the structure of the tree?
Sorry for the "long-winded question", but I hope you understand what I am asking.
If you look at the docs, they say:
A negative level -n consists of all parts of expr with depth n.
So negative levels are not counted from a reference point, but are defined based on the depth of subexpressions. z1*Cos[x1+y1] is of depth 4, so it's not returned when you ask for Level[..., {-2}].