I want to use the linear version of a somewhat big equation which is outputted by my Mathematica code - For simplicity I will here use the example equation:
Test = 3 x + x y + 8 y
Now, I want to use only the first order term, so that for x that will be 3 and for y that will be 8. I have tried to get those using
Coefficient[Test, x]
Coefficient[Test, y]
However these give me for x:
3+y
and for y:
8+x
Is there anyway I can use Mathematica to ignore the terms that depend in both xy simultaneously?
Related
Say that x and y are real numbers and y > 0. And say that I want to find for which values of A do (A + x + y > 0) and (A + x - y > 0) always hold, as long as x, y are in the domain.
How would I specify that on Wolfram Alpha? (Note: obviously these equations have no solution, but I just used it as an example.)
Or, if not on Wolfram, what software/website could I use?
I tried to write: solve for A: [input my first equation], y>0
but that didn't work, as it only gave integer solutions for when A, x, and y vary, instead of finding values of A such that it always holds no matter what x, y are.
https://www.wolframalpha.com/input?i=%28A+%2B+x+%2B+y+%3E+0%29+and+%28A+%2B+x+-+y+%3E+0%29+
[x>-A, -A - x<y<A + x]
I am am mathmatica notebook to find an analytical solution to the follow constrained optimization problem:
Max y^(1-b)(x^b(1-a(x/(x+1)))) s.t. M = Px+qy
x,y
I have tried the following code:
Maximize[{y^(1-b)(x^b(1-a(x/(x+1)))), M==Px+qy}, {x,y}]
and in returns the same function as an output. In the function a, b, M, P, and q are all parameters. I have also tried assigning the parameters arbitrary values to test to see if mathmatica is not sure how to deal with the parameters. I used to following code:
Maximize[{y^(1-0.5)(x^0.5(1-0.75(x/(x+1)))), 1000=5x+5y},{x,y}]
and it returns the same function. However, if I remove the constraint it will solve the optimization problem.
Maximize[{y^(1-0.5)(x^0.5(1-0.75(x/(x+1))))},{x,y}]
{7.2912*^59,{x->2.89727*^60,y->2.93582*^60}}
I am not sure what to do. After reading about constrained optimization problem the syntax appears to be correct. Sorry, it this question is really basic I am very new to mathmatica, also since I am using a notebook I could not past the output from the first two lines in.
The constraint is incorrectly specified, it should be 1000 == 5 x + 5 y. Maximize works better with exact numbers.
Maximize[{Rationalize[y^(1 - 0.5) (x^0.5 (1 - 0.75 (x/(x + 1))))],
1000 == 5 x + 5 y}, {x, y}] // N
(* {25.7537, {x -> 96.97, y -> 103.03}} *)
I'm to ask a question, which answers are solving this task:
Which right-angled triangles can be constructed by choosing three sides out of six segments of length being integers from 1 to 6
So, I'm thinking this is essential:
between(1,6,X),
between(1,6,Y),
between(1,6,Z),
Then we have to make sure it fits Pythagoras statement, so I'm trying this, adding to the above sentence:
(X^2 = Y^2 + Z^2 ;
Y^2 = X^2 + Z^2 ;
Z^2 = X^2 + Y^2)
Also I have been trying to replace X^2 with X*X, but it returns false every time. Why is that?
From my understanding, I need it to work like this:
Choose three sides from range 1-6, and make sure they fit Pythagoras statement. (Is triangle disparity also required here? I mean X>Y+Z,Y>X+Z,Z>X+Y ?
Check the prolog manual regarding the different comparators, etc. They mean and do various things. =:=/2 is specifically evaluates arithmetic expressions on either side and checks for equality of results. =/2 is not an equality operator; it performs prolog unification. It's important to know the difference. In your example, limiting all results to maximum of 6, then permutations of 3,4,5 are the only positive integer solutions to the right triangle.
?- between(1,6,X), between(1,6,Y), between(1,6,Z), Z^2 =:= X^2 + Y^2.
X = 3,
Y = 4,
Z = 5 ;
X = 4,
Y = 3,
Z = 5 ;
false.
This is an algorithm question that I've been struggling with. I figured I could get some insight here. I need to make the following function in Haskell:
Declare the type and define a function that takes two numbers as input and finds their product by addition. That is, add the first number, as many times as second number, to itself.
My problem is that this is basically just multiplying two numbers together, but it says that I need to do it with addition. Does anyone have any clue on how to do this?
This is all I can come up with (it's not right): (x + x) * y
Thank you
if a is the first number and b the second
sum $ take a $ cycle [b]
should do ot
mult (x, y):
sum = 0
for 1 to y:
sum = sum + x
return sum
This is just the algorithm. I do not know Haskell. So the lambda expression in the other answer may be more appropriate. Also, I use an intermediate variable.
PS: forget the previous embarrassing recursive algorithm
Work it out by induction.
We know the answer to one simple (the simplest) problem: multiplying anything by 0 yields 0. So we write:
mul x 0 = 0
Now, the inductive step: we can build a solution to a bigger problem, if we know a solution to the smaller problem; that way we can always reduce any big problem to the smallest problem, for which we know the solution. So, for any y, the solution for y+1 can be found by adding x to the solution for y: mul x (y+1) = x + (mul x y). In Haskell we can't write (y+1) on the left hand side, so we write equivalently:
mul x y = x + (mul x (y-1))
This function will keep adding x until y is zero.
Try this also
multiply::(Num a,Eq a) => a -> a -> a
multiply a 0 = 0
multiply a b = a + multiply a (b - 1)
main = print $ multiply 5 7
I'm trying to obtain the real part of the result of an operation which involves an undefined variable (let's say x).
How can I have Mathematica return x when I execute Re[x] if I know that x will never be a complex number? I think this involves telling Mathematica that x is a real, but I don't know how.
In my case the expression for which I want the real part is more complicated than a simple variable, but the concept will remain the same.
Some examples:
INPUT OUTPUT DESIRED RESULT
----- ------ --------------
Re[x] Re[x] x
Re[1] 1 1
Re[Sin[x]] Re[Sin[x]] Sin[x]
Re[1+x+I] 1 + Re[x] 1+x
Re[1 + x*I] 1-Im[x] 1
You can use for example the input Simplify[Re[x], x \[Element] Reals] which will give x as output.
Use ComplexExpand. It assumes that the variables are real unless you indicate otherwise. For example:
In[76]:= ComplexExpand[Re[x]]
Out[76]= x
In[77]:= ComplexExpand[Re[Sin[x]]]
Out[77]= Sin[x]
In[78]:= ComplexExpand[Re[1+x+I]]
Out[78]= 1+x
Two more possibilities:
Assuming[x \[Element] Reals, Refine[Re[x]]]
Refine[Re[x], x \[Element] Reals]
Both return x.
It can at times be useful to define UpValues for a symbol. This is far from robust, but it nevertheless can handle a number of cases.
Re[x] ^= x;
Im[x] ^= 0;
Re[x]
Re[1]
Re[1 + x + I]
Re[1 + x*I]
x
1
1 + x
1
Re[Sin[x]] does not evaluate as you desire, but one of the transformations used by FullSimplify does place it in a form that triggers Re[x]:
Re[Sin[x]] // FullSimplify
Sin[x]