Checking if two expressions are relatedChecking if two trigonometric expressions are equalWhy is ToRadicals[] not able to handle all cases? Is there a workaround?Can't solve equation having complex coefficientsCan Mathematica reliably determine the equality of any two constructible numbers?Complicated implicit plottingsymbolically solving two equations simultaneously involving trigonometric function of three variablesFinding an inverse functionSolving an equation on a particular domainHelp Simplifying ExpressionsCan Mathematica factor out constrained functions when it simplifies?
Is there precedent or are there procedures for a US president refusing to concede to an electoral defeat?
Where to draw the line between quantum mechanics theory and its interpretation(s)?
Correct way of drawing empty, half-filled and fully filled circles?
Why aren't nationalizations in Russia described as socialist?
Is the book wrong about the Nyquist Sampling Criterion?
Is there an age requirement to play in Adventurers League?
Where are the "shires" in the UK?
Can there be a single technologically advanced nation, in a continent full of non-technologically advanced nations?
Formatting Datetime.now()
Will 700 more planes a day fly because of the Heathrow expansion?
Is it normal for gliders not to have attitude indicators?
Why did the Apollo 13 crew extend the LM landing gear?
Why is my arithmetic with a long long int behaving this way?
Notation: What does the tilde bellow of the Expectation mean?
Should homeowners insurance cover the cost of the home?
Is there a word that describes the unjustified use of a more complex word?
Hostile Divisor Numbers
Is an HNN extension of a virtually torsion-free group virtually torsion-free?
Checking if two expressions are related
Does "Captain Marvel" contain spoilers for "Avengers: Infinity War"?
Why do these characters still seem to be the same age after the events of Endgame?
Can my 2 children, aged 10 and 12, who are US citizens, travel to the USA on expired American passports?
It is as simple as ABC
Which sphere is fastest?
Checking if two expressions are related
Checking if two trigonometric expressions are equalWhy is ToRadicals[] not able to handle all cases? Is there a workaround?Can't solve equation having complex coefficientsCan Mathematica reliably determine the equality of any two constructible numbers?Complicated implicit plottingsymbolically solving two equations simultaneously involving trigonometric function of three variablesFinding an inverse functionSolving an equation on a particular domainHelp Simplifying ExpressionsCan Mathematica factor out constrained functions when it simplifies?
$begingroup$
Is it possible in mathematica to check if two expressions may be able to be written in terms of the other? The expressions I had in mind were of the form $$alpha = frac1-x1-y,,,,,,, beta = frac1+x1+y$$ and was wondering if it's possible to find a relation $$alpha = f(beta)?$$
equation-solving simplifying-expressions expression-manipulation
asked 2 hours ago
CAFCAF
270110
$endgroup$
add a comment |
$begingroup$
Is it possible in mathematica to check if two expressions may be able to be written in terms of the other? The expressions I had in mind were of the form $$alpha = frac1-x1-y,,,,,,, beta = frac1+x1+y$$ and was wondering if it's possible to find a relation $$alpha = f(beta)?$$
equation-solving simplifying-expressions expression-manipulation
asked 2 hours ago
CAFCAF
270110
$endgroup$
add a comment |
$begingroup$
Is it possible in mathematica to check if two expressions may be able to be written in terms of the other? The expressions I had in mind were of the form $$alpha = frac1-x1-y,,,,,,, beta = frac1+x1+y$$ and was wondering if it's possible to find a relation $$alpha = f(beta)?$$
equation-solving simplifying-expressions expression-manipulation
asked 2 hours ago
CAFCAF
270110
$endgroup$
Is it possible in mathematica to check if two expressions may be able to be written in terms of the other? The expressions I had in mind were of the form $$alpha = frac1-x1-y,,,,,,, beta = frac1+x1+y$$ and was wondering if it's possible to find a relation $$alpha = f(beta)?$$
equation-solving simplifying-expressions expression-manipulation
equation-solving simplifying-expressions expression-manipulation
asked 2 hours ago
CAFCAF
270110
asked 2 hours ago
CAFCAF
270110
asked 2 hours ago
CAFCAF
270110
asked 2 hours ago
CAFCAF
270110
asked 2 hours ago
CAFCAF
270110
270110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It looks like the answer for the given system is No.
Defining the equations first:
eqns = a == (1-x)/(1-y), b == (1+x)/(1+y);
We can use Eliminate to try and suss out a relation which does not depend on x or y.
Eliminate[eqns, x, y]
True
However, Eliminate returns True. This means that a and b can be chosen independently of each other, disproving the existence of a relation of the form $alpha = f(beta)$ immediately. If that's not convincing, however, you can use Reduce to find out the precise conditions for which a solution for a exists:
Reduce[eqns, a, x, y]
(b == 1 && a == 1) || (-1 + a) (a - b) (-1 + b) != 0
So if a and b are both 1, or so long as the other expression is not 0, any pairing of a and b should be possible. You can use FindInstance to find a few, if you wish:
FindInstance[eqns /. a->4, b->2, x, y]
FindInstance[eqns /. a->5, b->2, x, y]
x -> 5, y -> 2
x -> 13/3, y -> 5/3
answered 1 hour ago
eyorbleeyorble
5,81311028
$endgroup$
add a comment |
$begingroup$
Solve $beta$ for $x=x(beta,y)$:
S = Solve[β == (1 + x)/(1 + y), x]
x -> -1 + β + y β
Use this list of solutions to express $alpha=alpha(beta,y)$:
α = FullSimplify[(1 - x)/(1 - y) /. S]
(-2 + β + y β)/(-1 + y)
Pick those solutions where $alpha$ depends only on $beta$ but not on $y$:
Select[α, FreeQ[y]]
No solutions found in this case.
answered 2 hours ago
RomanRoman
7,12511134
$endgroup$
add a comment |
$begingroup$
You can use FindInstance to generate a simple counter-example to the hypothesis that a is a function of b
a = (1 - x)/(1 - y);
b = (1 + x)/(1 + y);
FindInstance[a == 2, b == 1/2, x, y]
(* x -> -(1/3), y -> 1/3 *)
FindInstance[a != 2, b == 1/2, x, y]
(* x -> 1, y -> 3 *)
answered 58 mins ago
mikadomikado
7,0021929
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f197630%2fchecking-if-two-expressions-are-related%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It looks like the answer for the given system is No.
Defining the equations first:
eqns = a == (1-x)/(1-y), b == (1+x)/(1+y);
We can use Eliminate to try and suss out a relation which does not depend on x or y.
Eliminate[eqns, x, y]
True
However, Eliminate returns True. This means that a and b can be chosen independently of each other, disproving the existence of a relation of the form $alpha = f(beta)$ immediately. If that's not convincing, however, you can use Reduce to find out the precise conditions for which a solution for a exists:
Reduce[eqns, a, x, y]
(b == 1 && a == 1) || (-1 + a) (a - b) (-1 + b) != 0
So if a and b are both 1, or so long as the other expression is not 0, any pairing of a and b should be possible. You can use FindInstance to find a few, if you wish:
FindInstance[eqns /. a->4, b->2, x, y]
FindInstance[eqns /. a->5, b->2, x, y]
x -> 5, y -> 2
x -> 13/3, y -> 5/3
answered 1 hour ago
eyorbleeyorble
5,81311028
$endgroup$
add a comment |
$begingroup$
It looks like the answer for the given system is No.
Defining the equations first:
eqns = a == (1-x)/(1-y), b == (1+x)/(1+y);
We can use Eliminate to try and suss out a relation which does not depend on x or y.
Eliminate[eqns, x, y]
True
However, Eliminate returns True. This means that a and b can be chosen independently of each other, disproving the existence of a relation of the form $alpha = f(beta)$ immediately. If that's not convincing, however, you can use Reduce to find out the precise conditions for which a solution for a exists:
Reduce[eqns, a, x, y]
(b == 1 && a == 1) || (-1 + a) (a - b) (-1 + b) != 0
So if a and b are both 1, or so long as the other expression is not 0, any pairing of a and b should be possible. You can use FindInstance to find a few, if you wish:
FindInstance[eqns /. a->4, b->2, x, y]
FindInstance[eqns /. a->5, b->2, x, y]
x -> 5, y -> 2
x -> 13/3, y -> 5/3
answered 1 hour ago
eyorbleeyorble
5,81311028
$endgroup$
add a comment |
$begingroup$
It looks like the answer for the given system is No.
Defining the equations first:
eqns = a == (1-x)/(1-y), b == (1+x)/(1+y);
We can use Eliminate to try and suss out a relation which does not depend on x or y.
Eliminate[eqns, x, y]
True
However, Eliminate returns True. This means that a and b can be chosen independently of each other, disproving the existence of a relation of the form $alpha = f(beta)$ immediately. If that's not convincing, however, you can use Reduce to find out the precise conditions for which a solution for a exists:
Reduce[eqns, a, x, y]
(b == 1 && a == 1) || (-1 + a) (a - b) (-1 + b) != 0
So if a and b are both 1, or so long as the other expression is not 0, any pairing of a and b should be possible. You can use FindInstance to find a few, if you wish:
FindInstance[eqns /. a->4, b->2, x, y]
FindInstance[eqns /. a->5, b->2, x, y]
x -> 5, y -> 2
x -> 13/3, y -> 5/3
answered 1 hour ago
eyorbleeyorble
5,81311028
$endgroup$
It looks like the answer for the given system is No.
Defining the equations first:
eqns = a == (1-x)/(1-y), b == (1+x)/(1+y);
We can use Eliminate to try and suss out a relation which does not depend on x or y.
Eliminate[eqns, x, y]
True
However, Eliminate returns True. This means that a and b can be chosen independently of each other, disproving the existence of a relation of the form $alpha = f(beta)$ immediately. If that's not convincing, however, you can use Reduce to find out the precise conditions for which a solution for a exists:
Reduce[eqns, a, x, y]
(b == 1 && a == 1) || (-1 + a) (a - b) (-1 + b) != 0
So if a and b are both 1, or so long as the other expression is not 0, any pairing of a and b should be possible. You can use FindInstance to find a few, if you wish:
FindInstance[eqns /. a->4, b->2, x, y]
FindInstance[eqns /. a->5, b->2, x, y]
x -> 5, y -> 2
x -> 13/3, y -> 5/3
answered 1 hour ago
eyorbleeyorble
5,81311028
answered 1 hour ago
eyorbleeyorble
5,81311028
answered 1 hour ago
eyorbleeyorble
5,81311028
answered 1 hour ago
eyorbleeyorble
5,81311028
5,81311028
add a comment |
add a comment |
$begingroup$
Solve $beta$ for $x=x(beta,y)$:
S = Solve[β == (1 + x)/(1 + y), x]
x -> -1 + β + y β
Use this list of solutions to express $alpha=alpha(beta,y)$:
α = FullSimplify[(1 - x)/(1 - y) /. S]
(-2 + β + y β)/(-1 + y)
Pick those solutions where $alpha$ depends only on $beta$ but not on $y$:
Select[α, FreeQ[y]]
No solutions found in this case.
answered 2 hours ago
RomanRoman
7,12511134
$endgroup$
add a comment |
$begingroup$
Solve $beta$ for $x=x(beta,y)$:
S = Solve[β == (1 + x)/(1 + y), x]
x -> -1 + β + y β
Use this list of solutions to express $alpha=alpha(beta,y)$:
α = FullSimplify[(1 - x)/(1 - y) /. S]
(-2 + β + y β)/(-1 + y)
Pick those solutions where $alpha$ depends only on $beta$ but not on $y$:
Select[α, FreeQ[y]]
No solutions found in this case.
answered 2 hours ago
RomanRoman
7,12511134
$endgroup$
add a comment |
$begingroup$
Solve $beta$ for $x=x(beta,y)$:
S = Solve[β == (1 + x)/(1 + y), x]
x -> -1 + β + y β
Use this list of solutions to express $alpha=alpha(beta,y)$:
α = FullSimplify[(1 - x)/(1 - y) /. S]
(-2 + β + y β)/(-1 + y)
Pick those solutions where $alpha$ depends only on $beta$ but not on $y$:
Select[α, FreeQ[y]]
No solutions found in this case.
answered 2 hours ago
RomanRoman
7,12511134
$endgroup$
Solve $beta$ for $x=x(beta,y)$:
S = Solve[β == (1 + x)/(1 + y), x]
x -> -1 + β + y β
Use this list of solutions to express $alpha=alpha(beta,y)$:
α = FullSimplify[(1 - x)/(1 - y) /. S]
(-2 + β + y β)/(-1 + y)
Pick those solutions where $alpha$ depends only on $beta$ but not on $y$:
Select[α, FreeQ[y]]
No solutions found in this case.
answered 2 hours ago
RomanRoman
7,12511134
answered 2 hours ago
RomanRoman
7,12511134
answered 2 hours ago
RomanRoman
7,12511134
answered 2 hours ago
RomanRoman
7,12511134
7,12511134
add a comment |
add a comment |
$begingroup$
You can use FindInstance to generate a simple counter-example to the hypothesis that a is a function of b
a = (1 - x)/(1 - y);
b = (1 + x)/(1 + y);
FindInstance[a == 2, b == 1/2, x, y]
(* x -> -(1/3), y -> 1/3 *)
FindInstance[a != 2, b == 1/2, x, y]
(* x -> 1, y -> 3 *)
answered 58 mins ago
mikadomikado
7,0021929
$endgroup$
add a comment |
$begingroup$
You can use FindInstance to generate a simple counter-example to the hypothesis that a is a function of b
a = (1 - x)/(1 - y);
b = (1 + x)/(1 + y);
FindInstance[a == 2, b == 1/2, x, y]
(* x -> -(1/3), y -> 1/3 *)
FindInstance[a != 2, b == 1/2, x, y]
(* x -> 1, y -> 3 *)
answered 58 mins ago
mikadomikado
7,0021929
$endgroup$
add a comment |
$begingroup$
You can use FindInstance to generate a simple counter-example to the hypothesis that a is a function of b
a = (1 - x)/(1 - y);
b = (1 + x)/(1 + y);
FindInstance[a == 2, b == 1/2, x, y]
(* x -> -(1/3), y -> 1/3 *)
FindInstance[a != 2, b == 1/2, x, y]
(* x -> 1, y -> 3 *)
answered 58 mins ago
mikadomikado
7,0021929
$endgroup$
You can use FindInstance to generate a simple counter-example to the hypothesis that a is a function of b
a = (1 - x)/(1 - y);
b = (1 + x)/(1 + y);
FindInstance[a == 2, b == 1/2, x, y]
(* x -> -(1/3), y -> 1/3 *)
FindInstance[a != 2, b == 1/2, x, y]
(* x -> 1, y -> 3 *)
answered 58 mins ago
mikadomikado
7,0021929
answered 58 mins ago
mikadomikado
7,0021929
answered 58 mins ago
mikadomikado
7,0021929
answered 58 mins ago
mikadomikado
7,0021929
7,0021929
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f197630%2fchecking-if-two-expressions-are-related%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
