Are these expressions not equal? Mathematica output is ambiguousChecking if two trigonometric expressions are equalHow to group the powers of one variable?How do I simplify an embedded sub-expression without affecting other sub-expressions?Why are some equal expressions more equal than others?How can I get Mathematica to simplify my expression?Equality of Reciprocal with Negative ExponentMake Mathematica not simplify an expressionWSTP simplifying expression from CHow do I handle complicated algebraic manipulations?Smart way of simplifying an expression with square roots?
How to implement a feedback to keep the DC gain at zero for this conceptual passive filter?
Count the occurrence of each unique word in the file
What does "Scientists rise up against statistical significance" mean? (Comment in Nature)
How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?
Are paving bricks differently sized for sand bedding vs mortar bedding?
Not using 's' for he/she/it
What should you do when eye contact makes your subordinate uncomfortable?
How to explain what's wrong with this application of the chain rule?
How can "mimic phobia" be cured or prevented?
Lowest total scrabble score
Can I sign legal documents with a smiley face?
What is Cash Advance APR?
Is it safe to use olive oil to clean the ear wax?
copy and scale one figure (wheel)
The screen of my macbook suddenly broken down how can I do to recover
Strong empirical falsification of quantum mechanics based on vacuum energy density
"Spoil" vs "Ruin"
What should you do if you miss a job interview (deliberately)?
Terse Method to Swap Lowest for Highest?
Is there a name for this algorithm to calculate the concentration of a mixture of two solutions containing the same solute?
lightning-datatable row number error
Fear of getting stuck on one programming language / technology that is not used in my country
Is it better practice to read straight from sheet music rather than memorize it?
Which one is correct as adjective “protruding” or “protruded”?
Are these expressions not equal? Mathematica output is ambiguous
Checking if two trigonometric expressions are equalHow to group the powers of one variable?How do I simplify an embedded sub-expression without affecting other sub-expressions?Why are some equal expressions more equal than others?How can I get Mathematica to simplify my expression?Equality of Reciprocal with Negative ExponentMake Mathematica not simplify an expressionWSTP simplifying expression from CHow do I handle complicated algebraic manipulations?Smart way of simplifying an expression with square roots?
$begingroup$
The following plot indicates that the first expression equals the second. But how can I use Mathematica to show that is true:
Plot[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), P, 0, 1]
An attempt to simplify indicates the expressions are not equal:
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]
That gives the following answer:
(Log[-1 + 1/P] + Log[-(P/(-1 + P))])/Log[10]
plotting simplifying-expressions
asked yesterday
user120911user120911
73628
$endgroup$
add a comment |
$begingroup$
The following plot indicates that the first expression equals the second. But how can I use Mathematica to show that is true:
Plot[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), P, 0, 1]
An attempt to simplify indicates the expressions are not equal:
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]
That gives the following answer:
(Log[-1 + 1/P] + Log[-(P/(-1 + P))])/Log[10]
plotting simplifying-expressions
asked yesterday
user120911user120911
73628
$endgroup$
3
$begingroup$
Add the optionAssumptions -> 0 < P < 1toFullSimplifyand it will tell you they are the same on that domain. Without assumptions, Mathematica will try to solve the equation for every possible complex value ofPand the two equations are not generally equal due to branch cuts.
$endgroup$
– Sjoerd Smit
yesterday
1
$begingroup$
In addition to what Sjoerd said: EvaluatingFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/ Log[10]), P > 1]reveals that the expressions are not equal for arbitrary realP. So Mathematica would have lied if she had simplifiedFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]to0.
$endgroup$
– Henrik Schumacher
yesterday
1
$begingroup$
Always remember: Mathematica always assumes variables are complex-valued unless told otherwise. You did not tellFullSimplify[]what you know aboutP, so of course you get a general result.
$endgroup$
– J. M. is slightly pensive♦
yesterday
add a comment |
$begingroup$
The following plot indicates that the first expression equals the second. But how can I use Mathematica to show that is true:
Plot[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), P, 0, 1]
An attempt to simplify indicates the expressions are not equal:
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]
That gives the following answer:
(Log[-1 + 1/P] + Log[-(P/(-1 + P))])/Log[10]
plotting simplifying-expressions
asked yesterday
user120911user120911
73628
$endgroup$
The following plot indicates that the first expression equals the second. But how can I use Mathematica to show that is true:
Plot[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), P, 0, 1]
An attempt to simplify indicates the expressions are not equal:
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]
That gives the following answer:
(Log[-1 + 1/P] + Log[-(P/(-1 + P))])/Log[10]
plotting simplifying-expressions
plotting simplifying-expressions
asked yesterday
user120911user120911
73628
asked yesterday
user120911user120911
73628
asked yesterday
user120911user120911
73628
asked yesterday
user120911user120911
73628
asked yesterday
user120911user120911
73628
73628
3
$begingroup$
Add the optionAssumptions -> 0 < P < 1toFullSimplifyand it will tell you they are the same on that domain. Without assumptions, Mathematica will try to solve the equation for every possible complex value ofPand the two equations are not generally equal due to branch cuts.
$endgroup$
– Sjoerd Smit
yesterday
1
$begingroup$
In addition to what Sjoerd said: EvaluatingFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/ Log[10]), P > 1]reveals that the expressions are not equal for arbitrary realP. So Mathematica would have lied if she had simplifiedFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]to0.
$endgroup$
– Henrik Schumacher
yesterday
1
$begingroup$
Always remember: Mathematica always assumes variables are complex-valued unless told otherwise. You did not tellFullSimplify[]what you know aboutP, so of course you get a general result.
$endgroup$
– J. M. is slightly pensive♦
yesterday
add a comment |
3
$begingroup$
Add the optionAssumptions -> 0 < P < 1toFullSimplifyand it will tell you they are the same on that domain. Without assumptions, Mathematica will try to solve the equation for every possible complex value ofPand the two equations are not generally equal due to branch cuts.
$endgroup$
– Sjoerd Smit
yesterday
1
$begingroup$
In addition to what Sjoerd said: EvaluatingFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/ Log[10]), P > 1]reveals that the expressions are not equal for arbitrary realP. So Mathematica would have lied if she had simplifiedFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]to0.
$endgroup$
– Henrik Schumacher
yesterday
1
$begingroup$
Always remember: Mathematica always assumes variables are complex-valued unless told otherwise. You did not tellFullSimplify[]what you know aboutP, so of course you get a general result.
$endgroup$
– J. M. is slightly pensive♦
yesterday
3
3
$begingroup$
Add the option
Assumptions -> 0 < P < 1 to FullSimplify and it will tell you they are the same on that domain. Without assumptions, Mathematica will try to solve the equation for every possible complex value of P and the two equations are not generally equal due to branch cuts.$endgroup$
– Sjoerd Smit
yesterday
$begingroup$
Add the option
Assumptions -> 0 < P < 1 to FullSimplify and it will tell you they are the same on that domain. Without assumptions, Mathematica will try to solve the equation for every possible complex value of P and the two equations are not generally equal due to branch cuts.$endgroup$
– Sjoerd Smit
yesterday
1
1
$begingroup$
In addition to what Sjoerd said: Evaluating
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/ Log[10]), P > 1] reveals that the expressions are not equal for arbitrary real P. So Mathematica would have lied if she had simplified FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])] to 0.$endgroup$
– Henrik Schumacher
yesterday
$begingroup$
In addition to what Sjoerd said: Evaluating
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/ Log[10]), P > 1] reveals that the expressions are not equal for arbitrary real P. So Mathematica would have lied if she had simplified FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])] to 0.$endgroup$
– Henrik Schumacher
yesterday
1
1
$begingroup$
Always remember: Mathematica always assumes variables are complex-valued unless told otherwise. You did not tell
FullSimplify[] what you know about P, so of course you get a general result.$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
Always remember: Mathematica always assumes variables are complex-valued unless told otherwise. You did not tell
FullSimplify[] what you know about P, so of course you get a general result.$endgroup$
– J. M. is slightly pensive♦
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can ask Mathematica when this expression is zero, assuming we work on the reals:
Reduce[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]) == 0, P, Reals]
(* 0 < P < 1 *)
FullSimplify will confirm that result.
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), 0 < P < 1]
(* 0 *)
answered yesterday
SzabolcsSzabolcs
162k14444942
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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%2f193756%2fare-these-expressions-not-equal-mathematica-output-is-ambiguous%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can ask Mathematica when this expression is zero, assuming we work on the reals:
Reduce[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]) == 0, P, Reals]
(* 0 < P < 1 *)
FullSimplify will confirm that result.
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), 0 < P < 1]
(* 0 *)
answered yesterday
SzabolcsSzabolcs
162k14444942
$endgroup$
add a comment |
$begingroup$
You can ask Mathematica when this expression is zero, assuming we work on the reals:
Reduce[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]) == 0, P, Reals]
(* 0 < P < 1 *)
FullSimplify will confirm that result.
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), 0 < P < 1]
(* 0 *)
answered yesterday
SzabolcsSzabolcs
162k14444942
$endgroup$
add a comment |
$begingroup$
You can ask Mathematica when this expression is zero, assuming we work on the reals:
Reduce[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]) == 0, P, Reals]
(* 0 < P < 1 *)
FullSimplify will confirm that result.
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), 0 < P < 1]
(* 0 *)
answered yesterday
SzabolcsSzabolcs
162k14444942
$endgroup$
You can ask Mathematica when this expression is zero, assuming we work on the reals:
Reduce[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]) == 0, P, Reals]
(* 0 < P < 1 *)
FullSimplify will confirm that result.
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10]), 0 < P < 1]
(* 0 *)
answered yesterday
SzabolcsSzabolcs
162k14444942
answered yesterday
SzabolcsSzabolcs
162k14444942
answered yesterday
SzabolcsSzabolcs
162k14444942
answered yesterday
SzabolcsSzabolcs
162k14444942
162k14444942
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%2f193756%2fare-these-expressions-not-equal-mathematica-output-is-ambiguous%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
3
$begingroup$
Add the option
Assumptions -> 0 < P < 1toFullSimplifyand it will tell you they are the same on that domain. Without assumptions, Mathematica will try to solve the equation for every possible complex value ofPand the two equations are not generally equal due to branch cuts.$endgroup$
– Sjoerd Smit
yesterday
1
$begingroup$
In addition to what Sjoerd said: Evaluating
FullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/ Log[10]), P > 1]reveals that the expressions are not equal for arbitrary realP. So Mathematica would have lied if she had simplifiedFullSimplify[(-(Log[(1 - P)/P]/Log[10])) - (Log[-(P/(-1 + P))]/Log[10])]to0.$endgroup$
– Henrik Schumacher
yesterday
1
$begingroup$
Always remember: Mathematica always assumes variables are complex-valued unless told otherwise. You did not tell
FullSimplify[]what you know aboutP, so of course you get a general result.$endgroup$
– J. M. is slightly pensive♦
yesterday