Cancellation of common terms

Question

Can someone explain why the term A in case 2 does not get canceled out when case 1 does? I use "9.0 for Microsoft Windows (64-bit) (January 25, 2013)"

case1 = A*(E^(-λ) + α) == A;
Simplify[case1, A > 0]

gives 1 + E^λ α == E^λ

case2 = A*(E^(-2*λ) + α) == A;
Simplify[case2, A > 0]

gives A (E^(-2 λ) + α) == A

I get E^(-2*λ) + α == 1 for case2 in version 7. Perhaps try FullSimplify? — Mr.Wizard, Jan 25 '14 at 00:06
@Mr.Wizard And also in version 8. But not in version 9! Moreover, case1 is pretty weird too because E^-\[Lambda] + \[Alpha] == 1 (what v8 returns) is clearly simpler than 1 + E^λ α == E^λ (what v9 returns). A v9 bug perhaps? user11946: you can consider reporting this to support at wolfram.com since it seems that both results you mention are worse than what v8 (or v7) gives. — Szabolcs, Jan 25 '14 at 00:12
@user11946 Why don't you write to support about this problem and report back here with what they told you? — Szabolcs, Jan 25 '14 at 00:14
This has been observed a year ago in another question: FullSimplify does not work on this expression with no unknowns. There is an answer explaining how you can get what you need. — Artes, Jan 27 '14 at 23:24

score 5 · Answer 1 · edited Jan 27 '14 at 23:30

5

Below is the email response from Wolfram support.

Thank you for your email. It is not clear to me why Mathematica's behavior has changed between version 8 and 9 in this example with Simplify. As such, I have filed a report with our developers so that they can be aware of this behavior and make any changes as needed for future versions of Mathematica. In the meantime, I will also be sure to pass on any useful information that I get from our developers. Please let me know if you have any further questions.

Karl Isensee

Technical Support Wolfram Research, Inc.

edited Jan 27 '14 at 23:30

Artes

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answered Jan 27 '14 at 23:14

user11946

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See this answer to get an insight how to proceed. – Artes Jan 27 '14 at 23:27

Cancellation of common terms

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