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May
17
awarded  Popular Question
May
8
comment how to cancel arbitrary term in fractional formula
@MichaelE2 Oh, I mean divide R^2 (g m[1] - g m[2])/(J + m[1] R^2 + m[2] R^2) by p^2, not (p^2 (g m[1] - g m[2]))/(J + p^2 m[1] + p^2 m[2])
May
8
comment how to cancel arbitrary term in fractional formula
@MichaelE2 That is odd! I tried, it didn't work. I am using M10
May
8
comment how to cancel arbitrary term in fractional formula
@MichaelE2 Well, the reason that I want to divide numerator and denominator by $R^2$ is that I want the $J/R^2$, because $J/R^2$ has a special meaning. If $J/p^2$ has a special meaning, then I have to divide $p^2$.
May
7
comment how to cancel arbitrary term in fractional formula
It seems it only works with t[R^2]. I tried t[p^2], then nothing happens.
May
7
asked how to cancel arbitrary term in fractional formula
Apr
23
awarded  Popular Question
Mar
4
awarded  Popular Question
Feb
28
awarded  Nice Question
Dec
16
comment how to get continuation of square matrices with periodic diagonal band automatically?
@bills Thank you for suggesting FindLinearRecurrence. But I found that FindLinearRecurrence needs at least 5 elements in a list. So I think I just stick with Tally approach which is easy and understandable.
Dec
16
answered how to get continuation of square matrices with periodic diagonal band automatically?
Dec
15
awarded  Self-Learner
Dec
15
comment how to get continuation of square matrices with periodic diagonal band automatically?
What you gave is a manual version.
Dec
15
revised how to get continuation of square matrices with periodic diagonal band automatically?
added 260 characters in body
Dec
15
comment how to get continuation of square matrices with periodic diagonal band automatically?
Thank you! I know band function. But you misunderstood me. I want a function continuation which can automatically analyze any square matrix mat we feed to it, and give general version continuation[mat]?
Dec
15
asked how to get continuation of square matrices with periodic diagonal band automatically?
Dec
14
comment How to transform power of -1 into exponential
Thank you for pointing this out!
Dec
13
comment about the design of Part related to Association
@SimonWoods Well, My last question is actually a question about the transparency of Association. Since assoc[[1]] directly acts on values of assoc, then why doesn't assoc[[span]] gives collection of values? I admit I made a mistake. Like you said, original head should be preserved. Then, <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1;;2]] should give <|1,2|>, but this is not an association!It is a contradiction! So they must make it as <|"a" -> 1, "b" -> 2|>. Wow, maybe this is what the creators of Association thought in their mind!!! You enlightened me!:) What do you think?
Dec
13
revised How to transform power of -1 into exponential
added 54 characters in body
Dec
13
comment How to transform power of -1 into exponential
Thank you! What if I also want to transform -1 into E^[I π]?