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Re: FindROot and substitutions
*To*: mathgroup at smc.vnet.net
*Subject*: [mg89884] Re: FindROot and substitutions
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Tue, 24 Jun 2008 03:20:32 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <g3ihp3$fd5$1@smc.vnet.net> <g3kus2$7hn$1@smc.vnet.net> <g3nh71$cu$1@smc.vnet.net>
Aaron Fude wrote:
> Thanks.
>
> Just one follow up question. How was it that the semicolon told you
> that I didn't understand something? I thought it just meant "suppress
> output" or "compound expression".
>
> Thanks again.
It is good practice to quote the relevant part of the message to whom
you are responding to. Here I can just imagine that you are referring to
the following (note that I am not the original poster, so I speak for
myself about my interpretation of the original poster's thought!):
ï¿½ wrote:
> Aaron Fude wrote:
>> Hi,
>>
>> I have found a workaround for this problem, but I would like to
>> understand how Mathematica wants you to think so I'm asking the
>> question anyway.
>>
>> Why does "bad" not work, while "good" works?
>>
>> g := x z;
>
> ----------^
> You used a semicolon here, so I suppose that you do not understand the
> difference between SetDelayed and Set.
The above expression is not a compound expression.
SetDelayed does not produce any output.
Therefore the semicolon is not needed and does not change anything.
So someone may infer that you do not have a clear idea about about the
differences between immediate and delayed assignments, which are topics
of some importance to answer the rest of your original question.
Best regards,
-- Jean-Marc
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