All right, then, here it is:
Clear exposition presents information in a format that minimizes the
likelihood of misunderstanding. Adding technically unnecessary
parentheses to a mathematical expression can help to achieve that. This
is most especially true with the limited fonts generally used on usenet.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by glen herrmannsfeldt●May 17, 20072007-05-17

Jerry Avins wrote:
(snip)

>>> I think you mean F(s) = (a/s)*(s+c)/(s+c-b).

>> In math newsgroups, like sci.math, the two expressions are understood
>> to be equivalent. The 'space' implies multiplication (as in a
>> textbook), and unnecessary operators can make math newsgroup postings
>> much less readable.

Mathematica is the only programming language that I know of that
uses 'space' as a multiplication operator. There are some complications
to doing it, but mostly it works just fine.

>> It is generally understood that the "standard order of operations"
>> rules also apply (unless stated otherwise).

I learned Fortran before I learned the order of operations in math.
I don't know the history of the rules for mixing multiply and divide
in mathematical (not programming) expressions.
-- glen

Reply by ooo●May 17, 20072007-05-17

On Thu, 17 May 2007 00:18:11 -0400, Jerry Avins <jya@ieee.org> wrote:

>ooo wrote:
>> On Wed, 16 May 2007 21:40:25 -0400, Jerry Avins <jya@ieee.org> wrote:
>>=20
>> [snip]
>>> So when I write "The acceleration of gravity here is
>>> 22 miles/hour/second" what do I mean?
>>>
>>> Jerry
>>=20
>> What I would suppose you mean is
>>=20
>> 22 miles / (hour * second)
>>=20
>> just as
>>=20
>> 32 ft/sec/sec =3D 32 ft/(sec * sec) =3D 32 ft/sec^2
>>=20
>> or
>> a/b/c
>> =3D (a/b) / c
>> =3D (a/b) / (c/1)
>> =3D a / (b * c)
>> =20
>
>So /sec/sec and /sec*sec are synonyms? That's fine when one knows the=20
>context (acceleration), but too ambiguous for general written assertion.
>
>Jerry

> On Wed, 16 May 2007 21:40:25 -0400, Jerry Avins <jya@ieee.org> wrote:
>
> [snip]
>> So when I write "The acceleration of gravity here is
>> 22 miles/hour/second" what do I mean?
>>
>> Jerry
>
> What I would suppose you mean is
>
> 22 miles / (hour * second)
>
> just as
>
> 32 ft/sec/sec = 32 ft/(sec * sec) = 32 ft/sec^2
>
> or
> a/b/c
> = (a/b) / c
> = (a/b) / (c/1)
> = a / (b * c)
>

So /sec/sec and /sec*sec are synonyms? That's fine when one knows the
context (acceleration), but too ambiguous for general written assertion.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by ooo●May 16, 20072007-05-16

On Wed, 16 May 2007 21:40:25 -0400, Jerry Avins <jya@ieee.org> wrote:
[snip]

>So when I write "The acceleration of gravity here is
>22 miles/hour/second" what do I mean?
>
>Jerry

What I would suppose you mean is
22 miles / (hour * second)
just as
32 ft/sec/sec =3D 32 ft/(sec * sec) =3D 32 ft/sec^2
or
a/b/c
=3D (a/b) / c
=3D (a/b) / (c/1)
=3D a / (b * c)
=20

Reply by Jerry Avins●May 16, 20072007-05-16

ooo wrote:

> On Wed, 16 May 2007 18:34:04 -0400, Jerry Avins <jya@ieee.org> wrote:
>
> [snip]
>>>>> F(s) = a/s (s+c)/(s+c-b)
>>>> I think you mean F(s) = (a/s)*(s+c)/(s+c-b).
>>> In math newsgroups, like sci.math, the two expressions are understood
>>> to be equivalent. The 'space' implies multiplication (as in a
>>> textbook), and unnecessary operators can make math newsgroup postings
>>> much less readable.
>>>
>>> It is generally understood that the "standard order of operations"
>>> rules also apply (unless stated otherwise).
>> a
>> How does one distinguish F(s) = a/s*(s+c)/(s+c-b) = ------------------ ?
>> s*(s+c)/(s+c-b)
>>
>> Jerry
>
> The operators 'multiply' and 'divide' are usually assigned equal
> precedence. Where repeated operators of the same precedence are used
> (exponentiation being a common exception), evaluation proceeds left to
> right.
>
> Thus the expressions at issue above, namely
>
> a/s (s+c)/(s+c-b)
> (a/s)*(s+c)/(s+c-b) [This and the next line are yours]
> a/s*(s+c)/(s+c-b)
>
> are read (evaluated) left to right and are equivalent.
>
>
> This expression,
>
>> a
>> ------------------
>> s*(s+c)/(s+c-b)
>
> as you have written it (i.e. as a fraction, not as a one-line
> expression), is not equal to any of the three above. It could be
> written, among various ways, as any of
>
> a / ( s (s+c) / (s+c-b) )
>
> a/(s*(s+c)/(s+c-b))
>
> a/s (s+c-b) / (s+c)
>
> (a/s)*(s+c-b)/(s+c)
>
> a (s+c-b) / ( s (s+c) )

So when I write "The acceleration of gravity here is
22 miles/hour/second" what do I mean?
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by ooo●May 16, 20072007-05-16

On Wed, 16 May 2007 18:34:04 -0400, Jerry Avins <jya@ieee.org> wrote:
[snip]

>>>> F(s) =3D a/s (s+c)/(s+c-b)
>>> I think you mean F(s) =3D (a/s)*(s+c)/(s+c-b).
>>=20
>> In math newsgroups, like sci.math, the two expressions are understood
>> to be equivalent. The 'space' implies multiplication (as in a
>> textbook), and unnecessary operators can make math newsgroup postings
>> much less readable.
>>=20
>> It is generally understood that the "standard order of operations"
>> rules also apply (unless stated otherwise).
> a
>How does one distinguish F(s) =3D a/s*(s+c)/(s+c-b) =3D =

------------------ ?

> s*(s+c)/(s+c-b)
>
>Jerry

The operators 'multiply' and 'divide' are usually assigned equal
precedence. Where repeated operators of the same precedence are used
(exponentiation being a common exception), evaluation proceeds left to
right.
Thus the expressions at issue above, namely
a/s (s+c)/(s+c-b)
(a/s)*(s+c)/(s+c-b) [This and the next line are yours]
a/s*(s+c)/(s+c-b)
are read (evaluated) left to right and are equivalent.
This expression,

> a
> ------------------
> s*(s+c)/(s+c-b)

as you have written it (i.e. as a fraction, not as a one-line
expression), is not equal to any of the three above. It could be
written, among various ways, as any of
a / ( s (s+c) / (s+c-b) )
a/(s*(s+c)/(s+c-b))
a/s (s+c-b) / (s+c)
(a/s)*(s+c-b)/(s+c)
a (s+c-b) / ( s (s+c) )
=20

Reply by Jerry Avins●May 16, 20072007-05-16

ooo wrote:

> On Wed, 16 May 2007 14:28:11 -0400, Jerry Avins <jya@ieee.org> wrote:
>
>> Passerby wrote:
>>> On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt
>>> <gah@ugcs.caltech.edu> wrote:
>>>
>>>> Mike wrote:
>>>>
>>>> (snip)
>>>>
>>>> I don't understand the step from:
>>>>
>>>>> F(s)= a/s + b/(s+c)*F(s),
>>> Solve for F(s)
>>>
>>> F(s) [ 1 - b/(s+c) ] = a/s
>>>
>>> F(s) = a/s (s+c)/(s+c-b)
>> I think you mean F(s) = (a/s)*(s+c)/(s+c-b).
>
> In math newsgroups, like sci.math, the two expressions are understood
> to be equivalent. The 'space' implies multiplication (as in a
> textbook), and unnecessary operators can make math newsgroup postings
> much less readable.
>
> It is generally understood that the "standard order of operations"
> rules also apply (unless stated otherwise).

a
How does one distinguish F(s) = a/s*(s+c)/(s+c-b) = ------------------ ?
s*(s+c)/(s+c-b)
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by ooo●May 16, 20072007-05-16

On Wed, 16 May 2007 14:28:11 -0400, Jerry Avins <jya@ieee.org> wrote:

>Passerby wrote:
>> On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt
>> <gah@ugcs.caltech.edu> wrote:
>>=20
>>> Mike wrote:
>>>
>>> (snip)
>>>
>>> I don't understand the step from:
>>>
>>>> F(s)=3D a/s + b/(s+c)*F(s),
>>=20
>> Solve for F(s)
>>=20
>> F(s) [ 1 - b/(s+c) ] =3D a/s
>>=20
>> F(s) =3D a/s (s+c)/(s+c-b)
>
>I think you mean F(s) =3D (a/s)*(s+c)/(s+c-b).

In math newsgroups, like sci.math, the two expressions are understood
to be equivalent. The 'space' implies multiplication (as in a
textbook), and unnecessary operators can make math newsgroup postings
much less readable.
It is generally understood that the "standard order of operations"
rules also apply (unless stated otherwise).

>
> ...
>
>Jerry

Reply by Jerry Avins●May 16, 20072007-05-16

Don Stockbauer wrote:
...

> I was told that there would be no mathematics required in this group.

There isn't. A little high-school algebra is helpful, but no one will
send you away if you can't do it.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯