# Please comment on my solution of this equation using Laplace transform

Started by May 15, 2007
```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
>
> 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.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```
```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
>>=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
```
```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
>>>
>>> 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.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```
```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
```
```ooo wrote:
> 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.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```
```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

The ambiguities are resolved by well known rules (conventions) that
are followed by essentially everybody that understands basic math.

=46WIW, here are two relevant links:
<http://en.wikipedia.org/wiki/Order_of_operations>
<http://planetmath.org/?op=3Dgetobj&from=3Dobjects&id=3D3951>

You're welcome to the last word.

Bye.
=20
```
```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

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

```
```ooo wrote:

...

> You're welcome to the last word.

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.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```