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. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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

Started by ●May 15, 2007

Reply by ●May 16, 20072007-05-16

Reply by ●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) > >JerryThe 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 ●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 ●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? > >JerryWhat 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 ●May 17, 20072007-05-17

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. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by ●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. > >JerryThe 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

Reply by ●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 ●May 17, 20072007-05-17

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. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯