amy paradox
amy666
for a = 0
x = -c / b
for a =/= 0
x = [-b + sqrt(b^2 - 4ac)]/2a
we dont see the first formula as a part in the second ?
what if we fill in a = 0 in the second formula ?
amy's paradox.
have fun with it.
amy666
Neilist
<snip Tommy-esque math musings>
Hey "amy", if you're so smart, straighten out that idiot Tommy1729
regarding his WRONG definition of the noun "integral":
On August 28, 2007, 2:35 PM, Tommy1729 wrote:
"integers ...-2,-1,0,1,2,...
integrals 1,2,3,4,...
natural numbers 0,1,2,3,... "
kiran
amy, second formula is for quadratic equation means it works with
valid equation of the form ax^2+bx+c where a is not 0 in any case.
while first case written by u exists when equation is not quadratic.
hence first formula cannot be taken or derived as a case from second
formula.
amzoti
In your formula for x, what do you get if you let a = 0 --> (0/0); an
indeterminate form.
So, what can you do to resolve that, so you get the same as the first
equation?
Have fun with it!
~A
David W. Cantrell
>
> for a =/= 0, x = [-b + sqrt(b^2 - 4ac)]/2a
Hmm. The quadratic formula is normally written to indicate, explicitly,
that square roots of both signs are to be considered.
> we dont see the first formula as a part in the second ?
>
> what if we fill in a = 0 in the second formula ?
>
> amy's paradox.
It's no paradox. The common quadratic formula holds as long as we have an
actual (i.e., nondegenerate) quadratic.
OTOH, it would be nice to have a combined "linear & quadratic formula":
To obtain the root(s) of ax^2 + bx + c = 0, assuming that not both a and b
are zero, use
x = -2c/(b +/- sqrt(b^2 - 4ac))
with the proviso that, if a = 0, the only sign to used on the square root
is the same as the sign of b.
David W. Cantrell
Dave L. Renfro
> In your formula for x, what do you get if you let
> a = 0 --> (0/0); an indeterminate form.
>
> So, what can you do to resolve that, so you get the same
> as the first equation?
>
> Have fun with it!
I've seen the process of letting a --> 0 in the quadratic
equations in a number of old texts. Here's a few that
happen to be digitized at google-books. I searched for
"limit" and "roots" and "quadratic equation", with
a date restriction of 1850-1910:
http://books.google.com/books?q=limit+roots+%22quadratic+equation%22+date%3A1850-1910
Section 8 (pp. 14-16) of Smith/Gale's " The Elements of
Analytic Geometry" (1904)
http://tinyurl.com/yv59y8
Section 8 (pp. 7-8) of Newcomb's "Elements of Analytic
Geometry" (1885)
http://tinyurl.com/27fzcy
Section 275 (p. 227) of Hall/Knight's "Higher Algebra:
A Sequel to Elementary Algebra for Schools" (1894)
http://tinyurl.com/295ktm
Dave L. Renfro
amy666
and the conclusion is ... ?
b + sqrt(b^2 -4ac) / 2a
(hospital's rule with respect to a )
[1/2 * sqrt(1 / (b^2 -4ac) ) * -4c] / 2
= -c * sqrt( 1 / (b^2 - 4ac) )
= -c / b
it fits.
all that was needed was hopital's rule.
excercise finished.
it took quite long for a math forum ( ahem )
maybe intresting is a formula for both lineair and quadratic. (without limits)
this has been given by someone already however with restriction a and b =/= 0.
also extensions to higher degree should be investigated.
amy666
amy666
in the other post i used hospital's rule to find -c/b for the formula [-b +/- sqrt(b^2 - 4ac)] / 2a
it should also work for -2c/(b +/- sqrt(b^2 - 4ac))
in fact easier , there is only one a.
so we get -2c/2b = -c/b.
how about a "linear ,quadratic & cubic formula" ?
amy
Neilist
<snip>
> excercise finished.
>
> it took quite long for a math forum ( ahem )
Fuck off, Tommy1729.
Tommy1729, you've turned into a crank just like James Harris, even
using pseudonyms like Amy666 to praise yourself and to repost your
math garbage.
So, when are you going to read about NOUNS like "integral"? Are you
deathly afraid of learning how WRONG for so long you have been? As in
the following:
amzoti
I don't think most people actually cared much for the question - so it
is not that they don't know.
Trust me, there are some pretty smart folks on here - including Dave
Renfro - and I enjoy reading many of their postings. I can think of
about ten people that I really like on here and they are very helpful
to posters in general. We also have our fair share of cranks, trolls
and snake-oil sales(wo)men on here too.
I am sick of the TON of SPAM postings from google groups - but have no
other access to the usenet feeds at work.
I suppose I can get a pay account which uses a web feed with login to
get around the problem.
Anyway - hope the answers you received were adequate.
By the way - try resolving the limit without the use of L'Hopital's
and see if you can get without those silly tricks.
~A
amy666
well , surprise surprise , i agree.
renfro is good.
and so are about 9 others.
but 10 people is not much ...
and they dont often start a thread , they rather reply.
like e.g. robert israel.
hes good. but wont start a thread or introduce a new idea.
>
> I am sick of the TON of SPAM postings from google
> groups - but have no
> other access to the usenet feeds at work.
seems JSH and tommy are far from the biggest problems hmm.
>
> I suppose I can get a pay account which uses a web
> feed with login to
> get around the problem.
thats not a good idea.
this is one of the last spaces on internet where one can speak freely ( freely : no moderators and no cost )
besides then even less people who can do math will appear and more spam will occur ( since they know we can afford an account )
>
> Anyway - hope the answers you received were adequate.
>
> By the way - try resolving the limit without the use
> of L'Hopital's
> and see if you can get without those silly tricks.
its not silly.
hospital is one of the best methods to compute limits.
series expansions is another great way.
are you asking for a method without derivates ?
what do you have in mind ?
computing limits is usually calculus not ?
>
> ~A
amy
amzoti
>
> - Show quoted text -
Do it from first principles - that is, use the definition of the limit
and not this rule to resolve the limit.
I wish they wouldn't teach L'H rule (although it is an elegant
discovery) and have people proceed with the definition of the limit as
I think that is much more instructive in understanding what is going
on.
Also, there are many great posters - like Magadin, Dubuque, Ramsden,
quasi, Mario, Ullrich and many others - the list is greater than 10.
I think most of the others read - but do not reply given the
increasing amount of SPAM on usenet (an unfortunate thing really).
Usenet is truly one of the beautiful things about a networked
community - too bad these spamming sons-a-bitches are ruining it too.
I will NEVER follow a SPAM posting!
Bill Dubuque
>>
>> for a =/= 0, x = [-b + sqrt(b^2 - 4ac)]/2a
>>
>> what if we fill in a = 0 in the second formula ?
>
>
> x = 2c/(-b +/- sqrt(b^2 - 4ac))
Note that's simply the quadratic formula for reciprocal equation
i.e. 1/x = (-b +/- sqrt(b^2 - 4ac))/2c for c/xx + b/x + a = 0
Reciprocating puts the cases a = 0 and c = 0 in correspondence,
so the case a = 0 can be derived by reciprocating the case c = 0,
which avoids the naively perceived need to use some "limit" process.
--Bill Dubuque
amy666
as already posted ; what about an equation that solves all of linear , quadratic and cubic ?
btw you are mentioned here as a good mathematician.
better put this thread with your favorites ;-)
amy