mm-1002
===
Subject: Integral involving exponential function
Can anyone help me solve the following integral:
1/(e^(-x)+1) dx
According to Mathematica the answer is ln(e^x+1)+C but I
canÕt seem to
get to the answer.
===
Subject: Re: Integral involving exponential function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i15E0X016558;
>Can anyone help me solve the following integral:
>1/(e^(-x)+1) dx
>According to Mathematica the answer is ln(e^x+1)+C but I
canÕt seem
>get to the answer.
Hi Jennifer,
Multiply numerator and denominator of 1/(e^(-x)+1) by e^x to
get
e^x/(e^x+1) as the function to integrate.
Now notice that the function is of type fÕ(x)/f(x)
(with f(x) = e^x + 1) for which you get F(x) = log(f(x)) + C
as
antiderivative.
Best wishes
Torsten.
===
Subject: Re: Integral involving exponential function
> Can anyone help me solve the following integral:
> 1/(e^(-x)+1) dx
> According to Mathematica the answer is ln(e^x+1)+C but I
canÕt seem to
> get to the answer.
e^(-x) = 1/e^x
so 1/(e^(-x)+1) dx = 1/(1 + e^x) e^x dx
now let u = e^x+1, du = e^x dx
===
Subject: Re: Integral involving exponential function
> Can anyone help me solve the following integral:
> 1/(e^(-x)+1) dx
> According to Mathematica the answer is ln(e^x+1)+C but I
canÕt seem to
> get to the answer.
e^(-x) + 1 = ?
--
Paul Sperry
Columbia, SC (USA)
===
Subject: physics math
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i152q3231290;
Two charges are located on the x axis: q1 = +5.4 uC at x1 =
+4.2 cm,
and q2 = +4.7 uC at x2 = -4.1 cm. Two other charges are
located on the
y axis: q3 = +4.1 uC at y3 = +4.6 cm, and q4 = -7 uC at y4 =
+10 cm.
Find (a) the magnitude and (b) the direction of of the net
electric
\field at the origin. Specify the direction as an angle with
respect to
the +x direction.
problems IÕve
solved leading up to this.
===
Subject: Re: physics math
> Two charges are located on the x axis: q1 = +5.4 uC at x1 =
+4.2 cm,
> and q2 = +4.7 uC at x2 = -4.1 cm. Two other charges are
located on the
> y axis: q3 = +4.1 uC at y3 = +4.6 cm, and q4 = -7 uC at y4
= +10 cm.
> Find (a) the magnitude and (b) the direction of of the net
electric
> \field at the origin. Specify the direction as an angle with
respect to
> the +x direction.
problems IÕve
> solved leading up to this.
Do you know how to express the (scalar) electric potential at
one point
due to a point charge at another point?
Potentials are additive.
The gradient of the total potential at any point due to all
charges is
total electric vector \field at that point.
===
Subject: More on strange results
As I said I was using an iteration function x = - ln x which
ended up
alternating in the following complex conjugate pair:
x = -0.72281631256851 + 1.92926749633194i
x = -0.72281631256851 - 1.92926749633194i
My question is why does this happen? I know that the
iteration formula
in question will no behave nicely. More speci\fically \
IÕd like
to know
why it ends up alternating in this pair?
//Mattias
--
Sigblock empty. By choice.
===
Subject: i need help with 3 riddels
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i15E0Yc16595;
Break me and I will disappear, with no mess left behind and
no crash
to overhear.......what am I..?
I have three eyes but my root has two, I have all the time in
the
world, much more than you do.....what am I?
On the wall I may be found or from heaven IÕll come down, to
teach to
warn or to astound, IÕll help the lost if I am
found......what am I?
its only 4 letters or less...
any help???
===
Subject: Re: Permutation Question
 A question for any with time to burn about permutations.
 1. using the numerals 0,1,2,3,4,5,6,7 how many different 4
digit
> numbers can be made greater than 3400?
 please explain the process not just the answer if possible.
> It depends on whether a digit may appear more than once in
any one
> number. Since you refer to permutations, I will presume not.
 How many starting with 4,or 5, or 6, or 7?
 How many starting with 34, or 35, or 36 or 37?
 Add.
 a student) but I came up with 1080 being 7P3 x 4 for the
numbers
> starting with either 4,5,6 or 7 and 6P2 x 4 for the numbers
starting
> with 34,35,35 or 37. Have I got it right or did I go wrong
somewhere?
 I think your arithmetic is a bit off. Do you understand how
to
> calculate mPn? It is the product of n integers, starting
with m and
> successively decreasing by one, so, e.g., 7P3 = 7*6*5.
> I get (7*6*5)*4 =210*4 = 840
> and (6*5)*4 =30*4 = 120
> adding up to 960
True!!
For some reason I added another 120 to my answer. Let me know
if your
up for another one. IÕm currently a volunteer accountant in \
an
orphanage in El Salvador and IÕm trying to tutor the senor
kids but
IÕm very rusty on this math type.
===
Subject: Abstracting out the method, non-polynomial
factorization
Despite all the controvery over my method, basically what
IÕve done is
a balancing act--simple against complex--and hereÕs an
abstraction of
the technique:
Consider f_1(x) f_2(x) = g F(x), and gab = gc, where
(f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
where f_1(0) = f_2(0) = F(0) = G(0) = 0.
Here you have the factorization
gab = gc,
balanced against the factorization
f_1(x) f_2(x) = g F(x),
where the point is that itÕs *unknown* how factors of g \
split
between
f_1(x) and f_2(x).
So you can think of my method as a solution for an unknown
factorization.
Now I use the fact that if youÕre in the ring of algebraic
integers
then dividing g from both sides *must* give
ab = c,
whereas, if youÕre in some other ring, like the \
\field of
algebraic
numbers, then you have an *in\finity* of factorizations on the
left,
like
(sqrt(g) a)(b/sqrt(g)) = c.
Now then given that you have
(f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
itÕs forced that f_1(x) *should* have g as a factor in the
ring of
algebraic integers, and in many cases it does.
For instance, with f_1(x) = 3x, f_2(x) = x, a=1, b=1, c=1,
g=3, you
have
(3x + 3)(x + 1) = 3(x^2 + 2x + 1).
The problem for so many on the sci.math newsgroup has been
that you
can rather creatively \find functions f_1(x) and f_2(x), which
donÕt
have factors of g in the ring of algebraic integers.
There are two things to do when faced with such an issue:
1. Assume an error in the underlying reasoning, and check
2. If checks reveal no error in your reasoning, then realize
that the
conclusion is correct.
Notice that the abstracted method itself does not say
anything about
algebraic integers.
ItÕs just a mathematical tool.
Notice it doesnÕt even bother with whether or not you have a
polynomial or non-polynomial factorization.
When that tool is used with certain functions from
non-polynomial
factorizations it gives the result which has sparked so much
controversy on the sci.math newsgroup.
Now mathematicians can accept mathematics, or they can cling
to false
beliefs shown to be false by a rather basic tool which
balances a
simple factorization against a more complex one.
James Harris
===
Subject: Re: Abstracting out the method, non-polynomial
factorization
> Despite all the controvery over my method, basically what
IÕve done is
> a balancing act--simple against complex--and hereÕs an
abstraction of
> the technique:
> Consider f_1(x) f_2(x) = g F(x), and gab = gc, where
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> where f_1(0) = f_2(0) = F(0) = G(0) = 0.
> Here you have the factorization
> gab = gc,
> balanced against the factorization
> f_1(x) f_2(x) = g F(x),
> where the point is that itÕs *unknown* how factors of g
split between
> f_1(x) and f_2(x).
> So you can think of my method as a solution for an unknown
> factorization.
> Now I use the fact that if youÕre in the ring of algebraic
integers
> then dividing g from both sides *must* give
> ab = c,
> whereas, if youÕre in some other ring, like the \
\field of
algebraic
> numbers, then you have an *in\finity* of factorizations on
the left,
> like
> (sqrt(g) a)(b/sqrt(g)) = c.
> Now then given that you have
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> itÕs forced that f_1(x) *should* have g as a factor in the
ring of
> algebraic integers, and in many cases it does.
Whenever I see the word should in a proof, its like a \fire
alarm
going off.
You say f_1(x) *should* have g as a factor, but can you prove
f_1(x) **does** have g as a factor?
If you cant prove that, then you have no proof.
> For instance, with f_1(x) = 3x, f_2(x) = x, a=1, b=1, c=1,
g=3, you
> have
> (3x + 3)(x + 1) = 3(x^2 + 2x + 1).
> The problem for so many on the sci.math newsgroup has been
that you
> can rather creatively \find functions f_1(x) and f_2(x),
which donÕt
> have factors of g in the ring of algebraic integers.
The existence of functions f_1(x) and f_2(x) which donÕt \
have
factors
of g in the ring of algebraic integers proves your claim is
false by
counterexample. Since when is providing counterexamples a
problem?
> There are two things to do when faced with such an issue:
> 1. Assume an error in the underlying reasoning, and check
The best check is never done by the author. That is what peer
review
is for. If you dont like sci.math, submit it to a journal.
> 2. If checks reveal no error in your reasoning, then
realize that the
> conclusion is correct.
> Notice that the abstracted method itself does not say
anything about
> algebraic integers.
> ItÕs just a mathematical tool.
> Notice it doesnÕt even bother with whether or not you have \
a
> polynomial or non-polynomial factorization.
> When that tool is used with certain functions from
non-polynomial
> factorizations it gives the result which has sparked so much
> controversy on the sci.math newsgroup.
> Now mathematicians can accept mathematics, or they can
cling to false
> beliefs shown to be false by a rather basic tool which
balances a
> simple factorization against a more complex one.
> James Harris
===
Subject: Re: Abstracting out the method, non-polynomial
factorization
<snipNow I use the fact that if youÕre in the ring of
algebraic integers
>then dividing g from both sides *must* give
>ab = c,
>whereas, if youÕre in some other ring, like the \
\field of
algebraic
>numbers, then you have an *in\finity* of factorizations on the
left,
>like
>(sqrt(g) a)(b/sqrt(g)) = c.
Your fact about the ring of algebraic integers appears to be
the
conßation of two assertions.
Firstly, you assert that if a, b, c and g are algebraic
integers
such that gab = gc, then ab = c. This is correct, providing of
course that g is nonzero.
Secondly, you assert that if a, b and c are algebraic
integers such
that ab = c, then this is the only factorization of c in the
ring
of algebraic integers. That is clearly false. For example, if
y
is any algebraic integer, then the zeros of t^2 + yt + c are
cofactors of c in the ring of algebraic integers.
An easy way to see that your second assertion fails is to
note that
it is false in the ring of rational integers, and this is a
subring of the algebraic integers. In the ring of rational
integers, 12 has the distinct factorizations
12 = (1)(12) = (2)(6) = (3)(4).
It still has all of those factorizations in the algebraic
integers.
Of course, it also has a lot of new ones, such as
12 = (sqrt(3))(sqrt(48)) =
(sqrt(7)+sqrt(-5))(sqrt(7)-sqrt(-5)).
John Roberts-Jones
===
Subject: Re: Abstracting out the method, non-polynomial
factorization
> Despite all the controvery over my method, basically what
IÕve done is
> a balancing act--simple against complex--and hereÕs an
abstraction of
> the technique:
> Consider f_1(x) f_2(x) = g F(x), and gab = gc, where
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> where f_1(0) = f_2(0) = F(0) = G(0) = 0.
> Here you have the factorization
> gab = gc,
> balanced against the factorization
> f_1(x) f_2(x) = g F(x),
> where the point is that itÕs *unknown* how factors of g
split between
> f_1(x) and f_2(x).
> So you can think of my method as a solution for an unknown
> factorization.
> Now I use the fact that if youÕre in the ring of algebraic
integers
> then dividing g from both sides *must* give
> ab = c,
> whereas, if youÕre in some other ring, like the \
\field of
algebraic
> numbers, then you have an *in\finity* of factorizations on
the left,
> like
> (sqrt(g) a)(b/sqrt(g)) = c.
> Now then given that you have
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> itÕs forced that f_1(x) *should* have g as a factor in the
ring of
> algebraic integers, and in many cases it does.
Counterexample:
a=b=1, c=2, g=2, F(x) = x^2, G(x) = 2x
f_1(x) = 2x if x=0,1 mod 3, 7 if x=2 mod 3
f_2(x) = x if x=0,1 mod 3, (2(x^2+x+1)/9 -1) if x=2 mod 3
Then
(f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
for all x but f_1(x) does not have g (=2) as a factor.
- William Hughes
===
Subject: Re: Abstracting out the method, non-polynomial
factorization
> Despite all the controvery over my method, basically what
IÕve done is
> a balancing act--simple against complex--and hereÕs an
abstraction of
> the technique:
James Always in error, never in doubt! Harris. Hey schmuck,
alt.math.undergrad? Do you believe even children will be
impressed by
your loathsome brainfarting?
Hey stooopid loud troll James Harris, put up or shut up. James
Harris, King of the Primes! Where are your sceptor and crown,
delusional James Harris, your regal clothes? Is a $10,000
prize no
questions asked too small to justify your submission of two
little
prime numbers? Or are you a psychotic impotent gelding?
http://www.rsasecurity.com/rsalabs/challenges/factoring/
faq.html
http://www.rsasecurity.com/rsalabs/challenges/factoring/
numbers.html
http://www.crank.net/harris.html
ItÕs not every braying jackass that gets a whole page at
crank.net
--
Uncle Al
http://www.mazepath.com/uncleal/qz.pdf
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
===
Subject: Re: Abstracting out the method, non-polynomial
factorization
Some corrections...
> Despite all the controvery over my method, basically what
IÕve done is
> a balancing act--simple against complex--and hereÕs an
abstraction of
> the technique:
> Consider f_1(x) f_2(x) = g F(x), and gab = gc, where
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> where f_1(0) = f_2(0) = F(0) = G(0) = 0.
<deleted Now I use the fact that if youÕre in the ring of
algebraic integers
> then dividing g from both sides *must* give
> ab = c,
Actually that factorization is available but so are an
in\finity of
other unit factor factorizations like
(-a)(-b) = c
so I went for a strong condition and got it wrong, when the
proper
condition that works well is simply that the factorization is
available in the ring being considered.
> whereas, if youÕre in some other ring, like the \
\field of
algebraic
> numbers, then you have an *in\finity* of factorizations on
the left,
> like
> (sqrt(g) a)(b/sqrt(g)) = c.
Correcting following up from before, the issue isnÕt the
number of
factorizations but availability of a given factorization.
Here the proper point is that the factorization shown is NOT
available
in the ring of algebraic integers.
James Harris
===
Subject: Re: Abstracting out the method, non-polynomial
factorization
> Despite all the controvery over my method, basically what
IÕve done is
> a balancing act--simple against complex--and hereÕs an
abstraction of
> the technique:
> Consider f_1(x) f_2(x) = g F(x), and gab = gc, where
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> where f_1(0) = f_2(0) = F(0) = G(0) = 0.
> Here you have the factorization
> gab = gc,
> balanced against the factorization
> f_1(x) f_2(x) = g F(x),
> where the point is that itÕs *unknown* how factors of g
split between
> f_1(x) and f_2(x).
> So you can think of my method as a solution for an unknown
> factorization.
> Now I use the fact that if youÕre in the ring of algebraic
integers
> then dividing g from both sides *must* give
> ab = c,
> whereas, if youÕre in some other ring, like the \
\field of
algebraic
> numbers, then you have an *in\finity* of factorizations on
the left,
> like
> (sqrt(g) a)(b/sqrt(g)) = c.
> Now then given that you have
> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)
> itÕs forced that f_1(x) *should* have g as a factor in the
ring of
> algebraic integers, and in many cases it does.
> For instance, with f_1(x) = 3x, f_2(x) = x, a=1, b=1, c=1,
g=3, you
> have
> (3x + 3)(x + 1) = 3(x^2 + 2x + 1).
> The problem for so many on the sci.math newsgroup has been
that you
> can rather creatively \find functions f_1(x) and f_2(x),
which donÕt
> have factors of g in the ring of algebraic integers.
> There are two things to do when faced with such an issue:
> 1. Assume an error in the underlying reasoning, and check
> 2. If checks reveal no error in your reasoning, then
realize that the
> conclusion is correct.
3. Read the name of the original poster. If it is James
Harris, the
conclusion is false.
> James Often in error, but never in doubt! Harris
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: JSH: Abstracting the tool
You might have noticed that IÕve abstracted out the
mathematical tool
that IÕve been using, as well, itÕs fun to do \
so, and
probably very
useful.
The full abstraction now is at my blog:
http://mathforpro\fit.blogspot.com/
ItÕs a lot easier to correct there, versus making posts and
having to
go back when I make mistakes as I just did, with follow-ups.
Now discussion can be over the *tool* versus over where itÕs
used.
Later, discussion can focus on speci\fic examples, like the \
core
error that comes from interesting and intriguing properties of
algebraic integers: properties revealed by my tool.
My hope then is that in going over the tool thoroughly
progress can be
made in understanding what knowledge is gained by its
application.
James Harris
===
Subject: Re: Abstracting the tool
> You might have noticed that IÕve abstracted out the
mathematical tool
> that IÕve been using, as well, itÕs fun to \
do so, and
probably very
> useful.
> The full abstraction now is at my blog:
> http://mathforpro\fit.blogspot.com/
Since you have adverts on that page I suppose that is what
you mean by
mathforpro\fit.
How much money have you earned from the adverts so far?
Continually posting a link to that page is, I suppose, a form
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: What are all real values of x?
Perhaps I should clarify a bit,
The problem is properly written with (3-x) in paranthesis:
2/(3-x)=1/3-1/x and the question is
What are all real values of x?
After multiplying the expression by the least common
denominator of
3x(3-x) you get x^2+6x-9. So according to my study guide the
solutions
for x are two integers whos product is -9 and whose sum is 6.
The only
possibilty I can imagine would be 3 and 3 or -3 and -3 or -3
and 3. No
of the options work. So there is no real value of x for this
problem.
Correct? Or do I confuse something?
Dennis