who is right, wolfram or me?
Fjoka
(http://integrals.wolfram.com/index.jsp)
I entered x^2/(x-1) and pressed compute. Got this:
1/2(x-1)(x+3) + log(x-1).
when I integratd myself mechanically I got:
x^2/(x-1)=>x^3/3 + ln(x-1).
all good, for input of x=2 I got 2.666667 while wolfram told me exactly 2.5,
who is right?is the formula there
wrong?
Arturo Magidin
Fjoka <inv...@probably.com> wrote:
>hi, surfed into the wolfram integrator
>(http://integrals.wolfram.com/index.jsp)
>I entered x^2/(x-1) and pressed compute. Got this:
>1/2(x-1)(x+3) + log(x-1).
This is unclear. You mean you got (1/2)(x-1)(x+3) + log(x-1)
and as it clearly states, "log" is to be interpreted as the natural log.
Take the derivative; what do you get?
(1/2)(x+3) + (1/2)(x-1) + 1/(x-1) =
= (1/2)((x-1)(x+3) + (x-1)^2 + 2)/(x-1)
= (1/2)(x^2 + 2x - 3 + x^2 - 2x + 1 + 2)/(x-1)
= (1/2)(2x^2)/(x-1)
= x^2/(x-1).
Looks good, if you ignore the missing "+C".
>when I integratd myself mechanically I got:
>x^2/(x-1)=>x^3/3 + ln(x-1).
Take the derivative of x^3/ 3 + ln(x-1). I get
x^2 + 1/(x-1) = (x^2(x-1) + 1)/(x-1)
= (x^3 - x^2 + 1)/(x-1)
which does not look at all like x^2/(x-1). So you are
definitely wrong, even ignoring that you forgot the constant of
integration.
>all good, for input of x=2 I got 2.666667 while wolfram told me exactly 2.5,
This is nonsense.
>who is right?
Neither one. You got it completely wrong, and though the wolfram
integrator is almost right, it did not include the constant of
integration.
>is the formula there wrong?
Which formula? Wrong for what?
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Fjoka
"Arturo Magidin" <mag...@math.berkeley.edu> skrev i meddelandet
news:feg1kf$ml3$1...@agate.berkeley.edu...
integral of x^2/(x-1) is x^3/3+ln(x-1)
derivative of x^3/3+ln(x-1) is according to me:
3x^2/(3*(x-1))
which is x^2/(x-1)
which is the function.
on insertion of x=2, why don't I get the same results?
umm??
Arturo Magidin
>umm...
>integral of x^2/(x-1) is x^3/3+ln(x-1)
No, it's not. I did the derivative up there for you.
>derivative of x^3/3+ln(x-1) is according to me:
>3x^2/(3*(x-1))
You're wrong. The derivative of a sum is the sum of the
derivatives. So the derivative of
(x^3/3) + ln(x-1)
is
(x^3/3)' + (ln(x-1))'.
The derivative of x^3/3 is 3x^2/3 = x^2.
The derivative of ln(x-1) is (1/(x-1))(x-1)' = 1/(x-1).
So the derivative of (x^3/3) + ln(x-1) is
x^2 + (1/(x-1)).
And this is NOT equal to x^2/(x-1).
>which is x^2/(x-1)
>which is the function.
>on insertion of x=2, why don't I get the same results?
Because you are apparently incapable of differentiating properly,
integrating properly, or bothering to read what I wrote for you.
Fjoka
David Moran
that I'm 100% confident in.
Dave
OPPT
wrote:
Don't multipost. Your question was answered in the alt.math NG.
Your antiderivative is wrong.
Arturo Magidin
>something tells me you are not trustworthy*S*S*S*S*,
Fine, go with that. Do tell us how badly you failed Calculus when it
is all over, though.
bob
"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
news:feg85a$ofp$1...@agate.berkeley.edu...
Arturo is OK, and spot on.
But ask JSH for advice on solving this problem, see what he does.
(nothing!).
Fjoka
David C. Ullrich
wrote:
Something tells me that _you're_ an idiot.
Differentiation is not exactly rockket science, you know.
>
************************
David C. Ullrich
G.E. Ivey
"integral of x^2/(x-1) is x^3/3+ln(x-1)"
No, that's the integral of x^2+ 1/(x-1). The correct integral is just what Arturo said.
Fjoka
"G.E. Ivey" <georg...@gallaudet.edu> skrev i meddelandet
news:11111901.1192018054...@nitrogen.mathforum.org...
same.
Still, being nice to people is not expensive, and you gain a lot.
Fjoka
"David C. Ullrich" <ull...@math.okstate.edu> skrev i meddelandet
news:r38pg3pgr5o4bdk8m...@4ax.com...
Arturo Magidin
Fjoka <inv...@probably.com> wrote:
>umm..I'm retired*LOL*
Excellent! Plenty of time for you to learn proper usenet quoting
behavior, then...
Arturo Magidin
Fjoka <inv...@probably.com> wrote:
[...]
>> You asked for help- he gave you help. And because he didn't say "yes,
>> you are completely correct", that's your response? I guess you don't want
>> help, then.
>>
>> "integral of x^2/(x-1) is x^3/3+ln(x-1)"
>> No, that's the integral of x^2+ 1/(x-1). The correct integral is just
>> what Arturo said.
>yes, putting a little time on it, after a good nights sleep I concluded the
>same.
>Still, being nice to people is not expensive, and you gain a lot.
Excellent advice. How about taking it next time? Instead of wasting my
time and yours by not reading my response and simply repeating the
same mistake again, why not put that "little time on it" first? And
then, if you don't understand the answer, how about not saying that
someone is not "trustworthy", simply because you had not bothered to
think about the answer?
--
======================================================================
fjoka doka
> In article <Ek5Pi.235683$jo2.1...@reader1.news.saunalahti.fi>, Fjoka
> <inv...@probably.com> wrote:
>
> [...]
>
>>> You asked for help- he gave you help. And because he didn't say
>>> "yes,
>>> you are completely correct", that's your response? I guess you don't
>>> want help, then.
>>>
>>> "integral of x^2/(x-1) is x^3/3+ln(x-1)" No, that's the integral of
>>> x^2+ 1/(x-1). The correct integral is just
>>> what Arturo said.
>>yes, putting a little time on it, after a good nights sleep I concluded
>>the same.
>>Still, being nice to people is not expensive, and you gain a lot.
>
> Excellent advice. How about taking it next time? Instead of wasting my
> time and yours by not reading my response and simply repeating the same
> mistake again, why not put that "little time on it" first? And then, if
> you don't understand the answer, how about not saying that someone is
> not "trustworthy", simply because you had not bothered to think about
> the answer?
heh, right, well I DID look at it, but it made no sense(then).sleeping
and looking at it again it DID make sense. I guess that's wow I
function*S*. And about being trustworthy or not, if you belive everything
on the net is trustworthy, then next time you get one of those Nigerian
spams, do follow instructions. *S*
Stan Brown
> And about being trustworthy or not, if you belive everything
> on the net is trustworthy, then next time you get one of those Nigerian
> spams, do follow instructions. *S*
That's your idea of an excuse for your rudeness?
Hint: if you want help ever again, don't compare someone who tried to
help you -- who *would* have helped you if you'd bothered to read
what he wrote -- to a Nigerian scammer.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/
"If there's one thing I know, it's men. I ought to: it's
been my life work." -- Marie Dressler, in /Dinner at Eight/
David C. Ullrich
wrote:
Guffaw. You have problems with freshman calculus. You
get a professional mathematician to try to help you
for free. You show a remarkable inability to follow
his simple explanation, so he repeats himself, still
trying to help. You reply by saying you doubt that
he's trustworty. And now you have complaints with
_my_ behavior?
Never mind the "something tells me" part - you're
an idiot.
No, you can't believe everything you read on the internet.
But when it's _meth_ that shouldn't matter - you can
independently _verify_ what you read yourself.
>
************************
David C. Ullrich
G.E. Ivey
And why start a whole new thread to answer a statement made in the orginal?
Arturo Magidin
fjoka doka <n...@really.valid> wrote:
>Den Wed, 10 Oct 2007 17:00:37 +0000 skrev Arturo Magidin:
>
>> In article <Ek5Pi.235683$jo2.1...@reader1.news.saunalahti.fi>, Fjoka
>> <inv...@probably.com> wrote:
>>
>> [...]
>>
>>>> You asked for help- he gave you help. And because he didn't say
>>>> "yes,
>>>> you are completely correct", that's your response? I guess you don't
>>>> want help, then.
>>>>
>>>> "integral of x^2/(x-1) is x^3/3+ln(x-1)" No, that's the integral of
>>>> x^2+ 1/(x-1). The correct integral is just
>>>> what Arturo said.
>>>yes, putting a little time on it, after a good nights sleep I concluded
>>>the same.
>>>Still, being nice to people is not expensive, and you gain a lot.
>>
>> Excellent advice. How about taking it next time? Instead of wasting my
>> time and yours by not reading my response and simply repeating the same
>> mistake again, why not put that "little time on it" first? And then, if
>> you don't understand the answer, how about not saying that someone is
>> not "trustworthy", simply because you had not bothered to think about
>> the answer?
>
>heh, right, well I DID look at it, but it made no sense(then).sleeping
>and looking at it again it DID make sense. I guess that's wow I
>function*S*.
That's fine; but then ->why<-, instead of saying "I don't understand",
or instead of WAITING until after you slept on it, you decided that
the appropriate response was to question my "trustworthiness"?
To repeat the question you asked David Ullrich: is that how ->you<-
behave? If you don't understand what someone tells you because you
can't be bothered to think about it, do you reply by insulting them?
>And about being trustworthy or not, if you belive everything
>on the net is trustworthy, then next time you get one of those Nigerian
>spams, do follow instructions. *S*
You're an idiot.
TCL
> >who is right?
>
> Neither one.
wolfram's formula is correct -- any two indefinite integrals different by a constant are the same.
>You got it completely wrong, and though
> the wolfram
> integrator is almost right, it did not include the
> constant of
> integration.
>
> >is the formula there wrong?
>
> Which formula? Wrong for what?
>
> --
> ======================================================
> ================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes" by Bill
Arturo Magidin
TCL <tl...@cox.net> wrote:
>>
>> >who is right?
>>
>> Neither one.
>
>wolfram's formula is correct -- any two indefinite integrals different by a constant are the same.
Not quite. Any two ->antiderivatives<- differ by a constant. However,
there is only one indefinite integral, and the indefinite integral is
the FAMILY of antiderivatives. Omiting the constant of integration
yields a correct antiderivative, but it gives an incorrect indefinite
integral.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
TCL
> <30813455.1192166849...@nitrogen.math
> forum.org>,
> TCL <tl...@cox.net> wrote:
> >>
> >> >who is right?
> >>
> >> Neither one.
> >
> >wolfram's formula is correct -- any two indefinite
> integrals different by a constant are the same.
>
> Not quite. Any two ->antiderivatives<- differ by a
> constant. However,
> there is only one indefinite integral, and the
> indefinite integral is
> the FAMILY of antiderivatives. Omiting the constant
> of integration
> yields a correct antiderivative, but it gives an
> incorrect indefinite
> integral.
>
x^2/2 + C and (x^2-1)/2 +C are the same, because they
define the same family.
> --
> ======================================================
> ================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes" by Bill
Arturo Magidin
TCL <tl...@cox.net> wrote:
>> In article
>> <30813455.1192166849...@nitrogen.math
>> forum.org>,
>> TCL <tl...@cox.net> wrote:
>> >>
>> >> >who is right?
>> >>
>> >> Neither one.
>> >
>> >wolfram's formula is correct -- any two indefinite
>> integrals different by a constant are the same.
>>
>> Not quite. Any two ->antiderivatives<- differ by a
>> constant. However,
>> there is only one indefinite integral, and the
>> indefinite integral is
>> the FAMILY of antiderivatives. Omiting the constant
>> of integration
>> yields a correct antiderivative, but it gives an
>> incorrect indefinite
>> integral.
>>
>Consider indefinite integral of x. The two answers
>x^2/2 + C and (x^2-1)/2 +C are the same, because they
>define the same family.
I agree with the above, but it seems to have absolutely nothing to do
with my original statement or your later assertion.
I did not say that there is one and only one way to write down the
family of functions that is the indefinite integral. What I said was
that the wolfram integrator answer was wrong, because that answer was
a ->particular<- antiderivative, not the indefinite integral. You
replied by claiming the answer was correct "because" any two
"indefinite integrals [differ] by a constant". There is only ONE
indefinite integral, which consists of a family of functions; that
there may be more than one way to describe such a family is neither
here nor there, and is also irrelevant as to whether the wolfram
response was correct.
Consider your example. The two answers you give are two different ways
of describing the same family of functions, and they are both correct
descriptions of the indefinite integral of x^2/2. However, the wolfram
integrator answer was the parallel of just writing "x^2/2". and that
is wrong. The answer "x^2/2" does NOT describe a family of functions,
and as such is not the same as x^2/2 + C or any of its multiple
variants.
The answer that the Wolfram integrator gave was a ->particular<-
antiderivative, and as such is not the correct answer to the question
of what was the indefinite integral.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
Barry Schwarz
You wrote an insulting response. Now you are trying to rationalize
your behavior and simultaneously pretend it wasn't insulting. You
would do much better to simply admit you were wrong and apologize. You
are not likely to get much help from the regulars until you do.
Remove del for email