[Date Index]
[Thread Index]
[Author Index]
Re: Re: Bug of Integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg83050] Re: [mg83019] Re: Bug of Integrate
*From*: DrMajorBob <drmajorbob at bigfoot.com>
*Date*: Thu, 8 Nov 2007 06:17:32 -0500 (EST)
*References*: <fg4dfv$6c3$1@smc.vnet.net> <fg6pse$d44$1@smc.vnet.net> <28316690.1194448168488.JavaMail.root@m35>
*Reply-to*: drmajorbob at bigfoot.com
In the first place, your integral isn't the same for all functions x and
y. For instance:
Clear[x, y]
x[t_] = t^2;
y[t_] = t;
Integrate[Sqrt[1 + (y'[t]/x'[t])^2]*x'[t], {t, 0, 2 Pi}]
\[Pi] Sqrt[1 + 16 \[Pi]^2] + 1/4 ArcSinh[4 \[Pi]]
and even this is 0, not 2 Pi:
Clear[x, y]
x[t_] = Cos[t];
y[t_] = Sin[t];
Integrate[Sqrt[1 + (y'[t]/x'[t])^2]*x'[t], {t, 0, 2 Pi}]
0
I think you meant something like
Clear[x, y]
x[t_] = Cos[t];
y[t_] = Sin[t];
Integrate[Sqrt[x'[t]^2 + y'[t]^2], {t, 0, 2 Pi}]
2 \[Pi]
or (since x'[t] is negative from 0 to Pi/2):
-4 Integrate[Sqrt[1 + (y'[t]/x'[t])^2]*x'[t], {t, 0, Pi/2}]
2 \[Pi]
Bobby
On Wed, 07 Nov 2007 05:53:48 -0600, Miguel <misvrne at gmail.com> wrote:
> On 1 nov, 11:38, "David W.Cantrell" <DWCantr... at sigmaxi.net> wrote:
>> m... at inbox.ru wrote:
>> > On Oct 30, 2:26 am, "David W.Cantrell" <DWCantr... at sigmaxi.net> wrote:
>> [snip]
>> > > However, related to the above, version 5.2 does give an incorrect
>> > > result for a definite integral with a symbolic real limit. Whether
>> this
>> > > error still exists in version 6, I don't know:
>>
>> > > In[3]:= Assuming[Element[x,Reals],Integrate[3*Sign[Cos[t]],{t,0,x}]]
>>
>> > > Out[3]= 3 If[x > 0, x Abs[Cos[x]] Sec[x],
>> > > Integrate[Sign[Cos[t]], {t, 0, x}, Assumptions -> x <= 0]]
>>
>> > > The above is incorrect for x > Pi/2. A correct result would have
>> been
>>
>> > > 3 Sign[Cos[x]] (x - Pi Floor[x/Pi + 1/2])
>>
>> > > for all real x.
>>
>> > > David W. Cantrell
>>
>> > Note that your formula isn't correct for x = Pi/2 + Pi k. The correct
>> > expression for all real x is
>>
>> > In[1]:= Assuming[0 <= x < 2 Pi, Integrate[3 Sign[Cos[t]], {t, 0,
>> > x}]] /.
>> > x -> Mod[x, 2 Pi]
>>
>> > Out[1]= Piecewise[{{-3 Pi/2, Mod[x, 2 Pi] == 3 Pi/2}, {3 (Pi - Mod[x,
>> > 2 Pi]), Pi/2 < Mod[x, 2 Pi] < 3 Pi/2}, {-3 (2 Pi - Mod[x, 2 Pi]), 3
>> Pi/
>> > 2 < Mod[x, 2 Pi] < 2 Pi}, {3 Mod[x, 2 Pi], 0 < Mod[x, 2 Pi] <= Pi /2}}]
>>
>> Moments ago, I sent a message thanking Maxim for pointing out my error.
>> I
>> also mentioned a much shorter result which is correct for all real x:
>>
>> 3 ArcSin[Sin[x]]
>>
>> But perhaps it's worth mentioning that there is also an expression
>> which is correct for all real x which avoids using any functions such as
>> ArcSin or Sin, while still being shorter than his Piecewise expression:
>>
>> 3 (-1)^Floor[x/Pi + 1/2] (x - Pi Floor[x/Pi + 1/2])
>>
>> That result is "in the same spirit" as what I originally intended.
>>
>> David W. Cantrell- Ocultar texto de la cita -
>>
>> - Mostrar texto de la cita -
>
> By other hand, I think everybody knows the solution of the following
> problem:
> "Derive the formula for the circunference of a circle of radius "a" by
> computing the length of the arc x=a cost; y= a sint for 0<= t<=Pi
> "
>
> L=Integrate[Sqrt[1+(y'[t]/x'[t])^2]*x'[t],{t,0,2Pi}]
>
> I have tried to resolve with version 5.2 and version 6.0.1. The
> results have been differents.
>
> When the result is unknown, version 6.0.1 is reliable?
>
>
>
-- =
DrMajorBob at bigfoot.com
Prev by Date:
**Re: Bug of Integrate**
Next by Date:
**Re: Memory requirements in Mathematica 6**
Previous by thread:
**Re: Bug of Integrate**
Next by thread:
**[Q] Nonlinear Fitting in symbolic Integration ..?**
| |